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Preface v
Acknowledgments vi
A Note to the Student ix
1.1 Introduction 4
1.2 Systems of Units 4
1.3 Charge and Current 6
1.4 Voltage 9
1.5 Power and Energy 10
1.6 Circuit Elements 13
†1.7 Applications 15
1.7.1 TV Picture Tube
1.7.2 Electricity Bills
†1.8 Problem Solving 18
1.9 Summary 21
Review Questions 22
Problems 23
Comprehensive Problems 25
2.1 Introduction 28
2.2 Ohm’s Laws 28
†2.3 Nodes, Branches, and Loops 33
2.4 Kirchhoff’s Laws 35
2.5 Series Resistors and Voltage Division 41
2.6 Parallel Resistors and Current Division 42
†2.7 Wye-Delta Transformations 50
†2.8 Applications 54
2.8.1 Lighting Systems
2.8.2 Design of DC Meters
2.9 Summary 60
Review Questions 61
Problems 63
Comprehensive Problems 72
3.1 Introduction 76
3.2 Nodal Analysis 76
3.3 Nodal Analysis with Voltage Sources 82
3.4 Mesh Analysis 87
3.5 Mesh Analysis with Current Sources 92
†3.6 Nodal and Mesh Analyses by Inspection 95
3.7 Nodal Versus Mesh Analysis 99
3.8 Circuit Analysis with PSpice 100
†3.9 Applications: DC Transistor Circuits 102
3.10 Summary 107
Review Questions 107
Problems 109
Comprehensive Problems 117
4.1 Introduction 120
4.2 Linearity Property 120
4.3 Superposition 122
4.4 Source Transformation 127
4.5 Thevenin’s Theorem 131
4.6 Norton’s Theorem 137
†4.7 Derivations of Thevenin’s and Norton’s
Theorems 140
4.8 Maximum Power Transfer 142
4.9 Verifying Circuit Theorems
with PSpice 144
†4.10 Applications 147
4.10.1 Source Modeling
4.10.2 Resistance Measurement
4.11 Summary 153
Review Questions 153
Problems 154
Comprehensive Problems 162
5.1 Introduction 166
5.2 Operational Amplifiers 166
5.3 Ideal Op Amp 170
5.4 Inverting Amplifier 171
5.5 Noninverting Amplifier 174
5.6 Summing Amplifier 176
5.7 Difference Amplifier 177
5.8 Cascaded Op Amp Circuits 181
5.9 Op Amp Circuit Analysis
with PSpice 183
†5.10 Applications 185
5.10.1 Digital-to Analog Converter
5.10.2 Instrumentation Amplifiers
5.11 Summary 188
Review Questions 190
Problems 191
Comprehensive Problems 200
Contents
xi
Chapter 2 Basic Laws 27
Chapter 3 Methods of Analysis 75
PART 1 DC CIRCUITS 1
Chapter 1 Basic Concepts 3
Chapter 4 Circuit Theorems 119
Chapter 5 Operational Amplifiers 165
f51-cont.qxd 3/16/00 4:22 PM Page xi
6.1 Introduction 202
6.2 Capacitors 202
6.3 Series and Parallel Capacitors 208
6.4 Inductors 211
6.5 Series and Parallel Inductors 216
†6.6 Applications 219
6.6.1 Integrator
6.6.2 Differentiator
6.6.3 Analog Computer
6.7 Summary 225
Review Questions 226
Problems 227
Comprehensive Problems 235
7.1 Introduction 238
7.2 The Source-free RC Circuit 238
7.3 The Source-free RL Circuit 243
7.4 Singularity Functions 249
7.5 Step Response of an RC Circuit 257
7.6 Step Response of an RL Circuit 263
†7.7 First-order Op Amp Circuits 268
7.8 Transient Analysis with PSpice 273
†7.9 Applications 276
7.9.1 Delay Circuits
7.9.2 Photoflash Unit
7.9.3 Relay Circuits
7.9.4 Automobile Ignition Circuit
7.10 Summary 282
Review Questions 283
Problems 284
Comprehensive Problems 293
8.1 Introduction 296
8.2 Finding Initial and Final Values 296
8.3 The Source-Free Series RLC Circuit 301
8.4 The Source-Free Parallel RLC Circuit 308
8.5 Step Response of a Series RLC
Circuit 314
8.6 Step Response of a Parallel RLC
Circuit 319
8.7 General Second-Order Circuits 322
8.8 Second-Order Op Amp Circuits 327
8.9 PSpice Analysis of RLC Circuits 330
†8.10 Duality 332
†8.11 Applications 336
8.11.1 Automobile Ignition System
8.11.2 Smoothing Circuits
8.12 Summary 340
Review Questions 340
Problems 341
Comprehensive Problems 350
9.1 Introduction 354
9.2 Sinusoids 355
9.3 Phasors 359
9.4 Phasor Relationships for Circuit
Elements 367
9.5 Impedance and Admittance 369
9.6 Kirchhoff’s Laws in the Frequency
Domain 372
9.7 Impedance Combinations 373
†9.8 Applications 379
9.8.1 Phase-Shifters
9.8.2 AC Bridges
9.9 Summary 384
Review Questions 385
Problems 385
Comprehensive Problems 392
10.1 Introduction 394
10.2 Nodal Analysis 394
10.3 Mesh Analysis 397
10.4 Superposition Theorem 400
10.5 Source Transformation 404
10.6 Thevenin and Norton Equivalent
Circuits 406
10.7 Op Amp AC Circuits 411
10.8 AC Analysis Using PSpice 413
†10.9 Applications 416
10.9.1 Capacitance Multiplier
10.9.2 Oscillators
10.10 Summary 420
Review Questions 421
Problems 422
11.1 Introduction 434
11.2 Instantaneous and Average Power 434
11.3 Maximum Average Power Transfer 440
11.4 Effective or RMS Value 443
11.5 Apparent Power and Power Factor 447
11.6 Complex Power 449
†11.7 Conservation of AC Power 453
xii CONTENTS
Chapter 8 Second-Order Circuits 295
Chapter 10 Sinusoidal Steady-State Analysis 393
Chapter 11 AC Power Analysis 433
Chapter 6 Capacitors and Inductors 201
Chapter 7 First-Order Circuits 237
PART 2 AC CIRCUITS 351
Chapter 9 Sinusoids and Phasors 353
11.8 Power Factor Correction 457
†11.9 Applications 459
11.9.1 Power Measurement
11.9.2 Electricity Consumption Cost
11.10 Summary 464
Review Questions 465
Problems 466
Comprehensive Problems 474
12.1 Introduction 478
12.2 Balanced Three-Phase Voltages 479
12.3 Balanced Wye-Wye Connection 482
12.4 Balanced Wye-Delta Connection 486
12.5 Balanced Delta-Delta Connection 488
12.6 Balanced Delta-Wye Connection 490
12.7 Power in a Balanced System 494
†12.8 Unbalanced Three-Phase Systems 500
12.9 PSpice for Three-Phase Circuits 504
†12.10 Applications 508
12.10.1 Three-Phase Power Measurement
12.10.2 Residential Wiring
12.11 Summary 516
Review Questions 517
Problems 518
Comprehensive Problems 525
13.1 Introduction 528
13.2 Mutual Inductance 528
13.3 Energy in a Coupled Circuit 535
13.4 Linear Transformers 539
13.5 Ideal Transformers 545
13.6 Ideal Autotransformers 552
†13.7 Three-Phase Transformers 556
13.8 PSpice Analysis of Magnetically Coupled
Circuits 559
†13.9 Applications 563
13.9.1 Transformer as an Isolation Device
13.9.2 Transformer as a Matching Device
13.9.3 Power Distribution
13.10 Summary 569
Review Questions 570
Problems 571
Comprehensive Problems 582
14.1 Introduction 584
14.2 Transfer Function 584
†14.3 The Decibel Scale 588
14.4 Bode Plots 589
14.5 Series Resonance 600
14.6 Parallel Resonance 605
14.7 Passive Filters 608
14.7.1 Lowpass Filter
14.7.2 Highpass Filter
14.7.3 Bandpass Filter
14.7.4 Bandstop Filter
14.8 Active Filters 613
14.8.1 First-Order Lowpass Filter
14.8.2 First-Order Highpass Filter
14.8.3 Bandpass Filter
14.8.4 Bandreject (or Notch) Filter
†14.9 Scaling 619
14.9.1 Magnitude Scaling
14.9.2 Frequency Scaling
14.9.3 Magnitude and Frequency Scaling
14.10 Frequency Response Using
PSpice 622
†14.11 Applications 626
14.11.1 Radio Receiver
14.11.2 Touch-Tone Telephone
14.11.3 Crossover Network
14.12 Summary 631
Review Questions 633
Problems 633
Comprehensive Problems 640
15.1 Introduction 646
15.2 Definition of the Laplace
Transform 646
15.3 Properties of the Laplace
Transform 649
15.4 The Inverse Laplace Transform 659
15.4.1 Simple Poles
15.4.2 Repeated Poles
15.4.3 Complex Poles
15.5 Applicaton to Circuits 666
15.6 Transfer Functions 672
15.7 The Convolution Integral 677
†15.8 Application to Integrodifferential
Equations 685
†15.9 Applications 687
15.9.1 Network Stability
15.9.2 Network Synthesis
15.10 Summary 694
xiii
CONTENTS
PART 3 ADVANCEDCIRCUITANALYSIS 643
Chapter 15 The Laplace Transform 645
Chapter 12 Three-Phase Circuits 477
Chapter 13 Magnetically Coupled Circuits 527
Chapter 14 Frequency Response 583
Review Questions 696
Problems 696
Comprehensive Problems 705
16.1 Introduction 708
16.2 Trigonometric Fourier Series 708
16.3 Symmetry Considerations 717
16.3.1 Even Symmetry
16.3.2 Odd Symmetry
16.3.3 Half-Wave Symmetry
16.4 Circuit Applicatons 727
16.5 Average Power and RMS Values 730
16.6 Exponential Fourier Series 734
16.7 Fourier Analysis with PSpice 740
16.7.1 Discrete Fourier Transform
16.7.2 Fast Fourier Transform
†16.8 Applications 746
16.8.1 Spectrum Analyzers
16.8.2 Filters
16.9 Summary 749
Review Questions 751
Problems 751
Comprehensive Problems 758
17.1 Introduction 760
17.2 Definition of the Fourier Transform 760
17.3 Properties of the Fourier Transform 766
17.4 Circuit Applications 779
17.5 Parseval’s Theorem 782
17.6 Comparing the Fourier and Laplace
Transforms 784
†17.7 Applications 785
17.7.1 Amplitude Modulation
17.7.2 Sampling
17.8 Summary 789
Review Questions 790
Problems 790
Comprehensive Problems 794
18.1 Introduction 796
18.2 Impedance Parameters 796
18.3 Admittance Parameters 801
18.4 Hybrid Parameters 804
18.5 Transmission Parameters 809
†18.6 Relationships between Parameters 814
18.7 Interconnection of Networks 817
18.8 Computing Two-Port Parameters Using
PSpice 823
†18.9 Applications 826
18.9.1 Transistor Circuits
18.9.2 Ladder Network Synthesis
18.10 Summary 833
Review Questions 834
Problems 835
Comprehensive Problems 844
AppendixA Solution of Simultaneous Equations Using
Cramer’s Rule 845
Appendix B Complex Numbers 851
Appendix C Mathematical Formulas 859
Appendix D PSpice for Windows 865
Appendix E Answers to Odd-Numbered Problems 893
Selected Bibliography 929
Index 933
xiv CONTENTS
Chapter 16 The Fourier Series 707
Chapter 17 Fourier Transform 759
Chapter 18 Two-Port Networks 795
Features
In spite of the numerous textbooks on circuit analysis
available in the market, students often find the course
difficult to learn. The main objective of this book is
to present circuit analysis in a manner that is clearer,
more interesting, and easier to understand than earlier
texts. This objective is achieved in the following
ways:
• A course in circuit analysis is perhaps the first
exposure students have to electrical engineering.
We have included several features to help stu-
dents feel at home with the subject. Each chapter
opens with either a historical profile of some
electrical engineering pioneers to be mentioned in
the chapter or a career discussion on a subdisci-
pline of electrical engineering. An introduction
links the chapter with the previous chapters and
states the chapter’s objectives. The chapter ends
with a summary of the key points and formulas.
• All principles are presented in a lucid, logical,
step-by-step manner. We try to avoid wordiness
and superfluous detail that could hide concepts
and impede understanding the material.
• Important formulas are boxed as a means of
helping students sort what is essential from what
is not; and to ensure that students clearly get the
gist of the matter, key terms are defined and
highlighted.
• Marginal notes are used as a pedagogical aid. They
serve multiple uses—hints, cross-references, more
exposition, warnings, reminders, common mis-
takes, and problem-solving insights.
• Thoroughly worked examples are liberally given at
the end of every section. The examples are regard-
ed as part of the text and are explained clearly, with-
out asking the reader to fill in missing steps.
Thoroughly worked examples give students a good
understanding of the solution and the confidence to
solve problems themselves. Some of the problems
are solved in two or three ways to facilitate an
understanding and comparison of different
approaches.
• To give students practice opportunity, each illus-
trative example is immediately followed by a
practice problem with the answer. The students can
follow the example step-by-step to solve the prac-
tice problem without flipping pages or searching
the end of the book for answers. The practice prob-
lem is also intended to test students’ understanding
of the preceding example. It will reinforce their
grasp of the material before moving to the next
section.
• In recognition of ABET’s requirement on integrat-
ing computer tools, the use of PSpice is encouraged
in a student-friendly manner. Since the Windows
version of PSpice is becoming popular, it is used
instead of the MS-DOS version. PSpice is covered
early so that students can use it throughout the text.
Appendix D serves as a tutorial on PSpice for
Windows.
• The operational amplifier (op amp) as a basic ele-
ment is introduced early in the text.
• To ease the transition between the circuit course
and signals/systems courses, Fourier and Laplace
transforms are covered lucidly and thoroughly.
• The last section in each chapter is devoted to appli-
cations of the concepts covered in the chapter. Each
chapter has at least one or two practical problems or
devices. This helps students apply the concepts to
real-life situations.
• Ten multiple-choice review questions are provided
at the end of each chapter, with answers. These are
intended to cover the little “tricks” that the exam-
ples and end-of-chapter problems may not cover.
They serve as a self-test device and help students
determine how well they have mastered the chapter.
Organization
This book was written for a two-semester or three-semes-
ter course in linear circuit analysis. The book may
also be used for a one-semester course by a proper selec-
tion of chapters and sections. It is broadly divided into
three parts.
• Part 1, consisting of Chapters 1 to 8, is devoted to
dc circuits. It covers the fundamental laws and the-
orems, circuit techniques, passive and active ele-
ments.
• Part 2, consisting of Chapters 9 to 14, deals with ac
circuits. It introduces phasors, sinusoidal steady-
state analysis, ac power, rms values, three-phase
systems, and frequency response.
• Part 3, consisting of Chapters 15 to 18, is devoted
to advanced techniques for network analysis.
It provides a solid introduction to the Laplace
transform, Fourier series, the Fourier transform,
and two-port network analysis.
The material in three parts is more than suffi-
cient for a two-semester course, so that the instructor
PREFACE
v
F51-pref.qxd 3/17/00 10:11 AM Page v
must select which chapters/sections to cover. Sections
marked with the dagger sign (†) may be skipped,
explained briefly, or assigned as homework. They can
be omitted without loss of continuity. Each chapter has
plenty of problems, grouped according to the sections
of the related material, and so diverse that the instruc-
tor can choose some as examples and assign some as
homework. More difficult problems are marked with a
star (*). Comprehensive problems appear last; they are
mostly applications problems that require multiple
skills from that particular chapter.
The book is as self-contained as possible. At the
end of the book are some appendixes that review
solutions of linear equations, complex numbers, math-
ematical formulas, a tutorial on PSpice for Windows,
and answers to odd-numbered problems. Answers to
all the problems are in the solutions manual, which is
available from the publisher.
Prerequisites
As with most introductory circuit courses, the main
prerequisites are physics and calculus. Although famil-
iarity with complex numbers is helpful in the later part
of the book, it is not required.
Supplements
Solutions Manual—an Instructor’s Solutions Manual is
available to instructors who adopt the text. It contains
complete solutions to all the end-of-chapter problems.
Transparency Masters—over 200 important figures
are available as transparency masters for use as over-
heads.
Student CD-ROM—100 circuit files from the book are
presented as Electronics Workbench (EWB) files; 15–20
of these files are accessible using the free demo of Elec-
tronics Workbench. The students are able to experiment
with the files. For those who wish to fully unlock all 100
circuit files, EWB’s full version may be purchased from
Interactive Image Technologies for approximately
$79.00. The CD-ROM also contains a selection of prob-
lem-solving, analysis and design tutorials, designed to
further support important concepts in the text.
Problem-Solving Workbook—a paperback work-
book is for sale to students who wish to practice their
problem solving techniques. The workbook contains a
discussion of problem solving strategies and 150 addi-
tional problems with complete solutions provided.
Online Learning Center (OLC)—the Web site for
the book will serve as an online learning center for stu-
dents as a useful resource for instructors. The OLC
will provide access to:
300 test questions—for instructors only
Downloadable figures for overhead
presentations—for instructors only
Solutions manual—for instructors only
Web links to useful sites
Sample pages from the Problem-Solving
Workbook
PageOut Lite—a service provided to adopters
who want to create their own Web site. In
just a few minutes, instructors can change
the course syllabus into a Web site using
PageOut Lite.
The URL for the web site is www.mhhe.com.alexander.
Although the textbook is meant to be self-explanatory
and act as a tutor for the student, the personal contact
involved in teaching is not to be forgotten. The book
and supplements are intended to supply the instructor
with all the pedagogical tools necessary to effectively
present the material.
We wish to take the opportunity to thank the staff of
McGraw-Hill for their commitment and hard
work: Lynn Cox, Senior Editor; Scott Isenberg,
Senior Sponsoring Editor; Kelley Butcher, Senior
Developmental Editor; Betsy Jones, Executive
Editor; Catherine Fields, Sponsoring Editor;
Kimberly Hooker, Project Manager; and Michelle
Flomenhoft, Editorial Assistant. They got numerous
reviews, kept the book on track, and helped in many
ways. We really appreciate their inputs. We are
greatly in debt to Richard Mickey for taking the pain
ofchecking and correcting the entire manuscript. We
wish to record our thanks to Steven Durbin at Florida
State University and Daniel Moore at Rose Hulman
Institute of Technology for serving as accuracy
checkers of examples, practice problems, and end-
of-chapter problems. We also wish to thank the fol-
lowing reviewers for their constructive criticisms
and helpful comments.
Promod Vohra, Northern Illinois University
Moe Wasserman, Boston University
Robert J. Krueger, University of Wisconsin
Milwaukee
John O’Malley, University of Florida
vi PREFACE
ACKNOWLEDGMENTS
F51-pref.qxd 3/17/00 10:11 AM Page vi
Aniruddha Datta, Texas A&M University
John Bay, Virginia Tech
Wilhelm Eggimann, Worcester Polytechnic
Institute
A. B. Bonds, Vanderbilt University
Tommy Williamson, University of Dayton
Cynthia Finelli, Kettering University
John A. Fleming, Texas A&M University
Roger Conant, University of Illinois
at Chicago
Daniel J. Moore, Rose-Hulman Institute of
Technology
Ralph A. Kinney, Louisiana State University
Cecilia Townsend, North Carolina State
University
Charles B. Smith, University of Mississippi
H. Roland Zapp, Michigan State University
Stephen M. Phillips, Case Western University
Robin N. Strickland, University of Arizona
David N. Cowling, Louisiana Tech University
Jean-Pierre R. Bayard, California State
University
Jack C. Lee, University of Texas at Austin
E. L. Gerber, Drexel University
The first author wishes to express his apprecia-
tion to his department chair, Dr. Dennis Irwin, for his
outstanding support. In addition, he is extremely grate-
ful to Suzanne Vazzano for her help with the solutions
manual.
The second author is indebted to Dr. Cynthia
Hirtzel, the former dean of the college of engineering
at Temple University, and Drs.. Brian Butz, Richard
Klafter, and John Helferty, his departmental chairper-
sons at different periods, for their encouragement while
working on the manuscript. The secretarial support
provided by Michelle Ayers and Carol Dahlberg is
gratefully appreciated. Special thanks are due to Ann
Sadiku, Mario Valenti, Raymond Garcia, Leke and
Tolu Efuwape, and Ope Ola for helping in various
ways. Finally, we owe the greatest debt to our wives,
Paulette and Chris, without whose constant support and
cooperation this project would have been impossible.
Please address comments and corrections to the
publisher.
C. K. Alexander and M. N. O. Sadiku
PREFACE vii
F51-pref.qxd 3/17/00 10:11 AM Page vii
F51-pref.qxd 3/17/00 10:11 AM Page viii
This may be your first course in electrical engineer-
ing. Although electrical engineering is an exciting and
challenging discipline, the course may intimidate you.
This book was written to prevent that. A good textbook
and a good professor are an advantage—but you are
the one who does the learning. If you keep the follow-
ing ideas in mind, you will do very well in this course.
• This course is the foundation on which most
other courses in the electrical engineering cur-
riculum rest. For this reason, put in as much
effort as you can. Study the course regularly.
• Problem solving is an essential part of the learn-
ing process. Solve as many problems as you can.
Begin by solving the practice problem following
each example, and then proceed to the end-of-
chapter problems. The best way to learn is to
solve a lot of problems. An asterisk in front of a
problem indicates a challenging problem.
• Spice, a computer circuit analysis program, is
used throughout the textbook. PSpice, the per-
sonal computer version of Spice, is the popular
standard circuit analysis program at most uni-
versities. PSpice for Windows is described in
Appendix D. Make an effort to learn PSpice,
because you can check any circuit problem with
PSpice and be sure you are handing in a correct
problem solution.
• Each chapter ends with a section on how the
material covered in the chapter can be applied to
real-life situations. The concepts in this section
may be new and advanced to you. No doubt, you
will learn more of the details in other courses.
We are mainly interested in gaining a general
familiarity with these ideas.
• Attempt the review questions at the end of each
chapter. They will help you discover some
“tricks” not revealed in class or in the textbook.
A short review on finding determinants is cov-
ered in Appendix A, complex numbers in Appendix B,
and mathematical formulas in Appendix C. Answers to
odd-numbered problems are given in Appendix E.
Have fun!
C.K.A. and M.N.O.S.
A NOTE TO THE STUDENT
ix
F51-pref.qxd 3/17/00 10:11 AM Page ix
1
DC CIRCUITS
P A R T 1
C h a p t e r 1 Basic Concepts
C h a p t e r 2 Basic Laws
C h a p t e r 3 Methods of Analysis
C h a p t e r 4 Circuit Theorems
C h a p t e r 5 Operational Amplifier
C h a p t e r 6 Capacitors and Inductors
C h a p t e r 7 First-Order Circuits
C h a p t e r 8 Second-Order Circuits
2
3
C H A P T E R
BASIC CONCEPTS
1
It is engineering that changes the world.
—Isaac Asimov
Historical Profiles
Alessandro Antonio Volta (1745–1827), an Italian physicist, invented the electric
battery—which provided the first continuous flow of electricity—and the capacitor.
Born into a noble family in Como, Italy, Volta was performing electrical
experiments at age 18. His invention of the battery in 1796 revolutionized the use of
electricity. The publication of his work in 1800 marked the beginning of electric circuit
theory. Volta received many honors during his lifetime. The unit of voltage or potential
difference, the volt, was named in his honor.
Andre-Marie Ampere (1775–1836), a French mathematician and physicist, laid the
foundation of electrodynamics. He defined the electric current and developed a way to
measure it in the 1820s.
Born in Lyons, France, Ampere at age 12 mastered Latin in a few weeks, as he
was intensely interested in mathematics and many of the best mathematical works were
in Latin. He was a brilliant scientist and a prolific writer. He formulated the laws of
electromagnetics. He invented the electromagnet and the ammeter. The unit of electric
current, the ampere, was named after him.
4 PART 1 DC Circuits
1.1 INTRODUCTION
Electric circuit theory and electromagnetic theory are the two fundamen-
tal theories upon which all branches of electrical engineering are built.
Many branches of electrical engineering, such as power, electric ma-
chines, control, electronics, communications, and instrumentation, are
based on electric circuit theory. Therefore, the basic electric circuit the-
ory course is the most important course for an electrical engineering
student, and always an excellent starting point for a beginning student
in electrical engineering education. Circuit theory is also valuable to
students specializing in other branches of the physical sciences because
circuits are a good model for the study of energy systems in general, and
because of the applied mathematics, physics, and topology involved.
In electrical engineering, we are often interested in communicating
or transferring energy from one point to another. To do this requires an
interconnection of electrical devices. Such interconnection is referred to
as an electric circuit, and each component of the circuit is known as an
element.
An electric circuit is an interconnection of electrical elements.
A simple electric circuit is shown in Fig. 1.1. It consists of three
basic components: a battery, a lamp, and connecting wires. Such a simple
circuit can exist by itself; it has several applications, such as a torch light,
a search light, and so forth.
+
−
Current
Lamp
Battery
Figure1.1 A simple electric circuit.
A complicated real circuit is displayed in Fig. 1.2, representing the
schematic diagram for a radio receiver. Although it seems complicated,
this circuit can be analyzed using the techniques we cover in this book.
Our goal in this text is to learn various analytical techniques and computer
software applications for describing the behavior of a circuit like this.
Electric circuits are used in numerous electrical systems to accom-
plish different tasks. Our objective in this book is not the study of various
uses and applications of circuits. Rather our major concern is the anal-
ysis of the circuits. By the analysis of a circuit, we mean a study of the
behavior of the circuit: How does it respond to a given input? How do
the interconnected elements and devices in the circuit interact?
We commence our study by defining some basic concepts. These
concepts include charge, current, voltage, circuit elements, power, and
energy. Before defining these concepts, we must first establish a system
of units that we will use throughout the text.
1.2 SYSTEMS OF UNITS
As electrical engineers, we deal with measurable quantities. Our mea-
surement, however, must be communicated in a standard language that
virtually all professionals can understand, irrespective of the country
where the measurement is conducted. Such an international measure-
ment language is the International System of Units (SI), adopted by the
General Conference on Weights and Measures in 1960. In this system,
CHAPTER 1 Basic Concepts 5
2, 5, 6
C
Oscillator
E
B
R2
10 k
R3
10 k
R1 47
Y1
7 MHz
C6 5
L2
22.7 mH
(see text)
to
U1, Pin 8
R10
10 k
GAIN +
+
C16
100 mF
16 V
C11
100 mF
16 V
C10
1.0 mF
16 V
C9
1.0 mF
16 V
C15
0.47
16 V
C17
100 mF
16 V
+
−
12-V dc
Supply
Audio
Output
+
C18
0.1
R12
10
1
4
2
3
C14
0.0022
0.1
C13
U2A
1 ⁄2 TL072
U2B
1⁄2 TL072
R9
15 k
R5
100 k
R8
15 k
R6
100 k
5
6
R7
1 M
C12 0.0033
+
L3
1 mH
R11
47
C8
0.1
Q1
2N2222A
7
C3 0.1
L1
0.445 mH
Antenna C1
2200 pF
C2
2200 pF
1
8
7
U1
SBL-1
Mixer
3, 4
C7
532
C4
910
C5
910
R4
220
U3
LM386N
Audio power amp
5
4
6
3
2
+
+
−
+
−
+
−
+
8
Figure 1.2 Electric circuit of a radio receiver.
(Reproduced with permission from QST, August 1995, p. 23.)
there are six principal units from which the units of all other physical
quantities can be derived. Table 1.1 shows the six units, their symbols,
and the physical quantities they represent. The SI units are used through-
out this text.
One great advantage of the SI unit is that it uses prefixes based on
the power of 10 to relate larger and smaller units to the basic unit. Table
1.2 shows the SI prefixes and their symbols. For example, the following
are expressions of the same distance in meters (m):
600,000,000 mm 600,000 m 600 km
TABLE 1.2 The SI prefixes.
Multiplier Prefix Symbol
1018
exa E
1015
peta P
1012
tera T
109
giga G
106
mega M
103
kilo k
102
hecto h
10 deka da
10−1
deci d
10−2
centi c
10−3
milli m
10−6
micro µ
10−9
nano n
10−12
pico p
10−15
femto f
10−18
atto a
TABLE 1.1 The six basic SI units.
Quantity Basic unit Symbol
Length meter m
Mass kilogram kg
Time second s
Electric current ampere A
Thermodynamic temperature kelvin K
Luminous intensity candela cd
6 PART 1 DC Circuits
1.3 CHARGE AND CURRENT
The concept of electric charge is the underlying principle for explaining
all electrical phenomena. Also, the most basic quantity in an electric
circuit is the electric charge. We all experience the effect of electric
charge when we try to remove our wool sweater and have it stick to our
body or walk across a carpet and receive a shock.
Charge is an electrical property of the atomic particles of which
matter consists, measured in coulombs (C).
We know from elementary physics that all matter is made of fundamental
building blocks known as atoms and that each atom consists of electrons,
protons, and neutrons. We also know that the charge e on an electron is
negative and equal in magnitude to 1.602×10−19
C, while a proton carries
a positive charge of the same magnitude as the electron. The presence of
equal numbers of protons and electrons leaves an atom neutrally charged.
The following points should be noted about electric charge:
1. The coulomb is a large unit for charges. In 1 C of charge, there
are 1/(1.602 × 10−19
) = 6.24 × 1018
electrons. Thus realistic
or laboratory values of charges are on the order of pC, nC, or
µC.1
2. According to experimental observations, the only charges that
occur in nature are integral multiples of the electronic charge
e = −1.602 × 10−19
C.
3. The law of conservation of charge states that charge can neither
be created nor destroyed, only transferred. Thus the algebraic
sum of the electric charges in a system does not change.
We now consider the flow of electric charges. A unique feature of
electric charge or electricity is the fact that it is mobile; that is, it can
be transferred from one place to another, where it can be converted to
another form of energy.
Battery
I − −
− −
+ −
Figure1.3 Electric current due to flow
of electronic charge in a conductor.
A convention is a standard way of describing
something so that others in the profession can
understandwhatwemean. WewillbeusingIEEE
conventions throughout this book.
When a conducting wire (consisting of several atoms) is connected
to a battery (a source of electromotive force), the charges are compelled
to move; positive charges move in one direction while negative charges
move in the opposite direction. This motion of charges creates electric
current. It is conventional to take the current flow as the movement of
positive charges, that is, opposite to the flow of negative charges, as Fig.
1.3 illustrates. This convention was introduced by Benjamin Franklin
(1706–1790), the American scientist and inventor. Although we now
know that current in metallic conductors is due to negatively charged
electrons, we will follow the universally accepted convention that current
is the net flow of positive charges. Thus,
1However, a large power supply capacitor can store up to 0.5 C of charge.
CHAPTER 1 Basic Concepts 7
Electric current is the time rate of change of charge, measured in amperes (A).
Mathematically, the relationship between current i, charge q, and time t
is
i =
dq
dt
(1.1)
where current is measured in amperes (A), and
1 ampere = 1 coulomb/second
The charge transferred between time t0 and t is obtained by integrating
both sides of Eq. (1.1). We obtain
q =
 t
t0
i dt (1.2)
The way we define current as i in Eq. (1.1) suggests that current need not
be a constant-valued function. As many of the examples and problems in
this chapter and subsequent chapters suggest, there can be several types
of current; that is, charge can vary with time in several ways that may be
represented by different kinds of mathematical functions.
If the current does not change with time, but remains constant, we
call it a direct current (dc).
A direct current (dc) is a current that remains constant with time.
By convention the symbol I is used to represent such a constant current.
A time-varying current is represented by the symbol i. A com-
mon form of time-varying current is the sinusoidal current or alternating
current (ac).
An alternating current (ac) is a current that varies sinusoidally with time.
Such current is used in your household, to run the air conditioner, refrig-
erator, washing machine, and other electric appliances. Figure 1.4 shows
direct current and alternating current; these are the two most common
types of current. We will consider other types later in the book.
I
0 t
(a)
(b)
i
t
0
Figure1.4 Two common types of
current: (a) direct current (dc),
(b) alternating current (ac).
Once we define current as the movement of charge, we expect cur-
rent to have an associated direction of flow. As mentioned earlier, the
directionofcurrentflowisconventionallytakenasthedirectionofpositive
charge movement. Based on this convention, a current of 5 A may be
represented positively or negatively as shown in Fig. 1.5. In other words,
a negative current of −5 A flowing in one direction as shown in Fig.
1.5(b) is the same as a current of +5 A flowing in the opposite direction.
5 A
(a)
−5 A
(b)
Figure1.5 Conventional current flow:
(a) positive current flow, (b) negative current
flow.
8 PART 1 DC Circuits
E X A M P L E 1 . 1
How much charge is represented by 4,600 electrons?
Solution:
Each electron has −1.602 × 10−19
C. Hence 4,600 electrons will have
−1.602 × 10−19
C/electron × 4,600 electrons = −7.369 × 10−16
C
P R A C T I C E P R O B L E M 1 . 1
Calculate the amount of charge represented by two million protons.
Answer: +3.204 × 10−13
C.
E X A M P L E 1 . 2
The total charge entering a terminal is given by q = 5t sin 4πt mC. Cal-
culate the current at t = 0.5 s.
Solution:
i =
dq
dt
=
d
dt
(5t sin 4πt) mC/s = (5 sin 4πt + 20πt cos 4πt) mA
At t = 0.5,
i = 5 sin 2π + 10π cos 2π = 0 + 10π = 31.42 mA
P R A C T I C E P R O B L E M 1 . 2
If in Example 1.2, q = (10 − 10e−2t
) mC, find the current at t = 0.5 s.
Answer: 7.36 mA.
E X A M P L E 1 . 3
Determine the total charge entering a terminal between t = 1 s and t = 2 s
if the current passing the terminal is i = (3t2
− t) A.
Solution:
q =
 2
t=1
i dt =
 2
1
(3t2
− t) dt
=

t3
−
t2
2




2
1
= (8 − 2) −

1 −
1
2

= 5.5 C
P R A C T I C E P R O B L E M 1 . 3
The current flowing through an element is
i =

2 A, 0  t  1
2t2
A, t  1
Calculate the charge entering the element from t = 0 to t = 2 s.
Answer: 6.667 C.
CHAPTER 1 Basic Concepts 9
1.4 VOLTAGE
As explained briefly in the previous section, to move the electron in a
conductor in a particular direction requires some work or energy transfer.
Thisworkisperformedbyanexternalelectromotiveforce(emf), typically
represented by the battery in Fig. 1.3. This emf is also known as voltage
or potential difference. The voltage vab between two points a and b in
an electric circuit is the energy (or work) needed to move a unit charge
from a to b; mathematically,
vab =
dw
dq
(1.3)
where w is energy in joules (J) and q is charge in coulombs (C). The
voltage vab or simply v is measured in volts (V), named in honor of the
Italian physicist Alessandro Antonio Volta (1745–1827), who invented
the first voltaic battery. From Eq. (1.3), it is evident that
1 volt = 1 joule/coulomb = 1 newton meter/coulomb
Thus,
Voltage (or potential difference) is the energy required to move
a unit charge through an element, measured in volts (V).
Figure 1.6 shows the voltage across an element (represented by a
rectangular block) connected to points a and b. The plus (+) and minus
(−) signs are used to define reference direction or voltage polarity. The
vab can be interpreted in two ways: (1) point a is at a potential of vab
volts higher than point b, or (2) the potential at point a with respect to
point b is vab. It follows logically that in general
vab = −vba (1.4)
For example, in Fig. 1.7, we have two representations of the same vol-
tage. In Fig. 1.7(a), point a is +9 V above point b; in Fig. 1.7(b), point b is
−9 V above point a. We may say that in Fig. 1.7(a), there is a 9-V voltage
drop from a to b or equivalently a 9-V voltage rise from b to a. In other
words, a voltage drop from a to b is equivalent to a voltage rise from
b to a.
a
b
vab
+
−
Figure1.6 Polarity
of voltage vab.
9 V
(a)
a
b
+
−
−9 V
(b)
a
b
+
−
Figure1.7 Two equivalent
representations of the same
voltage vab: (a) point a is 9 V
above point b, (b) point b is
−9 V above point a.
Current and voltage are the two basic variables in electric circuits.
The common term signal is used for an electric quantity such as a current
or a voltage (or even electromagnetic wave) when it is used for conveying
information. Engineers prefer to call such variables signals rather than
mathematical functions of time because of their importance in commu-
nications and other disciplines. Like electric current, a constant voltage
is called a dc voltage and is represented by V, whereas a sinusoidally
time-varying voltage is called an ac voltage and is represented by v. A
dc voltage is commonly produced by a battery; ac voltage is produced by
an electric generator.
Keep in mind that electric current is always
through an element and that electric voltage is al-
ways across the element or between two points.
10 PART 1 DC Circuits
1.5 POWER AND ENERGY
Although current and voltage are the two basic variables in an electric
circuit, they are not sufficient by themselves. For practical purposes,
we need to know how much power an electric device can handle. We
all know from experience that a 100-watt bulb gives more light than a
60-watt bulb. We also know that when we pay our bills to the electric
utility companies, we are paying for the electric energy consumed over a
certain period of time. Thus power and energy calculations are important
in circuit analysis.
To relate power and energy to voltage and current, we recall from
physics that:
Power is the time rate of expending or absorbing energy, measured in watts (W).
We write this relationship as
p =
dw
dt
(1.5)
where p is power in watts (W), w is energy in joules (J), and t is time in
seconds (s). From Eqs. (1.1), (1.3), and (1.5), it follows that
p =
dw
dt
=
dw
dq
·
dq
dt
= vi (1.6)
or
p = vi (1.7)
The power p in Eq. (1.7) is a time-varying quantity and is called the
instantaneouspower. Thus, thepowerabsorbedorsuppliedbyanelement
is the product of the voltage across the element and the current through
it. If the power has a + sign, power is being delivered to or absorbed
by the element. If, on the other hand, the power has a − sign, power is
being supplied by the element. But how do we know when the power has
a negative or a positive sign?
Current direction and voltage polarity play a major role in deter-
mining the sign of power. It is therefore important that we pay attention
to the relationship between current i and voltage v in Fig. 1.8(a). The vol-
tage polarity and current direction must conform with those shown in Fig.
1.8(a) in order for the power to have a positive sign. This is known as
the passive sign convention. By the passive sign convention, current en-
ters through the positive polarity of the voltage. In this case, p = +vi or
vi  0 implies that the element is absorbing power. However, if p = −vi
or vi  0, as in Fig. 1.8(b), the element is releasing or supplying power.
p = +vi
(a)
v
+
−
p = −vi
(b)
v
+
−
i
i
Figure1.8 Reference
polarities for power using
the passive sign conven-
tion: (a) absorbing power,
(b) supplying power.
Passive sign convention is satisfied when the current enters through
the positive terminal of an element and p = +vi. If the current
enters through the negative terminal, p = −vi.
CHAPTER 1 Basic Concepts 11
When the voltage and current directions con-
form to Fig. 1.8(b), we have the active sign con-
vention and p = +vi.
Unless otherwise stated, we will follow the passive sign convention
throughout this text. For example, the element in both circuits of Fig. 1.9
has an absorbing power of +12 W because a positive current enters the
positive terminal in both cases. In Fig. 1.10, however, the element is
supplying power of −12 W because a positive current enters the negative
terminal. Of course, an absorbing power of +12 W is equivalent to a
supplying power of −12 W. In general,
Power absorbed = −Power supplied
(a)
4 V
3 A
(a)
+
−
3 A
4 V
3 A
(b)
+
−
Figure1.9 Two cases of an
element with an absorbing
power of 12 W:
(a) p = 4 × 3 = 12 W,
(b) p = 4 × 3 = 12 W.
3 A
(a)
4 V
3 A
(a)
+
−
3 A
4 V
3 A
(b)
+
−
Figure1.10 Two cases of
an element with a supplying
power of 12 W:
(a) p = 4 × (−3) = −12 W,
(b) p = 4 × (−3) = −12 W.
In fact, the law of conservation of energy must be obeyed in any
electric circuit. For this reason, the algebraic sum of power in a circuit,
at any instant of time, must be zero:

p = 0 (1.8)
This again confirms the fact that the total power supplied to the circuit
must balance the total power absorbed.
From Eq. (1.6), the energy absorbed or supplied by an element from
time t0 to time t is
w =
 t
t0
p dt =
 t
t0
vi dt (1.9)
Energy is the capacity to do work, measured in joules (J).
The electric power utility companies measure energy in watt-hours (Wh),
where
1 Wh = 3,600 J
E X A M P L E 1 . 4
An energy source forces a constant current of 2 A for 10 s to flow through
a lightbulb. If 2.3 kJ is given off in the form of light and heat energy,
calculate the voltage drop across the bulb.
12 PART 1 DC Circuits
Solution:
The total charge is
q = it = 2 × 10 = 20 C
The voltage drop is
v =
w
q
=
2.3 × 103
20
= 115 V
P R A C T I C E P R O B L E M 1 . 4
To move charge q from point a to point b requires −30 J. Find the voltage
drop vab if: (a) q = 2 C, (b) q = −6 C .
Answer: (a) −15 V, (b) 5 V.
E X A M P L E 1 . 5
Find the power delivered to an element at t = 3 ms if the current entering
its positive terminal is
i = 5 cos 60πt A
and the voltage is: (a) v = 3i, (b) v = 3 di/dt.
Solution:
(a) The voltage is v = 3i = 15 cos 60πt; hence, the power is
p = vi = 75 cos2
60πt W
At t = 3 ms,
p = 75 cos2
(60π × 3 × 10−3
) = 75 cos2
0.18π = 53.48 W
(b) We find the voltage and the power as
v = 3
di
dt
= 3(−60π)5 sin 60πt = −900π sin 60πt V
p = vi = −4500π sin 60πt cos 60πt W
At t = 3 ms,
p = −4500π sin 0.18π cos 0.18π W
= −14137.167 sin 32.4◦
cos 32.4◦
= −6.396 kW
P R A C T I C E P R O B L E M 1 . 5
Find the power delivered to the element in Example 1.5 at t = 5 ms if
the current remains the same but the voltage is: (a) v = 2i V, (b) v =

10 + 5
 t
0
i dt

V.
Answer: (a) 17.27 W, (b) 29.7 W.
CHAPTER 1 Basic Concepts 13
E X A M P L E 1 . 6
How much energy does a 100-W electric bulb consume in two hours?
Solution:
w = pt = 100 (W) × 2 (h) × 60 (min/h) × 60 (s/min)
= 720,000 J = 720 kJ
This is the same as
w = pt = 100 W × 2 h = 200 Wh
P R A C T I C E P R O B L E M 1 . 6
A stove element draws 15 A when connected to a 120-V line. How long
does it take to consume 30 kJ?
Answer: 16.67 s.
1.6 CIRCUIT ELEMENTS
As we discussed in Section 1.1, an element is the basic building block of
a circuit. An electric circuit is simply an interconnection of the elements.
Circuit analysis is the process of determining voltages across (or the
currents through) the elements of the circuit.
There are two types of elements found in electric circuits: passive
elements and active elements. An active element is capable of generating
energy while a passive element is not. Examples of passive elements
are resistors, capacitors, and inductors. Typical active elements include
generators, batteries, and operational amplifiers. Our aim in this section
is to gain familiarity with some important active elements.
The most important active elements are voltage or current sources
that generally deliver power to the circuit connected to them. There are
two kinds of sources: independent and dependent sources.
An ideal independent source is an active element that provides a specified voltage
or current that is completely independent of other circuit variables.
V
(b)
+
−
v
(a)
+
−
Figure1.11 Symbols for
independent voltage sources:
(a) used for constant or
time-varying voltage, (b) used for
constant voltage (dc).
In other words, an ideal independent voltage source delivers to the circuit
whatever current is necessary to maintain its terminal voltage. Physical
sources such as batteries and generators may be regarded as approxima-
tions to ideal voltage sources. Figure 1.11 shows the symbols for inde-
pendent voltage sources. Notice that both symbols in Fig. 1.11(a) and (b)
can be used to represent a dc voltage source, but only the symbol in Fig.
1.11(a) can be used for a time-varying voltage source. Similarly, an ideal
independent current source is an active element that provides a specified
current completely independent of the voltage across the source. That is,
the current source delivers to the circuit whatever voltage is necessary to
14 PART 1 DC Circuits
maintain the designated current. The symbol for an independent current
source is displayed in Fig. 1.12, where the arrow indicates the direction
of current i.
i
Figure1.12 Symbol
for independent
current source.
An ideal dependent (or controlled) source is an active element in which the source
quantity is controlled by another voltage or current.
Dependent sources are usually designated by diamond-shaped symbols,
as shown in Fig. 1.13. Since the control of the dependent source is ac-
hieved by a voltage or current of some other element in the circuit, and
the source can be voltage or current, it follows that there are four possible
types of dependent sources, namely:
1. A voltage-controlled voltage source (VCVS).
2. A current-controlled voltage source (CCVS).
3. A voltage-controlled current source (VCCS).
4. A current-controlled current source (CCCS).
(a) (b)
v +
− i
Figure1.13 Symbols for:
(a) dependent voltage source,
(b) dependent current source.
Dependent sources are useful in modeling elements such as transistors,
operational amplifiers and integrated circuits. An example of a current-
controlled voltage source is shown on the right-hand side of Fig. 1.14,
where the voltage 10i of the voltage source depends on the current i
through element C. Students might be surprised that the value of the
dependent voltage source is 10i V (and not 10i A) because it is a voltage
source. The key idea to keep in mind is that a voltage source comes
with polarities (+ −) in its symbol, while a current source comes with
an arrow, irrespective of what it depends on.
i
A B
C 10i
5 V
+
−
+
−
Figure1.14 The source on the right-hand
side is a current-controlled voltage source.
It should be noted that an ideal voltage source (dependent or in-
dependent) will produce any current required to ensure that the terminal
voltage is as stated, whereas an ideal current source will produce the
necessary voltage to ensure the stated current flow. Thus an ideal source
could in theory supply an infinite amount of energy. It should also be
noted that not only do sources supply power to a circuit, they can absorb
power from a circuit too. For a voltage source, we know the voltage but
not the current supplied or drawn by it. By the same token, we know the
current supplied by a current source but not the voltage across it.
E X A M P L E 1 . 7
Calculate the power supplied or absorbed by each element in Fig. 1.15.
p2
p3
I = 5 A
20 V
6 A
8 V 0.2I
12 V
+
−
+
−
+ −
p1 p4
Figure1.15 For Example 1.7.
Solution:
We apply the sign convention for power shown in Figs. 1.8 and 1.9. For
p1, the 5-A current is out of the positive terminal (or into the negative
terminal); hence,
p1 = 20(−5) = −100 W Supplied power
For p2 and p3, the current flows into the positive terminal of the element
in each case.
CHAPTER 1 Basic Concepts 15
p2 = 12(5) = 60 W Absorbed power
p3 = 8(6) = 48 W Absorbed power
For p4, we should note that the voltage is 8 V (positive at the top), the same
as the voltage for p3, since both the passive element and the dependent
source are connected to the same terminals. (Remember that voltage is
always measured across an element in a circuit.) Since the current flows
out of the positive terminal,
p4 = 8(−0.2I) = 8(−0.2 × 5) = −8 W Supplied power
We should observe that the 20-V independent voltage source and 0.2I
dependent current source are supplying power to the rest of the network,
while the two passive elements are absorbing power. Also,
p1 + p2 + p3 + p4 = −100 + 60 + 48 − 8 = 0
In agreement with Eq. (1.8), the total power supplied equals the total
power absorbed.
P R A C T I C E P R O B L E M 1 . 7
Compute the power absorbed or supplied by each component of the circuit
in Fig. 1.16.
8 A
5 V 3 V
2 V
3 A
I = 5 A
0.6I
+
−
+ −
+
−
+
−
+
−
p2
p1 p3 p4
Figure1.16 For Practice Prob. 1.7.
Answer: p1 = −40 W, p2 = 16 W, p3 = 9 W, p4 = 15 W.
†1.7 APPLICATIONS2
In this section, we will consider two practical applications of the concepts
developed in this chapter. The first one deals with the TV picture tube
and the other with how electric utilities determine your electric bill.
1.7.1 TV Picture Tube
One important application of the motion of electrons is found in both
the transmission and reception of TV signals. At the transmission end, a
TV camera reduces a scene from an optical image to an electrical signal.
Scanning is accomplished with a thin beam of electrons in an iconoscope
camera tube.
At the receiving end, the image is reconstructed by using a cath-
ode-ray tube (CRT) located in the TV receiver.3
The CRT is depicted in
2The dagger sign preceding a section heading indicates a section that may be skipped,
explained briefly, or assigned as homework.
3Modern TV tubes use a different technology.
16 PART 1 DC Circuits
Fig. 1.17. Unlike the iconoscope tube, which produces an electron beam
of constant intensity, the CRT beam varies in intensity according to the
incoming signal. The electron gun, maintained at a high potential, fires
the electron beam. The beam passes through two sets of plates for vertical
and horizontal deflections so that the spot on the screen where the beam
strikes can move right and left and up and down. When the electron beam
strikes the fluorescent screen, it gives off light at that spot. Thus the beam
can be made to “paint” a picture on the TV screen.
Vertical
deflection
plates
Horizontal
deflection
plates
Electron
trajectory
Bright spot on
fluorescent screen
Electron gun
Figure1.17 Cathode-ray tube.
(Source: D. E. Tilley, Contemporary College Physics [Menlo Park, CA:
Benjamin/Cummings, 1979], p. 319.)
E X A M P L E 1 . 8
The electron beam in a TV picture tube carries 1015
electrons per second.
As a design engineer, determine the voltage Vo needed to accelerate the
electron beam to achieve 4 W.
Solution:
The charge on an electron is
e = −1.6 × 10−19
C
If the number of electrons is n, then q = ne and
i =
dq
dt
= e
dn
dt
= (−1.6 × 10−19
)(1015
) = −1.6 × 10−4
A
The negative sign indicates that the electron flows in a direction opposite
to electron flow as shown in Fig. 1.18, which is a simplified diagram of
the CRT for the case when the vertical deflection plates carry no charge.
The beam power is
p = Voi or Vo =
p
i
=
4
1.6 × 10−4
= 25,000 V
Thus the required voltage is 25 kV.
i
q
Vo
Figure1.18 A simplified diagram of the
cathode-ray tube; for Example 1.8.
P R A C T I C E P R O B L E M 1 . 8
If an electron beam in a TV picture tube carries 1013
electrons/second and
is passing through plates maintained at a potential difference of 30 kV,
calculate the power in the beam.
Answer: 48 mW.
CHAPTER 1 Basic Concepts 17
1.7.2 Electricity Bills
The second application deals with how an electric utility company charges
their customers. The cost of electricity depends upon the amount of
energy consumed in kilowatt-hours (kWh). (Other factors that affect the
cost include demand and power factors; we will ignore these for now.)
However, even if a consumer uses no energy at all, there is a minimum
service charge the customer must pay because it costs money to stay
connected to the power line. As energy consumption increases, the cost
per kWh drops. It is interesting to note the average monthly consumption
of household appliances for a family of five, shown in Table 1.3.
TABLE 1.3 Typical average monthly consumption of household
appliances.
Appliance kWh consumed Appliance kWh consumed
Water heater 500 Washing machine 120
Freezer 100 Stove 100
Lighting 100 Dryer 80
Dishwasher 35 Microwave oven 25
Electric iron 15 Personal computer 12
TV 10 Radio 8
Toaster 4 Clock 2
E X A M P L E 1 . 9
A homeowner consumes 3,300 kWh in January. Determine the electricity
bill for the month using the following residential rate schedule:
Base monthly charge of $12.00.
First 100 kWh per month at 16 cents/kWh.
Next 200 kWh per month at 10 cents/kWh.
Over 200 kWh per month at 6 cents/kWh.
Solution:
We calculate the electricity bill as follows.
Base monthly charge = $12.00
First 100 kWh @ $0.16/kWh = $16.00
Next 200 kWh @ $0.10/kWh = $20.00
Remaining 100 kWh @ $0.06/kWh = $6.00
Total Charge = $54.00
Average cost =
$54
100 + 200 + 100
= 13.5 cents/kWh
18 PART 1 DC Circuits
P R A C T I C E P R O B L E M 1 . 9
Referring to the residential rate schedule in Example 1.9, calculate the
average cost per kWh if only 400 kWh are consumed in July when the
family is on vacation most of the time.
Answer: 13.5 cents/kWh.
†1.8 PROBLEM SOLVING
Although the problems to be solved during one’s career will vary in
complexity and magnitude, the basic principles to be followed remain
the same. The process outlined here is the one developed by the authors
over many years of problem solving with students, for the solution of
engineering problems in industry, and for problem solving in research.
We will list the steps simply and then elaborate on them.
1. Carefully Define the problem.
2. Present everything you know about the problem.
3. Establish a set of Alternative solutions and determine the one
that promises the greatest likelihood of success.
4. Attempt a problem solution.
5. Evaluate the solution and check for accuracy.
6. Has the problem been solved Satisfactorily? If so, present the
solution; if not, then return to step 3 and continue through the
process again.
1. Carefully Define the problem. This may be the most important
part of the process, because it becomes the foundation for all the rest of the
steps. In general, the presentation of engineering problems is somewhat
incomplete. You must do all you can to make sure you understand the
problem as thoroughly as the presenter of the problem understands it.
Time spent at this point clearly identifying the problem will save you
considerable time and frustration later. As a student, you can clarify a
problem statement in a textbook by asking your professor to help you
understand it better. A problem presented to you in industry may require
that you consult several individuals. At this step, it is important to develop
questionsthatneedtobeaddressedbeforecontinuingthesolutionprocess.
If you have such questions, you need to consult with the appropriate
individuals or resources to obtain the answers to those questions. With
those answers, you can now refine the problem, and use that refinement
as the problem statement for the rest of the solution process.
2. Present everything you know about the problem. You are now
ready to write down everything you know about the problem and its
possible solutions. This important step will save you time and frustration
later.
CHAPTER 1 Basic Concepts 19
3. Establish a set of Alternative solutions and determine the one
that promises the greatest likelihood of success. Almost every problem
will have a number of possible paths that can lead to a solution. It is highly
desirable to identify as many of those paths as possible. At this point, you
also need to determine what tools are available to you, such as Matlab
and other software packages that can greatly reduce effort and increase
accuracy. Again, we want to stress that time spent carefully defining the
problem and investigating alternative approaches to its solution will pay
big dividends later. Evaluating the alternatives and determining which
promises the greatest likelihood of success may be difficult but will be
well worth the effort. Document this process well since you will want to
come back to it if the first approach does not work.
4. Attempt a problem solution. Now is the time to actually begin
solving the problem. The process you follow must be well documented
in order to present a detailed solution if successful, and to evaluate the
process if you are not successful. This detailed evaluation may lead to
corrections that can then lead to a successful solution. It can also lead to
new alternatives to try. Many times, it is wise to fully set up a solution
before putting numbers into equations. This will help in checking your
results.
5. Evaluate the solution and check for accuracy. You now thor-
oughly evaluate what you have accomplished. Decide if you have an
acceptable solution, one that you want to present to your team, boss, or
professor.
6. Has the problem been solved Satisfactorily? If so, present the
solution; if not, then return to step 3 and continue through the process
again. Now you need to present your solution or try another alternative.
At this point, presenting your solution may bring closure to the process.
Often, however, presentation of a solution leads to further refinement of
the problem definition, and the process continues. Following this process
will eventually lead to a satisfactory conclusion.
Now let us look at this process for a student taking an electrical
and computer engineering foundations course. (The basic process also
applies to almost every engineering course.) Keep in mind that although
the steps have been simplified to apply to academic types of problems,
the process as stated always needs to be followed. We consider a simple
example.
Assume that we have been given the following circuit. The instruc-
tor asks us to solve for the current flowing through the 8-ohm resistor.
2 Ω 4 Ω
8 Ω
5 V 3 V
+
−
1. Carefully Define the problem. This is only a simple example,
but we can already see that we do not know the polarity on the 3-V
source. We have the following options. We can ask the professor what
20 PART 1 DC Circuits
the polarity should be. If we cannot ask, then we need to make a decision
on what to do next. If we have time to work the problem both ways, we
can solve for the current when the 3-V source is plus on top and then plus
on the bottom. If we do not have the time to work it both ways, assume a
polarity and then carefully document your decision. Let us assume that
the professor tells us that the source is plus on the bottom.
2. Present everything you know about the problem. Presenting all
that we know about the problem involves labeling the circuit clearly so
that we define what we seek.
Given the following circuit, solve for i8.
2 Ω 4 Ω
8 Ω
5 V 3 V
+
− +
−
i8Ω
Wenowcheckwiththeprofessor, ifreasonable, toseeiftheproblem
is properly defined.
3. Establish a set of Alternative solutions and determine the one
that promises the greatest likelihood of success. There are essentially
three techniques that can be used to solve this problem. Later in the text
you will see that you can use circuit analysis (using Kirchoff’s laws and
Ohm’s law), nodal analysis, and mesh analysis.
To solve for i8 using circuit analysis will eventually lead to a
solution, but it will likely take more work than either nodal or mesh
analysis. To solve for i8 using mesh analysis will require writing two
simultaneous equations to find the two loop currents indicated in the
following circuit. Using nodal analysis requires solving for only one
unknown. This is the easiest approach.
2 Ω 4 Ω
8 Ω
5 V 3 V
+
− +
−
i2
i1 i3
+ −
v1
+ −
v3
+
−
v2
Loop 1 Loop 2
v1
Therefore, we will solve for i8 using nodal analysis.
4. Attempt a problem solution. We first write down all of the
equations we will need in order to find i8.
i8 = i2, i2 =
v1
8
, i8 =
v1
8
v1 − 5
2
+
v1 − 0
8
+
v1 + 3
4
= 0
CHAPTER 1 Basic Concepts 21
Now we can solve for v1.
8

v1 − 5
2
+
v1 − 0
8
+
v1 + 3
4

= 0
leads to (4v1 − 20) + (v1) + (2v1 + 6) = 0
7v1 = +14, v1 = +2 V, i8 =
v1
8
=
2
8
= 0.25 A
5. Evaluate the solution and check for accuracy. We can now use
Kirchoff’s voltage law to check the results.
i1 =
v1 − 5
2
=
2 − 5
2
= −
3
2
= −1.5 A
i2 = i8 = 0.25 A
i3 =
v1 + 3
4
=
2 + 3
4
=
5
4
= 1.25 A
i1 + i2 + i3 = −1.5 + 0.25 + 1.25 = 0 (Checks.)
Applying KVL to loop 1,
−5 + v1 + v2 = −5 + (−i1 × 2) + (i2 × 8)
= −5 + (−(−1.5)2) + (0.25 × 8)
= −5 + 3 + 2 = 0 (Checks.)
Applying KVL to loop 2,
−v2 + v3 − 3 = −(i2 × 8) + (i3 × 4) − 3
= −(0.25 × 8) + (1.25 × 4) − 3
= −2 + 5 − 3 = 0 (Checks.)
So we now have a very high degree of confidence in the accuracy
of our answer.
6. Has the problem been solved Satisfactorily? If so, present the
solution; if not, then return to step 3 and continue through the process
again. This problem has been solved satisfactorily.
The current through the 8-ohm resistor is 0.25 amp flowing
down through the 8-ohm resistor.
1.9 SUMMARY
1. An electric circuit consists of electrical elements connected
together.
2. The International System of Units (SI) is the international mea-
surement language, which enables engineers to communicate their
results. From the six principal units, the units of other physical
quantities can be derived.
3. Current is the rate of charge flow.
i =
dq
dt
22 PART 1 DC Circuits
4. Voltage is the energy required to move 1 C of charge through an
element.
v =
dw
dq
5. Power is the energy supplied or absorbed per unit time. It is also the
product of voltage and current.
p =
dw
dt
= vi
6. According to the passive sign convention, power assumes a positive
sign when the current enters the positive polarity of the voltage
across an element.
7. An ideal voltage source produces a specific potential difference
across its terminals regardless of what is connected to it. An ideal
current source produces a specific current through its terminals
regardless of what is connected to it.
8. Voltage and current sources can be dependent or independent. A
dependent source is one whose value depends on some other circuit
variable.
9. Two areas of application of the concepts covered in this chapter are
the TV picture tube and electricity billing procedure.
REVIEW QUESTIONS
1.1 One millivolt is one millionth of a volt.
(a) True (b) False
1.2 The prefix micro stands for:
(a) 106
(b) 103
(c) 10−3
(d) 10−6
1.3 The voltage 2,000,000 V can be expressed in powers
of 10 as:
(a) 2 mV (b) 2 kV (c) 2 MV (d) 2 GV
1.4 A charge of 2 C flowing past a given point each
second is a current of 2 A.
(a) True (b) False
1.5 A 4-A current charging a dielectric material will
accumulate a charge of 24 C after 6 s.
(a) True (b) False
1.6 The unit of current is:
(a) Coulomb (b) Ampere
(c) Volt (d) Joule
1.7 Voltage is measured in:
(a) Watts (b) Amperes
(c) Volts (d) Joules per second
1.8 The voltage across a 1.1 kW toaster that produces a
current of 10 A is:
(a) 11 kV (b) 1100 V (c) 110 V (d) 11 V
1.9 Which of these is not an electrical quantity?
(a) charge (b) time (c) voltage
(d) current (e) power
1.10 The dependent source in Fig. 1.19 is:
(a) voltage-controlled current source
(b) voltage-controlled voltage source
(c) current-controlled voltage source
(d) current-controlled current source
vs
io
6io
+
−
Figure 1.19 For Review Question 1.10.
Answers: 1.1b, 1.2d, 1.3c, 1.4a, 1.5a, 1.6b, 1.7c, 1.8c, 1.9b, 1.10d.
CHAPTER 1 Basic Concepts 23
PROBLEMS
Section 1.3 Charge and Current
1.1 How many coulombs are represented by these
amounts of electrons:
(a) 6.482 × 1017
(b) 1.24 × 1018
(c) 2.46 × 1019
(d) 1.628 × 1020
1.2 Find the current flowing through an element if the
charge flow is given by:
(a) q(t) = (t + 2) mC
(b) q(t) = (5t2
+ 4t − 3) C
(c) q(t) = 10e−4t
pC
(d) q(t) = 20 cos 50πt nC
(e) q(t) = 5e−2t
sin 100t µC
1.3 Find the charge q(t) flowing through a device if the
current is:
(a) i(t) = 3 A, q(0) = 1 C
(b) i(t) = (2t + 5) mA, q(0) = 0
(c) i(t) = 20 cos(10t + π/6) µA, q(0) = 2 µC
(d) i(t) = 10e−30t
sin 40t A, q(0) = 0
1.4 The current flowing through a device is
i(t) = 5 sin 6πt A. Calculate the total charge flow
through the device from t = 0 to t = 10 ms.
1.5 Determine the total charge flowing into an element
for 0  t  2 s when the current entering its
positive terminal is i(t) = e−2t
mA.
1.6 The charge entering a certain element is shown in
Fig. 1.20. Find the current at:
(a) t = 1 ms (b) t = 6 ms (c) t = 10 ms
q(t) (mC)
t (ms)
0 2 4 6 8 10 12
80
Figure 1.20 For Prob. 1.6.
1.7 The charge flowing in a wire is plotted in Fig. 1.21.
Sketch the corresponding current.
q (C)
t (s)
50
−50
0
2 4 6 8
Figure 1.21 For Prob. 1.7.
1.8 The current flowing past a point in a device is shown
in Fig. 1.22. Calculate the total charge through the
point.
i (mA)
t (ms)
0 1 2
10
Figure 1.22 For Prob. 1.8.
1.9 The current through an element is shown in Fig.
1.23. Determine the total charge that passed through
the element at:
(a) t = 1 s (b) t = 3 s (c) t = 5 s
0 1 2 3 4 5
5
10
i (A)
t (s)
Figure 1.23 For Prob. 1.9.
Sections 1.4 and 1.5 Voltage, Power, and
Energy
1.10 A certain electrical element draws the current
i(t) = 10 cos 4t A at a voltage v(t) = 120 cos 4t V.
Find the energy absorbed by the element in 2 s.
1.11 The voltage v across a device and the current i
through it are
v(t) = 5 cos 2t V, i(t) = 10(1 − e−0.5t
) A
Calculate:
(a) the total charge in the device at t = 1 s
(b) the power consumed by the device at t = 1 s.
24 PART 1 DC Circuits
1.12 The current entering the positive terminal of a
device is i(t) = 3e−2t
A and the voltage across the
device is v(t) = 5 di/dt V.
(a) Find the charge delivered to the device between
t = 0 and t = 2 s.
(b) Calculate the power absorbed.
(c) Determine the energy absorbed in 3 s.
1.13 Figure 1.24 shows the current through and the
voltage across a device. Find the total energy
absorbed by the device for the period of 0  t  4 s.
0 2 4
50
i (mA)
t (s)
10
0 1 3 4
v (V)
t (s)
Figure 1.24 For Prob. 1.13.
Section 1.6 Circuit Elements
1.14 Figure 1.25 shows a circuit with five elements. If
p1 = −205 W, p2 = 60 W, p4 = 45 W, p5 = 30 W,
calculate the power p3 received or delivered by
element 3.
3
1
2 4
5
Figure 1.25 For Prob. 1.14.
1.15 Find the power absorbed by each of the elements in
Fig. 1.26.
I = 10 A 10 V
30 V
8 V
14 A
20 V 12 V
4 A
0.4I
+
−
+ −
+
−
+
−
+ −
p2
p1 p3
p4
p5
Figure 1.26 For Prob. 1.15.
1.16 Determine Io in the circuit of Fig. 1.27.
5 A
3 A
20 V 20 V 8 V
12 V
3 A
Io
+
−
+
−
+
−
+ −
Figure 1.27 For Prob. 1.16.
1.17 Find Vo in the circuit of Fig. 1.28.
6 A
6 A
1 A
3 A
3 A
Vo 5Io
Io = 2 A
28 V
12 V
+
−
+ −
28 V
+ −
+ −
30 V
–
+
+
−
Figure 1.28 For Prob. 1.17.
Section 1.7 Applications
1.18 It takes eight photons to strike the surface of a
photodetector in order to emit one electron. If
4 × 1011
photons/second strike the surface of the
photodetector, calculate the amount of current flow.
1.19 Find the power rating of the following electrical
appliances in your household:
(a) Lightbulb (b) Radio set
(c) TV set (d) Refrigerator
(e) Personal computer (f) PC printer
(g) Microwave oven (h) Blender
1.20 A 1.5-kW electric heater is connected to a 120-V
source.
(a) How much current does the heater draw?
(b) If the heater is on for 45 minutes, how much
energy is consumed in kilowatt-hours (kWh)?
(c) Calculate the cost of operating the heater for 45
minutes if energy costs 10 cents/kWh.
1.21 A 1.2-kW toaster takes roughly 4 minutes to heat
four slices of bread. Find the cost of operating the
toaster once per day for 1 month (30 days). Assume
energy costs 9 cents/kWh.
CHAPTER 1 Basic Concepts 25
1.22 A flashlight battery has a rating of 0.8 ampere-hours
(Ah) and a lifetime of 10 hours.
(a) How much current can it deliver?
(b) How much power can it give if its terminal
voltage is 6 V?
(c) How much energy is stored in the battery in
kWh?
1.23 A constant current of 3 A for 4 hours is required to
charge an automotive battery. If the terminal voltage
is 10 + t/2 V, where t is in hours,
(a) how much charge is transported as a result of the
charging?
(b) how much energy is expended?
(c) how much does the charging cost? Assume
electricity costs 9 cents/kWh.
1.24 A 30-W incandescent lamp is connected to a 120-V
source and is left burning continuously in an
otherwise dark staircase. Determine:
(a) the current through the lamp,
(b) the cost of operating the light for one non-leap
year if electricity costs 12 cents per kWh.
1.25 An electric stove with four burners and an oven is
used in preparing a meal as follows.
Burner 1: 20 minutes Burner 2: 40 minutes
Burner 3: 15 minutes Burner 4: 45 minutes
Oven: 30 minutes
If each burner is rated at 1.2 kW and the oven at
1.8 kW, and electricity costs 12 cents per kWh,
calculate the cost of electricity used in preparing the
meal.
1.26 PECO (the electric power company in Philadelphia)
charged a consumer $34.24 one month for using
215 kWh. If the basic service charge is $5.10, how
much did PECO charge per kWh?
COMPREHENSIVE PROBLEMS
1.27 A telephone wire has a current of 20 µA flowing
through it. How long does it take for a charge of
15 C to pass through the wire?
1.28 A lightning bolt carried a current of 2 kA and lasted
for 3 ms. How many coulombs of charge were
contained in the lightning bolt?
1.29 The power consumption for a certain household for
a day is shown in Fig. 1.29. Determine:
(a) the total energy consumed in kWh
(b) the average power per hour.
12 2 4 6 8 10 12
noon
2 4 6 8 10 12
p(t)
t
(hour)
400 W
1000 W
200 W
1200 W
400 W
Figure 1.29 For Prob. 1.29.
1.30 The graph in Fig. 1.30 represents the power drawn
by an industrial plant between 8:00 and 8:30 A.M.
Calculate the total energy in MWh consumed by the
plant.
8.00 8.05 8.10 8.15 8.20 8.25 8.30
5
4
3
8
p (MW)
t
Figure 1.30 For Prob. 1.30.
1.31 A battery may be rated in ampere-hours (Ah). An
lead-acid battery is rated at 160 Ah.
(a) What is the maximum current it can supply for
40 h?
(b) How many days will it last if it is discharged at
1 mA?
1.32 How much work is done by a 12-V automobile
battery in moving 5 × 1020
electrons from the
positive terminal to the negative terminal?
1.33 How much energy does a 10-hp motor deliver in 30
minutes? Assume that 1 horsepower = 746 W.
1.34 A 2-kW electric iron is connected to a 120-V line.
Calculate the current drawn by the iron.
27
C H A P T E R
BASIC LAWS
2
The chessboard is the world, the pieces are the phenomena of the universe,
the rules of the game are what we call the laws of Nature. The player
on the other side is hidden from us, we know that his play is always fair,
just, and patient. But also we know, to our cost, that he never overlooks
a mistake, or makes the smallest allowance for ignorance.
— Thomas Henry Huxley
Historical Profiles
Georg Simon Ohm (1787–1854), a German physicist, in 1826 experimentally deter-
mined the most basic law relating voltage and current for a resistor. Ohm’s work was
initially denied by critics.
Born of humble beginnings in Erlangen, Bavaria, Ohm threw himself into
electrical research. His efforts resulted in his famous law. He was awarded the Copley
Medal in 1841 by the Royal Society of London. In 1849, he was given the Professor
of Physics chair by the University of Munich. To honor him, the unit of resistance was
named the ohm.
Gustav Robert Kirchhoff (1824–1887), a German physicist, stated two basic laws
in 1847 concerning the relationship between the currents and voltages in an electrical
network. Kirchhoff’s laws, along with Ohm’s law, form the basis of circuit theory.
Born the son of a lawyer in Konigsberg, East Prussia, Kirchhoff entered
the University of Konigsberg at age 18 and later became a lecturer in Berlin. His
collaborative work in spectroscopy with German chemist Robert Bunsen led to the
discovery of cesium in 1860 and rubidium in 1861. Kirchhoff was also credited with
the Kirchhoff law of radiation. Thus Kirchhoff is famous among engineers, chemists,
and physicists.
28 PART 1 DC Circuits
2.1 INTRODUCTION
Chapter 1 introduced basic concepts such as current, voltage, and power
in an electric circuit. To actually determine the values of these variables
in a given circuit requires that we understand some fundamental laws that
govern electric circuits. These laws, known as Ohm’s law and Kirchhoff’s
laws, form the foundation upon which electric circuit analysis is built.
In this chapter, in addition to these laws, we shall discuss some
techniques commonly applied in circuit design and analysis. These tech-
niques include combining resistors in series or parallel, voltage division,
current division, and delta-to-wye and wye-to-delta transformations. The
application of these laws and techniques will be restricted to resistive cir-
cuits in this chapter. We will finally apply the laws and techniques to
real-life problems of electrical lighting and the design of dc meters.
2.2 OHM’S LAW
Materials in general have a characteristic behavior of resisting the flow
of electric charge. This physical property, or ability to resist current, is
known as resistance and is represented by the symbol R. The resistance
of any material with a uniform cross-sectional area A depends on A and
its length , as shown in Fig. 2.1(a). In mathematical form,
R = ρ

A
(2.1)
where ρ is known as the resistivity of the material in ohm-meters. Good
conductors, such as copper and aluminum, have low resistivities, while
insulators, such as mica and paper, have high resistivities. Table 2.1
presents the values of ρ for some common materials and shows which
materials are used for conductors, insulators, and semiconductors.
l
Cross-sectional
area A
(a)
Material with
resistivity r
v R
i
+
−
(b)
Figure 2.1 (a) Resistor, (b) Circuit symbol
for resistance.
TABLE 2.1 Resistivities of common materials.
Material Resistivity (·m) Usage
Silver 1.64 × 10−8
Conductor
Copper 1.72 × 10−8
Conductor
Aluminum 2.8 × 10−8
Conductor
Gold 2.45 × 10−8
Conductor
Carbon 4 × 10−5
Semiconductor
Germanium 47 × 10−2
Semiconductor
Silicon 6.4 × 102
Semiconductor
Paper 1010
Insulator
Mica 5 × 1011
Insulator
Glass 1012
Insulator
Teflon 3 × 1012
Insulator
The circuit element used to model the current-resisting behavior of
a material is the resistor. For the purpose of constructing circuits, resistors
are usually made from metallic alloys and carbon compounds. The circuit
CHAPTER 2 Basic Laws 29
symbol for the resistor is shown in Fig. 2.1(b), where R stands for the
resistance of the resistor. The resistor is the simplest passive element.
Georg Simon Ohm (1787–1854), a German physicist, is credited
with finding the relationship between current and voltage for a resistor.
This relationship is known as Ohm’s law.
Ohm’s law states that the voltage v across a resistor is directly proportional
to the current i flowing through the resistor.
That is,
v ∝ i (2.2)
Ohm defined the constant of proportionality for a resistor to be the resis-
tance, R. (The resistance is a material property which can change if the
internal or external conditions of the element are altered, e.g., if there are
changes in the temperature.) Thus, Eq. (2.2) becomes
v = iR (2.3)
which is the mathematical form of Ohm’s law. R in Eq. (2.3) is measured
in the unit of ohms, designated . Thus,
The resistance R of an element denotes its ability to resist the flow
of electric current; it is measured in ohms ().
We may deduce from Eq. (2.3) that
R =
v
i
(2.4)
so that
1  = 1 V/A
To apply Ohm’s law as stated in Eq. (2.3), we must pay careful
attention to the current direction and voltage polarity. The direction of
current i and the polarity of voltage v must conform with the passive sign
convention, as shown in Fig. 2.1(b). This implies that current flows from
a higher potential to a lower potential in order for v = iR. If current
flows from a lower potential to a higher potential, v = −iR.
(a)
(b)
R = 0
i
R = ∞
i = 0
v = 0
+
−
v
+
−
Figure 2.2 (a) Short circuit (R = 0),
(b)Opencircuit(R = ∞).
Since the value of R can range from zero to infinity, it is important
that we consider the two extreme possible values of R. An element with
R = 0 is called a short circuit, as shown in Fig. 2.2(a). For a short circuit,
v = iR = 0 (2.5)
showing that the voltage is zero but the current could be anything. In
practice, a short circuit is usually a connecting wire assumed to be a
perfect conductor. Thus,
30 PART 1 DC Circuits
A short circuit is a circuit element with resistance approaching zero.
Similarly, an element with R = ∞ is known as an open circuit, as shown
in Fig. 2.2(b). For an open circuit,
i = lim
R→∞
v
R
= 0 (2.6)
indicating that the current is zero though the voltage could be anything.
Thus,
An open circuit is a circuit element with resistance approaching infinity.
(a)
(b)
Figure2.3 Fixed resistors: (a) wire-
wound type, (b) carbon film type.
(Courtesy of Tech America.)
(a) (b)
Figure2.4 Circuit symbol for: (a) a variable
resistor in general, (b) a potentiometer.
A resistor is either fixed or variable. Most resistors are of the fixed
type, meaning their resistance remains constant. The two common types
of fixed resistors (wirewound and composition) are shown in Fig. 2.3.
The composition resistors are used when large resistance is needed. The
circuit symbol in Fig. 2.1(b) is for a fixed resistor. Variable resistors
have adjustable resistance. The symbol for a variable resistor is shown
in Fig. 2.4(a). A common variable resistor is known as a potentiometer
or pot for short, with the symbol shown in Fig. 2.4(b). The pot is a
three-terminal element with a sliding contact or wiper. By sliding the
wiper, the resistances between the wiper terminal and the fixed terminals
vary. Like fixed resistors, variable resistors can either be of wirewound or
composition type, as shown in Fig. 2.5. Although resistors like those in
Figs.2.3and2.5areusedincircuitdesigns, todaymostcircuitcomponents
including resistors are either surface mounted or integrated, as typically
shown in Fig. 2.6.
(a) (b)
Figure2.5 Variable resistors: (a) composition type, (b) slider pot.
(Courtesy of Tech America.)
Figure2.6 Resistors in a thick-film circuit.
(Source: G. Daryanani, Principles of Active
Network Synthesis and Design [New York:
John Wiley, 1976], p. 461c.)
It should be pointed out that not all resistors obey Ohm’s law. A
resistor that obeys Ohm’s law is known as a linear resistor. It has a con-
stant resistance and thus its current-voltage characteristic is as illustrated
in Fig. 2.7(a): its i-v graph is a straight line passing through the ori-
gin. A nonlinear resistor does not obey Ohm’s law. Its resistance varies
with current and its i-v characteristic is typically shown in Fig. 2.7(b).
CHAPTER 2 Basic Laws 31
Examples of devices with nonlinear resistance are the lightbulb and the
diode. Although all practical resistors may exhibit nonlinear behavior
under certain conditions, we will assume in this book that all elements
actually designated as resistors are linear.
Slope = R
(a)
v
i
Slope = R
(b)
v
i
Figure 2.7 The i-v characteristic of:
(a) a linear resistor,
(b) a nonlinear resistor.
A useful quantity in circuit analysis is the reciprocal of resistance
R, known as conductance and denoted by G:
G =
1
R
=
i
v
(2.7)
The conductance is a measure of how well an element will conduct
electric current. The unit of conductance is the mho (ohm spelled back-
ward) or reciprocal ohm, with symbol

, the inverted omega. Although
engineers often use the mhos, in this book we prefer to use the siemens
(S), the SI unit of conductance:
1 S = 1

= 1 A/V (2.8)
Thus,
Conductance is the ability of an element to conduct electric current; it is
measured in mhos (

) or siemens (S).
The same resistance can be expressed in ohms or siemens. For
example, 10  is the same as 0.1 S. From Eq. (2.7), we may write
i = Gv (2.9)
The power dissipated by a resistor can be expressed in terms of R.
Using Eqs. (1.7) and (2.3),
p = vi = i2
R =
v2
R
(2.10)
The power dissipated by a resistor may also be expressed in terms of G
as
p = vi = v2
G =
i2
G
(2.11)
We should note two things from Eqs. (2.10) and (2.11):
1. The power dissipated in a resistor is a nonlinear function of
either current or voltage.
2. Since R and G are positive quantities, the power dissipated in
a resistor is always positive. Thus, a resistor always absorbs
power from the circuit. This confirms the idea that a resistor is
a passive element, incapable of generating energy.
E X A M P L E 2 . 1
An electric iron draws 2 A at 120 V. Find its resistance.
32 PART 1 DC Circuits
Solution:
From Ohm’s law,
R =
v
i
=
120
2
= 60 
P R A C T I C E P R O B L E M 2 . 1
The essential component of a toaster is an electrical element (a resistor)
that converts electrical energy to heat energy. How much current is drawn
by a toaster with resistance 12  at 110 V?
Answer: 9.167 A.
E X A M P L E 2 . 2
In the circuit shown in Fig. 2.8, calculate the current i, the conductance
G, and the power p.
30 V
i
+
− 5 kΩ v
+
−
Figure2.8 For Example 2.2.
Solution:
The voltage across the resistor is the same as the source voltage (30 V)
because the resistor and the voltage source are connected to the same pair
of terminals. Hence, the current is
i =
v
R
=
30
5 × 103
= 6 mA
The conductance is
G =
1
R
=
1
5 × 103
= 0.2 mS
We can calculate the power in various ways using either Eqs. (1.7), (2.10),
or (2.11).
p = vi = 30(6 × 10−3
) = 180 mW
or
p = i2
R = (6 × 10−3
)2
5 × 103
= 180 mW
or
p = v2
G = (30)2
0.2 × 10−3
= 180 mW
P R A C T I C E P R O B L E M 2 . 2
For the circuit shown in Fig. 2.9, calculate the voltage v, the conductance
G, and the power p.
2 mA
i
10 kΩ v
+
−
Figure2.9 For Practice Prob. 2.2
Answer: 20 V, 100 µS, 40 mW.
CHAPTER 2 Basic Laws 33
E X A M P L E 2 . 3
A voltage source of 20 sin πt V is connected across a 5-k resistor. Find
the current through the resistor and the power dissipated.
Solution:
i =
v
R
=
20 sin πt
5 × 103
= 4 sin πt mA
Hence,
p = vi = 80 sin2
πt mW
P R A C T I C E P R O B L E M 2 . 3
A resistor absorbs an instantaneous power of 20 cos2
t mW when con-
nected to a voltage source v = 10 cos t V. Find i and R.
Answer: 2 cos t mA, 5 k.
†2.3 NODES, BRANCHES, AND LOOPS
Since the elements of an electric circuit can be interconnected in several
ways, weneedtounderstandsomebasicconceptsofnetworktopology. To
differentiate between a circuit and a network, we may regard a network as
an interconnection of elements or devices, whereas a circuit is a network
providing one or more closed paths. The convention, when addressing
network topology, is to use the word network rather than circuit. We
do this even though the words network and circuit mean the same thing
when used in this context. In network topology, we study the properties
relating to the placement of elements in the network and the geometric
configuration of the network. Such elements include branches, nodes,
and loops.
A branch represents a single element such as a voltage source or a resistor.
In other words, a branch represents any two-terminal element. The circuit
in Fig. 2.10 has five branches, namely, the 10-V voltage source, the 2-A
current source, and the three resistors.
10 V 2 A
a b
c
5 Ω
+
− 2 Ω 3 Ω
Figure2.10 Nodes, branches, and loops.
b
c
a
10 V
5 Ω
2 Ω
3 Ω 2 A
+
−
Figure 2.11 The three-node circuit of Fig. 2.10
is redrawn.
A node is the point of connection between two or more branches.
A node is usually indicated by a dot in a circuit. If a short circuit (a
connecting wire) connects two nodes, the two nodes constitute a single
node. The circuit in Fig. 2.10 has three nodes a, b, and c. Notice that
the three points that form node b are connected by perfectly conducting
wires and therefore constitute a single point. The same is true of the four
points forming node c. We demonstrate that the circuit in Fig. 2.10 has
only three nodes by redrawing the circuit in Fig. 2.11. The two circuits in
34 PART 1 DC Circuits
Figs. 2.10 and 2.11 are identical. However, for the sake of clarity, nodes
b and c are spread out with perfect conductors as in Fig. 2.10.
A loop is any closed path in a circuit.
A loop is a closed path formed by starting at a node, passing through a
set of nodes, and returning to the starting node without passing through
any node more than once. A loop is said to be independent if it contains a
branch which is not in any other loop. Independent loops or paths result
in independent sets of equations.
For example, the closed path abca containing the 2- resistor in
Fig. 2.11 is a loop. Another loop is the closed path bcb containing the
3- resistor and the current source. Although one can identify six loops
in Fig. 2.11, only three of them are independent.
A network with b branches, n nodes, and l independent loops will
satisfy the fundamental theorem of network topology:
b = l + n − 1 (2.12)
As the next two definitions show, circuit topology is of great value
to the study of voltages and currents in an electric circuit.
Two or more elements are in series if they are cascaded or connected sequentially
and consequently carry the same current.
Two or more elements are in parallel if they are connected to the same two nodes
and consequently have the same voltage across them.
Elements are in series when they are chain-connected or connected se-
quentially, end to end. For example, two elements are in series if they
share one common node and no other element is connected to that com-
mon node. Elements in parallel are connected to the same pair of termi-
nals. Elements may be connected in a way that they are neither in series
nor in parallel. In the circuit shown in Fig. 2.10, the voltage source and
the 5- resistor are in series because the same current will flow through
them. The 2- resistor, the 3- resistor, and the current source are in
parallel because they are connected to the same two nodes (b and c)
and consequently have the same voltage across them. The 5- and 2-
resistors are neither in series nor in parallel with each other.
E X A M P L E 2 . 4
Determine the number of branches and nodes in the circuit shown in Fig.
2.12. Identify which elements are in series and which are in parallel.
Solution:
Since there are four elements in the circuit, the circuit has four branches:
10 V, 5 , 6 , and 2 A. The circuit has three nodes as identified in
CHAPTER 2 Basic Laws 35
Fig. 2.13. The 5- resistor is in series with the 10-V voltage source
because the same current would flow in both. The 6- resistor is in
parallel with the 2-A current source because both are connected to the
same nodes 2 and 3.
5 Ω
6 Ω 2 A
10 V +
−
Figure2.12 For Example 2.4.
1 2
5 Ω
6 Ω 2 A
10 V +
−
3
Figure2.13 The three nodes in the circuit
of Fig. 2.12.
P R A C T I C E P R O B L E M 2 . 4
How many branches and nodes does the circuit in Fig. 2.14 have? Identify
the elements that are in series and in parallel.
Answer: Five branches and three nodes are identified in Fig. 2.15. The
1- and 2- resistors are in parallel. The 4- resistor and 10-V source
are also in parallel.
5 Ω
1 Ω 2 Ω 4 Ω
10 V
+
−
Figure2.14 For Practice Prob. 2.4.
3 Ω
3
1 Ω 2 Ω 4 Ω
10 V
+
−
1 2
Figure2.15 Answer for Practice Prob. 2.4.
2.4 KIRCHHOFF’S LAWS
Ohm’s law by itself is not sufficient to analyze circuits. However, when
it is coupled with Kirchhoff’s two laws, we have a sufficient, powerful
set of tools for analyzing a large variety of electric circuits. Kirchhoff’s
laws were first introduced in 1847 by the German physicist Gustav Robert
Kirchhoff (1824–1887). These laws are formally known as Kirchhoff’s
current law (KCL) and Kirchhoff’s voltage law (KVL).
Kirchhoff’s first law is based on the law of conservation of charge,
which requires that the algebraic sum of charges within a system cannot
change.
36 PART 1 DC Circuits
Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering
a node (or a closed boundary) is zero.
Mathematically, KCL implies that
N

n=1
in = 0 (2.13)
where N is the number of branches connected to the node and in is the
nth current entering (or leaving) the node. By this law, currents entering
a node may be regarded as positive, while currents leaving the node may
be taken as negative or vice versa.
To prove KCL, assume a set of currents ik(t), k = 1, 2, . . . , flow
into a node. The algebraic sum of currents at the node is
iT (t) = i1(t) + i2(t) + i3(t) + · · · (2.14)
Integrating both sides of Eq. (2.14) gives
qT (t) = q1(t) + q2(t) + q3(t) + · · · (2.15)
where qk(t) =

ik(t) dt and qT (t) =

iT (t) dt. But the law of conser-
vation of electric charge requires that the algebraic sum of electric charges
at the node must not change; that is, the node stores no net charge. Thus
qT (t) = 0 → iT (t) = 0, confirming the validity of KCL.
i1
i5
i4
i3
i2
Figure2.16 Currents at
a node illustrating KCL. Consider the node in Fig. 2.16. Applying KCL gives
i1 + (−i2) + i3 + i4 + (−i5) = 0 (2.16)
since currents i1, i3, and i4 are entering the node, while currents i2 and
i5 are leaving it. By rearranging the terms, we get
i1 + i3 + i4 = i2 + i5 (2.17)
Equation (2.17) is an alternative form of KCL:
The sum of the currents entering a node is equal to the sum
of the currents leaving the node.
Note that KCL also applies to a closed boundary. This may be
regarded as a generalized case, because a node may be regarded as a
closed surface shrunk to a point. In two dimensions, a closed boundary
is the same as a closed path. As typically illustrated in the circuit of
Fig. 2.17, the total current entering the closed surface is equal to the total
current leaving the surface.
Closed boundary
Figure 2.17 Applying KCL to a closed
boundary.
Twosources(orcircuitsingeneral)aresaidtobe
equivalent if they have the same i-v relationship
at a pair of terminals.
A simple application of KCL is combining current sources in par-
allel. The combined current is the algebraic sum of the current supplied
by the individual sources. For example, the current sources shown in Fig.
2.18(a) can be combined as in Fig. 2.18(b). The combined or equivalent
current source can be found by applying KCL to node a.
IT + I2 = I1 + I3
CHAPTER 2 Basic Laws 37
or
IT = I1 − I2 + I3 (2.18)
A circuit cannot contain two different currents, I1 and I2, in series, unless
I1 = I2; otherwise KCL will be violated.
Kirchhoff’s second law is based on the principle of conservation of
energy:
a
(a)
(b)
I1 I2 I3
b
a
IS = I1 – I2 + I3
b
IT
IT
Figure2.18 Current sources in parallel:
(a) original circuit, (b) equivalent circuit.
Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages
around a closed path (or loop) is zero.
Expressed mathematically, KVL states that
M

m=1
vm = 0 (2.19)
where M is the number of voltages in the loop (or the number of branches
in the loop) and vm is the mth voltage.
KVLcanbeappliedintwoways: bytakingeithera
clockwiseoracounterclockwisetriparoundthe
loop. Either way, the algebraic sum of voltages
around the loop is zero.
To illustrate KVL, consider the circuit in Fig. 2.19. The sign on
each voltage is the polarity of the terminal encountered first as we travel
around the loop. We can start with any branch and go around the loop
either clockwise or counterclockwise. Suppose we start with the voltage
source and go clockwise around the loop as shown; then voltages would
be −v1, +v2, +v3, −v4, and +v5, in that order. For example, as we reach
branch 3, the positive terminal is met first; hence we have +v3. For branch
4, we reach the negative terminal first; hence, −v4. Thus, KVL yields
−v1 + v2 + v3 − v4 + v5 = 0 (2.20)
Rearranging terms gives
v2 + v3 + v5 = v1 + v4 (2.21)
which may be interpreted as
Sum of voltage drops = Sum of voltage rises (2.22)
This is an alternative form of KVL. Notice that if we had traveled coun-
terclockwise, the result would have been +v1, −v5, +v4, −v3, and −v2,
which is the same as before except that the signs are reversed. Hence,
Eqs. (2.20) and (2.21) remain the same.
v4
v1
+
− +
−
v3
v2
v5
+ − + −
+
−
Figure 2.19 A single-loop circuit
illustrating KVL.
When voltage sources are connected in series, KVL can be applied
to obtain the total voltage. The combined voltage is the algebraic sum
of the voltages of the individual sources. For example, for the voltage
sources shown in Fig. 2.20(a), the combined or equivalent voltage source
in Fig. 2.20(b) is obtained by applying KVL.
−Vab + V1 + V2 − V3 = 0
38 PART 1 DC Circuits
or
Vab = V1 + V2 − V3 (2.23)
To avoid violating KVL, a circuit cannot contain two different voltages
V1 and V2 in parallel unless V1 = V2.
V1
V2
V3
a
b
(a)
VS = V1 + V2 − V3
a
b
(b)
+
−
+
−
+
−
Vab
+
−
Vab
+
−
+
−
Figure 2.20 Voltage sources in series:
(a) original circuit, (b) equivalent circuit.
E X A M P L E 2 . 5
For the circuit in Fig. 2.21(a), find voltages v1 and v2.
(a)
20 V +
− 3 Ω
v2
2 Ω
v1
+ −
+
−
(b)
20 V +
− 3 Ω
v2
2 Ω
v1
+ −
+
−
i
Figure2.21 For Example 2.5.
Solution:
To find v1 and v2, we apply Ohm’s law and Kirchhoff’s voltage law.
Assume that current i flows through the loop as shown in Fig. 2.21(b).
From Ohm’s law,
v1 = 2i, v2 = −3i (2.5.1)
Applying KVL around the loop gives
−20 + v1 − v2 = 0 (2.5.2)
Substituting Eq. (2.5.1) into Eq. (2.5.2), we obtain
−20 + 2i + 3i = 0 or 5i = 20 ⇒ i = 4 A
Substituting i in Eq. (2.5.1) finally gives
v1 = 8 V, v2 = −12 V
CHAPTER 2 Basic Laws 39
P R A C T I C E P R O B L E M 2 . 5
Find v1 and v2 in the circuit of Fig. 2.22.
10 V +
− 8 V
+
−
4 Ω
v1
2 Ω
v2
+ −
+ −
Figure2.22 For Practice Prob. 2.5
Answer: 12 V, −6 V.
E X A M P L E 2 . 6
Determine vo and i in the circuit shown in Fig. 2.23(a).
4 Ω
(a)
12 V
2vo
i
4 V
i
+ −
+
− +
−
4 Ω
(b)
12 V
2vo
4 V
+ −
+
− +
−
6 Ω
vo
6 Ω
vo
+ − + −
Figure2.23 For Example 2.6.
Solution:
We apply KVL around the loop as shown in Fig. 2.23(b). The result is
−12 + 4i + 2vo − 4 + 6i = 0 (2.6.1)
Applying Ohm’s law to the 6- resistor gives
vo = −6i (2.6.2)
Substituting Eq. (2.6.2) into Eq. (2.6.1) yields
−16 + 10i − 12i = 0 ⇒ i = −8 A
and vo = 48 V.
P R A C T I C E P R O B L E M 2 . 6
Find vx and vo in the circuit of Fig. 2.24.
35 V 2vx
+
−
+
−
10 Ω
vx
5 Ω
vo
+ −
+ −
Figure2.24 For Practice Prob. 2.6.
Answer: 10 V, −5 V.
40 PART 1 DC Circuits
E X A M P L E 2 . 7
Find current io and voltage vo in the circuit shown in Fig. 2.25.
a
0.5io 3 A
io
4 Ω
vo
+
−
Figure2.25 For Example 2.7.
Solution:
Applying KCL to node a, we obtain
3 + 0.5io = io ⇒ io = 6 A
For the 4- resistor, Ohm’s law gives
vo = 4io = 24 V
P R A C T I C E P R O B L E M 2 . 7
Find vo and io in the circuit of Fig. 2.26.
io
4
6 A
io
2 Ω 8 Ω vo
+
−
Figure2.26 For Practice Prob. 2.7.
Answer: 8 V, 4 A.
E X A M P L E 2 . 8
Find the currents and voltages in the circuit shown in Fig. 2.27(a).
8 Ω
30 V +
−
(a)
v1 i2
i3
i1
a
6 Ω
v3
3 Ω
v2
+ −
+
−
+
−
8 Ω
30 V +
−
(b)
v1 i2
i3
i1
a
6 Ω
v3
3 Ω
v2
+ −
+
−
+
−
Loop 2
Loop 1
Figure2.27 For Example 2.8.
Solution:
We apply Ohm’s law and Kirchhoff’s laws. By Ohm’s law,
v1 = 8i1, v2 = 3i2, v3 = 6i3 (2.8.1)
Since the voltage and current of each resistor are related by Ohm’s
law as shown, we are really looking for three things: (v1, v2, v3) or
(i1, i2, i3). At node a, KCL gives
i1 − i2 − i3 = 0 (2.8.2)
Applying KVL to loop 1 as in Fig. 2.27(b),
−30 + v1 + v2 = 0
CHAPTER 2 Basic Laws 41
We express this in terms of i1 and i2 as in Eq. (2.8.1) to obtain
−30 + 8i1 + 3i2 = 0
or
i1 =
(30 − 3i2)
8
(2.8.3)
Applying KVL to loop 2,
−v2 + v3 = 0 ⇒ v3 = v2 (2.8.4)
as expected since the two resistors are in parallel. We express v1 and v2
in terms of i1 and i2 as in Eq. (2.8.1). Equation (2.8.4) becomes
6i3 = 3i2 ⇒ i3 =
i2
2
(2.8.5)
Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives
30 − 3i2
8
− i2 −
i2
2
= 0
or i2 = 2 A. From the value of i2, we now use Eqs. (2.8.1) to (2.8.5) to
obtain
i1 = 3 A, i3 = 1 A, v1 = 24 V, v2 = 6 V, v3 = 6 V
P R A C T I C E P R O B L E M 2 . 8
Find the currents and voltages in the circuit shown in Fig. 2.28.
5 V 3 V
+
−
i2
i3
i1
8 Ω
v2
+
−
2 Ω
v1
4 Ω
v3
+
−
+ − + −
Figure2.28 For Practice Prob. 2.8.
Answer: v1 = 3 V, v2 = 2 V, v3 = 5 V, i1 = 1.5 A, i2 = 0.25 A,
i3 =1.25 A.
2.5 SERIES RESISTORS AND VOLTAGE DIVISION
v +
−
R1
v1
R2
v2
i
+ − + −
a
b
Figure2.29 A single-loop circuit
with two resistors in series.
The need to combine resistors in series or in parallel occurs so frequently
that it warrants special attention. The process of combining the resistors
is facilitated by combining two of them at a time. With this in mind,
consider the single-loop circuit of Fig. 2.29. The two resistors are in
series, since the same current i flows in both of them. Applying Ohm’s
law to each of the resistors, we obtain
v1 = iR1, v2 = iR2 (2.24)
If we apply KVL to the loop (moving in the clockwise direction), we have
−v + v1 + v2 = 0 (2.25)
42 PART 1 DC Circuits
Combining Eqs. (2.24) and (2.25), we get
v = v1 + v2 = i(R1 + R2) (2.26)
or
i =
v
R1 + R2
(2.27)
Notice that Eq. (2.26) can be written as
v = iReq (2.28)
implying that the two resistors can be replaced by an equivalent resistor
Req; that is,
Req = R1 + R2 (2.29)
Thus, Fig. 2.29 can be replaced by the equivalent circuit in Fig. 2.30. The
two circuits in Figs. 2.29 and 2.30 are equivalent because they exhibit the
same voltage-current relationships at the terminals a-b. An equivalent
circuit such as the one in Fig. 2.30 is useful in simplifying the analysis
of a circuit. In general,
v
Req
v
+
−
i
+ −
a
b
Figure2.30 Equivalent circuit
of the Fig. 2.29 circuit.
The equivalent resistance of any number of resistors connected in series
is the sum of the individual resistances.
Resistors in series behave as a single resistor
whose resistance is equal to the sum of the re-
sistances of the individual resistors.
For N resistors in series then,
Req = R1 + R2 + · · · + RN =
N

n=1
Rn (2.30)
To determine the voltage across each resistor in Fig. 2.29, we sub-
stitute Eq. (2.26) into Eq. (2.24) and obtain
v1 =
R1
R1 + R2
v, v2 =
R2
R1 + R2
v (2.31)
Notice that the source voltage v is divided among the resistors in direct
proportion to their resistances; the larger the resistance, the larger the
voltage drop. This is called the principle of voltage division, and the
circuit in Fig. 2.29 is called a voltage divider. In general, if a voltage
divider has N resistors (R1, R2, . . . , RN ) in series with the source voltage
v, the nth resistor (Rn) will have a voltage drop of
vn =
Rn
R1 + R2 + · · · + RN
v (2.32)
2.6 PARALLEL RESISTORS AND CURRENT DIVISION
Consider the circuit in Fig. 2.31, where two resistors are connected in
parallel and therefore have the same voltage across them. From Ohm’s
law,
v = i1R1 = i2R2
CHAPTER 2 Basic Laws 43
or
i1 =
v
R1
, i2 =
v
R2
(2.33)
Applying KCL at node a gives the total current i as
i = i1 + i2 (2.34)
Substituting Eq. (2.33) into Eq. (2.34), we get
i =
v
R1
+
v
R2
= v

1
R1
+
1
R2

=
v
Req
(2.35)
where Req is the equivalent resistance of the resistors in parallel:
1
Req
=
1
R1
+
1
R2
(2.36)
or
1
Req
=
R1 + R2
R1R2
or
Req =
R1R2
R1 + R2
(2.37)
Thus,
The equivalent resistance of two parallel resistors is equal to the product
of their resistances divided by their sum.
It must be emphasized that this applies only to two resistors in parallel.
From Eq. (2.37), if R1 = R2, then Req = R1/2.
Node b
Node a
v +
− R1 R2
i1 i2
i
Figure2.31 Two resistors in parallel.
We can extend the result in Eq. (2.36) to the general case of a circuit
with N resistors in parallel. The equivalent resistance is
1
Req
=
1
R1
+
1
R2
+ · · · +
1
RN
(2.38)
Note that Req is always smaller than the resistance of the smallest resistor
in the parallel combination. If R1 = R2 = · · · = RN = R, then
Req =
R
N
(2.39)
For example, if four 100- resistors are connected in parallel, their equiv-
alent resistance is 25 .
Conductances in parallel behave as a single con-
ductance whose value is equal to the sum of the
individual conductances.
It is often more convenient to use conductance rather than resistance
when dealing with resistors in parallel. From Eq. (2.38), the equivalent
conductance for N resistors in parallel is
Geq = G1 + G2 + G3 + · · · + GN (2.40)
where Geq = 1/Req, G1 = 1/R1, G2 = 1/R2, G3 = 1/R3, . . . , GN =
1/RN . Equation (2.40) states:
44 PART 1 DC Circuits
The equivalent conductance of resistors connected in parallel is the sum
of their individual conductances.
This means that we may replace the circuit in Fig. 2.31 with that in
Fig. 2.32. Notice the similarity between Eqs. (2.30) and (2.40). The
equivalent conductance of parallel resistors is obtained the same way
as the equivalent resistance of series resistors. In the same manner, the
equivalent conductance of resistors in series is obtained just the same way
as the resistance of resistors in parallel. Thus the equivalent conductance
Geq of N resistors in series (such as shown in Fig. 2.29) is
1
Geq
=
1
G1
+
1
G2
+
1
G3
+ · · · +
1
GN
(2.41)
b
a
v +
− Req or Geq
v
i
Figure 2.32 Equivalent circuit to
Fig. 2.31.
Given the total current i entering node a in Fig. 2.31, how do we
obtain current i1 and i2? We know that the equivalent resistor has the
same voltage, or
v = iReq =
iR1R2
R1 + R2
(2.42)
Combining Eqs. (2.33) and (2.42) results in
i1 =
R2 i
R1 + R2
, i2 =
R1 i
R1 + R2
(2.43)
which shows that the total current i is shared by the resistors in inverse
proportion to their resistances. This is known as the principle of current
division, and the circuit in Fig. 2.31 is known as a current divider. Notice
that the larger current flows through the smaller resistance.
As an extreme case, suppose one of the resistors in Fig. 2.31 is zero,
say R2 = 0; that is, R2 is a short circuit, as shown in Fig. 2.33(a). From
Eq. (2.43), R2 = 0 implies that i1 = 0, i2 = i. This means that the
entire current i bypasses R1 and flows through the short circuit R2 = 0,
the path of least resistance. Thus when a circuit is short circuited, as
shown in Fig. 2.33(a), two things should be kept in mind:
R2 = 0
(a)
R1
i
i1 = 0 i2 = i
R2 = ∞
(b)
R1
i
i1 = i
i2 = 0
Figure 2.33 (a) A shorted circuit,
(b) an open circuit.
1. The equivalent resistance Req = 0. [See what happens when
R2 = 0 in Eq. (2.37).]
2. The entire current flows through the short circuit.
As another extreme case, suppose R2 = ∞, that is, R2 is an open
circuit, as shown in Fig. 2.33(b). The current still flows through the path
of least resistance, R1. By taking the limit of Eq. (2.37) as R2 → ∞, we
obtain Req = R1 in this case.
If we divide both the numerator and denominator by R1R2, Eq.
(2.43) becomes
i1 =
G1
G1 + G2
i (2.44a)
i2 =
G2
G1 + G2
i (2.44b)
CHAPTER 2 Basic Laws 45
Thus, in general, if a current divider has N conductors (G1, G2, . . . , GN )
in parallel with the source current i, the nth conductor (Gn) will have
current
in =
Gn
G1 + G2 + · · · + GN
i (2.45)
In general, it is often convenient and possible to combine resistors
in series and parallel and reduce a resistive network to a single equivalent
resistance Req. Such an equivalent resistance is the resistance between
the designated terminals of the network and must exhibit the same i-v
characteristics as the original network at the terminals.
E X A M P L E 2 . 9
Find Req for the circuit shown in Fig. 2.34.
2 Ω
5 Ω
Req
4 Ω
8 Ω
1 Ω
6 Ω 3 Ω
Figure2.34 For Example 2.9.
Solution:
6 Ω
Req
4 Ω
(a)
8 Ω
2 Ω
2 Ω
2.4 Ω
Req
4 Ω
(b)
8 Ω
Figure 2.35 Equivalent circuits for
Example 2.9.
To get Req, we combine resistors in series and in parallel. The 6- and
3- resistors are in parallel, so their equivalent resistance is
6  3  =
6 × 3
6 + 3
= 2 
(The symbol is used to indicate a parallel combination.) Also, the 1-
and 5- resistors are in series; hence their equivalent resistance is
1  + 5  = 6 
Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig.
2.35(a), we notice that the two 2- resistors are in series, so the equivalent
resistance is
2  + 2  = 4 
This 4- resistor is now in parallel with the 6- resistor in Fig. 2.35(a);
their equivalent resistance is
4  6  =
4 × 6
4 + 6
= 2.4 
The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig.
2.35(b), the three resistors are in series. Hence, the equivalent resistance
for the circuit is
Req = 4  + 2.4  + 8  = 14.4 
P R A C T I C E P R O B L E M 2 . 9
By combining the resistors in Fig. 2.36, find Req.
5 Ω
4 Ω
6 Ω
Req
2 Ω
1 Ω
3 Ω 4 Ω
3 Ω
Figure2.36 For Practice Prob. 2.9.
Answer: 6 .
46 PART 1 DC Circuits
E X A M P L E 2 . 1 0
Calculate the equivalent resistance Rab in the circuit in Fig. 2.37.
a
b
b b
c d
6 Ω
12 Ω
5 Ω
4 Ω
10 Ω 1 Ω 1 Ω
Rab
3 Ω
Figure2.37 For Example 2.10.
Solution:
The 3- and 6- resistors are in parallel because they are connected to
the same two nodes c and b. Their combined resistance is
3  6  =
3 × 6
3 + 6
= 2  (2.10.1)
Similarly, the 12- and 4- resistors are in parallel since they are con-
nected to the same two nodes d and b. Hence
12  4  =
12 × 4
12 + 4
= 3  (2.10.2)
Also the 1- and 5- resistors are in series; hence, their equivalent
resistance is
1  + 5  = 6  (2.10.3)
With these three combinations, we can replace the circuit in Fig. 2.37 with
that in Fig. 2.38(a). In Fig. 2.38(a), 3- in parallel with 6- gives 2-, as
calculated in Eq. (2.10.1). This 2- equivalent resistance is now in series
with the 1- resistance to give a combined resistance of 1 +2  = 3 .
Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In
Fig. 2.38(b), we combine the 2- and 3- resistors in parallel to get
(a)
b
b
d
b
c
3 Ω 6 Ω
2 Ω
10 Ω 1 Ω
a
b
(b)
b b
c
3 Ω
2 Ω
10 Ω
a
b
Figure 2.38 Equivalent circuits for
Example 2.10.
2  3  =
2 × 3
2 + 3
= 1.2 
This 1.2- resistor is in series with the 10- resistor, so that
Rab = 10 + 1.2 = 11.2 
P R A C T I C E P R O B L E M 2 . 1 0
Find Rab for the circuit in Fig. 2.39.
1 Ω
9 Ω
18 Ω
20 Ω
20 Ω
2 Ω
5 Ω
8 Ω
a
b
Rab
Figure2.39 For Practice Prob. 2.10.
Answer: 11 .
CHAPTER 2 Basic Laws 47
E X A M P L E 2 . 1 1
Find the equivalent conductance Geq for the circuit in Fig. 2.40(a).
12 S
8 S
6 S
(a)
5 S
Geq
20 S
6 S
(b)
5 S
Geq
(c)
Req
Ω
1
5
Ω
1
6 Ω
1
8 Ω
1
12
Figure2.40 For Example 2.11: (a) original
circuit, (b) its equivalent circuit, (c) same
circuit as in (a) but resistors are expressed in
ohms.
Solution:
The 8-S and 12-S resistors are in parallel, so their conductance is
8 S + 12 S = 20 S
This 20-S resistor is now in series with 5 S as shown in Fig. 2.40(b) so
that the combined conductance is
20 × 5
20 + 5
= 4 S
This is in parallel with the 6-S resistor. Hence
Geq = 6 + 4 = 10 S
We should note that the circuit in Fig. 2.40(a) is the same as that in
Fig. 2.40(c). While the resistors in Fig. 2.40(a) are expressed in siemens,
they are expressed in ohms in Fig. 2.40(c). To show that the circuits are
the same, we find Req for the circuit in Fig. 2.40(c).
Req =
1
6





1
5
+
1
8




1
12

=
1
6





1
5
+
1
20

=
1
6




1
4
=
1
6
× 1
4
1
6
+ 1
4
=
1
10

Geq =
1
Req
= 10 S
This is the same as we obtained previously.
P R A C T I C E P R O B L E M 2 . 1 1
Calculate Geq in the circuit of Fig. 2.41.
4 S
6 S
8 S
2 S
12 Ω
Geq
Figure2.41 For Practice Prob. 2.11.
Answer: 4 S.
E X A M P L E 2 . 1 2
Find io and vo in the circuit shown in Fig. 2.42(a). Calculate the power
dissipated in the 3- resistor.
Solution:
The 6- and 3- resistors are in parallel, so their combined resistance is
6  3  =
6 × 3
6 + 3
= 2
48 PART 1 DC Circuits
Thus our circuit reduces to that shown in Fig. 2.42(b). Notice that vo is
not affected by the combination of the resistors because the resistors are
in parallel and therefore have the same voltage vo. From Fig. 2.42(b), we
can obtain vo in two ways. One way is to apply Ohm’s law to get
i =
12
4 + 2
= 2 A
and hence, vo = 2i = 2 × 2 = 4 V. Another way is to apply voltage
division, since the 12 V in Fig. 2.42(b) is divided between the 4- and
2- resistors. Hence,
vo =
2
2 + 4
(12 V) = 4 V
a
b
(a)
12 V
4 Ω
i io
6 Ω 3 Ω
vo
+
−
a
b
(b)
12 V
4 Ω
i
+
− 2 Ω
vo
+
−
+
−
Figure2.42 For Example 2.12: (a) original
circuit, (b) its equivalent circuit.
Similarly, io can be obtained in two ways. One approach is to apply
Ohm’s law to the 3- resistor in Fig. 2.42(a) now that we know vo; thus,
vo = 3io = 4 ⇒ io =
4
3
A
Another approach is to apply current division to the circuit in Fig. 2.42(a)
now that we know i, by writing
io =
6
6 + 3
i =
2
3
(2 A) =
4
3
A
The power dissipated in the 3- resistor is
po = voio = 4

4
3

= 5.333 W
P R A C T I C E P R O B L E M 2 . 1 2
Find v1 and v2 in the circuit shown in Fig. 2.43. Also calculate i1 and i2
and the power dissipated in the 12- and 40- resistors.
15 V
i1
+
− 40 Ω
v2
+
−
10 Ω
12 Ω
v1
6 Ω
i2
+ −
Figure2.43 For Practice Prob. 2.12.
Answer: v1 = 5 V, i1 = 416.7 mA, p1 = 2.083 W, v2 = 10 V,
i2 = 250 mA, p2 = 2.5 W.
E X A M P L E 2 . 1 3
For the circuit shown in Fig. 2.44(a), determine: (a) the voltage vo, (b)
the power supplied by the current source, (c) the power absorbed by each
resistor.
Solution:
(a) The 6-k and 12-k resistors are in series so that their combined
value is 6 + 12 = 18 k. Thus the circuit in Fig. 2.44(a) reduces to that
CHAPTER 2 Basic Laws 49
shown in Fig. 2.44(b). We now apply the current division technique to
find i1 and i2.
i1 =
18,000
9000 + 18,000
(30 mA) = 20 mA
i2 =
9000
9000 + 18,000
(30 A) = 10 mA
Notice that the voltage across the 9-k and 18-k resistors is the same,
and vo = 9,000i1 = 18,000i2 = 180 V, as expected.
(a)
30 mA 9 kΩ
vo
+
−
12 kΩ
6 kΩ
(b)
30 mA 9 kΩ
vo
+
−
18 kΩ
i1
io i2
Figure 2.44 For Example 2.13:
(a) original circuit,
(b) its equivalent circuit.
(b) Power supplied by the source is
po = voio = 180(30) mW = 5.4 W
(c) Power absorbed by the 12-k resistor is
p = iv = i2(i2R) = i2
2 R = (10 × 10−3
)2
(12,000) = 1.2 W
Power absorbed by the 6-k resistor is
p = i2
2 R = (10 × 10−3
)2
(6000) = 0.6 W
Power absorbed by the 9-k resistor is
p =
v2
o
R
=
(180)2
9000
= 3.6 W
or
p = voi1 = 180(20) mW = 3.6 W
Notice that the power supplied (5.4 W) equals the power absorbed (1.2+
0.6 + 3.6 = 5.4 W). This is one way of checking results.
P R A C T I C E P R O B L E M 2 . 1 3
For the circuit shown in Fig. 2.45, find: (a) v1 and v2, (b) the power dis-
sipated in the 3-k and 20-k resistors, and (c) the power supplied by
the current source.
10 mA
3 kΩ 5 kΩ 20 kΩ
1 kΩ
v1
+
−
v2
+
−
Figure2.45 For Practice Prob. 2.13.
Answer: (a) 15 V, 20 V, (b) 75 mW, 20 mW, (c) 200 mW.
50 PART 1 DC Circuits
†2.7 WYE-DELTA TRANSFORMATIONS
Situations often arise in circuit analysis when the resistors are neither in
parallel nor in series. For example, consider the bridge circuit in Fig.
2.46. How do we combine resistors R1 through R6 when the resistors
are neither in series nor in parallel? Many circuits of the type shown in
Fig. 2.46 can be simplified by using three-terminal equivalent networks.
These are the wye (Y) or tee (T) network shown in Fig. 2.47 and the
delta () or pi (#) network shown in Fig. 2.48. These networks occur by
themselves or as part of a larger network. They are used in three-phase
networks, electrical filters, and matching networks. Our main interest
here is in how to identify them when they occur as part of a network and
how to apply wye-delta transformation in the analysis of that network.
vs
+
−
R1
R4
R2
R5
R3
R6
Figure2.46 The bridge network.
1 3
2 4
R3
R2
R1
(a)
1 3
2 4
R3
R2
R1
(b)
Figure2.47 Two forms of the same network: (a) Y, (b) T.
1 3
2 4
Rc
(a)
1 3
2 4
(b)
Ra
Rb
Rc
Ra
Rb
Figure2.48 Two forms of the
same network: (a) , (b) #.
Delta to Wye Conversion
Suppose it is more convenient to work with a wye network in a place
where the circuit contains a delta configuration. We superimpose a wye
network on the existing delta network and find the equivalent resistances
in the wye network. To obtain the equivalent resistances in the wye
network, we compare the two networks and make sure that the resistance
between each pair of nodes in the  (or #) network is the same as the
resistance between the same pair of nodes in the Y (or T) network. For
terminals 1 and 2 in Figs. 2.47 and 2.48, for example,
R12(Y) = R1 + R3
R12() = Rb (Ra + Rc)
(2.46)
Setting R12(Y)= R12() gives
R12 = R1 + R3 =
Rb(Ra + Rc)
Ra + Rb + Rc
(2.47a)
Similarly,
R13 = R1 + R2 =
Rc(Ra + Rb)
Ra + Rb + Rc
(2.47b)
R34 = R2 + R3 =
Ra(Rb + Rc)
Ra + Rb + Rc
(2.47c)
Subtracting Eq. (2.47c) from Eq. (2.47a), we get
R1 − R2 =
Rc(Rb − Ra)
Ra + Rb + Rc
(2.48)
CHAPTER 2 Basic Laws 51
Adding Eqs. (2.47b) and (2.48) gives
R1 =
RbRc
Ra + Rb + Rc
(2.49)
and subtracting Eq. (2.48) from Eq. (2.47b) yields
R2 =
RcRa
Ra + Rb + Rc
(2.50)
Subtracting Eq. (2.49) from Eq. (2.47a), we obtain
R3 =
RaRb
Ra + Rb + Rc
(2.51)
We do not need to memorize Eqs. (2.49) to (2.51). To transform a 
network to Y, we create an extra node n as shown in Fig. 2.49 and follow
this conversion rule:
Each resistor in the Y network is the product of the resistors in the two adjacent 
branches, divided by the sum of the three  resistors.
R3
Ra
Rb
R1 R2
Rc
b
n
a
c
Figure2.49 Superposition of Y and 
networks as an aid in transforming one to
the other.
Wye to Delta Conversion
To obtain the conversion formulas for transforming a wye network to an
equivalent delta network, we note from Eqs. (2.49) to (2.51) that
R1R2 + R2R3 + R3R1 =
RaRbRc(Ra + Rb + Rc)
(Ra + Rb + Rc)2
=
RaRbRc
Ra + Rb + Rc
(2.52)
Dividing Eq. (2.52) by each of Eqs. (2.49) to (2.51) leads to the following
equations:
Ra =
R1R2 + R2R3 + R3R1
R1
(2.53)
Rb =
R1R2 + R2R3 + R3R1
R2
(2.54)
Rc =
R1R2 + R2R3 + R3R1
R3
(2.55)
From Eqs. (2.53) to (2.55) and Fig. 2.49, the conversion rule for Y to 
is as follows:
52 PART 1 DC Circuits
Each resistor in the  network is the sum of all possible products of Y resistors
taken two at a time, divided by the opposite Y resistor.
The Y and  networks are said to be balanced when
R1 = R2 = R3 = RY , Ra = Rb = Rc = R (2.56)
Under these conditions, conversion formulas become
RY =
R
3
or R = 3RY (2.57)
One may wonder why RY is less than R. Well, we notice that the Y-
connection is like a “series” connection while the -connection is like a
“parallel” connection.
Note that in making the transformation, we do not take anything out
of the circuit or put in anything new. We are merely substituting different
but mathematically equivalent three-terminal network patterns to create
a circuit in which resistors are either in series or in parallel, allowing us
to calculate Req if necessary.
E X A M P L E 2 . 1 4
Convert the  network in Fig. 2.50(a) to an equivalent Y network.
c
b
a
10 Ω 15 Ω
(a)
Rb Ra
Rc
25 Ω
c
b
a
5 Ω
3 Ω
7.5 Ω
R2
R1
R3
(b)
Figure2.50 For Example 2.14: (a) original  network, (b) Y equivalent network.
Solution:
Using Eqs. (2.49) to (2.51), we obtain
CHAPTER 2 Basic Laws 53
R1 =
RbRc
Ra + Rb + Rc
=
25 × 10
25 + 10 + 15
=
250
50
= 5 
R2 =
RcRa
Ra + Rb + Rc
=
25 × 15
50
= 7.5 
R3 =
RaRb
Ra + Rb + Rc
=
15 × 10
50
= 3 
The equivalent Y network is shown in Fig. 2.50(b).
P R A C T I C E P R O B L E M 2 . 1 4
Transform the wye network in Fig. 2.51 to a delta network.
20 Ω
R2
b
a
c
10 Ω
R1
R3 40 Ω
Figure2.51 For Practice Prob. 2.14.
Answer: Ra = 140 , Rb = 70 , Rc = 35 .
E X A M P L E 2 . 1 5
Obtain the equivalent resistance Rab for the circuit in Fig. 2.52 and use it
to find current i.
a a
i
b
b
c n
120 V
5 Ω
30 Ω
12.5 Ω
15 Ω
10 Ω
20 Ω
+
−
Figure2.52 For Example 2.15.
Solution:
Inthiscircuit, therearetwoYnetworksandonenetwork. Transforming
just one of these will simplify the circuit. If we convert the Y network
comprising the 5-, 10-, and 20- resistors, we may select
R1 = 10 , R2 = 20 , R3 = 5 
Thus from Eqs. (2.53) to (2.55) we have
Ra =
R1R2 + R2R3 + R3R1
R1
=
10 × 20 + 20 × 5 + 5 × 10
10
=
350
10
= 35 
Rb =
R1R2 + R2R3 + R3R1
R2
=
350
20
= 17.5 
Rc =
R1R2 + R2R3 + R3R1
R3
=
350
5
= 70 
With the Y converted to , the equivalent circuit (with the voltage
source removed for now) is shown in Fig. 2.53(a). Combining the three
pairs of resistors in parallel, we obtain
54 PART 1 DC Circuits
70 30 =
70 × 30
70 + 30
= 21 
12.5 17.5 =
12.5 × 17.5
12.5 + 17.5
= 7.2917 
15 35 =
15 × 35
15 + 35
= 10.5 
so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we find
Rab = (7.292 + 10.5) 21 =
17.792 × 21
17.792 + 21
= 9.632 
Then
i =
vs
Rab
=
120
9.632
= 12.458 A
a
b
30 Ω
70 Ω
17.5 Ω
35 Ω
12.5 Ω
15 Ω
(a)
a
b
21 Ω
(b)
7.292 Ω
10.5 Ω
Figure2.53 Equivalent circuits to Fig. 2.52, with the voltage removed.
P R A C T I C E P R O B L E M 2 . 1 5
For the bridge network in Fig. 2.54, find Rab and i.
24 Ω
100 V
i
30 Ω
10 Ω
50 Ω
13 Ω
20 Ω
+
−
b
a
Figure2.54 For Practice Prob. 2.15.
Answer: 40 , 2.5 A.
†2.8 APPLICATIONS
Resistors are often used to model devices that convert electrical energy
into heat or other forms of energy. Such devices include conducting
wire, lightbulbs, electric heaters, stoves, ovens, and loudspeakers. In this
CHAPTER 2 Basic Laws 55
section, we will consider two real-life problems that apply the concepts
developed in this chapter: electrical lighting systems and design of dc
meters.
So far, we have assumed that connecting wires
are perfect conductors (i.e., conductors of zero
resistance). In real physical systems, however,
the resistance of the connecting wire may be ap-
preciably large, and the modeling of the system
must include that resistance.
2.8.1 Lighting Systems
Lighting systems, such as in a house or on a Christmas tree, often consist
of N lamps connected either in parallel or in series, as shown in Fig.
2.55. Each lamp is modeled as a resistor. Assuming that all the lamps are
identical and Vo is the power-line voltage, the voltage across each lamp
is Vo for the parallel connection and Vo/N for the series connection. The
series connection is easy to manufacture but is seldom used in practice,
for at least two reasons. First, it is less reliable; when a lamp fails, all the
lamps go out. Second, it is harder to maintain; when a lamp is bad, one
must test all the lamps one by one to detect the faulty one.
Vo
+
−
Power
plug
1 2 3 N
Lamp
(a)
Vo
+
−
1
2
3
N
(b)
Figure2.55 (a) Parallel connection of lightbulbs, (b) series connection of lightbulbs.
E X A M P L E 2 . 1 6
Three lightbulbs are connected to a 9-V battery as shown in Fig. 2.56(a).
Calculate: (a) the total current supplied by the battery, (b) the current
through each bulb, (c) the resistance of each bulb.
(a)
9 V
10 W
15 W
20 W
(b)
9 V
+
−
+
−
+
−
I1
I2
V3
V2
V1 R1
I
R3
R2
Figure 2.56 (a) Lighting system with three bulbs, (b) resistive circuit equivalent
model.
56 PART 1 DC Circuits
Solution:
(a) The total power supplied by the battery is equal to the total power
absorbed by the bulbs, that is,
p = 15 + 10 + 20 = 45 W
Since p = V I, then the total current supplied by the battery is
I =
p
V
=
45
9
= 5 A
(b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b). Since
R1 (20-W bulb) is in parallel with the battery as well as the series com-
bination of R2 and R3,
V1 = V2 + V3 = 9 V
The current through R1 is
I1 =
p1
V1
=
20
9
= 2.222 A
By KCL, the current through the series combination of R2 and R3
is
I2 = I − I1 = 5 − 2.222 = 2.778 A
(c) Since p = I2
R,
R1 =
p1
I2
1
=
20
2.2222
= 4.05 
R2 =
p2
I2
2
=
15
2.7772
= 1.945 
R3 =
p3
I2
3
=
10
2.7772
= 1.297 
P R A C T I C E P R O B L E M 2 . 1 6
Refer to Fig. 2.55 and assume there are 10 lightbulbs, each with a power
rating of 40 W. If the voltage at the plug is 110 V for the parallel and
series connections, calculate the current through each bulb for both cases.
Answer: 0.364 A (parallel), 3.64 A (series).
+
+
−
−
Vin
Vout
a
b
c
Max
Min
Figure2.57 The potentiometer
controlling potential levels.
2.8.2 Design of DC Meters
By their nature, resistors are used to control the flow of current. We take
advantage of this property in several applications, such as in a poten-
tiometer (Fig. 2.57). The word potentiometer, derived from the words
potential and meter, implies that potential can be metered out. The po-
tentiometer (or pot for short) is a three-terminal device that operates on
the principle of voltage division. It is essentially an adjustable voltage
divider. As a voltage regulator, it is used as a volume or level control on
radios, TVs, and other devices. In Fig. 2.57,
Vout = Vbc =
Rbc
Rac
Vin (2.58)
where Rac = Rab + Rbc. Thus, Vout decreases or increases as the sliding
contact of the pot moves toward c or a, respectively.
CHAPTER 2 Basic Laws 57
Another application where resistors are used to control current flow
is in the analog dc meters—the ammeter, voltmeter, and ohmmeter, which
measure current, voltage, and resistance, respectively. Each of these me-
ters employs the d’Arsonval meter movement, shown in Fig. 2.58. The
movement consists essentially of a movable iron-core coil mounted on
a pivot between the poles of a permanent magnet. When current flows
through the coil, it creates a torque which causes the pointer to deflect.
The amount of current through the coil determines the deflection of the
pointer, which is registered on a scale attached to the meter movement.
For example, if the meter movement is rated 1 mA, 50 , it would take
1 mA to cause a full-scale deflection of the meter movement. By introduc-
ing additional circuitry to the d’Arsonval meter movement, an ammeter,
voltmeter, or ohmmeter can be constructed.
Aninstrumentcapableofmeasuringvoltage,cur-
rent, and resistance is called a multimeter or a
volt-ohm meter (VOM).
Aloadisacomponentthatisreceivingenergy(an
energysink),asopposedtoageneratorsupplying
energy (an energy source). More about loading
will be discussed in Section 4.9.1.
Consider Fig. 2.59, where an analog voltmeter and ammeter are
connected to an element. The voltmeter measures the voltage across a
load and is therefore connected in parallel with the element. As shown
in Fig. 2.60(a), the voltmeter consists of a d’Arsonval movement in par-
allel with a resistor whose resistance Rm is deliberately made very large
(theoretically, infinite), to minimize the current drawn from the circuit.
To extend the range of voltage that the meter can measure, series multi-
plier resistors are often connected with the voltmeters, as shown in Fig.
2.60(b). The multiple-range voltmeter in Fig. 2.60(b) can measure volt-
age from 0 to 1 V, 0 to 10 V, or 0 to 100 V, depending on whether the
switch is connected to R1, R2, or R3, respectively.
Let us calculate the multiplier resistor Rn for the single-range volt-
meter in Fig. 2.60(a), or Rn = R1, R2, or R3 for the multiple-range
voltmeter in Fig. 2.60(b). We need to determine the value of Rn to be
connected in series with the internal resistance Rm of the voltmeter. In
any design, we consider the worst-case condition. In this case, the worst
case occurs when the full-scale current Ifs = Im flows through the meter.
This should also correspond to the maximum voltage reading or the full-
scale voltage Vfs. Since the multiplier resistance Rn is in series with the
scale
pointer
spring
permanent magnet
rotating coil
stationary iron core
spring
N
S
Figure2.58 A d’Arsonval meter movement.
V
A
V
I
+
−
Voltmeter
Ammeter
Element
Figure2.59 Connection of a
voltmeter and an ammeter to an
element.
58 PART 1 DC Circuits
Probes V
+
−
R1
R2
R3
1 V
10 V
100 V
Switch
Im
(b)
Rn
Im
Multiplier
Probes V
+
−
(a)
Rm
Meter
Rm
Meter
Figure2.60 Voltmeters: (a) single-range type, (b) multiple-range type.
internal resistance Rm,
Vfs = Ifs(Rn + Rm) (2.59)
From this, we obtain
Rn =
Vfs
Ifs
− Rm (2.60)
Im
I
Probes
(a)
Rn
In
(b)
R1
R2
R3
10 mA
100 mA
1 A
Switch
Im
I
Probes
Rm
Meter
Rm
Meter
Figure 2.61 Ammeters: (a) single-range type,
(b) multiple-range type.
Similarly, the ammeter measures the current through the load and
is connected in series with it. As shown in Fig. 2.61(a), the ammeter con-
sists of a d’Arsonval movement in parallel with a resistor whose resistance
Rm is deliberately made very small (theoretically, zero) to minimize the
voltage drop across it. To allow multiple range, shunt resistors are often
connected in parallel with Rm as shown in Fig. 2.61(b). The shunt resis-
tors allow the meter to measure in the range 0–10 mA, 0–100 mA, or
0–1 A, depending on whether the switch is connected to R1, R2, or
R3, respectively.
Now our objective is to obtain the multiplier shunt Rn for the single-
range ammeter in Fig. 2.61(a), or Rn = R1, R2, or R3 for the multiple-
range ammeter in Fig. 2.61(b). We notice that Rm and Rn are in parallel
and that at full-scale reading I = Ifs = Im + In, where In is the current
through the shunt resistor Rn. Applying the current division principle
yields
Im =
Rn
Rn + Rm
Ifs
CHAPTER 2 Basic Laws 59
or
Rn =
Im
Ifs − Im
Rm (2.61)
The resistance Rx of a linear resistor can be measured in two ways.
An indirect way is to measure the current I that flows through it by
connecting an ammeter in series with it and the voltage V across it by
connecting a voltmeter in parallel with it, as shown in Fig. 2.62(a). Then
Rx =
V
I
(2.62)
The direct method of measuring resistance is to use an ohmmeter. An
ohmmeterconsistsbasicallyofad’Arsonvalmovement, avariableresistor
or potentiometer, and a battery, as shown in Fig. 2.62(b). Applying KVL
to the circuit in Fig. 2.62(b) gives
E = (R + Rm + Rx)Im
or
Rx =
E
Im
− (R + Rm) (2.63)
The resistor R is selected such that the meter gives a full-scale deflection,
that is, Im = Ifs when Rx = 0. This implies that
E = (R + Rm)Ifs (2.64)
Substituting Eq. (2.64) into Eq. (2.63) leads to
Rx =

Ifs
Im
− 1

(R + Rm) (2.65)
Im
R
E Rx
Ohmmeter
(b)
(a)
V
A
+
−
V
Rx
I
Rm
Figure2.62 Two ways of measuring
resistance: (a) using an ammeter and a
voltmeter, (b) using an ohmmeter.
As mentioned, the types of meters we have discussed are known as
analog meters and are based on the d’Arsonval meter movement. Another
type of meter, called a digital meter, is based on active circuit elements
such as op amps. For example, a digital multimeter displays measure-
ments of dc or ac voltage, current, and resistance as discrete numbers,
instead of using a pointer deflection on a continuous scale as in an ana-
log multimeter. Digital meters are what you would most likely use in a
modern lab. However, the design of digital meters is beyond the scope
of this book.
E X A M P L E 2 . 1 7
Following the voltmeter setup of Fig. 2.60, design a voltmeter for the fol-
lowing multiple ranges:
(a) 0–1 V (b) 0–5 V (c) 0–50 V (d) 0–100 V
Assume that the internal resistance Rm = 2 k and the full-scale current
Ifs = 100 µA.
Solution:
We apply Eq. (2.60) and assume that R1, R2, R3, and R4 correspond with
ranges 0–1 V, 0–5 V, 0–50 V, and 0–100 V, respectively.
(a) For range 0–1 V,
R1 =
1
100 × 10−6
− 2000 = 10,000 − 2000 = 8 k
60 PART 1 DC Circuits
(b) For range 0–5 V,
R2 =
5
100 × 10−6
− 2000 = 50,000 − 2000 = 48 k
(c) For range 0–50 V,
R3 =
50
100 × 10−6
− 2000 = 500,000 − 2000 = 498 k
(d) For range 0–100 V,
R4 =
100 V
100 × 10−6
− 2000 = 1,000,000 − 2000 = 998 k
Note that the ratio of the total resistance (Rn+Rm) to the full-scale voltage
Vfs is constant and equal to 1/Ifs for the four ranges. This ratio (given in
ohms per volt, or /V) is known as the sensitivity of the voltmeter. The
larger the sensitivity, the better the voltmeter.
P R A C T I C E P R O B L E M 2 . 1 7
Following the ammeter setup of Fig. 2.61, design an ammeter for the fol-
lowing multiple ranges:
(a) 0–1 A (b) 0–100 mA (c) 0–10 mA
Take the full-scale meter current as Im = 1 mA and the internal resistance
of the ammeter as Rm = 50 .
Answer: Shunt resistors: 0.05 , 0.505 , 5.556 .
2.9 SUMMARY
1. A resistor is a passive element in which the voltage v across it is
directly proportional to the current i through it. That is, a resistor is
a device that obeys Ohm’s law,
v = iR
where R is the resistance of the resistor.
2. A short circuit is a resistor (a perfectly conducting wire) with zero
resistance (R = 0). An open circuit is a resistor with infinite resis-
tance (R = ∞).
3. The conductance G of a resistor is the reciprocal of its resistance:
G =
1
R
4. A branch is a single two-terminal element in an electric circuit. A
node is the point of connection between two or more branches. A
loop is a closed path in a circuit. The number of branches b, the
number of nodes n, and the number of independent loops l in a
network are related as
b = l + n − 1
CHAPTER 2 Basic Laws 61
5. Kirchhoff’s current law (KCL) states that the currents at any node
algebraically sum to zero. In other words, the sum of the currents
entering a node equals the sum of currents leaving the node.
6. Kirchhoff’s voltage law (KVL) states that the voltages around a
closed path algebraically sum to zero. In other words, the sum of
voltage rises equals the sum of voltage drops.
7. Two elements are in series when they are connected sequentially,
end to end. When elements are in series, the same current flows
through them (i1 = i2). They are in parallel if they are connected to
the same two nodes. Elements in parallel always have the same
voltage across them (v1 = v2).
8. When two resistors R1 (= 1/G1) and R2 (= 1/G2) are in series,
their equivalent resistance Req and equivalent conductance Geq are
Req = R1 + R2, Geq =
G1G2
G1 + G2
9. When two resistors R1 (= 1/G1) and R2 (= 1/G2) are in parallel,
their equivalent resistance Req and equivalent conductance Geq are
Req =
R1R2
R1 + R2
, Geq = G1 + G2
10. The voltage division principle for two resistors in series is
v1 =
R1
R1 + R2
v, v2 =
R2
R1 + R2
v
11. The current division principle for two resistors in parallel is
i1 =
R2
R1 + R2
i, i2 =
R1
R1 + R2
i
12. The formulas for a delta-to-wye transformation are
R1 =
RbRc
Ra + Rb + Rc
, R2 =
RcRa
Ra + Rb + Rc
R3 =
RaRb
Ra + Rb + Rc
13. The formulas for a wye-to-delta transformation are
Ra =
R1R2 + R2R3 + R3R1
R1
, Rb =
R1R2 + R2R3 + R3R1
R2
Rc =
R1R2 + R2R3 + R3R1
R3
14. The basic laws covered in this chapter can be applied to the prob-
lems of electrical lighting and design of dc meters.
REVIEW QUESTIONS
2.1 The reciprocal of resistance is:
(a) voltage (b) current
(c) conductance (d) coulombs
2.2 An electric heater draws 10 A from a 120-V line.
The resistance of the heater is:
(a) 1200  (b) 120 
(c) 12  (d) 1.2
62 PART 1 DC Circuits
2.3 The voltage drop across a 1.5-kW toaster that draws
12 A of current is:
(a) 18 kV (b) 125 V
(c) 120 V (d) 10.42 V
2.4 The maximum current that a 2W, 80 k resistor can
safely conduct is:
(a) 160 kA (b) 40 kA
(c) 5 mA (d) 25 µA
2.5 A network has 12 branches and 8 independent loops.
How many nodes are there in the network?
(a) 19 (b) 17 (c) 5 (d) 4
2.6 The current I in the circuit in Fig. 2.63 is:
(a) −0.8 A (b) −0.2 A
(c) 0.2 A (d) 0.8 A
3 V 5 V
+
−
+
−
4 Ω I
6 Ω
Figure 2.63 For Review Question 2.6.
2.7 The current Io in Fig. 2.64 is:
(a) −4 A (b) −2 A (c) 4 A (d) 16 A
10 A
4 A
2 A
Io
Figure 2.64 For Review Question 2.7.
2.8 In the circuit in Fig. 2.65, V is:
(a) 30 V (b) 14 V (c) 10 V (d) 6 V
+
−
+
−
+ −
+ −
10 V
12 V 8 V
V
Figure 2.65 For Review Question 2.8.
2.9 Which of the circuits in Fig. 2.66 will give you
Vab = 7 V?
3 V
a
b
5 V
1 V
(a)
+
−
+ −
+ −
3 V
a
b
5 V
1 V
(b)
+
−
+
−
+ −
3 V
a
5 V
1 V
(c)
+
−
+ −
+
− b
3 V
a
5 V
1 V
(d)
+
−
+
−
+
− b
Figure 2.66 For Review Question 2.9.
2.10 The equivalent resistance of the circuit in Fig. 2.67
is:
(a) 4 k (b) 5 k (c) 8 k (d) 14 k
2 kΩ 3 kΩ
Req
6 kΩ 3 kΩ
Figure 2.67 For Review Question 2.10.
Answers: 2.1c, 2.2c, 2.3b, 2.4c, 2.5c, 2.6b, 2.7a, 2.8d, 2.9d, 2.10a.
CHAPTER 2 Basic Laws 63
PROBLEMS
Section 2.2 Ohm’s Law
2.1 The voltage across a 5-k resistor is 16 V. Find the
current through the resistor.
2.2 Find the hot resistance of a lightbulb rated 60 W,
120 V.
2.3 When the voltage across a resistor is 120 V, the
current through it is 2.5 mA. Calculate its
conductance.
2.4 (a) Calculate current i in Fig. 2.68 when the switch
is in position 1.
(b) Find the current when the switch is in position 2.
+
−
150 Ω
100 Ω
3 V
1 2
i
Figure 2.68 For Prob. 2.4.
Section 2.3 Nodes, Branches, and Loops
2.5 For the network graph in Fig. 2.69, find the number
of nodes, branches, and loops.
Figure 2.69 For Prob. 2.5.
2.6 In the network graph shown in Fig. 2.70, determine
the number of branches and nodes.
Figure 2.70 For Prob. 2.6.
2.7 Determine the number of branches and nodes in the
circuit in Fig. 2.71.
+
−
6 Ω
3 Ω
2 Ω
5 Ω
10 V
5i
4 Ω i
Figure 2.71 For Prob. 2.7.
Section 2.4 Kirchhoff’s Laws
2.8 Use KCL to obtain currents i1, i2, and i3 in the
circuit shown in Fig. 2.72.
8 mA
9 mA
12 mA
i1
i3
i2
Figure 2.72 For Prob. 2.8.
2.9 Find i1, i2, and i3 in the circuit in Fig. 2.73.
i2
i3
10 A i1 3 A
2 A
1 A
Figure 2.73 For Prob. 2.9.
64 PART 1 DC Circuits
2.10 Determine i1 and i2 in the circuit in Fig. 2.74.
3 A
4 A –2 A
i2
i1
Figure 2.74 For Prob. 2.10.
2.11 Determine v1 through v4 in the circuit in Fig. 2.75.
12 V
+
−
10 V
−
+
+ −
8 V
+
−
v1
+
−
v3
+
−
v2
−
+
v4
− +
6 V
Figure 2.75 For Prob. 2.11.
2.12 In the circuit in Fig. 2.76, obtain v1, v2, and v3.
+
−
20 V
+
−
v1
+
−
v3
25 V 10 V
15 V
v2
+ −
+
− + − + −
Figure 2.76 For Prob. 2.12.
2.13 Find v1 and v2 in the circuit in Fig. 2.77.
6 V v1
+
−
+
−
v1
v2
+ − + −
+ −
+ −
12 V 10 V
Figure 2.77 For Prob. 2.13.
2.14 Obtain v1 through v3 in the circuit of Fig. 2.78.
24 V
12 V
10 V
v3
v2
+
−
+
−
+
−
+
−
+
−
v1
+ −
Figure 2.78 For Prob. 2.14.
2.15 Find I and Vab in the circuit of Fig. 2.79.
5 Ω
3 Ω
+
−
+
−
+
−
Vab
30 V 8 V
b
a
+
−
10 V
I
Figure 2.79 For Prob. 2.15.
2.16 From the circuit in Fig. 2.80, find I, the power
dissipated by the resistor, and the power supplied by
each source.
−8 V
10 V
12 V 3 Ω
+
−
+ −
+ −
I
Figure 2.80 For Prob. 2.16.
CHAPTER 2 Basic Laws 65
2.17 Determine io in the circuit of Fig. 2.81.
36 V +
−
4 Ω
+
− 5io
io
Figure 2.81 For Prob. 2.17.
2.18 Calculate the power dissipated in the 5- resistor in
the circuit of Fig. 2.82.
–
+
45 V
1 Ω
5 Ω
3Vo
+ −
Vo
+
−
Figure 2.82 For Prob. 2.18.
2.19 Find Vo in the circuit in Fig. 2.83 and the power
dissipated by the controlled source.
10 A
6 Ω 2Vo
+ −
4 Ω
Vo
Figure 2.83 For Prob. 2.19.
2.20 For the circuit in Fig. 2.84, find Vo/Vs in terms of
α, R1, R2, R3, and R4. If R1 = R2 = R3 = R4,
what value of α will produce |Vo/Vs| = 10?
Vo
+
−
+
−
R4
R3
R1
R2 aIo
Vs
Io
Figure 2.84 For Prob. 2.20.
2.21 For the network in Fig. 2.85, find the current,
voltage, and power associated with the 20-k
resistor.
0.01Vo
Vo
+
−
20 kΩ
5 kΩ
10 kΩ
5 mA
Figure 2.85 For Prob. 2.21.
Sections 2.5 and 2.6 Series and Parallel
Resistors
2.22 For the circuit in Fig. 2.86, find i1 and i2.
4 kΩ
6 kΩ
20 mA
i1 i2
Figure 2.86 For Prob. 2.22.
2.23 Find v1 and v2 in the circuit in Fig. 2.87.
24 V
3 kΩ
9 kΩ
v1
v2
+
−
+ −
+
−
Figure 2.87 For Prob. 2.23.
2.24 Find v1, v2, and v3 in the circuit in Fig. 2.88.
40 V
14 Ω
15 Ω
v1
v2
+
−
+ −
+
−
10 Ω
v3
+
−
Figure 2.88 For Prob. 2.24.
2.25 Calculate v1, i1, v2, and i2 in the circuit of Fig. 2.89.
3 Ω
4 Ω 6 Ω
i1
i2
v1
v2
+
−
+ −
+
−
12 V
Figure 2.89 For Prob. 2.25.
66 PART 1 DC Circuits
2.26 Find i, v, and the power dissipated in the 6-
resistor in Fig. 2.90.
9 A 4 Ω
8 Ω
6 Ω
i
+
−
v
Figure 2.90 For Prob. 2.26.
2.27 In the circuit in Fig. 2.91, find v, i, and the power
absorbed by the 4- resistor.
20 V 6 Ω
10 Ω
5 Ω
+
−
4 Ω
+
−
v
i
Figure 2.91 For Prob. 2.27.
2.28 Find i1 through i4 in the circuit in Fig. 2.92.
20 A
10 Ω
40 Ω
i4
i3
20 Ω
30 Ω
i2
i1
Figure 2.92 For Prob. 2.28.
2.29 Obtain v and i in the circuit in Fig. 2.93.
9 A 2 S
1 S
4 S 6 S
3 S
+
−
v
i
Figure 2.93 For Prob. 2.29.
2.30 Determine i1, i2, v1, and v2 in the ladder network in
Fig. 2.94. Calculate the power dissipated in the 2-
resistor.
28 V 13 Ω
15 Ω
6 Ω
+
−
8 Ω 2 Ω
4 Ω
−
10 Ω
12 Ω
+
v1
−
+
v2
i1 i2
Figure 2.94 For Prob. 2.30.
2.31 Calculate Vo and Io in the circuit of Fig. 2.95.
50 V
30 Ω
70 Ω
+
−
5 Ω
20 Ω
+
−
Vo
Io
Figure 2.95 For Prob. 2.31.
2.32 Find Vo and Io in the circuit of Fig. 2.96.
4 V 6 Ω
3 Ω
1 Ω
+
−
Vo
8 Ω
2 Ω
+
−
Io
Figure 2.96 For Prob. 2.32.
2.33 In the circuit of Fig. 2.97, find R if Vo = 4V.
20 V 6 Ω
16 Ω
+
−
+
−
Vo
R
Figure 2.97 For Prob. 2.33.
CHAPTER 2 Basic Laws 67
2.34 Find I and Vs in the circuit of Fig. 2.98 if the current
through the 3- resistor is 2 A.
4 Ω
2 A
10 Ω
+
−
2 Ω
3 Ω
6 Ω
V
s
I
Figure 2.98 For Prob. 2.34.
2.35 Find the equivalent resistance at terminals a-b for
each of the networks in Fig. 2.99.
R
(a) (b) (c)
a
b
3R R
R
R
(d) (e)
a
b
R
R
R
R
a
b
R
R
R
R
a b
R 2R 3R
a
b
Figure 2.99 For Prob. 2.35.
2.36 For the ladder network in Fig. 2.100, find I and Req.
10 V 6 Ω
2 Ω
+
−
3 Ω 1 Ω
2 Ω
4 Ω
I
Req
Figure 2.100 For Prob. 2.36.
2.37 If Req = 50  in the circuit in Fig. 2.101, find R.
Req
30 Ω
10 Ω
60 Ω
R
12 Ω 12 Ω 12 Ω
Figure 2.101 For Prob. 2.37.
2.38 Reduce each of the circuits in Fig. 2.102 to a single
resistor at terminals a-b.
8 Ω
5 Ω
20 Ω
30 Ω
a b
(a)
5 Ω
4 Ω
8 Ω
5 Ω
10 Ω
4 Ω
2 Ω
3 Ω
a b
(b)
Figure 2.102 For Prob. 2.38.
2.39 Calculate the equivalent resistance Rab at terminals
a-b for each of the circuits in Fig. 2.103.
40 Ω
10 Ω
5 Ω
20 Ω
(a)
a
b
30 Ω
80 Ω
60 Ω
(b)
a
b
10 Ω
20 Ω
Figure 2.103 For Prob. 2.39.
68 PART 1 DC Circuits
2.40 Obtain the equivalent resistance at the terminals a-b
for each of the circuits in Fig. 2.104.
11 Ω
10 Ω
20 Ω
6 Ω
5 Ω
4 Ω
9 Ω
8 Ω
4 Ω
5 Ω
15 Ω
(b)
a
b
(a)
a
b
10 Ω
20 Ω
60 Ω 30 Ω
Figure 2.104 For Prob. 2.40.
2.41 Find Req at terminals a-b for each of the circuits in
Fig. 2.105.
(a)
a
b
40 Ω
70 Ω
30 Ω
60 Ω
20 Ω
(b)
a
b
6 Ω
40 Ω
60 Ω
30 Ω
20 Ω
50 Ω
80 Ω
10 Ω
70 Ω
4 Ω
8 Ω
Figure 2.105 For Prob. 2.41.
2.42 Find the equivalent resistance Rab in the circuit of
Fig. 2.106.
a
d e
f
b
c
6 Ω
3 Ω
5 Ω
20 Ω
10 Ω 8 Ω
Figure 2.106 For Prob. 2.42.
Section 2.7 Wye-Delta Transformations
2.43 Convert the circuits in Fig. 2.107 from Y to .
10 Ω 10 Ω
10 Ω
b
a
c
(a)
20 Ω
30 Ω
50 Ω
a
(b)
b
c
Figure 2.107 For Prob. 2.43.
2.44 Transform the circuits in Fig. 2.108 from  to Y.
12 Ω
12 Ω 12 Ω
(a)
a b
c
60 Ω
30 Ω 10 Ω
(b)
a b
c
Figure 2.108 For Prob. 2.44.
CHAPTER 2 Basic Laws 69
2.45 What value of R in the circuit of Fig. 2.109 would
cause the current source to deliver 800 mW to the
resistors?
30 mA
R R
R
R
R
Figure 2.109 For Prob. 2.45.
2.46 Obtain the equivalent resistance at the terminals a-b
for each of the circuits in Fig. 2.110.
(a)
b
a
30 Ω
10 Ω
10 Ω
20 Ω
20 Ω
10 Ω
20 Ω
10 Ω
30 Ω
25 Ω
(b)
b
a
15 Ω
5 Ω
Figure 2.110 For Prob. 2.46.
2.47
∗
Find the equivalent resistance Rab in each of the
circuits of Fig. 2.111. Each resistor is 100 .
(a)
b
a
(b)
b
a
Figure 2.111 For Prob. 2.47.
2.48
∗
Obtain the equivalent resistance Rab in each of the
circuits of Fig. 2.112. In (b), all resistors have a
value of 30 .
(b)
40 Ω
50 Ω
10 Ω
60 Ω
30 Ω
20 Ω
(a)
b
a
80 Ω
30 Ω
a
b
Figure 2.112 For Prob. 2.48.
2.49 Calculate Io in the circuit of Fig. 2.113.
20 Ω
40 Ω
60 Ω
50 Ω
10 Ω
20 Ω
24 V +
−
Io
Figure 2.113 For Prob. 2.49.
∗An asterisk indicates a challenging problem.
70 PART 1 DC Circuits
2.50 Determine V in the circuit of Fig. 2.114.
100 V
30 Ω
15 Ω 10 Ω
16 Ω
35 Ω 12 Ω 20 Ω
+
− V
+
−
Figure 2.114 For Prob. 2.50.
2.51
∗
Find Req and I in the circuit of Fig. 2.115.
2 Ω
4 Ω
12 Ω
6 Ω 1 Ω
8 Ω 2 Ω
3 Ω
10 Ω
5 Ω
4 Ω
20 V +
−
Req
I
Figure 2.115 For Prob. 2.51.
Section 2.8 Applications
2.52 The lightbulb in Fig. 2.116 is rated 120 V, 0.75 A.
Calculate Vs to make the lightbulb operate at the
rated conditions.
+
−
40 Ω
Vs 80 Ω
Bulb
Figure 2.116 For Prob. 2.52.
2.53 Three lightbulbs are connected in series to a 100-V
battery as shown in Fig. 2.117. Find the current I
through the bulbs.
30 W 40 W 50 W
100 V +
−
I
Figure 2.117 For Prob. 2.53.
2.54 If the three bulbs of Prob. 2.53 are connected in
parallel to the 100-V battery, calculate the current
through each bulb.
2.55 As a design engineer, you are asked to design a
lighting system consisting of a 70-W power supply
and two lightbulbs as shown in Fig. 2.118. You must
select the two bulbs from the following three
available bulbs.
R1 = 80 , cost = $0.60 (standard size)
R2 = 90 , cost = $0.90 (standard size)
R3 = 100 , cost = $0.75 (nonstandard size)
The system should be designed for minimum cost
such that I = 1.2 A ± 5 percent.
I
Rx Ry
70 W
Power
Supply
+
−
Figure 2.118 For Prob. 2.55.
2.56 If an ammeter with an internal resistance of 100 
and a current capacity of 2 mA is to measure 5 A,
determine the value of the resistance needed.
Calculate the power dissipated in the shunt resistor.
2.57 The potentiometer (adjustable resistor) Rx in Fig.
2.119 is to be designed to adjust current ix from 1 A
to 10 A. Calculate the values of R and Rx to achieve
this.
+
−
ix R
Rx
ix
110 V
Figure 2.119 For Prob. 2.57.
2.58 A d’Arsonval meter with an internal resistance of 1
k requires 10 mA to produce full-scale deflection.
Calculate the value of a series resistance needed to
measure 50 V of full scale.
CHAPTER 2 Basic Laws 71
2.59 A 20-k/V voltmeter reads 10 V full scale.
(a) What series resistance is required to make the
meter read 50 V full scale?
(b) What power will the series resistor dissipate
when the meter reads full scale?
2.60 (a) Obtain the voltage vo in the circuit of Fig.
2.120(a).
(b) Determine the voltage v
o measured when a
voltmeter with 6-k internal resistance is
connected as shown in Fig. 2.120(b).
(c) The finite resistance of the meter introduces an
error into the measurement. Calculate the
percent error as




vo − v
o
vo



 × 100%
(d) Find the percent error if the internal resistance
were 36 k.
+
−
2 mA
1 kΩ
5 kΩ 4 kΩ vo
(a)
(b)
2 mA
+
−
1 kΩ
5 kΩ 4 kΩ Voltmeter
vo
Figure 2.120 For Prob. 2.60.
2.61 (a) Find the current i in the circuit of Fig. 2.121(a).
(b) An ammeter with an internal resistance of 1  is
inserted in the network to measure i
as shown in
Fig. 2.121(b). What is i
?
(c) Calculate the percent error introduced by the
meter as




i − i
i



 × 100%
+
−
i
4 V
16 Ω
40 Ω 60 Ω
(a)
+
−
i'
4 V
16 Ω
40 Ω 60 Ω
(b)
Ammeter
Figure 2.121 For Prob. 2.61.
2.62 A voltmeter is used to measure Vo in the circuit in
Fig. 2.122. The voltmeter model consists of an ideal
voltmeter in parallel with a 100-k resistor. Let
Vs = 40 V, Rs = 10 k, and R1 = 20 k. Calculate
Vo with and without the voltmeter when
(a) R2 = 1 k (b) R2 = 10 k
(c) R2 = 100 k
+
−
+
−
V
100 kΩ
Vo
Vs
Rs
R1
R2
Figure 2.122 For Prob. 2.62.
2.63 An ammeter model consists of an ideal ammeter in
series with a 20- resistor. It is connected with a
current source and an unknown resistor Rx as shown
in Fig. 2.123. The ammeter reading is noted. When
a potentiometer R is added and adjusted until the
ammeter reading drops to one half its previous
reading, then R = 65 . What is the value of Rx?
I
A
R
Rx
20 Ω
Ammeter
model
Figure 2.123 For Prob. 2.63.
2.64 The circuit in Fig. 2.124 is to control the speed of a
motor such that the motor draws currents 5 A, 3 A,
72 PART 1 DC Circuits
and 1 A when the switch is at high, medium, and
low positions, respectively. The motor can be
modeled as a load resistance of 20 m. Determine
the series dropping resistances R1, R2, and R3.
6 V
High
Medium
Low
10-A, 0.01-Ω fuse
R1
R2
R3
Motor
Figure 2.124 For Prob. 2.64.
2.65 An ohmmeter is constructed with a 2-V battery and
0.1-mA (full-scale) meter with 100- internal
resistance.
(a) Calculate the resistance of the (variable) resistor
required to be in series with the meter and the
battery.
(b) Determine the unknown resistance across the
terminals of the ohmmeter that will cause the
meter to deflect half scale.
COMPREHENSIVE PROBLEMS
2.66 An electric heater connected to a 120-V source
consists of two identical 0.4- elements made of
Nichrome wire. The elements provide low heat
when connected in series and high heat when
connected in parallel. Find the power at low and
high heat settings.
2.67 Suppose your circuit laboratory has the following
standard commercially available resistors in large
quantities:
1.8  20  300  24 k 56 k
Using series and parallel combinations and a
minimum number of available resistors, how would
you obtain the following resistances for an
electronic circuit design?
(a) 5  (b) 311.8 
(c) 40 k (d) 52.32 k
2.68 In the circuit in Fig. 2.125, the wiper divides the
potentiometer resistance between αR and (1 − α)R,
0 ≤ α ≤ 1. Find vo/vs.
vo
+
−
+
− R
R
aR
vs
Figure 2.125 For Prob. 2.68.
2.69 An electric pencil sharpener rated 240 mW, 6 V is
connected to a 9-V battery as shown in Fig. 2.126.
Calculate the value of the series-dropping resistor
Rx needed to power the sharpener.
9 V
Switch Rx
Figure 2.126 For Prob. 2.69.
2.70 A loudspeaker is connected to an amplifier as shown
in Fig. 2.127. If a 10- loudspeaker draws the
maximum power of 12 W from the amplifier,
determine the maximum power a 4- loudspeaker
will draw.
Amplifier
Loudspeaker
Figure 2.127 For Prob. 2.70.
2.71 In a certain application, the circuit in Fig. 2.128
must be designed to meet these two criteria:
(a) Vo/Vs = 0.05 (b) Req = 40 k
CHAPTER 2 Basic Laws 73
If the load resistor 5 k is fixed, find R1 and R2 to
meet the criteria.
Vs
+
−
+
−
5 kΩ
Vo
R2
R1
Req
Figure 2.128 For Prob. 2.71.
2.72 The pin diagram of a resistance array is shown in
Fig. 2.129. Find the equivalent resistance between
the following:
(a) 1 and 2 (b) 1 and 3 (c) 1 and 4
20 Ω 20 Ω
40 Ω
10 Ω
10 Ω
1 2
3
4
80 Ω
Figure 2.129 For Prob. 2.72.
2.73 Two delicate devices are rated as shown in Fig.
2.130. Find the values of the resistors R1 and R2
needed to power the devices using a 24-V battery.
Device 1
Device 2
24 V
R1
R2
60-mA, 2-Ω fuse
9 V, 45 mW
24 V, 480 mW
Figure 2.130 For Prob. 2.73.
75
C H A P T E R
METHODS OF ANALYSIS
3
Scientists study the world as it is, engineers create the world that never
has been.
—Theodore von Karman
Enhancing Your Career
Career in Electronics One area of application for electric
circuit analysis is electronics. The term electronics was orig-
inally used to distinguish circuits of very low current levels.
This distinction no longer holds, as power semiconductor de-
vices operate at high levels of current. Today, electronics is
regarded as the science of the motion of charges in a gas, vac-
uum, or semiconductor. Modern electronics involves tran-
sistors and transistor circuits. The earlier electronic circuits
were assembled from components. Many electronic circuits
are now produced as integrated circuits, fabricated in a semi-
conductor substrate or chip.
Electronic circuits find applications in many areas,
such as automation, broadcasting, computers, and instru-
mentation. The range of devices that use electronic circuits
is enormous and is limited only by our imagination. Radio,
television, computers, and stereo systems are but a few.
An electrical engineer usually performs diverse func-
tions and is likely to use, design, or construct systems that
incorporate some form of electronic circuits. Therefore, an
understanding of the operation and analysis of electronics
is essential to the electrical engineer. Electronics has
become a specialty distinct from other disciplines within
electrical engineering. Because the field of electronics
is ever advancing, an electronics engineer must update
his/her knowledge from time to time. The best way to do
this is by being a member of a professional organization
such as the Institute of Electrical and Electronics Engineers
Troubleshooting an electronic circuit board. Source: T. J. Mal-
oney, Modern Industrial Electronics, 3rd ed. Englewood Cliffs, NJ:
Prentice Hall, 1996, p. 408.
(IEEE). With a membership of over 300,000, the IEEE is
the largest professional organization in the world. Members
benefit immensely from the numerous magazines, journals,
transactions, and conference/symposium proceedings pub-
lished yearly by IEEE. You should consider becoming an
IEEE member.
76 PART 1 DC Circuits
3.1 INTRODUCTION
Having understood the fundamental laws of circuit theory (Ohm’s law and
Kirchhoff’s laws), we are now prepared to apply these laws to develop
two powerful techniques for circuit analysis: nodal analysis, which is
based on a systematic application of Kirchhoff’s current law (KCL), and
mesh analysis, which is based on a systematic application of Kirchhoff’s
voltage law (KVL). The two techniques are so important that this chapter
should be regarded as the most important in the book. Students are
therefore encouraged to pay careful attention.
With the two techniques to be developed in this chapter, we can
analyze almost any circuit by obtaining a set of simultaneous equations
that are then solved to obtain the required values of current or voltage.
One method of solving simultaneous equations involves Cramer’s rule,
whichallowsustocalculatecircuitvariablesasaquotientofdeterminants.
The examples in the chapter will illustrate this method; Appendix A also
briefly summarizes the essentials the reader needs to know for applying
Cramer’s rule.
Also in this chapter, we introduce the use of PSpice for Windows, a
circuit simulation computer software program that we will use throughout
thetext. Finally, weapplythetechniqueslearnedinthischaptertoanalyze
transistor circuits.
3.2 NODAL ANALYSIS
Nodal analysis provides a general procedure for analyzing circuits using
node voltages as the circuit variables. Choosing node voltages instead
of element voltages as circuit variables is convenient and reduces the
number of equations one must solve simultaneously.
To simplify matters, we shall assume in this section that circuits do
not contain voltage sources. Circuits that contain voltage sources will be
analyzed in the next section.
Nodal analysis is also known as the node-voltage
method.
In nodal analysis, we are interested in finding the node voltages.
Given a circuit with n nodes without voltage sources, the nodal analysis
of the circuit involves taking the following three steps.
Steps to Determine Node Voltages:
1. Select a node as the reference node. Assign voltages
v1, v2, . . . , vn−1 to the remaining n − 1 nodes. The voltages are
referenced with respect to the reference node.
2. Apply KCL to each of the n − 1 nonreference nodes. Use Ohm’s
law to express the branch currents in terms of node voltages.
3. Solve the resulting simultaneous equations to obtain the unknown
node voltages.
We shall now explain and apply these three steps.
The first step in nodal analysis is selecting a node as the reference
or datum node. The reference node is commonly called the ground since
it is assumed to have zero potential. A reference node is indicated by
CHAPTER 3 Methods of Analysis 77
any of the three symbols in Fig. 3.1. The type of ground in Fig. 3.1(b) is
called a chassis ground and is used in devices where the case, enclosure,
or chassis acts as a reference point for all circuits. When the potential of
the earth is used as reference, we use the earth ground in Fig. 3.1(a) or
(c). We shall always use the symbol in Fig. 3.1(b).
The number of nonreference nodes is equal to
the number of independent equations that we
will derive.
Once we have selected a reference node, we assign voltage desig-
nations to nonreference nodes. Consider, for example, the circuit in Fig.
3.2(a). Node 0 is the reference node (v = 0), while nodes 1 and 2 are
assigned voltages v1 and v2, respectively. Keep in mind that the node
voltages are defined with respect to the reference node. As illustrated in
Fig. 3.2(a), each node voltage is the voltage rise from the reference node
to the corresponding nonreference node or simply the voltage of that node
with respect to the reference node.
(a) (b) (c)
Figure3.1 Common symbols for
indicating a reference node.
As the second step, we apply KCL to each nonreference node in the
circuit. To avoid putting too much information on the same circuit, the
circuit in Fig. 3.2(a) is redrawn in Fig. 3.2(b), where we now add i1, i2,
and i3 as the currents through resistors R1, R2, and R3, respectively. At
node 1, applying KCL gives
I1 = I2 + i1 + i2 (3.1)
At node 2,
I2 + i2 = i3 (3.2)
We now apply Ohm’s law to express the unknown currents i1, i2, and i3
in terms of node voltages. The key idea to bear in mind is that, since
resistance is a passive element, by the passive sign convention, current
must always flow from a higher potential to a lower potential.
Current flows from a higher potential to a lower potential in a resistor.
(a)
(b)
1 2
v1
i1
i2 i2
i3
v2
I2
0
R3
v2
+
−
R3
R1
v1
+
−
R1
I1
I2
R2
R2
I1
Figure 3.2 Typicalcircuitfornodal
analysis.
We can express this principle as
i =
vhigher − vlower
R
(3.3)
Note that this principle is in agreement with the way we defined resistance
in Chapter 2 (see Fig. 2.1). With this in mind, we obtain from Fig. 3.2(b),
i1 =
v1 − 0
R1
or i1 = G1v1
i2 =
v1 − v2
R2
or i2 = G2(v1 − v2)
i3 =
v2 − 0
R3
or i3 = G3v2
(3.4)
Substituting Eq. (3.4) in Eqs. (3.1) and (3.2) results, respectively, in
I1 = I2 +
v1
R1
+
v1 − v2
R2
(3.5)
I2 +
v1 − v2
R2
=
v2
R3
(3.6)
78 PART 1 DC Circuits
In terms of the conductances, Eqs. (3.5) and (3.6) become
I1 = I2 + G1v1 + G2(v1 − v2) (3.7)
I2 + G2(v1 − v2) = G3v2 (3.8)
The third step in nodal analysis is to solve for the node voltages. If
we apply KCL to n−1 nonreference nodes, we obtain n−1 simultaneous
equations such as Eqs. (3.5) and (3.6) or (3.7) and (3.8). For the circuit
of Fig. 3.2, we solve Eqs. (3.5) and (3.6) or (3.7) and (3.8) to obtain the
node voltages v1 and v2 using any standard method, such as the substitu-
tion method, the elimination method, Cramer’s rule, or matrix inversion.
To use either of the last two methods, one must cast the simultaneous
equations in matrix form. For example, Eqs. (3.7) and (3.8) can be cast
in matrix form as

G1 + G2
−G2
−G2
G2 + G3
 
v1
v2

=

I1 − I2
I2

(3.9)
which can be solved to get v1 and v2. Equation 3.9 will be generalized
in Section 3.6. The simultaneous equations may also be solved using
calculators such as HP48 or with software packages such as Matlab,
Mathcad, Maple, and Quattro Pro.
AppendixAdiscusseshowtouseCramer’srule.
E X A M P L E 3 . 1
Calculate the node voltages in the circuit shown in Fig. 3.3(a).
Solution:
Consider Fig. 3.3(b), where the circuit in Fig. 3.3(a) has been prepared for
nodal analysis. Notice how the currents are selected for the application
of KCL. Except for the branches with current sources, the labeling of the
currents is arbitrary but consistent. (By consistent, we mean that if, for
example, we assume that i2 enters the 4 resistor from the left-hand side,
i2 must leave the resistor from the right-hand side.) The reference node
is selected, and the node voltages v1 and v2 are now to be determined.
At node 1, applying KCL and Ohm’s law gives
i1 = i2 + i3 ⇒ 5 =
v1 − v2
4
+
v1 − 0
2
Multiplying each term in the last equation by 4, we obtain
20 = v1 − v2 + 2v1
or
3v1 − v2 = 20 (3.1.1)
2
1
5 A
10 A
2 Ω 6 Ω
4 Ω
(a)
5 A
10 A
2 Ω 6 Ω
4 Ω
(b)
i1 = 5 i1 = 5
i4 = 10
i2
i3
i2 i5
v2
v1
Figure 3.3 For Example 3.1: (a) original
circuit, (b) circuit for analysis.
At node 2, we do the same thing and get
i2 + i4 = i1 + i5 ⇒
v1 − v2
4
+ 10 = 5 +
v2 − 0
6
Multiplying each term by 12 results in
3v1 − 3v2 + 120 = 60 + 2v2
or
−3v1 + 5v2 = 60 (3.1.2)
CHAPTER 3 Methods of Analysis 79
Now we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve
the equations using any method and obtain the values of v1 and v2.
METHOD 1 Using the elimination technique, we add Eqs. (3.1.1) and
(3.1.2).
4v2 = 80 ⇒ v2 = 20 V
Substituting v2 = 20 in Eq. (3.1.1) gives
3v1 − 20 = 20 ⇒ v1 =
40
3
= 13.33 V
METHOD 2 To use Cramer’s rule, we need to put Eqs. (3.1.1) and
(3.1.2) in matrix form as

3
−3
−1
5
 
v1
v2

=

20
60

(3.1.3)
The determinant of the matrix is
=




3
−3
−1
5



 = 15 − 3 = 12
We now obtain v1 and v2 as
v1 =
1
=




20
60
−1
5




=
100 + 60
12
= 13.33 V
v2 =
2
=




3
−3
20
60




=
180 + 60
12
= 20 V
giving us the same result as did the elimination method.
If we need the currents, we can easily calculate them from the values
of the nodal voltages.
i1 = 5 A, i2 =
v1 − v2
4
= −1.6667 A, i3 =
v1
2
= 6.666
i4 = 10 A, i5 =
v2
6
= 3.333 A
The fact that i2 is negative shows that the current flows in the direction
opposite to the one assumed.
P R A C T I C E P R O B L E M 3 . 1
Obtain the node voltages in the circuit in Fig. 3.4.
1 A 4 A
6 Ω
2 Ω 7 Ω
1 2
Figure3.4 For Practice Prob. 3.1.
Answer: v1 = −2 V, v2 = −14 V.
80 PART 1 DC Circuits
E X A M P L E 3 . 2
Determine the voltages at the nodes in Fig. 3.5(a).
Solution:
The circuit in this example has three nonreference nodes, unlike the pre-
vious example which has two nonreference nodes. We assign voltages to
the three nodes as shown in Fig. 3.5(b) and label the currents.
4 Ω
4 Ω
2 Ω 8 Ω
ix
1 3
2
0
3 A 2ix
(a)
ix ix
i3
4 Ω
4 Ω
2 Ω 8 Ω
i1
v1
v2
i2 i2
i1
v3
3 A
3 A
2ix
(b)
Figure3.5 For Example 3.2: (a) original circuit, (b) circuit for analysis.
At node 1,
3 = i1 + ix ⇒ 3 =
v1 − v3
4
+
v1 − v2
2
Multiplying by 4 and rearranging terms, we get
3v1 − 2v2 − v3 = 12 (3.2.1)
At node 2,
ix = i2 + i3 ⇒
v1 − v2
2
=
v2 − v3
8
+
v2 − 0
4
Multiplying by 8 and rearranging terms, we get
−4v1 + 7v2 − v3 = 0 (3.2.2)
At node 3,
i1 + i2 = 2ix ⇒
v1 − v3
4
+
v2 − v3
8
=
2(v1 − v2)
2
Multiplying by 8, rearranging terms, and dividing by 3, we get
2v1 − 3v2 + v3 = 0 (3.2.3)
We have three simultaneous equations to solve to get the node voltages
v1, v2, and v3. We shall solve the equations in two ways.
METHOD 1 Using the elimination technique, we add Eqs. (3.2.1) and
(3.2.3).
5v1 − 5v2 = 12
CHAPTER 3 Methods of Analysis 81
or
v1 − v2 =
12
5
= 2.4 (3.2.4)
Adding Eqs. (3.2.2) and (3.2.3) gives
−2v1 + 4v2 = 0 ⇒ v1 = 2v2 (3.2.5)
Substituting Eq. (3.2.5) into Eq. (3.2.4) yields
2v2 − v2 = 2.4 ⇒ v2 = 2.4, v1 = 2v2 = 4.8 V
From Eq. (3.2.3), we get
v3 = 3v2 − 2v1 = 3v2 − 4v2 = −v2 = −2.4 V
Thus,
v1 = 4.8 V, v2 = 2.4 V, v3 = −2.4 V
METHOD 2 To use Cramer’s rule, we put Eqs. (3.2.1) to (3.2.3) in
matrix form.


3
−4
2
−2
7
−3
−1
−1
1




v1
v2
v3

 =


12
0
0


From this, we obtain
v1 =
1
, v2 =
2
, v3 =
3
where , 1, 2, and 3 are the determinants to be calculated as follows.
As explained in Appendix A, to calculate the determinant of a 3 by 3
matrix, we repeat the first two rows and cross multiply.
−
−
− +
+
+
= 21 − 12 + 4 + 14 − 9 − 8 = 10
3
3
7
7
−4
−2
−2
−4
=
= −4
−2
−1
−1
−1
−1
−1
−1
−3
−3
1
1
2
2
3
7

Similarly, we obtain
= 84 + 0 + 0 − 0 − 36 − 0 = 48
−
−
− +
+
+
7
7
−2
−2
=
0
0
0
12
12
−3
1
−1
−1
1
−1
−1
82 PART 1 DC Circuits
= 0 + 0 − 24 − 0 − 0 + 48 = 24
−
−
− +
+
+
3
3
−4
−4
−1
−1
1
−1
−1
=
0
0
0
12
12
2
2
= 0 + 144 + 0 − 168 − 0 − 0 = −24
−
−
− +
+
+
3
3
7
7
−4
−2
−2
−4
=
0
0
0
12
12
−3
2
3
Thus, we find
v1 =
1
=
48
10
= 4.8 V, v2 =
2
=
24
10
= 2.4 V
v3 =
3
=
−24
10
= −2.4 V
as we obtained with Method 1.
P R A C T I C E P R O B L E M 3 . 2
Find the voltages at the three nonreference nodes in the circuit of Fig. 3.6.
10 A
2 Ω
3 Ω
4 Ω 6 Ω
ix
4ix
1 3
2
Figure3.6 For Practice Prob. 3.2.
Answer: v1 = 80 V, v2 = −64 V, v3 = 156 V.
3.3 NODAL ANALYSIS WITH VOLTAGE SOURCES
We now consider how voltage sources affect nodal analysis. We use the
circuitinFig.3.7forillustration. Considerthefollowingtwopossibilities.
CASE 1 If a voltage source is connected between the reference node
and a nonreference node, we simply set the voltage at the nonreference
node equal to the voltage of the voltage source. In Fig. 3.7, for example,
v1 = 10 V (3.10)
Thus our analysis is somewhat simplified by this knowledge of the voltage
at this node.
CHAPTER 3 Methods of Analysis 83
10 V
5 V
4 Ω
8 Ω 6 Ω
2 Ω
v1 v3
v2
i3
i1
i2
i4
Supernode
+
−
+ −
Figure3.7 A circuit with a supernode.
CASE 2 If the voltage source (dependent or independent) is connected
between two nonreference nodes, the two nonreference nodes form a gen-
eralized node or supernode; we apply both KCL and KVL to determine
the node voltages.
Asupernodemayberegardedasaclosedsurface
enclosing the voltage source and its two nodes.
A supernode is formed by enclosing a (dependent or independent) voltage
source connected between two nonreference nodes and any
elements connected in parallel with it.
In Fig. 3.7, nodes 2 and 3 form a supernode. (We could have more than
two nodes forming a single supernode. For example, see the circuit in
Fig. 3.14.) We analyze a circuit with supernodes using the same three
steps mentioned in the previous section except that the supernodes are
treated differently. Why? Because an essential component of nodal
analysis is applying KCL, which requires knowing the current through
each element. There is no way of knowing the current through a voltage
source in advance. However, KCL must be satisfied at a supernode like
any other node. Hence, at the supernode in Fig. 3.7,
i1 + i4 = i2 + i3 (3.11a)
or
v1 − v2
2
+
v1 − v3
4
=
v2 − 0
8
+
v3 − 0
6
(3.11b)
To apply Kirchhoff’s voltage law to the supernode in Fig. 3.7, we redraw
the circuit as shown in Fig. 3.8. Going around the loop in the clockwise
direction gives
−v2 + 5 + v3 = 0 ⇒ v2 − v3 = 5 (3.12)
From Eqs. (3.10), (3.11b), and (3.12), we obtain the node voltages.
+ −
v2 v3
5 V
+ +
− −
Figure 3.8 Applying KVL to a supernode.
84 PART 1 DC Circuits
Note the following properties of a supernode:
1. The voltage source inside the supernode provides a constraint
equation needed to solve for the node voltages.
2. A supernode has no voltage of its own.
3. A supernode requires the application of both KCL and KVL.
E X A M P L E 3 . 3
For the circuit shown in Fig. 3.9, find the node voltages.
+
−
2 A
2 V
7 A
4 Ω
10 Ω
2 Ω
v1 v2
Figure3.9 For Example 3.3.
Solution:
The supernode contains the 2-V source, nodes 1 and 2, and the 10- re-
sistor. Applying KCL to the supernode as shown in Fig. 3.10(a) gives
2 = i1 + i2 + 7
Expressing i1 and i2 in terms of the node voltages
2 =
v1 − 0
2
+
v2 − 0
4
+ 7 ⇒ 8 = 2v1 + v2 + 28
or
v2 = −20 − 2v1 (3.3.1)
To get the relationship between v1 and v2, we apply KVL to the circuit
in Fig. 3.10(b). Going around the loop, we obtain
−v1 − 2 + v2 = 0 ⇒ v2 = v1 + 2 (3.3.2)
From Eqs. (3.3.1) and (3.3.2), we write
v2 = v1 + 2 = −20 − 2v1
or
3v1 = −22 ⇒ v1 = −7.333 V
and v2 = v1 + 2 = −5.333 V. Note that the 10- resistor does not make
any difference because it is connected across the supernode.
2 A
2 A
7 A
7 A
2 Ω 4 Ω
v2
v1
i1 i2
1 2
(a)
+
−
(b)
2 V
1 2
+
+
− −
v1 v2
Figure3.10 Applying: (a) KCL to the supernode, (b) KVL to the loop.
CHAPTER 3 Methods of Analysis 85
P R A C T I C E P R O B L E M 3 . 3
Find v and i in the circuit in Fig. 3.11.
7 V
3 V
4 Ω
3 Ω 2 Ω 6 Ω
+
−
+
−
i
v
+
−
Figure3.11 For Practice Prob. 3.3.
Answer: −0.2 V, 1.4 A.
E X A M P L E 3 . 4
Find the node voltages in the circuit of Fig. 3.12.
20 V
2 Ω 4 Ω
6 Ω
3 Ω
1 Ω
vx
3vx
+ − + −
10 A
1 4
3
2
+ −
Figure3.12 For Example 3.4.
Solution:
Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCL to
the two supernodes as in Fig. 3.13(a). At supernode 1-2,
i3 + 10 = i1 + i2
Expressing this in terms of the node voltages,
v3 − v2
6
+ 10 =
v1 − v4
3
+
v1
2
or
5v1 + v2 − v3 − 2v4 = 60 (3.4.1)
At supernode 3-4,
i1 = i3 + i4 + i5 ⇒
v1 − v4
3
=
v3 − v2
6
+
v4
1
+
v3
4
or
4v1 + 2v2 − 5v3 − 16v4 = 0 (3.4.2)
86 PART 1 DC Circuits
10 A
3 Ω
6 Ω
2 Ω 4 Ω 1 Ω
(a)
i1
i2
i3
i4
i5
v1
v2 v3
v4
vx
+ −
(b)
+ − + −
+ −
20 V
3 Ω
6 Ω
i3
v1 v2 v3 v4
vx
Loop 1 Loop 2
Loop 3
3vx
+ + +
+
− − − −
i1
i3
Figure3.13 Applying: (a) KCL to the two supernodes, (b) KVL to the loops.
We now apply KVL to the branches involving the voltage sources
as shown in Fig. 3.13(b). For loop 1,
−v1 + 20 + v2 = 0 ⇒ v1 − v2 = 20 (3.4.3)
For loop 2,
−v3 + 3vx + v4 = 0
But vx = v1 − v4 so that
3v1 − v3 − 2v4 = 0 (3.4.4)
For loop 3,
vx − 3vx + 6i3 − 20 = 0
But 6i3 = v3 − v2 and vx = v1 − v4. Hence
−2v1 − v2 + v3 + 2v4 = 20 (3.4.5)
We need four node voltages, v1, v2, v3, and v4, and it requires
only four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although
the fifth equation is redundant, it can be used to check results. We can
eliminate one node voltage so that we solve three simultaneous equations
instead of four. From Eq. (3.4.3), v2 = v1 − 20. Substituting this into
Eqs. (3.4.1) and (3.4.2), respectively, gives
6v1 − v3 − 2v4 = 80 (3.4.6)
and
6v1 − 5v3 − 16v4 = 40 (3.4.7)
Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as


3
6
6
−1
−1
−5
−2
−2
−16




v1
v3
v4

 =


0
80
40


CHAPTER 3 Methods of Analysis 87
Using Cramer’s rule,
=






3
6
6
−1
−1
−5
−2
−2
−16






= −18, 1 =






0
80
40
−1
−1
−5
−2
−2
−16






= −480
3 =






3
6
6
0
80
40
−2
−2
−16






= −3120, 4 =






3
6
6
−1
−1
−5
0
80
40






= 840
Thus, we arrive at the node voltages as
v1 =
1
=
−480
−18
= 26.667 V, v3 =
3
=
−3120
−18
= 173.333 V
v4 =
4
=
840
−18
= −46.667 V
and v2 = v1 −20 = 6.667 V. We have not used Eq. (3.4.5); it can be used
to cross check results.
P R A C T I C E P R O B L E M 3 . 4
Find v1, v2, and v3 in the circuit in Fig. 3.14 using nodal analysis.
2 Ω 4 Ω 3 Ω
6 Ω
i
v1
v2
v3
+ −
+
−
10 V 5i
Figure3.14 For Practice Prob. 3.4.
Answer: v1 = 3.043 V, v2 = −6.956 V, v3 = 0.6522 V.
3.4 MESH ANALYSIS
Mesh analysis provides another general procedure for analyzing circuits,
using mesh currents as the circuit variables. Using mesh currents instead
of element currents as circuit variables is convenient and reduces the
number of equations that must be solved simultaneously. Recall that a
loop is a closed path with no node passed more than once. A mesh is a
loop that does not contain any other loop within it.
Meshanalysisisalsoknownasloopanalysisorthe
mesh-current method.
Nodal analysis applies KCL to find unknown voltages in a given
circuit, while mesh analysis applies KVL to find unknown currents. Mesh
analysis is not quite as general as nodal analysis because it is only ap-
plicable to a circuit that is planar. A planar circuit is one that can be
drawn in a plane with no branches crossing one another; otherwise it is
nonplanar. A circuit may have crossing branches and still be planar if it
can be redrawn such that it has no crossing branches. For example, the
88 PART 1 DC Circuits
circuit in Fig. 3.15(a) has two crossing branches, but it can be redrawn
as in Fig. 3.15(b). Hence, the circuit in Fig. 3.15(a) is planar. However,
the circuit in Fig. 3.16 is nonplanar, because there is no way to redraw
it and avoid the branches crossing. Nonplanar circuits can be handled
using nodal analysis, but they will not be considered in this text.
(a)
1 A
(b)
1 A
1 Ω
1 Ω 3 Ω
2 Ω
4 Ω
5 Ω
8 Ω 7 Ω
6 Ω
2 Ω
4 Ω
7 Ω
8 Ω
5 Ω 6 Ω 3 Ω
Figure3.15 (a) A planar circuit with crossing
branches, (b) the same circuit redrawn with no
crossing branches.
5 A
1 Ω
5 Ω
4 Ω
6 Ω
10 Ω
11 Ω
12 Ω
13 Ω
9 Ω
8 Ω
3 Ω
2 Ω
7 Ω
Figure3.16 A nonplanar circuit.
To understand mesh analysis, we should first explain more about
what we mean by a mesh.
A mesh is a loop which does not contain any other loops within it.
In Fig. 3.17, for example, paths abefa and bcdeb are meshes, but path
abcdefa is not a mesh. The current through a mesh is known as mesh
current. In mesh analysis, we are interested in applying KVL to find the
mesh currents in a given circuit.
Although path abcdefa is a loop and not a mesh,
KVL still holds. This is the reason for loosely
using the terms loop analysis and mesh analysis to
mean the same thing.
+
−
+
−
I1 R1 R2
R3
i1
i2
I2
I3
V1 V2
a b c
d
e
f
Figure3.17 A circuit with two meshes.
In this section, we will apply mesh analysis to planar circuits that
do not contain current sources. In the next sections, we will consider
circuits with current sources. In the mesh analysis of a circuit with n
meshes, we take the following three steps.
CHAPTER 3 Methods of Analysis 89
Steps to Determine Mesh Currents:
1. Assign mesh currents i1, i2, . . . , in to the n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s law to express
the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh
currents.
To illustrate the steps, consider the circuit in Fig. 3.17. The first
step requires that mesh currents i1 and i2 are assigned to meshes 1 and
2. Although a mesh current may be assigned to each mesh in an arbi-
trary direction, it is conventional to assume that each mesh current flows
clockwise.
The direction of the mesh current is arbitrary—
(clockwise or counterclockwise)—and does not
affect the validity of the solution.
As the second step, we apply KVL to each mesh. Applying KVL
to mesh 1, we obtain
−V1 + R1i1 + R3(i1 − i2) = 0
or
(R1 + R3)i1 − R3i2 = V1 (3.13)
For mesh 2, applying KVL gives
R2i2 + V2 + R3(i2 − i1) = 0
or
−R3i1 + (R2 + R3)i2 = −V2 (3.14)
Note in Eq. (3.13) that the coefficient of i1 is the sum of the resistances in
the first mesh, while the coefficient of i2 is the negative of the resistance
common to meshes 1 and 2. Now observe that the same is true in Eq.
(3.14). This can serve as a shortcut way of writing the mesh equations.
We will exploit this idea in Section 3.6.
The shortcut way will not apply if one mesh cur-
rentisassumedclockwiseandtheotherassumed
anticlockwise, although this is permissible.
The third step is to solve for the mesh currents. Putting Eqs. (3.13)
and (3.14) in matrix form yields

R1 + R3
−R3
−R3
R2 + R3
 
i1
i2

=

V1
−V2

(3.15)
which can be solved to obtain the mesh currents i1 and i2. We are at liberty
to use any technique for solving the simultaneous equations. According
to Eq. (2.12), if a circuit has n nodes, b branches, and l independent
loops or meshes, then l = b − n + 1. Hence, l independent simultaneous
equations are required to solve the circuit using mesh analysis.
Notice that the branch currents are different from the mesh currents
unless the mesh is isolated. To distinguish between the two types of
currents, we use i for a mesh current and I for a branch current. The
current elements I1, I2, and I3 are algebraic sums of the mesh currents.
It is evident from Fig. 3.17 that
I1 = i1, I2 = i2, I3 = i1 − i2 (3.16)
90 PART 1 DC Circuits
E X A M P L E 3 . 5
For the circuit in Fig. 3.18, find the branch currents I1, I2, and I3 using
mesh analysis.
+
−
+
−
15 V
10 V
5 Ω 6 Ω
10 Ω
4 Ω
I1
i1
I2
i2
I3
Figure3.18 For Example 3.5.
Solution:
We first obtain the mesh currents using KVL. For mesh 1,
−15 + 5i1 + 10(i1 − i2) + 10 = 0
or
3i1 − 2i2 = 1 (3.5.1)
For mesh 2,
6i2 + 4i2 + 10(i2 − i1) − 10 = 0
or
i1 = 2i2 − 1 (3.5.2)
METHOD 1 Using the substitution method, we substitute Eq. (3.5.2)
into Eq. (3.5.1), and write
6i2 − 3 − 2i2 = 1 ⇒ i2 = 1 A
From Eq. (3.5.2), i1 = 2i2 − 1 = 2 − 1 = 1 A. Thus,
I1 = i1 = 1 A, I2 = i2 = 1 A, I3 = i1 − i2 = 0
METHOD 2 To use Cramer’s rule, we cast Eqs. (3.5.1) and (3.5.2) in
matrix form as

3
−1
−2
2
 
i1
i2

=

1
1

We obtain the determinants
=




3
−1
−2
2



 = 6 − 2 = 4
1 =




1
1
−2
2



 = 2 + 2 = 4, 2 =




3
−1
1
1



 = 3 + 1 = 4
Thus,
i1 =
1
= 1 A, i2 =
2
= 1 A
as before.
P R A C T I C E P R O B L E M 3 . 5
Calculate the mesh currents i1 and i2 in the circuit of Fig. 3.19.
12 V 8 V
2 Ω
4 Ω 3 Ω
12 Ω
9 Ω
i1
i2
+
−
+
−
Figure3.19 For Practice Prob. 3.5.
Answer: i1 = 2
3
A, i2 = 0 A.
CHAPTER 3 Methods of Analysis 91
E X A M P L E 3 . 6
Use mesh analysis to find the current io in the circuit in Fig. 3.20.
+
−
+
−
24 V
12 Ω
4 Ω
10 Ω 24 Ω
i1
i1
i3
i2
i2
io
4io
A
Figure3.20 For Example 3.6.
Solution:
We apply KVL to the three meshes in turn. For mesh 1,
−24 + 10(i1 − i2) + 12(i1 − i3) = 0
or
11i1 − 5i2 − 6i3 = 12 (3.6.1)
For mesh 2,
24i2 + 4(i2 − i3) + 10(i2 − i1) = 0
or
−5i1 + 19i2 − 2i3 = 0 (3.6.2)
For mesh 3,
4io + 12(i3 − i1) + 4(i3 − i2) = 0
But at node A, io = i1 − i2, so that
4(i1 − i2) + 12(i3 − i1) + 4(i3 − i2) = 0
or
−i1 − i2 + 2i3 = 0 (3.6.3)
In matrix form, Eqs. (3.6.1) to (3.6.3) become


11
−5
−1
−5
19
−1
−6
−2
2




i1
i2
i3

 =


12
0
0


We obtain the determinants as
= 418 − 30 − 10 − 114 − 22 − 50 = 192
−
−
− +
+
+
19
19
−5
−5
=
−5
−1
11
−5
11
−1

−6
−2
2
−6
−2
= 456 − 24 = 432
−
−
− +
+
+
19
19
−5
−5
=
0
0
12
0
12
−1
−6
−2
2
−6
−2
1
= 24 + 120 = 144
−
−
− +
+
+
0
0
12
12
=
−5
−1
11
−5
11
0
−6
−2
2
−6
−2
2
92 PART 1 DC Circuits
= 60 + 228 = 288
−
−
− +
+
+
0
19
19 0
12
12
=
−5
−5
−5
−1 −1
11
−5
11
0
3
We calculate the mesh currents using Cramer’s rule as
i1 =
1
=
432
192
= 2.25 A, i2 =
2
=
144
192
= 0.75 A
i3 =
3
=
288
192
= 1.5 A
Thus, io = i1 − i2 = 1.5 A.
P R A C T I C E P R O B L E M 3 . 6
Using mesh analysis, find io in the circuit in Fig. 3.21.
+
−
–
+
20 V
4 Ω 8 Ω
2 Ω
6 Ω
i1 i2
i3
10io
io
Figure3.21 For Practice Prob. 3.6.
Answer: −5 A.
3.5 MESH ANALYSIS WITH CURRENT SOURCES
Applying mesh analysis to circuits containing current sources (dependent
or independent) may appear complicated. But it is actually much easier
than what we encountered in the previous section, because the presence
of the current sources reduces the number of equations. Consider the
following two possible cases.
+
− 5 A
10 V
4 Ω 3 Ω
6 Ω
i1 i2
Figure3.22 A circuit with a current source.
CASE 1 When a current source exists only in one mesh: Consider the
circuit in Fig. 3.22, for example. We set i2 = −5 A and write a mesh
equation for the other mesh in the usual way, that is,
−10 + 4i1 + 6(i1 − i2) = 0 ⇒ i1 = −2 A (3.17)
CASE 2 When a current source exists between two meshes: Consider
the circuit in Fig. 3.23(a), for example. We create a supermesh by ex-
cluding the current source and any elements connected in series with it,
as shown in Fig. 3.23(b). Thus,
A supermesh results when two meshes have a (dependent or independent)
current source in common.
CHAPTER 3 Methods of Analysis 93
(b)
20 V 4 Ω
6 Ω 10 Ω
i1 i2
+
−
+
−
6 A
20 V
6 Ω 10 Ω
2 Ω
4 Ω
i1
i1
i2
i2
0
(a)
Exclude these
elements
Figure 3.23 (a) Two meshes having a current source in common, (b) a supermesh, created by excluding the current source.
As shown in Fig. 3.23(b), we create a supermesh as the periphery of
the two meshes and treat it differently. (If a circuit has two or more
supermeshes that intersect, they should be combined to form a larger
supermesh.) Why treat the supermesh differently? Because mesh analy-
sis applies KVL—which requires that we know the voltage across each
branch—and we do not know the voltage across a current source in ad-
vance. However, a supermesh must satisfy KVL like any other mesh.
Therefore, applying KVL to the supermesh in Fig. 3.23(b) gives
−20 + 6i1 + 10i2 + 4i2 = 0
or
6i1 + 14i2 = 20 (3.18)
We apply KCL to a node in the branch where the two meshes intersect.
Applying KCL to node 0 in Fig. 3.23(a) gives
i2 = i1 + 6 (3.19)
Solving Eqs. (3.18) and (3.19), we get
i1 = −3.2 A, i2 = 2.8 A (3.20)
Note the following properties of a supermesh:
1. The current source in the supermesh is not completely ignored;
it provides the constraint equation necessary to solve for the
mesh currents.
2. A supermesh has no current of its own.
3. A supermesh requires the application of both KVL and KCL.
E X A M P L E 3 . 7
For the circuit in Fig. 3.24, find i1 to i4 using mesh analysis.
Solution:
Notethatmeshes1and2formasupermeshsincetheyhaveanindependent
current source in common. Also, meshes 2 and 3 form another supermesh
94 PART 1 DC Circuits
+
− 10 V
6 Ω 8 Ω
2 Ω
4 Ω
i1
i2 i3 i4
2 Ω
5 A
i1
i2
i2 i3
io
P
Q
3io
Figure3.24 For Example 3.7.
because they have a dependent current source in common. The two
supermeshes intersect and form a larger supermesh as shown. Applying
KVL to the larger supermesh,
2i1 + 4i3 + 8(i3 − i4) + 6i2 = 0
or
i1 + 3i2 + 6i3 − 4i4 = 0 (3.7.1)
For the independent current source, we apply KCL to node P:
i2 = i1 + 5 (3.7.2)
For the dependent current source, we apply KCL to node Q:
i2 = i3 + 3io
But io = −i4, hence,
i2 = i3 − 3i4 (3.7.3)
Applying KVL in mesh 4,
2i4 + 8(i4 − i3) + 10 = 0
or
5i4 − 4i3 = −5 (3.7.4)
From Eqs. (3.7.1) to (3.7.4),
i1 = −7.5 A, i2 = −2.5 A, i3 = 3.93 A, i4 = 2.143 A
P R A C T I C E P R O B L E M 3 . 7
Use mesh analysis to determine i1, i2, and i3 in Fig. 3.25.
+
− 3 A
6 V
1 Ω
2 Ω 2 Ω
8 Ω
4 Ω
i1
i3
i2
Figure3.25 For Practice Prob. 3.7.
Answer: i1 = 3.474 A, i2 = 0.4737 A, i3 = 1.1052 A.
CHAPTER 3 Methods of Analysis 95
†3.6 NODAL AND MESH ANALYSES BY INSPECTION
This section presents a generalized procedure for nodal or mesh analysis.
It is a shortcut approach based on mere inspection of a circuit.
When all sources in a circuit are independent current sources, we
do not need to apply KCL to each node to obtain the node-voltage equa-
tions as we did in Section 3.2. We can obtain the equations by mere
inspection of the circuit. As an example, let us reexamine the circuit in
Fig. 3.2, shown again in Fig. 3.26(a) for convenience. The circuit has
two nonreference nodes and the node equations were derived in Section
3.2 as

G1 + G2
−G2
−G2
G2 + G3
 
v1
v2

=

I1 − I2
I2

(3.21)
Observe that each of the diagonal terms is the sum of the conductances
connected directly to node 1 or 2, while the off-diagonal terms are the
negatives of the conductances connected between the nodes. Also, each
term on the right-hand side of Eq. (3.21) is the algebraic sum of the
currents entering the node.
I1
v1
G1 G3
G2
I2
v2
(a)
(b)
i1 i3
V1 V2
+
−
+
−
R1 R2
R3
Figure 3.26 (a) The circuit in Fig. 3.2,
(b) the circuit in Fig. 3.17.
In general, if a circuit with independent current sources has N
nonreference nodes, the node-voltage equations can be written in terms
of the conductances as





G11
G21
.
.
.
GN1
G12
G22
.
.
.
GN2
. . .
. . .
.
.
.
. . .
G1N
G2N
.
.
.
GNN









v1
v2
.
.
.
vN



 =




i1
i2
.
.
.
iN



 (3.22)
or simply
Gv = i (3.23)
where
Gkk = Sum of the conductances connected to node k
Gkj = Gjk = Negative of the sum of the conductances directly
connecting nodes k and j, k = j
vk = Unknown voltage at node k
ik = Sum of all independent current sources directly
connected to node k, with currents entering the node
treated as positive
G is called the conductance matrix, v is the output vector; and i is the
input vector. Equation (3.22) can be solved to obtain the unknown node
voltages. Keep in mind that this is valid for circuits with only independent
current sources and linear resistors.
Similarly, wecanobtainmesh-currentequationsbyinspectionwhen
a linear resistive circuit has only independent voltage sources. Consider
the circuit in Fig. 3.17, shown again in Fig. 3.26(b) for convenience. The
circuit has two nonreference nodes and the node equations were derived
in Section 3.4 as

R1 + R3
−R3
−R3
R2 + R3
 
i1
i2

=

v1
−v2

(3.24)
96 PART 1 DC Circuits
We notice that each of the diagonal terms is the sum of the resistances in
the related mesh, while each of the off-diagonal terms is the negative of
the resistance common to meshes 1 and 2. Each term on the right-hand
side of Eq. (3.24) is the algebraic sum taken clockwise of all independent
voltage sources in the related mesh.
In general, if the circuit has N meshes, the mesh-current equations
can be expressed in terms of the resistances as




R11
R21
.
.
.
RN1
R12
R22
.
.
.
RN2
. . .
. . .
.
.
.
. . .
R1N
R2N
.
.
.
RNN








i1
i2
.
.
.
iN



 =




v1
v2
.
.
.
vN



 (3.25)
or simply
Ri = v (3.26)
where
Rkk = Sum of the resistances in mesh k
Rkj = Rjk = Negative of the sum of the resistances in common with
meshes k and j, k = j
ik = Unknown mesh current for mesh k in the clockwise
direction
vk = Sum taken clockwise of all independent voltage sources
in mesh k, with voltage rise treated as positive
R is called the resistance matrix, i is the output vector; and v is the input
vector. We can solve Eq. (3.25) to obtain the unknown mesh currents.
E X A M P L E 3 . 8
Write the node-voltage matrix equations for the circuit in Fig. 3.27 by
inspection.
3 A 1 A 4 A
2 A
10 Ω
5 Ω
1 Ω
8 Ω 8 Ω
v1 v2 v3 v4
4 Ω 2 Ω
Figure3.27 For Example 3.8.
CHAPTER 3 Methods of Analysis 97
Solution:
The circuit in Fig. 3.27 has four nonreference nodes, so we need four
node equations. This implies that the size of the conductance matrix G,
is 4 by 4. The diagonal terms of G, in siemens, are
G11 =
1
5
+
1
10
= 0.3, G22 =
1
5
+
1
8
+
1
1
= 1.325
G33 =
1
8
+
1
8
+
1
4
= 0.5, G44 =
1
8
+
1
2
+
1
1
= 1.625
The off-diagonal terms are
G12 = −
1
5
= −0.2, G13 = G14 = 0
G21 = −0.2, G23 = −
1
8
= −0.125, G24 = −
1
1
= −1
G31 = 0, G32 = −0.125, G34 = −
1
8
= −0.125
G41 = 0, G42 = −1, G43 = −0.125
The input current vector i has the following terms, in amperes:
i1 = 3, i2 = −1 − 2 = −3, i3 = 0, i4 = 2 + 4 = 6
Thus the node-voltage equations are




0.3
−0.2
0
0
−0.2
1.325
−0.125
−1
0
−0.125
0.5
−0.125
0
−1
−0.125
1.625








v1
v2
v3
v4



 =




3
−3
0
6




which can be solved to obtain the node voltages v1, v2, v3, and v4.
P R A C T I C E P R O B L E M 3 . 8
By inspection, obtain the node-voltage equations for the circuit in Fig.
3.28.
1 A
2 A
3 A
10 Ω
1 Ω
5 Ω
4 Ω
2 Ω
v1
v2
v3 v4
Figure3.28 For Practice Prob. 3.8.
Answer:




1.3
−0.2
−1
0
−0.2
0.2
0
0
−1
0
1.25
−0.25
0
0
−0.25
0.75








v1
v2
v3
v4



 =




0
3
−1
3




E X A M P L E 3 . 9
Byinspection, writethemesh-currentequationsforthecircuitinFig.3.29.
98 PART 1 DC Circuits
+
−
+ −
+
−
+
−
10 V
4 V
2 Ω
2 Ω
5 Ω
2 Ω
4 Ω
3 Ω
3 Ω
1 Ω 1 Ω
4 Ω
i1
i2
i3
i4 i5 6 V
12 V
Figure3.29 For Example 3.9.
Solution:
We have five meshes, so the resistance matrix is 5 by 5. The diagonal
terms, in ohms, are:
R11 = 5 + 2 + 2 = 9, R22 = 2 + 4 + 1 + 1 + 2 = 10
R33 = 2 + 3 + 4 = 9, R44 = 1 + 3 + 4 = 8, R55 = 1 + 3 = 4
The off-diagonal terms are:
R12 = −2, R13 = −2, R14 = 0 = R15
R21 = −2, R23 = −4, R24 = −1, R25 = −1
R31 = −2, R32 = −4, R34 = 0 = R35
R41 = 0, R42 = −1, R43 = 0, R45 = −3
R51 = 0, R52 = −1, R53 = 0, R54 = −3
The input voltage vector v has the following terms in volts:
v1 = 4, v2 = 10 − 4 = 6
v3 = −12 + 6 = −6, v4 = 0, v5 = −6
Thus the mesh-current equations are:






9
−2
−2
0
0
−2
10
−4
−1
−1
−2
−4
9
0
0
0
−1
0
8
−3
0
−1
0
−3
4












i1
i2
i3
i4
i5






=






4
6
−6
0
−6






From this, we can obtain mesh currents i1, i2, i3, i4, and i5.
P R A C T I C E P R O B L E M 3 . 9
By inspection, obtain the mesh-current equations for the circuit in Fig.
3.30.
CHAPTER 3 Methods of Analysis 99
+
−
+
−
24 V
12 V
10 V
50 Ω
40 Ω
i1
i2 i3
i4 i5
10 Ω
30 Ω
20 Ω
60 Ω
80 Ω
+
−
Figure3.30 For Practice Prob. 3.9.
Answer:






170
−40
0
−80
0
−40
80
−30
−10
0
0
−30
50
0
−20
−80
−10
0
90
0
0
0
−20
0
80












i1
i2
i3
i4
i5






=






24
0
−12
10
−10






3.7 NODAL VERSUS MESH ANALYSIS
Both nodal and mesh analyses provide a systematic way of analyzing a
complex network. Someone may ask: Given a network to be analyzed,
how do we know which method is better or more efficient? The choice
of the better method is dictated by two factors.
The first factor is the nature the particular network. Networks that
containmanyseries-connectedelements, voltagesources, orsupermeshes
are more suitable for mesh analysis, whereas networks with parallel-
connected elements, current sources, or supernodes are more suitable for
nodal analysis. Also, a circuit with fewer nodes than meshes is better
analyzed using nodal analysis, while a circuit with fewer meshes than
nodes is better analyzed using mesh analysis. The key is to select the
method that results in the smaller number of equations.
The second factor is the information required. If node voltages are
required, it may be expedient to apply nodal analysis. If branch or mesh
currents are required, it may be better to use mesh analysis.
It is helpful to be familiar with both methods of analysis, for at least
two reasons. First, one method can be used to check the results from the
other method, if possible. Second, since each method has its limitations,
only one method may be suitable for a particular problem. For example,
100 PART 1 DC Circuits
mesh analysis is the only method to use in analyzing transistor circuits,
as we shall see in Section 3.9. But mesh analysis cannot easily be used to
solve an op amp circuit, as we shall see in Chapter 5, because there is no
direct way to obtain the voltage across the op amp itself. For nonplanar
networks, nodal analysis is the only option, because mesh analysis only
applies to planar networks. Also, nodal analysis is more amenable to
solution by computer, as it is easy to program. This allows one to analyze
complicated circuits that defy hand calculation. A computer software
package based on nodal analysis is introduced next.
3.8 CIRCUIT ANALYSIS WITH PSPICE
PSpice is a computer software circuit analysis program that we will grad-
ually learn to use throught the course of this text. This section illustrates
how to use PSpice for Windows to analyze the dc circuits we have studied
so far.
Appendix D provides a tutorial on using PSpice
for Windows.
The reader is expected to review Sections D.1 through D.3 of Ap-
pendix D before proceeding in this section. It should be noted that PSpice
is only helpful in determining branch voltages and currents when the nu-
merical values of all the circuit components are known.
E X A M P L E 3 . 1 0
Use PSpice to find the node voltages in the circuit of Fig. 3.31.
+
− 3 A
120 V
20 Ω
30 Ω 40 Ω
10 Ω
1 2 3
0
Figure3.31 For Example 3.10.
Solution:
The first step is to draw the given circuit using Schematics. If one follows
the instructions given in Appendix sections D.2 and D.3, the schematic in
Fig. 3.32 is produced. Since this is a dc analysis, we use voltage source
VDC and current source IDC. The pseudocomponent VIEWPOINTS are
added to display the required node voltages. Once the circuit is drawn and
saved as exam310.sch, we run PSpice by selecting Analysis/Simulate.
The circuit is simulated and the results are displayed on VIEWPOINTS
and also saved in output file exam310.out. The output file includes the
following:
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) 120.0000 (2) 81.2900 (3) 89.0320
indicating that V1 = 120 V, V2 = 81.29 V, V3 = 89.032 V.
+
−
R1 R3
20 10
120 V V1 R2 R4
30 40 I1 3 A
IDC
0
1 2 3
120.0000 81.2900 89.0320
Figure 3.32 For Example 3.10; the schematic of the circuit in Fig. 3.31.
CHAPTER 3 Methods of Analysis 101
P R A C T I C E P R O B L E M 3 . 1 0
For the circuit in Fig. 3.33, use PSpice to find the node voltages.
+
−
2 A
200 V
30 Ω 60 Ω 50 Ω
100 Ω
25 Ω
1 2 3
0
Figure3.33 For Practice Prob. 3.10.
Answer: V1 = −40 V, V2 = 57.14 V, V3 = 200 V.
E X A M P L E 3 . 1 1
In the circuit in Fig. 3.34, determine the currents i1, i2, and i3.
Solution:
The schematic is shown in Fig. 3.35. (The schematic in Fig. 3.35 includes
the output results, implying that it is the schematic displayed on the screen
after the simulation.) Notice that the voltage-controlled voltage source
E1 in Fig. 3.35 is connected so that its input is the voltage across the 4-
resistor; its gain is set equal to 3. In order to display the required currents,
we insert pseudocomponent IPROBES in the appropriate branches. The
schematic is saved as exam311.sch and simulated by selecting Analy-
sis/Simulate. The results are displayed on IPROBES as shown in Fig.
3.35 and saved in output file exam311.out. From the output file or the
IPROBES, we obtain i1 = i2 = 1.333 A and i3 = 2.667 A.
+
−
+
−
24 V
1 Ω
i1 i2 i3
+
−
4 Ω 2 Ω
2 Ω 8 Ω 4 Ω
3vo
vo
Figure3.34 For Example 3.11.
102 PART 1 DC Circuits
+
−
24 V V1
R1
4
R2 2 R3 8 R4 4
1.333E+00 1.333E+00 2.667E+00
0
R6
1
R5
2
E E1
+
−
− +
Figure3.35 The schematic of the circuit in Fig. 3.34.
P R A C T I C E P R O B L E M 3 . 1 1
Use PSpice to determine currents i1, i2, and i3 in the circuit of Fig. 3.36.
+
−
2 A
10 V
2 Ω
i1
i1
i2
4 Ω
1 Ω 2 Ω
i3
Figure3.36 For Practice Prob. 3.11.
Answer: i1 = −0.4286 A, i2 = 2.286 A, i3 = 2 A.
†3.9 APPLICATIONS: DC TRANSISTOR CIRCUITS
Most of us deal with electronic products on a routine basis and have
some experience with personal computers. A basic component for the
integrated circuits found in these electronics and computers is the ac-
tive, three-terminal device known as the transistor. Understanding the
transistor is essential before an engineer can start an electronic circuit
design.
Figure 3.37 depicts various kinds of transistors commercially avail-
able. There are two basic types of transistors: bipolar junction transis-
tors (BJTs) and field-effect transistors (FETs). Here, we consider only
the BJTs, which were the first of the two and are still used today. Our
objective is to present enough detail about the BJT to enable us to apply
the techniques developed in this chapter to analyze dc transistor circuits.
TherearetwotypesofBJTs: npnandpnp, withtheircircuitsymbols
as shown in Fig. 3.38. Each type has three terminals, designated as emit-
ter (E), base (B), and collector (C). For the npn transistor, the currents and
CHAPTER 3 Methods of Analysis 103
Figure3.37 Various types of transistors.
(Courtesy of Tech America.)
n
n
p
Base
Collector
Emitter
E
B
C
(a)
p
p
n
Base
Collector
Emitter
E
B
C
(b)
Figure3.38 Two types of BJTs and
their circuit symbols: (a) npn, (b) pnp.
B
C
E
+
+
+
−
−
−
VCB
VCE
VBE
B
C
E
IB
IC
IE
(a)
(b)
Figure3.39 The terminal
variables of an npn transistor:
(a) currents, (b) voltages.
voltages of the transistor are specified as in Fig. 3.39. Applying KCL to
Fig. 3.39(a) gives
IE = IB + IC (3.27)
whereIE, IC, andIB areemitter, collector, andbasecurrents, respectively.
Similarly, applying KVL to Fig. 3.39(b) gives
VCE + VEB + VBC = 0 (3.28)
where VCE, VEB, and VBC are collector-emitter, emitter-base, and base-
collector voltages. The BJT can operate in one of three modes: active,
cutoff, and saturation. When transistors operate in the active mode, typ-
ically VBE  0.7 V,
IC = αIE (3.29)
where α is called the common-base current gain. In Eq. (3.29), α denotes
the fraction of electrons injected by the emitter that are collected by the
collector. Also,
IC = βIB (3.30)
where β is known as the common-emitter current gain. The α and β are
characteristic properties of a given transistor and assume constant values
for that transistor. Typically, α takes values in the range of 0.98 to 0.999,
while β takes values in the range 50 to 1000. From Eqs. (3.27) to (3.30),
it is evident that
IE = (1 + β)IB (3.31)
and
β =
α
1 − α
(3.32)
104 PART 1 DC Circuits
These equations show that, in the active mode, the BJT can be modeled as
a dependent current-controlled current source. Thus, in circuit analysis,
the dc equivalent model in Fig. 3.40(b) may be used to replace the npn
transistor in Fig. 3.40(a). Since β in Eq. (3.32) is large, a small base
current controls large currents in the output circuit. Consequently, the
bipolar transistor can serve as an amplifier, producing both current gain
and voltage gain. Such amplifiers can be used to furnish a considerable
amount of power to transducers such as loudspeakers or control motors.
B C
E
IB IC
VBE
VCE
+
−
+
−
B
C
E
IB
(a) (b)
VBE
VCE
+
+
−
−
bIB
Figure 3.40 (a) An npn transistor, (b) its dc equivalent
model.
In fact, transistor circuits provide motivation to
study dependent sources.
It should be observed in the following examples that one cannot
directly analyze transistor circuits using nodal analysis because of the
potential difference between the terminals of the transistor. Only when
the transistor is replaced by its equivalent model can we apply nodal
analysis.
E X A M P L E 3 . 1 2
Find IB, IC, and vo in the transistor circuit of Fig. 3.41. Assume that the
transistor operates in the active mode and that β = 50.
Solution:
For the input loop, KVL gives
−4 + IB(20 × 103
) + VBE = 0
IC
+
−
+
+
−
−
+
−
4 V
6 V
20 kΩ
IB
VBE
vo
Output
loop
Input
loop
100 Ω
Figure3.41 For Example 3.12.
CHAPTER 3 Methods of Analysis 105
Since VBE = 0.7 V in the active mode,
IB =
4 − 0.7
20 × 103
= 165 µA
But
IC = βIB = 50 × 165 µA = 8.25 mA
For the output loop, KVL gives
−vo − 100IC + 6 = 0
or
vo = 6 − 100IC = 6 − 0.825 = 5.175 V
Note that vo = VCE in this case.
P R A C T I C E P R O B L E M 3 . 1 2
For the transistor circuit in Fig. 3.42, let β = 100 and VBE = 0.7 V. De-
termine vo and VCE.
+
−
+
+
+
−
−
−
+
−
5 V
12 V
10 kΩ
500 Ω
VBE
VCE
200 Ω vo
Figure3.42 For Practice Prob. 3.12.
Answer: 2.876 V, 1.984 V.
E X A M P L E 3 . 1 3
For the BJT circuit in Fig. 3.43, β = 150 and VBE = 0.7 V. Find vo.
Solution:
We can solve this problem in two ways. One way is by direct analysis of
the circuit in Fig. 3.43. Another way is by replacing the transistor with
its equivalent circuit.
2 V
100 kΩ
+
−
+
−
16 V
200 kΩ
1 kΩ
+
−
vo
Figure3.43 For Example 3.13.
METHOD 1 We can solve the problem as we solved the problem in
the previous example. We apply KVL to the input and output loops as
shown in Fig. 3.44(a). For loop 1,
2 = 100 × 103
I1 + 200 × 103
I2 (3.13.1)
For loop 2,
VBE = 0.7 = 200 × 103
I2 ⇒ I2 = 3.5 µA (3.13.2)
For loop 3,
−vo − 1000IC + 16 = 0
106 PART 1 DC Circuits
or
vo = 16 − 1000IC (3.13.3)
From Eqs. (3.13.1) and (3.13.2),
I1 =
2 − 0.7
100 × 103
= 13 µA, IB = I1 − I2 = 9.5 µA
IC = βIB = 150 × 9.5 µA = 1.425 mA
Substituting for IC in Eq. (3.13.3),
vo = 16 − 1.425 = 14.575 V
+
−
Vo
+
−
+
−
I1 IB
IC
I2
1 kΩ
100 kΩ
200 kΩ
2 V
16 V
16 V
2 V
Loop 1 Loop 2
Loop 3
(a)
(b)
+
−
+
−
0.7 V
100 kΩ
200 kΩ
1 kΩ
150IB
IB
I1
I2
B C
E
+
−
Vo
Figure 3.44 Solution of the problem in Example 3.13: (a) method 1,
(b) method 2.
METHOD 2 We can modify the circuit in Fig. 3.43 by replacing the
transistor by its equivalent model in Fig. 3.40(b). The result is the circuit
shown in Fig. 3.44(b). Notice that the locations of the base (B), emitter
(E), and collector (C) remain the same in both the original circuit in Fig.
3.43 and its equivalent circuit in Fig. 3.44(b). From the output loop,
vo = 16 − 1000(150IB)
But
IB = I1 − I2 =
2 − 0.7
100 × 103
−
0.7
200 × 103
= (13 − 3.5) µA = 9.5 µA
and so
vo = 16 − 1000(150 × 9.5 × 10−6
) = 14.575 V
CHAPTER 3 Methods of Analysis 107
P R A C T I C E P R O B L E M 3 . 1 3
The transistor circuit in Fig. 3.45 has β = 80 and VBE = 0.7 V. Find vo
and io.
1 V
10 V
30 kΩ
20 kΩ
20 kΩ
+
−
io
VBE
+
−
vo
+
−
+
−
Figure 3.45 For Practice Prob. 3.13.
Answer: −3 V, −150 µA.
3.10 SUMMARY
1. Nodal analysis is the application of Kirchhoff’s current law at the
nonreference nodes. (It is applicable to both planar and nonplanar
circuits.) We express the result in terms of the node voltages.
Solving the simultaneous equations yields the node voltages.
2. A supernode consists of two nonreference nodes connected by a
(dependent or independent) voltage source.
3. Mesh analysis is the application of Kirchhoff’s voltage law around
meshes in a planar circuit. We express the result in terms of mesh
currents. Solving the simultaneous equations yields the mesh
currents.
4. A supermesh consists of two meshes that have a (dependent or
independent) current source in common.
5. Nodal analysis is normally used when a circuit has fewer node
equations than mesh equations. Mesh analysis is normally used
when a circuit has fewer mesh equations than node equations.
6. Circuit analysis can be carried out using PSpice.
7. DC transistor circuits can be analyzed using the techniques cover-
ed in this chapter.
REVIEW QUESTIONS
3.1 At node 1 in the circuit in Fig. 3.46, applying KCL
gives:
(a) 2 +
12 − v1
3
=
v1
6
+
v1 − v2
4
(b) 2 +
v1 − 12
3
=
v1
6
+
v2 − v1
4
(c) 2 +
12 − v1
3
=
0 − v1
6
+
v1 − v2
4
(d) 2 +
v1 − 12
3
=
0 − v1
6
+
v2 − v1
4
108 PART 1 DC Circuits
2 A
v1
1 2
v2
12 V +
−
3 Ω 4 Ω
6 Ω 6 Ω
8 Ω
Figure 3.46 For Review Questions 3.1 and 3.2.
3.2 In the circuit in Fig. 3.46, applying KCL at node 2
gives:
(a)
v2 − v1
4
+
v2
8
=
v2
6
(b)
v1 − v2
4
+
v2
8
=
v2
6
(c)
v1 − v2
4
+
12 − v2
8
=
v2
6
(d)
v2 − v1
4
+
v2 − 12
8
=
v2
6
3.3 For the circuit in Fig. 3.47, v1 and v2 are related as:
(a) v1 = 6i + 8 + v2 (b) v1 = 6i − 8 + v2
(c) v1 = −6i + 8 + v2 (d) v1 = −6i − 8 + v2
12 V +
− 4 Ω
6 Ω 8 V
v2
v1
i
+
−
Figure 3.47 For Review Questions 3.3 and 3.4.
3.4 In the circuit in Fig. 3.47, the voltage v2 is:
(a) −8 V (b) −1.6 V
(c) 1.6 V (d) 8 V
3.5 The current i in the circuit in Fig. 3.48 is:
(a) −2.667 A (b) −0.667 A
(c) 0.667 A (d) 2.667 A
10 V +
− 6 V
+
−
4 Ω
i
2 Ω
Figure 3.48 For Review Questions 3.5 and 3.6.
3.6 The loop equation for the circuit in Fig. 3.48 is:
(a) −10 + 4i + 6 + 2i = 0
(b) 10 + 4i + 6 + 2i = 0
(c) 10 + 4i − 6 + 2i = 0
(d) −10 + 4i − 6 + 2i = 0
3.7 In the circuit in Fig. 3.49, current i1 is:
(a) 4 A (b) 3 A (c) 2 A (d) 1 A
i1 i2
2 A
20 V +
−
2 Ω 1 Ω
3 Ω 4 Ω
v
+
−
Figure 3.49 For Review Questions 3.7 and 3.8.
3.8 The voltage v across the current source in the circuit
of Fig. 3.49 is:
(a) 20 V (b) 15 V (c) 10 V (d) 5 V
3.9 The PSpice part name for a current-controlled
voltage source is:
(a) EX (b) FX (c) HX (d) GX
3.10 Which of the following statements are not true of the
pseudocomponent IPROBE:
(a) It must be connected in series.
(b) It plots the branch current.
(c) It displays the current through the branch in
which it is connected.
(d) It can be used to display voltage by connecting it
in parallel.
(e) It is used only for dc analysis.
(f) It does not correspond to a particular circuit
element.
Answers: 3.1a, 3.2c, 3.3b, 3.4d, 3.5c, 3.6a, 3.7d, 3.8b, 3.9c, 3.10b,d.
CHAPTER 3 Methods of Analysis 109
PROBLEMS
Sections 3.2 and 3.3 Nodal Analysis
3.1 Determine v1, v2, and the power dissipated in all the
resistors in the circuit of Fig. 3.50.
10 A
6 A 8 Ω
4 Ω
v1 v2
2 Ω
Figure 3.50 For Prob. 3.1.
3.2 For the circuit in Fig. 3.51, obtain v1 and v2.
3 A
6 A
5 Ω
10 Ω
2 Ω
v1 v2
4 Ω
Figure 3.51 For Prob. 3.2.
3.3 Find the currents i1 through i4 and the voltage vo in
the circuit in Fig. 3.52.
10 A 2 A 60 Ω
30 Ω
20 Ω
10 Ω
i1 i2 i3 i4
vo
Figure 3.52 For Prob. 3.3.
3.4 Given the circuit in Fig. 3.53, calculate the currents
i1 through i4.
4 A 5 A
2 A
5 Ω
10 Ω
10 Ω
5 Ω
i1 i2 i3 i4
Figure 3.53 For Prob. 3.4.
3.5 Obtain vo in the circuit of Fig. 3.54.
30 V +
−
2 kΩ
20 V +
−
5 kΩ
4 kΩ vo
+
−
Figure 3.54 For Prob. 3.5.
3.6 Use nodal analysis to obtain vo in the circuit in Fig.
3.55.
6 Ω 2 Ω
12 V
10 V
+
−
+ −
4 Ω
i3
i2
vo
i1
Figure 3.55 For Prob. 3.6.
3.7 Using nodal analysis, find vo in the circuit of Fig.
3.56.
3 V
4vo
+
−
2 Ω
vo
+
− 1 Ω
3 Ω 5 Ω
+
−
Figure 3.56 For Prob. 3.7.
3.8 Calculate vo in the circuit in Fig. 3.57.
12 V 2vo
+
− 8 Ω
6 Ω
+
−
3 Ω
vo
+ −
Figure 3.57 For Prob. 3.8.
3.9 Find io in the circuit in Fig. 3.58.
110 PART 1 DC Circuits
2 Ω 4 Ω
8 Ω
1 Ω
4 A 2io
io
Figure 3.58 For Prob. 3.9.
3.10 Solve for i1 and i2 in the circuit in Fig. 3.22 (Section
3.5) using nodal analysis.
3.11 Use nodal analysis to find currents i1 and i2 in the
circuit of Fig. 3.59.
24 V +
− 40 Ω 20 Ω
5 A
10 Ω 20 Ω 30 Ω
i2
i1
Figure 3.59 For Prob. 3.11.
3.12 Calculate v1 and v2 in the circuit in Fig. 3.60 using
nodal analysis.
8 Ω 4 Ω 3 A
2 Ω 2 V
v2
v1
+ −
Figure 3.60 For Prob. 3.12.
3.13 Using nodal analysis, find vo in the circuit of Fig.
3.61.
2 Ω
5 A
8 Ω
+
−
+
−
4 Ω 20 V
vo
+
−
1 Ω
40 V
Figure 3.61 For Prob. 3.13.
3.14 Apply nodal analysis to find io and the power
dissipated in each resistor in the circuit of Fig. 3.62.
5 S
6 S
2 A
io
4 A
3 S
10 V
+ −
Figure 3.62 For Prob. 3.14.
3.15 Determine voltages v1 through v3 in the circuit of
Fig. 3.63 using nodal analysis.
1 S 13 V
2 S
v1 v2
2vo
v3
8 S
2 A 4 S
vo
+
−
+
−
+ −
Figure 3.63 For Prob. 3.15.
3.16 Using nodal analysis, find current io in the circuit of
Fig. 3.64.
60 V
io
3io
10 Ω
8 Ω
2 Ω
+
−
4 Ω
Figure 3.64 For Prob. 3.16.
3.17 Determine the node voltages in the circuit in Fig.
3.65 using nodal analysis.
5 A
2
3
1
2 Ω
2 Ω
10 V
+
−
8 Ω
4 Ω
Figure 3.65 For Prob. 3.17.
CHAPTER 3 Methods of Analysis 111
3.18 For the circuit in Fig. 3.66, find v1 and v2 using
nodal analysis.
3 mA
v2
v1
2 kΩ
4 kΩ
1 kΩ vo
3vo
+
−
+
−
Figure 3.66 For Prob. 3.18.
3.19 Determine v1 and v2 in the circuit in Fig. 3.67.
3 A
v2
5vo
v1
8 Ω
1 Ω
4 Ω
12 V
2 Ω
vo
+
−
–
+
+ −
Figure 3.67 For Prob. 3.19.
3.20 Obtain v1 and v2 in the circuit of Fig. 3.68.
2 A
5 A 10 Ω
v1 v2
5 Ω
8 V
+
−
Figure 3.68 For Prob. 3.20.
3.21 Find vo and io in the circuit in Fig. 3.69.
20 V 2 Ω
2 Ω
1 Ω
+
−
40 V +
−
10 V +
−
4 Ω vo
+
−
io
Figure 3.69 For Prob. 3.21.
3.22
∗
Use nodal analysis to determine voltages v1, v2, and
v3 in the circuit in Fig. 3.70.
2 S
2 A 4 S 2 S 4 A
io
1 S
4 S
1 S
v1
3io
v2 v3
Figure 3.70 For Prob. 3.22.
3.23 Using nodal analysis, find vo and io in the circuit of
Fig. 3.71.
+
−
100 V 80 Ω vo
+
−
10 Ω 20 Ω
40 Ω
120 V
+
− 2io
4vo
+
−
io
Figure 3.71 For Prob. 3.23.
3.24 Find the node voltages for the circuit in Fig. 3.72.
4 Ω
1 A 1 Ω 4 Ω 10 V
io
1 Ω
2 Ω
v1
2vo
4io
v2 v3
+
−
vo
+
−
+ −
Figure 3.72 For Prob. 3.24.
∗An asterisk indicates a challenging problem.
112 PART 1 DC Circuits
3.25
∗
Obtain the node voltages v1, v2, and v3 in the circuit
of Fig. 3.73.
10 kΩ
4 mA
5 kΩ
v1
20 V
10 V v2
v3
12 V
+
−
+
− + −
Figure 3.73 For Prob. 3.25.
Sections 3.4 and 3.5 Mesh Analysis
3.26 Which of the circuits in Fig. 3.74 is planar? For the
planar circuit, redraw the circuits with no crossing
branches.
2 Ω
6 Ω
5 Ω
2 A
(a)
4 Ω
3 Ω
1 Ω
(b)
12 V +
− 2 Ω
3 Ω
5 Ω
4 Ω
1 Ω
Figure 3.74 For Prob. 3.26.
3.27 Determine which of the circuits in Fig. 3.75 is
planar and redraw it with no crossing branches.
10 V +
−
3 Ω
5 Ω
2 Ω
7 Ω
4 Ω
(a)
1 Ω
6 Ω
7 Ω
6 Ω
1 Ω 3 Ω
4 A
(b)
8 Ω
2 Ω
5 Ω 4 Ω
Figure 3.75 For Prob. 3.27.
3.28 Rework Prob. 3.5 using mesh analysis.
3.29 Rework Prob. 3.6 using mesh analysis.
3.30 Solve Prob. 3.7 using mesh analysis.
3.31 Solve Prob. 3.8 using mesh analysis.
3.32 For the bridge network in Fig. 3.76, find io using
mesh analysis.
30 V +
−
2 kΩ
2 kΩ
6 kΩ 6 kΩ
4 kΩ
4 kΩ
io
Figure 3.76 For Prob. 3.32.
3.33 Apply mesh analysis to find i in Fig. 3.77.
CHAPTER 3 Methods of Analysis 113
+
−
+ −
10 Ω
2 Ω
5 Ω
1 Ω
8 V
6 V
i1
i2 i3
i
4 Ω
Figure 3.77 For Prob. 3.33.
3.34 Use mesh analysis to find vab and io in the circuit in
Fig. 3.78.
+
−
20 Ω
20 Ω
30 Ω
30 Ω
20 Ω
80 V
+
−
80 V
30 Ω vab
+
−
io
Figure 3.78 For Prob. 3.34.
3.35 Use mesh analysis to obtain io in the circuit of Fig.
3.79.
3 A
12 V
+
−
4 Ω
io 1 Ω
6 V
2 Ω
5 Ω
+ −
Figure 3.79 For Prob. 3.35.
3.36 Find current i in the circuit in Fig. 3.80.
4 A
30 V
i
+
− 3 Ω 1 Ω
2 Ω 6 Ω
4 Ω 8 Ω
Figure 3.80 For Prob. 3.36.
3.37 Find vo and io in the circuit of Fig. 3.81.
16 V
2io
3 Ω
1 Ω 2 Ω
2 Ω +
−
io
vo
Figure 3.81 For Prob. 3.37.
3.38 Use mesh analysis to find the current io in the circuit
in Fig. 3.82.
3io
10 Ω
4 Ω
60 V +
−
io
8 Ω
2 Ω
Figure 3.82 For Prob. 3.38.
3.39 Apply mesh analysis to find vo in the circuit in Fig.
3.83.
20 V
5 A
2 Ω 8 Ω
1 Ω
40 V
vo
+
−
+
−
4 Ω
Figure 3.83 For Prob. 3.39.
3.40 Use mesh analysis to find i1, i2, and i3 in the circuit
of Fig. 3.84.
114 PART 1 DC Circuits
12 V +
−
8 Ω
4 Ω +
−
2 Ω
vo
2vo
i2
i3
i1
3 A
+
−
Figure 3.84 For Prob. 3.40.
3.41 Rework Prob. 3.11 using mesh analysis.
3.42
∗
In the circuit of Fig. 3.85, solve for i1, i2, and i3.
4 A 2 Ω
1 A
i3
i1
i2
6 Ω
12 Ω 4 Ω
8 V
+ −
10 V
+ −
Figure 3.85 For Prob. 3.42.
3.43 Determine v1 and v2 in the circuit of Fig. 3.86.
12 V
2 Ω 2 Ω
2 Ω
+
−
2 Ω
v1
2 Ω
v2
+
−
+ −
Figure 3.86 For Prob. 3.43.
3.44 Find i1, i2, and i3 in the circuit in Fig. 3.87.
10 Ω
10 Ω
120 V 30 Ω
30 Ω
30 Ω
i3
i2
i1
+
−
Figure 3.87 For Prob. 3.44.
3.45 Rework Prob. 3.23 using mesh analysis.
3.46 Calculate the power dissipated in each resistor in the
circuit in Fig. 3.88.
10 V
0.5io
4 Ω 8 Ω
1 Ω 2 Ω
+
−
io
Figure 3.88 For Prob. 3.46.
3.47 Calculate the current gain io/is in the circuit of Fig.
3.89.
5vo
20 Ω 10 Ω
40 Ω
io
is 30 Ω
vo
+
−
–
+
Figure 3.89 For Prob. 3.47.
3.48 Find the mesh currents i1, i2, and i3 in the network
of Fig. 3.90.
4 kΩ 8 kΩ 2 kΩ
100 V 4 mA 2i1 40 V
+
−
+
−
i1 i2 i3
Figure 3.90 For Prob. 3.48.
3.49 Find vx and ix in the circuit shown in Fig. 3.91.
ix
2 Ω
vx 4ix
+
−
5 Ω
50 V
3 A
+
−
+
−
vx
4
10 Ω
Figure 3.91 For Prob. 3.49.
CHAPTER 3 Methods of Analysis 115
3.50 Find vo and io in the circuit of Fig. 3.92.
+
−
+
−
io
+ −
2 A
100 V 40 Ω
10 Ω
50 Ω 10 Ω
vo
0.2vo
4io
Figure 3.92 For Prob. 3.50.
Section 3.6 Nodal and Mesh Analyses by
Inspection
3.51 Obtain the node-voltage equations for the circuit in
Fig. 3.93 by inspection. Determine the node
voltages v1 and v2.
3 A 5 A
4 Ω 2 Ω
v1 v2
1 Ω
6 A
Figure 3.93 For Prob. 3.51.
3.52 By inspection, write the node-voltage equations for
the circuit in Fig. 3.94 and obtain the node voltages.
4 A 3 S 2 A 1 A
v1 1 S 2 S
5 S
v2
v3
Figure 3.94 For Prob. 3.52.
3.53 For the circuit shown in Fig. 3.95, write the
node-voltage equations by inspection.
2 kΩ 2 kΩ 10 mA
20 mA
v1 4 kΩ 4 kΩ
1 kΩ
5 mA
v2
v3
Figure 3.95 For Prob. 3.53.
3.54 Write the node-voltage equations of the circuit in
Fig. 3.96 by inspection.
I1
v1 v3
G4 G5
G2 G3
G1
v2
I2
Figure 3.96 For Prob. 3.54.
3.55 Obtain the mesh-current equations for the circuit in
Fig. 3.97 by inspection. Calculate the power
absorbed by the 8- resistor.
+
−
+
−
+
−
12 A 20 V
2 Ω 2 Ω
i1 i2 i3
8 V
4 Ω 8 Ω 5 Ω
Figure 3.97 For Prob. 3.55.
3.56 By inspection, write the mesh-current equations for
the circuit in Fig. 3.98.
116 PART 1 DC Circuits
+
−
+
−
+
−
10 V
4 Ω
5 Ω 2 Ω 4 Ω
i1 i2 i3
8 V 4 V
i4
1 Ω
Figure 3.98 For Prob. 3.56.
3.57 Write the mesh-current equations for the circuit in
Fig. 3.99.
+
−
+ −
+
−
+
−
6 V 4 V
1 Ω 1 Ω
3 Ω
1 Ω
i1 i2
i4
i3
2 V 3 V
2 Ω
4 Ω
5 Ω
Figure 3.99 For Prob. 3.57.
3.58 By inspection, obtain the mesh-current equations for
the circuit in Fig. 3.100.
+
−
+ −
+
−
i1
i3
V1
V3
V2
V4
i2
i4
R1 R2 R3
R4
R5
R6
R7
R8
+
−
Figure 3.100 For Prob. 3.58.
Section 3.8 Circuit Analysis with PSpice
3.59 Use PSpice to solve Prob. 3.44.
3.60 Use PSpice to solve Prob. 3.22.
3.61 Rework Prob. 3.51 using PSpice.
3.62 Find the nodal voltages v1 through v4 in the circuit
in Fig. 3.101 using PSpice.
+
−
+ −
8 A
20 V
1 Ω
v1
2 Ω
4 Ω
10 Ω 12 Ω
v2
v3
io
6io
v4
Figure 3.101 For Prob. 3.62.
3.63 Use PSpice to solve the problem in Example 3.4.
3.64 If the Schematics Netlist for a network is as follows,
draw the network.
R_R1 1 2 2K
R_R2 2 0 4K
R_R3 3 0 8K
R_R4 3 4 6K
R_R5 1 3 3K
V_VS 4 0 DC 100
I_IS 0 1 DC 4
F_F1 1 3 VF_F1 2
VF_F1 5 0 0V
E_E1 3 2 1 3 3
3.65 The following program is the Schematics Netlist of
a particular circuit. Draw the circuit and determine
the voltage at node 2.
R_R1 1 2 20
R_R2 2 0 50
R_R3 2 3 70
R_R4 3 0 30
V_VS 1 0 20V
I_IS 2 0 DC 2A
Section 3.9 Applications
3.66 Calculate vo and io in the circuit of Fig. 3.102.
+
−
+
−
3 mV vo
+
−
4 kΩ
50io
io
vo
100
20 kΩ
Figure 3.102 For Prob. 3.66.
3.67 For the simplified transistor circuit of Fig. 3.103,
calculate the voltage vo.
CHAPTER 3 Methods of Analysis 117
+
−
+
−
i
2 kΩ
5 kΩ
1 kΩ
30 mV vo
400i
Figure 3.103 For Prob. 3.67.
3.68 For the circuit in Fig. 3.104, find the gain vo/vs.
+
−
–
+
+
−
+
−
500 Ω 400 Ω
2 kΩ 200 Ω
vs vo
v1 60v1
Figure 3.104 For Prob. 3.68.
3.69
∗
Determine the gain vo/vs of the transistor amplifier
circuit in Fig. 3.105.
2 kΩ
100 Ω
200 Ω
vs 40io
io
vo 10 kΩ
vo
1000
+
−
+
−
+
−
Figure 3.105 For Prob. 3.69.
3.70 For the simple transistor circuit of Fig. 3.106, let
β = 75, VBE = 0.7 V. What value of vi is required
to give a collector-emitter voltage of 2 V?
+
−
5 V
2 kΩ
vi
40 kΩ
Figure 3.106 For Prob. 3.70.
3.71 Calculate vs for the transistor in Fig. 3.107 given
that vo = 4 V, β = 150, VBE = 0.7 V.
+
−
18 V
1 kΩ
vs
10 kΩ
+
−
500 Ω vo
Figure 3.107 For Prob. 3.71.
3.72 For the transistor circuit of Fig. 3.108, find IB , VCE,
and vo. Take β = 200, VBE = 0.7 V.
+
−
9 V
5 kΩ
3 V
6 kΩ
+
−
400 Ω vo
VCE
+
−
IB
2 kΩ
Figure 3.108 For Prob. 3.72.
3.73 Find IB and VC for the circuit in Fig. 3.109. Let
β = 100, VBE = 0.7 V.
+
−
IB
12 V
4 kΩ
VC
10 kΩ
5 kΩ
Figure 3.109 For Prob. 3.73.
COMPREHENSIVE PROBLEMS
3.74
∗
Rework Example 3.11 with hand calculation.
119
C H A P T E R
CIRCUIT THEOREMS
4
Our schools had better get on with what is their overwhelmingly most
important task: teaching their charges to express themselves clearly and
with precision in both speech and writing; in other words, leading them
toward mastery of their own language. Failing that, all their instruction
in mathematics and science is a waste of time.
—Joseph Weizenbaum, M.I.T.
Enhancing Your Career
Enhancing Your Communication Skill Taking a course
in circuit analysis is one step in preparing yourself for a
career in electrical engineering. Enhancing your commu-
nication skill while in school should also be part of that
preparation, as a large part of your time will be spent com-
municating.
People in industry have complained again and again
that graduating engineers are ill-prepared in written and
oral communication. An engineer who communicates ef-
fectively becomes a valuable asset.
You can probably speak or write easily and quickly.
But how effectively do you communicate? The art of ef-
fective communication is of the utmost importance to your
success as an engineer.
For engineers in industry, communication is key to
promotability. Consider the result of a survey of U.S. cor-
porations that asked what factors influence managerial pro-
motion. Thesurveyincludesalistingof22personalqualities
and their importance in advancement. You may be surprised
to note that “technical skill based on experience” placed
fourth from the bottom. Attributes such as self-confidence,
ambition, flexibility, maturity, ability to make sound deci-
sions, getting things done with and through people, and ca-
pacity for hard work all ranked higher. At the top of the list
was “ability to communicate.” The higher your professional
career progresses, the more you will need to communicate.
Therefore, you should regard effective communication as an
important tool in your engineering tool chest.
Learning to communicate effectively is a lifelong
task you should always work toward. The best time to begin
is while still in school. Continually look for opportunities
to develop and strengthen your reading, writing, listening,
Ability to
work hard
Working and
getting along
with people
Maturity
Appearance
Problem-solving
skills
Self-
determination
College
education
Effective
communication
Ability to communicate effectively is regarded by many as the most
important step to an executive promotion.
(Adapted from J. Sherlock, A Guide to Technical Communication.
Boston, MA: Allyn and Bacon, 1985, p. 7.)
and speaking skills. You can do this through classroom
presentations, team projects, active participation in student
organizations, and enrollment in communication courses.
The risks are less now than later in the workplace.
120 PART 1 DC Circuits
4.1 INTRODUCTION
A major advantage of analyzing circuits using Kirchhoff’s laws as we did
in Chapter 3 is that we can analyze a circuit without tampering with its
original configuration. A major disadvantage of this approach is that, for
a large, complex circuit, tedious computation is involved.
The growth in areas of application of electric circuits has led to an
evolution from simple to complex circuits. To handle the complexity,
engineers over the years have developed some theorems to simplify cir-
cuit analysis. Such theorems include Thevenin’s and Norton’s theorems.
Since these theorems are applicable to linear circuits, we first discuss the
concept of circuit linearity. In addition to circuit theorems, we discuss the
concepts of superposition, source transformation, and maximum power
transfer in this chapter. The concepts we develop are applied in the last
section to source modeling and resistance measurement.
4.2 LINEARITY PROPERTY
Linearity is the property of an element describing a linear relationship
between cause and effect. Although the property applies to many circuit
elements, we shall limit its applicability to resistors in this chapter. The
property is a combination of both the homogeneity (scaling) property and
the additivity property.
The homogeneity property requires that if the input (also called the
excitation) is multiplied by a constant, then the output (also called the
response) is multiplied by the same constant. For a resistor, for example,
Ohm’s law relates the input i to the output v,
v = iR (4.1)
If the current is increased by a constant k, then the voltage increases
correspondingly by k, that is,
kiR = kv (4.2)
The additivity property requires that the response to a sum of inputs
is the sum of the responses to each input applied separately. Using the
voltage-current relationship of a resistor, if
v1 = i1R (4.3a)
and
v2 = i2R (4.3b)
then applying (i1 + i2) gives
v = (i1 + i2)R = i1R + i2R = v1 + v2 (4.4)
We say that a resistor is a linear element because the voltage-current
relationship satisfies both the homogeneity and the additivity properties.
In general, a circuit is linear if it is both additive and homogeneous.
A linear circuit consists of only linear elements, linear dependent sources,
and independent sources.
CHAPTER 4 Circuit Theorems 121
A linear circuit is one whose output is linearly related
(or directly proportional) to its input.
Throughout this book we consider only linear circuits. Note that since
p = i2
R = v2
/R (making it a quadratic function rather than a linear
one), the relationship between power and voltage (or current) is nonlinear.
Therefore, the theorems covered in this chapter are not applicable to
power.
To understand the linearity principle, consider the linear circuit
shown in Fig. 4.1. The linear circuit has no independent sources inside
it. It is excited by a voltage source vs, which serves as the input. The
circuit is terminated by a load R. We may take the current i through R as
the output. Suppose vs = 10 V gives i = 2 A. According to the linearity
principle, vs = 1 V will give i = 0.2 A. By the same token, i = 1 mA
must be due to vs = 5 mV.
vs R
i
+
− Linear circuit
Figure 4.1 A linear circuit with input vs and
output i.
E X A M P L E 4 . 1
For the circuit in Fig. 4.2, find io when vs = 12 V and vs = 24 V.
+
−
vs
vx
3vx
i1 i2
2 Ω 8 Ω
4 Ω
6 Ω
4 Ω
–
+
+ − io
Figure4.2 For Example 4.1.
Solution:
Applying KVL to the two loops, we obtain
12i1 − 4i2 + vs = 0 (4.1.1)
− 4i1 + 16i2 − 3vx − vs = 0 (4.1.2)
But vx = 2i1. Equation (4.1.2) becomes
−10i1 + 16i2 − vs = 0 (4.1.3)
Adding Eqs. (4.1.1) and (4.1.3) yields
2i1 + 12i2 = 0 ⇒ i1 = −6i2
Substituting this in Eq. (4.1.1), we get
−76i2 + vs = 0 ⇒ i2 =
vs
76
When vs = 12 V,
io = i2 =
12
76
A
When vs = 24 V,
io = i2 =
24
76
A
showing that when the source value is doubled, io doubles.
P R A C T I C E P R O B L E M 4 . 1
For the circuit in Fig. 4.3, find vo when is = 15 and is = 30 A.
is
6 Ω
4 Ω
2 Ω
+
−
vo
Figure4.3 For Practice Prob. 4.1.
Answer: 10 V, 20 V.
122 PART 1 DC Circuits
E X A M P L E 4 . 2
Assume Io = 1 A and use linearity to find the actual value of Io in the
circuit in Fig. 4.4.
Io
I4 I2
I3
V2
6 Ω 2 Ω
2
5 Ω
7 Ω
I1
V1
3 Ω
1
4 Ω
Is = 15 A
Figure4.4 For Example 4.2.
Solution:
If Io = 1 A, then V1 = (3 + 5)Io = 8 V and I1 = V1/4 = 2 A. Applying
KCL at node 1 gives
I2 = I1 + Io = 3 A
V2 = V1 + 2I2 = 8 + 6 = 14 V, I3 =
V2
7
= 2 A
Applying KCL at node 2 gives
I4 = I3 + I2 = 5 A
Therefore, Is = 5 A. This shows that assuming Io = 1 gives Is = 5 A;
the actual source current of 15 A will give Io = 3 A as the actual value.
P R A C T I C E P R O B L E M 4 . 2
Assume that Vo = 1 V and use linearity to calculate the actual value of
Vo in the circuit of Fig. 4.5.
10 V
12 Ω
8 Ω
5 Ω
+
−
+
−
Vo
Figure4.5 For Practice Prob. 4.2
Answer: 4 V.
4.3 SUPERPOSITION
If a circuit has two or more independent sources, one way to determine
the value of a specific variable (voltage or current) is to use nodal or mesh
analysis as in Chapter 3. Another way is to determine the contribution of
each independent source to the variable and then add them up. The latter
approach is known as the superposition.
CHAPTER 4 Circuit Theorems 123
The idea of superposition rests on the linearity property. Superpositionisnotlimitedtocircuitanalysisbut
isapplicableinmanyfieldswherecauseandeffect
bear a linear relationship to one another.
The superposition principle states that the voltage across (or current through) an
element in a linear circuit is the algebraic sum of the voltages across (or currents
through) that element due to each independent source acting alone.
The principle of superposition helps us to analyze a linear circuit with
more than one independent source by calculating the contribution of each
independent source separately. However, to apply the superposition prin-
ciple, we must keep two things in mind:
1. We consider one independent source at a time while all other
independent sources are turned off. This implies that we
replace every voltage source by 0 V (or a short circuit), and
every current source by 0 A (or an open circuit). This way we
obtain a simpler and more manageable circuit.
Other terms such as killed, made inactive, dead-
ened, or set equal to zero are often used to con-
vey the same idea.
2. Dependent sources are left intact because they are controlled
by circuit variables.
With these in mind, we apply the superposition principle in three steps:
Steps to Apply Superposition Principle:
1. Turn off all independent sources except one source. Find the
output (voltage or current) due to that active source using nodal or
mesh analysis.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the
contributions due to the independent sources.
Analyzing a circuit using superposition has one major disadvan-
tage: it may very likely involve more work. If the circuit has three
independent sources, we may have to analyze three simpler circuits each
providing the contribution due to the respective individual source. How-
ever, superposition does help reduce a complex circuit to simpler circuits
through replacement of voltage sources by short circuits and of current
sources by open circuits.
Keep in mind that superposition is based on linearity. For this
reason, it is not applicable to the effect on power due to each source,
because the power absorbed by a resistor depends on the square of the
voltage or current. If the power value is needed, the current through (or
voltage across) the element must be calculated first using superposition.
For example, when current i1 flows through re-
sistorR,thepowerisp1 =Ri2
1,andwhencurrent
i2 flows through R, the power is p2 = Ri2
2. If cur-
rent i1 + i2 flows through R, the power absorb-
ed is p3 = R(i1 + i2)2
= Ri2
1 + Ri2
2 + 2Ri1i2 =
p1 + p2. Thus, the power relation is nonlinear.
E X A M P L E 4 . 3
Use the superposition theorem to find v in the circuit in Fig. 4.6.
6 V v 3 A
8 Ω
4 Ω
+
−
+
−
Figure4.6 For Example 4.3.
Solution:
Since there are two sources, let
v = v1 + v2
where v1 and v2 are the contributions due to the 6-V voltage source and
124 PART 1 DC Circuits
the 3-A current source, respectively. To obtain v1, we set the current
source to zero, as shown in Fig. 4.7(a). Applying KVL to the loop in Fig.
4.7(a) gives
12i1 − 6 = 0 ⇒ i1 = 0.5 A
Thus,
v1 = 4i1 = 2 V
We may also use voltage division to get v1 by writing
v1 =
4
4 + 8
(6) = 2 V
To get v2, we set the voltage source to zero, as in Fig. 4.7(b). Using
current division,
i3 =
8
4 + 8
(3) = 2 A
Hence,
v2 = 4i3 = 8 V
And we find
v = v1 + v2 = 2 + 8 = 10 V
+
−
6 V i1
8 Ω
v1
4 Ω
(a)
+
−
3 A
8 Ω
v2
i2
i3
4 Ω
(b)
+
−
Figure4.7 For Example 4.3:
(a) calculating v1, (b) calculating v2.
P R A C T I C E P R O B L E M 4 . 3
Using the superposition theorem, find vo in the circuit in Fig. 4.8.
3 Ω 5 Ω
2 Ω 8 A 20 V
+
−
+
−
vo
Figure4.8 For Practice Prob. 4.3.
Answer: 12 V.
E X A M P L E 4 . 4
Find io in the circuit in Fig. 4.9 using superposition.
4 A
20 V
3 Ω
5 Ω
1 Ω
2 Ω
4 Ω
+ −
5io
io
+ −
Figure4.9 For Example 4.4.
Solution:
The circuit in Fig. 4.9 involves a dependent source, which must be left
intact. We let
io = i
o + i
o (4.4.1)
where i
o and i
o are due to the 4-A current source and 20-V voltage source
respectively. To obtain i
o, we turn off the 20-V source so that we have
the circuit in Fig. 4.10(a). We apply mesh analysis in order to obtain i
o.
For loop 1,
i1 = 4 A (4.4.2)
CHAPTER 4 Circuit Theorems 125
4 A
3 Ω
5 Ω
1 Ω
2 Ω
4 Ω
+ −
i1 i3
i′
o
5i′
o
0
(a)
3 Ω
5 Ω
1 Ω
2 Ω
4 Ω
+ −
i′′
o
5i′′
o
(b)
20 V
+ −
i1
i2
i3
i5
i4
Figure4.10 For Example 4.4: Applying superposition to (a) obtain i
0, (b) obtain i
0 .
For loop 2,
−3i1 + 6i2 − 1i3 − 5i
o = 0 (4.4.3)
For loop 3,
−5i1 − 1i2 + 10i3 + 5i
o = 0 (4.4.4)
But at node 0,
i3 = i1 − i
o = 4 − i
o (4.4.5)
Substituting Eqs. (4.4.2) and (4.4.5) into Eqs. (4.4.3) and (4.4.4) gives
two simultaneous equations
3i2 − 2i
o = 8 (4.4.6)
i2 + 5i
o = 20 (4.4.7)
which can be solved to get
i
o =
52
17
A (4.4.8)
To obtain i
o , we turn off the 4-A current source so that the circuit
becomes that shown in Fig. 4.10(b). For loop 4, KVL gives
6i4 − i5 − 5i
o = 0 (4.4.9)
and for loop 5,
−i4 + 10i5 − 20 + 5i
o = 0 (4.4.10)
But i5 = −i
o . Substituting this in Eqs. (4.4.9) and (4.4.10) gives
6i4 − 4i
o = 0 (4.4.11)
i4 + 5i
o = −20 (4.4.12)
which we solve to get
126 PART 1 DC Circuits
i
o = −
60
17
A (4.4.13)
Now substituting Eqs. (4.4.8) and (4.4.13) into Eq. (4.4.1) gives
io = −
8
17
= −0.4706 A
P R A C T I C E P R O B L E M 4 . 4
Use superposition to find vx in the circuit in Fig. 4.11.
vx
20 Ω
0.1vx
4 Ω
10 V 2 A
+
−
Figure4.11 For Practice Prob. 4.4.
Answer: vx = 12.5 V.
E X A M P L E 4 . 5
For the circuit in Fig. 4.12, use the superposition theorem to find i.
+ −
+
−
24 V 8 Ω
4 Ω
3 Ω 3 A
12 V
4 Ω
i
Figure4.12 For Example 4.5.
Solution:
In this case, we have three sources. Let
i = i1 + i2 + i3
where i1, i2, and i3 are due to the 12-V, 24-V, and 3-A sources respectively.
To get i1, consider the circuit in Fig. 4.13(a). Combining 4  (on the right-
hand side) in series with 8  gives 12 . The 12  in parallel with 4 
gives 12 × 4/16 = 3 . Thus,
i1 =
12
6
= 2 A
To get i2, consider the circuit in Fig. 4.13(b). Applying mesh analysis,
16ia − 4ib + 24 = 0 ⇒ 4ia − ib = −6 (4.5.1)
7ib − 4ia = 0 ⇒ ia =
7
4
ib (4.5.2)
Substituting Eq. (4.5.2) into Eq. (4.5.1) gives
i2 = ib = −1
To get i3, consider the circuit in Fig. 4.13(c). Using nodal analysis,
3 =
v2
8
+
v2 − v1
4
⇒ 24 = 3v2 − 2v1 (4.5.3)
v2 − v1
4
=
v1
4
+
v1
3
⇒ v2 =
10
3
v1 (4.5.4)
Substituting Eq. (4.5.4) into Eq. (4.5.3) leads to v1 = 3 and
CHAPTER 4 Circuit Theorems 127
8 Ω
4 Ω 4 Ω
3 Ω
12 V +
−
3 Ω
3 Ω
12 V +
−
(a)
8 Ω
24 V
4 Ω 4 Ω
3 Ω
(b)
+ −
ib
ia
8 Ω
4 Ω 4 Ω
3 Ω 3 A
v1
v2
(c)
i2
i2 i2
i1
Figure4.13 For Example 4.5.
i3 =
v1
3
= 1 A
Thus,
i = i1 + i2 + i3 = 2 − 1 + 1 = 2 A
P R A C T I C E P R O B L E M 4 . 5
Find i in the circuit in Fig. 4.14 using the superposition principle.
16 V
8 Ω
2 Ω
4 A
6 Ω
+
− 12 V
+
−
i
Figure4.14 For Practice Prob. 4.5.
Answer: 0.75 A.
4.4 SOURCE TRANSFORMATION
We have noticed that series-parallel combination and wye-delta transfor-
mation help simplify circuits. Source transformation is another tool for
simplifying circuits. Basic to these tools is the concept of equivalence.
128 PART 1 DC Circuits
We recall that an equivalent circuit is one whose v-i characteristics are
identical with the original circuit.
In Section 3.6, we saw that node-voltage (or mesh-current) equa-
tions can be obtained by mere inspection of a circuit when the sources are
all independent current (or all independent voltage) sources. It is there-
fore expedient in circuit analysis to be able to substitute a voltage source
in series with a resistor for a current source in parallel with a resistor, or
vice versa, as shown in Fig. 4.15. Either substitution is known as a source
transformation.
+
−
vs
R
a
b
is R
a
b
Figure4.15 Transformation of independent sources.
A source transformation is the process of replacing a voltage source vs
in series with a resistor R by a current source is in parallel
with a resistor R, or vice versa.
The two circuits in Fig. 4.15 are equivalent—provided they have the same
voltage-current relation at terminals a-b. It is easy to show that they are
indeed equivalent. If the sources are turned off, the equivalent resistance
at terminals a-b in both circuits is R. Also, when terminals a-b are short-
circuited, the short-circuit current flowing from a to b is isc = vs/R in
the circuit on the left-hand side and isc = is for the circuit on the right-
hand side. Thus, vs/R = is in order for the two circuits to be equivalent.
Hence, source transformation requires that
vs = isR or is =
vs
R
(4.5)
Source transformation also applies to dependent sources, provided
we carefully handle the dependent variable. As shown in Fig. 4.16, a
dependent voltage source in series with a resistor can be transformed to
a dependent current source in parallel with the resistor or vice versa.
vs
R
a
b
is R
a
b
+
−
Figure4.16 Transformation of dependent sources.
Like the wye-delta transformation we studied in Chapter 2, a source
transformation does not affect the remaining part of the circuit. When
CHAPTER 4 Circuit Theorems 129
applicable, source transformation is a powerful tool that allows circuit
manipulations to ease circuit analysis. However, we should keep the
following points in mind when dealing with source transformation.
1. Note from Fig. 4.15 (or Fig. 4.16) that the arrow of the current
source is directed toward the positive terminal of the voltage
source.
2. Note from Eq. (4.5) that source transformation is not possible
when R = 0, which is the case with an ideal voltage source.
However, for a practical, nonideal voltage source, R = 0.
Similarly, an ideal current source with R = ∞ cannot be
replaced by a finite voltage source. More will be said on ideal
and nonideal sources in Section 4.10.1.
E X A M P L E 4 . 6
Use source transformation to find vo in the circuit in Fig. 4.17. 2 Ω 3 Ω
12 V
8 Ω
4 Ω 3 A +
−
+
−
vo
Figure4.17 For Example 4.6.
Solution:
We first transform the current and voltage sources to obtain the circuit in
Fig. 4.18(a). Combining the 4- and 2- resistors in series and trans-
forming the 12-V voltage source gives us Fig. 4.18(b). We now combine
the 3- and 6- resistors in parallel to get 2-. We also combine the
2-A and 4-A current sources to get a 2-A source. Thus, by repeatedly
applying source transformations, we obtain the circuit in Fig. 4.18(c).
4 Ω 2 Ω
4 A
8 Ω 3 Ω
12 V +
−
(a)
+
−
vo
4 A
8 Ω
6 Ω 3 Ω
2 A
(b)
2 A
8 Ω 2 Ω
(c)
i
+
−
vo
+
−
vo
Figure4.18 For Example 4.6.
We use current division in Fig. 4.18(c) to get
i =
2
2 + 8
(2) = 0.4
and
vo = 8i = 8(0.4) = 3.2 V
130 PART 1 DC Circuits
Alternatively, since the 8- and 2- resistors in Fig. 4.18(c) are in
parallel, they have the same voltage vo across them. Hence,
vo = (8 2)(2 A) =
8 × 2
10
(2) = 3.2 V
P R A C T I C E P R O B L E M 4 . 6
Find io in the circuit of Fig. 4.19 using source transformation.
4 Ω
5 A
5 V
7 Ω 3 A
3 Ω
1 Ω
6 Ω
− +
io
Figure4.19 For Practice Prob. 4.6.
Answer: 1.78 A.
E X A M P L E 4 . 7
Find vx in Fig. 4.20 using source transformation.
4 Ω
2 Ω
0.25vx
2 Ω
6 V 18 V
+
−
+
−
vx
+
−
Figure4.20 For Example 4.7.
Solution:
The circuit in Fig. 4.20 involves a voltage-controlled dependent current
source. We transform this dependent current source as well as the 6-V
independent voltage source as shown in Fig. 4.21(a). The 18-V volt-
age source is not transformed because it is not connected in series with
any resistor. The two 2- resistors in parallel combine to give a 1-
resistor, which is in parallel with the 3-A current source. The current is
transformed to a voltage source as shown in Fig. 4.21(b). Notice that the
terminals for vx are intact. Applying KVL around the loop in Fig. 4.21(b)
gives
−3 + 5i + vx + 18 = 0 (4.7.1)
18 V
3 A
4 Ω
2 Ω
2 Ω
+ −
+
−
(a)
18 V
3 V
4 Ω
1 Ω
vx
vx
+
−
+ −
+
−
+
−
(b)
i
+
−
vx
Figure4.21 For Example 4.7: Applying source transformation to the circuit in Fig. 4.20.
CHAPTER 4 Circuit Theorems 131
Applying KVL to the loop containing only the 3-V voltage source, the
1- resistor, and vx yields
−3 + 1i + vx = 0 ⇒ vx = 3 − i (4.7.2)
Substituting this into Eq. (4.7.1), we obtain
15 + 5i + 3 − i = 0 ⇒ i = −4.5 A
Alternatively, we may apply KVL to the loop containing vx, the 4-
resistor, the voltage-controlled dependent voltage source, and the 18-V
voltage source in Fig. 4.21(b). We obtain
−vx + 4i + vx + 18 = 0 ⇒ i = −4.5 A
Thus, vx = 3 − i = 7.5 V.
P R A C T I C E P R O B L E M 4 . 7
Use source transformation to find ix in the circuit shown in Fig. 4.22.
2ix
5 Ω
4 A 10 Ω
–
+
ix
Figure4.22 For Practice Prob. 4.7.
Answer: 1.176 A.
4.5 THEVENIN’S THEOREM
It often occurs in practice that a particular element in a circuit is variable
(usually called the load) while other elements are fixed. As a typical
example, a household outlet terminal may be connected to different ap-
pliances constituting a variable load. Each time the variable element is
changed, the entire circuit has to be analyzed all over again. To avoid this
problem, Thevenin’s theorem provides a technique by which the fixed
part of the circuit is replaced by an equivalent circuit.
According to Thevenin’s theorem, the linear circuit in Fig. 4.23(a)
can be replaced by that in Fig. 4.23(b). (The load in Fig. 4.23 may be a
single resistor or another circuit.) The circuit to the left of the terminals
a-b in Fig. 4.23(b) is known as the Thevenin equivalent circuit; it was
developed in 1883 by M. Leon Thevenin (1857–1926), a French telegraph
engineer.
Linear
two-terminal
circuit
Load
I a
b
V
+
−
(a)
Load
I a
b
V
+
−
(b)
+
−
VTh
RTh
Figure4.23 Replacing a linear two-terminal
circuit by its Thevenin equivalent: (a) original
circuit, (b) the Thevenin equivalent circuit.
Thevenin’s theorem states that a linear two-terminal circuit can be replaced
by an equivalent circuit consisting of a voltage source VTh in series with
a resistor RTh, where VTh is the open-circuit voltage at the terminals
and RTh is the input or equivalent resistance at the terminals when
the independent sources are turned off.
The proof of the theorem will be given later, in Section 4.7. Our
major concern right now is how to find the Thevenin equivalent voltage
132 PART 1 DC Circuits
VTh and resistance RTh. To do so, suppose the two circuits in Fig. 4.23
are equivalent. Two circuits are said to be equivalent if they have the
same voltage-current relation at their terminals. Let us find out what will
make the two circuits in Fig. 4.23 equivalent. If the terminals a-b are
made open-circuited (by removing the load), no current flows, so that
the open-circuit voltage across the terminals a-b in Fig. 4.23(a) must be
equal to the voltage source VTh in Fig. 4.23(b), since the two circuits are
equivalent. Thus VTh is the open-circuit voltage across the terminals as
shown in Fig. 4.24(a); that is,
VTh = voc (4.6)
Linear
two-terminal
circuit
a
b
voc
+
−
(a)
VTh = voc
Linear circuit with
all independent
sources set equal
to zero
a
b
Rin
(b)
RTh = Rin
Figure4.24 Finding VTh and RTh.
Again, withtheloaddisconnectedandterminalsa-b open-circuited,
we turn off all independent sources. The input resistance (or equivalent
resistance) of the dead circuit at the terminals a-b in Fig. 4.23(a) must
be equal to RTh in Fig. 4.23(b) because the two circuits are equivalent.
Thus, RTh is the input resistance at the terminals when the independent
sources are turned off, as shown in Fig. 4.24(b); that is,
RTh = Rin (4.7)
To apply this idea in finding the Thevenin resistance RTh, we need
to consider two cases.
CASE 1 If the network has no dependent sources, we turn off all in-
dependent sources. RTh is the input resistance of the network looking
between terminals a and b, as shown in Fig. 4.24(b).
CASE 2 If the network has dependent sources, we turn off all inde-
pendent sources. As with superposition, dependent sources are not to be
turned off because they are controlled by circuit variables. We apply a
voltage source vo at terminals a and b and determine the resulting cur-
rent io. Then RTh = vo/io, as shown in Fig. 4.25(a). Alternatively, we
may insert a current source io at terminals a-b as shown in Fig. 4.25(b)
and find the terminal voltage vo. Again RTh = vo/io. Either of the two
approaches will give the same result. In either approach we may assume
any value of vo and io. For example, we may use vo = 1 V or io = 1 A,
or even use unspecified values of vo or io.
vo
Circuit with
all independent
sources set equal
to zero
a
b
(a)
RTh =
+
−
vo
io
io
io
vo
Circuit with
all independent
sources set equal
to zero
a
b
(b)
RTh =
vo
io
+
−
Figure 4.25 Finding RTh when circuit
has dependent sources.
Laterwewillseethatanalternativewayoffinding
RTh is RTh = voc/isc.
It often occurs that RTh takes a negative value. In this case, the
negative resistance (v = −iR) implies that the circuit is supplying power.
CHAPTER 4 Circuit Theorems 133
This is possible in a circuit with dependent sources; Example 4.10 will
illustrate this.
Thevenin’s theorem is very important in circuit analysis. It helps
simplify a circuit. A large circuit may be replaced by a single independent
voltage source and a single resistor. This replacement technique is a
powerful tool in circuit design.
As mentioned earlier, a linear circuit with a variable load can be re-
placed by the Thevenin equivalent, exclusive of the load. The equivalent
network behaves the same way externally as the original circuit. Con-
sider a linear circuit terminated by a load RL, as shown in Fig. 4.26(a).
The current IL through the load and the voltage VL across the load are
easily determined once the Thevenin equivalent of the circuit at the load’s
terminals is obtained, as shown in Fig. 4.26(b). From Fig. 4.26(b), we
obtain
IL =
VTh
RTh + RL
(4.8a)
VL = RLIL =
RL
RTh + RL
VTh (4.8b)
Note from Fig. 4.26(b) that the Thevenin equivalent is a simple voltage
divider, yielding VL by mere inspection.
Linear
circuit
a
b
(a)
RL
IL
a
b
(b)
RL
IL
+
−
VTh
RTh
Figure4.26 A circuit with a load:
(a) original circuit, (b) Thevenin
equivalent.
E X A M P L E 4 . 8
Find the Thevenin equivalent circuit of the circuit shown in Fig. 4.27, to
the left of the terminals a-b. Then find the current through RL = 6, 16,
and 36 .
RL
32 V 2 A
4 Ω 1 Ω
12 Ω
+
−
a
b
Figure4.27 For Example 4.8.
Solution:
We find RTh by turning off the 32-V voltage source (replacing it with
a short circuit) and the 2-A current source (replacing it with an open
circuit). The circuit becomes what is shown in Fig. 4.28(a). Thus,
RTh = 4 12 + 1 =
4 × 12
16
+ 1 = 4 
32 V 2 A
4 Ω 1 Ω
12 Ω
+
− VTh
VTh
+
−
(b)
4 Ω 1 Ω
12 Ω
(a)
RTh
i1 i2
Figure4.28 For Example 4.8: (a) finding RTh, (b) finding VTh.
To find VTh, consider the circuit in Fig. 4.28(b). Applying mesh
analysis to the two loops, we obtain
−32 + 4i1 + 12(i1 − i2) = 0, i2 = −2 A
134 PART 1 DC Circuits
Solving for i1, we get i1 = 0.5 A. Thus,
VTh = 12(i1 − i2) = 12(0.5 + 2.0) = 30 V
Alternatively, it is even easier to use nodal analysis. We ignore the 1-
resistor since no current flows through it. At the top node, KCL gives
32 − VTh
4
+ 2 =
VTh
12
or
96 − 3VTh + 24 = VTh ⇒ VTh = 30 V
as obtained before. We could also use source transformation to find VTh.
The Thevenin equivalent circuit is shown in Fig. 4.29. The current
through RL is
IL =
VTh
RTh + RL
=
30
4 + RL
When RL = 6,
IL =
30
10
= 3 A
When RL = 16,
IL =
30
20
= 1.5 A
When RL = 36,
IL =
30
40
= 0.75 A
RL
30 V
4 Ω
+
−
a
b
iL
Figure4.29 The Thevenin
equivalent circuit for Example 4.8.
P R A C T I C E P R O B L E M 4 . 8
Using Thevenin’s theorem, find the equivalent circuit to the left of the
terminals in the circuit in Fig. 4.30. Then find i.
12 V 2 A
6 Ω 6 Ω
4 Ω 1 Ω
+
−
a
b
i
Figure4.30 For Practice Prob. 4.8.
Answer: VTh = 6 V, RTh = 3 , i = 1.5 A.
E X A M P L E 4 . 9
Find the Thevenin equivalent of the circuit in Fig. 4.31.
CHAPTER 4 Circuit Theorems 135
Solution:
This circuit contains a dependent source, unlike the circuit in the previ-
ous example. To find RTh, we set the independent source equal to zero
but leave the dependent source alone. Because of the presence of the
dependent source, however, we excite the network with a voltage source
vo connected to the terminals as indicated in Fig. 4.32(a). We may set
vo = 1 V to ease calculation, since the circuit is linear. Our goal is to find
the current io through the terminals, and then obtain RTh = 1/io. (Al-
ternatively, we may insert a 1-A current source, find the corresponding
voltage vo, and obtain RTh = vo/1.)
5 A
2 Ω
2vx
2 Ω
6 Ω
4 Ω
a
b
− +
+
−
vx
Figure4.31 For Example 4.9.
2 Ω
2vx
2 Ω
6 Ω
4 Ω
a
b
− +
+
− vo = 1 V
io
(a)
i1
i2
(b)
5 A
2 Ω
2vx
2 Ω
6 Ω
4 Ω
a
b
− +
voc
+
−
i3
i1 i2
i3
+
−
vx
+
−
vx
Figure4.32 Finding RTh and VTh for Example 4.9.
Applying mesh analysis to loop 1 in the circuit in Fig. 4.32(a) results
in
−2vx + 2(i1 − i2) = 0 or vx = i1 − i2
But −4i2 = vx = i1 − i2; hence,
i1 = −3i2 (4.9.1)
For loops 2 and 3, applying KVL produces
4i2 + 2(i2 − i1) + 6(i2 − i3) = 0 (4.9.2)
6(i3 − i2) + 2i3 + 1 = 0 (4.9.3)
Solving these equations gives
i3 = −
1
6
A
But io = −i3 = 1/6 A. Hence,
RTh =
1 V
io
= 6 
To get VTh, we find voc in the circuit of Fig. 4.32(b). Applying mesh
analysis, we get
136 PART 1 DC Circuits
i1 = 5 (4.9.4)
− 2vx + 2(i3 − i2) = 0 ⇒ vx = i3 − i2 (4.9.5)
4(i2 − i1) + 2(i2 − i3) + 6i2 = 0
or
12i2 − 4i1 − 2i3 = 0 (4.9.6)
But 4(i1 − i2) = vx. Solving these equations leads to i2 = 10/3. Hence,
VTh = voc = 6i2 = 20 V
The Thevenin equivalent is as shown in Fig. 4.33.
20 V
6 Ω
a
b
+
−
Figure4.33 The Thevenin
equivalent of the circuit in
Fig. 4.31.
P R A C T I C E P R O B L E M 4 . 9
Find the Thevenin equivalent circuit of the circuit in Fig. 4.34 to the left
of the terminals.
6 V
3 Ω
5 Ω
4 Ω
a
b
1.5Ix
+
−
Ix
Figure4.34 For Practice Prob. 4.9.
Answer: VTh = 5.33 V, RTh = 0.44 .
E X A M P L E 4 . 1 0
Determine the Thevenin equivalent of the circuit in Fig. 4.35(a).
2ix 4 Ω 2 Ω
a
b
ix
vo
(a)
2ix io
4 Ω 2 Ω
a
b
ix
(b)
Figure4.35 For Example 4.10.
Solution:
Since the circuit in Fig. 4.35(a) has no independent sources, VTh = 0 V.
To find RTh, it is best to apply a current source io at the terminals as shown
in Fig. 4.35(b). Applying nodal analysis gives
io + ix = 2ix +
vo
4
(4.10.1)
But
ix =
0 − vo
2
= −
vo
2
(4.10.2)
Substituting Eq. (4.10.2) into Eq. (4.10.1) yields
io = ix +
vo
4
= −
vo
2
+
vo
4
= −
vo
4
or vo = −4io
Thus,
RTh =
vo
io
= −4 
The negative value of the resistance tells us that, according to the passive
sign convention, the circuit in Fig. 4.35(a) is supplying power. Of course,
the resistors in Fig. 4.35(a) cannot supply power (they absorb power); it
CHAPTER 4 Circuit Theorems 137
is the dependent source that supplies the power. This is an example of
how a dependent source and resistors could be used to simulate negative
resistance.
P R A C T I C E P R O B L E M 4 . 1 0
Obtain the Thevenin equivalent of the circuit in Fig. 4.36.
5 Ω 15 Ω
a
b
10 Ω
4vx
+ −
+
−
vx
Figure 4.36 For Practice Prob. 4.10.
Answer: VTh = 0 V, RTh = −7.5 .
4.6 NORTON’S THEOREM
In 1926, about 43 years after Thevenin published his theorem, E. L.
Norton, an American engineer at Bell Telephone Laboratories, proposed
a similar theorem.
Norton’s theorem states that a linear two-terminal circuit can be replaced
by an equivalent circuit consisting of a current source IN in parallel with
a resistor RN, where IN is the short-circuit current through the terminals
and RN is the input or equivalent resistance at the terminals when the
independent sources are turned off.
Thus, the circuit in Fig. 4.37(a) can be replaced by the one in Fig. 4.37(b).
Linear
two-terminal
circuit
a
b
(a)
(b)
RN
a
b
IN
Figure4.37 (a) Original circuit,
(b) Norton equivalent circuit.
The proof of Norton’s theorem will be given in the next section. For
now, we are mainly concerned with how to get RN and IN . We find RN
in the same way we find RTh. In fact, from what we know about source
transformation, the Thevenin and Norton resistances are equal; that is,
RN = RTh (4.9)
TofindtheNortoncurrentIN , wedeterminetheshort-circuitcurrent
flowing from terminal a to b in both circuits in Fig. 4.37. It is evident
that the short-circuit current in Fig. 4.37(b) is IN . This must be the same
short-circuit current from terminal a to b in Fig. 4.37(a), since the two
circuits are equivalent. Thus,
IN = isc (4.10)
shown in Fig. 4.38. Dependent and independent sources are treated the
same way as in Thevenin’s theorem. Linear
two-terminal
circuit
a
b
isc = IN
Figure 4.38 Finding Norton
current IN .
Observe the close relationship between Norton’s and Thevenin’s
theorems: RN = RTh as in Eq. (4.9), and
IN =
VTh
RTh
(4.11)
138 PART 1 DC Circuits
This is essentially source transformation. For this reason, source trans-
formation is often called Thevenin-Norton transformation.
TheTheveninandNortonequivalentcircuitsare
related by a source transformation.
Since VTh, IN , and RTh are related according to Eq. (4.11), to de-
termine the Thevenin or Norton equivalent circuit requires that we find:
• The open-circuit voltage voc across terminals a and b.
• The short-circuit current isc at terminals a and b.
• The equivalent or input resistance Rin at terminals a and b when
all independent sources are turned off.
We can calculate any two of the three using the method that takes the
least effort and use them to get the third using Ohm’s law. Example 4.11
will illustrate this. Also, since
VTh = voc (4.12a)
IN = isc (4.12b)
RTh =
voc
isc
= RN (4.12c)
the open-circuit and short-circuit tests are sufficient to find any Thevenin
or Norton equivalent.
E X A M P L E 4 . 1 1
Find the Norton equivalent circuit of the circuit in Fig. 4.39.
2 A
8 Ω
8 Ω
5 Ω
4 Ω
12 V
a
b
+
−
Figure4.39 For Example 4.11.
Solution:
We find RN in the same way we find RTh in the Thevenin equivalent cir-
cuit. Set the independent sources equal to zero. This leads to the circuit
in Fig. 4.40(a), from which we find RN . Thus,
RN = 5 (8 + 4 + 8) = 5 20 =
20 × 5
25
= 4 
To find IN , we short-circuit terminals a and b, as shown in Fig. 4.40(b).
We ignore the 5- resistor because it has been short-circuited. Applying
mesh analysis, we obtain
i1 = 2 A, 20i2 − 4i1 − 12 = 0
From these equations, we obtain
i2 = 1 A = isc = IN
Alternatively, we may determine IN from VTh/RTh. We obtain VTh
as the open-circuit voltage across terminals a and b in Fig. 4.40(c). Using
mesh analysis, we obtain
i3 = 2 A
25i4 − 4i3 − 12 = 0 ⇒ i4 = 0.8 A
and
voc = VTh = 5i4 = 4 V
CHAPTER 4 Circuit Theorems 139
2 A
5 Ω
4 Ω
12 V
a
b
+
−
isc = IN
(b)
2 A 5 Ω
4 Ω
12 V
a
b
+
−
(c)
8 Ω
5 Ω
a
b
4 Ω
(a)
RN
VTh = voc
+
−
i1
i3
i4
i2
8 Ω 8 Ω
8 Ω
8 Ω
8 Ω
Figure4.40 For Example 4.11; finding: (a) RN , (b) IN = isc, (c) VTh = voc.
Hence,
IN =
VTh
RTh
=
4
4
= 1 A
as obtained previously. This also serves to confirm Eq. (4.7) that RTh =
voc/isc = 4/1 = 4 . Thus, the Norton equivalent circuit is as shown in
Fig. 4.41.
1 A 4 Ω
a
b
Figure4.41 Norton equiva-
lent of the circuit in Fig. 4.39.
P R A C T I C E P R O B L E M 4 . 1 1
Find the Norton equivalent circuit for the circuit in Fig. 4.42.
4 A
15 V 6 Ω
a
b
3 Ω
+
−
3 Ω
Figure4.42 For Practice Prob. 4.11.
Answer: RN = 3 , IN = 4.5 A.
E X A M P L E 4 . 1 2
Using Norton’s theorem, find RN and IN of the circuit in Fig. 4.43 at ter-
minals a-b.
Solution:
To find RN , we set the independent voltage source equal to zero and con-
nect a voltage source of vo = 1 V (or any unspecified voltage vo) to the
140 PART 1 DC Circuits
terminals. WeobtainthecircuitinFig.4.44(a). Weignorethe4-resistor
because it is short-circuited. Also due to the short circuit, the 5- resistor,
the voltage source, and the dependent current source are all in parallel.
Hence, ix = vo/5 = 1/5 = 0.2. At node a, −io = ix + 2ix = 3ix = 0.6,
and
RN =
vo
io
=
1
−0.6
= −1.67 
5 Ω
2Ix
10 V
4 Ω
a
b
+
−
ix
Figure4.43 For Example 4.12.
TofindIN , weshort-circuitterminalsa andb andfindthecurrentisc,
as indicated in Fig. 4.44(b). Note from this figure that the 4- resistor, the
10-V voltage source, the 5- resistor, and the dependent current source
are all in parallel. Hence,
ix =
10 − 0
5
= 2 A
At node a, KCL gives
isc = ix + 2ix = 2 + 4 = 6 A
Thus,
IN = 6 A
5 Ω
2ix
vo = 1 V
io
4 Ω
a
b
+
−
(a)
5 Ω
2ix
isc = IN
4 Ω
a
b
(b)
10 V
+
−
ix ix
Figure4.44 For Example 4.12: (a) finding RN , (b) finding IN .
P R A C T I C E P R O B L E M 4 . 1 2
Find the Norton equivalent circuit of the circuit in Fig. 4.45.
10 A
2vx
6 Ω 2 Ω
a
b
−
+
+
−
vx
Figure4.45 For Practice Prob. 4.12.
Answer: RN = 1 , IN = 10 A.
†4.7 DERIVATIONS OF THEVENIN’S AND NORTON’S
THEOREMS
In this section, we will prove Thevenin’s and Norton’s theorems using
the superposition principle.
CHAPTER 4 Circuit Theorems 141
Consider the linear circuit in Fig. 4.46(a). It is assumed that the cir-
cuit contains resistors, and dependent and independent sources. We have
access to the circuit via terminals a and b, through which current from
an external source is applied. Our objective is to ensure that the voltage-
current relation at terminals a and b is identical to that of the Thevenin
equivalent in Fig. 4.46(b). For the sake of simplicity, suppose the linear
circuit in Fig. 4.46(a) contains two independent voltage sources vs1 and
vs2 and two independent current sources is1 and is2. We may obtain any
circuit variable, such as the terminal voltage v, by applying superposition.
That is, we consider the contribution due to each independent source in-
cluding the external source i. By superposition, the terminal voltage v
is
i
Linear
circuit
a
b
(a)
i
a
b
(b)
v
+
−
v
+
−
VTh
+
−
RTh
Figure4.46 Derivation of
Thevenin equivalent: (a) a
current-driven circuit, (b) its
Thevenin equivalent.
v = A0i + A1vs1 + A2vs2 + A3is1 + A4is2 (4.13)
where A0, A1, A2, A3, and A4 are constants. Each term on the right-hand
side of Eq. (4.13) is the contribution of the related independent source;
that is, A0i is the contribution to v due to the external current source i,
A1vs1 is the contribution due to the voltage source vs1, and so on. We
may collect terms for the internal independent sources together as B0, so
that Eq. (4.13) becomes
v = A0i + B0 (4.14)
where B0 = A1vs1 + A2vs2 + A3is1 + A4is2. We now want to evaluate
the values of constants A0 and B0. When the terminals a and b are open-
circuited, i = 0 and v = B0. Thus B0 is the open-circuit voltage voc,
which is the same as VTh, so
B0 = VTh (4.15)
When all the internal sources are turned off, B0 = 0. The circuit can then
be replaced by an equivalent resistance Req, which is the same as RTh,
and Eq. (4.14) becomes
v = A0i = RThi ⇒ A0 = RTh (4.16)
Substituting the values of A0 and B0 in Eq. (4.14) gives
v = RThi + VTh (4.17)
which expresses the voltage-current relation at terminals a and b of the
circuit in Fig. 4.46(b). Thus, the two circuits in Fig. 4.46(a) and 4.46(b)
are equivalent.
When the same linear circuit is driven by a voltage source v as
shown in Fig. 4.47(a), the current flowing into the circuit can be obtained
by superposition as
v
Linear
circuit
a
b
(a)
v
a
b
(b)
IN
RN
+
−
+
−
i
i
Figure4.47 Derivation of Norton
equivalent: (a) a voltage-driven
circuit, (b) its Norton equivalent.
i = C0v + D0 (4.18)
where C0v is the contribution to i due to the external voltage source v and
D0 contains the contributions to i due to all internal independent sources.
When the terminals a-b are short-circuited, v = 0 so that i = D0 = −isc,
where isc is the short-circuit current flowing out of terminal a, which is
the same as the Norton current IN , i.e.,
D0 = −IN (4.19)
142 PART 1 DC Circuits
When all the internal independent sources are turned off, D0 = 0 and the
circuit can be replaced by an equivalent resistance Req (or an equivalent
conductance Geq = 1/Req), which is the same as RTh or RN . Thus Eq.
(4.19) becomes
i =
v
RTh
− IN (4.20)
This expresses the voltage-current relation at terminals a-b of the circuit
in Fig. 4.47(b), confirming that the two circuits in Fig. 4.47(a) and 4.47(b)
are equivalent.
4.8 MAXIMUM POWER TRANSFER
In many practical situations, a circuit is designed to provide power to a
load. While for electric utilities, minimizing power losses in the process
of transmission and distribution is critical for efficiency and economic
reasons, there are other applications in areas such as communications
where it is desirable to maximize the power delivered to a load. We now
address the problem of delivering the maximum power to a load when
given a system with known internal losses. It should be noted that this
will result in significant internal losses greater than or equal to the power
delivered to the load.
The Thevenin equivalent is useful in finding the maximum power a
linear circuit can deliver to a load. We assume that we can adjust the load
resistance RL. If the entire circuit is replaced by its Thevenin equivalent
except for the load, as shown in Fig. 4.48, the power delivered to the load
is
p = i2
RL =

VTh
RTh + RL
2
RL (4.21)
For a given circuit, VTh and RTh are fixed. By varying the load resistance
RL, the power delivered to the load varies as sketched in Fig. 4.49. We
notice from Fig. 4.49 that the power is small for small or large values of
RL but maximum for some value of RL between 0 and ∞. We now want
to show that this maximum power occurs when RL is equal to RTh. This
is known as the maximum power theorem.
RL
VTh
RTh
+
−
a
b
i
Figure4.48 The circuit used for
maximum power transfer.
p
RL
RTh
0
pmax
Figure 4.49 Power delivered to the load
as a function of RL.
Maximum power is transferred to the load when the load resistance equals the
Thevenin resistance as seen from the load (RL = RTh).
To prove the maximum power transfer theorem, we differentiate
p in Eq. (4.21) with respect to RL and set the result equal to zero. We
obtain
dp
dRL
= V 2
Th

(RTh + RL)2
− 2RL(RTh + RL)
(RTh + RL)4

= V 2
Th

(RTh + RL − 2RL)
(RTh + RL)3

= 0
CHAPTER 4 Circuit Theorems 143
This implies that
0 = (RTh + RL − 2RL) = (RTh − RL) (4.22)
which yields
RL = RTh (4.23)
showing that the maximum power transfer takes place when the load
resistance RL equals the Thevenin resistance RTh. We can readily confirm
that Eq. (4.23) gives the maximum power by showing that d2
p/dR2
L  0.
Thesourceandloadaresaidtobematched when
RL = RTh.
The maximum power transferred is obtained by substituting Eq.
(4.23) into Eq. (4.21), for
pmax =
V 2
Th
4RTh
(4.24)
Equation (4.24) applies only when RL = RTh. When RL = RTh, we
compute the power delivered to the load using Eq. (4.21).
E X A M P L E 4 . 1 3
Find the value of RL for maximum power transfer in the circuit of Fig.
4.50. Find the maximum power.
12 V 2 A
6 Ω 3 Ω 2 Ω
12 Ω RL
+
−
a
b
Figure4.50 For Example 4.13.
Solution:
We need to find the Thevenin resistance RTh and the Thevenin voltage
VTh across the terminals a-b. To get RTh, we use the circuit in Fig. 4.51(a)
and obtain
RTh = 2 + 3 + 6 12 = 5 +
6 × 12
18
= 9 
6 Ω 3 Ω 2 Ω
12 Ω
RTh
12 V 2 A
6 Ω 3 Ω 2 Ω
12 Ω
+
− VTh
+
−
(a) (b)
i1 i2
Figure4.51 For Example 4.13: (a) finding RTh, (b) finding VTh.
144 PART 1 DC Circuits
To get VTh, we consider the circuit in Fig. 4.51(b). Applying mesh anal-
ysis,
−12 + 18i1 − 12i2 = 0, i2 = −2 A
Solving for i1, we get i1 = −2/3. Applying KVL around the outer loop
to get VTh across terminals a-b, we obtain
−12 + 6i1 + 3i2 + 2(0) + VTh = 0 ⇒ VTh = 22 V
For maximum power transfer,
RL = RTh = 9 
and the maximum power is
pmax =
V 2
Th
4RL
=
222
4 × 9
= 13.44 W
P R A C T I C E P R O B L E M 4 . 1 3
Determine the value of RL that will draw the maximum power from the
rest of the circuit in Fig. 4.52. Calculate the maximum power.
9 V
4 Ω
2 Ω
RL
1 Ω
3vx
+
−
+
−
+ −
vx
Figure4.52 For Practice Prob. 4.13.
Answer: 4.22 , 2.901 W.
4.9 VERIFYING CIRCUIT THEOREMS WITH PSPICE
In this section, we learn how to use PSpice to verify the theorems covered
in this chapter. Specifically, we will consider using dc sweep analysis to
find the Thevenin or Norton equivalent at any pair of nodes in a circuit
and the maximum power transfer to a load. The reader is advised to read
Section D.3 of Appendix D in preparation for this section.
To find the Thevenin equivalent of a circuit at a pair of open ter-
minals using PSpice, we use the schematic editor to draw the circuit and
insert an independent probing current source, say, Ip, at the terminals.
The probing current source must have a part name ISRC. We then per-
form a DC Sweep on Ip, as discussed in Section D.3. Typically, we may
let the current through Ip vary from 0 to 1 A in 0.1-A increments. After
simulating the circuit, we use Probe to display a plot of the voltage across
Ip versus the current through Ip. The zero intercept of the plot gives us
the Thevenin equivalent voltage, while the slope of the plot is equal to
the Thevenin resistance.
To find the Norton equivalent involves similar steps except that we
insert a probing independent voltage source (with a part name VSRC),
say, Vp, at the terminals. We perform a DC Sweep on Vp and let Vp
vary from 0 to 1 V in 0.1-V increments. A plot of the current through
CHAPTER 4 Circuit Theorems 145
Vp versus the voltage across Vp is obtained using the Probe menu after
simulation. The zero intercept is equal to the Norton current, while the
slope of the plot is equal to the Norton conductance.
To find the maximum power transfer to a load using PSpice involves
performing a dc parametric sweep on the component value of RL in Fig.
4.48 and plotting the power delivered to the load as a function of RL.
According to Fig. 4.49, the maximum power occurs when RL = RTh.
This is best illustrated with an example, and Example 4.15 provides one.
We use VSRC and ISRC as part names for the independent voltage
and current sources.
E X A M P L E 4 . 1 4
Consider the circuit is in Fig. 4.31 (see Example 4.9). Use PSpice to find
the Thevenin and Norton equivalent circuits.
Solution:
(a) To find the Thevenin resistance RTh and Thevenin voltage VTh at the
terminals a-b in the circuit in Fig. 4.31, we first use Schematics to draw
the circuit as shown in Fig. 4.53(a). Notice that a probing current source
I2 is inserted at the terminals. Under Analysis/Setput, we select DC
Sweep. In the DC Sweep dialog box, we select Linear for the Sweep
Type and Current Source for the Sweep Var. Type. We enter I2 under the
Name box, 0 as Start Value, 1 as End Value, and 0.1 as Increment. After
simulation, we add trace V(I2:−) from the Probe menu and obtain the
plot shown in Fig. 4.53(b). From the plot, we obtain
VTh = Zero intercept = 20 V, RTh = Slope =
26 − 20
1
= 6 
These agree with what we got analytically in Example 4.9.
R2 R4
2 2
GAIN=2
E1
R4 4 R3 6 I2
I1
0
+
−
(a) (b)
26 V
24 V
22 V
20 V
0 A 0.2 A 0.4 A 0.6 A 0.8 A 1.0 A
= V(I2:-)
+
−
Figure4.53 For Example 4.14: (a) schematic and (b) plot for finding RTh and VTh.
(b) To find the Norton equivalent, we modify the schematic in Fig. 4.53(a)
by replaying the probing current source with a probing voltage source V1.
The result is the schematic in Fig. 4.54(a). Again, in the DC Sweep dialog
box, we select Linear for the Sweep Type and Voltage Source for the Sweep
Var. Type. We enter V1 under Name box, 0 as Start Value, 1 as End Value,
146 PART 1 DC Circuits
and 0.1 as Increment. When the Probe is running, we add trace I(V1) and
obtain the plot in Fig. 4.54(b). From the plot, we obtain
IN = Zero intercept = 3.335 A
GN = Slope =
3.335 − 3.165
1
= 0.17 S
R2 R1
2 2
GAIN=2
E1
R4 4 R3 6 V1
I1
0
+
−
(a) (b)
3.4 A
3.3 A
3.2 A
3.1 A
0 V 0.2 V 0.4 V 0.6 V 0.8 V 1.0 V
I(V1) V_V1
+
−
+
−
Figure4.54 For Example 4.14: (a) schematic and (b) plot for finding GN and IN .
P R A C T I C E P R O B L E M 4 . 1 4
Rework Practice Prob. 4.9 using PSpice.
Answer: VTh = 5.33 V, RTh = 0.44 .
E X A M P L E 4 . 1 5
Refer to the circuit in Fig. 4.55. Use PSpice to find the maximum power
transfer to RL.
RL
1 V
1 kΩ
+
−
Figure 4.55 For Example 4.15.
Solution:
We need to perform a dc sweep on RL to determine when the power across
it is maximum. We first draw the circuit using Schematics as shown in
Fig. 4.56. Once the circuit is drawn, we take the following three steps to
further prepare the circuit for a dc sweep.
{RL}
DC=1 V +
−
0
R1
R2
1k
V1
PARAMETERS:
RL 2k
Figure 4.56 Schematic for the circuit in
Fig. 4.55.
The first step involves defining the value of RL as a parameter, since
we want to vary it. To do this:
1. DCLICKL the value 1k of R2 (representing RL) to open up
the Set Attribute Value dialog box.
2. Replace 1k with {RL} and click OK to accept the change.
Note that the curly brackets are necessary.
The second step is to define parameter. To achieve this:
1. Select Draw/Get New Part/Libraries · · ·/special.slb.
2. Type PARAM in the PartName box and click OK.
3. DRAG the box to any position near the circuit.
4. CLICKL to end placement mode.
CHAPTER 4 Circuit Theorems 147
5. DCLICKL to open up the PartName: PARAM dialog box.
6. CLICKL on NAME1 = and enter RL (with no curly brackets)
in the Value box, and CLICKL Save Attr to accept change.
7. CLICKL on VALUE1 = and enter 2k in the Value box, and
CLICKL Save Attr to accept change.
8. Click OK.
The value 2k in item 7 is necessary for a bias point calculation; it
cannot be left blank.
The third step is to set up the DC Sweep to sweep the parameter.
To do this:
1. Select Analysis/Setput to bring up the DC Sweep dialog box.
2. For the Sweep Type, select Linear (or Octave for a wide range
of RL).
3. For the Sweep Var. Type, select Global Parameter.
4. Under the Name box, enter RL.
5. In the Start Value box, enter 100.
6. In the End Value box, enter 5k.
7. In the Increment box, enter 100.
8. Click OK and Close to accept the parameters.
250 uW
150 uW
200 uW
100 uW
50 uW
0 2.0 K 4.0 K 6.0 K
-V(R2:2)*I(R2)
RL
Figure 4.57 For Example 4.15: the plot
of power across PL.
After taking these steps and saving the circuit, we are ready to sim-
ulate. Select Analysis/Simulate. If there are no errors, we select Add
TraceintheProbemenuandtype−V(R2:2)∗
I(R2)intheTraceCommand
box. [The negative sign is needed since I(R2) is negative.] This gives the
plot of the power delivered to RL as RL varies from 100  to 5 k. We
can also obtain the power absorbed by RL by typing V(R2:2)∗
V(R2:2)/RL
in the Trace Command box. Either way, we obtain the plot in Fig. 4.57.
It is evident from the plot that the maximum power is 250 µW. Notice
that the maximum occurs when RL = 1 k, as expected analytically.
P R A C T I C E P R O B L E M 4 . 1 5
Find the maximum power transferred to RL if the 1-k resistor in Fig.
4.55 is replaced by a 2-k resistor.
Answer: 125 µW.
†4.10 APPLICATIONS
In this section we will discuss two important practical applications of
the concepts covered in this chapter: source modeling and resistance
measurement.
4.10.1 Source Modeling
Source modeling provides an example of the usefulness of the Thevenin
or the Norton equivalent. An active source such as a battery is often
characterized by its Thevenin or Norton equivalent circuit. An ideal
voltage source provides a constant voltage irrespective of the current
148 PART 1 DC Circuits
drawn by the load, while an ideal current source supplies a constant
current regardless of the load voltage. As Fig. 4.58 shows, practical
voltage and current sources are not ideal, due to their internal resistances
or source resistances Rs and Rp. They become ideal as Rs → 0 and
Rp → ∞. To show that this is the case, consider the effect of the load
on voltage sources, as shown in Fig. 4.59(a). By the voltage division
principle, the load voltage is
vL =
RL
Rs + RL
vs (4.25)
As RL increases, the load voltage approaches a source voltage vs, as
illustrated in Fig. 4.59(b). From Eq. (4.25), we should note that:
vs
Rs
+
−
(a)
is
Rp
(b)
Figure4.58 (a) Practical
voltage source, (b) practical
current source.
1. The load voltage will be constant if the internal resistance Rs
of the source is zero or, at least, Rs  RL. In other words, the
smaller Rs is compared to RL, the closer the voltage source is
to being ideal.
2. When the load is disconnected (i.e., the source is open-
circuited so that RL → ∞), voc = vs. Thus, vs may be
regarded as the unloaded source voltage. The connection of
the load causes the terminal voltage to drop in magnitude; this
is known as the loading effect.
RL
vs
Rs
+
− vL
+
−
(a) (b)
vL
RL
0
vs
Practical source
Ideal source
Figure 4.59 (a) Practical voltage source connected to a load RL,
(b) load voltage decreases as RL decreases.
The same argument can be made for a practical current source when
connected to a load as shown in Fig. 4.60(a). By the current division
principle,
iL =
Rp
Rp + RL
is (4.26)
Figure 4.60(b) shows the variation in the load current as the load re-
sistance increases. Again, we notice a drop in current due to the load
(loading effect), and load current is constant (ideal current source) when
the internal resistance is very large (i.e., Rp → ∞ or, at least, Rp  RL).
RL
(a)
is Rp
IL
(b)
IL
RL
0
is
Practical source
Ideal source
Figure4.60 (a) Practical current
source connected to a load RL,
(b) load current decreases as RL
increases.
Sometimes, we need to know the unloaded source voltage vs and
the internal resistance Rs of a voltage source. To find vs and Rs, we follow
the procedure illustrated in Fig. 4.61. First, we measure the open-circuit
voltage voc as in Fig. 4.61(a) and set
vs = voc (4.27)
CHAPTER 4 Circuit Theorems 149
Then, we connect a variable load RL across the terminals as in Fig.
4.61(b). We adjust the resistance RL until we measure a load voltage
of exactly one-half of the open-circuit voltage, vL = voc/2, because now
RL = RTh = Rs. At that point, we disconnect RL and measure it. We
set
Rs = RL (4.28)
For example, a car battery may have vs = 12 V and Rs = 0.05 .
Signal
source
(a)
voc
+
−
Signal
source
(b)
vL
+
−
RL
Figure4.61 (a) Measuring voc, (b) measuring vL.
E X A M P L E 4 . 1 6
The terminal voltage of a voltage source is 12 V when connected to a 2-W
load. When the load is disconnected, the terminal voltage rises to 12.4 V.
(a) Calculate the source voltage vs and internal resistance Rs. (b) Deter-
mine the voltage when an 8- load is connected to the source.
Rs
(a)
(b)
RL
vs
Rs iL
+
− vL
+
−
8 Ω
12 V −
+ v
+
−
2.4 Ω
Figure4.62 For Example 4.16.
Solution:
(a)WereplacethesourcebyitsTheveninequivalent. Theterminalvoltage
when the load is disconnected is the open-circuit voltage,
vs = voc = 12.4 V
When the load is connected, as shown in Fig. 4.62(a), vL = 12 V and
pL = 2 W. Hence,
pL =
vL2
RL
⇒ RL =
v2
L
pL
=
122
2
= 72 
The load current is
iL =
vL
RL
=
12
72
=
1
6
A
The voltage across Rs is the difference between the source voltage vs and
the load voltage vL, or
12.4 − 12 = 0.4 = RsiL, Rs =
0.4
IL
= 2.4 
(b) Now that we have the Thevenin equivalent of the source, we connect
the 8- load across the Thevenin equivalent as shown in Fig. 4.62(b).
Using voltage division, we obtain
v =
8
8 + 2.4
(12) = 9.231 V
150 PART 1 DC Circuits
P R A C T I C E P R O B L E M 4 . 1 6
The measured open-circuit voltage across a certain amplifier is 9 V. The
voltage drops to 8 V when a 20- loudspeaker is connected to the am-
plifier. Calculate the voltage when a 10- loudspeaker is used instead.
Answer: 7.2 V.
4.10.2 Resistance Measurement
Although the ohmmeter method provides the simplest way to measure re-
sistance, more accurate measurement may be obtained using the Wheat-
stone bridge. While ohmmeters are designed to measure resistance in
low, mid, or high range, a Wheatstone bridge is used to measure resis-
tance in the mid range, say, between 1  and 1 M. Very low values of
resistances are measured with a milliohmmeter, while very high values
are measured with a Megger tester.
Historical note: The bridge was invented by
Charles Wheatstone (1802–1875), a British
professor who also invented the telegraph, as
Samuel Morse did independently in the United
States.
v
R1 R3
R2 Rx
+
−
Galvanometer
v1
+
−
+
−
v2
Figure 4.63 The Wheatstone bridge; Rx is
the resistance to be measured.
The Wheatstone bridge (or resistance bridge) circuit is used in a
number of applications. Here we will use it to measure an unknown re-
sistance. The unknown resistance Rx is connected to the bridge as shown
in Fig. 4.63. The variable resistance is adjusted until no current flows
through the galvanometer, which is essentially a d’Arsonval movement
operating as a sensitive current-indicating device like an ammeter in the
microamp range. Under this condition v1 = v2, and the bridge is said
to be balanced. Since no current flows through the galvanometer, R1
and R2 behave as though they were in series; so do R3 and Rx. The fact
that no current flows through the galvanometer also implies that v1 = v2.
Applying the voltage division principle,
v1 =
R2
R1 + R2
v = v2 =
Rx
R3 + Rx
v (4.29)
Hence, no current flows through the galvanometer when
R2
R1 + R2
=
Rx
R3 + Rx
⇒ R2R3 = R1Rx
or
Rx =
R3
R1
R2 (4.30)
If R1 = R3, and R2 is adjusted until no current flows through the gal-
vanometer, then Rx = R2.
How do we find the current through the galvanometer when the
Wheatstone bridge is unbalanced? We find the Thevenin equivalent (VTh
and RTh) with respect to the galvanometer terminals. If Rm is the resis-
tance of the galvanometer, the current through it under the unbalanced
condition is
I =
VTh
RTh + Rm
(4.31)
Example 4.18 will illustrate this.
CHAPTER 4 Circuit Theorems 151
E X A M P L E 4 . 1 7
In Fig. 4.63, R1 = 500  and R3 = 200 . The bridge is balanced when
R2 is adjusted to be 125 . Determine the unknown resistance Rx.
Solution:
Using Eq. (4.30),
Rx =
R3
R1
R2 =
200
500
125 = 50 
P R A C T I C E P R O B L E M 4 . 1 7
A Wheatstone bridge has R1 = R3 = 1 k. R2 is adjusted until no cur-
rent flows through the galvanometer. At that point, R2 = 3.2 k. What
is the value of the unknown resistance?
Answer: 3.2 k.
E X A M P L E 4 . 1 8
The circuit in Fig. 4.64 represents an unbalanced bridge. If the galvano-
meter has a resistance of 40 , find the current through the galvanometer.
220 V
400 Ω
600 Ω
+
− G
3 kΩ
1 kΩ
40 Ω
a b
Figure4.64 Unbalanced bridge of Example 4.18.
Solution:
We first need to replace the circuit by its Thevenin equivalent at termi-
nals a and b. The Thevenin resistance is found using the circuit in Fig.
4.65(a). Notice that the 3-k and 1-k resistors are in parallel; so are the
400- and 600- resistors. The two parallel combinations form a series
combination with respect to terminals a and b. Hence,
RTh = 3000 1000 + 400 600
=
3000 × 1000
3000 + 1000
+
400 × 600
400 + 600
= 750 + 240 = 990 
To find the Thevenin voltage, we consider the circuit in Fig. 4.65(b).
Using the voltage division principle,
v1 =
1000
1000 + 3000
(220) = 55 V, v2 =
600
600 + 400
(220) = 132 V
Applying KVL around loop ab gives
−v1 + VTh + v2 = 0 or VTh = v1 − v2 = 55 − 132 = −77 V
152 PART 1 DC Circuits
220 V
400 Ω
600 Ω
+
−
3 kΩ
1 kΩ
a b
+ −
VTh
(b)
VTh
40 Ω
+
−
(c)
400 Ω
600 Ω
3 kΩ
1 kΩ
a b
RTh
(a)
RTh a
b
G
IG
+
−
v1
+
−
v2
Figure 4.65 For Example 4.18: (a) Finding RTh, (b) finding VTh, (c) determining the current through the
galvanometer.
Having determined the Thevenin equivalent, we find the current through
the galvanometer using Fig. 4.65(c).
IG =
VTh
RTh + Rm
=
−77
990 + 40
= −74.76 mA
The negative sign indicates that the current flows in the direction opposite
to the one assumed, that is, from terminal b to terminal a.
P R A C T I C E P R O B L E M 4 . 1 8
Obtain the current through the galvanometer, having a resistance of 14 ,
in the Wheatstone bridge shown in Fig. 4.66.
14 Ω
60 Ω
16 V
40 Ω
20 Ω 30 Ω
G
Figure4.66 For Practice Prob. 4.18.
Answer: 64 mA.
CHAPTER 4 Circuit Theorems 153
4.11 SUMMARY
1. A linear network consists of linear elements, linear dependent
sources, and linear independent sources.
2. Network theorems are used to reduce a complex circuit to a simpler
one, thereby making circuit analysis much simpler.
3. The superposition principle states that for a circuit having multiple
independent sources, the voltage across (or current through) an
element is equal to the algebraic sum of all the individual voltages
(or currents) due to each independent source acting one at a time.
4. Source transformation is a procedure for transforming a voltage
source in series with a resistor to a current source in parallel with a
resistor, or vice versa.
5. Thevenin’s and Norton’s theorems allow us to isolate a portion of a
network while the remaining portion of the network is replaced by an
equivalent network. The Thevenin equivalent consists of a voltage
source VTh in series with a resistor RTh, while the Norton equivalent
consists of a current source IN in parallel with a resistor RN . The
two theorems are related by source transformation.
RN = RTh, IN =
VTh
RTh
6. For a given Thevenin equivalent circuit, maximum power transfer
occurs when RL = RTh, that is, when the load resistance is equal to
the Thevenin resistance.
7. PSpice can be used to verify the circuit theorems covered in this
chapter.
8. Source modeling and resistance measurement using the Wheatstone
bridge provide applications for Thevenin’s theorem.
REVIEW QUESTIONS
4.1 The current through a branch in a linear network is
2 A when the input source voltage is 10 V. If the
voltage is reduced to 1 V and the polarity is
reversed, the current through the branch is:
(a) −2 (b) −0.2 (c) 0.2
(d) 2 (e) 20
4.2 For superposition, it is not required that only one
independent source be considered at a time; any
number of independent sources may be considered
simultaneously.
(a) True (b) False
4.3 The superposition principle applies to power
calculation.
(a) True (b) False
4.4 Refer to Fig. 4.67. The Thevenin resistance at
terminals a and b is:
(a) 25  (b) 20 
(c) 5  (d) 4 
50 V 20 Ω
+
−
5 Ω
a
b
Figure 4.67 For Review Questions 4.4 to 4.6.
4.5 The Thevenin voltage across terminals a and b of
the circuit in Fig. 4.67 is:
(a) 50 V (b) 40 V
(c) 20 V (d) 10 V
154 PART 1 DC Circuits
4.6 The Norton current at terminals a and b of the
circuit in Fig. 4.67 is:
(a) 10 A (b) 2.5 A
(c) 2 A (d) 0 A
4.7 The Norton resistance RN is exactly equal to the
Thevenin resistance RTh.
(a) True (b) False
4.8 Which pair of circuits in Fig. 4.68 are equivalent?
(a) a and b (b) b and d
(c) a and c (d) c and d
+
−
20 V
5 Ω
(a)
4 A
5 Ω
(b)
5 Ω
(c)
+
−
20 V 5 Ω
(d)
4 A
Figure 4.68 For Review Question 4.8.
4.9 A load is connected to a network. At the terminals
to which the load is connected, RTh = 10  and
VTh = 40 V. The maximum power supplied to the
load is:
(a) 160 W (b) 80 W
(c) 40 W (d) 1 W
4.10 The source is supplying the maximum power to the
load when the load resistance equals the source
resistance.
(a) True (b) False
Answers: 4.1b, 4.2a, 4.3b, 4.4d, 4.5b, 4.6a, 4.7a, 4.8c, 4.9c, 4.10b.
PROBLEMS
Section 4.2 Linearity Property
4.1 Calculate the current io in the current of Fig. 4.69.
What does this current become when the input
voltage is raised to 10 V?
+
−
io
1 Ω 5 Ω
3 Ω
8 Ω
1 V
Figure 4.69 For Prob. 4.1.
4.2 Find vo in the circuit of Fig. 4.70. If the source
current is reduced to 1 µA, what is vo?
5 Ω 4 Ω
6 Ω
8 Ω
1 A 2 Ω
+
−
vo
Figure 4.70 For Prob. 4.2.
4.3 (a) In the circuit in Fig. 4.71, calculate vo and io
when vs = 1 V.
(b) Find vo and io when vs = 10 V.
(c) What are vo and io when each of the 1-
resistors is replaced by a 10- resistor and
vs = 10 V?
+
−
1 Ω
1 Ω
1 Ω 1 Ω
vs
1 Ω
io
+
−
vo
Figure 4.71 For Prob. 4.3.
4.4 Use linearity to determine io in the circuit of Fig.
4.72.
2 Ω
3 Ω
4 Ω
6 Ω 9 A
io
Figure 4.72 For Prob. 4.4.
CHAPTER 4 Circuit Theorems 155
4.5 For the circuit in Fig. 4.73, assume vo = 1 V, and
use linearity to find the actual value of vo.
2 Ω 3 Ω
4 Ω
6 Ω
vo 2 Ω
6 Ω
15 V +
−
Figure 4.73 For Prob. 4.5.
Section 4.3 Superposition
4.6 Apply superposition to find i in the circuit of Fig.
4.74.
20 V 5 A
6 Ω
4 Ω
+
−
i
Figure 4.74 For Prob. 4.6.
4.7 Given the circuit in Fig. 4.75, calculate ix and the
power dissipated by the 10- resistor using
superposition.
12 Ω
4 A
10 Ω 40 Ω
15 V −
+
ix
Figure 4.75 For Prob. 4.7.
4.8 For the circuit in Fig. 4.76, find the terminal voltage
Vab using superposition.
4 V 2 A
a
b
10 Ω
3Vab
+ −
+
− Vab
+
−
Figure 4.76 For Prob. 4.8.
4.9 Use superposition principle to find i in Fig. 4.77.
6 Ω
4 A
2 Ω 3 Ω
12 V −
+
i
Figure 4.77 For Prob. 4.9.
4.10 Determine vo in the circuit of Fig. 4.78 using the
superposition principle.
12 V
5 Ω
6 Ω
2 A
4 Ω
12 Ω
3 Ω
+
− 19 V
+
−
+ −
vo
Figure 4.78 For Prob. 4.10.
4.11 Apply the superposition principle to find vo in the
circuit of Fig. 4.79.
+
−
6 Ω
2 Ω
3 Ω
1 A
2 A
20 V
4 Ω
+
−
vo
Figure 4.79 For Prob. 4.11.
4.12 For the circuit in Fig. 4.80, use superposition to find
i. Calculate the power delivered to the 3- resistor.
20 V
2 A
3 Ω
2 Ω
1 Ω
4 Ω
16 V
−
+
i
+
−
Figure 4.80 For Probs. 4.12 and 4.45.
156 PART 1 DC Circuits
4.13 Given the circuit in Fig. 4.81, use superposition to
get io.
12 V
3 Ω
4 Ω
4 A
2 Ω
5 Ω
10 Ω
+
− 2 A
io
Figure 4.81 For Probs. 4.13 and 4.23.
4.14 Use superposition to obtain vx in the circuit of Fig.
4.82. Check your result using PSpice.
90 V 6 A
30 Ω 10 Ω 20 Ω
60 Ω 30 Ω
+
− 40 V
+
−
+ −
vx
Figure 4.82 For Prob. 4.14.
4.15 Find vx in Fig. 4.83 by superposition.
2 Ω
1 Ω
5ix
2 A 4 Ω
10 V +
−
ix
+ −
vx
Figure 4.83 For Prob. 4.15.
4.16 Use superposition to solve for ix in the circuit of
Fig. 4.84.
8 Ω
2 Ω 6 A 4 A
− +
ix
4ix
+
−
vx
Figure 4.84 For Prob. 4.16.
Section 4.4 Source Transformation
4.17 Find i in Prob. 4.9 using source transformation.
4.18 Apply source transformation to determine vo and io
in the circuit in Fig. 4.85.
12 V 2 A
6 Ω
3 Ω
+
−
io
+
−
vo
Figure 4.85 For Prob. 4.18.
4.19 For the circuit in Fig. 4.86, use source
transformation to find i.
5 Ω 10 Ω
4 Ω
5 Ω
2 A 20 V
+
−
i
Figure 4.86 For Prob. 4.19.
4.20 Obtain vo in the circuit of Fig. 4.87 using source
transformation. Check your result using PSpice.
3 A
9 Ω
2 Ω
2 A
30 V
5 Ω
4 Ω 6 A
+ −
+ −
vo
Figure 4.87 For Prob. 4.20.
4.21 Use source transformation to solve Prob. 4.14.
4.22 Apply source transformation to find vx in the circuit
of Fig. 4.88.
CHAPTER 4 Circuit Theorems 157
50 V 8 A
10 Ω 12 Ω 20 Ω
40 Ω
+
− 40 V
+
−
a b
+ −
vx
Figure 4.88 For Probs. 4.22 and 4.32.
4.23 Given the circuit in Fig. 4.81, use source
transformation to find io.
4.24 Use source transformation to find vo in the circuit of
Fig. 4.89.
4 kΩ
1 kΩ
3 mA
2 kΩ
3vo
− +
+
−
vo
Figure 4.89 For Prob. 4.24.
4.25 Determine vx in the circuit of Fig. 4.90 using source
transformation.
+
−
3 Ω 6 Ω
2vx
8 Ω
12 V
+
−
+ −
vx
Figure 4.90 For Prob. 4.25.
4.26 Use source transformation to find ix in the circuit of
Fig. 4.91.
10 Ω
15 Ω
0.5ix
40 Ω
60 V +
− 50 Ω
ix
Figure 4.91 For Prob. 4.26.
Sections 4.5 and 4.6 Thevenin’s and Norton’s
Theorems
4.27 Determine RTh and VTh at terminals 1-2 of each of
the circuits in Fig. 4.92.
10 Ω
+
−
20 V 40 Ω
(a)
(b)
1
2
2 A 30 Ω 30 V
+
−
60 Ω
1
2
Figure 4.92 For Probs. 4.27 and 4.37.
4.28 Find the Thevenin equivalent at terminals a-b of the
circuit in Fig. 4.93.
20 Ω
10 Ω
3 A
40 V 40 Ω
a
b
+
−
Figure 4.93 For Probs. 4.28 and 4.39.
4.29 Use Thevenin’s theorem to find vo in Prob. 4.10.
4.30 Solve for the current i in the circuit of Fig. 4.94
using Thevenin’s theorem. (Hint: Find the Thevenin
equivalent across the 12- resistor.)
12 Ω
30 V
40 Ω
+
−
10 Ω
50 V +
−
i
Figure 4.94 For Prob. 4.30.
4.31 For Prob. 4.8, obtain the Thevenin equivalent at
terminals a-b.
158 PART 1 DC Circuits
4.32 Given the circuit in Fig. 4.88, obtain the Thevenin
equivalent at terminals a-b and use the result to
get vx.
4.33
∗
For the circuit in Fig. 4.95, find the Thevenin
equivalent between terminals a and b.
20 Ω
20 Ω
10 Ω
10 Ω
5 A 10 Ω
20 V
30 V +
−
−
+
10 Ω
a b
Figure 4.95 For Prob. 4.33.
4.34 Find the Thevenin equivalent looking into terminals
a-b of the circuit in Fig. 4.96 and solve for ix.
20 V 2 A
10 Ω 6 Ω
10 Ω
+
− 5 Ω
a b
ix
Figure 4.96 For Prob. 4.34.
4.35 For the circuit in Fig. 4.97, obtain the Thevenin
equivalent as seen from terminals:
(a) a-b (b) b-c
4 Ω
24 V
5 Ω
2 Ω
1 Ω
3 Ω
2 A
a
b
c
+
−
Figure 4.97 For Prob. 4.35.
4.36 Find the Norton equivalent of the circuit in Fig. 4.98.
*An asterisk indicates a challenging problem.
4 A 4 Ω
a
b
6 Ω
6 Ω
Figure 4.98 For Prob. 4.36.
4.37 Obtain RN and IN at terminals 1 and 2 of each of the
circuits in Fig. 4.92.
4.38 Determine the Norton equivalent at terminals a-b
for the circuit in Fig. 4.99.
2 A
a
b
4 Ω
2 Ω
10io
+ −
io
Figure 4.99 For Prob. 4.38.
4.39 Find the Norton equivalent looking into terminals
a-b of the circuit in Fig. 4.93.
4.40 Obtain the Norton equivalent of the circuit in Fig.
4.100 to the left of terminals a-b. Use the result to
find current i.
2 A
a
b
4 A
4 Ω 5 Ω
12 V
6 Ω
+ −
i
Figure 4.100 For Prob. 4.40.
4.41 Given the circuit in Fig. 4.101, obtain the Norton
equivalent as viewed from terminals:
(a) a-b (b) c-d
120 V
c
a b
d
6 A 2 Ω
3 Ω
4 Ω
6 Ω
+
−
Figure 4.101 For Prob. 4.41.
CHAPTER 4 Circuit Theorems 159
4.42 For the transistor model in Fig. 4.102, obtain the
Thevenin equivalent at terminals a-b.
6 V 20io
3 kΩ
2 kΩ
+
−
a
b
io
Figure 4.102 For Prob. 4.42.
4.43 Find the Norton equivalent at terminals a-b of the
circuit in Fig. 4.103.
2 Ω
6 Ω
0.25vo
3 Ω
18 V +
−
vo
+
−
a
b
Figure 4.103 For Prob. 4.43.
4.44
∗
Obtain the Norton equivalent at terminals a-b of the
circuit in Fig. 4.104.
2 V 80I
8 kΩ
50 kΩ
+
−
a
b
0.01Vab
+
−
I
+
−
Vab
Figure 4.104 For Prob. 4.44.
4.45 Use Norton’s theorem to find current i in the circuit
of Fig. 4.80.
4.46 Obtain the Thevenin and Norton equivalent circuits
at the terminals a-b for the circuit in Fig. 4.105.
50 V
3 Ω 2 Ω
10 Ω
+
−
a
b
0.5vx
6 Ω
+
−
vx
Figure 4.105 For Probs. 4.46 and 4.65.
4.47 The network in Fig. 4.106 models a bipolar
transistor common-emitter amplifier connected to a
load. Find the Thevenin resistance seen by the load.
vs
R1
bib
RL
+
− R2
ib
Figure 4.106 For Prob. 4.47.
4.48 Determine the Thevenin and Norton equivalents at
terminals a-b of the circuit in Fig. 4.107.
8 A
10 Ω 20 Ω
50 Ω 40 Ω
a b
Figure 4.107 For Probs. 4.48 and 4.66.
4.49
∗
For the circuit in Fig. 4.108, find the Thevenin and
Norton equivalent circuits at terminals a-b.
+ −
18 V
3 A
4 Ω 6 Ω
5 Ω
a b
2 A
10 V
+ −
Figure 4.108 For Probs. 4.49 and 4.67.
4.50
∗
Obtain the Thevenin and Norton equivalent circuits
at terminals a-b of the circuit in Fig. 4.109.
160 PART 1 DC Circuits
12 V
6 Ω
2 Ω
6 Ω
2 Ω
6 Ω
+
− 12 V
2 Ω
+
−
12 V
+
−
a
b
Figure 4.109 For Prob. 4.50.
4.51
∗
Find the Thevenin equivalent of the circuit in Fig.
4.110.
10 Ω
20 Ω
40 Ω
+ −
io
0.1io
2vo
+
−
vo
b
a
Figure 4.110 For Prob. 4.51.
4.52 Find the Norton equivalent for the circuit in Fig.
4.111.
0.5vo
10 Ω
+
−
vo 20 V
Figure 4.111 For Prob. 4.52.
4.53 Obtain the Thevenin equivalent seen at terminals
a-b of the circuit in Fig. 4.112.
10ix
4 Ω
2 Ω
1 Ω
+
−
ix
a
b
Figure 4.112 For Prob. 4.53.
Section 4.8 Maximum Power Transfer
4.54 Find the maximum power that can be delivered to
the resistor R in the circuit in Fig. 4.113.
R
3 Ω
2 Ω
5 Ω
20 V 6 A
+
−
− +
10 V
Figure 4.113 For Prob. 4.54.
4.55 Refer to Fig. 4.114. For what value of R is the
power dissipated in R maximum? Calculate that power.
6 Ω
30 V
4 Ω
8 Ω
R
12 Ω
+ −
Figure 4.114 For Prob. 4.55.
4.56
∗
Compute the value of R that results in maximum
power transfer to the 10- resistor in Fig. 4.115.
Find the maximum power.
+
−
+
−
R
20 Ω
10 Ω
8 V
12 V
Figure 4.115 For Prob. 4.56.
4.57 Find the maximum power transferred to resistor R
in the circuit of Fig. 4.116.
R
40 kΩ 30 kΩ
100 V +
− 3vo
22 kΩ
10 kΩ
+
−
vo
Figure 4.116 For Prob. 4.57.
CHAPTER 4 Circuit Theorems 161
4.58 For the circuit in Fig. 4.117, what resistor connected
across terminals a-b will absorb maximum power
from the circuit? What is that power?
8 V 120vo
3 kΩ 10 kΩ
40 kΩ
1 kΩ
+
−
a
b
–
+
+
−
vo
Figure 4.117 For Prob. 4.58.
4.59 (a) For the circuit in Fig. 4.118, obtain the Thevenin
equivalent at terminals a-b.
(b) Calculate the current in RL = 8 .
(c) Find RL for maximum power deliverable to RL.
(d) Determine that maximum power.
6 Ω
4 Ω
2 A
20 V
4 A 2 Ω
a
b
RL
+ −
Figure 4.118 For Prob. 4.59.
4.60 For the bridge circuit shown in Fig. 4.119, find the
load RL for maximum power transfer and the
maximum power absorbed by the load.
R3
RL
R4
R1
R2
vs
+
−
Figure 4.119 For Prob. 4.60.
4.61 For the circuit in Fig. 4.120, determine the value of
R such that the maximum power delivered to the
load is 3 mW.
R
R
R
2 V
3 V
RL
+
−
1 V +
−
+
−
Figure 4.120 For Prob. 4.61.
Section 4.9 Verifying Circuit Theorems with
PSpice
4.62 Solve Prob. 4.28 using PSpice.
4.63 Use PSpice to solve Prob. 4.35.
4.64 Use PSpice to solve Prob. 4.42.
4.65 Obtain the Thevenin equivalent of the circuit in Fig.
4.105 using PSpice.
4.66 Use PSpice to find the Thevenin equivalent circuit at
terminals a-b of the circuit in Fig. 4.107.
4.67 For the circuit in Fig. 4.108, use PSpice to find the
Thevenin equivalent at terminals a-b.
Section 4.10 Applications
4.68 A battery has a short-circuit current of 20 A and an
open-circuit voltage of 12 V. If the battery is
connected to an electric bulb of resistance 2 ,
calculate the power dissipated by the bulb.
4.69 The following results were obtained from
measurements taken between the two terminals of a
resistive network.
Terminal Voltage 12 V 0 V
Terminal Current 0 V 1.5 A
Find the Thevenin equivalent of the network.
4.70 A black box with a circuit in it is connected to a
variable resistor. An ideal ammeter (with zero
resistance) and an ideal voltmeter (with infinite
resistance) are used to measure current and voltage
as shown in Fig. 4.121. The results are shown in the
table below.
Black
box
V
A
i
R
Figure 4.121 For Prob. 4.70.
162 PART 1 DC Circuits
(a) Find i when R = 4 .
(b) Determine the maximum power from the box.
R() V (V) i(A)
2 3 1.5
8 8 1.0
14 10.5 0.75
4.71 A transducer is modeled with a current source Is and
a parallel resistance Rs. The current at the terminals
of the source is measured to be 9.975 mA when an
ammeter with an internal resistance of 20  is used.
(a) If adding a 2-k resistor across the source
terminals causes the ammeter reading to fall to
9.876 mA, calculate Is and Rs.
(b) What will the ammeter reading be if the
resistance between the source terminals is
changed to 4 k?
4.72 The Wheatstone bridge circuit shown in Fig. 4.122
is used to measure the resistance of a strain gauge.
The adjustable resistor has a linear taper with a
maximum value of 100 . If the resistance of the
strain gauge is found to be 42.6 , what fraction of
the full slider travel is the slider when the bridge is
balanced?
4 kΩ
100 Ω
2 kΩ
+
−
vs
Rs
Rx
G
Figure 4.122 For Prob. 4.72.
4.73 (a) In the Wheatstone bridge circuit of Fig. 4.123,
select the values of R1 and R3 such that the
bridge can measure Rx in the range of 0–10 .
R3
Rx
R1
V +
−
G
50 Ω
Figure 4.123 For Prob. 4.73.
(b) Repeat for the range of 0–100 .
4.74
∗
Consider the bridge circuit of Fig. 4.124. Is the
bridge balanced? If the 10-k resistor is replaced
by an 18-k resistor, what resistor connected
between terminals a-b absorbs the maximum
power? What is this power?
220 V
2 kΩ
3 kΩ 6 kΩ
5 kΩ 10 kΩ
+
− a b
Figure 4.124 For Prob. 4.74.
COMPREHENSIVE PROBLEMS
4.75 The circuit in Fig. 4.125 models a common-emitter
transistor amplifier. Find ix using source
transformation.
vs
Rs
+
− bix
Ro
ix
Figure 4.125 For Prob. 4.75.
4.76 An attenuator is an interface circuit that reduces the
voltage level without changing the output resistance.
(a) By specifying Rs and Rp of the interface circuit
in Fig. 4.126, design an attenuator that will meet
the following requirements:
Vo
Vg
= 0.125, Req = RTh = Rg = 100 
(b) Using the interface designed in part (a),
calculate the current through a load of
RL = 50  when Vg = 12 V.
CHAPTER 4 Circuit Theorems 163
V
g
Rg Rs
RL
Req
+
− Rp Vo
+
−
Attenuator
Load
Figure 4.126 For Prob. 4.76.
4.77
∗
A dc voltmeter with a sensitivity of 20 k/V is used
to find the Thevenin equivalent of a linear network.
Readings on two scales are as follows:
(a) 0–10 V scale: 4 V (b) 0–50 V scale: 5 V
Obtain the Thevenin voltage and the Thevenin
resistance of the network.
4.78
∗
A resistance array is connected to a load resistor R
and a 9-V battery as shown in Fig. 4.127.
(a) Find the value of R such that Vo = 1.8 V.
(b) Calculate the value of R that will draw the
maximum current. What is the maximum
current?
60 Ω 10 Ω
10 Ω
8 Ω 8 Ω
R
10 Ω 40 Ω
9 V
+ −
3
4
1
2
+ −
Vo
Figure 4.127 For Prob. 4.78.
4.79 A common-emitter amplifier circuit is shown in Fig.
4.128. Obtain the Thevenin equivalent to the left of
points B and E.
RL
Rc
E
4 kΩ
6 kΩ
12 V
B +
−
Figure 4.128 For Prob. 4.79.
4.80
∗
For Practice Prob. 4.17, determine the current
through the 40- resistor and the power dissipated
by the resistor.
165
C H A P T E R
OPERATIONAL AMPLIFIERS
5
If A is success in life, then A equals X plus Y plus Z. Work is X, Y is
play and Z is keeping your mouth shut.
—Albert Einstein
Enhancing Your Career
Career in Electronic Instrumentation Engineering in-
volves applying physical principles to design devices for
the benefit of humanity. But physical principles cannot be
understood without measurement. In fact, physicists often
say that physics is the science that measures reality. Just
as measurements are a tool for understanding the physical
world, instruments are tools for measurement. The opera-
tional amplifier introduced in this chapter is a building block
of modern electronic instrumentation. Therefore, mastery
of operational amplifier fundamentals is paramount to any
practical application of electronic circuits.
Electronic instruments are used in all fields of sci-
ence and engineering. They have proliferated in science and
technology to the extent that it would be ridiculous to have
a scientific or technical education without exposure to elec-
tronic instruments. For example, physicists, physiologists,
chemists, and biologists must learn to use electronic instru-
ments. For electrical engineering students in particular, the
skill in operating digital and analog electronic instruments
is crucial. Such instruments include ammeters, voltmeters,
ohmmeters, oscilloscopes, spectrum analyzers, and signal
generators.
Beyond developing the skill for operating the instru-
ments, some electrical engineers specialize in designing and
constructing electronic instruments. These engineers derive
pleasure in building their own instruments. Most of them
Electronic Instrumentation used in medical research.
Source: Geoff Tompkinson/Science Photo Library.
invent and patent their inventions. Specialists in electronic
instruments find employment in medical schools, hospitals,
research laboratories, aircraft industries, and thousands of
other industries where electronic instruments are routinely
used.
166 PART 1 DC Circuits
5.1 INTRODUCTION
Having learned the basic laws and theorems for circuit analysis, we are
now ready to study an active circuit element of paramount importance:
the operational amplifier, or op amp for short. The op amp is a versatile
circuit building block.
The term operational amplifier was introduced
in 1947 by John Ragazzini and his colleagues, in
theirworkonanalogcomputersfortheNational
Defense Research Council during World War II.
Thefirstopampsusedvacuumtubesratherthan
transistors.
The op amp is an electronic unit that behaves like a
voltage-controlled voltage source.
An op amp may also be regarded as a voltage
amplifier with very high gain. It can also be used in making a voltage- or current-controlled current
source. An op amp can sum signals, amplify a signal, integrate it, or
differentiate it. The ability of the op amp to perform these mathematical
operations is the reason it is called an operational amplifier. It is also
the reason for the widespread use of op amps in analog design. Op
amps are popular in practical circuit designs because they are versatile,
inexpensive, easy to use, and fun to work with.
We begin by discussing the ideal op amp and later consider the
nonideal op amp. Using nodal analysis as a tool, we consider ideal op
amp circuits such as the inverter, voltage follower, summer, and difference
amplifier. We will analyze op amp circuits with PSpice. Finally, we learn
how an op amp is used in digital-to-analog converters and instrumentation
amplifiers.
5.2 OPERATIONAL AMPLIFIERS
An operational amplifier is designed so that it performs some mathemat-
ical operations when external components, such as resistors and capaci-
tors, are connected to its terminals. Thus,
An op amp is an active circuit element designed to perform mathematical operations
of addition, subtraction, multiplication, division, differentiation, and integration.
The op amp is an electronic device consisting of a complex arrange-
ment of resistors, transistors, capacitors, and diodes. A full discussion
of what is inside the op amp is beyond the scope of this book. It will
suffice to treat the op amp as a circuit building block and simply study
what takes place at its terminals.
Figure 5.1 A typical operational amplifier.
(Courtesy of Tech America.)
Op amps are commercially available in integrated circuit packages
in several forms. Figure 5.1 shows a typical op amp package. A typical
one is the eight-pin dual in-line package (or DIP), shown in Fig. 5.2(a).
Pin or terminal 8 is unused, and terminals 1 and 5 are of little concern to
us. The five important terminals are:
ThepindiagraminFig.5.2(a)correspondstothe
741 general-purpose op amp made by Fairchild
Semiconductor.
1. The inverting input, pin 2.
2. The noninverting input, pin 3.
3. The output, pin 6.
CHAPTER 5 Operational Amplifiers 167
4. The positive power supply V +
, pin 7.
5. The negative power supply V −
, pin 4.
The circuit symbol for the op amp is the triangle in Fig. 5.2(b); as shown,
the op amp has two inputs and one output. The inputs are marked with
minus (−) and plus (+) to specify inverting and noninverting inputs,
respectively. An input applied to the noninverting terminal will appear
with the same polarity at the output, while an input applied to the inverting
terminal will appear inverted at the output.
+
−
1
Balance
2
Inverting input
3
Noninverting input
4
V− 5 Balance
6 Output
7 V+
8 No connection
(a) (b)
2
Inverting input
3
Noninverting input
4
V−
V+
1 5
Offset Null
6 Output
7
Figure5.2 A typical op amp: (a) pin configuration, (b) circuit symbol.
As an active element, the op amp must be powered by a voltage
supply as typically shown in Fig. 5.3. Although the power supplies are
often ignored in op amp circuit diagrams for the sake of simplicity, the
power supply currents must not be overlooked. By KCL,
io = i1 + i2 + i+ + i− (5.1)
7
4
6
Vcc
Vcc
+
−
+
−
io
i1
i2
i+
i−
2
3
Figure5.3 Powering the op amp.
v1
v2
vo
+
−
+
−
vd Ri
Ro
Avd
Figure 5.4 The equivalent circuit of the non-
ideal op amp.
The equivalent circuit model of an op amp is shown in Fig. 5.4. The
output section consists of a voltage-controlled source in series with the
output resistance Ro. It is evident from Fig. 5.4 that the input resistance
Ri is the Thevenin equivalent resistance seen at the input terminals, while
the output resistance Ro is the Thevenin equivalent resistance seen at the
output. The differential input voltage vd is given by
vd = v2 − v1 (5.2)
where v1 is the voltage between the inverting terminal and ground and v2
is the voltage between the noninverting terminal and ground. The op amp
senses the difference between the two inputs, multiplies it by the gain A,
and causes the resulting voltage to appear at the output. Thus, the output
vo is given by
vo = Avd = A(v2 − v1) (5.3)
A is called the open-loop voltage gain because it is the gain of the op amp
without any external feedback from output to input. Table 5.1 shows
typical values of voltage gain A, input resistance Ri, output resistance
Ro, and supply voltage VCC.
Sometimes, voltage gain is expressed in decibels
(dB), as discussed in Chapter 14.
A dB = 20 log10 A
The concept of feedback is crucial to our understanding of op amp
circuits. A negative feedback is achieved when the output is fed back to
the inverting terminal of the op amp. As Example 5.1 shows, when there
168 PART 1 DC Circuits
TABLE 5.1 Typical ranges for op amp
parameters.
Parameter Typical range Ideal values
Open-loop gain, A 105
to 108
∞
Input resistance, Ri 106
to 1013
 ∞ 
Output resistance, Ro 10 to 100  0 
Supply voltage, Vcc 5 to 24 V
is a feedback path from output to input, the ratio of the output voltage to
the input voltage is called the closed-loop gain. As a result of the negative
feedback, it can be shown that the closed-loop gain is almost insensitive
to the open-loop gain A of the op amp. For this reason, op amps are used
in circuits with feedback paths.
A practical limitation of the op amp is that the magnitude of its
output voltage cannot exceed |VCC|. In other words, the output voltage
is dependent on and is limited by the power supply voltage. Figure 5.5
illustrates that the op amp can operate in three modes, depending on the
differential input voltage vd:
1. Positive saturation, vo = VCC.
2. Linear region, −VCC ≤ vo = Avd ≤ VCC.
3. Negative saturation, vo = −VCC.
Positive saturation
Negative saturation
vd
vo
VCC
−VCC
0
Figure5.5 Op amp output voltage vo as a
function of the differential input voltage vd .
If we attempt to increase vd beyond the linear range, the op amp becomes
saturated and yields vo = VCC or vo = −VCC. Throughout this book,
we will assume that our op amps operate in the linear mode. This means
that the output voltage is restricted by
−VCC ≤ vo ≤ VCC (5.4)
Although we shall always operate the op amp in the linear region, the
possibility of saturation must be borne in mind when one designs with
op amps, to avoid designing op amp circuits that will not work in the
laboratory.
E X A M P L E 5 . 1
A 741 op amp has an open-loop voltage gain of 2 × 105
, input resistance
of 2 M, and output resistance of 50 . The op amp is used in the circuit
of Fig. 5.6(a). Find the closed-loop gain vo/vs. Determine current i when
vs = 2 V.
Solution:
Using the op amp model in Fig. 5.4, we obtain the equivalent circuit of
Fig. 5.6(a) as shown in Fig. 5.6(b). We now solve the circuit in Fig. 5.6(b)
by using nodal analysis. At node 1, KCL gives
vs − v1
10 × 103
=
v1
2000 × 103
+
v1 − vo
20 × 103
CHAPTER 5 Operational Amplifiers 169
10 kΩ
20 kΩ
vs
i
vo
+
−
+
−
1
O
(a) (b)
+
−
741
10 kΩ
20 kΩ
vs
i
Ro = 50 Ω
Ri = 2 MΩ
+
−
1 O
+
− Avd
v1 vo
−
+
vd
Figure5.6 For Example 5.1: (a) original circuit, (b) the equivalent circuit.
Multiplying through by 2000 × 103
, we obtain
200vs = 301v1 − 100vo
or
2vs  3v1 − vo ⇒ v1 =
2vs + vo
3
(5.1.1)
At node O,
v1 − vo
20 × 103
=
vo − Avd
50
But vd = −v1 and A = 200,000. Then
v1 − vo = 400(vo + 200,000v1) (5.1.2)
Substituting v1 from Eq. (5.1.1) into Eq. (5.1.2) gives
0  26,667,067vo + 53,333,333vs
vo
vs
= −1.9999699
This is closed-loop gain, because the 20-k feedback resistor closes the
loop between the output and input terminals. When vs = 2 V, vo =
−3.9999398 V. From Eq. (5.1.1), we obtain v1 = 20.066667 µV. Thus,
i =
v1 − vo
20 × 103
= 0.1999 mA
It is evident that working with a nonideal op amp is tedious, as we are
dealing with very large numbers.
P R A C T I C E P R O B L E M 5 . 1
If the same 741 op amp in Example 5.1 is used in the circuit of Fig. 5.7,
calculate the closed-loop gain vo/vs. Find io when vs = 1 V.
40 kΩ
20 kΩ
5 kΩ
vs
vo
io
+
− +
−
+
−
741
Figure5.7 For Practice Prob. 5.1.
Answer: 9.0041, −362 mA.
170 PART 1 DC Circuits
5.3 IDEAL OP AMP
To facilitate the understanding of op amp circuits, we will assume ideal
op amps. An op amp is ideal if it has the following characteristics:
1. Infinite open-loop gain, A  ∞.
2. Infinite input resistance, Ri  ∞.
3. Zero output resistance, Ro  0.
An ideal op amp is an amplifier with infinite open-loop gain, infinite input
resistance, and zero output resistance.
Although assuming an ideal op amp provides only an approxi-
mate analysis, most modern amplifiers have such large gains and input
impedances that the approximate analysis is a good one. Unless stated
otherwise, we will assume from now on that every op amp is ideal.
For circuit analysis, the ideal op amp is illustrated in Fig. 5.8, which
is derived from the nonideal model in Fig. 5.4. Two important character-
istics of the ideal op amp are:
i2 = 0
i1 = 0
v1
v2 = v1
+
−
vo
+
−
vd
+
−
+
−
+
−
Figure5.8 Ideal op amp model.
1. The currents into both input terminals are zero:
i1 = 0, i2 = 0 (5.5)
This is due to infinite input resistance. An infinite resistance
between the input terminals implies that an open circuit exists
there and current cannot enter the op amp. But the output
current is not necessarily zero according to Eq. (5.1).
2. The voltage across the input terminals is negligibly small; i.e.,
vd = v2 − v1  0 (5.6)
or
v1 = v2 (5.7)
Thus, an ideal op amp has zero current into its two input
terminals and negligibly small voltage between the two input
terminals. Equations (5.5) and (5.7) are extremely important
and should be regarded as the key handles to analyzing op amp
circuits.
The two characteristics can be exploited by
noting that for voltage calculations the input
port behaves as a short circuit, while for current
calculations the input port behaves as an open
circuit.
E X A M P L E 5 . 2
Rework Practice Prob. 5.1 using the ideal op amp model.
Solution:
We may replace the op amp in Fig. 5.7 by its equivalent model in Fig.
5.9 as we did in Example 5.1. But we do not really need to do this. We
CHAPTER 5 Operational Amplifiers 171
just need to keep Eqs. (5.5) and (5.7) in mind as we analyze the circuit in
Fig. 5.7. Thus, the Fig. 5.7 circuit is presented as in Fig. 5.9. Notice that
v2 = vs (5.2.1)
Since i1 = 0, the 40-k and 5-k resistors are in series because the same
current flows through them. v1 is the voltage across the 5-k resistor.
Hence, using the voltage division principle,
v1 =
5
5 + 40
vo =
vo
9
(5.2.2)
According to Eq. (5.7),
v2 = v1 (5.2.3)
Substituting Eqs. (5.2.1) and (5.2.2) into Eq. (5.2.3) yields the closed-loop
gain,
vs =
vo
9
⇒
vo
vs
= 9 (5.2.4)
which is very close to the value of 8.99955796 obtained with the nonideal
model in Practice Prob. 5.1. This shows that negligibly small error results
from assuming ideal op amp characteristics.
40 kΩ
20 kΩ
5 kΩ
vs
i1 = 0
i2 = 0
i0
+
−
v1
v2
O
+
−
+
−
vo
Figure5.9 For Example 5.2.
At node O,
io =
vo
40 + 5
+
vo
20
mA (5.2.5)
From Eq. (5.2.4), when vs = 1 V, vo = 9 V. Substituting for vo = 9 V
in Eq. (5.2.5) produces
io = 0.2 + 0.45 = 0.65 mA
This, again, is close to the value of 0.649 mA obtained in Practice Prob.
5.1 with the nonideal model.
P R A C T I C E P R O B L E M 5 . 2
Repeat Example 5.1 using the ideal op amp model.
Answer: −2 , 0.2 mA.
Throughoutthisbook,weassumethatanopamp
operates in the linear range. Keep in mind the
voltage constraint on the op amp in this mode.
A key feature of the inverting amplifier is that
boththeinputsignalandthefeedbackareapplied
at the inverting terminal of the op amp.
5.4 INVERTING AMPLIFIER
Inthisandthefollowingsections, weconsidersomeusefulopampcircuits
that often serve as modules for designing more complex circuits. The first
ofsuchopampcircuitsistheinvertingamplifiershowninFig.5.10. Inthis
circuit, the noninverting input is grounded, vi is connected to the inverting
input through R1, and the feedback resistor Rf is connected between the
inverting input and output. Our goal is to obtain the relationship between
the input voltage vi and the output voltage vo. Applying KCL at node 1,
i1 = i2 ⇒
vi − v1
R1
=
v1 − vo
Rf
(5.8)
172 PART 1 DC Circuits
But v1 = v2 = 0 for an ideal op amp, since the noninverting terminal is
grounded. Hence,
vi
R1
= −
vo
Rf
or
vo = −
Rf
R1
vi (5.9)
The voltage gain is Av = vo/vi = −Rf /R1. The designation of the
circuit in Fig. 5.10 as an inverter arises from the negative sign. Thus,
An inverting amplifier reverses the polarity of the input signal while amplifying it.
Notice that the gain is the feedback resistance divided by the input
resistance which means that the gain depends only on the external ele-
ments connected to the op amp. In view of Eq. (5.9), an equivalent circuit
for the inverting amplifier is shown in Fig. 5.11. The inverting amplifier
is used, for example, in a current-to-voltage converter.
Note there are two types of gains: the one here
is the closed-loop voltage gain Av, while the op
amp itself has an open-loop voltage gain A.
R1
Rf
vi
vo
+
−
+
− v2
v1
0 A
0 V
+
−
+
−
1
i1
i2
Figure5.10 The inverting amplifier.
–
+
+
−
vi
+
−
vo
R1
Rf
R1
vi
Figure5.11 An equivalent circuit
for the inverter in Fig. 5.10.
E X A M P L E 5 . 3
Refer to the op amp in Fig. 5.12. If vi = 0.5 V, calculate: (a) the output
voltage vo, and (b) the current in the 10 k resistor.
10 kΩ
25 kΩ
vi vo
+
−
+
−
+
−
Figure5.12 For Example 5.3.
Solution:
(a) Using Eq. (5.9),
vo
vi
= −
Rf
R1
= −
25
10
= −2.5
vo = −2.5vi = −2.5(0.5) = −1.25 V
(b) The current through the 10-k resistor is
i =
vi − 0
R1
=
0.5 − 0
10 × 103
= 50 µA
CHAPTER 5 Operational Amplifiers 173
P R A C T I C E P R O B L E M 5 . 3
Find the output of the op amp circuit shown in Fig. 5.13. Calculate the
current through the feedback resistor.
5 kΩ
15 kΩ
40 mV vo
+
−
+
−
+
−
Figure5.13 For Practice Prob. 5.3.
Answer: −120 mV, 8 µA.
E X A M P L E 5 . 4
Determine vo in the op amp circuit shown in Fig. 5.14.
20 kΩ
40 kΩ
6 V vo
+
−
2 V +
−
+
−
a
b +
−
Figure5.14 For Example 5.4.
Solution:
Applying KCL at node a,
va − vo
40
=
6 − va
20
va − vo = 12 − 2va ⇒ vo = 3va − 12
But va = vb = 2 V for an ideal op amp, because of the zero voltage drop
across the input terminals of the op amp. Hence,
vo = 6 − 12 = −6 V
Notice that if vb = 0 = va, then vo = −12, as expected from Eq. (5.9).
P R A C T I C E P R O B L E M 5 . 4
Two kinds of current-to-voltage converters (also known as transresistance
amplifiers) are shown in Fig. 5.15.
(a) Show that for the converter in Fig. 5.15(a),
vo
is
= −R
(b) Show that for the converter in Fig. 5.15(b),
vo
is
= −R1

1 +
R3
R1
+
R3
R2

Answer: Proof.
R
is vo
+
−
(a)
+
−
R1
is
R2
vo
+
−
(b)
R3
+
−
Figure5.15 For Practice Prob. 5.4.
174 PART 1 DC Circuits
R1
Rf
vo
+
−
v1
v2
vi
+
−
i2
i1
+
−
Figure5.16 The noninverting amplifier.
5.5 NONINVERTING AMPLIFIER
Another important application of the op amp is the noninverting amplifier
shown in Fig. 5.16. In this case, the input voltage vi is applied directly at
the noninverting input terminal, and resistor R1 is connected between the
ground and the inverting terminal. We are interested in the output voltage
and the voltage gain. Application of KCL at the inverting terminal gives
i1 = i2 ⇒
0 − v1
R1
=
v1 − vo
Rf
(5.10)
But v1 = v2 = vi. Equation (5.10) becomes
−vi
R1
=
vi − vo
Rf
or
vo =

1 +
Rf
R1

vi (5.11)
The voltage gain is Av = vo/vi = 1 + Rf /R1, which does not have a
negative sign. Thus, the output has the same polarity as the input.
A noninverting amplifier is an op amp circuit designed
to provide a positive voltage gain.
Again we notice that the gain depends only on the external resistors.
Notice that if feedback resistor Rf = 0 (short circuit) or R1 = ∞
(open circuit) or both, the gain becomes 1. Under these conditions (Rf =
0 and R1 = ∞), the circuit in Fig. 5.16 becomes that shown in Fig. 5.17,
which is called a voltage follower (or unity gain amplifier) because the
output follows the input. Thus, for a voltage follower
vo = vi (5.12)
Such a circuit has a very high input impedance and is therefore useful as an
intermediate-stage (or buffer) amplifier to isolate one circuit from another,
as portrayed in Fig. 5.18. The voltage follower minimizes interaction
between the two stages and eliminates interstage loading.
vo = vi
+
−
vi
+
−
+
−
Figure5.17 The voltage
follower.
+
−
vi
+
−
vo
+
−
First
stage
Second
stage
Figure5.18 A voltage follower used to
isolate two cascaded stages of a circuit.
CHAPTER 5 Operational Amplifiers 175
E X A M P L E 5 . 5
For the op amp circuit in Fig. 5.19, calculate the output voltage vo.
4 kΩ
10 kΩ
6 V vo
+
−
+
− 4 V +
−
a
b +
−
Figure5.19 For Example 5.9.
Solution:
We may solve this in two ways: using superposition and using nodal
analysis.
METHOD 1 Using superposition, we let
vo = vo1 + vo2
where vo1 is due to the 6-V voltage source, and vo2 is due to the 4-V input.
To get vo1, we set the 4-V source equal to zero. Under this condition, the
circuit becomes an inverter. Hence Eq. (5.9) gives
vo1 = −
10
4
(6) = −15 V
To get vo2, we set the 6-V source equal to zero. The circuit becomes a
noninverting amplifier so that Eq. (5.11) applies.
vo2 =

1 +
10
4

4 = 14 V
Thus,
vo = vo1 + vo2 = −15 + 14 = −1 V
METHOD 2 Applying KCL at node a,
6 − va
4
=
va − vo
10
But va = vb = 4, and so
6 − 4
4
=
4 − vo
10
⇒ 5 = 4 − vo
or vo = −1 V, as before.
P R A C T I C E P R O B L E M 5 . 5
Calculate vo in the circuit in Fig. 5.20.
5 kΩ
4 kΩ
2 kΩ
3 V vo
+
−
+
− 8 kΩ
+
−
Figure5.20 For Practice Prob. 5.5.
Answer: 7 V.
176 PART 1 DC Circuits
5.6 SUMMING AMPLIFIER
Besides amplification, the op amp can perform addition and subtraction.
The addition is performed by the summing amplifier covered in this sec-
tion; the subtraction is performed by the difference amplifier covered in
the next section.
A summing amplifier is an op amp circuit that combines several inputs and produces
an output that is the weighted sum of the inputs.
i1
i2
i3
v1
v2
v3
i
i
+
−
vo
0
0
R1 Rf
R2
R3
a
+
−
Figure5.21 The summing amplifier.
The summing amplifier, shown in Fig. 5.21, is a variation of the
inverting amplifier. It takes advantage of the fact that the inverting con-
figuration can handle many inputs at the same time. We keep in mind that
the current entering each op amp input is zero. Applying KCL at node a
gives
i = i1 + i2 + i3 (5.13)
But
i1 =
v1 − va
R1
, i2 =
v2 − va
R2
i3 =
v3 − va
R3
, i =
va − vo
Rf
(5.14)
We note that va = 0 and substitute Eq. (5.14) into Eq. (5.13). We get
vo = −

Rf
R1
v1 +
Rf
R2
v2 +
Rf
R3
v3

(5.15)
indicating that the output voltage is a weighted sum of the inputs. For
this reason, the circuit in Fig. 5.21 is called a summer. Needless to say,
the summer can have more than three inputs.
E X A M P L E 5 . 6
Calculate vo and io in the op amp circuit in Fig. 5.22.
10 kΩ
5 kΩ
2.5 kΩ
2 kΩ
io
vo
a
b
1 V
+
−
+
− +
−
2 V
+
−
Figure5.22 For Example 5.6.
CHAPTER 5 Operational Amplifiers 177
Solution:
This is a summer with two inputs. Using Eq. (5.15),
vo = −

10
5
(2) +
10
2.5
(1)

= −(4 + 4) = −8 V
The current io is the sum of the currents through the 10-k and 2-k
resistors. Both of these resistors have voltage vo = −8 V across them,
since va = vb = 0. Hence,
io =
vo − 0
10
+
vo − 0
2
mA = −0.8 − 0.4 = −1.2 mA
P R A C T I C E P R O B L E M 5 . 6
Find vo and io in the op amp circuit shown in Fig. 5.23.
vo
io
+
−
20 kΩ
10 kΩ
6 kΩ
8 kΩ
4 kΩ
+
−
+
−
+
−
1.2 V
2 V
1.5 V
+
−
Figure5.23 For Practice Prob. 5.6.
Answer: −3.8 V, −1.425 mA.
5.7 DIFFERENCE AMPLIFIER
Difference (or differential) amplifiers are used in various applications
where there is need to amplify the difference between two input signals.
They are first cousins of the instrumentation amplifier, the most useful
and popular amplifier, which we will discuss in Section 5.10.
A difference amplifier is a device that amplifies the difference between two inputs
but rejects any signals common to the two inputs.
Thedifferenceamplifierisalsoknownasthesub-
tractor, for reasons to be shown later.
Consider the op amp circuit shown in Fig. 5.24. Keep in mind that
zero currents enter the op amp terminals. Applying KCL to node a,
v1 − va
R1
=
va − vo
R2
or
vo =

R2
R1
+ 1

va −
R2
R1
v1 (5.16)
178 PART 1 DC Circuits
v1
v2
+
−
vo
0
0
+
−
R1
R2
R3
R4
va
vb
+
−
+
−
Figure5.24 Difference amplifier.
Applying KCL to node b,
v2 − vb
R3
=
vb − 0
R4
or
vb =
R4
R3 + R4
v2 (5.17)
But va = vb. Substituting Eq. (5.17) into Eq. (5.16) yields
vo =

R2
R1
+ 1

R4
R3 + R4
v2 −
R2
R1
v1
or
vo =
R2
R1
(1 + R1/R2)
(1 + R3/R4)
v2 −
R2
R1
v1 (5.18)
Since a difference amplifier must reject a signal common to the two inputs,
the amplifier must have the property that vo = 0 when v1 = v2. This
property exists when
R1
R2
=
R3
R4
(5.19)
Thus, when the op amp circuit is a difference amplifier, Eq. (5.18) be-
comes
vo =
R2
R1
(v2 − v1) (5.20)
If R2 = R1 and R3 = R4, the difference amplifier becomes a subtractor,
with the output
vo = v2 − v1 (5.21)
E X A M P L E 5 . 7
Design an op amp circuit with inputs v1 and v2 such that vo = −5v1 +3v2.
Solution:
The circuit requires that
vo = 3v2 − 5v1 (5.7.1)
This circuit can be realized in two ways.
CHAPTER 5 Operational Amplifiers 179
DESIGN 1 If we desire to use only one op amp, we can use the op
amp circuit of Fig. 5.24. Comparing Eq. (5.7.1) with Eq. (5.18),
R2
R1
= 5 ⇒ R2 = 5R1 (5.7.2)
Also,
5
(1 + R1/R2)
(1 + R3/R4)
= 3 ⇒
6
5
1 + R3/R4
=
3
5
or
2 = 1 +
R3
R4
⇒ R3 = R4 (5.7.3)
If we choose R1 = 10 k and R3 = 20 k, then R2 = 50 k and
R4 = 20 k.
vo
5R1
R6
R1
va +
−
+
−
v1
v2
3R3
5R1
Figure5.25 For Example 5.7.
DESIGN 2 If we desire to use more than one op amp, we may cascade
an inverting amplifier and a two-input inverting summer, as shown in Fig.
5.25. For the summer,
vo = −va − 5v1 (5.7.4)
and for the inverter,
va = −3v2 (5.7.5)
Combining Eqs. (5.7.4) and (5.7.5) gives
vo = 3v2 − 5v1
which is the desired result. In Fig. 5.25, we may select R1 = 10 k and
R2 = 20 k or R1 = R2 = 10 k.
P R A C T I C E P R O B L E M 5 . 7
Design a difference amplifier with gain 4.
Answer: Typical: R1 = R3 = 10 k, R2 = R4 = 40 k.
E X A M P L E 5 . 8
An instrumentation amplifier shown in Fig. 5.26 is an amplifier of low-
level signals used in process control or measurement applications and
commercially available in single-package units. Show that
vo =
R2
R1

1 +
2R3
R4

(v2 − v1)
Solution:
We recognize that the amplifier A3 in Fig. 5.26 is a difference amplifier.
Thus, from Eq. (5.20),
vo =
R2
R1
(vo2 − vo1) (5.8.1)
180 PART 1 DC Circuits
vo1
vo2
v1
v2
0
0
vo
+
−
+
−
+
−
A1
A2
A3
R3
R4
R1
R1
R2
R2
R3
va
vb
+
−
+
−
i
Figure5.26 Instrumentation amplifier; for Example 5.8.
Since the op amps A1 and A2 draw no current, current i flows through
the three resistors as though they were in series. Hence,
vo1 − vo2 = i(R3 + R4 + R3) = i(2R3 + R4) (5.8.2)
But
i =
va − vb
R4
and va = v1, vb = v2. Therefore,
i =
v1 − v2
R4
(5.8.3)
Inserting Eqs. (5.8.2) and (5.8.3) into Eq. (5.8.1) gives
vo =
R2
R1

1 +
2R3
R4

(v2 − v1)
as required. We will discuss the instrumentation amplifier in detail in
Section 5.10.
P R A C T I C E P R O B L E M 5 . 8
Obtain io in the instrumentation amplifier circuit of Fig. 5.27.
+
−
+
−
+
−
io
20 kΩ
20 kΩ
40 kΩ
10 kΩ
40 kΩ
8.00 V
8.01 V
Figure5.27 Instrumentation amplifier; for Practice Prob. 5.8.
Answer: 2 µA.
CHAPTER 5 Operational Amplifiers 181
5.8 CASCADED OP AMP CIRCUITS
As we know, op amp circuits are modules or building blocks for designing
complex circuits. It is often necessary in practical applications to connect
op amp circuits in cascade (i.e., head to tail) to achieve a large overall
gain. In general, two circuits are cascaded when they are connected in
tandem, one behind another in a single file.
A cascade connection is a head-to-tail arrangement of two or more op amp circuits
such that the output of one is the input of the next.
When op amp circuits are cascaded, each circuit in the string is
called a stage; the original input signal is increased by the gain of the
individual stage. Op amp circuits have the advantage that they can be
cascaded without changing their input-output relationships. This is due
to the fact that each (ideal) op amp circuit has infinite input resistance and
zero output resistance. Figure 5.28 displays a block diagram representa-
tion of three op amp circuits in cascade. Since the output of one stage is
the input to the next stage, the overall gain of the cascade connection is
the product of the gains of the individual op amp circuits, or
A = A1A2A3 (5.22)
Although the cascade connection does not affect the op amp input-output
relationships, care must be exercised in the design of an actual op amp
circuit to ensure that the load due to the next stage in the cascade does
not saturate the op amp.
Stage 1
v2 = A1v1
+
−
v1
+
−
+
−
A1
Stage 2
A2
v3 = A2v2 vo = A3v3
+
−
Stage 3
A3
Figure5.28 A three-stage cascaded connection.
E X A M P L E 5 . 9
Find vo and io in the circuit in Fig. 5.29.
10 kΩ
12 kΩ
4 kΩ
20 mV
vo
+
−
+
−
3 kΩ
a
b
io
+
−
+
−
Figure5.29 For Example 5.9.
Solution:
This circuit consists of two noninverting amplifiers cascaded. At the
output of the first op amp,
va =

1 +
12
3

(20) = 100 mV
At the output of the second op amp,
vo =

1 +
10
4

va = (1 + 2.5)100 = 350 mV
182 PART 1 DC Circuits
The required current io is the current through the 10-k resistor.
io =
vo − vb
10
mA
But vb = va = 100 mV. Hence,
io =
(350 − 100) × 10−3
10 × 103
= 25 µA
P R A C T I C E P R O B L E M 5 . 9
Determine vo and io in the op amp circuit in Fig. 5.30.
6 kΩ
4 kΩ
4 V vo
+
−
+
−
io
+
−
+
−
Figure5.30 For Practice Prob. 5.9.
Answer: 10 V, 1 mA.
E X A M P L E 5 . 1 0
If v1 = 1 V and v2 = 2 V, find vo in the op amp circuit of Fig. 5.31.
+
−
+
−
+
−
A
B
C
5 kΩ
15 kΩ
v1
10 kΩ
2 kΩ
4 kΩ
8 kΩ
6 kΩ
v2
vo
a
b
Figure5.31 For Example 5.10.
Solution:
The circuit consists of two inverters A and B and a summer C as shown
in Fig. 5.31. We first find the outputs of the inverters.
va = −
6
2
(v1) = −3(1) = −3 V, vb = −
8
4
(v2) = −2(2) = −4 V
CHAPTER 5 Operational Amplifiers 183
These become the inputs to the summer so that the output is obtained as
vo = −

10
5
va +
10
15
vb

= −

2(−3) +
2
3
(−4)

= 8.333 V
P R A C T I C E P R O B L E M 5 . 1 0
If v1 = 2 V and v2 = 1.5 V, find vo in the op amp circuit of Fig. 5.32.
+
−
+
−
+
−
+
−
+
−
10 kΩ
v1
v2
vo
50 kΩ
20 kΩ
30 kΩ
60 kΩ
Figure5.32 Practice Prob. 5.10.
Answer: 9 V.
5.9 OP AMP CIRCUIT ANALYSIS WITH PSPICE
PSpice for Windows does not have a model for an ideal op amp, although
one may create one as a subcircuit using the Create Subcircuit line in the
Tools menu. Rather than creating an ideal op amp, we will use one of the
four nonideal, commercially available op amps supplied in the PSpice
library eval.slb. The op amp models have the part names LF411, LM111,
LM324, and uA471, as shown in Fig. 5.33. Each of them can be obtained
from Draw/Get New Part/libraries · · ·/eval.lib or by simply selecting
Draw/Get New Part and typing the part name in the PartName dialog
box, as usual. Note that each of them requires dc supplies, without which
the op amp will not work. The dc supplies should be connected as shown
in Fig. 5.3.
+
−
LM324
2
3
1
4 U1A
11
V+
V−
+
−
LM111
3
2
7
V+
V−
U2
8 5
6
1
4
G
BB ⁄S
(c) Five–connection
op amp subcircuit
(b) Op amp subcircuit
+
−
uA741
2
3
U3
4
V+
V−
+
−
LF411
2
3
6
7
5
1
U4
4
V+
V−
7
5
1
6
052
051
B2
B1
(d) Five–connection
op amp subcircuit
(a) JFET–input op amp
subcircuit
Figure5.33 Nonideal op amp model available in PSpice.
184 PART 1 DC Circuits
E X A M P L E 5 . 1 1
Use PSpice to solve the op amp circuit for Example 5.1.
Solution:
Using Schematics, we draw the circuit in Fig. 5.6(a) as shown in Fig.
5.34. Notice that the positive terminal of the voltage source vs is con-
nected to the inverting terminal (pin 2) via the 10-k resistor, while the
noninverting terminal (pin 3) is grounded as required in Fig. 5.6(a). Also,
notice how the op amp is powered; the positive power supply terminal
V+ (pin 7) is connected to a 15-V dc voltage source, while the negative
power supply terminal V− (pin 4) is connected to −15 V. Pins 1 and 5 are
left floating because they are used for offset null adjustment, which does
not concern us in this chapter. Besides adding the dc power supplies to
the original circuit in Fig. 5.6(a), we have also added pseudocomponents
VIEWPOINT and IPROBE to respectively measure the output voltage vo
at pin 6 and the required current i through the 20-k resistor.
+
−
uA741
2
3
U1
4
V+
V−
7
5
1
6
052
051
+
−
+
−
+
−
20 K
R2
1.999E–04
V3
–15 V 0
15 V
V2
10 K
R1
VS 2 V
0
–3.9983
Figure5.34 Schematic for Example 5.11.
After saving the schematic, we simulate the circuit by selecting
Analysis/Simulate and have the results displayed on VIEWPOINT and
IPROBE. From the results, the closed-loop gain is
vo
vs
=
−3.9983
2
= −1.99915
and i = 0.1999 mA, in agreement with the results obtained analytically
in Example 5.1.
P R A C T I C E P R O B L E M 5 . 1 1
Rework Practice Prob. 5.1 using PSpice.
Answer: 9.0027, 0.6502 mA.
CHAPTER 5 Operational Amplifiers 185
†5.10 APPLICATIONS
The op amp is a fundamental building block in modern electronic instru-
mentation. It is used extensively in many devices, along with resistors
and other passive elements. Its numerous practical applications include
instrumentation amplifiers, digital-to-analog converters, analog comput-
ers, level shifters, filters, calibration circuits, inverters, summers, inte-
grators, differentiators, subtractors, logarithmic amplifiers, comparators,
gyrators, oscillators, rectifiers, regulators, voltage-to-current converters,
current-to-voltage converters, and clippers. Some of these we have al-
ready considered. We will consider two more applications here: the
digital-to-analog converter and the instrumentation amplifier.
5.10.1 Digital-to-Analog Converter
The digital-to-analog converter (DAC) transforms digital signals into ana-
logform. Atypicalexampleofafour-bitDACisillustratedinFig.5.35(a).
The four-bit DAC can be realized in many ways. A simple realization is
the binary weighted ladder, shown in Fig. 5.35(b). The bits are weights
according to the magnitude of their place value, by descending value of
Rf /Rn so that each lesser bit has half the weight of the next higher. This
is obviously an inverting summing amplifier. The output is related to the
inputs as shown in Eq. (5.15). Thus,
−Vo =
Rf
R1
V1 +
Rf
R2
V2 +
Rf
R3
V3 +
Rf
R4
V4 (5.23)
Input V1 is called the most significant bit (MSB), while input V4 is the
least significant bit (LSB). Each of the four binary inputs V1, . . . , V4 can
assume only two voltage levels: 0 or 1 V. By using the proper input
and feedback resistor values, the DAC provides a single output that is
proportional to the inputs.
Analog
output
Digital
input
(0000–1111)
Four-bit
DAC
(a)
+
−
V1 V2 V3 V4
R1 R2 R3 R4
Rf
V
o
MSB LSB
(b)
Figure 5.35 Four-bit DAC: (a) block diagram,
(b) binary weighted ladder type.
In practice, the voltage levels may be typically 0
and ± 5 V.
E X A M P L E 5 . 1 2
In the op amp circuit of Fig. 5.35(b), let Rf = 10 k, R1 = 10 k,
R2 = 20 k, R3 = 40 k, and R4 = 80 k. Obtain the analog output
for binary inputs [0000], [0001], [0010], . . . , [1111].
Solution:
Substituting the given values of the input and feedback resistors in Eq.
(5.23) gives
−Vo =
Rf
R1
V1 +
Rf
R2
V2 +
Rf
R3
V3 +
Rf
R4
V4
= V1 + 0.5V2 + 0.25V3 + 0.125V4
Using this equation, a digital input [V1V2V3V4] = [0000] produces an
analog output of −Vo = 0 V; [V1V2V3V4] = [0001] gives −Vo =
0.125 V. Similarly,
186 PART 1 DC Circuits
[V1V2V3V4] = [0010] ⇒ −Vo = 0.25 V
[V1V2V3V4] = [0011] ⇒ −Vo = 0.25 + 0.125 = 0.375 V
[V1V2V3V4] = [0100] ⇒ −Vo = 0.5 V
.
.
.
[V1V2V3V4] = [1111] ⇒ −Vo = 1 + 0.5 + 0.25 + 0.125
= 1.875 V
Table 5.2 summarizes the result of the digital-to-analog conversion. Note
that we have assumed that each bit has a value of 0.125 V. Thus, in
this system, we cannot represent a voltage between 1.000 and 1.125,
for example. This lack of resolution is a major limitation of digital-to-
analog conversions. For greater accuracy, a word representation with a
greater number of bits is required. Even then a digital representation of
an analog voltage is never exact. In spite of this inexact representation,
digital representation has been used to accomplish remarkable things such
as audio CDs and digital photography.
TABLE 5.2 Input and output values
of the four-bit DAC.
Binary input Output
[V1V2V3V4] Decimal value −Vo
0000 0 0
0001 1 0.125
0010 2 0.25
0011 3 0.375
0100 4 0.5
0101 5 0.625
0110 6 0.75
0111 7 0.875
1000 8 1.0
1001 9 0.125
1010 10 0.25
1011 11 1.375
1011 12 1.5
1100 13 1.625
1101 14 1.75
1111 15 1.875
P R A C T I C E P R O B L E M 5 . 1 2
A three-bit DAC is shown in Fig. 5.36.
(a) Determine |Vo| for [V1V2V3] = [010].
(b) Find |Vo| if [V1V2V3] = [110].
CHAPTER 5 Operational Amplifiers 187
(c) If |Vo| = 1.25 V is desired, what should be [V1V2V3] ?
(d) To get |Vo| = 1.75 V, what should be [V1V2V3] ?
+
−
10 kΩ
20 kΩ
40 kΩ
10 kΩ
v1
v2
v3
vo
Figure 5.36 Three-bit DAC; for Practice
Prob. 5.12.
Answer: 0.5 V, 1.5 V, [101], [111].
5.10.2 Instrumentation Amplifiers
One of the most useful and versatile op amp circuits for precision mea-
surement and process control is the instrumentation amplifier (IA), so
called because of its widespread use in measurement systems. Typical
applications of IAs include isolation amplifiers, thermocouple amplifiers,
and data acquisition systems.
The instrumentation amplifier is an extension of the difference am-
plifier in that it amplifies the difference between its input signals. As
shown in Fig. 5.26 (see Example 5.8), an instrumentation amplifier typ-
ically consists of three op amps and seven resistors. For convenience,
the amplifier is shown again in Fig. 5.37(a), where the resistors are made
equal except for the external gain-setting resistor RG, connected between
the gain set terminals. Figure 5.37(b) shows its schematic symbol. Ex-
ample 5.8 showed that
vo = Av(v2 − v1) (5.24)
where the voltage gain is
Av = 1 +
2R
RG
(5.25)
As shown in Fig. 5.38, the instrumentation amplifier amplifies small dif-
ferential signal voltages superimposed on larger common-mode voltages.
Since the common-mode voltages are equal, they cancel each other.
The IA has three major characteristics:
+
−
+
−
+
−
1
2
3
R
R
R
R
R
R
RG
v1
v2
vo
Inverted input
Gain set
Gain set
Noninverting input
Output
(a) (b)
+
−
Figure5.37 (a) The instrumentation amplifier with an external resistance to adjust the gain, (b) schematic diagram.
188 PART 1 DC Circuits
+
−
RG
Small differential signals riding on larger
common–mode signals
Instrumentation amplifier Amplified differential signal,
No common-mode signal
Figure 5.38 The IA rejects common voltages but amplifies small signal voltages.
(Source: T. L. Floyd, Electronic Devices, 2nd ed., Englewood Cliffs, NJ: Prentice Hall, 1996, p. 795.)
1. The voltage gain is adjusted by one external resistor RG.
2. The input impedance of both inputs is very high and does not
vary as the gain is adjusted.
3. The output vo depends on the difference between the inputs v1
and v2, not on the voltage common to them (common-mode
voltage).
Due to the widespread use of IAs, manufacturers have developed
these amplifiers on single-package units. A typical example is the
LH0036, developed by National Semiconductor. The gain can be var-
ied from 1 to 1,000 by an external resistor whose value may vary from
100  to 10 k.
E X A M P L E 5 . 1 3
In Fig. 5.37, let R = 10 k, v1 = 2.011 V, and v2 = 2.017 V. If RG is ad-
justed to 500 , determine: (a) the voltage gain, (b) the output voltage vo.
Solution:
(a) The voltage gain is
Av = 1 +
2R
RG
= 1 +
2 × 10,000
500
= 41
(b) The output voltage is
vo = Av(v2 − v1) = 41(2.017 − 2.011) = 41(6) mV = 246 mV
P R A C T I C E P R O B L E M 5 . 1 3
Determine the value of the external gain-setting resistor RG required for
the IA in Fig. 5.37 to produce a gain of 142 when R = 25 k.
Answer: 354.6 .
5.11 SUMMARY
1. The op amp is a high-gain amplifier that has high input resistance
and low output resistance.
CHAPTER 5 Operational Amplifiers 189
2. Table 5.3 summarizes the op amp circuits considered in this
chapter. The expression for the gain of each amplifier circuit holds
whether the inputs are dc, ac, or time-varying in general.
TABLE 5.3 Summary of basic op amp circuits.
Op amp circuit Name/output-input relationship
+
−
R2
R1
vi
vo
Inverting amplifier
vo = −
R2
R1
vi
vo
R1
+
−
vi
R2 Noninverting amplifier
vo =

1 +
R2
R1

vi
+
− vo
vi
Voltage follower
vo = vi
v1
v2
v3
vo
R1
R2
R3
Rf
+
−
Summer
vo = −

Rf
R1
v1 +
Rf
R2
v2 +
Rf
R3
v3

+
−
R2
R1
v1
R1 R2
v2
vo
Difference amplifier
vo =
R2
R1
(v2 − v1)
3. An ideal op amp has an infinite input resistance, a zero output
resistance, and an infinite gain.
4. For an ideal op amp, the current into each of its two input terminals
is zero, and the voltage across its input terminals is negligibly small.
5. In an inverting amplifier, the output voltage is a negative multiple of
the input.
6. In a noninverting amplifier, the output is a positive multiple of the
input.
7. In a voltage follower, the output follows the input.
8. In a summing amplifier, the output is the weighted sum of the
inputs.
190 PART 1 DC Circuits
9. In a difference amplifier, the output is proportional to the difference
of the two inputs.
10. Op amp circuits may be cascaded without changing their
input-output relationships.
11. PSpice can be used to analyze an op amp circuit.
12. Typical applications of the op amp considered in this chapter
include the digital-to-analog converter and the instrumentation
amplifier.
REVIEW QUESTIONS
5.1 The two input terminals of an op amp are labeled as:
(a) high and low.
(b) positive and negative.
(c) inverting and noninverting.
(d) differential and nondifferential.
5.2 For an ideal op amp, which of the following
statements are not true?
(a) The differential voltage across the input
terminals is zero.
(b) The current into the input terminals is zero.
(c) The current from the output terminal is zero.
(d) The input resistance is zero.
(e) The output resistance is zero.
5.3 For the circuit in Fig. 5.39, voltage vo is:
(a) −6 V (b) −5 V
(c) −1.2 V (d) −0.2 V
+
−
+
−
2 kΩ
ix
vo
1 V
10 kΩ
3 kΩ
+
−
Figure 5.39 For Reivew Questions 5.3 and 5.4.
5.4 For the circuit in Fig. 5.39, current ix is:
(a) 0.6 A (b) 0.5 A
(c) 0.2 A (d) 1/12 A
5.5 If vs = 0 in the circuit of Fig. 5.40, current io is:
(a) −10 A (b) −2.5 A
(c) 10/12 A (d) 10/14 A
+
−
+
− +
−
4 kΩ
io
vo
10 V
8 kΩ
2 kΩ
+
−
vs
a
Figure 5.40 For Review Questions 5.5 to 5.7.
5.6 If vs = 8 V in the circuit of Fig. 5.40, the output
voltage is:
(a) −44 V (b) −8 V
(c) 4 V (d) 7 V
5.7 Refer to Fig. 5.40. If vs = 8 V, voltage va is:
(a) −8 V (b) 0 V
(c) 10/3 V (d) 8 V
5.8 The power absorbed by the 4-k resistor in Fig.
5.41 is:
(a) 9 mW (b) 4 mW
(c) 2 mW (d) 1 mW
+
−
6 V 2 kΩ vo
+
−
4 kΩ
+
−
Figure 5.41 For Review Question 5.8.
5.9 Which of these amplifiers is used in a
digital-to-analog converter?
(a) noninverter (b) voltage follower
(c) summer (d) difference amplifier
CHAPTER 5 Operational Amplifiers 191
5.10 Difference amplifiers are used in:
(a) instrumentation amplifiers
(b) voltage followers
(c) voltage regulators
(d) buffers
(e) summing amplifiers
(f) subtracting amplifiers
Answers: 5.1c, 5.2c,d, 5.3b, 5.4b, 5.5a, 5.6c, 5.7d, 5.8b, 5.9c, 5.10a,f.
PROBLEMS
Section 5.2 Operational Amplifiers
5.1 The equivalent model of a certain op amp is shown
in Fig. 5.42. Determine:
(a) the input resistance
(b) the output resistance
(c) the voltage gain in dB.
60 Ω
+
−
vd +
−
1.5 MΩ 8 × 10vd
Figure 5.42 For Prob. 5.1.
5.2 The open-loop gain of an op amp is 100,000.
Calculate the output voltage when there are inputs of
+10 µV on the inverting terminal and + 20 µV on
the noninverting terminal.
5.3 Determine the output voltage when −20 µV is
applied to the inverting terminal of an op amp and
+30 µV to its noninverting terminal. Assume that
the op amp has an open-loop gain of 200,000.
5.4 The output voltage of an op amp is −4 V when the
noninverting input is 1 mV. If the open-loop gain of
the op amp is 2 × 106
, what is the inverting input?
5.5 For the op amp circuit of Fig. 5.43, the op amp has
an open-loop gain of 100,000, an input resistance of
10 k, and an output resistance of 100 . Find the
voltage gain vo/vi using the nonideal model of the
op amp.
+
−
+
−
vo
vi
+
−
Figure 5.43 For Prob. 5.5.
5.6 Using the same parameters for the 741 op amp in
Example 5.1, find vo in the op amp circuit of Fig.
5.44.
+
−
+ −
vo
741
1 mV
Figure 5.44 For Prob. 5.6.
5.7 The op amp in Fig. 5.45 has Ri = 100 k,
Ro = 100 , A = 100,000. Find the differential
voltage vd and the output voltage vo.
+
−
+
−
10 kΩ 100 kΩ
vo
vd
+
−
1 mV
+
−
Figure 5.45 For Prob. 5.7.
Section 5.3 Ideal Op Amp
5.8 Obtain vo for each of the op amp circuits in Fig. 5.46.
2 kΩ
(a)
vo
+
−
1 mA
2 V
10 kΩ
(b)
vo
+
−
1 V 2 kΩ
+
−
+
−
+
− +
−
Figure 5.46 For Prob. 5.8.
5.9 Determine vo for each of the op amp circuits in Fig.
5.47.
192 PART 1 DC Circuits
+
−
+
− 4 V
2 kΩ
vo
1 mA
+
− 1 V
2 kΩ vo
+
−
+
−
3 V
+
−
+
−
Figure 5.47 For Prob. 5.9.
5.10 Find the gain vo/vs of the circuit in Fig. 5.48.
10 kΩ
10 kΩ
vo
+
−
+
−
20 kΩ
+
−
vs
Figure 5.48 For Prob. 5.10.
5.11 Find vo and io in the circuit in Fig. 5.49.
+
−
5 kΩ
2 kΩ
+
−
io
+
−
8 kΩ
10 kΩ 4 kΩ
3 V
vo
Figure 5.49 For Prob. 5.11.
5.12 Refer to the op amp circuit in Fig. 5.50. Determine
the power supplied by the voltage source.
1.2 V
2 kΩ
4 kΩ
+
−
vo
1 kΩ
4 kΩ
+
−
Figure 5.50 For Prob. 5.12.
5.13 Find vo and io in the circuit of Fig. 5.51.
50 kΩ
vo
+
−
+
−
1 V
100 kΩ
90 kΩ
10 kΩ
io
10 kΩ
+
−
Figure 5.51 For Prob. 5.13.
5.14 Determine the output voltage vo in the circuit of Fig.
5.52.
5 kΩ vo
+
−
2 mA
20 kΩ
10 kΩ
10 kΩ
+
−
Figure 5.52 For Prob. 5.14.
Section 5.4 Inverting Amplifier
5.15 (a) For the circuit shown in Fig. 5.53, show that the
gain is
vo
vi
= −
1
R

R1 + R2 +
R1R2
R3

(b) Evaluate the gain when R = 10 k,
R1 = 100 k, R2 = 50 k, R3 = 25 k.
CHAPTER 5 Operational Amplifiers 193
vo
vi
R
R1 R2
R3
+
−
Figure 5.53 For Prob. 5.15.
5.16 Calculate the gain vo/vi when the switch in Fig.
5.54 is in:
(a) position 1 (b) position 2 (c) position 3
10 kΩ
1
vo
+
−
5 kΩ
+
−
vi
2 MΩ
80 kΩ
12 kΩ
2
3
+
−
Figure 5.54 For Prob. 5.16.
5.17 Calculate the gain vo/vi of the op amp circuit in Fig.
5.55.
20 kΩ vo
+
−
+
−
vi
10 kΩ 50 kΩ
1 MΩ
+
−
Figure 5.55 For Prob. 5.17.
5.18 Determine io in the circuit of Fig. 5.56.
1 V
5 kΩ
4 kΩ
io
+
−
4 kΩ 10 kΩ
2 kΩ
+
−
Figure 5.56 For Prob. 5.18.
5.19 In the circuit in Fig. 5.57, calculate vo if vs = 0.
+
− +
−
9 V
4 kΩ 4 kΩ
2 kΩ
8 kΩ
vo
+
−
vs
+
−
Figure 5.57 For Prob. 5.19.
5.20 Repeat the previous problem if vs = 3 V.
5.21 Design an inverting amplifier with a gain of −15.
Section 5.5 Noninverting Amplifier
5.22 Find va and vo in the op amp circuit of Fig. 5.58.
2 V
vo
+ −
3 V
va
+
−
Figure 5.58 For Prob. 5.22.
5.23 Refer to Fig. 5.59.
(a) Determine the overall gain vo/vi of the circuit.
(b) What value of vi will result in
vo = 15 cos 120πt?
+
− 2 kΩ
+
−
1 MΩ
vo
8 kΩ
20 kΩ
vi
+
−
Figure 5.59 For Prob. 5.23.
194 PART 1 DC Circuits
5.24 Find io in the op amp circuit of Fig. 5.60.
+
− 0.4 V 20 kΩ
10 kΩ
io
50 kΩ
+
−
Figure 5.60 For Prob. 5.24.
5.25 In the circuit shown in Fig. 5.61, find ix and the
power absorbed by the 20- resistor.
+
−
1.2 V 30 kΩ 20 kΩ
ix
60 kΩ
+
−
Figure 5.61 For Prob. 5.25.
5.26 For the circuit in Fig. 5.62, find ix.
+
−
+
−
6 kΩ
6 kΩ
3 kΩ
4 mA
vo
12 kΩ
ix
Figure 5.62 For Prob. 5.26.
5.27 Calculate ix and vo in the circuit of Fig. 5.63. Find
the power dissipated by the 60-k resistor.
+
−
vo
+
− 30 kΩ
60 kΩ
ix
4 mV
20 kΩ
50 kΩ
10 kΩ
+
−
Figure 5.63 For Prob. 5.27.
5.28 Refer to the op amp circuit in Fig. 5.64. Calculate ix
and the power dissipated by the 3-k resistor.
+
−
4 kΩ 2 kΩ
ix
1 mA 3 kΩ
1 kΩ
Figure 5.64 For Prob. 5.28.
5.29 Design a noninverting amplifier with a gain of 10.
Section 5.6 Summing Amplifier
5.30 Determine the output of the summing amplifier in
Fig. 5.65.
30 kΩ
10 kΩ
1 V
20 kΩ
2 V
30 kΩ
3 V
+ −
+
−
+
−
vo
+
−
+
−
Figure 5.65 For Prob. 5.30.
5.31 Calculate the output voltage due to the summing
amplifier shown in Fig. 5.66.
50 kΩ
25 kΩ
10 mV
20 kΩ
20 mV
50 kΩ
100 mV
+ −
+ −
10 kΩ
50 mV
+
−
+
−
vo
+
−
+
−
Figure 5.66 For Prob. 5.31.
5.32 An averaging amplifier is a summer that provides an
output equal to the average of the inputs. By using
CHAPTER 5 Operational Amplifiers 195
proper input and feedback resistor values, one can
get
−vout = 1
4
(v1 + v2 + v3 + v4)
Using a feedback resistor of 10 k, design an
averaging amplifier with four inputs.
5.33 A four-input summing amplifier has
R1 = R2 = R3 = R4 = 12 k. What value of
feedback resistor is needed to make it an averaging
amplifier?
5.34 Show that the output voltage vo of the circuit in Fig.
5.67 is
vo =
(R3 + R4)
R3(R1 + R2)
(R2v1 + R1v2)
R4
R3
R1
R2
vo
v1
v2
+
−
Figure 5.67 For Prob. 5.34.
5.35 Design an op amp circuit to perform the following
operation:
vo = 3v1 − 2v2
All resistances must be ≤ 100 k.
5.36 Using only two op amps, design a circuit to solve
−vout =
v1 − v2
3
+
v3
2
Section 5.7 Difference Amplifier
5.37 Find vo and io in the differential amplifier of Fig.
5.68.
10 V
io
2 kΩ
1 kΩ
4 kΩ
3 kΩ 5 kΩ vo
+
−
+
−
8 V
+
−
+
−
Figure 5.68 For Prob. 5.37.
5.38 The circuit in Fig. 5.69 is a differential amplifier
driven by a bridge. Find vo.
20 kΩ
80 kΩ
20 kΩ
80 kΩ
vo
+ 5 mV
40 kΩ
10 kΩ
60 kΩ
30 kΩ
+
−
Figure 5.69 For Prob. 5.38.
5.39 Design a difference amplifier to have a gain of 2 and
a common mode input resistance of 10 k at each
input.
5.40 Design a circuit to amplify the difference between
two inputs by 2.
(a) Use only one op amp.
(b) Use two op amps.
5.41 Using two op amps, design a subtractor.
5.42
∗
The ordinary difference amplifier for fixed-gain
operation is shown in Fig. 5.70(a). It is simple and
reliable unless gain is made variable. One way of
providing gain adjustment without losing simplicity
and accuracy is to use the circuit in Fig. 5.70(b).
Another way is to use the circuit in Fig. 5.70(c).
Show that:
(a) for the circuit in Fig. 5.70(a),
vo
vi
=
R2
R1
(b) for the circuit in Fig. 5.70(b),
vo
vi
=
R2
R1
1
1 +
R1
2RG
(c) for the circuit in Fig. 5.70(c),
vo
vi
=
R2
R1

1 +
R2
2RG

∗An asterisk indicates a challenging problem.
196 PART 1 DC Circuits
R1
R2
R2
R2
R1
(a)
vo
R2
RG
(b)
R1
2
R1
2
R2
2
R2
2
R2
2
R2
2
R1
2
R1
2
vi
(c)
R1
R1
RG
+
−
+
−
+
−
+
−
vi
+
−
vi
+
− +
−
vo
+
−
vo
+
−
Figure 5.70 For Prob. 5.42.
Section 5.8 Cascaded Op Amp Circuits
5.43 The individual gains of the stages in a multistage
amplifier are shown in Fig. 5.71.
(a) Calculate the overall voltage gain vo/vi.
(b) Find the voltage gain that would be needed in a
fourth stage which would make the overall gain
to be 60 dB when added.
vo
vi –20 –12.5 +0.8
Figure 5.71 For Prob. 5.43.
5.44 In a certain electronic device, a three-stage amplifier
is desired, whose overall voltage gain is 42 dB. The
individual voltage gains of the first two stages are to
be equal, while the gain of the third is to be
one-fourth of each of the first two. Calculate the
voltage gain of each.
5.45 Refer to the circuit in Fig. 5.72. Calculate io if:
(a) vs = 12 mV (b) vs = 10 cos 377t mV.
vs
+
−
2 kΩ
io
12 kΩ
6 kΩ
12 kΩ
4 kΩ
+
−
+
−
Figure 5.72 For Prob. 5.45.
5.46 Calculate io in the op amp circuit of Fig. 5.73.
0.6 V +
− 4 kΩ
io
1 kΩ
10 kΩ
5 kΩ
2 kΩ
+
−
+
−
3 kΩ
Figure 5.73 For Prob. 5.46.
5.47 Find the voltage gain vo/vs of the circuit in Fig. 5.74.
vs
+
−
10 kΩ
5 kΩ
20 kΩ
vo
+
−
+
− +
−
Figure 5.74 For Prob. 5.47.
5.48 Calculate the current gain io/is of the op amp circuit
in Fig. 5.75.
CHAPTER 5 Operational Amplifiers 197
is
10 kΩ
3 kΩ
4 kΩ
+
−
+
−
5 kΩ
3 kΩ 2 kΩ
io
Figure 5.75 For Prob. 5.48.
5.49 Find vo in terms of v1 and v2 in the circuit in Fig.
5.76.
R3
R2
R1
R4
+
−
+
−
v1
v2
vo
R5
Figure 5.76 For Prob. 5.49.
5.50 Obtain the closed-loop voltage gain vo/vi of the
circuit in Fig. 5.77.
Rf
R2
R1
+
−
+
−
vi
R3
vo
R4
+
−
+
−
Figure 5.77 For Prob. 5.50.
5.51 Determine the gain vo/vi of the circuit in Fig. 5.78.
+
−
+ +
−
−
R3
R2
R1
R4
R5
R6
vo
vi
+
−
Figure 5.78 For Prob. 5.51.
5.52 For the circuit in Fig. 5.79, find vo.
+
−
+
−
+
−
+
−
+
−
25 kΩ
10 kΩ
40 kΩ 100 kΩ
20 kΩ
6 V
4 V
2 V
20 kΩ
vo
+
−
Figure 5.79 For Prob. 5.52.
5.53 Obtain the output vo in the circuit of Fig. 5.80.
+
−
+
−
+
−
80 kΩ 80 kΩ
20 kΩ
0.4 V
40 kΩ
20 kΩ
vo
+
−
+
−
0.2 V
+
−
Figure 5.80 For Prob. 5.53.
5.54 Find vo in the circuit in Fig. 5.81, assuming that
Rf = ∞ (open circuit).
198 PART 1 DC Circuits
10 mV +
−
15 kΩ
6 kΩ
5 kΩ
Rf
+
−
+
−
1 kΩ
2 kΩ
+
−
vo
Figure 5.81 For Probs. 5.54 and 5.55.
5.55 Repeat the previous problem if Rf = 10 k.
5.56 Determine vo in the op amp circuit of Fig. 5.82.
+
−
30 kΩ
A
C
40 kΩ
10 kΩ
1 V
20 kΩ
60 kΩ
+
−
10 kΩ
2 V
+
−
20 kΩ
B
3 V
+
−
10 kΩ
4 V
10 kΩ
vo
+
−
+
−
+
−
10 kΩ
Figure 5.82 For Prob. 5.56.
5.57 Find the load voltage vL in the circuit of Fig. 5.83.
+
−
+
−
+
−
100 kΩ 250 kΩ
0.4 V 2 kΩ
+
−
+
−
+
−
vL
20 kΩ
Figure 5.83 For Prob. 5.57.
5.58 Determine the load voltage vL in the circuit of Fig.
5.84.
50 kΩ
10 kΩ
5 kΩ
1.8 V
4 kΩ vL
+
−
+
−
+
−
+
−
Figure 5.84 For Prob. 5.58.
5.59 Find io in the op amp circuit of Fig. 5.85.
+
−
+
−
+
−
100 kΩ 32 kΩ
10 kΩ
20 kΩ
1.6 kΩ
0.6 V
+
−
0.4 V
+
−
+
−
io
Figure 5.85 For Prob. 5.59.
Section 5.9 Op Amp Circuit Analysis with
PSpice
5.60 Rework Example 5.11 using the nonideal op amp
LM324 instead of uA741.
5.61 Solve Prob. 5.18 using PSpice and op amp uA741.
5.62 Solve Prob. 5.38 using PSpice and op amp LM324.
5.63 Use PSpice to obtain vo in the circuit of Fig. 5.86.
+
−
20 kΩ 30 kΩ
10 kΩ
1 V +
−
+
−
+
−
40 kΩ
2 V +
−
+
−
vo
+
−
Figure 5.86 For Prob. 5.63.
5.64 Determine vo in the op amp circuit of Fig. 5.87
using PSpice.
CHAPTER 5 Operational Amplifiers 199
vo
+
−
20 kΩ
5 V
1 V
10 kΩ
+
−
+
−
20 kΩ 10 kΩ 40 kΩ
+
−
100 kΩ
+
−
Figure 5.87 For Prob. 5.64.
5.65 Use PSpice to solve Prob. 5.56, assuming that the op
amps are uA741.
5.66 Use PSpice to verify the results in Example 5.9.
Assume nonideal op amps LM324.
Section 5.10 Applications
5.67 A five-bit DAC covers a voltage range of 0 to 7.75 V.
Calculate how much voltage each bit is worth.
5.68 Design a six-bit digital-to-analog converter.
(a) If |Vo| = 1.1875 V is desired, what should
[V1V2V3V4V5V6] be?
(b) Calculate |Vo| if [V1V2V3V4V5V6] = [011011].
(c) What is the maximum value |Vo| can assume?
5.69
∗
A four-bit R-2R ladder DAC is presented in Fig. 5.88.
(a) Show that the output voltage is given by
−Vo = Rf

V1
2R
+
V2
4R
+
V3
8R
+
V4
16R

(b) If Rf = 12 k and R = 10 k, find |Vo| for
[V1V2V3V4] = [1011] and [V1V2V3V4] = [0101].
R
R
R
R
V
o
+
−
V1
V2
V3
V4
2R
2R
2R
2R
Rf
Figure 5.88 For Prob. 5.69.
5.70 If RG = 100  and R = 20 k, calculate the
voltage gain of the IA in Fig. 5.37.
5.71 Assuming a gain of 200 for an IA, find its output
voltage for:
(a) v1 = 0.402 V and v2 = 0.386 V
(b) v1 = 1.002 V and v2 = 1.011 V.
5.72 Figure 5.89 displays a two-op-amp instrumentation
amplifier. Derive an expression for vo in terms of v1
and v2. How can this amplifier be used as a
subtractor?
v2
v1
vo
R4
R3
R2
R1
+
−
+
−
Figure 5.89 For Prob. 5.72.
5.73
∗
Figure 5.90 shows an instrumentation amplifier
driven by a bridge. Obtain the gain vo/vi of the
amplifier.
25 kΩ
10 kΩ
10 kΩ
500 kΩ
vo
25 kΩ
2 kΩ
30 kΩ
20 kΩ
vi
80 kΩ
40 kΩ
500 kΩ
+
−
+
−
+
−
Figure 5.90 For Prob. 5.73.
200 PART 1 DC Circuits
COMPREHENSIVE PROBLEMS
5.74 A gain of 6 (+ or −, it does not matter) is required
in an audio system. Design an op amp circuit to
provide the gain with an input resistance of 2 k.
5.75 The op amp circuit in Fig. 5.91 is a current amplifier.
Find the current gain io/is of the amplifier.
+
−
20 kΩ
4 kΩ
5 kΩ 2 kΩ
is
io
Figure 5.91 For Prob. 5.75.
5.76 A noninverting current amplifier is portrayed in Fig.
5.92. Calculate the gain io/is. Take R1 = 8 k and
R2 = 1 k.
+
−
R1
R2
R2
is
io
Figure 5.92 For Prob. 5.76.
5.77 Refer to the bridge amplifier shown in Fig. 5.93.
Determine the voltage gain vo/vi.
+
−
60 kΩ
vi
vo
RL
+
−
+
−
+
−
50 kΩ
20 kΩ
30 kΩ
Figure 5.93 For Prob. 5.77.
5.78
∗
A voltage-to-current converter is shown in Fig. 5.94,
which means that iL = Avi if R1R2 = R3R4. Find
the constant term A.
+
−
R3
R1
iL
R2
vi
RL
R4
+
−
Figure 5.94 For Prob. 5.78.
201
C H A P T E R
CAPACITORS AND INDUCTORS
6
The important thing about a problem is not its solution, but the strength
we gain in finding the solution.
—Anonymous
Historical Profiles
Michael Faraday (1791–1867), an English chemist and physicist, was probably the
greatest experimentalist who ever lived.
Born near London, Faraday realized his boyhood dream by working with the
great chemist Sir Humphry Davy at the Royal Institution, where he worked for 54 years.
He made several contributions in all areas of physical science and coined such words
as electrolysis, anode, and cathode. His discovery of electromagnetic induction in
1831 was a major breakthrough in engineering because it provided a way of generating
electricity. The electric motor and generator operate on this principle. The unit of
capacitance, the farad, was named in his honor.
Joseph Henry (1797–1878), an American physicist, discovered inductance and con-
structed an electric motor.
Born in Albany, New York, Henry graduated from Albany Academy and taught
philosophy at Princeton University from 1832 to 1846. He was the first secretary of the
Smithsonian Institution. He conducted several experiments on electromagnetism and
developed powerful electromagnets that could lift objects weighing thousands of pounds.
Interestingly, Joseph Henry discovered electromagnetic induction before Faraday
but failed to publish his findings. The unit of inductance, the henry, was named after him.
202 PART 1 DC Circuits
6.1 INTRODUCTION
So far we have limited our study to resistive circuits. In this chapter, we
shall introduce two new and important passive linear circuit elements:
the capacitor and the inductor. Unlike resistors, which dissipate energy,
capacitors and inductors do not dissipate but store energy, which can be
retrieved at a later time. For this reason, capacitors and inductors are
called storage elements.
In contrast to a resistor, which spends or dis-
sipates energy irreversibly, an inductor or ca-
pacitor stores or releases energy (i.e., has a
memory).
The application of resistive circuits is quite limited. With the in-
troduction of capacitors and inductors in this chapter, we will be able to
analyze more important and practical circuits. Be assured that the circuit
analysis techniques covered in Chapters 3 and 4 are equally applicable to
circuits with capacitors and inductors.
We begin by introducing capacitors and describing how to combine
them in series or in parallel. Later, we do the same for inductors. As
typical applications, we explore how capacitors are combined with op
amps to form integrators, differentiators, and analog computers.
6.2 CAPACITORS
A capacitor is a passive element designed to store energy in its electric
field. Besides resistors, capacitors are the most common electrical com-
ponents. Capacitors are used extensively in electronics, communications,
computers, and power systems. For example, they are used in the tuning
circuits of radio receivers and as dynamic memory elements in computer
systems.
A capacitor is typically constructed as depicted in Fig. 6.1.
Metal plates,
each with area A
d
Dielectric with permittivity e
Figure6.1 A typical capacitor.
A capacitor consists of two conducting plates separated
by an insulator (or dielectric).
In many practical applications, the plates may be aluminum foil while the
dielectric may be air, ceramic, paper, or mica.
When a voltage source v is connected to the capacitor, as in Fig.
6.2, the source deposits a positive charge q on one plate and a negative
charge −q on the other. The capacitor is said to store the electric charge.
The amount of charge stored, represented by q, is directly proportional
to the applied voltage v so that
q = Cv (6.1)
where C, the constant of proportionality, is known as the capacitance
of the capacitor. The unit of capacitance is the farad (F), in honor of
the English physicist Michael Faraday (1791–1867). From Eq. (6.1), we
may derive the following definition.
Alternatively,capacitanceistheamountofcharge
stored per plate for a unit voltage difference in a
capacitor.
−
−
−
−q
+q
+
+
+
+
+
+
−
+
v
Figure6.2 A capacitor
with applied voltage v.
Capacitance is the ratio of the charge on one plate of a capacitor to the voltage
difference between the two plates, measured in farads (F).
Note from Eq. (6.1) that 1 farad = 1 coulomb/volt.
CHAPTER 6 Capacitors and Inductors 203
Although the capacitance C of a capacitor is the ratio of the charge
q per plate to the applied voltage v, it does not depend on q or v. It
depends on the physical dimensions of the capacitor. For example, for
the parallel-plate capacitor shown in Fig. 6.1, the capacitance is given by
C =
A
d
(6.2)
where A is the surface area of each plate, d is the distance between the
plates, and  is the permittivity of the dielectric material between the
plates. Although Eq. (6.2) applies to only parallel-plate capacitors, we
may infer from it that, in general, three factors determine the value of the
capacitance:
Capacitorvoltageratingandcapacitancearetyp-
ically inversely rated due to the relationships in
Eqs. (6.1) and (6.2). Arcing occurs if d is small
and V is high.
1. The surface area of the plates—the larger the area, the greater
the capacitance.
2. The spacing between the plates—the smaller the spacing, the
greater the capacitance.
3. The permittivity of the material—the higher the permittivity,
the greater the capacitance.
Capacitors are commercially available in different values and types.
Typically, capacitors have values in the picofarad (pF) to microfarad (µF)
range. They are described by the dielectric material they are made of and
by whether they are of fixed or variable type. Figure 6.3 shows the circuit
symbols for fixed and variable capacitors. Note that according to the
passive sign convention, current is considered to flow into the positive
terminal of the capacitor when the capacitor is being charged, and out of
the positive terminal when the capacitor is discharging.
i i
C
v
+ −
C
v
+ −
Figure6.3 Circuit symbols for capacitors:
(a) fixed capacitor, (b) variable capacitor.
Figure6.4showscommontypesoffixed-valuecapacitors. Polyester
capacitors are light in weight, stable, and their change with temperature is
predictable. Instead of polyester, other dielectric materials such as mica
and polystyrene may be used. Film capacitors are rolled and housed in
metal or plastic films. Electrolytic capacitors produce very high capaci-
tance. Figure 6.5 shows the most common types of variable capacitors.
The capacitance of a trimmer (or padder) capacitor or a glass piston capac-
itor is varied by turning the screw. The trimmer capacitor is often placed
in parallel with another capacitor so that the equivalent capacitance can
be varied slightly. The capacitance of the variable air capacitor (meshed
plates) is varied by turning the shaft. Variable capacitors are used in radio
(a) (b) (c)
Figure 6.4 Fixed capacitors: (a) polyester capacitor, (b) ceramic capacitor, (c) electrolytic capacitor.
(Courtesy of Tech America.)
204 PART 1 DC Circuits
receivers allowing one to tune to various stations. In addition, capacitors
are used to block dc, pass ac, shift phase, store energy, start motors, and
suppress noise.
(a)
(b)
Figure6.5 Variable capacitors:
(a) trimmer capacitor, (b) filmtrim
capacitor.
(Courtesy of Johanson.)
To obtain the current-voltage relationship of the capacitor, we take
the derivative of both sides of Eq. (6.1). Since
i =
dq
dt
(6.3)
differentiating both sides of Eq. (6.1) gives
i = C
dv
dt
(6.4)
According to Eq. (6.4), for a capacitor to carry
current, its voltage must vary with time. Hence,
for constant voltage, i = 0 .
This is the current-voltage relationship for a capacitor, assuming the pos-
itive sign convention. The relationship is illustrated in Fig. 6.6 for a
capacitor whose capacitance is independent of voltage. Capacitors that
satisfy Eq. (6.4) are said to be linear. For a nonlinear capacitor, the
plot of the current-voltage relationship is not a straight line. Although
some capacitors are nonlinear, most are linear. We will assume linear
capacitors in this book.
Slope = C
dv⁄dt
0
i
Figure6.6 Current-voltage
relationship of a capacitor.
The voltage-current relation of the capacitor can be obtained by
integrating both sides of Eq. (6.4). We get
v =
1
C
 t
−∞
i dt (6.5)
or
v =
1
C
 t
t0
i dt + v(t0) (6.6)
where v(t0) = q(t0)/C is the voltage across the capacitor at time t0.
Equation (6.6) shows that capacitor voltage depends on the past history
of the capacitor current. Hence, the capacitor has memory—a property
that is often exploited.
The instantaneous power delivered to the capacitor is
p = vi = Cv
dv
dt
(6.7)
The energy stored in the capacitor is therefore
w =
 t
−∞
p dt = C
 t
−∞
v
dv
dt
dt = C
 t
−∞
v dv =
1
2
Cv2




t
t=−∞
(6.8)
We note that v(−∞) = 0, because the capacitor was uncharged at t =
−∞. Thus,
w =
1
2
Cv2
(6.9)
Using Eq. (6.1), we may rewrite Eq. (6.9) as
w =
q2
2C
(6.10)
CHAPTER 6 Capacitors and Inductors 205
Equation (6.9) or (6.10) represents the energy stored in the electric field
that exists between the plates of the capacitor. This energy can be re-
trieved, since an ideal capacitor cannot dissipate energy. In fact, the word
capacitor is derived from this element’s capacity to store energy in an
electric field.
We should note the following important properties of a capacitor:
1. Note from Eq. (6.4) that when the voltage across a capacitor is
not changing with time (i.e., dc voltage), the current through
the capacitor is zero. Thus,
A capacitor is an open circuit to dc.
However, if a battery (dc voltage) is connected across a
capacitor, the capacitor charges.
2. The voltage on the capacitor must be continuous.
The voltage on a capacitor cannot change abruptly.
The capacitor resists an abrupt change in the voltage across it.
According to Eq. (6.4), a discontinuous change in voltage
requires an infinite current, which is physically impossible.
For example, the voltage across a capacitor may take the form
shown in Fig. 6.7(a), whereas it is not physically possible for
the capacitor voltage to take the form shown in Fig. 6.7(b)
because of the abrupt change. Conversely, the current through
a capacitor can change instantaneously.
3. The ideal capacitor does not dissipate energy. It takes power
from the circuit when storing energy in its field and returns
previously stored energy when delivering power to the circuit.
4. A real, nonideal capacitor has a parallel-model leakage
resistance, as shown in Fig. 6.8. The leakage resistance may be
as high as 100 M and can be neglected for most practical
applications. For this reason, we will assume ideal capacitors
in this book.
An alternative way of looking at this is using Eq.
(6.9), whichindicatesthatenergyisproportional
to voltage squared. Since injecting or extracting
energy can only be done over some finite time,
voltage cannot change instantaneously across a
capacitor.
v
t
(a)
v
t
(b)
Figure6.7 Voltage across a capacitor:
(a) allowed, (b) not allowable; an abrupt
change is not possible.
Leakage resistance
Capacitance
Figure 6.8 Circuit model of a
nonideal capacitor.
E X A M P L E 6 . 1
(a) Calculate the charge stored on a 3-pF capacitor with 20 V across it.
(b) Find the energy stored in the capacitor.
Solution:
(a) Since q = Cv,
q = 3 × 10−12
× 20 = 60 pC
(b) The energy stored is
w =
1
2
Cv2
=
1
2
× 3 × 10−12
× 400 = 600 pJ
206 PART 1 DC Circuits
P R A C T I C E P R O B L E M 6 . 1
What is the voltage across a 3-µF capacitor if the charge on one plate is
0.12 mC? How much energy is stored?
Answer: 40 V, 2.4 mJ.
E X A M P L E 6 . 2
The voltage across a 5-µF capacitor is
v(t) = 10 cos 6000t V
Calculate the current through it.
Solution:
By definition, the current is
i(t) = C
dv
dt
= 5 × 10−6 d
dt
(10 cos 6000t)
= −5 × 10−6
× 6000 × 10 sin 6000t = −0.3 sin 6000t A
P R A C T I C E P R O B L E M 6 . 2
If a 10-µF capacitor is connected to a voltage source with
v(t) = 50 sin 2000t V
determine the current through the capacitor.
Answer: cos 2000t A.
E X A M P L E 6 . 3
Determine the voltage across a 2-µF capacitor if the current through it is
i(t) = 6e−3000t
mA
Assume that the initial capacitor voltage is zero.
Solution:
Since v =
1
C
 t
0
i dt + v(0) and v(0) = 0,
v =
1
2 × 10−6
 t
0
6e−3000t
dt·10−3
=
3 × 103
−3000
e−3000t




t
0
= (1 − e−3000t
) V
P R A C T I C E P R O B L E M 6 . 3
The current through a 100-µF capacitor is i(t) = 50 sin 120πt mA. Cal-
culate the voltage across it at t = 1 ms and t = 5 ms. Take v(0) = 0.
Answer: −93.137 V, −1.736 V.
CHAPTER 6 Capacitors and Inductors 207
E X A M P L E 6 . 4
Determinethecurrentthrougha200-µFcapacitorwhosevoltageisshown
in Fig. 6.9.
v(t)
0
4
3
2
1
50
−50
t
Figure6.9 For Example 6.4.
Solution:
The voltage waveform can be described mathematically as
v(t) =







50t V 0  t  1
100 − 50t V 1  t  3
−200 + 50t V 3  t  4
0 otherwise
Since i = C dv/dt and C = 200 µF, we take the derivative of v to obtain
i(t) = 200 × 10−6
×







50 0  t  1
−50 1  t  3
50 3  t  4
0 otherwise
=







10 mA 0  t  1
−10 mA 1  t  3
10 mA 3  t  4
0 otherwise
Thus the current waveform is as shown in Fig. 6.10.
i (mA)
0
4
3
2
1
10
−10
t
Figure6.10 For Example 6.4.
P R A C T I C E P R O B L E M 6 . 4
An initially uncharged 1-mF capacitor has the current shown in Fig. 6.11
across it. Calculate the voltage across it at t = 2 ms and t = 5 ms.
i (mA)
0
6
4
2
100
t (ms)
Figure6.11 For Practice Prob. 6.4.
Answer: 100 mV, 400 mV.
E X A M P L E 6 . 5
Obtain the energy stored in each capacitor in Fig. 6.12(a) under dc con-
ditions.
Solution:
Under dc conditions, we replace each capacitor with an open circuit, as
shown in Fig. 6.12(b). The current through the series combination of the
2-k and 4-k resistors is obtained by current division as
i =
3
3 + 2 + 4
(6 mA) = 2 mA
Hence, the voltages v1 and v2 across the capacitors are
v1 = 2000i = 4 V v2 = 4000i = 8 V
208 PART 1 DC Circuits
v1
+ −
v2
+
−
6 mA 3 kΩ
5 kΩ
4 kΩ
2 kΩ
2 mF
4 mF
(a)
6 mA 3 kΩ
5 kΩ
4 kΩ
2 kΩ
(b)
i
Figure6.12 For Example 6.5.
and the energies stored in them are
w1 =
1
2
C1v2
1 =
1
2
(2 × 10−3
)(4)2
= 16 mJ
w2 =
1
2
C2v2
2 =
1
2
(4 × 10−3
)(8)2
= 128 mJ
P R A C T I C E P R O B L E M 6 . 5
Under dc conditions, find the energy stored in the capacitors in Fig. 6.13.
10 V +
− 6 kΩ
1 kΩ
20 mF
10 mF
3 kΩ
Figure6.13 For Practice Prob. 6.5.
Answer: 405 µJ, 90 µJ.
6.3 SERIES AND PARALLEL CAPACITORS
We know from resistive circuits that series-parallel combination is a pow-
erful tool for reducing circuits. This technique can be extended to series-
parallel connections of capacitors, which are sometimes encountered. We
desire to replace these capacitors by a single equivalent capacitor Ceq.
i C1
(a)
i1
C2 C3 CN
iN
v
i
(b)
Ceq v
+
−
+
−
i2 i3
Figure6.14 (a) Parallel-connected N
capacitors, (b) equivalent circuit for the parallel
capacitors.
In order to obtain the equivalent capacitor Ceq of N capacitors in
parallel, consider the circuit in Fig. 6.14(a). The equivalent circuit is in
Fig. 6.14(b). Note that the capacitors have the same voltage v across
them. Applying KCL to Fig. 6.14(a),
i = i1 + i2 + i3 + · · · + iN (6.11)
But ik = Ck dv/dt. Hence,
i = C1
dv
dt
+ C2
dv
dt
+ C3
dv
dt
+ · · · + CN
dv
dt
=

N

k=1
Ck
dv
dt
= Ceq
dv
dt
(6.12)
CHAPTER 6 Capacitors and Inductors 209
where
Ceq = C1 + C2 + C3 + · · · + CN (6.13)
The equivalent capacitance of N parallel-connected capacitors is the
sum of the individual capacitances.
We observe that capacitors in parallel combine in the same manner as
resistors in series.
v
C1
(a)
C2 C3 CN
v
(b)
Ceq v
v1 v2 v3 vN
+
−
+
−
i
i
+ −
+ −
+ − + −
+
−
Figure6.15 (a) Series-connected N
capacitors, (b) equivalent circuit for the series
capacitor.
We now obtain Ceq of N capacitors connected in series by compar-
ing the circuit in Fig. 6.15(a) with the equivalent circuit in Fig. 6.15(b).
Note that the same current i flows (and consequently the same charge)
through the capacitors. Applying KVL to the loop in Fig. 6.15(a),
v = v1 + v2 + v3 + · · · + vN (6.14)
But vk =
1
Ck
 t
t0
i(t) dt + vk(t0). Therefore,
v =
1
C1
 t
t0
i(t) dt + v1(t0) +
1
C2
 t
t0
i(t) dt + v2(t0)
+ · · · +
1
CN
 t
t0
i(t) dt + vN (t0)
=
1
C1
+
1
C2
+ · · · +
1
CN
 t
t0
i(t) dt + v1(t0) + v2(t0)
+ · · · + vN (t0)
=
1
Ceq
 t
t0
i(t) dt + v(t0)
(6.15)
where
1
Ceq
=
1
C1
+
1
C2
+
1
C3
+ · · · +
1
CN
(6.16)
The initial voltage v(t0) across Ceq is required by KVL to be the sum of
the capacitor voltages at t0. Or according to Eq. (6.15),
v(t0) = v1(t0) + v2(t0) + · · · + vN (t0)
Thus, according to Eq. (6.16),
The equivalent capacitance of series-connected capacitors is the reciprocal of the
sum of the reciprocals of the individual capacitances.
Note that capacitors in series combine in the same manner as resistors in
parallel. For N = 2 (i.e., two capacitors in series), Eq. (6.16) becomes
1
Ceq
=
1
C1
+
1
C2
210 PART 1 DC Circuits
or
Ceq =
C1C2
C1 + C2
(6.17)
E X A M P L E 6 . 6
Find the equivalent capacitance seen between terminals a and b of the
circuit in Fig. 6.16.
a
b
Ceq
5 mF
20 mF 20 mF
6 mF
60 mF
Figure6.16 For Example 6.6.
Solution:
The 20-µF and 5-µF capacitors are in series; their equivalent capacitance
is
20 × 5
20 + 5
= 4 µF
This 4-µF capacitor is in parallel with the 6-µF and 20-µF capacitors;
their combined capacitance is
4 + 6 + 20 = 30 µF
This 30-µF capacitor is in series with the 60-µF capacitor. Hence, the
equivalent capacitance for the entire circuit is
Ceq =
30 × 60
30 + 60
= 20 µF
P R A C T I C E P R O B L E M 6 . 6
Find the equivalent capacitance seen at the terminals of the circuit in Fig.
6.17.
Answer: 40 µF.
Ceq
120 mF
20 mF
70 F
60 mF
50 mF
Figure6.17 For Practice Prob. 6.6.
E X A M P L E 6 . 7
For the circuit in Fig. 6.18, find the voltage across each capacitor.
CHAPTER 6 Capacitors and Inductors 211
Solution:
We first find the equivalent capacitance Ceq, shown in Fig. 6.19. The two
parallel capacitors in Fig. 6.18 can be combined to get 40+20 = 60 mF.
This 60-mF capacitor is in series with the 20-mF and 30-mF capacitors.
Thus,
Ceq =
1
1
60
+ 1
30
+ 1
20
mF = 10 mF
The total charge is
q = Ceqv = 10 × 10−3
× 30 = 0.3 C
This is the charge on the 20-mF and 30-mF capacitors, because they are
in series with the 30-V source. (A crude way to see this is to imagine that
charge acts like current, since i = dq/dt.) Therefore,
v1 =
q
C1
=
0.3
20 × 10−3
= 15 V v2 =
q
C2
=
0.3
30 × 10−3
= 10 V
Having determined v1 and v2, we now use KVL to determine v3 by
v3 = 30 − v1 − v2 = 5 V
20 mF
40 mF
30 mF
20 mF
30 V +
−
v1 v2
v3
+
−
+ − + −
Figure6.18 For Example 6.7.
Ceq
30 V +
−
q
+
−
Figure6.19 Equivalent
circuit for Fig. 6.18.
Alternatively, since the 40-mF and 20-mF capacitors are in parallel,
they have the same voltage v3 and their combined capacitance is 40 +
20 = 60 mF. This combined capacitance is in series with the 20-mF and
30-mF capacitors and consequently has the same charge on it. Hence,
v3 =
q
60 mF
=
0.3
60 × 10−3
= 5 V
P R A C T I C E P R O B L E M 6 . 7
Find the voltage across each of the capacitors in Fig. 6.20.
30 mF
20 mF
60 mF
40 mF
60 V +
−
v1 v3
v2 v4
+ − + −
+
−
+
−
Figure6.20 For Practice Prob. 6.7.
Answer: v1 = 30 V, v2 = 30 V, v3 = 10 V, v4 = 20 V.
6.4 INDUCTORS
An inductor is a passive element designed to store energy in its magnetic
field. Inductors find numerous applications in electronic and power sys-
tems. They are used in power supplies, transformers, radios, TVs, radars,
and electric motors.
Any conductor of electric current has inductive properties and may
be regarded as an inductor. But in order to enhance the inductive effect,
a practical inductor is usually formed into a cylindrical coil with many
turns of conducting wire, as shown in Fig. 6.21.
212 PART 1 DC Circuits
An inductor consists of a coil of conducting wire.
If current is allowed to pass through an inductor, it is found that the voltage
across the inductor is directly proportional to the time rate of change of
the current. Using the passive sign convention,
v = L
di
dt
(6.18)
where L is the constant of proportionality called the inductance of the
inductor. The unit of inductance is the henry (H), named in honor of the
American inventor Joseph Henry (1797–1878). It is clear from Eq. (6.18)
that 1 henry equals 1 volt-second per ampere.
Length, l
Cross-sectional area, A
Core material
Number of turns, N
Figure6.21 Typical form of an inductor.
In view of Eq. (6.18), for an inductor to have
voltageacrossitsterminals,itscurrentmustvary
with time. Hence, v = 0 for constant current
through the inductor.
Inductance is the property whereby an inductor exhibits opposition to the change
of current flowing through it, measured in henrys (H).
The inductance of an inductor depends on its physical dimension
and construction. Formulas for calculating the inductance of inductors
of different shapes are derived from electromagnetic theory and can be
found in standard electrical engineering handbooks. For example, for the
inductor (solenoid) shown in Fig. 6.21,
L =
N2
µA

(6.19)
where N is the number of turns,  is the length, A is the cross-sectional
area, and µ is the permeability of the core. We can see from Eq. (6.19)
that inductance can be increased by increasing the number of turns of
coil, using material with higher permeability as the core, increasing the
cross-sectional area, or reducing the length of the coil.
Like capacitors, commercially available inductors come in different
values and types. Typical practical inductors have inductance values
ranging from a few microhenrys (µH), as in communication systems,
to tens of henrys (H) as in power systems. Inductors may be fixed or
variable. The core may be made of iron, steel, plastic, or air. The terms
coil and choke are also used for inductors. Common inductors are shown
in Fig. 6.22. The circuit symbols for inductors are shown in Fig. 6.23,
following the passive sign convention.
(a)
(b)
(c)
Figure6.22 Various types of inductors:
(a) solenoidal wound inductor, (b) toroidal
inductor, (c) chip inductor.
(Courtesy of Tech America.)
Equation (6.18) is the voltage-current relationship for an inductor.
Figure 6.24 shows this relationship graphically for an inductor whose
inductance is independent of current. Such an inductor is known as a
linear inductor. For a nonlinear inductor, the plot of Eq. (6.18) will not
be a straight line because its inductance varies with current. We will
assume linear inductors in this textbook unless stated otherwise.
The current-voltage relationship is obtained from Eq. (6.18) as
di =
1
L
v dt
CHAPTER 6 Capacitors and Inductors 213
Integrating gives
i =
1
L
 t
−∞
v(t) dt (6.20)
i i i
(a)
v L
+
−
(b)
v L
+
−
(c)
v L
+
−
Figure6.23 Circuit symbols for inductors:
(a) air-core, (b) iron-core, (c) variable
iron-core.
or
i =
1
L
 t
t0
v(t) dt + i(t0) (6.21)
where i(t0) is the total current for −∞  t  t0 and i(−∞) = 0. The
idea of making i(−∞) = 0 is practical and reasonable, because there
must be a time in the past when there was no current in the inductor.
The inductor is designed to store energy in its magnetic field. The
energy stored can be obtained from Eqs. (6.18) and (6.20). The power
delivered to the inductor is
p = vi = L
di
dt
i (6.22)
The energy stored is
w =
 t
−∞
p dt =
 t
−∞
L
di
dt
i dt
= L
 t
−∞
i di =
1
2
Li2
(t) −
1
2
Li2
(−∞)
(6.23)
Since i(−∞) = 0,
w =
1
2
Li2
(6.24)
Slope = L
di ⁄dt
0
v
Figure6.24 Voltage-current
relationship of an inductor.
We should note the following important properties of an inductor.
i
t
(a)
i
t
(b)
Figure6.25 Current through an inductor:
(a) allowed, (b) not allowable; an abrupt
change is not possible.
1. Note from Eq. (6.18) that the voltage across an inductor is zero
when the current is constant. Thus,
An inductor acts like a short circuit to dc.
2. An important property of the inductor is its opposition to the
change in current flowing through it.
The current through an inductor cannot change instantaneously.
According to Eq. (6.18), a discontinuous change in the current
through an inductor requires an infinite voltage, which is not
physically possible. Thus, an inductor opposes an abrupt
change in the current through it. For example, the current
through an inductor may take the form shown in Fig. 6.25(a),
whereas the inductor current cannot take the form shown in
Fig. 6.25(b) in real-life situations due to the discontinuities.
However, the voltage across an inductor can change abruptly.
214 PART 1 DC Circuits
3. Like the ideal capacitor, the ideal inductor does not dissipate
energy. The energy stored in it can be retrieved at a later time.
The inductor takes power from the circuit when storing energy
and delivers power to the circuit when returning previously
stored energy.
4. A practical, nonideal inductor has a significant resistive
component, as shown in Fig. 6.26. This is due to the fact that
the inductor is made of a conducting material such as copper,
which has some resistance. This resistance is called the
winding resistance Rw, and it appears in series with the
inductance of the inductor. The presence of Rw makes it both
an energy storage device and an energy dissipation device.
Since Rw is usually very small, it is ignored in most cases. The
nonideal inductor also has a winding capacitance Cw due to
the capacitive coupling between the conducting coils. Cw is
very small and can be ignored in most cases, except at high
frequencies. We will assume ideal inductors in this book.
Since an inductor is often made of a highly con-
ducting wire, it has a very small resistance.
L Rw
Cw
Figure6.26 Circuit model
for a practical inductor.
E X A M P L E 6 . 8
The current through a 0.1-H inductor is i(t) = 10te−5t
A. Find the voltage
across the inductor and the energy stored in it.
Solution:
Since v = L di/dt and L = 0.1 H,
v = 0.1
d
dt
(10te−5t
) = e−5t
+ t(−5)e−5t
= e−5t
(1 − 5t) V
The energy stored is
w =
1
2
Li2
=
1
2
(0.1)100t2
e−10t
= 5t2
e−10t
J
P R A C T I C E P R O B L E M 6 . 8
If the current through a 1-mH inductor is i(t) = 20 cos 100t mA, find the
terminal voltage and the energy stored.
Answer: −2 sin 100t mV, 0.2 cos2
100t µJ.
E X A M P L E 6 . 9
Find the current through a 5-H inductor if the voltage across it is
v(t) =
30t2
, t  0
0, t  0
Also find the energy stored within 0  t  5 s.
Solution:
Since i =
1
L
 t
t0
v(t) dt + i(t0) and L = 5 H,
i =
1
5
 t
0
30t2
dt + 0 = 6 ×
t3
3
= 2t3
A
CHAPTER 6 Capacitors and Inductors 215
The power p = vi = 60t5
, and the energy stored is then
w =

p dt =
 5
0
60t5
dt = 60
t6
6




5
0
= 156.25 kJ
Alternatively, we can obtain the energy stored using Eq. (6.13), by writing
w

5
0
=
1
2
Li2
(5) −
1
2
Li(0) =
1
2
(5)(2 × 53
)2
− 0 = 156.25 kJ
as obtained before.
P R A C T I C E P R O B L E M 6 . 9
The terminal voltage of a 2-H inductor is v = 10(1−t) V. Find the current
flowing through it at t = 4 s and the energy stored in it within 0  t 
4 s. Assume i(0) = 2 A.
Answer: −18 A, 320 J.
E X A M P L E 6 . 1 0
Consider the circuit in Fig. 6.27(a). Under dc conditions, find: (a) i, vC,
and iL, (b) the energy stored in the capacitor and inductor.
12 V
1 F
+
−
4 Ω
5 Ω
1 Ω
2 H
i
iL
vC
+
−
vC
+
−
(a)
(a)
12 V +
−
4 Ω
5 Ω
1 Ω
i
iL
(b)
Figure6.27 For Example 6.10.
Solution:
(a) Under dc conditions, we replace the capacitor with an open circuit
and the inductor with a short circuit, as in Fig. 6.27(b). It is evident from
Fig. 6.27(b) that
i = iL =
12
1 + 5
= 2 A
The voltage vC is the same as the voltage across the 5- resistor. Hence,
vC = 5i = 10 V
(b) The energy in the capacitor is
wC =
1
2
Cv2
C =
1
2
(1)(102
) = 50 J
and that in the inductor is
wL =
1
2
Li2
L =
1
2
(2)(22
) = 4 J
P R A C T I C E P R O B L E M 6 . 1 0
Determine vC, iL, and the energy stored in the capacitor and inductor in
the circuit of Fig. 6.28 under dc conditions.
4 A 2 F
3 Ω 1 Ω
0.25 H
iL
vC
+
−
Figure6.28 For Practice Prob. 6.10.
Answer: 3 V, 3 A, 9 J, 1.125 J.
216 PART 1 DC Circuits
6.5 SERIES AND PARALLEL INDUCTORS
Now that the inductor has been added to our list of passive elements, it is
necessary to extend the powerful tool of series-parallel combination. We
need to know how to find the equivalent inductance of a series-connected
or parallel-connected set of inductors found in practical circuits.
Consider a series connection of N inductors, as shown in Fig.
6.29(a), with the equivalent circuit shown in Fig. 6.29(b). The inductors
have the same current through them. Applying KVL to the loop,
v = v1 + v2 + v3 + · · · + vN (6.25)
L1
(a)
L2 L3 LN
(b)
Leq
i
i
v
+
−
v
+
−
+ −
v1
+ −
v2
+ −
v3
+ −
vN
. . .
Figure6.29 (a) A series connection of N
inductors, (b) equivalent circuit for the
series inductors.
Substituting vk = Lk di/dt results in
v = L1
di
dt
+ L2
di
dt
+ L3
di
dt
+ · · · + LN
di
dt
= (L1 + L2 + L3 + · · · + LN )
di
dt
=

N

k=1
Lk
di
dt
= Leq
di
dt
(6.26)
where
Leq = L1 + L2 + L3 + · · · + LN (6.27)
Thus,
The equivalent inductance of series-connected inductors is the
sum of the individual inductances.
Inductors in series are combined in exactly the same way as resistors in
series.
(a)
(b)
Leq
i
v
+
−
v
+
−
L1 L2 L3 LN
i
i1 i2 i3 iN
Figure6.30 (a) A parallel connection of N
inductors, (b) equivalent circuit for the parallel
inductors.
We now consider a parallel connection of N inductors, as shown
in Fig. 6.30(a), with the equivalent circuit in Fig. 6.30(b). The inductors
have the same voltage across them. Using KCL,
i = i1 + i2 + i3 + · · · + iN (6.28)
But ik =
1
Lk
 t
t0
v dt + ik(t0); hence,
i =
1
L1
 t
t0
v dt + i1(t0) +
1
L2
 t
t0
v dt + i2(t0)
+ · · · +
1
LN
 t
t0
v dt + iN (t0)
=
1
L1
+
1
L2
+ · · · +
1
LN
 t
t0
v dt + i1(t0) + i2(t0)
+ · · · + iN (t0)
=

N

k=1
1
Lk
 t
t0
v dt +
N

k=1
ik(t0) =
1
Leq
 t
t0
v dt + i(t0)
(6.29)
CHAPTER 6 Capacitors and Inductors 217
where
1
Leq
=
1
L1
+
1
L2
+
1
L3
+ · · · +
1
LN
(6.30)
The initial current i(t0) through Leq at t = t0 is expected by KCL to be
the sum of the inductor currents at t0. Thus, according to Eq. (6.29),
i(t0) = i1(t0) + i2(t0) + · · · + iN (t0)
According to Eq. (6.30),
The equivalent inductance of parallel inductors is the reciprocal of the sum of the
reciprocals of the individual inductances.
Note that the inductors in parallel are combined in the same way as resis-
tors in parallel.
For two inductors in parallel (N = 2), Eq. (6.30) becomes
1
Leq
=
1
L1
+
1
L2
or Leq =
L1L2
L1 + L2
(6.31)
It is appropriate at this point to summarize the most important character-
istics of the three basic circuit elements we have studied. The summary
is given in Table 6.1.
TABLE 6.1 Important characteristics of the basic elements.†
Relation Resistor (R) Capacitor (C) Inductor (L)
v-i: v = iR v =
1
C
 t
t0
i dt + v(t0) v = L
di
dt
i-v: i = v/R i = C
dv
dt
i =
1
L
 t
t0
i dt + i(t0)
p or w: p = i2
R =
v2
R
w =
1
2
Cv2
w =
1
2
Li2
Series: Req = R1 + R2 Ceq =
C1C2
C1 + C2
Leq = L1 + L2
Parallel: Req =
R1R2
R1 + R2
Ceq = C1 + C2 Leq =
L1L2
L1 + L2
At dc: Same Open circuit Short circuit
Circuit variable
that cannot
change abruptly: Not applicable v i
†Passive sign convention is assumed.
E X A M P L E 6 . 1 1
Find the equivalent inductance of the circuit shown in Fig. 6.31.
218 PART 1 DC Circuits
4 H 20 H
8 H 10 H
12 H
7 H
Leq
Figure6.31 For Example 6.11.
Solution:
The 10-H, 12-H, and 20-H inductors are in series; thus, combining them
gives a 42-H inductance. This 42-H inductor is in parallel with the 7-H
inductor so that they are combined, to give
7 × 42
7 + 42
= 6 H
This 6-H inductor is in series with the 4-H and 8-H inductors. Hence,
Leq = 4 + 6 + 8 = 18 H
P R A C T I C E P R O B L E M 6 . 1 1
Calculate the equivalent inductance for the inductive ladder network in
Fig. 6.32.
20 mH 100 mH 40 mH
30 mH 20 mH
40 mH
50 mH
Leq
Figure6.32 For Practice Prob. 6.11.
Answer: 25 mH.
E X A M P L E 6 . 1 2
For the circuit in Fig. 6.33, i(t) = 4(2 − e−10t
) mA. If i2(0) = −1 mA,
find: (a) i1(0); (b) v(t), v1(t), and v2(t); (c) i1(t) and i2(t).
4 H
12 H
4 H
v
+
−
v2
v1
+
+ −
−
i
i1 i2
Figure6.33 For Example 6.12.
Solution:
(a) From i(t) = 4(2 − e−10t
) mA, i(0) = 4(2 − 1) = 4 mA. Since i =
i1 + i2,
i1(0) = i(0) − i2(0) = 4 − (−1) = 5 mA
(b) The equivalent inductance is
Leq = 2 + 4  12 = 2 + 3 = 5 H
Thus,
v(t) = Leq
di
dt
= 5(4)(−1)(−10)e−10t
mV = 200e−10t
mV
and
v1(t) = 2
di
dt
= 2(−4)(−10)e−10t
mV = 80e−10t
mV
Since v = v1 + v2,
v2(t) = v(t) − v1(t) = 120e−10t
mV
CHAPTER 6 Capacitors and Inductors 219
(c) The current i1 is obtained as
i1(t) =
1
4
 t
0
v2 dt + i1(0) =
120
4
 t
0
e−10t
dt + 5 mA
= −3e−10t

t
0
+ 5 mA = −3e−10t
+ 3 + 5 = 8 − 3e−10t
mA
Similarly,
i2(t) =
1
12
 t
0
v2 dt + i2(0) =
120
12
 t
0
e−10t
dt − 1 mA
= −e−10t

t
0
− 1 mA = −e−10t
+ 1 − 1 = −e−10t
mA
Note that i1(t) + i2(t) = i(t).
P R A C T I C E P R O B L E M 6 . 1 2
In the circuit of Fig. 6.34, i1(t) = 0.6e−2t
A. If i(0) = 1.4 A, find:
(a) i2(0); (b) i2(t) and i(t); (c) v(t), v1(t), and v2(t).
3 H
6 H
8 H
v
+
−
v2
+
−
i
i1
i2
+ −
v1
Figure 6.34 For Practice Prob. 6.12.
Answer: (a) 0.8 A, (b) (−0.4 + 1.2e−2t
) A, (−0.4 + 1.8e−2t
) A,
(c) −7.2e−2t
V, −28.8e−2t
V, −36e−2t
V.
†6.6 APPLICATIONS
Circuit elements such as resistors and capacitors are commercially avail-
able in either discrete form or integrated-circuit (IC) form. Unlike ca-
pacitors and resistors, inductors with appreciable inductance are difficult
to produce on IC substrates. Therefore, inductors (coils) usually come
in discrete form and tend to be more bulky and expensive. For this rea-
son, inductors are not as versatile as capacitors and resistors, and they
are more limited in applications. However, there are several applications
in which inductors have no practical substitute. They are routinely used
in relays, delays, sensing devices, pick-up heads, telephone circuits, ra-
dio and TV receivers, power supplies, electric motors, microphones, and
loudspeakers, to mention a few.
Capacitors and inductors possess the following three special prop-
erties that make them very useful in electric circuits:
1. The capacity to store energy makes them useful as temporary
voltage or current sources. Thus, they can be used for
generating a large amount of current or voltage for a short
period of time.
2. Capacitors oppose any abrupt change in voltage, while
inductors oppose any abrupt change in current. This property
220 PART 1 DC Circuits
makes inductors useful for spark or arc suppression and for
converting pulsating dc voltage into relatively smooth dc
voltage.
3. Capacitors and inductors are frequency sensitive. This
property makes them useful for frequency discrimination.
The first two properties are put to use in dc circuits, while the third
one is taken advantage of in ac circuits. We will see how useful these
properties are in later chapters. For now, consider three applications
involving capacitors and op amps: integrator, differentiator, and analog
computer.
6.6.1 Integrator
Important op amp circuits that use energy-storage elements include inte-
grators and differentiators. These op amp circuits often involve resistors
and capacitors; inductors (coils) tend to be more bulky and expensive.
The op amp integrator is used in numerous applications, especially
in analog computers, to be discussed in Section 6.6.3.
An integrator is an op amp circuit whose output is proportional
to the integral of the input signal.
R1
Rf
i1 v1
i2
vi
+
−
vo
+
−
v2
0 A
0 V
+
−
+
−
(a)
R
a
C
iR
iC
vi
+
−
vo
+
−
+
−
(b)
1
Figure6.35 Replacing the feedback resistor
in the inverting amplifier in (a) produces an
integrator in (b).
If the feedback resistor Rf in the familiar inverting amplifier of
Fig. 6.35(a) is replaced by a capacitor, we obtain an ideal integrator, as
shown in Fig. 6.35(b). It is interesting that we can obtain a mathematical
representation of integration this way. At node a in Fig. 6.35(b),
iR = iC (6.32)
But
iR =
vi
R
, iC = −C
dvo
dt
Substituting these in Eq. (6.32), we obtain
vi
R
= −C
dvo
dt
(6.33a)
dvo = −
1
RC
vi dt (6.33b)
Integrating both sides gives
vo(t) − vo(0) = −
1
RC
 t
0
vi(t) dt (6.34)
To ensure that vo(0) = 0, it is always necessary to discharge the integra-
tor’s capacitor prior to the application of a signal. Assuming vo(0) = 0,
vo = −
1
RC
 t
0
vi(t) dt (6.35)
which shows that the circuit in Fig. 6.35(b) provides an output voltage
proportional to the integral of the input. In practice, the op amp integrator
CHAPTER 6 Capacitors and Inductors 221
requires a feedback resistor to reduce dc gain and prevent saturation. Care
must be taken that the op amp operates within the linear range so that it
does not saturate.
E X A M P L E 6 . 1 3
If v1 = 10 cos 2t mV and v2 = 0.5t mV, find vo in the op amp circuit in
Fig. 6.36. Assume that the voltage across the capacitor is initially zero.
vo
v1
v2
2 mF
3 MΩ
100 kΩ
+
−
Figure6.36 For Example 6.13.
Solution:
This is a summing integrator, and
vo = −
1
R1C

v1 dt −
1
R2C

v2 dt
= −
1
3 × 106 × 2 × 10−6
 t
0
10 cos 2t dt
−
1
100 × 103 × 2 × 10−6
 t
0
0.5t dt
= −
1
6
10
2
sin 2t −
1
0.2
0.5t2
2
= −0.833 sin 2t − 1.25t2
mV
P R A C T I C E P R O B L E M 6 . 1 3
The integrator in Fig. 6.35 has R = 25 k, C = 10 µF. Determine the
output voltage when a dc voltage of 10 mV is applied at t = 0. Assume
that the op amp is initially nulled.
Answer: −40t mV.
6.6.2 Differentiator
A differentiator is an op amp circuit whose output is proportional to
the rate of change of the input signal.
R
a
C
iC
iR
vi
+
−
vo
+
−
+
−
Figure6.37 An op amp differentiator.
In Fig. 6.35(a), if the input resistor is replaced by a capacitor, the
resulting circuit is a differentiator, shown in Fig. 6.37. Applying KCL at
node a,
iR = iC (6.36)
But
iR = −
vo
R
, iC = C
dvi
dt
Substituting these in Eq. (6.36) yields
vo = −RC
dvi
dt
(6.37)
showing that the output is the derivative of the input. Differentiator cir-
cuits are electronically unstable because any electrical noise within the
222 PART 1 DC Circuits
circuit is exaggerated by the differentiator. For this reason, the differen-
tiator circuit in Fig. 6.37 is not as useful and popular as the integrator. It
is seldom used in practice.
E X A M P L E 6 . 1 4
Sketch the output voltage for the circuit in Fig. 6.38(a), given the input
voltage in Fig. 6.38(b). Take vo = 0 at t = 0.
vo
vi
+
−
(a)
(b)
+
−
0.2 mF
5 kΩ
vi
8
6
4
2
0
4
t (ms)
+
−
Figure6.38 For Example 6.14.
Solution:
This is a differentiator with
RC = 5 × 103
× 0.2 × 10−6
= 10−3
s
For 0  t  4 ms, we can express the input voltage in Fig. 6.38(b) as
vi =
2t 0  t  2 ms
8 − 2t 2  t  4 ms
This is repeated for 4  t  8. Using Eq. (6.37), the output is obtained
as
vo = −RC
dvi
dt
=
−2 mV 0  t  2 ms
2 mV 2  t  4 ms
Thus, the output is as sketched in Fig. 6.39.
vi (mV)
8
6
4
2
2
0
−2
t (ms)
Figure6.39 Output of the circuit in Fig. 6.38(a).
P R A C T I C E P R O B L E M 6 . 1 4
The differentiator in Fig. 6.37 has R = 10 k and C = 2 µF. Given that
vi = 3t V, determine the output vo.
Answer: −60 mV.
6.6.3 Analog Computer
Op amps were initially developed for electronic analog computers. Ana-
log computers can be programmed to solve mathematical models of me-
chanical or electrical systems. These models are usually expressed in
terms of differential equations.
To solve simple differential equations using the analog computer
requires cascading three types of op amp circuits: integrator circuits,
summing amplifiers, and inverting/noninverting amplifiers for negative/
CHAPTER 6 Capacitors and Inductors 223
positive scaling. The best way to illustrate how an analog computer solves
a differential equation is with an example.
Suppose we desire the solution x(t) of the equation
a
d2
x
dt2
+ b
dx
dt
+ cx = f (t), t  0 (6.38)
where a, b, and c are constants, and f (t) is an arbitrary forcing function.
The solution is obtained by first solving the highest-order derivative term.
Solving for d2
x/dt2
yields
d2
x
dt2
=
f (t)
a
−
b
a
dx
dt
−
c
a
x (6.39)
To obtain dx/dt, the d2
x/dt2
term is integrated and inverted. Finally, to
obtain x, the dx/dt term is integrated and inverted. The forcing function
is injected at the proper point. Thus, the analog computer for solving Eq.
(6.38) is implemented by connecting the necessary summers, inverters,
and integrators. A plotter or oscilloscope may be used to view the output
x, or dx/dt, or d2
x/dt2
, depending on where it is connected in the
system.
Although the above example is on a second-order differential equa-
tion, any differential equation can be simulated by an analog computer
comprising integrators, inverters, and inverting summers. But care must
be exercised in selecting the values of the resistors and capacitors, to
ensure that the op amps do not saturate during the solution time
interval.
The analog computers with vacuum tubes were built in the 1950s
and 1960s. Recently their use has declined. They have been superseded
by modern digital computers. However, we still study analog computers
for two reasons. First, the availability of integrated op amps has made
it possible to build analog computers easily and cheaply. Second, un-
derstanding analog computers helps with the appreciation of the digital
computers.
E X A M P L E 6 . 1 5
Design an analog computer circuit to solve the differential equation:
d2
vo
dt2
+ 2
dvo
dt
+ vo = 10 sin 4t t  0
subject to vo(0) = −4, v
o(0) = 1, where the prime refers to the time
derivative.
Solution:
We first solve for the second derivative as
d2
vo
dt2
= 10 sin 4t − 2
dvo
dt
− vo (6.15.1)
Solving this requires some mathematical operations, including summing,
scaling, and integration. Integrating both sides of Eq. (6.15.1) gives
dvo
dt
= −
 t
0
−10 sin 4t + 2
dvo
dt
+ vo + v
o(0) (6.15.2)
224 PART 1 DC Circuits
where v
o(0) = 1. We implement Eq. (6.15.2) using the summing inte-
grator shown in Fig. 6.40(a). The values of the resistors and capacitors
have been chosen so that RC = 1 for the term
−
1
RC
 t
0
vo dt
Other terms in the summing integrator of Eq. (6.15.2) are implemented
accordingly. The initial condition dvo(0)/dt = 1 is implemented by
connecting a 1-V battery with a switch across the capacitor as shown in
Fig. 6.40(a).
(a)
1 mF
1 MΩ
1 V
0.6 MΩ
1 MΩ
dvo
dt
dvo
dt
dvo
dt
t = 0
–10 sin (4t)
vo
1 mF
1 MΩ 1 V
0.5 MΩ
1 MΩ
t = 0
10 sin (4t)
vo
(b)
1 mF
4 V
1 MΩ
1 MΩ
dvo
dt
t = 0
−vo
vo
1 MΩ
(c)
1 mF
4 V
1 MΩ
1 MΩ
t = 0
vo
1 MΩ
+
−
+
−
+
−
+
−
+
−
+
−
+
−
+ −
+ −
+ −
+ −
Figure6.40 For Example 6.15.
The next step is to obtain vo by integrating dvo/dt and inverting
the result,
vo = −
 t
0
−
dvo
dt
dt + v(0) (6.15.3)
This is implemented with the circuit in Fig. 6.40(b) with the battery giving
the initial condition of −4 V. We now combine the two circuits in Fig.
6.40 (a) and (b) to obtain the complete circuit shown in Fig. 6.40(c). When
theinputsignal10 sin 4t isapplied, weopentheswitchesatt = 0toobtain
the output waveform vo, which may be viewed on an oscilloscope.
CHAPTER 6 Capacitors and Inductors 225
P R A C T I C E P R O B L E M 6 . 1 5
Design an analog computer circuit to solve the differential equation:
d2
vo
dt2
+ 3
dvo
dt
+ 2vo = 4 cos 10t t  0
subject to vo(0) = 2, v
o(0) = 0.
Answer: See Fig. 6.41, where RC = 1 s.
d2
v
dt2
d2
v
dt2
cos (10t)
2 V
t = 0
v
+
−
C
R
R
2
R
C
R
R
R
R
3
R
4
+
−
+
−
+
−
+
−
+
−
R
R
Figure6.41 For Practice Prob. 6.15.
6.7 SUMMARY
1. The current through a capacitor is directly proportional to the time
rate of change of the voltage across it.
i = C
dv
dt
The current through a capacitor is zero unless the voltage is
changing. Thus, a capacitor acts like an open circuit to a dc source.
2. The voltage across a capacitor is directly proportional to the time
integral of the current through it.
v =
1
C
 t
−∞
i dt =
1
C
 t
t0
i dt + i(t0)
The voltage across a capacitor cannot change instantly.
3. Capacitors in series and in parallel are combined in the same way as
conductances.
226 PART 1 DC Circuits
4. The voltage across an inductor is directly proportional to the time
rate of change of the current through it.
v = L
di
dt
The voltage across the inductor is zero unless the current is chang-
ing. Thus an inductor acts like a short circuit to a dc source.
5. The current through an inductor is directly proportional to the time
integral of the voltage across it.
i =
1
L
 t
−∞
v dt =
1
L
 t
t0
v dt + v(t0)
The current through an inductor cannot change instantly.
6. Inductors in series and in parallel are combined in the same way
resistors in series and in parallel are combined.
7. At any given time t, the energy stored in a capacitor is 1
2
Cv2
, while
the energy stored in an inductor is 1
2
Li2
.
8. Three application circuits, the integrator, the differentiator, and the
analog computer, can be realized using resistors, capacitors, and op
amps.
REVIEW QUESTIONS
6.1 What charge is on a 5-F capacitor when it is
connected across a 120-V source?
(a) 600 C (b) 300 C
(c) 24 C (d) 12 C
6.2 Capacitance is measured in:
(a) coulombs (b) joules
(c) henrys (d) farads
6.3 When the total charge in a capacitor is doubled, the
energy stored:
(a) remains the same (b) is halved
(c) is doubled (d) is quadrupled
6.4 Can the voltage waveform in Fig. 6.42 be associated
with a capacitor?
(a) Yes (b) No
0
2
1
10
−10
t
v(t)
Figure 6.42 For Review Question 6.4.
6.5 The total capacitance of two 40-mF series-connected
capacitors in parallel with a 4-mF capacitor is:
(a) 3.8 mF (b) 5 mF (c) 24 mF
(d) 44 mF (e) 84 mF
6.6 In Fig. 6.43, if i = cos 4t and v = sin 4t, the
element is:
(a) a resistor (b) a capacitor (c) an inductor
v +
−
i
Element
Figure 6.43 For Review Question 6.6.
6.7 A 5-H inductor changes its current by 3 A in 0.2 s.
The voltage produced at the terminals of the
inductor is:
(a) 75 V (b) 8.888 V
(c) 3 V (d) 1.2 V
6.8 If the current through a 10-mH inductor increases
from zero to 2 A, how much energy is stored in the
inductor?
(a) 40 mJ (b) 20 mJ
(c) 10 mJ (d) 5 mJ
CHAPTER 6 Capacitors and Inductors 227
6.9 Inductors in parallel can be combined just like
resistors in parallel.
(a) True (b) False
6.10 For the circuit in Fig. 6.44, the voltage divider
formula is:
(a) v1 =
L1 + L2
L1
vs (b) v1 =
L1 + L2
L2
vs
(c) v1 =
L2
L1 + L2
vs (d) v1 =
L1
L1 + L2
vs
vs
+
−
v2
v1
L1
L2
+
−
+ −
Figure 6.44 For Review Question 6.10.
Answers: 6.1a, 6.2d, 6.3d, 6.4b, 6.5c, 6.6b, 6.7a, 6.8b, 6.9a, 6.10d.
PROBLEMS
Section 6.2 Capacitors
6.1 If the voltage across a 5-F capacitor is 2te−3t
V, find
the current and the power.
6.2 A 40-µF capacitor is charged to 120 V and is then
allowed to discharge to 80 V. How much energy is
lost?
6.3 In 5 s, the voltage across a 40-mF capacitor changes
from 160 V to 220 V. Calculate the average current
through the capacitor.
6.4 A current of 6 sin 4t A flows through a 2-F
capacitor. Find the voltage v(t) across the capacitor
given that v(0) = 1 V.
6.5 If the current waveform in Fig. 6.45 is applied to a
20-µF capacitor, find the voltage v(t) across the
capacitor. Assume that v(0) = 0.
0
2
1
4
t
i(t)
Figure 6.45 For Prob. 6.5.
6.6 The voltage waveform in Fig. 6.46 is applied across
a 30-µF capacitor. Draw the current waveform
through it.
v(t) V
0
6 8 10 12
4
2
10
−10
t (ms)
Figure 6.46 For Prob. 6.6.
6.7 At t = 0, the voltage across a 50-mF capacitor is
10 V. Calculate the voltage across the capacitor for
t  0 when current 4t mA flows through it.
6.8 The current through a 0.5-F capacitor is
6(1 − e−t
) A. Determine the voltage and power at
t = 2 s. Assume v(0) = 0.
6.9 If the voltage across a 2-F capacitor is as shown in
Fig. 6.47, find the current through the capacitor.
v(t) (V)
0
5
3 4 5 6 7
2
1
10
t (s)
Figure 6.47 For Prob. 6.9.
6.10 The current through an initially uncharged 4-µF
capacitor is shown in Fig. 6.48. Find the voltage
across the capacitor for 0  t  3.
228 PART 1 DC Circuits
0
2 3
1
40
−40
t (s)
i(t) (mA)
Figure 6.48 For Prob. 6.10.
6.11 A voltage of 60 cos 4πt V appears across the
terminals of a 3-mF capacitor. Calculate the current
through the capacitor and the energy stored in it
from t = 0 to t = 0.125 s.
6.12 Find the voltage across the capacitors in the circuit
of Fig. 6.49 under dc conditions.
3 Ω
60 V
20 Ω
10 Ω 50 Ω
v2
v1
C1 C2
+
−
+
−
+
−
Figure 6.49 For Prob. 6.12.
Section 6.3 Series and Parallel Capacitors
6.13 What is the total capacitance of four 30-mF
capacitors connected in:
(a) parallel (b) series
6.14 Two capacitors (20 µF and 30 µF) are connected to
a 100-V source. Find the energy stored in each
capacitor if they are connected in:
(a) parallel (b) series
6.15 Determine the equivalent capacitance for each of the
circuits in Fig. 6.50.
4 F
4 F
6 F
3 F
12 F
(a)
6 F
4 F 2 F
5 F
(b)
2 F
3 F
(c)
6 F
3 F
4 F
Figure 6.50 For Prob. 6.15.
6.16 Find Ceq for the circuit in Fig. 6.51.
30 mF
5 mF 40 mF
15 mF
20 mF
Ceq
Figure 6.51 For Prob. 6.16.
6.17 Calculate the equivalent capacitance for the circuit
in Fig. 6.52. All capacitances are in mF.
4
8
1
15
5
6 6
2
3
Ceq
Figure 6.52 For Prob. 6.17.
6.18 Determine the equivalent capacitance at terminals
a-b of the circuit in Fig. 6.53.
6 mF 4 mF
5 mF
3 mF 12 mF
2 mF
a
b
Figure 6.53 For Prob. 6.18.
CHAPTER 6 Capacitors and Inductors 229
6.19 Obtain the equivalent capacitance of the circuit in
Fig. 6.54.
40 mF
20 mF
a b
35 mF 5 mF
10 mF
15 mF 15 mF
10 mF
Figure 6.54 For Prob. 6.19.
6.20 For the circuit in Fig. 6.55, determine:
(a) the voltage across each capacitor,
(b) the energy stored in each capacitor.
2 mF
6 mF
4 mF
3 mF
+
−
120 V
Figure 6.55 For Prob. 6.20.
6.21 Repeat Prob. 6.20 for the circuit in Fig. 6.56.
60 mF 20 mF
14 mF 80 mF
30 mF
+
−
90 V
Figure 6.56 For Prob. 6.21.
6.22 (a) Show that the voltage-division rule for two
capacitors in series as in Fig. 6.57(a) is
v1 =
C2
C1 + C2
vs, v2 =
C1
C1 + C2
vs
assuming that the initial conditions are zero.
C1
is C2
(b)
C1
vs
v1
v2 C2
(a)
+
−
+
−
+ − i1 i2
Figure 6.57 For Prob. 6.22.
(b) For two capacitors in parallel as in Fig. 6.57(b),
show that the current-division rule is
i1 =
C1
C1 + C2
is, i2 =
C2
C1 + C2
is
assuming that the initial conditions are zero.
6.23 Three capacitors, C1 = 5 µF, C2 = 10 µF, and
C3 = 20 µF, are connected in parallel across a
150-V source. Determine:
(a) the total capacitance,
(b) the charge on each capacitor,
(c) the total energy stored in the parallel
combination.
6.24 The three capacitors in the previous problem are
placed in series with a 200-V source. Compute:
(a) the total capacitance,
(b) the charge on each capacitor,
(c) the total energy stored in the series combination.
6.25
∗
Obtain the equivalent capacitance of the network
shown in Fig. 6.58.
30 mF
20 mF
10 mF
50 mF
40 mF
Figure 6.58 For Prob. 6.25.
6.26 Determine Ceq for each circuit in Fig. 6.59.
∗An asterisk indicates a challenging problem.
230 PART 1 DC Circuits
C
C
C
C
C
Ceq
(a)
C
C
C
C
Ceq
(b)
Figure 6.59 For Prob. 6.26.
6.27 Assuming that the capacitors are initially uncharged,
find vo(t) in the circuit in Fig. 6.60.
is +
−
vo(t)
6 mF
3 mF
is (mA)
0
2
1
60
t (s)
Figure 6.60 For Prob. 6.27.
6.28 If v(0) = 0, find v(t), i1(t), and i2(t) in the circuit in
Fig. 6.61.
i1
is
i2
v
6 mF 4 mF
is (mA)
5
3 4
1 2
20
0
−20
t
+
−
Figure 6.61 For Prob. 6.28.
6.29 For the circuit in Fig. 6.62, let v = 10e−3t
V and
v1(0) = 2 V. Find:
(a) v2(0) (b) v1(t) and v2(t)
(c) i(t), i1(t), and i2(t)
50 mF
30 mF
20 mF
v
+
−
v2
v1
i
−
+
+
−
i1 i2
Figure 6.62 For Prob. 6.29.
Section 6.4 Inductors
6.30 The current through a 10-mH inductor is 6e−t/2
A.
Find the voltage and the power at t = 3 s.
6.31 The current in a coil increases uniformly from 0.4 to
1 A in 2 s so that the voltage across the coil is
60 mV. Calculate the inductance of the coil.
6.32 The current through a 0.25-mH inductor is
12 cos 2t A. Determine the terminal voltage and the
power.
6.33 The current through a 12-mH inductor is
4 sin 100t A. Find the voltage, and also the energy
stored in the inductor for 0  t  π/200 s.
6.34 The current through a 40-mH inductor is
i(t) =
0, t  0
te−2t
A, t  0
Find the voltage v(t).
6.35 The voltage across a 2-H inductor is 20(1 − e−2t
) V.
If the initial current through the inductor is 0.3 A,
find the current and the energy stored in the inductor
at t = 1 s.
6.36 If the voltage waveform in Fig. 6.63 is applied
across the terminals of a 5-H inductor, calculate the
current through the inductor. Assume i(0) = −1 A.
v(t) (V)
5
4
2
1 3
10
0
t
Figure 6.63 For Prob. 6.36.
6.37 The current in an 80-mH inductor increases from 0
to 60 mA. How much energy is stored in the
inductor?
6.38 A voltage of (4 + 10 cos 2t) V is applied to a 5-H
inductor. Find the current i(t) through the inductor
if i(0) = −1 A.
CHAPTER 6 Capacitors and Inductors 231
6.39 If the voltage waveform in Fig. 6.64 is applied to a
10-mH inductor, find the inductor current i(t).
Assume i(0) = 0.
v(t)
0
2
1
5
–5
t
Figure 6.64 For Prob. 6.39.
6.40 Find vC, iL, and the energy stored in the capacitor
and inductor in the circuit of Fig. 6.65 under dc
conditions.
5 Ω
2 Ω
4 Ω
2 F
3 A 0.5 H
vC
+
−
iL
Figure 6.65 For Prob. 6.40.
6.41 For the circuit in Fig. 6.66, calculate the value of R
that will make the energy stored in the capacitor the
same as that stored in the inductor under dc
conditions.
R
2 Ω
5 A 4 mH
160 mF
Figure 6.66 For Prob. 6.41.
6.42 Under dc conditions, find the voltage across the
capacitors and the current through the inductors in
the circuit of Fig. 6.67.
6 Ω
4 Ω
30 V +
− C1
L1
C2 L2
Figure 6.67 For Prob. 6.42.
Section 6.5 Series and Parallel Inductors
6.43 Find the equivalent inductance for each circuit in
Fig. 6.68.
5 H 1 H
4 H 4 H
6 H
(a)
1 H 2 H
6 H 4 H
12 H
(b)
6 H
2 H
4 H
3 H
(c)
Figure 6.68 For Prob. 6.43.
6.44 Obtain Leq for the inductive circuit of Fig. 6.69. All
inductances are in mH.
6
5
4
12
10
3
Figure 6.69 For Prob. 6.44.
232 PART 1 DC Circuits
6.45 Determine Leq at terminals a-b of the circuit in Fig.
6.70.
60 mH
20 mH
30 mH
25 mH
10 mH
a b
Figure 6.70 For Prob. 6.45.
6.46 Find Leq at the terminals of the circuit in Fig. 6.71.
8 mH
6 mH
8 mH
12 mH
4 mH
6 mH
5 mH
8 mH
10 mH
a
b
Figure 6.71 For Prob. 6.46.
6.47 Find the equivalent inductance looking into the
terminals of the circuit in Fig. 6.72.
9 H
6 H
4 H
3 H
12 H
10 H
a b
Figure 6.72 For Prob. 6.47.
6.48 Determine Leq in the circuit in Fig. 6.73.
L
L
L
L
L
L
Leq
Figure 6.73 For Prob. 6.48.
6.49 Find Leq in the circuit in Fig. 6.74.
L
L
L
L L
Leq
L
L
L
Figure 6.74 For Prob. 6.49.
6.50
∗
Determine Leq that may be used to represent the
inductive network of Fig. 6.75 at the terminals.
3 H
4 H
5 H
Leq
+ −
i
a
b
dt
di
2
Figure 6.75 For Prob. 6.50.
6.51 The current waveform in Fig. 6.76 flows through a
3-H inductor. Sketch the voltage across the inductor
over the interval 0  t  6 s.
i(t)
0
2
3 4 5 6
2
1 t
Figure 6.76 For Prob. 6.51.
CHAPTER 6 Capacitors and Inductors 233
6.52 (a) For two inductors in series as in Fig. 6.77(b),
show that the current-division principle is
v1 =
L1
L1 + L2
vs, v2 =
L2
L1 + L2
vs
assuming that the initial conditions are zero.
(b) For two inductors in parallel as in Fig. 6.77(b),
show that the current-division principle is
i1 =
L2
L1 + L2
is, i2 =
L1
L1 + L2
is
assuming that the initial conditions are zero.
vs
+
−
+
−
v2
+ −
v1
L1
L2
(a)
(a)
is L1 L2
(b)
i1 i2
Figure 6.77 For Prob. 6.52.
6.53 In the circuit of Fig. 6.78, let is(t) = 6e−2t
mA,
t ≥ 0 and i1(0) = 4 mA. Find:
(a) i2(0),
(b) i1(t) and i2(t), t  0,
(c) v1(t) and v2(t), t  0,
(d) the energy in each inductor at t = 0.5 s.
is(t)
i1 i2
20 mH
30 mH
10 mH
+ −
v1
v2
+
−
Figure 6.78 For Prob. 6.53.
6.54 The inductors in Fig. 6.79 are initially charged and
are connected to the black box at t = 0. If
i1(0) = 4 A, i2(0) = −2 A, and v(t) = 50e−200t
mV,
t ≥ 0, find:
(a) the energy initially stored in each inductor,
(b) the total energy delivered to the black box from
t = 0 to t = ∞,
(c) i1(t) and i2(t), t ≥ 0,
(d) i(t), t ≥ 0.
i1 i2
20 H
5 H
v
+
−
Black box
i(t)
t = 0
Figure 6.79 For Prob. 6.54.
6.55 Find i and v in the circuit of Fig. 6.80 assuming that
i(0) = 0 = v(0).
40 mH
60 mH
20 mH
16 mH
v
+
−
12 sin 4t mV +
−
i
Figure 6.80 For Prob. 6.55.
Section 6.6 Applications
6.56 An op amp integrator has R = 50 k and
C = 0.04 µF. If the input voltage is
vi = 10 sin 50t mV, obtain the output voltage.
6.57 A 10-V dc voltage is applied to an integrator with
R = 50 k, C = 100 µF at t = 0. How long will it
take for the op amp to saturate if the saturation
voltages are +12 V and −12 V? Assume that the
initial capacitor voltage was zero.
6.58 An op amp integrator with R = 4 M and
C = 1 µF has the input waveform shown in Fig.
6.81. Plot the output waveform.
vi (mV)
0
20
10
–10
–20
3 4 5 6
2
1 t (ms)
Figure 6.81 For Prob. 6.58.
6.59 Using a single op amp, a capacitor, and resistors of
100 k or less, design a circuit to implement
vo = −50
 t
0
vi(t) dt
Assume vo = 0 at t = 0.
234 PART 1 DC Circuits
6.60 Show how you would use a single op amp to
generate
vo = −
 t
0
(v1 + 4v2 + 10v3) dt
If the integrating capacitor is C = 2 µF, obtain other
component values.
6.61 At t = 1.5 ms, calculate vo due to the cascaded
integrators in Fig. 6.82. Assume that the integrators
are reset to 0 V at t = 0.
1 V
2 mF
10 kΩ
20 kΩ
vo
+
−
0.5 mF
+
−
+
−
+
−
Figure 6.82 For Prob. 6.61.
6.62 Show that the circuit in Fig. 6.83 is a noninverting
integrator.
vo
vi
+
−
+
−
R
R
R
C
R
+
−
Figure 6.83 For Prob. 6.62.
6.63 The triangular waveform in Fig. 6.84(a) is applied to
the input of the op amp differentiator in Fig. 6.84(b).
Plot the output.
(a)
vi(t)
0
10
3 4
2
1 t (ms)
–10
vo
vi
+
−
+
−
20 kΩ
0.01 mF
(b)
+
−
Figure 6.84 For Prob. 6.63.
6.64 An op amp differentiator has R = 250 k and
C = 10 µF. The input voltage is a ramp
r(t) = 12t mV. Find the output voltage.
6.65 A voltage waveform has the following
characteristics: a positive slope of 20 V/s for 5 ms
followed by a negative slope of 10 V/s for 10 ms. If
the waveform is applied to a differentiator with
R = 50 k, C = 10 µF, sketch the output voltage
waveform.
6.66
∗
The output vo of the op amp circuit of Fig. 6.85(a) is
shown in Fig. 6.85(b). Let Ri = Rf = 1 M and
C = 1 µF. Determine the input voltage waveform
and sketch it.
(b)
(a)
0
4
3 4
2
1 t (ms)
−4
vo
vi
vo
Ri
C
Rf
+
−
+
−
+
−
Figure 6.85 For Prob. 6.66.
6.67 Design an analog computer to simulate
d2
vo
dt2
+ 2
dvo
dt
+ vo = 10 sin 2t
where v0(0) = 2 and v
0(0) = 0.
CHAPTER 6 Capacitors and Inductors 235
6.68 Design an analog computer to solve the differential
equation
di(t)
dt
+ 3i(t) = 2 t  0
and assume that i(0) = 1 mA.
6.69 Figure 6.86 presents an analog computer designed to
solve a differential equation. Assuming f (t) is
known, set up the equation for f (t).
vo(t)
−f(t)
1 mF
1 mF
1 MΩ
1 MΩ
1 MΩ
100 kΩ
200 kΩ
500 kΩ
100 kΩ
+
−
+
−
+
−
+
−
Figure 6.86 For Prob. 6.69.
COMPREHENSIVE PROBLEMS
6.70 Your laboratory has available a large number of
10-µF capacitors rated at 300 V. To design a
capacitor bank of 40-µF rated at 600 V, how many
10-µF capacitors are needed and how would you
connect them?
6.71 When a capacitor is connected to a dc source, its
voltage rises from 20 V to 36 V in 4 µs with an
average charging current of 0.6 A. Determine the
value of the capacitance.
6.72 A square-wave generator produces the voltage
waveform shown in Fig. 6.87(a). What kind of a
circuit component is needed to convert the voltage
waveform to the triangular current waveform shown
in Fig. 6.87(b)? Calculate the value of the
component, assuming that it is initially uncharged.
v (V)
0
5
−5
3 4
2
1 t (ms)
(a)
(b)
i (A)
4
3 4
2
1
0 t (ms)
Figure 6.87 For Prob. 6.72.
6.73 In an electric power plant substation, a capacitor
bank is made of 10 capacitor strings connected in
parallel. Each string consists of eight 1000-µF
capacitors connected in series, with each capacitor
charged to 100 V.
(a) Calculate the total capacitance of the bank.
(b) Determine the total energy stored in the bank.
237
C H A P T E R
FIRST-ORDER CIRCUITS
7
I often say that when you can measure what you are speaking about, and
express it in numbers, you know something about it; but when you cannot
express it in numbers, your knowledge is of a meager and unsatisfactory
kind; it may be the beginning of knowledge, but you have scarcely, in
your thoughts, advanced to the stage of a science, whatever the matter
may be.
—Lord Kelvin
Enhancing Your Career
Careers in Computer Engineering Electrical engineer-
ing education has gone through drastic changes in recent
decades. Most departments have come to be known as De-
partment of Electrical and Computer Engineering, empha-
sizing the rapid changes due to computers. Computers oc-
cupy a prominent place in modern society and education.
They have become commonplace and are helping to change
the face of research, development, production, business,
and entertainment. The scientist, engineer, doctor, attor-
ney, teacher, airline pilot, businessperson—almost anyone
benefits from a computer’s abilities to store large amounts
of information and to process that information in very short
periods of time. The internet, a computer communication
network, is becoming essential in business, education, and
library science. Computer usage is growing by leaps and
bounds.
Three major disciplines study computer systems:
computer science, computer engineering, and information
management science. Computer engineering has grown so
fast and wide that it is divorcing itself from electrical en-
gineering. But, in many schools of engineering, computer
engineering is still an integral part of electrical engineering.
An education in computer engineering should provide
breadth in software, hardware design, and basic modeling
techniques. It should include courses in data structures, dig-
ital systems, computer architecture, microprocessors, inter-
facing, software engineering, and operating systems. Elec-
trical engineers who specialize in computer engineering find
Computer design of very large scale integrated (VLSI) circuits.
Source: M. E. Hazen, Fundamentals of DC and AC Circuits,
Philadelphia: Saunders, 1990, p. 20A4.
jobs in computer industries and in numerous fields where
computers are being used. Companies that produce soft-
ware are growing rapidly in number and size and providing
employment for those who are skilled in programming. An
excellent way to advance one’s knowledge of computers is
to join the IEEE Computer Society, which sponsors diverse
magazines, journals, and conferences.
238 PART 1 DC Circuits
7.1 INTRODUCTION
Now that we have considered the three passive elements (resistors, ca-
pacitors, and inductors) and one active element (the op amp) individually,
we are prepared to consider circuits that contain various combinations of
two or three of the passive elements. In this chapter, we shall examine
two types of simple circuits: a circuit comprising a resistor and capaci-
tor and a circuit comprising a resistor and an inductor. These are called
RC and RL circuits, respectively. As simple as these circuits are, they
find continual applications in electronics, communications, and control
systems, as we shall see.
We carry out the analysis of RC and RL circuits by applying Kirch-
hoff’s laws, as we did for resistive circuits. The only difference is that
applying Kirchhoff’s laws to purely resistive circuits results in algebraic
equations, while applying the laws to RC and RL circuits produces dif-
ferential equations, which are more difficult to solve than algebraic equa-
tions. The differential equations resulting from analyzing RC and RL
circuits are of the first order. Hence, the circuits are collectively known
as first-order circuits.
A first-order circuit is characterized by a first-order differential equation.
In addition to there being two types of first-order circuits (RC
and RL), there are two ways to excite the circuits. The first way is by
initial conditions of the storage elements in the circuits. In these so-
called source-free circuits, we assume that energy is initially stored in
the capacitive or inductive element. The energy causes current to flow in
the circuit and is gradually dissipated in the resistors. Although source-
free circuits are by definition free of independent sources, they may have
dependent sources. The second way of exciting first-order circuits is by
independent sources. In this chapter, the independent sources we will
consider are dc sources. (In later chapters, we shall consider sinusoidal
and exponential sources.) The two types of first-order circuits and the
two ways of exciting them add up to the four possible situations we will
study in this chapter.
Finally, we consider four typical applications of RC and RL cir-
cuits: delay and relay circuits, a photoflash unit, and an automobile igni-
tion circuit.
7.2 THE SOURCE-FREE RC CIRCUIT
A source-free RC circuit occurs when its dc source is suddenly discon-
nected. The energy already stored in the capacitor is released to the
resistors.
v
+
−
iR
iC
R
C
Figure 7.1 A source-free
RC circuit.
A circuit response is the manner in which the
circuit reacts to an excitation.
Consider a series combination of a resistor and an initially charged
capacitor, as shown in Fig. 7.1. (The resistor and capacitor may be the
equivalent resistance and equivalent capacitance of combinations of re-
sistors and capacitors.) Our objective is to determine the circuit response,
which, for pedagogic reasons, we assume to be the voltage v(t) across
CHAPTER 7 First-Order Circuits 239
the capacitor. Since the capacitor is initially charged, we can assume that
at time t = 0, the initial voltage is
v(0) = V0 (7.1)
with the corresponding value of the energy stored as
w(0) =
1
2
CV 2
0 (7.2)
Applying KCL at the top node of the circuit in Fig. 7.1,
iC + iR = 0 (7.3)
By definition, iC = C dv/dt and iR = v/R. Thus,
C
dv
dt
+
v
R
= 0 (7.4a)
or
dv
dt
+
v
RC
= 0 (7.4b)
This is a first-order differential equation, since only the first derivative of
v is involved. To solve it, we rearrange the terms as
dv
v
= −
1
RC
dt (7.5)
Integrating both sides, we get
ln v = −
t
RC
+ ln A
where ln A is the integration constant. Thus,
ln
v
A
= −
t
RC
(7.6)
Taking powers of e produces
v(t) = Ae−t/RC
But from the initial conditions, v(0) = A = V0. Hence,
v(t) = V0e−t/RC
(7.7)
This shows that the voltage response of the RC circuit is an exponential
decay of the initial voltage. Since the response is due to the initial energy
stored and the physical characteristics of the circuit and not due to some
external voltage or current source, it is called the natural response of the
circuit.
The natural response of a circuit refers to the behavior (in terms of voltages and
currents) of the circuit itself, with no external sources of excitation.
The natural response depends on the nature of
the circuit alone, with no external sources. In
fact, the circuit has a response only because of
the energy initially stored in the capacitor.
The natural response is illustrated graphically in Fig. 7.2. Note that at
t = 0, we have the correct initial condition as in Eq. (7.1). As t increases,
the voltage decreases toward zero. The rapidity with which the voltage
decreases is expressed in terms of the time constant, denoted by the lower
case Greek letter tau, τ.
240 PART 1 DC Circuits
The time constant of a circuit is the time required for the response to decay by a
factor of 1/e or 36.8 percent of its initial value.1
Voe−t ⁄ t
t t
0.368Vo
Vo
v
0
Figure 7.2 The voltage response of the RC
circuit.
t 2t 3t 4t 5t t (s)
0
v
Vo
0.37
0.25
0.75
1.0
0.50
Tangent at t = 0
Figure7.3 Graphical determination of the
time constant τ from the response curve.
This implies that at t = τ, Eq. (7.7) becomes
V0e−τ/RC
= V0e−1
= 0.368V0
or
τ = RC (7.8)
In terms of the time constant, Eq. (7.7) can be written as
v(t) = V0e−t/τ
(7.9)
With a calculator it is easy to show that the value of v(t)/V0 is as
shown in Table 7.1. It is evident from Table 7.1 that the voltage v(t) is less
than 1 percent of V0 after 5τ (five time constants). Thus, it is customary
to assume that the capacitor is fully discharged (or charged) after five
time constants. In other words, it takes 5τ for the circuit to reach its final
state or steady state when no changes take place with time. Notice that
for every time interval of τ, the voltage is reduced by 36.8 percent of its
previous value, v(t + τ) = v(t)/e = 0.368v(t), regardless of the value
of t.
TABLE 7.1 Values of
v(t)/V0 = e−t/τ
.
t v(t)/V0
τ 0.36788
2τ 0.13534
3τ 0.04979
4τ 0.01832
5τ 0.00674
Observe from Eq. (7.8) that the smaller the time constant, the more
rapidly the voltage decreases, that is, the faster the response. This is
illustrated in Fig. 7.4. A circuit with a small time constant gives a fast
response in that it reaches the steady state (or final state) quickly due to
quick dissipation of energy stored, whereas a circuit with a large time
1The time constant may be viewed from another perspective. Evaluating the derivative of
v(t) in Eq. (7.7) at t = 0, we obtain
d
dt

v
V0
 



t = 0
= −
1
τ
e−t/τ




t=0
= −
1
τ
Thus the time constant is the initial rate of decay, or the time taken for v/V0 to decay from
unity to zero, assuming a constant rate of decay. This initial slope interpretation of the
time constant is often used in the laboratory to find τ graphically from the response curve
displayed on an oscilloscope. To find τ from the response curve, draw the tangent to the
curve, as shown in Fig. 7.3. The tangent intercepts with the time axis at t = τ.
CHAPTER 7 First-Order Circuits 241
0 t
1
3 4 5
1 2
v
Vo
e−t ⁄t
=
t = 0.5
t = 1
t = 2
Figure7.4 Plot of v/V0 = e−t/τ for various values of the time constant.
constant gives a slow response because it takes longer to reach steady
state. At any rate, whether the time constant is small or large, the circuit
reaches steady state in five time constants.
With the voltage v(t) in Eq. (7.9), we can find the current iR(t),
iR(t) =
v(t)
R
=
V0
R
e−t/τ
(7.10)
The power dissipated in the resistor is
p(t) = viR =
V 2
0
R
e−2t/τ
(7.11)
The energy absorbed by the resistor up to time t is
wR(t) =
 t
0
p dt =
 t
0
V 2
0
R
e−2t/τ
dt
= −
τV 2
0
2R
e−2t/τ




t
0
=
1
2
CV 2
0 (1 − e−2t/τ
), τ = RC
(7.12)
Notice that as t → ∞, wR(∞) → 1
2
CV 2
0 , which is the same as wC(0),
the energy initially stored in the capacitor. The energy that was initially
stored in the capacitor is eventually dissipated in the resistor.
In summary:
The Key to Working with a Source-free RC Circuit is
Finding:
1. The initial voltage v(0) = V0 across the capacitor.
2. The time constant τ.
Thetimeconstantisthesameregardlessofwhat
the output is defined to be.
With these two items, we obtain the response as the capacitor voltage
vC(t) = v(t) = v(0)e−t/τ
. Once the capacitor voltage is first obtained,
other variables (capacitor current iC, resistor voltage vR, and resistor
current iR) can be determined. In finding the time constant τ = RC, R is
often the Thevenin equivalent resistance at the terminals of the capacitor;
that is, we take out the capacitor C and find R = RTh at its terminals.
When a circuit contains a single capacitor and
several resistors and dependent sources, the
Thevenin equivalent can be found at the termi-
nals of the capacitor to form a simple RC circuit.
Also, onecanuseThevenin’stheoremwhensev-
eral capacitors can be combined to form a single
equivalent capacitor.
242 PART 1 DC Circuits
E X A M P L E 7 . 1
In Fig. 7.5, let vC(0) = 15 V. Find vC, vx, and ix for t  0.
5 Ω
8 Ω
12 Ω
vC vx
ix
+
−
+
−
0.1 F
Figure7.5 For Example 7.1.
Solution:
We first need to make the circuit in Fig. 7.5 conform with the standard
RC circuit in Fig. 7.1. We find the equivalent resistance or the Thevenin
resistance at the capacitor terminals. Our objective is always to first obtain
capacitor voltage vC. From this, we can determine vx and ix.
The 8- and 12- resistors in series can be combined to give a
20- resistor. This 20- resistor in parallel with the 5- resistor can be
combined so that the equivalent resistance is
Req =
20 × 5
20 + 5
= 4 
Hence, the equivalent circuit is as shown in Fig. 7.6, which is analogous
to Fig. 7.1. The time constant is
τ = ReqC = 4(0.1) = 0.4 s
Thus,
v = v(0)e−t/τ
= 15e−t/0.4
V, vC = v = 15e−2.5t
V
From Fig. 7.5, we can use voltage division to get vx; so
vx =
12
12 + 8
v = 0.6(15e−2.5t
) = 9e−2.5t
V
Finally,
ix =
vx
12
= 0.75e−2.5t
A
v
+
−
Req 0.1 F
Figure7.6 Equivalent circuit
for the circuit in Fig. 7.5.
P R A C T I C E P R O B L E M 7 . 1
Refer to the circuit in Fig. 7.7. Let vC(0) = 30 V. Determine vC, vx, and
io for t ≥ 0.
12 Ω
8 Ω
vC
F
6 Ω
io
+
−
vx
+
−
1
3
Figure7.7 For Practice Prob. 7.1.
Answer: 30e−0.25t
V, 10e−0.25t
V, −2.5e−0.25t
A.
E X A M P L E 7 . 2
The switch in the circuit in Fig. 7.8 has been closed for a long time, and
it is opened at t = 0. Find v(t) for t ≥ 0. Calculate the initial energy
stored in the capacitor.
3 Ω
20 V
+
−
v
9 Ω
t = 0
1 Ω
20 mF
+
−
Figure7.8 For Example 7.2.
Solution:
For t  0, the switch is closed; the capacitor is an open circuit to dc, as
represented in Fig. 7.9(a). Using voltage division
CHAPTER 7 First-Order Circuits 243
vC(t) =
9
9 + 3
(20) = 15 V, t  0
Since the voltage across a capacitor cannot change instantaneously, the
voltage across the capacitor at t = 0−
is the same at t = 0, or
vC(0) = V0 = 15 V
9 Ω
1 Ω
vC(0)
3 Ω
9 Ω
1 Ω
+
−
+
−
20 V
(a)
(b)
+
−
Vo = 15 V 20 mF
Figure 7.9 For Example 7.2: (a) t  0,
(b) t  0.
For t  0, the switch is opened, and we have the RC circuit shown
in Fig. 7.9(b). [Notice that the RC circuit in Fig. 7.9(b) is source free;
the independent source in Fig. 7.8 is needed to provide V0 or the initial
energy in the capacitor.] The 1- and 9- resistors in series give
Req = 1 + 9 = 10 
The time constant is
τ = ReqC = 10 × 20 × 10−3
= 0.2 s
Thus, the voltage across the capacitor for t ≥ 0 is
v(t) = vC(0)e−t/τ
= 15e−t/0.2
V
or
v(t) = 15e−5t
V
The initial energy stored in the capacitor is
wC(0) =
1
2
Cv2
C(0) =
1
2
× 20 × 10−3
× 152
= 2.25 J
P R A C T I C E P R O B L E M 7 . 2
If the switch in Fig. 7.10 opens at t = 0, find v(t) for t ≥ 0 and wC(0).
Answer: 8e−2t
V, 5.33 J.
6 Ω
+
−
24 V
+
−
v 12 Ω 4 Ω
t = 0
F
1
6
Figure7.10 For Practice Prob. 7.2.
7.3 THE SOURCE-FREE RL CIRCUIT
vL
+
−
R
L
i
vR
+
−
Figure 7.11 A source-
free RL circuit.
Consider the series connection of a resistor and an inductor, as shown in
Fig. 7.11. Our goal is to determine the circuit response, which we will
assume to be the current i(t) through the inductor. We select the inductor
current as the response in order to take advantage of the idea that the
inductor current cannot change instantaneously. At t = 0, we assume
that the inductor has an initial current I0, or
i(0) = I0 (7.13)
with the corresponding energy stored in the inductor as
w(0) =
1
2
LI2
0 (7.14)
244 PART 1 DC Circuits
Applying KVL around the loop in Fig. 7.11,
vL + vR = 0 (7.15)
But vL = L di/dt and vR = iR. Thus,
L
di
dt
+ Ri = 0
or
di
dt
+
R
L
i = 0 (7.16)
Rearranging terms and integrating gives
 i(t)
I0
di
i
= −
 t
0
R
L
dt
ln i




i(t)
I0
= −
Rt
L




t
0
⇒ ln i(t) − ln I0 = −
Rt
L
+ 0
or
ln
i(t)
I0
= −
Rt
L
(7.17)
Taking the powers of e, we have
i(t) = I0e−Rt/L
(7.18)
This shows that the natural response of the RL circuit is an exponential
decay of the initial current. The current response is shown in Fig. 7.12.
It is evident from Eq. (7.18) that the time constant for the RL circuit is
τ =
L
R
(7.19)
with τ again having the unit of seconds. Thus, Eq. (7.18) may be written
as
i(t) = I0e−t/τ
(7.20)
Tangent at t = 0
Ioe−t ⁄t
t t
0.368Io
Io
i(t)
0
Figure 7.12 The current response of the RL
circuit.
The smaller the time constant τ of a circuit, the
faster the rate of decay of the response. The
larger the time constant, the slower the rate of
decayoftheresponse. Atanyrate, theresponse
decays to less than 1 percent of its initial value
(i.e., reaches steady state) after 5τ.
Figure 7.12 shows an initial slope interpretation
may be given to τ.
With the current in Eq. (7.20), we can find the voltage across the
resistor as
vR(t) = iR = I0Re−t/τ
(7.21)
The power dissipated in the resistor is
p = vRi = I2
0 Re−2t/τ
(7.22)
The energy absorbed by the resistor is
wR(t) =
 t
0
p dt =
 t
0
I2
0 Re−2t/τ
dt = −
1
2
τI2
0 Re−2t/τ




t
0
, τ =
L
R
or
wR(t) =
1
2
LI2
0 (1 − e−2t/τ
) (7.23)
Note that as t → ∞, wR(∞) → 1
2
LI2
0 , which is the same as wL(0), the
initial energy stored in the inductor as in Eq. (7.14). Again, the energy
initially stored in the inductor is eventually dissipated in the resistor.
CHAPTER 7 First-Order Circuits 245
In summary:
The Key to Working with a Source-free RL Circuit is
to Find:
1. The initial current i(0) = I0 through the inductor.
2. The time constant τ of the circuit.
With the two items, we obtain the response as the inductor current iL(t) =
i(t) = i(0)e−t/τ
. Once we determine the inductor current iL, other vari-
ables (inductor voltage vL, resistor voltage vR, and resistor current iR)
can be obtained. Note that in general, R in Eq. (7.19) is the Thevenin
resistance at the terminals of the inductor.
When a circuit has a single inductor and several
resistors and dependent sources, the Thevenin
equivalent can be found at the terminals of the
inductor to form a simple RL circuit. Also, one
can use Thevenin’s theorem when several induc-
torscanbecombinedtoformasingleequivalent
inductor.
E X A M P L E 7 . 3
Assuming that i(0) = 10 A, calculate i(t) and ix(t) in the circuit in Fig.
7.13.
2 Ω
4 Ω
0.5 H
+
−
i
3i
ix
Figure7.13 For Example 7.3.
Solution:
There are two ways we can solve this problem. One way is to obtain the
equivalent resistance at the inductor terminals and then use Eq. (7.20).
The other way is to start from scratch by using Kirchhoff’s voltage law.
Whichever approach is taken, it is always better to first obtain the inductor
current.
METHOD 1 The equivalent resistance is the same as the Thevenin
resistance at the inductor terminals. Because of the dependent source,
we insert a voltage source with vo = 1 V at the inductor terminals a-b,
as in Fig. 7.14(a). (We could also insert a 1-A current source at the ter-
minals.) Applying KVL to the two loops results in
2(i1 − i2) + 1 = 0 ⇒ i1 − i2 = −
1
2
(7.3.1)
6i2 − 2i1 − 3i1 = 0 ⇒ i2 =
5
6
i1 (7.3.2)
Substituting Eq. (7.3.2) into Eq. (7.3.1) gives
4 Ω
2 Ω
vo = 1 V +
−
+
−
io
i1 i2 3i
(a)
a
b
4 Ω
2 Ω +
−
i1 i2 3i
(b)
0.5 H
Figure7.14 Solving the circuit in Fig. 7.13.
246 PART 1 DC Circuits
i1 = −3 A, io = −i1 = 3 A
Hence,
Req = RTh =
vo
io
=
1
3

The time constant is
τ =
L
Req
=
1
2
1
3
=
3
2
s
Thus, the current through the inductor is
i(t) = i(0)e−t/τ
= 10e−(2/3)t
A, t  0
METHOD 2 We may directly apply KVL to the circuit as in Fig.
7.14(b). For loop 1,
1
2
di1
dt
+ 2(i1 − i2) = 0
or
di1
dt
+ 4i1 − 4i2 = 0 (7.3.3)
For loop 2,
6i2 − 2i1 − 3i1 = 0 ⇒ i2 =
5
6
i1 (7.3.4)
Substituting Eq. (7.3.4) into Eq. (7.3.3) gives
di1
dt
+
2
3
i1 = 0
Rearranging terms,
di1
i1
= −
2
3
dt
Since i1 = i, we may replace i1 with i and integrate:
ln i




i(t)
i(0)
= −
2
3
t




t
0
or
ln
i(t)
i(0)
= −
2
3
t
Taking the powers of e, we finally obtain
i(t) = i(0)e−(2/3)t
= 10e−(2/3)t
A, t  0
which is the same as by Method 1.
The voltage across the inductor is
v = L
di
dt
= 0.5(10)

−
2
3

e−(2/3)t
= −
10
3
e−(2/3)t
V
Since the inductor and the 2- resistor are in parallel,
ix(t) =
v
2
= −1.667e−(2/3)t
A, t  0
CHAPTER 7 First-Order Circuits 247
P R A C T I C E P R O B L E M 7 . 3
Find i and vx in the circuit in Fig. 7.15. Let i(0) = 5 A.
1 Ω
5 Ω
3 Ω
+
− 2vx
H
i + −
vx
1
6
Figure7.15 For Practice Prob. 7.3.
Answer: 5e−53t
A, −15e−53t
V.
E X A M P L E 7 . 4
The switch in the circuit of Fig. 7.16 has been closed for a long time. At
t = 0, the switch is opened. Calculate i(t) for t  0.
2 Ω 4 Ω
+
− 40 V 16 Ω
12 Ω 2 H
t = 0
i(t)
Figure7.16 For Example 7.4.
Solution:
When t  0, the switch is closed, and the inductor acts as a short circuit
to dc. The 16- resistor is short-circuited; the resulting circuit is shown
in Fig. 7.17(a). To get i1 in Fig. 7.17(a), we combine the 4- and 12-
resistors in parallel to get
4 × 12
4 + 12
= 3 
Hence,
i1 =
40
2 + 3
= 8 A
We obtain i(t) from i1 in Fig. 7.17(a) using current division, by writing
i(t) =
12
12 + 4
i1 = 6 A, t  0
Since the current through an inductor cannot change instantaneously,
i(0) = i(0−
) = 6 A
When t  0, the switch is open and the voltage source is discon-
nected. We now have the RL circuit in Fig. 7.17(b). Combining the re-
sistors, we have
Req = (12 + 4) 16 = 8 
The time constant is
τ =
L
Req
=
2
8
=
1
4
s
Thus,
i(t) = i(0)e−t/τ
= 6e−4t
A
4 Ω
12 Ω
2 Ω
+
−
i1
2 H
i(t)
40 V
i(t)
(a)
16 Ω
12 Ω
4 Ω
(b)
Figure 7.17 Solving the circuit of Fig. 7.16: (a)
for t  0, (b) for t  0.
248 PART 1 DC Circuits
P R A C T I C E P R O B L E M 7 . 4
For the circuit in Fig. 7.18, find i(t) for t  0.
5 Ω
5 A
12 Ω 8 Ω
2 H
t = 0
i(t)
Figure7.18 For Practice Prob. 7.4.
Answer: 2e−2t
A, t  0.
E X A M P L E 7 . 5
In the circuit shown in Fig. 7.19, find io, vo, and i for all time, assuming
that the switch was open for a long time.
10 V 6 Ω 2 H
t = 0
i
io
+ −
vo
3 Ω
2 Ω
+
−
Figure7.19 For Example 7.5.
Solution:
Itisbettertofirstfindtheinductorcurrenti andthenobtainotherquantities
from it.
2 Ω 3 Ω
+
−
10 V 6 Ω
i
io
+ −
vo
+ −
vo
(a)
(b)
6 Ω
3 Ω
2 H
i
io
vL
+
−
Figure 7.20 The circuit in Fig. 7.19 for:
(a) t  0, (b) t  0.
For t  0, the switch is open. Since the inductor acts like a short
circuit to dc, the 6- resistor is short-circuited, so that we have the circuit
shown in Fig. 7.20(a). Hence, io = 0, and
i(t) =
10
2 + 3
= 2 A, t  0
vo(t) = 3i(t) = 6 V, t  0
Thus, i(0) = 2.
For t  0, the switch is closed, so that the voltage source is short-
circuited. We now have a source-free RL circuit as shown in Fig. 7.20(b).
At the inductor terminals,
RTh = 3 6 = 2 
so that the time constant is
τ =
L
RTh
= 1 s
Hence,
i(t) = i(0)e−t/τ
= 2e−t
A, t  0
Since the inductor is in parallel with the 6- and 3- resistors,
vo(t) = −vL = −L
di
dt
= −2(−2e−t
) = 4e−t
V, t  0
and
io(t) =
vL
6
= −
2
3
e−t
A, t  0
CHAPTER 7 First-Order Circuits 249
Thus, for all time,
io(t) =



0 A, t  0
−
2
3
e−t
A, t  0
, vo(t) =

6 V, t  0
4e−t
V, t  0
i(t) =

2 A, t  0
2e−t
A, t ≥ 0
We notice that the inductor current is continuous at t = 0, while the
current through the 6- resistor drops from 0 to −2/3 at t = 0, and the
voltage across the 3- resistor drops from 6 to 4 at t = 0. We also notice
that the time constant is the same regardless of what the output is defined
to be. Figure 7.21 plots i and io.
t
2
i(t)
2
3
−
io(t)
Figure7.21 A plot of i and i0.
P R A C T I C E P R O B L E M 7 . 5
Determine i, io, and vo for all t in the circuit shown in Fig. 7.22. Assume
that the switch was closed for a long time.
1 H
4 Ω 2 Ω
3 Ω
6 A
i
t = 0
io
vo
+
−
Figure7.22 For Practice Prob. 7.5.
Answer: i =

4 A, t  0
4e−2t
A, t ≥ 0
, io =

2 A, t  0
−(4/3)e−2t
A, t  0
,
vo =

4 V, t  0
−(8/3)e−2t
V, t  0
7.4 SINGULARITY FUNCTIONS
Before going on with the second half of this chapter, we need to digress
and consider some mathematical concepts that will aid our understanding
of transient analysis. A basic understanding of singularity functions will
help us make sense of the response of first-order circuits to a sudden
application of an independent dc voltage or current source.
Singularity functions (also called switching functions) are very use-
fulincircuitanalysis. Theyserveasgoodapproximationstotheswitching
signals that arise in circuits with switching operations. They are helpful in
the neat, compact description of some circuit phenomena, especially the
step response of RC or RL circuits to be discussed in the next sections.
By definition,
Singularity functions are functions that either are discontinuous or have
discontinuous derivatives.
The three most widely used singularity functions in circuit analysis
are the unit step, the unit impulse, and the unit ramp functions.
250 PART 1 DC Circuits
The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.
In mathematical terms,
u(t) =

0, t  0
1, t  0
(7.24)
0 t
1
u(t)
Figure 7.23 The unit step
function.
The unit step function is undefined at t = 0, where it changes abruptly
from 0 to 1. It is dimensionless, like other mathematical functions such as
sine and cosine. Figure 7.23 depicts the unit step function. If the abrupt
change occurs at t = t0 (where t0  0) instead of t = 0, the unit step
function becomes
u(t − t0) =

0, t  t0
1, t  t0
(7.25)
which is the same as saying that u(t) is delayed by t0 seconds, as shown in
Fig. 7.24(a). To get Eq. (7.25) from Eq. (7.24), we simply replace every
t by t − t0. If the change is at t = −t0, the unit step function becomes
u(t + t0) =

0, t  −t0
1, t  −t0
(7.26)
meaning that u(t) is advanced by t0 seconds, as shown in Fig. 7.24(b).
0 t
1
u(t − to)
to
(a)
0 t
u(t + to)
−to
(b)
1
Figure7.24 (a) The unit step
function delayed by t0, (b) the unit
step advanced by t0.
Alternatively,wemayderiveEqs.(7.25)and(7.26)
from Eq. (7.24) by writing u[f(t)] = 1, f(t)  0,
where f(t) may be t − t0 or t + t0.
We use the step function to represent an abrupt change in voltage
or current, like the changes that occur in the circuits of control systems
and digital computers. For example, the voltage
v(t) =

0, t  t0
V0, t  t0
(7.27)
may be expressed in terms of the unit step function as
v(t) = V0u(t − t0) (7.28)
If we let t0 = 0, then v(t) is simply the step voltage V0u(t). A voltage
source of V0u(t) is shown in Fig. 7.25(a); its equivalent circuit is shown
in Fig. 7.25(b). It is evident in Fig. 7.25(b) that terminals a-b are short-
circuited (v = 0) for t  0 and that v = V0 appears at the terminals
for t  0. Similarly, a current source of I0u(t) is shown in Fig. 7.26(a),
while its equivalent circuit is in Fig. 7.26(b). Notice that for t  0, there
is an open circuit (i = 0), and that i = I0 flows for t  0.
+
−
(a)
Vou(t) +
−
(b)
Vo
b
a
b
a
t = 0
=
Figure7.25 (a) Voltage source of V0u(t), (b) its equivalent circuit.
CHAPTER 7 First-Order Circuits 251
(a)
Iou(t)
(b)
Io
b
a
b
a
t = 0 i
=
Figure7.26 (a) Current source of I0u(t), (b) its equivalent circuit.
The derivative of the unit step function u(t) is the unit impulse
function δ(t), which we write as
δ(t) =
d
dt
u(t) =



0, t  0
Undefined, t = 0
0, t  0
(7.29)
The unit impulse function—also known as the delta function—is shown
in Fig. 7.27.
The unit impulse function δ(t) is zero everywhere except at t = 0, where
it is undefined.
Impulsive currents and voltages occur in electric circuits as a result of
switching operations or impulsive sources. Although the unit impulse
functionisnotphysicallyrealizable(justlikeidealsources, idealresistors,
etc.), it is a very useful mathematical tool.
The unit impulse may be regarded as an applied or resulting shock.
It may be visualized as a very short duration pulse of unit area. This may
be expressed mathematically as
 0+
0−
δ(t) dt = 1 (7.30)
where t = 0−
denotes the time just before t = 0 and t = 0+
is the time
just after t = 0. For this reason, it is customary to write 1 (denoting
unit area) beside the arrow that is used to symbolize the unit impulse
function, as in Fig. 7.27. The unit area is known as the strength of the
impulse function. When an impulse function has a strength other than
unity, the area of the impulse is equal to its strength. For example, an
impulse function 10δ(t) has an area of 10. Figure 7.28 shows the impulse
functions 5δ(t + 2), 10δ(t), and −4δ(t − 3).
0 t
(1)
d(t)
Figure7.27 The unit
impulse function.
5d(t + 2)
10d(t)
−4d(t − 3)
1
0 2 3 t
−1
−2
Figure7.28 Three impulse functions.
To illustrate how the impulse function affects other functions, let
us evaluate the integral
 b
a
f (t)δ(t − t0) dt (7.31)
where a  t0  b. Since δ(t − t0) = 0 except at t = t0, the integrand is
252 PART 1 DC Circuits
zero except at t0. Thus,
 b
a
f (t)δ(t − t0) dt =
 b
a
f (t0)δ(t − t0) dt
= f (t0)
 b
a
δ(t − t0) dt = f (t0)
or
 b
a
f (t)δ(t − t0) dt = f (t0) (7.32)
This shows that when a function is integrated with the impulse function,
we obtain the value of the function at the point where the impulse occurs.
This is a highly useful property of the impulse function known as the
sampling or sifting property. The special case of Eq. (7.31) is for t0 = 0.
Then Eq. (7.32) becomes
 0+
0−
f (t)δ(t) dt = f (0) (7.33)
Integrating the unit step function u(t) results in the unit ramp func-
tion r(t); we write
r(t) =
 t
−∞
u(t) dt = tu(t) (7.34)
or
r(t) =

0, t ≤ 0
t, t ≥ 0
(7.35)
The unit ramp function is zero for negative values of t and has a unit slope for
positive values of t.
Figure 7.29 shows the unit ramp function. In general, a ramp is a function
that changes at a constant rate.
0 t
1
r(t)
1
Figure 7.29 The unit ramp
function.
The unit ramp function may be delayed or advanced as shown in
Fig. 7.30. For the delayed unit ramp function,
r(t − t0) =

0, t ≤ t0
t − t0, t ≥ t0
(7.36)
and for the advanced unit ramp function,
r(t + t0) =

0, t ≤ −t0
t − t0, t ≥ −t0
(7.37)
(a)
0 t
−to + 1
−to
1
r(t + to)
r(t − to)
(b)
0 t
to + 1
to
1
Figure7.30 The unit ramp
function: (a) delayed by t0,
(b) advanced by t0.
We should keep in mind that the three singularity functions (im-
pulse, step, and ramp) are related by differentiation as
δ(t) =
du(t)
dt
, u(t) =
dr(t)
dt
(7.38)
or by integration as
u(t) =
 t
−∞
δ(t) dt, r(t) =
 t
−∞
u(t) dt (7.39)
CHAPTER 7 First-Order Circuits 253
Although there are many more singularity functions, we are only inter-
ested in these three (the impulse function, the unit step function, and the
ramp function) at this point.
E X A M P L E 7 . 6
Express the voltage pulse in Fig. 7.31 in terms of the unit step. Calculate
its derivative and sketch it.
0 t
10
v(t)
3 4 5
1 2
Figure7.31 For Example 7.6.
Solution:
The type of pulse in Fig. 7.31 is called the gate function. It may be re-
garded as a step function that switches on at one value of t and switches
off at another value of t. The gate function shown in Fig. 7.31 switches
on at t = 2 s and switches off at t = 5 s. It consists of the sum of two
unit step functions as shown in Fig. 7.32(a). From the figure, it is evident
that
v(t) = 10u(t − 2) − 10u(t − 5) = 10[u(t − 2) − u(t − 5)]
Taking the derivative of this gives
dv
dt
= 10[δ(t − 2) − δ(t − 5)]
which is shown in Fig. 7.32(b). We can obtain Fig. 7.32(b) directly from
Fig. 7.31 by simply observing that there is a sudden increase by 10 V at
t = 2 s leading to 10δ(t − 2). At t = 5 s, there is a sudden decrease by
10 V leading to −10 V δ(t − 5).
Gate functions are used along with switches to
pass or block another signal.
0 t
2
1
10
10u(t − 2) −10u(t − 5)
(a)
1 2
0
3 4 5 t
10
−10
+
(b)
10
3 4 5 t
1 2
0
−10
dv
dt
Figure7.32 (a) Decomposition of the pulse in Fig. 7.31, (b) derivative of the pulse in Fig. 7.31.
254 PART 1 DC Circuits
P R A C T I C E P R O B L E M 7 . 6
Express the current pulse in Fig. 7.33 in terms of the unit step. Find its
integral and sketch it.
Answer: 10[u(t)−2u(t −2)+u(t −4)], 10[r(t)−2r(t −2)+r(t −4)].
See Fig. 7.34.
0
t
10
−10
i(t)
2 4
Figure7.33 For Practice Prob. 7.6.
2
0 4 t
20
i dt
∫
Figure7.34 Integral of i(t) in Fig. 7.33.
E X A M P L E 7 . 7
Express the sawtooth function shown in Fig. 7.35 in terms of singularity
functions.
0 t
10
v(t)
2
Figure 7.35 For Example 7.7.
Solution:
There are three ways of solving this problem. The first method is by mere
observation of the given function, while the other methods involve some
graphical manipulations of the function.
METHOD 1 By looking at the sketch of v(t) in Fig. 7.35, it is not
hard to notice that the given function v(t) is a combination of singularity
functions. So we let
v(t) = v1(t) + v2(t) + · · · (7.7.1)
The function v1(t) is the ramp function of slope 5, shown in Fig. 7.36(a);
that is,
v1(t) = 5r(t) (7.7.2)
0 t
10
v1(t)
2 0 t
10
v1 + v2
2
0
t
−10
v2(t)
2
+
(a)
(b)
(c)
=
Figure7.36 Partial decomposition of v(t) in Fig. 7.35.
CHAPTER 7 First-Order Circuits 255
Since v1(t) goes to infinity, we need another function at t = 2 s in order
to get v(t). We let this function be v2, which is a ramp function of slope
−5, as shown in Fig. 7.36(b); that is,
v2(t) = −5r(t − 2) (7.7.3)
Adding v1 and v2 gives us the signal in Fig. 7.36(c). Obviously, this is
not the same as v(t) in Fig. 7.35. But the difference is simply a constant
10 units for t  2 s. By adding a third signal v3, where
v3 = −10u(t − 2) (7.7.4)
we get v(t), as shown in Fig. 7.37. Substituting Eqs. (7.7.2) through
(7.7.4) into Eq. (7.7.1) gives
v(t) = 5r(t) − 5r(t − 2) − 10u(t − 2)
0 t
10
v1 + v2
2
+
(c)
(a)
=
0 t
10
v(t)
2
0
t
−10
v3(t)
2
(b)
Figure7.37 Complete decomposition of v(t) in Fig. 7.35.
METHOD 2 A close observation of Fig. 7.35 reveals that v(t) is a mul-
tiplication of two functions: a ramp function and a gate function. Thus,
v(t) = 5t[u(t) − u(t − 2)]
= 5tu(t) − 5tu(t − 2)
= 5r(t) − 5(t − 2 + 2)u(t − 2)
= 5r(t) − 5(t − 2)u(t − 2) − 10u(t − 2)
= 5r(t) − 5r(t − 2) − 10u(t − 2)
the same as before.
METHOD 3 This method is similar to Method 2. We observe from
Fig. 7.35 that v(t) is a multiplication of a ramp function and a unit step
function, as shown in Fig. 7.38. Thus,
v(t) = 5r(t)u(−t + 2)
If we replace u(−t) by 1 − u(t), then we can replace u(−t + 2) by
1 − u(t − 2). Hence,
v(t) = 5r(t)[1 − u(t − 2)]
which can be simplified as in Method 2 to get the same result.
256 PART 1 DC Circuits
0 t
10
5r(t)
2
×
0 t
u(−t + 2)
2
1
Figure7.38 Decomposition of v(t) in Fig. 7.35.
P R A C T I C E P R O B L E M 7 . 7
Refer to Fig. 7.39. Express i(t) in terms of singularity functions.
i(t) (A)
1
0
2 3 t (s)
2
−2
Figure7.39 For Practice Prob. 7.7.
Answer: 2u(t) − 2r(t) + 4r(t − 2) − 2r(t − 3).
E X A M P L E 7 . 8
Given the signal
g(t) =



3, t  0
−2, 0  t  1
2t − 4, t  1
express g(t) in terms of step and ramp functions.
Solution:
The signal g(t) may be regarded as the sum of three functions specified
within the three intervals t  0, 0  t  1, and t  1.
For t  0, g(t) may be regarded as 3 multiplied by u(−t), where
u(−t) = 1 for t  0 and 0 for t  0. Within the time interval 0  t  1,
the function may be considered as −2 multiplied by a gated function
[u(t) − u(t − 1)]. For t  1, the function may be regarded as 2t − 4
multiplied by the unit step function u(t − 1). Thus,
g(t) = 3u(−t) − 2[u(t) − u(t − 1)] + (2t − 4)u(t − 1)
= 3u(−t) − 2u(t) + (2t − 4 + 2)u(t − 1)
= 3u(−t) − 2u(t) + 2(t − 1)u(t − 1)
= 3u(−t) − 2u(t) + 2r(t − 1)
One may avoid the trouble of using u(−t) by replacing it with 1 − u(t).
Then
CHAPTER 7 First-Order Circuits 257
g(t) = 3[1 − u(t)] − 2u(t) + 2r(t − 1) = 3 − 5u(t) + 2r(t − 1)
Alternatively, we may plot g(t) and apply Method 1 from Example 7.7.
P R A C T I C E P R O B L E M 7 . 8
If
h(t) =







0, t  0
4, 0  t  2
6 − t, 2  t  6
0, t  6
express h(t) in terms of the singularity functions.
Answer: 4u(t) − r(t − 2) + r(t − 6).
E X A M P L E 7 . 9
Evaluate the following integrals involving the impulse function:
 10
0
(t2
+ 4t − 2)δ(t − 2) dt
 ∞
−∞
(δ(t − 1)e−t
cos t + δ(t + 1)e−t
sin t)dt
Solution:
For the first integral, we apply the sifting property in Eq. (7.32).
 10
0
(t2
+ 4t − 2)δ(t − 2)dt = (t2
+ 4t − 2)|t=2 = 4 + 8 − 2 = 10
Similarly, for the second integral,
 ∞
−∞
(δ(t − 1)e−t
cos t + δ(t + 1)e−t
sin t)dt
= e−t
cos t|t=1 + e−t
sin t|t=−1
= e−1
cos 1 + e1
sin(−1) = 0.1988 − 2.2873 = −2.0885
P R A C T I C E P R O B L E M 7 . 9
Evaluate the following integrals:
 ∞
−∞
(t3
+ 5t2
+ 10)δ(t + 3) dt,
 10
0
δ(t − π) cos 3t dt
Answer: 28, −1.
7.5 STEP RESPONSE OF AN RC CIRCUIT
When the dc source of an RC circuit is suddenly applied, the voltage
or current source can be modeled as a step function, and the response is
known as a step response.
258 PART 1 DC Circuits
The step response of a circuit is its behavior when the excitation is the step
function, which may be a voltage or a current source.
The step response is the response of the circuit due to a sudden application
of a dc voltage or current source.
R
C
t = 0
+
−
Vsu(t)
Vs
+
−
v
(a)
R
C
+
−
+
−
v
(b)
Figure 7.40 An RC circuit with
voltage step input.
Consider the RC circuit in Fig. 7.40(a) which can be replaced by the
circuit in Fig. 7.40(b), where Vs is a constant, dc voltage source. Again,
we select the capacitor voltage as the circuit response to be determined.
We assume an initial voltage V0 on the capacitor, although this is not
necessary for the step response. Since the voltage of a capacitor cannot
change instantaneously,
v(0−
) = v(0+
) = V0 (7.40)
where v(0−
) is the voltage across the capacitor just before switching and
v(0+
) is its voltage immediately after switching. Applying KCL, we have
C
dv
dt
+
v − Vsu(t)
R
= 0
or
dv
dt
+
v
RC
=
Vs
RC
u(t) (7.41)
where v is the voltage across the capacitor. For t  0, Eq. (7.41) becomes
dv
dt
+
v
RC
=
Vs
RC
(7.42)
Rearranging terms gives
dv
dt
= −
v − Vs
RC
or
dv
v − Vs
= −
dt
RC
(7.43)
Integrating both sides and introducing the initial conditions,
ln(v − Vs)




v(t)
V0
= −
t
RC




t
0
ln(v(t) − Vs) − ln(V0 − Vs) = −
t
RC
+ 0
or
ln
v − Vs
V0 − Vs
= −
t
RC
(7.44)
Taking the exponential of both sides
v − Vs
V0 − Vs
= e−t/τ
, τ = RC
v − Vs = (V0 − Vs)e−t/τ
or
v(t) = Vs + (V0 − Vs)e−t/τ
, t  0 (7.45)
CHAPTER 7 First-Order Circuits 259
Thus,
v(t) =

V0, t  0
Vs + (V0 − Vs)e−t/τ
, t  0
(7.46)
This is known as the complete response of the RC circuit to a sudden
application of a dc voltage source, assuming the capacitor is initially
charged. The reason for the term “complete” will become evident a little
later. Assuming that Vs  V0, a plot of v(t) is shown in Fig. 7.41.
0 t
Vs
v(t)
Vo
Figure7.41 Response of an
RC circuit with initially charged
capacitor.
If we assume that the capacitor is uncharged initially, we set V0 = 0
in Eq. (7.46) so that
v(t) =

0, t  0
Vs(1 − e−t/τ
), t  0
(7.47)
which can be written alternatively as
v(t) = Vs(1 − e−t/τ
)u(t) (7.48)
This is the complete step response of the RC circuit when the capacitor
is initially uncharged. The current through the capacitor is obtained from
Eq. (7.47) using i(t) = C dv/dt. We get
i(t) = C
dv
dt
=
C
τ
Vse−t/τ
, τ = RC, t  0
or
i(t) =
Vs
R
e−t/τ
u(t) (7.49)
Figure 7.42 shows the plots of capacitor voltage v(t) and capacitor current
i(t).
0 t
Vs
v(t)
0 t
i(t)
Vs
R
(a)
(b)
Figure7.42 Step response of an
RC circuit with initially uncharged
capacitor: (a) voltage response,
(b) current response.
Rather than going through the derivations above, there is a sys-
tematic approach—or rather, a short-cut method—for finding the step
response of an RC or RL circuit. Let us reexamine Eq. (7.45), which is
more general than Eq. (7.48). It is evident that v(t) has two components.
Thus, we may write
v = vf + vn (7.50)
where
vf = Vs (7.51)
and
vn = (V0 − Vs)e−t/τ
(7.52)
We know that vn is the natural response of the circuit, as discussed in
Section 7.2. Since this part of the response will decay to almost zero
after five time constants, it is also called the transient response because it
is a temporary response that will die out with time. Now, vf is known as
the forced response because it is produced by the circuit when an external
“force” is applied (a voltage source in this case). It represents what the
circuit is forced to do by the input excitation. It is also known as the
steady-state response, because it remains a long time after the circuit is
excited.
260 PART 1 DC Circuits
The natural response or transient response is the circuit’s temporary response
that will die out with time.
The forced response or steady-state response is the behavior of the circuit
a long time after an external excitation is applied.
The complete response of the circuit is the sum of the natural response
and the forced response. Therefore, we may write Eq. (7.45) as
v(t) = v(∞) + [v(0) − v(∞)]e−t/τ
(7.53)
where v(0) is the initial voltage at t = 0+
and v(∞) is the final or steady-
state value. Thus, to find the step response of an RC circuit requires three
things:
This is the same as saying that the complete re-
sponse is the sum of the transient response and
the steady-state response.
Once we know x(0), x(∞), and τ, almost all the
circuit problems in this chapter can be solved
using the formula
x(t) = x(∞)+ [x(0) − x(∞)] e-t/τ
1. The initial capacitor voltage v(0).
2. The final capacitor voltage v(∞).
3. The time constant τ.
We obtain item 1 from the given circuit for t  0 and items 2 and 3 from
the circuit for t  0. Once these items are determined, we obtain the
response using Eq. (7.53). This technique equally applies to RL circuits,
as we shall see in the next section.
Note that if the switch changes position at time t = t0 instead of at
t = 0, there is a time delay in the response so that Eq. (7.53) becomes
v(t) = v(∞) + [v(t0) − v(∞)]e−(t−t0)/τ
(7.54)
where v(t0) is the initial value at t = t+
0 . Keep in mind that Eq. (7.53) or
(7.54) applies only to step responses, that is, when the input excitation is
constant.
E X A M P L E 7 . 1 0
The switch in Fig. 7.43 has been in position A for a long time. At t = 0,
the switch moves to B. Determine v(t) for t  0 and calculate its value
at t = 1 s and 4 s.
3 kΩ
24 V 30 V
v
5 kΩ 0.5 mF
4 kΩ
+
−
+
−
t = 0
A B
+
−
Figure7.43 For Example 7.10.
CHAPTER 7 First-Order Circuits 261
Solution:
For t  0, the switch is at position A. Since v is the same as the voltage
across the 5-k resistor, the voltage across the capacitor just before t = 0
is obtained by voltage division as
v(0−
) =
5
5 + 3
(24) = 15 V
Using the fact that the capacitor voltage cannot change instantaneously,
v(0) = v(0−
) = v(0+
) = 15 V
For t  0, the switch is in position B. The Thevenin resistance connected
to the capacitor is RTh = 4 k, and the time constant is
τ = RThC = 4 × 103
× 0.5 × 10−3
= 2 s
Since the capacitor acts like an open circuit to dc at steady state, v(∞) =
30 V. Thus,
v(t) = v(∞) + [v(0) − v(∞)]e−t/τ
= 30 + (15 − 30)e−t/2
= (30 − 15e−0.5t
) V
At t = 1,
v(1) = 30 − 15e−0.5
= 20.902 V
At t = 4,
v(4) = 30 − 15e−2
= 27.97 V
P R A C T I C E P R O B L E M 7 . 1 0
Find v(t) for t  0 in the circuit in Fig. 7.44. Assume the switch has been
open for a long time and is closed at t = 0. Calculate v(t) at t = 0.5.
2 Ω
10 V 50 V
v
6 Ω
+
−
+
−
t = 0
F
1
3
+
−
Figure7.44 For Practice Prob. 7.10.
Answer: −5 + 15e−2t
V, 0.5182 V.
E X A M P L E 7 . 1 1
In Fig. 7.45, the switch has been closed for a long time and is opened at
t = 0. Find i and v for all time.
10 Ω
30u(t) V 10 V
v
20 Ω
+
−
+
−
i t = 0
F
1
4
+
−
Figure7.45 For Example 7.11.
262 PART 1 DC Circuits
Solution:
The resistor current i can be discontinuous at t = 0, while the capacitor
voltage v cannot. Hence, it is always better to find v and then obtain i
from v.
By definition of the unit step function,
30u(t) =

0, t  0
30, t  0
For t  0, the switch is closed and 30u(t) = 0, so that the 30u(t)
voltage source is replaced by a short circuit and should be regarded as
contributing nothing to v. Since the switch has been closed for a long
time, the capacitor voltage has reached steady state and the capacitor acts
like an open circuit. Hence, the circuit becomes that shown in Fig. 7.46(a)
for t  0. From this circuit we obtain
v = 10 V, i = −
v
10
= −1 A
Since the capacitor voltage cannot change instantaneously,
v(0) = v(0−
) = 10 V
10 Ω
10 V
+
−
v
20 Ω +
−
i
(a)
10 Ω
30 V
+
−
v
20 Ω
+
−
i
(b)
F
1
4
Figure 7.46 Solution of Example 7.11:
(a) for t  0, (b) for t  0.
For t  0, the switch is opened and the 10-V voltage source is
disconnected from the circuit. The 30u(t) voltage source is now opera-
tive, so the circuit becomes that shown in Fig. 7.46(b). After a long time,
the circuit reaches steady state and the capacitor acts like an open circuit
again. We obtain v(∞) by using voltage division, writing
v(∞) =
20
20 + 10
(30) = 20 V
The Thevenin resistance at the capacitor terminals is
RTh = 10 20 =
10 × 20
30
=
20
3

and the time constant is
τ = RThC =
20
3
·
1
4
=
5
3
s
Thus,
v(t) = v(∞) + [v(0) − v(∞)]e−t/τ
= 20 + (10 − 20)e−(3/5)t
= (20 − 10e−0.6t
) V
To obtain i, we notice from Fig. 7.46(b) that i is the sum of the currents
through the 20- resistor and the capacitor; that is,
i =
v
20
+ C
dv
dt
= 1 − 0.5e−0.6t
+ 0.25(−0.6)(−10)e−0.6t
= (1 + e−0.6t
) A
Notice from Fig. 7.46(b) that v + 10i = 30 is satisfied, as expected.
Hence,
v =

10 V, t  0
(20 − 10e−0.6t
) V, t ≥ 0
CHAPTER 7 First-Order Circuits 263
i =

−1 A, t  0
(1 + e−0.6t
) A, t  0
Notice that the capacitor voltage is continuous while the resistor current
is not.
P R A C T I C E P R O B L E M 7 . 1 1
The switch in Fig. 7.47 is closed at t = 0. Find i(t) and v(t) for all time.
Note that u(−t) = 1 for t  0 and 0 for t  0. Also, u(−t) = 1 − u(t).
5 Ω
+
−
20u(−t) V 10 Ω
0.2 F 3 A
v
i t = 0
+
−
Figure7.47 For Practice Prob. 7.11.
Answer: i(t) =

0, t  0
−2(1 + e−1.5t
) A, t  0
,
v =

20 V, t  0
10(1 + e−1.5t
) V, t  0
7.6 STEP RESPONSE OF AN RL CIRCUIT
Consider the RL circuit in Fig. 7.48(a), which may be replaced by the
circuit in Fig. 7.48(b). Again, our goal is to find the inductor current i as
the circuit response. Rather than apply Kirchhoff’s laws, we will use the
simple technique in Eqs. (7.50) through (7.53). Let the response be the
sum of the natural current and the forced current,
i = in + if (7.55)
We know that the natural response is always a decaying exponential, that
is,
in = Ae−t/τ
, τ =
L
R
(7.56)
where A is a constant to be determined.
R
Vs
t = 0
i
i
+
−
+
−
v(t)
L
(a)
R
Vsu(t) +
−
+
−
v(t)
L
(b)
Figure 7.48 An RL circuit with a
step input voltage.
The forced response is the value of the current a long time after
the switch in Fig. 7.48(a) is closed. We know that the natural response
essentially dies out after five time constants. At that time, the inductor
becomes a short circuit, and the voltage across it is zero. The entire source
voltage Vs appears across R. Thus, the forced response is
if =
Vs
R
(7.57)
264 PART 1 DC Circuits
Substituting Eqs. (7.56) and (7.57) into Eq. (7.55) gives
i = Ae−t/τ
+
Vs
R
(7.58)
We now determine the constant A from the initial value of i. Let I0 be
the initial current through the inductor, which may come from a source
other than Vs. Since the current through the inductor cannot change
instantaneously,
i(0+
) = i(0−
) = I0 (7.59)
Thus at t = 0, Eq. (7.58) becomes
I0 = A +
Vs
R
From this, we obtain A as
A = I0 −
Vs
R
Substituting for A in Eq. (7.58), we get
i(t) =
Vs
R
+

I0 −
Vs
R

e−t/τ
(7.60)
This is the complete response of the RL circuit. It is illustrated in Fig.
7.49. The response in Eq. (7.60) may be written as
i(t) = i(∞) + [i(0) − i(∞)]e−t/τ
(7.61)
where i(0) and i(∞) are the initial and final values of i. Thus, to find the
step response of an RL circuit requires three things:
0 t
i(t)
Vs
R
Io
Figure7.49 Total response
of the RL circuit with initial
inductor current I0.
1. The initial inductor current i(0) at t = 0+
.
2. The final inductor current i(∞).
3. The time constant τ.
We obtain item 1 from the given circuit for t  0 and items 2 and 3 from
the circuit for t  0. Once these items are determined, we obtain the
response using Eq. (7.61). Keep in mind that this technique applies only
for step responses.
Again, if the switching takes place at time t = t0 instead of t = 0,
Eq. (7.61) becomes
i(t) = i(∞) + [i(t0) − i(∞)]e−(t−t0)/τ
(7.62)
If I0 = 0, then
i(t) =



0, t  0
Vs
R
(1 − e−t/τ
), t  0
(7.63a)
or
i(t) =
Vs
R
(1 − e−t/τ
)u(t) (7.63b)
This is the step response of the RL circuit. The voltage across the inductor
is obtained from Eq. (7.63) using v = L di/dt. We get
CHAPTER 7 First-Order Circuits 265
v(t) = L
di
dt
= Vs
L
τR
e−t/τ
, τ =
L
R
, t  0
or
v(t) = Vse−t/τ
u(t) (7.64)
Figure 7.50 shows the step responses in Eqs. (7.63) and (7.64).
0 t
v(t)
0 t
i(t)
Vs
R
(a) (b)
Vs
Figure7.50 Step responses of an RL circuit with no initial
inductor current: (a) current response, (b) voltage response.
E X A M P L E 7 . 1 2
Find i(t) in the circuit in Fig. 7.51 for t  0. Assume that the switch has
been closed for a long time.
2 Ω 3 Ω
+
−
10 V
i
t = 0
H
1
3
Figure7.51 For Example 7.12.
Solution:
When t  0, the 3- resistor is short-circuited, and the inductor acts
like a short circuit. The current through the inductor at t = 0−
(i.e., just
before t = 0) is
i(0−
) =
10
2
= 5 A
Since the inductor current cannot change instantaneously,
i(0) = i(0+
) = i(0−
) = 5 A
When t  0, the switch is open. The 2- and 3- resistors are in series,
so that
i(∞) =
10
2 + 3
= 2 A
The Thevenin resistance across the inductor terminals is
RTh = 2 + 3 = 5 
For the time constant,
τ =
L
RTh
=
1
3
5
=
1
15
s
Thus,
i(t) = i(∞) + [i(0) − i(∞)]e−t/τ
= 2 + (5 − 2)e−15t
= 2 + 3e−15t
A, t  0
266 PART 1 DC Circuits
Check: In Fig. 7.51, for t  0, KVL must be satisfied; that is,
10 = 5i + L
di
dt
5i + L
di
dt
= [10 + 15e−15t
] +
1
3
(3)(−15)e−15t
= 10
This confirms the result.
P R A C T I C E P R O B L E M 7 . 1 2
The switch in Fig. 7.52 has been closed for a long time. It opens at t = 0.
Find i(t) for t  0.
1.5 H
10 Ω
5 Ω 3 A
t = 0
i
Figure7.52 For Practice Prob. 7.12.
Answer: (2 + e−10t
) A, t  0.
E X A M P L E 7 . 1 3
At t = 0, switch 1 in Fig. 7.53 is closed, and switch 2 is closed 4 s later.
Find i(t) for t  0. Calculate i for t = 2 s and t = 5 s.
4 Ω 6 Ω
+
−
+
−
40 V
10 V
2 Ω 5 H
i
t = 0
t = 4
S1
S2
P
Figure7.53 For Example 7.13.
Solution:
We need to consider the three time intervals t ≤ 0, 0 ≤ t ≤ 4, and t ≥ 4
separately. For t  0, switches S1 and S2 are open so that i = 0. Since
the inductor current cannot change instantly,
i(0−
) = i(0) = i(0+
) = 0
For 0 ≤ t ≤ 4, S1 is closed so that the 4- and 6- resistors are in
series. Hence, assuming for now that S1 is closed forever,
i(∞) =
40
4 + 6
= 4 A, RTh = 4 + 6 = 10 
τ =
L
RTh
=
5
10
=
1
2
s
CHAPTER 7 First-Order Circuits 267
Thus,
i(t) = i(∞) + [i(0) − i(∞)]e−t/τ
= 4 + (0 − 4)e−2t
= 4(1 − e−2t
) A, 0 ≤ t ≤ 4
For t ≥ 4, S2 is closed; the 10-V voltage source is connected, and
the circuit changes. This sudden change does not affect the inductor
current because the current cannot change abruptly. Thus, the initial
current is
i(4) = i(4−
) = 4(1 − e−8
)  4 A
To find i(∞), let v be the voltage at node P in Fig. 7.53. Using KCL,
40 − v
4
+
10 − v
2
=
v
6
⇒ v =
180
11
V
i(∞) =
v
6
=
30
11
= 2.727 A
The Thevenin resistance at the inductor terminals is
RTh = 4 2 + 6 =
4 × 2
6
+ 6 =
22
3

and
τ =
L
RTh
=
5
22
3
=
15
22
s
Hence,
i(t) = i(∞) + [i(4) − i(∞)]e−(t−4)/τ
, t ≥ 4
We need (t − 4) in the exponential because of the time delay. Thus,
i(t) = 2.727 + (4 − 2.727)e−(t−4)/τ
, τ =
15
22
= 2.727 + 1.273e−1.4667(t−4)
, t ≥ 4
Putting all this together,
i(t) =



0, t ≤ 0
4(1 − e−2t
), 0 ≤ t ≤ 4
2.727 + 1.273e−1.4667(t−4)
, t ≥ 4
At t = 2,
i(2) = 4(1 − e−4
) = 3.93 A
At t = 5,
i(5) = 2.727 + 1.273e−1.4667
= 3.02 A
P R A C T I C E P R O B L E M 7 . 1 3
Switch S1 in Fig. 7.54 is closed at t = 0, and switch S2 is closed at t =
2 s. Calculate i(t) for all t. Find i(1) and i(3).
268 PART 1 DC Circuits
Answer:
i(t) =



0, t  0
2(1 − e−9t
), 0  t  2
3.6 − 1.6e−5(t−2)
, t  2
i(1) = 1.9997 A, i(3) = 3.589 A.
10 Ω
15 Ω
20 Ω
6 A 5 H
t = 0
S1
t = 2
S2
i(t)
Figure7.54 For Practice Prob. 7.13.
†7.7 FIRST-ORDER OP AMP CIRCUITS
An op amp circuit containing a storage element will exhibit first-order
behavior. Differentiators and integrators treated in Section 6.6 are exam-
ples of first-order op amp circuits. Again, for practical reasons, inductors
are hardly ever used in op amp circuits; therefore, the op amp circuits we
consider here are of the RC type.
As usual, we analyze op amp circuits using nodal analysis. Some-
times, the Thevenin equivalent circuit is used to reduce the op amp circuit
to one that we can easily handle. The following three examples illustrate
the concepts. The first one deals with a source-free op amp circuit, while
the other two involve step responses. The three examples have been care-
fully selected to cover all possible RC types of op amp circuits, depending
on the location of the capacitor with respect to the op amp; that is, the
capacitor can be located in the input, the output, or the feedback loop.
E X A M P L E 7 . 1 4
For the op amp circuit in Fig. 7.55(a) , find vo for t  0, given that v(0) =
3 V. Let Rf = 80 k, R1 = 20 k, and C = 5 µF.
vo
v +
−
R1
Rf
(a)
+ −
3
2
1 1
vo (0+
)
3 V +
−
(b)
+ −
3
2
vo
v +
−
(c)
80 kΩ
80 kΩ
20 kΩ
20 kΩ
1 A
C
−
+
C
+
−
+
−
+
−
Figure7.55 For Example 7.14.
CHAPTER 7 First-Order Circuits 269
Solution:
This problem can be solved in two ways:
METHOD 1 Consider the circuit in Fig. 7.55(a). Let us derive the
appropriate differential equation using nodal analysis. If v1 is the voltage
at node 1, at that node, KCL gives
0 − v1
R1
= C
dv
dt
(7.14.1)
Since nodes 2 and 3 must be at the same potential, the potential at node
2 is zero. Thus, v1 − 0 = v or v1 = v and Eq. (7.14.1) becomes
dv
dt
+
v
CR1
= 0 (7.14.2)
This is similar to Eq. (7.4b) so that the solution is obtained the same way
as in Section 7.2, i.e.,
v(t) = V0e−t/τ
, τ = R1C (7.14.3)
where V0 is the initial voltage across the capacitor. But v(0) = 3 = V0
and τ = 20 × 103
× 5 × 10−6
= 0.1. Hence,
v(t) = 3e−10t
(7.14.4)
Applying KCL at node 2 gives
C
dv
dt
=
0 − vo
Rf
or
vo = −Rf C
dv
dt
(7.14.5)
Now we can find v0 as
vo = −80 × 103
× 5 × 10−6
(−30e−10t
) = 12e−10t
V, t  0
METHOD 2 Let us now apply the short-cut method from Eq. (7.53).
We need to find vo(0+
), vo(∞), and τ. Since v(0+
) = v(0−
) = 3 V, we
apply KCL at node 2 in the circuit of Fig. 7.55(b) to obtain
3
20,000
+
0 − vo(0+
)
80,000
= 0
or vo(0+
) = 12 V. Since the circuit is source free, v(∞) = 0 V. To find τ,
we need the equivalent resistance Req across the capacitor terminals. If
we remove the capacitor and replace it by a 1-A current source, we have
the circuit shown in Fig. 7.55(c). Applying KVL to the input loop yields
20,000(1) − v = 0 ⇒ v = 20 kV
Then
Req =
v
1
= 20 k
and τ = ReqC = 0.1. Thus,
vo(t) = vo(∞) + [vo(0) − vo(∞)]e−t/τ
= 0 + (12 − 0)e−10t
= 12e−10t
V, t  0
as before.
270 PART 1 DC Circuits
P R A C T I C E P R O B L E M 7 . 1 4
For the op amp circuit in Fig. 7.56, find vo for t  0 if v(0) = 4 V. Assume
that Rf = 50 k, R1 = 10 k, and C = 10 µF.
vo
+
−
R1
Rf
v
+ −
C
+
−
Figure7.56 For Practice Prob. 7.14.
Answer: −4e−2t
V, t  0.
E X A M P L E 7 . 1 5
Determine v(t) and vo(t) in the circuit of Fig. 7.57.
vo
v1
+
−
3 V
v
+ −
1 mF
50 kΩ
20 kΩ
20 kΩ
+
−
t = 0
10 kΩ
+
−
Figure7.57 For Example 7.15.
Solution:
This problem can be solved in two ways, just like the previous example.
However, we will apply only the second method. Since what we are
looking for is the step response, we can apply Eq. (7.53) and write
v(t) = v(∞) + [v(0) − v(∞)]e−t/τ
, t  0 (7.15.1)
where we need only find the time constant τ, the initial value v(0), and
the final value v(∞). Notice that this applies strictly to the capacitor
voltage due a step input. Since no current enters the input terminals of
the op amp, the elements on the feedback loop of the op amp constitute
an RC circuit, with
τ = RC = 50 × 103
× 10−6
= 0.05 (7.15.2)
For t  0, the switch is open and there is no voltage across the capacitor.
Hence, v(0) = 0. For t  0, we obtain the voltage at node 1 by voltage
division as
v1 =
20
20 + 10
3 = 2 V (7.15.3)
Since there is no storage element in the input loop, v1 remains constant
for all t. At steady state, the capacitor acts like an open circuit so that the
op amp circuit is a noninverting amplifier. Thus,
vo(∞) =

1 +
50
20

v1 = 3.5 × 2 = 7 V (7.15.4)
But
v1 − vo = v (7.15.5)
CHAPTER 7 First-Order Circuits 271
so that
v(∞) = 2 − 7 = −5 V
Substituting τ, v(0), and v(∞) into Eq. (7.15.1) gives
v(t) = −5 + [0 − (−5)]e−20t
= 5(e−20t
− 1) V, t  0 (7.15.6)
From Eqs. (7.15.3), (7.15.5), and (7.15.6), we obtain
vo(t) = v1(t) − v(t) = 7 − 5e−20t
V, t  0 (7.15.7)
P R A C T I C E P R O B L E M 7 . 1 5
Find v(t) and vo(t) in the op amp circuit of Fig. 7.58.
vo
+
−
4 mV
v
+ −
1 mF
100 kΩ
+
−
t = 0
10 kΩ
+
−
Figure7.58 For Practice Prob. 7.15.
Answer: 40(1 − e−10t
) mV, 40(e−10t
− 1) mV.
E X A M P L E 7 . 1 6
Find the step response vo(t) for t  0 in the op amp circuit of Fig. 7.59.
Let vi = 2u(t) V, R1 = 20 k, Rf = 50 k, R2 = R3 = 10 k,
C = 2 µF.
vi vo
+
− C
+
−
R1
Rf
R2
R3
+
−
Figure7.59 For Example 7.16.
Solution:
Notice that the capacitor in Example 7.14 is located in the input loop,
while the capacitor in Example 7.15 is located in the feedback loop. In
this example, the capacitor is located in the output of the op amp. Again,
we can solve this problem directly using nodal analysis. However, using
the Thevenin equivalent circuit may simplify the problem.
We temporarily remove the capacitor and find the Thevenin equiv-
alent at its terminals. To obtain VTh, consider the circuit in Fig. 7.60(a).
Since the circuit is an inverting amplifier,
Vab = −
Rf
R1
vi
By voltage division,
VTh =
R3
R2 + R3
Vab = −
R3
R2 + R3
Rf
R1
vi
272 PART 1 DC Circuits
vi
+
−
R1
Rf
R2
R3
+
−
Vab VTh
+
−
a
b
(a) (b)
RTh
Ro
R2
R3
+
−
Figure7.60 Obtaining VTh and RTh across the capacitor in Fig. 7.59.
To obtain RTh, consider the circuit in Fig. 7.60(b), where Ro is the
output resistance of the op amp. Since we are assuming an ideal op amp,
Ro = 0, and
RTh = R2 R3 =
R2R3
R2 + R3
Substituting the given numerical values,
VTh = −
R3
R2 + R3
Rf
R1
vi = −
10
20
50
20
2u(t) = −2.5u(t)
RTh =
R2R3
R2 + R3
= 5 k
The Thevenin equivalent circuit is shown in Fig. 7.61, which is similar
to Fig. 7.40. Hence, the solution is similar to that in Eq. (7.48); that is,
vo(t) = −2.5(1 − e−t/τ
) u(t)
where τ = RThC = 5 × 103
× 2 × 10−6
= 0.01. Thus, the step response
for t  0 is
vo(t) = 2.5(e−100t
− 1) u(t) V
5 kΩ
+
−
−2.5u(t) 2 mF
Figure 7.61 Theveninequivalentcircuitof
the circuit in Fig. 7.59.
P R A C T I C E P R O B L E M 7 . 1 6
Obtain the step response vo(t) for the circuit of Fig. 7.62. Let vi = 2u(t)
V, R1 = 20 k, Rf = 40 k, R2 = R3 = 10 k, C = 2 µF.
Rf
+
−
R1
R2
R3
vo
vi
+
−
C
+
−
Figure7.62 For Practice Prob. 7.16.
Answer: 6(1 − e−50t
)u(t) V.
CHAPTER 7 First-Order Circuits 273
7.8 TRANSIENT ANALYSIS WITH PSPICE
As we discussed in Section 7.5, the transient response is the tem-
porary response of the circuit that soon disappears. PSpice can be used
to obtain the transient response of a circuit with storage elements. Sec-
tion D.4 in Appendix D provides a review of transient analysis using
PSpice for Windows. It is recommended that you read Section D.4 before
continuing with this section.
PSpice uses “transient” to mean “function of
time.” Therefore, the transient response in
PSpice may not actually die out as expected.
If necessary, dc PSpice analysis is first carried out to determine the
initial conditions. Then the initial conditions are used in the transient
PSpice analysis to obtain the transient responses. It is recommended
but not necessary that during this dc analysis, all capacitors should be
open-circuited while all inductors should be short-circuited.
E X A M P L E 7 . 1 7
Use PSpice to find the response i(t) for t  0 in the circuit of Fig. 7.63. 4 Ω
2 Ω
6 A 3 H
t = 0
i(t)
Figure7.63 For Example 7.17.
Solution:
Solving this problem by hand gives i(0) = 0, i(∞) = 2 A, RTh = 6, τ =
3/6 = 0.5 s, so that
i(t) = i(∞) + [i(0) − i(∞)]e−t/τ
= 2(1 − e−2t
), t  0
To use PSpice, we first draw the schematic as shown in Fig. 7.64.
We recall from Appendix D that the part name for a close switch is
Sw−tclose. We do not need to specify the initial condition of the in-
ductor because PSpice will determine that from the circuit. By select-
ing Analysis/Setup/Transient, we set Print Step to 25 ms and Final
Step to 5τ = 2.5 s. After saving the circuit, we simulate by selecting
Analysis/Simulate. In the Probe menu, we select Trace/Add and
display −I(L1) as the current through the inductor. Figure 7.65 shows the
plot of i(t), which agrees with that obtained by hand calculation.
R2
2
6 A 3 H
IDC
R1 L1
tClose = 0
1 2
U1 4
0
Figure 7.64 The schematic of the circuit in
Fig. 7.63.
1.5 A
0.5 A
2.0 A
1.0 A
0 A
0 s 1.0 s 2.0 s 3.0 s
-I(L1)
Time
Figure 7.65 For Example 7.17; the response
of the circuit in Fig. 7.63.
274 PART 1 DC Circuits
Note that the negative sign on I(L1) is needed because the current
enters through the upper terminal of the inductor, which happens to be the
negative terminal after one counterclockwise rotation. A way to avoid
the negative sign is to ensure that current enters pin 1 of the inductor. To
obtain this desired direction of positive current flow, the initially horizon-
tal inductor symbol should be rotated counterclockwise 270◦
and placed
in the desired location.
P R A C T I C E P R O B L E M 7 . 1 7
For the circuit in Fig. 7.66, use PSpice to find v(t) for t  0.
3 Ω
+
−
12 V 6 Ω 0.5 F
+
−
v(t)
t = 0
Figure7.66 For Practice Prob. 7.17.
Answer: v(t) = 8(1 − e−t
) V, t  0. The response is similar in shape
to that in Fig. 7.65.
E X A M P L E 7 . 1 8
In the circuit in Fig. 7.67, determine the response v(t).
12 Ω
+
−
30 V 3 Ω
6 Ω
6 Ω
0.1 F
4 A
+ −
v(t)
t = 0 t = 0
Figure7.67 For Example 7.18.
Solution:
There are two ways of solving this problem using PSpice.
METHOD 1 One way is to first do the dc PSpice analysis to determine
the initial capacitor voltage. The schematic of the revelant circuit is in
Fig. 7.68(a). Two pseudocomponent VIEWPOINTs are inserted to mea-
sure the voltages at nodes 1 and 2. When the circuit is simulated, we
obtain the displayed values in Fig. 7.68(a) as V1 = 0 V and V2 = 8 V.
Thus the initial capacitor voltage is v(0) = V1 − V2 = −8 V. The PSpice
transient analysis uses this value along with the schematic in Fig. 7.68(b).
Once the circuit in Fig. 7.68(b) is drawn, we insert the capacitor initial
voltage as IC = −8. We select Analysis/Setup/Transient and set Print
Step to 0.1 s and Final Step to 4τ = 4 s. After saving the circuit, we select
Analysis/Simulate to simulate the circuit. In the Probe menu, we select
CHAPTER 7 First-Order Circuits 275
Trace/Add and display V(R2:2) - V(R3:2) or V(C1:1) - V(C1:2) as the
capacitor voltage v(t). The plot of v(t) is shown in Fig. 7.69. This agrees
with the result obtained by hand calculation, v(t) = 10 − 18e−t
.
0.0000 8.0000
6 4A
0.1
1
R3 3
R4 I1
2
6
R2
0
C1
(a)
6
12
R2 6
R3
30 V
0
R1
(b)
+
−
0.1
C1
V1
Figure7.68 (a) Schematic for dc analysis to get v(0),
(b) schematic for transient analysis used in getting the
response v(t).
5 V
-5 V
10 V
0 V
-10 V
0 s 1.0 s 2.0 s 3.0 s 4.0 s
V(R2:2) - V(R3:2)
Time
Figure7.69 Response v(t) for the circuit in Fig. 7.67.
METHOD 2 We can simulate the circuit in Fig. 7.67 directly, since
PSpice can handle the open and close switches and determine the initial
conditions automatically. Using this approach, the schematic is drawn
as shown in Fig. 7.70. After drawing the circuit, we select Analysis/
Setup/Transient and set Print Step to 0.1 s and Final Step to 4τ = 4 s.
We save the circuit, then select Analysis/Simulate to simulate the circuit.
In the Probe menu, we select Trace/Add and display V(R2:2) - V(R3:2)
as the capacitor voltage v(t). The plot of v(t) is the same as that shown
in Fig. 7.69.
R1
6
30 V 4 A
R2 6
R3 3
R4 I1
tClose = 0
1 2
12 U1
1 2
U2
0
+
−
tOpen = 0
0.1
C1
V1
Figure7.70 For Example 7.18.
276 PART 1 DC Circuits
P R A C T I C E P R O B L E M 7 . 1 8
The switch in Fig. 7.71 was open for a long time but closed at t = 0. If
i(0) = 10 A, find i(t) for t  0 by hand and also by PSpice.
5 Ω
30 Ω
12 A 2 H
t = 0
6 Ω
i(t)
Figure7.71 For Practice Prob. 7.18.
Answer: i(t) = 6 + 4e−5t
A. The plot of i(t) obtained by PSpice
analysis is shown in Fig. 7.72.
9 A
10 A
7 A
8 A
6 A
0 s 0.5 s 1.0 s
I(L1)
Time
Figure7.72 For Practice Prob. 7.18.
†7.9 APPLICATIONS
The various devices in which RC and RL circuits find applications in-
clude filtering in dc power supplies, smoothing circuits in digital com-
munications, differentiators, integrators, delay circuits, and relay circuits.
Some of these applications take advantage of the short or long time con-
stants of the RC or RL circuits. We will consider four simple applications
here. The first two are RC circuits, the last two are RL circuits.
7.9.1 Delay Circuits
An RC circuit can be used to provide various time delays. Figure 7.73
shows such a circuit. It basically consists of an RC circuit with the
capacitor connected in parallel with a neon lamp. The voltage source can
provide enough voltage to fire the lamp. When the switch is closed, the
capacitor voltage increases gradually toward 110 V at a rate determined
R1
R2
110 V C 0.1 mF
S
+
−
70 V
Neon
lamp
Figure7.73 An RC delay circuit.
CHAPTER 7 First-Order Circuits 277
by the circuit’s time constant, (R1 + R2)C. The lamp will act as an open
circuit and not emit light until the voltage across it exceeds a particular
level, say 70 V. When the voltage level is reached, the lamp fires (goes
on), and the capacitor discharges through it. Due to the low resistance of
the lamp when on, the capacitor voltage drops fast and the lamp turns off.
The lamp acts again as an open circuit and the capacitor recharges. By
adjusting R2, we can introduce either short or long time delays into the
circuit and make the lamp fire, recharge, and fire repeatedly every time
constant τ = (R1 + R2)C, because it takes a time period τ to get the
capacitor voltage high enough to fire or low enough to turn off.
The warning blinkers commonly found on road construction sites
are one example of the usefulness of such an RC delay circuit.
E X A M P L E 7 . 1 9
Consider the circuit in Fig. 7.73, and assume that R1 = 1.5 M, 0  R 
2.5 M. (a) Calculate the extreme limits of the time constant of the cir-
cuit. (b) How long does it take for the lamp to glow for the first time after
the switch is closed? Let R2 assume its largest value.
Solution:
(a) The smallest value for R2 is 0 , and the corresponding time constant
for the circuit is
τ = (R1 + R2)C = (1.5 × 106
+ 0) × 0.1 × 10−6
= 0.15 s
The largest value for R2 is 2.5 M, and the corresponding time constant
for the circuit is
τ = (R1 + R2)C = (1.5 + 2.5) × 106
× 0.1 × 10−6
= 0.4 s
Thus, by proper circuit design, the time constant can be adjusted to in-
troduce a proper time delay in the circuit.
(b) Assuming that the capacitor is initially uncharged, vC(0) = 0, while
vC(∞) = 110. But
vC(t) = vC(∞) + [vC(0) − vC(∞)]e−t/τ
= 110[1 − e−t/τ
]
where τ = 0.4 s, as calculated in part (a). The lamp glows when vC =
70 V. If vC(t) = 70 V at t = t0, then
70 = 110[1 − e−t0/τ
] ⇒
7
11
= 1 − e−t0/τ
or
e−t0/τ
=
4
11
⇒ et0/τ
=
11
4
Taking the natural logarithm of both sides gives
t0 = τ ln
11
4
= 0.4 ln 2.75 = 0.4046 s
A more general formula for finding t0 is
t0 = τ ln
v(0) − v(∞)
v(t0) − v(∞)
278 PART 1 DC Circuits
The lamp will fire repeatedly every τ seconds if and only if t0  τ. In
this example, that condition is not satisfied.
P R A C T I C E P R O B L E M 7 . 1 9
The RC circuit in Fig. 7.74 is designed to operate an alarm which acti-
vates when the current through it exceeds 120 µA. If 0 ≤ R ≤ 6 k, find
the range of the time delay that the circuit can cause.
10 kΩ R
9 V 80 mF 4 kΩ
S
+
−
Alarm
Figure7.74 For Practice Prob. 7.19.
Answer: Between 47.23 ms and 124 ms.
7.9.2 Photoflash Unit
R1
+
−
High
voltage
dc supply
R2
C v
vs
1
2
i
+
−
Figure7.75 Circuit for a flash unit providing
slow charge in position 1 and fast discharge in
position 2.
An electronic flash unit provides a common example of an RC circuit.
This application exploits the ability of the capacitor to oppose any abrupt
change in voltage. Figure 7.75 shows a simplified circuit. It consists
essentially of a high-voltage dc supply, a current-limiting large resistor
R1, and a capacitor C in parallel with the flashlamp of low resistance R2.
When the switch is in position 1, the capacitor charges slowly due to the
large time constant (τ1 = R1C). As shown in Fig. 7.76, the capacitor
voltage rises gradually from zero to Vs, while its current decreases grad-
ually from I1 = Vs/R1 to zero. The charging time is approximately five
times the time constant,
tcharge = 5R1C (7.65)
With the switch in position 2, the capacitor voltage is discharged. The low
resistance R2 of the photolamp permits a high discharge current with peak
I2 = Vs/R2 in a short duration, as depicted in Fig. 7.76(b). Discharging
takes place in approximately five times the time constant,
0 t
Vs
v
0
(a) (b)
−I2
I1
i
Figure7.76 (a) Capacitor voltage showing slow charge and fast discharge,
(b) capacitor current showing low charging current I1 = Vs/R1 and high discharge
current I2 = Vs/R2.
CHAPTER 7 First-Order Circuits 279
tdischarge = 5R2C (7.66)
Thus, the simple RC circuit of Fig. 7.75 provides a short-duration, high-
current pulse. Such a circuit also finds applications in electric spot weld-
ing and the radar transmitter tube.
E X A M P L E 7 . 2 0
An electronic flashgun has a current-limiting 6-k resistor and 2000-µF
electrolytic capacitor charged to 240 V. If the lamp resistance is 12 ,
find: (a) the peak charging current, (b) the time required for the capaci-
tor to fully charge, (c) the peak discharging current, (d) the total energy
stored in the capacitor, and (e) the average power dissipated by the lamp.
Solution:
(a) The peak charging current is
I1 =
Vs
R1
=
240
6 × 103
= 40 mA
(b) From Eq. (7.65),
tcharge = 5R1C = 5 × 6 × 103
× 2000 × 10−6
= 60 s = 1 minute
(c) The peak discharging current is
I2 =
Vs
R2
=
240
12
= 20 A
(d) The energy stored is
W =
1
2
CV 2
s =
1
2
× 2000 × 10−6
× 2402
= 57.6 J
(e) The energy stored in the capacitor is dissipated across the lamp during
the discharging period. From Eq. (7.66),
tdischarge = 5R2C = 5 × 12 × 2000 × 10−6
= 0.12 s
Thus, the average power dissipated is
p =
W
tdischarge
=
57.6
0.12
= 480 W
P R A C T I C E P R O B L E M 7 . 2 0
The flash unit of a camera has a 2-mF capacitor charged to 80 V.
(a) How much charge is on the capacitor?
(b) What is the energy stored in the capacitor?
(c) If the flash fires in 0.8 ms, what is the average current through the
flashtube?
(d) How much power is delivered to the flashtube?
(e) After a picture has been taken, the capacitor needs to be recharged by
a power unit which supplies a maximum of 5 mA. How much time does
it take to charge the capacitor?
Answer: (a) 0.16 C, (b) 6.4 J, (c) 200 A, (d) 8 kW, (e) 32 s.
280 PART 1 DC Circuits
7.9.3 Relay Circuits
Amagneticallycontrolledswitchiscalledarelay. Arelayisessentiallyan
electromagneticdeviceusedtoopenorcloseaswitchthatcontrolsanother
circuit. Figure 7.77(a) shows a typical relay circuit. The coil circuit is an
RL circuit like that in Fig. 7.77(b), where R and L are the resistance and
inductance of the coil. When switch S1 in Fig. 7.77(a) is closed, the coil
circuit is energized. The coil current gradually increases and produces
a magnetic field. Eventually the magnetic field is sufficiently strong to
pull the movable contact in the other circuit and close switch S2. At this
point, the relay is said to be pulled in. The time interval td between the
closure of switches S1 and S2 is called the relay delay time.
Relays were used in the earliest digital circuits and are still used
for switching high-power circuits.
S2
Coil
Magnetic field
S1
S1
Vs
(a) (b)
Vs
R
L
Figure7.77 A relay circuit.
E X A M P L E 7 . 2 1
The coil of a certain relay is operated by a 12-V battery. If the coil has a
resistance of 150  and an inductance of 30 mH and the current needed
to pull in is 50 mA, calculate the relay delay time.
Solution:
The current through the coil is given by
i(t) = i(∞) + [i(0) − i(∞)]e−t/τ
where
i(0) = 0, i(∞) =
12
150
= 80 mA
τ =
L
R
=
30 × 10−3
150
= 0.2 ms
Thus,
i(t) = 80[1 − e−t/τ
] mA
If i(td) = 50 mA, then
50 = 80[1 − e−td /τ
] ⇒
5
8
= 1 − e−td /τ
CHAPTER 7 First-Order Circuits 281
or
e−td /τ
=
3
8
⇒ etd /τ
=
8
3
By taking the natural logarithm of both sides, we get
td = τ ln
8
3
= 0.2 ln
8
3
ms = 0.1962 ms
P R A C T I C E P R O B L E M 7 . 2 1
A relay has a resistance of 200  and an inductance of 500 mH. The relay
contacts close when the current through the coil reaches 350 mA. What
time elapses between the application of 110 V to the coil and contact
closure?
Answer: 2.529 ms.
7.9.4 Automobile Ignition Circuit
The ability of inductors to oppose rapid change in current makes them
useful for arc or spark generation. An automobile ignition system takes
advantage of this feature.
R
Vs v
+
−
i
Spark
plug
Air gap
L
Figure 7.78 Circuit for an automobile ignition
system.
The gasoline engine of an automobile requires that the fuel-air
mixture in each cylinder be ignited at proper times. This is achieved
by means of a spark plug (Fig. 7.78), which essentially consists of a
pair of electrodes separated by an air gap. By creating a large voltage
(thousands of volts) between the electrodes, a spark is formed across the
air gap, thereby igniting the fuel. But how can such a large voltage be
obtained from the car battery, which supplies only 12 V? This is achieved
by means of an inductor (the spark coil) L. Since the voltage across the
inductor is v = L di/dt, we can make di/dt large by creating a large
change in current in a very short time. When the ignition switch in Fig.
7.78 is closed, the current through the inductor increases gradually and
reaches the final value of i = Vs/R, where Vs = 12 V. Again, the time
taken for the inductor to charge is five times the time constant of the
circuit (τ = L/R),
tcharge = 5
L
R
(7.67)
Since at steady state, i is constant, di/dt = 0 and the inductor voltage
v = 0. When the switch suddenly opens, a large voltage is developed
across the inductor (due to the rapidly collapsing field) causing a spark
or arc in the air gap. The spark continues until the energy stored in the
inductor is dissipated in the spark discharge. In laboratories, when one
is working with inductive circuits, this same effect causes a very nasty
shock, and one must exercise caution.
E X A M P L E 7 . 2 2
A solenoid with resistance 4  and inductance 6 mH is used in an auto-
mobile ignition circuit similar to that in Fig. 7.78. If the battery supplies
12 V, determine: the final current through the solenoid when the switch
282 PART 1 DC Circuits
is closed, the energy stored in the coil, and the voltage across the air gap,
assuming that the switch takes 1 µs to open.
Solution:
The final current through the coil is
I =
Vs
R
=
12
4
= 3 A
The energy stored in the coil is
W =
1
2
LI2
=
1
2
× 6 × 10−3
× 32
= 27 mJ
The voltage across the gap is
V = L
)I
)t
= 6 × 10−3
×
3
1 × 10−6
= 18 kV
P R A C T I C E P R O B L E M 7 . 2 2
The spark coil of an automobile ignition system has a 20-mH inductance
and a 5- resistance. With a supply voltage of 12 V, calculate: the time
needed for the coil to fully charge, the energy stored in the coil, and the
voltage developed at the spark gap if the switch opens in 2 µs.
Answer: 20 ms, 57.6 mJ, and 24 kV.
7.10 SUMMARY
1. The analysis in this chapter is applicable to any circuit that can be
reduced to an equivalent circuit comprising a resistor and a single
energy-storage element (inductor or capacitor). Such a circuit is
first-order because its behavior is described by a first-order differen-
tial equation. When analyzing RC and RL circuits, one must always
keep in mind that the capacitor is an open circuit to steady-state dc
conditions while the inductor is a short circuit to steady-state dc
conditions.
2. The natural response is obtained when no independent source is
present. It has the general form
x(t) = x(0)e−t/τ
where x represents current through (or voltage across) a resistor, a
capacitor, or an inductor, and x(0) is the initial value of x. The
natural response is also called the transient response because it is the
temporary response that vanishes with time.
3. The time constant τ is the time required for a response to decay to
1/e of its initial value. For RC circuits, τ = RC and for RL circuits,
τ = L/R.
4. The singularity functions include the unit step, the unit ramp func-
tion, and the unit impulse functions. The unit step function u(t) is
u(t) =

0, t  0
1, t  0
CHAPTER 7 First-Order Circuits 283
The unit impulse function is
δ(t) =



0, t  0
Undefined, t = 0
0, t  0
The unit ramp function is
r(t) =

0, t ≤ 0
t, t ≥ 0
5. The forced (or steady-state) response is the behavior of the circuit
after an independent source has been applied for a long time.
6. The total or complete response consists of the natural response and
the forced response.
7. The step response is the response of the circuit to a sudden applica-
tion of a dc current or voltage. Finding the step response of a first-
order circuit requires the initial value x(0+
), the final value x(∞),
and the time constant τ. With these three items, we obtain the step
response as
x(t) = x(∞) + [x(0+
) − x(∞)]e−t/τ
A more general form of this equation is
x(t) = x(∞) + [x(t+
0 ) − x(∞)]e−(t−t0)/τ
Or we may write it as
Instantaneous value = Final + [Initial − Final]e−(t−t0)/τ
8. PSpice is very useful for obtaining the transient response of a circuit.
9. Four practical applications of RC and RL circuits are: a delay
circuit, a photoflash unit, a relay circuit, and an automobile ignition
circuit.
REVIEW QUESTIONS
7.1 An RC circuit has R = 2  and C = 4 F. The time
constant is:
(a) 0.5 s (b) 2 s (c) 4 s
(d) 8 s (e) 15 s
7.2 The time constant for an RL circuit with R = 2 
and L = 4 H is:
(a) 0.5 s (b) 2 s (c) 4 s
(d) 8 s (e) 15 s
7.3 A capacitor in an RC circuit with R = 2  and
C = 4 F is being charged. The time required for the
capacitor voltage to reach 63.2 percent of its
steady-state value is:
(a) 2 s (b) 4 s (c) 8 s
(d) 16 s (e) none of the above
7.4 An RL circuit has R = 2  and L = 4 H. The time
needed for the inductor current to reach 40 percent
of its steady-state value is:
(a) 0.5 s (b) 1 s (c) 2 s
(d) 4 s (e) none of the above
7.5 In the circuit of Fig. 7.79, the capacitor voltage just
before t = 0 is:
(a) 10 V (b) 7 V (c) 6 V
(d) 4 V (e) 0 V
v(t)
10 V
2 Ω
3 Ω
+
−
+
−
t = 0
7 F
Figure 7.79 For Review Questions 7.5 and 7.6.
284 PART 1 DC Circuits
7.6 In the circuit of Fig. 7.79, v(∞) is:
(a) 10 V (b) 7 V (c) 6 V
(d) 4 V (e) 0 V
7.7 For the circuit of Fig. 7.80, the inductor current just
before t = 0 is:
(a) 8 A (b) 6 A (c) 4 A
(d) 2 A (e) 0 A
10 A
3 Ω
2 Ω
5 H
i(t)
t = 0
Figure 7.80 For Review Questions 7.7 and 7.8.
7.8 In the circuit of Fig. 7.80, i(∞) is:
(a) 8 A (b) 6 A (c) 4 A
(d) 2 A (e) 0 A
7.9 If vs changes from 2 V to 4 V at t = 0, we may
express vs as:
(a) δ(t) V (b) 2u(t) V
(c) 2u(−t) + 4u(t) V (d) 2 + 2u(t) V
(e) 4u(t) − 2 V
7.10 The pulse in Fig. 7.110(a) can be expressed in terms
of singularity functions as:
(a) 2u(t) + 2u(t − 1) V (b) 2u(t) − 2u(t − 1) V
(c) 2u(t) − 4u(t − 1) V (d) 2u(t) + 4u(t − 1) V
Answers: 7.1d, 7.2b, 7.3c, 7.4b, 7.5d, 7.6a, 7.7c, 7.8e, 7.9c,d, 7.10b.
PROBLEMS
Section 7.2 The Source-Free RC Circuit
7.1 Show that Eq. (7.9) can be obtained by working with
the current i in the RC circuit rather than working
with the voltage v.
7.2 Find the time constant for the RC circuit in Fig.
7.81.
+
− 80 Ω
120 Ω 12 Ω
50 V 0.5 mF
Figure 7.81 For Prob. 7.2.
7.3 Determine the time constant of the circuit in Fig.
7.82.
4 kΩ
12 kΩ 3 mF
1 mF
5 kΩ
Figure 7.82 For Prob. 7.3.
7.4 Obtain the time constant of the circuit in Fig. 7.83.
+
− R2
R1
vs
C2
C1
Figure 7.83 For Prob. 7.4.
7.5 The switch in Fig. 7.84 has been in position a for a
long time, until t = 4 s when it is moved to position
b and left there. Determine v(t) at t = 10 s.
v(t)
24 V 20 Ω
80 Ω
+
−
+
−
0.1 F
t = 4
a b
Figure 7.84 For Prob. 7.5.
7.6 If v(0) = 20 V in the circuit in Fig. 7.85, obtain v(t)
for t  0.
10 Ω
8 Ω
0.5 V 0.1 F
+
−
v
+
−
Figure 7.85 For Prob. 7.6.
CHAPTER 7 First-Order Circuits 285
7.7 For the circuit in Fig. 7.86, if
v = 10e−4t
V and i = 0.2e−4t
A, t  0
(a) Find R and C.
(b) Determine the time constant.
(c) Calculate the initial energy in the capacitor.
(d) Obtain the time it takes to dissipate 50 percent
of the initial energy.
R v
i
C
+
−
Figure 7.86 For Prob. 7.7.
7.8 In the circuit of Fig. 7.87, v(0) = 20 V. Find v(t) for
t  0.
2 Ω
0.25 F
8 Ω
6 Ω 3 Ω
8 Ω
+
−
v
Figure 7.87 For Prob. 7.8.
7.9 Given that i(0) = 3 A, find i(t) for t  0 in the
circuit in Fig. 7.88.
i
10 mF
10 Ω
4 Ω
15 Ω
Figure 7.88 For Prob. 7.9.
Section 7.3 The Source-Free RL Circuit
7.10 Derive Eq. (7.20) by working with voltage v across
the inductor of the RL circuit instead of working
with the current i.
7.11 The switch in the circuit in Fig. 7.89 has been closed
for a long time. At t = 0, the switch is opened.
Calculate i(t) for t  0.
3 Ω
+
−
12 V 4 Ω
i
t = 0
2 H
Figure 7.89 For Prob. 7.11.
7.12 For the circuit shown in Fig. 7.90, calculate the time
constant.
70 Ω 2 mH
+
−
20 V 80 Ω 20 Ω
30 Ω
Figure 7.90 For Prob. 7.12.
7.13 What is the time constant of the circuit in Fig. 7.91?
10 kΩ
10 mH
30 kΩ 6 kΩ
20 mH
Figure 7.91 For Prob. 7.13.
7.14 Determine the time constant for each of the circuits
in Fig. 7.92.
L
R1
R2
R3
(a)
R1 R2
L2
L1
R3
(b)
Figure 7.92 For Prob. 7.14.
7.15 Consider the circuit of Fig. 7.93. Find vo(t) if
i(0) = 2 A and v(t) = 0.
286 PART 1 DC Circuits
vo(t)
v(t)
1 Ω
3 Ω +
−
+
−
i(t)
H
1
4
Figure 7.93 For Prob. 7.15.
7.16 For the circuit in Fig. 7.94, determine vo(t) when
i(0) = 1 A and v(t) = 0.
vo(t)
v(t) 3 Ω
+
−
+
−
i(t)
2 Ω
0.4 H
Figure 7.94 For Prob. 7.16.
7.17 In the circuit of Fig. 7.95, find i(t) for t  0 if
i(0) = 2 A.
40 Ω
10 Ω 0.5i
6 H
i
Figure 7.95 For Prob. 7.17.
7.18 For the circuit in Fig. 7.96,
v = 120e−50t
V
and
i = 30e−50t
A, t  0
(a) Find L and R.
(b) Determine the time constant.
(c) Calculate the initial energy in the inductor.
(d) What fraction of the initial energy is dissipated
in 10 ms?
R
i
+
−
v
L
Figure 7.96 For Prob. 7.18.
7.19 In the circuit in Fig. 7.97, find the value of R for
which energy stored in the inductor will be 1 J.
40 Ω R
+
−
60 V 2 H
80 Ω
Figure 7.97 For Prob. 7.19.
7.20 Find i(t) and v(t) for t  0 in the circuit of Fig.
7.98 if i(0) = 10 A.
5 Ω 20 Ω
1 Ω
2 H +
−
v(t)
i(t)
Figure 7.98 For Prob. 7.20.
7.21 Consider the circuit in Fig. 7.99. Given that
vo(0) = 2 V, find vo and vx for t  0.
3 Ω
1 Ω 2 Ω vo
+
−
vx H
1
3
+
−
Figure 7.99 For Prob. 7.21.
Section 7.4 Singularity Functions
7.22 Express the following signals in terms of singularity
functions.
(a) v(t) =

0, t  0
−5, t  0
(b) i(t) =





0, t  1
−10, 1  t  3
10, 3  t  5
0, t  5
CHAPTER 7 First-Order Circuits 287
(c) x(t) =





t − 1, 1  t  2
1, 2  t  3
4 − t, 3  t  4
0, Otherwise
(d) y(t) =



2, t  0
−5, 0  t  1
0, t  1
7.23 Express the signals in Fig. 7.100 in terms of
singularity functions.
0 t
1
−1
v1(t)
1
−1
(a)
0
1 2 t
−1
−2
v4(t)
(d)
0 2 4 6 t
2
4
v3(t)
(c)
0 2 4 t
2
v2(t)
(b)
Figure 7.100 For Prob. 7.23.
7.24 Sketch the waveform that is represented by
v(t) = u(t) + u(t − 1) − 3u(t − 2) + 2u(t − 3)
7.25 Sketch the waveform represented by
i(t) = r(t) + r(t − 1) − u(t − 2) − r(t − 2)
+ r(t − 3) + u(t − 4)
7.26 Evaluate the following integrals involving the
impulse functions:
(a)
 ∞
−∞
4t2
δ(t − 1) dt
(b)
 ∞
−∞
4t2
cos 2πtδ(t − 0.5) dt
7.27 Evaluate the following integrals:
(a)
 ∞
−∞
e−4t2
δ(t − 2) dt
(b)
 ∞
−∞
[5δ(t) + e−t
δ(t) + cos 2πtδ(t)]dt
7.28 The voltage across a 10-mH inductor is
20δ(t − 2) mV. Find the inductor current, assuming
that the inductor is initially uncharged.
7.29 Find the solution of the following first-order
differential equations subject to the specified initial
conditions.
(a) 5 dv/dt + 3v = 0, v(0) = −2
(b) 4 dv/dt − 6v = 0, v(0) = 5
7.30 Solve for v in the following differential equations,
subject to the stated initial condition.
(a) dv/dt + v = u(t), v(0) = 0
(b) 2 dv/dt − v = 3u(t), v(0) = −6
Section 7.5 Step Response of an RC Circuit
7.31 Calculate the capacitor voltage for t  0 and t  0
for each of the circuits in Fig. 7.101.
+
−
1 Ω
4 Ω
20 V
12 V
+
−
t = 0
v 2 F
(a)
(b)
3 Ω
2 A
4 Ω
+ −
+
− t = 0
2 F
v
Figure 7.101 For Prob. 7.31.
7.32 Find the capacitor voltage for t  0 and t  0 for
each of the circuits in Fig. 7.102.
288 PART 1 DC Circuits
3 Ω 2 Ω
+
− 3 F
+
−
v
12 V 4 V +
−
t = 0
(a)
(b)
4 Ω
2 Ω 5 F
6 A
+
−
v
t = 0
Figure 7.102 For Prob. 7.32.
7.33 For the circuit in Fig. 7.103, find v(t) for t  0.
1 F
+
−
v
6 Ω
30 Ω
12 V
t = 0
+
−
Figure 7.103 For Prob. 7.33.
7.34 (a) If the switch in Fig. 7.104 has been open for a
long time and is closed at t = 0, find vo(t).
(b) Suppose that the switch has been closed for a
long time and is opened at t = 0. Find vo(t).
3 F
+
−
vo
2 Ω
4 Ω
12 V +
−
t = 0
Figure 7.104 For Prob. 7.34.
7.35 Consider the circuit in Fig. 7.105. Find i(t) for
t  0 and t  0.
3 F
40 Ω 30 Ω
50 Ω
0.5i
80 V +
−
t = 0
i
Figure 7.105 For Prob. 7.35.
7.36 The switch in Fig. 7.106 has been in position a for a
long time. At t = 0, it moves to position b.
Calculate i(t) for all t  0.
2 F
6 Ω
3 Ω
30 V +
− 12 V +
−
i
t = 0
a
b
Figure 7.106 For Prob. 7.36.
7.37 Find the step responses v(t) and i(t) to
vs = 5u(t) V in the circuit of Fig. 7.107.
v(t)
vs 4 Ω
12 Ω
+
−
+
−
0.5 F
7 Ω
i(t)
Figure 7.107 For Prob. 7.37.
7.38 Determine v(t) for t  0 in the circuit in Fig. 7.108
if v(0) = 0.
3u(t − 1) A 3u(t) A
8 Ω
2 Ω
+ −
0.1 F
v
Figure 7.108 For Prob. 7.38.
7.39 Find v(t) and i(t) in the circuit of Fig. 7.109.
CHAPTER 7 First-Order Circuits 289
v
u(−t) A 10 Ω
+
−
0.1 F
20 Ω
i
Figure 7.109 For Prob. 7.39.
7.40 If the waveform in Fig. 7.110(a) is applied to the
circuit of Fig. 7.110(b), find v(t). Assume v(0) = 0.
v
is 4 Ω
+
−
0.5 F
6 Ω
(b)
0 1 t (s)
2
is (A)
(a)
Figure 7.110 For Prob. 7.40 and Review Question 7.10.
7.41
∗
In the circuit in Fig. 7.111, find ix for t  0. Let
R1 = R2 = 1 k, R3 = 2 k, and C = 0.25 mF.
R2
R1
30 mA
t = 0
R3
ix
C
Figure 7.111 For Prob. 7.41.
Section 7.6 Step Response of an RL Circuit
7.42 Rather than applying the short-cut technique used in
Section 7.6, use KVL to obtain Eq. (7.60).
7.43 For the circuit in Fig. 7.112, find i(t) for t  0.
40 Ω
20 V 5 H
i
+
−
t = 0
10 Ω
Figure 7.112 For Prob. 7.43.
7.44 Determine the inductor current i(t) for both t  0
and t  0 for each of the circuits in Fig. 7.113.
4 Ω
6 A 2 Ω 3 H
i
t = 0
(b)
25 V 4 H
i
(a)
+
− t = 0
2 Ω
3 Ω
Figure 7.113 For Prob. 7.44.
7.45 Obtain the inductor current for both t  0 and t  0
in each of the circuits in Fig. 7.114.
∗An asterisk indicates a challenging problem.
290 PART 1 DC Circuits
4 Ω
2 A
2 Ω 3 Ω
6 Ω
12 Ω
3.5 H
i
4 Ω
(a)
10 V 2 H
i
(b)
+
−
24 V +
−
t = 0
t = 0
Figure 7.114 For Prob. 7.45.
7.46 Find v(t) for t  0 and t  0 in the circuit in Fig.
7.115.
8 Ω
4io
3 Ω
0.5 H
2 Ω
20 V +
−
24 V +
−
t = 0
+
−
io
+
−
v
Figure 7.115 For Prob. 7.46.
7.47 For the network shown in Fig. 7.116, find v(t) for
t  0.
6 Ω
12 Ω
2 A 0.5 H
20 Ω
5 Ω
+
−
v
+
− 20 V
t = 0
Figure 7.116 For Prob. 7.47.
7.48
∗
Find i1(t) and i2(t) for t  0 in the circuit of Fig.
7.117.
6 Ω
5 A
2.5 H
5 Ω 20 Ω
4 H
i1 i2
t = 0
Figure 7.117 For Prob. 7.48.
7.49 Rework Prob. 7.15 if i(0) = 10 A and
v(t) = 20u(t) V.
7.50 Determine the step response vo(t) to vs = 18u(t) in
the circuit of Fig. 7.118.
3 Ω
6 Ω
vs
1.5 H
4 Ω
+
− +
−
vo
Figure 7.118 For Prob. 7.50.
7.51 Find v(t) for t  0 in the circuit of Fig. 7.119 if the
initial current in the inductor is zero.
5 Ω 20 Ω
4u(t) 8 H
+
−
v
Figure 7.119 For Prob. 7.51.
7.52 In the circuit in Fig. 7.120, is changes from 5 A to
10 A at t = 0; that is, is = 5u(−t) + 10u(t). Find v
and i.
4 Ω
is 0.5 H
+
−
v
i
Figure 7.120 For Prob. 7.52.
7.53 For the circuit in Fig. 7.121, calculate i(t) if
i(0) = 0.
CHAPTER 7 First-Order Circuits 291
3 Ω 6 Ω
+
−
u(t − 1) V u(t) V
2 H
i
+
−
Figure 7.121 For Prob. 7.53.
7.54 Obtain v(t) and i(t) in the circuit of Fig. 7.122.
5 Ω
+
−
10u(−t) V 20 Ω 0.5 H
i
+
−
v
Figure 7.122 For Prob. 7.54.
7.55 Find vo(t) for t  0 in the circuit of Fig. 7.123.
6 Ω
2 Ω
3 Ω
+
−
+
−
vo
t = 0
4 H
10 V
Figure 7.123 For Prob. 7.55.
7.56 If the input pulse in Fig. 7.124(a) is applied to the
circuit in Fig. 7.124(b), determine the response i(t).
5 Ω
+
−
vs 20 Ω 2 H
i
(b)
(a)
0 t (s)
vs (V)
10
1
Figure 7.124 For Prob. 7.56.
Section 7.7 First-order Op Amp Circuits
7.57 Find the output current io for t  0 in the op amp
circuit of Fig. 7.125. Let v(0) = −4 V.
10 kΩ
10 kΩ
v
20 kΩ
io
+ −
2 mF
+
−
Figure 7.125 For Prob. 7.57.
7.58 If v(0) = 5 V, find vo(t) for t  0 in the op amp
circuit in Fig. 7.126. Let R = 10 k and C = 1 µF.
R
R
R v
vo
+
−
C
+
−
Figure 7.126 For Prob. 7.58.
7.59 Obtain vo for t  0 in the circuit of Fig. 7.127.
10 kΩ
10 kΩ
+
− vo
+
−
25 mF
t = 0
4 V
+
−
Figure 7.127 For Prob. 7.59.
7.60 For the op amp circuit in Fig. 7.128, find vo(t) for
t  0.
292 PART 1 DC Circuits
20 kΩ 100 kΩ
10 kΩ
+
−
vo
+
−
25 mF
t = 0
4 V +
−
Figure 7.128 For Prob. 7.60.
7.61 Determine vo for t  0 when vs = 20 mV in the op
amp circuit of Fig. 7.129.
20 kΩ
+
−
vo
vs 5 mF
t = 0
+
−
Figure 7.129 For Prob. 7.61.
7.62 For the op amp circuit in Fig. 7.130, find io for t  2.
10 kΩ
10 kΩ 20 kΩ
+
− 100 mF
t = 2
4 V
io
+
−
Figure 7.130 For Prob. 7.62.
7.63 Find io in the op amp circuit in Fig. 7.131. Assume
that v(0) = −2 V, R = 10 k, and C = 10 µF.
R
+
−
v
3u(t)
io
+ −
C
+
−
Figure 7.131 For Prob. 7.63.
7.64 For the op amp circuit of Fig. 7.132, let R1 = 10 k,
Rf = 20 k, C = 20 µF, and v(0) = 1 V. Find vo.
Rf
R1
+
− vo
+
−
4u(t)
v
+ −
C
+
−
Figure 7.132 For Prob. 7.64.
7.65 Determine vo(t) for t  0 in the circuit of Fig.
7.133. Let is = 10u(t) µA and assume that the
capacitor is initially uncharged.
10 kΩ
50 kΩ vo
+
−
is
2 mF
+
−
Figure 7.133 For Prob. 7.65.
7.66 In the circuit of Fig. 7.134, find vo and io, given that
vs = 4u(t) V and v(0) = 1 V.
vo
vs
2 mF
10 kΩ
20 kΩ
+ −
v
+
−
io
+
−
Figure 7.134 For Prob. 7.66.
CHAPTER 7 First-Order Circuits 293
Section 7.8 Transient Analysis with PSpice
7.67 Repeat Prob. 7.40 using PSpice.
7.68 The switch in Fig. 7.135 opens at t = 0. Use PSpice
to determine v(t) for t  0.
5 Ω
4 Ω
5 A 6 Ω 20 Ω +
− 30 V
t = 0 + −
v
100 mF
Figure 7.135 For Prob. 7.68.
7.69 The switch in Fig. 7.136 moves from position a to b
at t = 0. Use PSpice to find i(t) for t  0.
4 Ω
6 Ω
3 Ω
+
−
108 V 6 Ω 2 H
i(t)
t = 0
a
b
Figure 7.136 For Prob. 7.69.
7.70 Repeat Prob. 7.56 using PSpice.
Section 7.9 Applications
7.71 In designing a signal-switching circuit, it was found
that a 100-µF capacitor was needed for a time
constant of 3 ms. What value resistor is necessary
for the circuit?
7.72 A simple relaxation oscillator circuit is shown in
Fig. 7.137. The neon lamp fires when its voltage
reaches 75 V and turns off when its voltage drops to
30 V. Its resistance is 120  when on and infinitely
high when off.
(a) For how long is the lamp on each time the
capacitor discharges?
(b) What is the time interval between light flashes?
120 V
4 MΩ
Neon lamp
6 mF
+
−
Figure 7.137 For Prob. 7.72.
7.73 Figure 7.138 shows a circuit for setting the length of
time voltage is applied to the electrodes of a welding
machine. The time is taken as how long it takes the
capacitor to charge from 0 to 8 V. What is the time
range covered by the variable resistor?
100 kΩ to 1 MΩ
12 V 2 mF
Welding
control
unit
Electrode
Figure 7.138 For Prob. 7.73.
7.74 A 120-V dc generator energizes a motor whose coil
has an inductance of 50 H and a resistance of 100 .
A field discharge resistor of 400  is connected in
parallel with the motor to avoid damage to the
motor, as shown in Fig. 7.139. The system is at
steady state. Find the current through the discharge
resistor 100 ms after the breaker is tripped.
+
−
120 V 400 Ω
Circuit breaker
Motor
Figure 7.139 For Prob. 7.74.
COMPREHENSIVE PROBLEMS
7.75 The circuit in Fig. 7.140(a) can be designed as an
approximate differentiator or an integrator,
depending on whether the output is taken across the
resistor or the capacitor, and also on the time
constant τ = RC of the circuit and the width T of
the input pulse in Fig. 7.140(b). The circuit is a
differentiator if τ  T , say τ  0.1T, or an
integrator if τ  T , say τ  10T .
(a) What is the minimum pulse width that will allow
a differentiator output to appear across the
capacitor?
294 PART 1 DC Circuits
(b) If the output is to be an integrated form of the
input, what is the maximum value the pulse
width can assume?
300 kΩ
+
− 200 pF
vi
(a)
0 T t
Vm
vi
(b)
Figure 7.140 For Prob. 7.75.
7.76 An RL circuit may be used as a differentiator if the
output is taken across the inductor and τ  T (say
τ  0.1T ), where T is the width of the input pulse.
If R is fixed at 200 k, determine the maximum
value of L required to differentiate a pulse with
T = 10 µs.
7.77 An attenuator probe employed with oscilloscopes
was designed to reduce the magnitude of the input
voltage vi by a factor of 10. As shown in Fig. 7.141,
the oscilloscope has internal resistance Rs and
capacitance Cs, while the probe has an internal
resistance Rp. If Rp is fixed at 6 M, find Rs and
Cs for the circuit to have a time constant of 15 µs.
vo
vi
Probe Scope
Rp
Cs
+
−
+
−
Rs
Figure 7.141 For Prob. 7.77.
7.78 The circuit in Fig. 7.142 is used by a biology student
to study “frog kick.” She noticed that the frog
kicked a little when the switch was closed but
kicked violently for 5 s when the switch was
opened. Model the frog as a resistor and calculate
its resistance. Assume that it takes 10 mA for the
frog to kick violently.
50 Ω
2 H
+
−
12 V
Switch
Frog
Figure 7.142 For Prob. 7.78.
7.79 To move a spot of a cathode-ray tube across the
screen requires a linear increase in the voltage
across the deflection plates, as shown in Fig. 7.143.
Given that the capacitance of the plates is 4 nF,
sketch the current flowing through the plates.
Rise time = 2 ms Drop time = 5 ms
t
10
v (V)
(not to scale)
Figure 7.143 For Prob. 7.79.
295
C H A P T E R
SECOND-ORDER CIRCUITS
8
“Engineering is not only a learned profession, it is also a learning pro-
fession, one whose practitioners first become and then remain students
throughout their active careers.”
—William L. Everitt
Enhancing Your Career
To increase your engineering career opportunities after grad-
uation, develop a strong fundamental understanding in a
broad set of engineering areas. When possible, this might
best be accomplished by working toward a graduate degree
immediately upon receiving your undergraduate degree.
Each degree in engineering represents certain skills
the students acquire. At the Bachelor degree level, you learn
the language of engineering and the fundamentals of engi-
neering and design. At the Master’s level, you acquire the
ability to do advanced engineering projects and to commu-
nicate your work effectively both orally and in writing. The
Ph.D. represents a thorough understanding of the fundamen-
tals of electrical engineering and a mastery of the skills nec-
essary both for working at the frontiers of an engineering
area and for communicating one’s effort to others.
If you have no idea what career you should pursue af-
ter graduation, a graduate degree program will enhance your
ability to explore career options. Since your undergraduate
degree will only provide you with the fundamentals of en-
gineering, a Master’s degree in engineering supplemented
by business courses benefits more engineering students than
does getting a Master’s of Business Administration (MBA).
The best time to get your MBA is after you have been a prac-
ticing engineer for some years and decide your career path
would be enhanced by strengthening your business skills.
Engineers should constantly educate themselves,
formally and informally, taking advantage of all means of
education. Perhaps there is no better way to enhance your
career than to join a professional society such as IEEE and
be an active member.
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your own niche. (Courtesy of IEEE.)
296 PART 1 DC Circuits
8.1 INTRODUCTION
In the previous chapter we considered circuits with a single storage ele-
ment (a capacitor or an inductor). Such circuits are first-order because
the differential equations describing them are first-order. In this chap-
ter we will consider circuits containing two storage elements. These are
known as second-order circuits because their responses are described by
differential equations that contain second derivatives.
Typical examples of second-order circuits are RLC circuits, in
which the three kinds of passive elements are present. Examples of such
circuits are shown in Fig. 8.1(a) and (b). Other examples are RC and RL
circuits, as shown in Fig. 8.1(c) and (d). It is apparent from Fig. 8.1 that
a second-order circuit may have two storage elements of different type or
the same type (provided elements of the same type cannot be represented
by an equivalent single element). An op amp circuit with two storage
elements may also be a second-order circuit. As with first-order circuits,
a second-order circuit may contain several resistors and dependent and
independent sources.
A second-order circuit is characterized by a second-order differential equation. It
consists of resistors and the equivalent of two energy storage elements.
Our analysis of second-order circuits will be similar to that used for
first-order. We will first consider circuits that are excited by the initial
conditions of the storage elements. Although these circuits may contain
dependent sources, they are free of independent sources. These source-
free circuits will give natural responses as expected. Later we will con-
sider circuits that are excited by independent sources. These circuits will
give both the natural response and the forced response. We consider
only dc independent sources in this chapter. The case of sinusoidal and
exponential sources is deferred to later chapters.
vs
R
R
L
C
+
−
(a)
is C L
R
(b)
vs
R1 R2
+
−
(c)
is C2
C1
(d)
L1 L2
Figure8.1 Typical examples of
second-order circuits: (a) series
RLC circuit, (b) parallel RLC
circuit, (c) RL circuit, (d) RC
circuit.
We begin by learning how to obtain the initial conditions for the cir-
cuit variables and their derivatives, as this is crucial to analyzing second-
order circuits. Then we consider series and parallel RLC circuits such as
shown in Fig. 8.1 for the two cases of excitation: by initial conditions of
the energy storage elements and by step inputs. Later we examine other
types of second-order circuits, including op amp circuits. We will con-
sider PSpice analysis of second-order circuits. Finally, we will consider
the automobile ignition system and smoothing circuits as typical appli-
cations of the circuits treated in this chapter. Other applications such as
resonant circuits and filters will be covered in Chapter 14.
8.2 FINDING INITIAL AND FINAL VALUES
Perhaps the major problem students face in handling second-order circuits
is finding the initial and final conditions on circuit variables. Students are
CHAPTER 8 Second-Order Circuits 297
usually comfortable getting the initial and final values of v and i but often
have difficulty finding the initial values of their derivatives: dv/dt and
di/dt. For this reason, this section is explicitly devoted to the subtleties
of getting v(0), i(0), dv(0)/dt, di(0)/dt, i(∞), and v(∞). Unless
otherwise stated in this chapter, v denotes capacitor voltage, while i is
the inductor current.
There are two key points to keep in mind in determining the initial
conditions.
First—as always in circuit analysis—we must carefully handle the
polarity of voltage v(t) across the capacitor and the direction of the cur-
rent i(t) through the inductor. Keep in mind that v and i are defined
strictly according to the passive sign convention (see Figs. 6.3 and 6.23).
One should carefully observe how these are defined and apply them ac-
cordingly.
Second, keep in mind that the capacitor voltage is always continu-
ous so that
v(0+
) = v(0−
) (8.1a)
and the inductor current is always continuous so that
i(0+
) = i(0−
) (8.1b)
where t = 0−
denotes the time just before a switching event and t = 0+
is
the time just after the switching event, assuming that the switching event
takes place at t = 0.
Thus, in finding initial conditions, we first focus on those variables
that cannot change abruptly, capacitor voltage and inductor current, by
applying Eq. (8.1). The following examples illustrate these ideas.
E X A M P L E 8 . 1
The switch in Fig. 8.2 has been closed for a long time. It is open at t = 0.
Find: (a) i(0+
), v(0+
), (b) di(0+
)dt, dv(0+
)/dt, (c) i(∞), v(∞).
12 V
4 Ω 0.25 H
+
− 0.1 F
i
v
+
−
2 Ω
t = 0
Figure8.2 For Example 8.1.
Solution:
(a) If the switch is closed a long time before t = 0, it means that the circuit
has reached dc steady state at t = 0. At dc steady state, the inductor acts
like a short circuit, while the capacitor acts like an open circuit, so we
have the circuit in Fig. 8.3(a) at t = 0−
. Thus,
12 V
4 Ω 0.25 H
+
− 0.1 F
i
(b)
12 V
4 Ω
+
−
i
v
+
−
2 Ω
(a)
12 V
4 Ω
+
−
i
v
+
−
(c)
+ −
vL
v
+
−
Figure8.3 Equivalent circuit of that in Fig. 8.2 for: (a) t = 0−, (b) t = 0+, (c) t → ∞.
298 PART 1 DC Circuits
i(0−
) =
12
4 + 2
= 2 A, v(0−
) = 2i(0−
) = 4 V
As the inductor current and the capacitor voltage cannot change abruptly,
i(0+
) = i(0−
) = 2 A, v(0+
) = v(0−
) = 4 V
(b) At t = 0+
, the switch is open; the equivalent circuit is as shown in Fig.
8.3(b). The same current flows through both the inductor and capacitor.
Hence,
iC(0+
) = i(0+
) = 2 A
Since C dv/dt = iC, dv/dt = iC/C, and
dv(0+
)
dt
=
iC(0+
)
C
=
2
0.1
= 20 V/s
Similarly, since L di/dt = vL, di/dt = vL/L. We now obtain vL by
applying KVL to the loop in Fig. 8.3(b). The result is
−12 + 4i(0+
) + vL(0+
) + v(0+
) = 0
or
vL(0+
) = 12 − 8 − 4 = 0
Thus,
di(0+
)
dt
=
vL(0+
)
L
=
0
0.25
= 0 A/s
(c) For t  0, the circuit undergoes transience. But as t → ∞, the circuit
reaches steady state again. The inductor acts like a short circuit and the
capacitor like an open circuit, so that the circuit becomes that shown in
Fig. 8.3(c), from which we have
i(∞) = 0 A, v(∞) = 12 V
P R A C T I C E P R O B L E M 8 . 1
The switch in Fig. 8.4 was open for a long time but closed at t = 0. De-
termine: (a) i(0+
), v(0+
), (b) di(0+
)dt, dv(0+
)/dt, (c) i(∞), v(∞).
10 Ω
24 V
v
+
−
2 Ω +
−
i
t = 0
0.4 H
F
1
20
Figure8.4 For Practice Prob. 8.1.
Answer: (a) 2 A, 4 V, (b) 50 A/s, 0 V/s, (c) 12 A, 24 V.
CHAPTER 8 Second-Order Circuits 299
E X A M P L E 8 . 2
In the circuit of Fig. 8.5, calculate: (a) iL(0+
), vC(0+
), vR(0+
),
(b) diL(0+
)dt, dvC(0+
)/dt, dvR(0+
)/dt, (c) iL(∞), vC(∞), vR(∞).
3u(t) A
4 Ω
20 V
0.6 H
vC
+
−
vR
+
−
2 Ω
+
−
iL
F
1
2
Figure8.5 For Example 8.2.
Solution:
(a) For t  0, 3u(t) = 0. At t = 0−
, since the circuit has reached steady
state, the inductor can be replaced by a short circuit, while the capacitor
is replaced by an open circuit as shown in Fig. 8.6(a). From this figure
we obtain
iL(0−
) = 0, vR(0−
) = 0, vC(0−
) = −20 V (8.2.1)
Although the derivatives of these quantities at t = 0−
are not required, it
is evident that they are all zero, since the circuit has reached steady state
and nothing changes.
3 A
4 Ω
20 V
0.6 H
vR
+
−
2 Ω
+
−
iL
iC
vL
(b)
a b
4 Ω
20 V
vC
+
−
vR
+
−
2 Ω
+
−
iL
(a)
vo
vC
+
−
+ −
+
−
F
1
2
Figure8.6 The circuit in Fig. 8.5 for: (a) t = 0−, (b) t = 0+.
For t  0, 3u(t) = 3, so that the circuit is now equivalent to that
in Fig. 8.6(b). Since the inductor current and capacitor voltage cannot
change abruptly,
iL(0+
) = iL(0−
) = 0, vC(0+
) = vC(0−
) = −20 V (8.2.2)
Although the voltage across the 4- resistor is not required, we will use
it to apply KVL and KCL; let it be called vo. Applying KCL at node a
in Fig. 8.6(b) gives
3 =
vR(0+
)
2
+
vo(0+
)
4
(8.2.3)
300 PART 1 DC Circuits
Applying KVL to the middle mesh in Fig. 8.6(b) yields
−vR(0+
) + vo(0+
) + vC(0+
) + 20 = 0 (8.2.4)
Since vC(0+) = −20 V from Eq. (8.2.2), Eq. (8.2.4) implies that
vR(0+
) = vo(0+
) (8.2.5)
From Eqs. (8.2.3) and (8.2.5), we obtain
vR(0+
) = vo(0+
) = 4 V (8.2.6)
(b) Since L diL/dt = vL,
diL(0+
)
dt
=
vL(0+
)
L
But applying KVL to the right mesh in Fig. 8.6(b) gives
vL(0+
) = vC(0+
) + 20 = 0
Hence,
diL(0+
)
dt
= 0 (8.2.7)
Similarly, since C dvC/dt = iC, then dvC/dt = iC/C. We apply KCL
at node b in Fig. 8.6(b) to get iC:
vo(0+
)
4
= iC(0+
) + iL(0+
) (8.2.8)
Since vo(0+
) = 4 and iL(0+
) = 0, iC(0+
) = 4/4 = 1 A. Then
dvC(0+
)
dt
=
iC(0+
)
C
=
1
0.5
= 2 V/s (8.2.9)
To get dvR(0+
)/dt, we apply KCL to node a and obtain
3 =
vR
2
+
vo
4
Taking the derivative of each term and setting t = 0+
gives
0 = 2
dvR(0+
)
dt
+
dvo(0+
)
dt
(8.2.10)
We also apply KVL to the middle mesh in Fig. 8.6(b) and obtain
−vR + vC + 20 + vo = 0
Again, taking the derivative of each term and setting t = 0+
yields
−
dvR(0+
)
dt
+
dvC(0+
)
dt
+
dvo(0+
)
dt
= 0
Substituting for dvC(0+
)/dt = 2 gives
dvR(0+
)
dt
= 2 +
dvo(0+
)
dt
(8.2.11)
From Eqs. (8.2.10) and (8.2.11), we get
CHAPTER 8 Second-Order Circuits 301
dvR(0+
)
dt
=
2
3
V/s
We can find diR(0+
)/dt although it is not required. Since vR = 5iR,
diR(0+
)
dt
=
1
5
dvR(0+
)
dt
=
1
5
2
3
=
2
15
A/s
(c) As t → ∞, the circuit reaches steady state. We have the equivalent
circuit in Fig. 8.6(a) except that the 3-A current source is now operative.
By current division principle,
iL(∞) =
2
2 + 4
3 A = 1 A
vR(∞) =
4
2 + 4
3 A × 2 = 4 V, vC(∞) = −20 V
(8.2.12)
P R A C T I C E P R O B L E M 8 . 2
For the circuit in Fig. 8.7, find: (a) iL(0+
), vC(0+
), vR(0+
),
(b) diL(0+
)/dt, dvC(0+
)/dt, dvR(0+
)/dt, (c) iL(∞), vC(∞), vR(∞).
2u(t) A 3 A
5 Ω
2 H
iC iL
vC
+
−
iR
vL
vR
+ −
+
−
F
1
5
Figure8.7 For Practice Prob. 8.2.
Answer: (a) −3 A, 0, 0, (b) 0, 10 V/s, 0, (c) −1 A, 10 V, 10 V.
8.3 THE SOURCE-FREE SERIES RLC CIRCUIT
An understanding of the natural response of the series RLC circuit is a
necessary background for future studies in filter design and communica-
tions networks.
Consider the series RLC circuit shown in Fig. 8.8. The circuit is
being excited by the energy initially stored in the capacitor and inductor.
The energy is represented by the initial capacitor voltage V0 and initial
inductor current I0. Thus, at t = 0,
v(0) =
1
C
 0
−∞
i dt = V0 (8.2a)
i(0) = I0 (8.2b)
i
R L
Io
Vo C
+
−
Figure 8.8 A source-free series
RLC circuit.
Applying KVL around the loop in Fig. 8.8,
Ri + L
di
dt
+
1
C
 t
−∞
i dt = 0 (8.3)
302 PART 1 DC Circuits
To eliminate the integral, we differentiate with respect to t and rearrange
terms. We get
d2
i
dt2
+
R
L
di
dt
+
i
LC
= 0 (8.4)
This is a second-order differential equation and is the reason for calling
the RLC circuits in this chapter second-order circuits. Our goal is to solve
Eq. (8.4). To solve such a second-order differential equation requires that
we have two initial conditions, such as the initial value of i and its first
derivative or initial values of some i and v. The initial value of i is given
in Eq. (8.2b). We get the initial value of the derivative of i from Eqs.
(8.2a) and (8.3); that is,
Ri(0) + L
di(0)
dt
+ V0 = 0
or
di(0)
dt
= −
1
L
(RI0 + V0) (8.5)
With the two initial conditions in Eqs. (8.2b) and (8.5), we can now
solve Eq. (8.4). Our experience in the preceding chapter on first-order
circuits suggests that the solution is of exponential form. So we let
i = Aest
(8.6)
where A and s are constants to be determined. Substituting Eq. (8.6) into
Eq. (8.4) and carrying out the necessary differentiations, we obtain
As2
est
+
AR
L
sest
+
A
LC
est
= 0
or
Aest

s2
+
R
L
s +
1
LC

= 0 (8.7)
Since i = Aest
is the assumed solution we are trying to find, only the
expression in parentheses can be zero:
s2
+
R
L
s +
1
LC
= 0 (8.8)
See Appendix C.1 for the formula to find the
roots of a quadratic equation.
This quadratic equation is known as the characteristic equation of the
differential Eq. (8.4), since the roots of the equation dictate the character
of i. The two roots of Eq. (8.8) are
s1 = −
R
2L
+


R
2L
2
−
1
LC
(8.9a)
s2 = −
R
2L
−


R
2L
2
−
1
LC
(8.9b)
A more compact way of expressing the roots is
s1 = −α +

α2 − ω2
0, s2 = −α −

α2 − ω2
0 (8.10)
CHAPTER 8 Second-Order Circuits 303
where
α =
R
2L
, ω0 =
1
√
LC
(8.11)
The roots s1 and s2 are called natural frequencies, measured in
nepers per second (Np/s), because they are associated with the natural
response of the circuit; ω0 is known as the resonant frequency or strictly as
the undamped natural frequency, expressed in radians per second (rad/s);
and α is the neper frequency or the damping factor, expressed in nepers
per second. In terms of α and ω0, Eq. (8.8) can be written as
s2
+ 2αs + ω2
0 = 0 (8.8a)
The variables s and ω are important quantities we will be discussing
throughout the rest of the text.
The neper (Np) is a dimensionless unit named
after John Napier (1550–1617), a Scottish math-
ematician.
The ratio α/ω0 is known as the damping ratio ζ.
The two values of s in Eq. (8.10) indicate that there are two possible
solutions for i, each of which is of the form of the assumed solution in
Eq. (8.6); that is,
i1 = A1es1t
, i2 = A2es2t
(8.12)
Since Eq. (8.4) is a linear equation, any linear combination of the two
distinct solutions i1 and i2 is also a solution of Eq. (8.4). A complete or
total solution of Eq. (8.4) would therefore require a linear combination
of i1 and i2. Thus, the natural response of the series RLC circuit is
i(t) = A1es1t
+ A2es2t
(8.13)
where the constants A1 and A2 are determined from the initial values i(0)
and di(0)/dt in Eqs. (8.2b) and (8.5).
From Eq. (8.10), we can infer that there are three types of solutions:
1. If α  ω0, we have the overdamped case.
2. If α = ω0, we have the critically damped case.
3. If α  ω0, we have the underdamped case.
We will consider each of these cases separately.
The response is overdamped when the roots of
the circuit’s characteristic equation are unequal
and real, critically damped when the roots are
equal and real, and underdamped when the roots
are complex.
Overdamped Case (α  ω0)
From Eqs. (8.9) and (8.10), α  ω0 when C  4L/R2
. When this hap-
pens, both roots s1 and s2 are negative and real. The response is
i(t) = A1es1t
+ A2es2t
(8.14)
which decays and approaches zero as t increases. Figure 8.9(a) illustrates
a typical overdamped response.
Critically Damped Case (α = ω0)
When α = ω0, C = 4L/R2
and
s1 = s2 = −α = −
R
2L
(8.15)
For this case, Eq. (8.13) yields
i(t) = A1e−αt
+ A2e−αt
= A3e−αt
304 PART 1 DC Circuits
where A3 = A1 + A2. This cannot be the solution, because the two
initial conditions cannot be satisfied with the single constant A3. What
then could be wrong? Our assumption of an exponential solution is
incorrect for the special case of critical damping. Let us go back to Eq.
(8.4). When α = ω0 = R/2L, Eq. (8.4) becomes
d2
i
dt2
+ 2α
di
dt
+ α2
i = 0
or
d
dt

di
dt
+ αi

+ α

di
dt
+ αi

= 0 (8.16)
If we let
f =
di
dt
+ αi (8.17)
then Eq. (8.16) becomes
df
dt
+ αf = 0
which is a first-order differential equation with solution f = A1e−αt
,
where A1 is a constant. Equation (8.17) then becomes
di
dt
+ αi = A1e−αt
or
eαt di
dt
+ eαt
αi = A1 (8.18)
This can be written as
d
dt
(eαt
i) = A1 (8.19)
Integrating both sides yields
eαt
i = A1t + A2
or
i = (A1t + A2)eαt
(8.20)
where A2 is another constant. Hence, the natural response of the critically
damped circuit is a sum of two terms: a negative exponential and a
negative exponential multiplied by a linear term, or
i(t) = (A2 + A1t)e−αt
(8.21)
A typical critically damped response is shown in Fig. 8.9(b). In fact, Fig.
8.9(b) is a sketch of i(t) = te−αt
, which reaches a maximum value of
e−1
/α at t = 1/α, one time constant, and then decays all the way to zero.
t
i(t)
0
e–t
(c)
t
1
a
i(t)
0
(b)
t
i(t)
0
(a)
2p
vd
Figure 8.9 (a) Overdamped response,
(b) critically damped response,
(c) underdamped response.
Underdamped Case (α  ω0)
For α  ω0, C  4L/R2
. The roots may be written as
s1 = −α +

−(ω2
0 − α2) = −α + jωd (8.22a)
s2 = −α −

−(ω2
0 − α2) = −α − jωd (8.22b)
CHAPTER 8 Second-Order Circuits 305
where j =
√
−1 and ωd =
√
ω2
0 − α2
, which is called the damping
frequency. Both ω0 and ωd are natural frequencies because they help
determine the natural response; while ω0 is often called the undamped
natural frequency, ωd is called the damped natural frequency. The natural
response is
i(t) = A1e−(α−jωd )t
+ A2e−(α+jωd )t
= e−αt
(A1ejωd t
+ A2e−jωd t
)
(8.23)
Using Euler’s identities,
ejθ
= cos θ + j sin θ, e−jθ
= cos θ − j sin θ (8.24)
we get
i(t) = e−αt
[A1(cos ωdt + j sin ωdt) + A2(cos ωdt − j sin ωdt)]
= e−αt
[(A1 + A2) cos ωdt + j(A1 − A2) sin ωdt]
(8.25)
Replacing constants (A1 + A2) and j(A1 − A2) with constants B1 and
B2, we write
i(t) = e−αt
(B1 cos ωdt + B2 sin ωdt) (8.26)
With the presence of sine and cosine functions, it is clear that the natural
response for this case is exponentially damped and oscillatory in nature.
The response has a time constant of 1/α and a period of T = 2π/ωd. Fig-
ure 8.9(c) depicts a typical underdamped response. [Figure 8.9 assumes
for each case that i(0) = 0.]
Once the inductor current i(t) is found for the RLC series circuit as
shown above, other circuit quantities such as individual element voltages
can easily be found. For example, the resistor voltage is vR = Ri, and the
inductor voltage is vL = L di/dt. The inductor current i(t) is selected
as the key variable to be determined first in order to take advantage of Eq.
(8.1b).
We conclude this section by noting the following interesting, pe-
culiar properties of an RLC network:
R = 0 produces a perfectly sinusoidal response.
Thisresponsecannotbepracticallyaccomplished
with L and C because of the inherent losses in
them. See Figs. 6.8 and 6.26. An electronic de-
vice called an oscillator can produce a perfectly
sinusoidal response.
Examples 8.5 and 8.7 demonstrate the effect of
varying R.
Theresponseofasecond-ordercircuitwithtwo
storage elements of the same type, as in Fig.
8.1(c) and (d), cannot be oscillatory.
1. The behavior of such a network is captured by the idea of
damping, which is the gradual loss of the initial stored energy,
as evidenced by the continuous decrease in the amplitude of
the response. The damping effect is due to the presence of
resistance R. The damping factor α determines the rate at
which the response is damped. If R = 0, then α = 0, and we
have an LC circuit with 1/
√
LC as the undamped natural
frequency. Since α  ω0 in this case, the response is not only
undamped but also oscillatory. The circuit is said to be loss-
less, because the dissipating or damping element (R) is absent.
By adjusting the value of R, the response may be made
undamped, overdamped, critically damped, or underdamped.
2. Oscillatory response is possible due to the presence of the two
types of storage elements. Having both L and C allows the
flow of energy back and forth between the two. The damped
oscillation exhibited by the underdamped response is known as
ringing. It stems from the ability of the storage elements L and
C to transfer energy back and forth between them.
306 PART 1 DC Circuits
3. Observe from Fig. 8.9 that the waveforms of the responses
differ. In general, it is difficult to tell from the waveforms the
difference between the overdamped and critically damped
responses. The critically damped case is the borderline
between the underdamped and overdamped cases and it decays
the fastest. With the same initial conditions, the overdamped
case has the longest settling time, because it takes the longest
time to dissipate the initial stored energy. If we desire the
fastest response without oscillation or ringing, the critically
damped circuit is the right choice.
Whatthismeansinmostpracticalcircuitsisthat
weseekanoverdampedcircuitthatisascloseas
possible to a critically damped circuit.
E X A M P L E 8 . 3
In Fig. 8.8, R = 40 , L = 4 H, and C = 1/4 F. Calculate the char-
acteristic roots of the circuit. Is the natural response overdamped, under-
damped, or critically damped?
Solution:
We first calculate
α =
R
2L
=
40
2(4)
= 5, ω0 =
1
√
LC
=
1

4 × 1
4
= 1
The roots are
s1,2 = −α ±

α2 − ω2
0 = −5 ±
√
25 − 1
or
s1 = −0.101, s2 = −9.899
Since α  ω0, we conclude that the response is overdamped. This is also
evident from the fact that the roots are real and negative.
P R A C T I C E P R O B L E M 8 . 3
If R = 10 , L = 5 H, and C = 2 mF in Fig. 8.8, find α, ω0, s1, and s2.
What type of natural response will the circuit have?
Answer: 1, 10, −1 ± j9.95, underdamped.
E X A M P L E 8 . 4
Find i(t) in the circuit in Fig. 8.10. Assume that the circuit has reached
steady state at t = 0−
.
t = 0
10 V
4 Ω
0.5 H
0.02 F v
+
−
3 Ω
+
−
6 Ω
i
Figure8.10 For Example 8.4.
Solution:
For t  0, the switch is closed. The capacitor acts like an open circuit
while the inductor acts like a shunted circuit. The equivalent circuit is
shown in Fig. 8.11(a). Thus, at t = 0,
i(0) =
10
4 + 6
= 1 A, v(0) = 6i(0) = 6 V
where i(0) is the initial current through the inductor and v(0) is the initial
voltage across the capacitor.
CHAPTER 8 Second-Order Circuits 307
0.5 H
0.02 F
9 Ω
i
(b)
10 V
4 Ω
v
+
−
6 Ω
+
−
i
(a)
v
+
−
Figure8.11 The circuit in Fig. 8.10: (a) for t  0, (b) for t  0.
For t  0, the switch is opened and the voltage source is dis-
connected. The equivalent circuit is shown in Fig. 8.11(b), which is a
source-free series RLC circuit. Notice that the 3- and 6- resistors,
which are in series in Fig. 8.10 when the switch is opened, have been
combined to give R = 9  in Fig. 8.11(b). The roots are calculated as
follows:
α =
R
2L
=
9
2
1
2
 = 9, ω0 =
1
√
LC
=
1

1
2
× 1
50
= 10
s1,2 = −α ±

α2 − ω2
0 = −9 ±
√
81 − 100
or
s1,2 = −9 ± j4.359
Hence, the response is underdamped (α  ω); that is,
i(t) = e−9t
(A1 cos 4.359t + A2 sin 4.359t) (8.4.1)
We now obtain A1 and A2 using the initial conditions. At t = 0,
i(0) = 1 = A1 (8.4.2)
From Eq. (8.5),
di
dt




t=0
= −
1
L
[Ri(0) + v(0)] = −2[9(1) − 6] = −6 A/s (8.4.3)
Note that v(0) = V0 = −6 V is used, because the polarity of v in Fig.
8.11(b) is opposite that in Fig. 8.8. Taking the derivative of i(t) in Eq.
(8.4.1),
di
dt
= −9e−9t
(A1 cos 4.359t + A2 sin 4.359t)
+ e−9t
(4.359)(−A1 sin 4.359t + A2 cos 4.359t)
Imposing the condition in Eq. (8.4.3) at t = 0 gives
−6 = −9(A1 + 0) + 4.359(−0 + A2)
But A1 = 1 from Eq. (8.4.2). Then
−6 = −9 + 4.359A2 ⇒ A2 = 0.6882
308 PART 1 DC Circuits
Substituting the values of A1 and A2 in Eq. (8.4.1) yields the com-
plete solution as
i(t) = e−9t
(cos 4.359t + 0.6882 sin 4.359t) A
P R A C T I C E P R O B L E M 8 . 4
The circuit in Fig. 8.12 has reached steady state at t = 0−
. If the make-
before-break switch moves to position b at t = 0, calculate i(t) for t  0.
t = 0
a b
50 V
10 Ω
1 H
+
− 5 Ω
i(t)
F
1
9
Figure8.12 For Practice Prob. 8.4.
Answer: e−2.5t
(5 cos 1.6583t − 7.5378 sin 1.6583t) A.
8.4 THE SOURCE-FREE PARALLEL RLC CIRCUIT
Parallel RLC circuits find many practical applications, notably in com-
munications networks and filter designs.
v
R L C
I0
v
+
−
v
+
−
V0
+
−
Figure 8.13 A source-free parallel RLC
circuit.
Consider the parallel RLC circuit shown in Fig. 8.13. Assume
initial inductor current I0 and initial capacitor voltage V0,
i(0) = I0 =
1
L
 0
∞
v(t) dt (8.27a)
v(0) = V0 (8.27b)
Since the three elements are in parallel, they have the same voltage v
across them. According to passive sign convention, the current is entering
each element; that is, the current through each element is leaving the top
node. Thus, applying KCL at the top node gives
v
R
+
1
L
 t
−∞
v dt + C
dv
dt
= 0 (8.28)
Taking the derivative with respect to t and dividing by C results in
d2
v
dt2
+
1
RC
dv
dt
+
1
LC
v = 0 (8.29)
We obtain the characteristic equation by replacing the first derivative by
s and the second derivative by s2
. By following the same reasoning
used in establishing Eqs. (8.4) through (8.8), the characteristic equation
is obtained as
s2
+
1
RC
s +
1
LC
= 0 (8.30)
The roots of the characteristic equation are
s1,2 = −
1
2RC
±


1
2RC
2
−
1
LC
CHAPTER 8 Second-Order Circuits 309
or
s1,2 = −α ±

α2 − ω2
0 (8.31)
where
α =
1
2RC
, ω0 =
1
√
LC
(8.32)
The names of these terms remain the same as in the preceding section,
as they play the same role in the solution. Again, there are three possible
solutions, depending on whether α  ω0, α = ω0, or α  ω0. Let us
consider these cases separately.
Overdamped Case (α  ω0)
From Eq. (8.32), α  ω0 when L  4R2
C. The roots of the characteristic
equation are real and negative. The response is
v(t) = A1es1t
+ A2es2t
(8.33)
Critically Damped Case (α = ω0)
For α = ω, L = 4R2
C. The roots are real and equal so that the response
is
v(t) = (A1 + A2t)e−αt
(8.34)
Underdamped Case (α  ω0)
When α  ω0, L  4R2
C. In this case the roots are complex and may
be expressed as
s1,2 = −α ± jωd (8.35)
where
ωd =

ω2
0 − α2 (8.36)
The response is
v(t) = e−αt
(A1 cos ωdt + A2 sin ωdt) (8.37)
The constants A1 and A2 in each case can be determined from the
initial conditions. We need v(0) and dv(0)/dt. The first term is known
from Eq. (8.27b). We find the second term by combining Eqs. (8.27) and
(8.28), as
V0
R
+ I0 + C
dv(0)
dt
= 0
or
dv(0)
dt
= −
(V0 + RI0)
RC
(8.38)
The voltage waveforms are similar to those shown in Fig. 8.9 and will
depend on whether the circuit is overdamped, underdamped, or critically
damped.
310 PART 1 DC Circuits
Having found the capacitor voltage v(t) for the parallel RLC circuit
as shown above, we can readily obtain other circuit quantities such as
individual element currents. For example, the resistor current is iR =
v/R and the capacitor voltage is vC = C dv/dt. We have selected
the capacitor voltage v(t) as the key variable to be determined first in
order to take advantage of Eq. (8.1a). Notice that we first found the
inductor current i(t) for the RLC series circuit, whereas we first found
the capacitor voltage v(t) for the parallel RLC circuit.
E X A M P L E 8 . 5
In the parallel circuit of Fig. 8.13, find v(t) for t  0, assuming v(0) =
5 V, i(0) = 0, L = 1 H, and C = 10 mF. Consider these cases:
R = 1.923 , R = 5 , and R = 6.25 .
Solution:
CASE 1 If R = 1.923 ,
α =
1
2RC
=
1
2 × 1.923 × 10 × 10−3
= 26
ω0 =
1
√
LC
=
1
√
1 × 10 × 10−3
= 10
Since α  ω0 in this case, the response is overdamped. The roots of the
characteristic equation are
s1,2 = −α ±

α2 − ω2
0 = −2, −50
and the corresponding response is
v(t) = A1e−2t
+ A2e−50t
(8.5.1)
We now apply the initial conditions to get A1 and A2.
v(0) = 5 = A1 + A2 (8.5.2)
dv(0)
dt
= −
v(0) + Ri(0)
RC
= −
5 + 0
1.923 × 10 × 10−3
= 260
But differentiating Eq. (8.5.1),
dv
dt
= −2A1e−2t
− 50A2e−50t
At t = 0,
260 = −2A1 − 50A2 (8.5.3)
From Eqs. (8.5.2) and (8.5.3), we obtain A1 = 10.625 and A2 = −5.625.
Substituting A1 and A2 in Eq. (8.5.1) yields
v(t) = 10.625e−2t
− 5.625e−50t
(8.5.4)
CASE 2 When R = 5 ,
α =
1
2RC
=
1
2 × 5 × 10 × 10−3
= 10
CHAPTER 8 Second-Order Circuits 311
while ω0 = 10 remains the same. Since α = ω0 = 10, the response is
critically damped. Hence, s1 = s2 = −10, and
v(t) = (A1 + A2t)e−10t
(8.5.5)
To get A1 and A2, we apply the initial conditions
v(0) = 5 = A1 (8.5.6)
dv(0)
dt
= −
v(0) + Ri(0)
RC
= −
5 + 0
5 × 10 × 10−3
= 100
But differentiating Eq. (8.5.5),
dv
dt
= (−10A1 − 10A2t + A2)e−10t
At t = 0,
100 = −10A1 + A2 (8.5.7)
From Eqs. (8.5.6) and (8.5.7), A1 = 5 and A2 = 150. Thus,
v(t) = (5 + 150t)e−10t
V (8.5.8)
CASE 3 When R = 6.25 ,
α =
1
2RC
=
1
2 × 6.25 × 10 × 10−3
= 8
while ω0 = 10 remains the same. As α  ω0 in this case, the response
is underdamped. The roots of the characteristic equation are
s1,2 = −α ±

α2 − ω2
0 = −8 ± j6
Hence,
v(t) = (A1 cos 6t + A2 sin 6t)e−8t
(8.5.9)
We now obtain A1 and A2, as
v(0) = 5 = A1 (8.5.10)
dv(0)
dt
= −
v(0) + Ri(0)
RC
= −
5 + 0
6.25 × 10 × 10−3
= 80
But differentiating Eq. (8.5.9),
dv
dt
= (−8A1 cos 6t − 8A2 sin 6t − 6A1 sin 6t + 6A2 cos 6t)e−8t
At t = 0,
80 = −8A1 + 6A2 (8.5.11)
From Eqs. (8.5.10) and (8.5.11), A1 = 5 and A2 = 20. Thus,
v(t) = (5 cos 6t + 20 sin 6t)e−8t
(8.5.12)
Notice that by increasing the value of R, the degree of damping
decreases and the responses differ. Figure 8.14 plots the three cases.
312 PART 1 DC Circuits
0.2 0.4 0.6
1
2
3
4
5
6
7
8
9
10
0.8 1
–1
t (s)
v(t) V
Overdamped
Critically damped
Underdamped
Figure8.14 For Example 8.5: responses for three degrees of damping.
P R A C T I C E P R O B L E M 8 . 5
In Fig. 8.13, let R = 2 , L = 0.4 H, C = 25 mF, v(0) = 0, i(0) =
3 A. Find v(t) for t  0.
Answer: −120te−10t
V.
E X A M P L E 8 . 6
Find v(t) for t  0 in the RLC circuit of Fig. 8.15.
40 V
0.4 H
50 Ω 20 mF
30 Ω
+
−
i
t = 0 v
+
−
Figure8.15 For Example 8.6.
Solution:
When t  0, the switch is open; the inductor acts like a short circuit while
the capacitor behaves like an open circuit. The initial voltage across the
capacitor is the same as the voltage across the 50- resistor; that is,
v(0) =
50
30 + 50
(40) =
5
8
× 40 = 25 V (8.6.1)
The initial current through the inductor is
CHAPTER 8 Second-Order Circuits 313
i(0) = −
40
30 + 50
= −0.5 A
The direction of i is as indicated in Fig. 8.15 to conform with the direction
of I0 in Fig. 8.13, which is in agreement with the convention that current
flows into the positive terminal of an inductor (see Fig. 6.23). We need
to express this in terms of dv/dt, since we are looking for v.
dv(0)
dt
= −
v(0) + Ri(0)
RC
= −
25 − 50 × 0.5
50 × 20 × 10−6
= 0 (8.6.2)
When t  0, the switch is closed. The voltage source along with
the 30- resistor is separated from the rest of the circuit. The parallel
RLC circuit acts independently of the voltage source, as illustrated in
Fig. 8.16. Next, we determine that the roots of the characteristic equation
are
α =
1
2RC
=
1
2 × 50 × 20 × 10−6
= 500
ω0 =
1
√
LC
=
1
√
0.4 × 20 × 10−6
= 354
s1,2 = −α ±

α2 − ω2
0
= −500 ±
√
250,000 − 124,997.6 = −500 ± 354
or
s1 = −854, s2 = −146
Since α  ω0, we have the overdamped response
v(t) = A1e−854t
+ A2e−164t
(8.6.3)
At t = 0, we impose the condition in Eq. (8.6.1),
v(0) = 25 = A1 + A2 ⇒ A2 = 25 − A1 (8.6.4)
Taking the derivative of v(t) in Eq. (8.6.3),
dv
dt
= −854A1e−854t
− 164A2e−164t
Imposing the condition in Eq. (8.6.2),
40 V
0.4 H
50 Ω 20 mF
30 Ω
+
−
Figure8.16 The circuit in Fig. 8.15 when t  0. The
parallel RLC circuit on the left-hand side acts inde-
pendently of the circuit on the right-hand side of the
junction.
314 PART 1 DC Circuits
dv(0)
dt
= 0 = −854A1 − 164A2
or
0 = 854A1 + 164A2 (8.6.5)
Solving Eqs. (8.6.4) and (8.6.5) gives
A1 = −5.16, A2 = 30.16
Thus, the complete solution in Eq. (8.6.3) becomes
v(t) = −5.16e−854t
+ 30.16e−164t
V
P R A C T I C E P R O B L E M 8 . 6
Refer to the circuit in Fig. 8.17. Find v(t) for t  0.
2 A 4 mF
20 Ω 10 H
t = 0
v
+
−
Figure8.17 For Practice Prob. 8.6.
Answer: 66.67(e−10t
− e−2.5t
) V.
8.5 STEP RESPONSE OF A SERIES RLC CIRCUIT
As we learned in the preceding chapter, the step response is obtained by
the sudden application of a dc source. Consider the series RLC circuit
shown in Fig. 8.18. Applying KVL around the loop for t  0,
L
di
dt
+ Ri + v = Vs (8.39)
But
i = C
dv
dt
Substituting for i in Eq. (8.39) and rearranging terms,
d2
v
dt2
+
R
L
dv
dt
+
v
LC
=
Vs
LC
(8.40)
which has the same form as Eq. (8.4). More specifically, the coefficients
are the same (and that is important in determining the frequency param-
eters) but the variable is different. (Likewise, see Eq. (8.47).) Hence, the
characteristic equation for the series RLC circuit is not affected by the
presence of the dc source.
Vs
R L
C
+
−
i
t = 0
v
+
−
Figure 8.18 Step voltage applied to a series
RLC circuit.
ThesolutiontoEq.(8.40)hastwocomponents: thenaturalresponse
vn(t) and the forced response vf (t); that is,
v(t) = vn(t) + vf (t) (8.41)
The natural response is the solution when we set Vs = 0 in Eq. (8.40)
and is the same as the one obtained in Section 8.3. The natural response
vn for the overdamped, underdamped, and critically damped cases are:
CHAPTER 8 Second-Order Circuits 315
vn(t) = A1es1t
+ A2es2t
(Overdamped) (8.42a)
vn(t) = (A1 + A2t)e−αt
(Critically damped) (8.42b)
vn(t) = (A1 cos ωdt + A2 sin ωdt)e−αt
(Underdamped) (8.42c)
The forced response is the steady state or final value of v(t). In the
circuit in Fig. 8.18, the final value of the capacitor voltage is the same as
the source voltage Vs. Hence,
vf (t) = v(∞) = Vs (8.43)
Thus, the complete solutions for the overdamped, underdamped, and
critically damped cases are:
v(t) = Vs + A1es1t
+ A2es2t
(Overdamped) (8.44a)
v(t) = Vs + (A1 + A2t)e−αt
(Critically damped) (8.44b)
v(t) = Vs + (A1 cos ωdt + A2 sin ωdt)e−αt
(Underdamped) (8.44c)
The values of the constants A1 and A2 are obtained from the initial con-
ditions: v(0) and dv(0)/dt. Keep in mind that v and i are, respectively,
the voltage across the capacitor and the current through the inductor.
Therefore, Eq. (8.44) only applies for finding v. But once the capaci-
tor voltage vC = v is known, we can determine i = C dv/dt, which is
the same current through the capacitor, inductor, and resistor. Hence,
the voltage across the resistor is vR = iR, while the inductor voltage is
vL = L di/dt.
Alternatively, the complete response for any variable x(t) can be
found directly, because it has the general form
x(t) = xf (t) + xn(t) (8.45)
where the xf = x(∞) is the final value and xn(t) is the natural response.
The final value is found as in Section 8.2. The natural response has the
same form as in Eq. (8.42), and the associated constants are determined
from Eq. (8.44) based on the values of x(0) and dx(0)/dt.
E X A M P L E 8 . 7
For the circuit in Fig. 8.19, find v(t) and i(t) for t  0. Consider these
cases: R = 5 , R = 4 , andR = 1 .
24 V
R 1 H
+
− 0.5 F 1 Ω
i
t = 0
v
+
−
Figure8.19 For Example 8.7.
Solution:
CASE 1 When R = 5 . For t  0, the switch is closed. The capa-
citor behaves like an open circuit while the inductor acts like a short cir-
cuit. The initial current through the inductor is
i(0) =
24
5 + 1
= 4 A
and the initial voltage across the capacitor is the same as the voltage
across the 1- resistor; that is,
316 PART 1 DC Circuits
v(0) = 1i(0) = 4 V
For t  0, the switch is opened, so that we have the 1- resistor
disconnected. What remains is the series RLC circuit with the voltage
source. The characteristic roots are determined as follows.
α =
R
2L
=
5
2 × 1
= 2.5, ω0 =
1
√
LC
=
1
√
1 × 0.25
= 2
s1,2 = −α ±

α2 − ω2
0 = −1, −4
Since α  ω0, we have the overdamped natural response. The total
response is therefore
v(t) = vf + (A1e−t
+ A2e−4t
)
where vf is the forced or steady-state response. It is the final value of the
capacitor voltage. In Fig. 8.19, vf = 24 V. Thus,
v(t) = 24 + (A1e−t
+ A2e−4t
) (8.7.1)
We now need to find A1 and A2 using the initial conditions.
v(0) = 4 = 24 + A1 + A2
or
−20 = A1 + A2 (8.7.2)
The current through the inductor cannot change abruptly and is the same
current through the capacitor at t = 0+
because the inductor and capacitor
are now in series. Hence,
i(0) = C
dv(0)
dt
= 4 ⇒
dv(0)
dt
=
4
C
=
4
0.25
= 16
Before we use this condition, we need to take the derivative of v in Eq.
(8.7.1).
dv
dt
= −A1e−t
− 4A2e−4t
(8.7.3)
At t = 0,
dv(0)
dt
= 16 = −A1 − 4A2 (8.7.4)
From Eqs. (8.7.2) and (8.7.4), A1 = −64/3 and A2 = 4/3. Substituting
A1 and A2 in Eq. (8.7.1), we get
v(t) = 24 +
4
3
(−16e−t
+ e−4t
) V (8.7.5)
Since the inductor and capacitor are in series for t  0, the inductor
current is the same as the capacitor current. Hence,
i(t) = C
dv
dt
Multiplying Eq. (8.7.3) by C = 0.25 and substituting the values of A1
and A2 gives
CHAPTER 8 Second-Order Circuits 317
i(t) =
4
3
(4e−t
− e−4t
) A (8.7.6)
Note that i(0) = 4 A, as expected.
CASE 2 WhenR = 4 . Again, theinitialcurrentthroughtheinductor
is
i(0) =
24
4 + 1
= 4.5 A
and the initial capacitor voltage is
v(0) = 1i(0) = 4.5 V
For the characteristic roots,
α =
R
2L
=
4
2 × 1
= 2
while ω0 = 2 remains the same. In this case, s1 = s2 = −α = −2,
and we have the critically damped natural response. The total response
is therefore
v(t) = vf + (A1 + A2t)e−2t
and, as vf = 24 V,
v(t) = 24 + (A1 + A2t)e−2t
(8.7.7)
To find A1 and A2, we use the initial conditions. We write
v(0) = 4.5 = 24 + A1 ⇒ A1 = −19.5 (8.7.8)
Since i(0) = C dv(0)/dt = 4.5 or
dv(0)
dt
=
4.5
C
= 18
From Eq. (8.7.7),
dv
dt
= (−2A1 − 2tA2 + A2)e−2t
(8.7.9)
At t = 0,
dv(0)
dt
= 18 = −2A1 + A2 (8.7.10)
From Eqs. (8.7.8) and (8.7.10), A1 = −19.5 and A2 = 57. Thus, Eq.
(8.7.7) becomes
v(t) = 24 + (−19.5 + 57t)e−2t
V (8.7.11)
The inductor current is the same as the capacitor current, that is,
i(t) = C
dv
dt
Multiplying Eq. (8.7.9) by C = 0.25 and substituting the values of A1
and A2 gives
i(t) = (4.5 − 28.5t)e−2t
A (8.7.12)
Note that i(0) = 4.5 A, as expected.
318 PART 1 DC Circuits
CASE 3 When R = 1 . The initial inductor current is
i(0) =
24
1 + 1
= 12 A
and the initial voltage across the capacitor is the same as the voltage
across the 1- resistor,
v(0) = 1i(0) = 12 V
α =
R
2L
=
1
2 × 1
= 0.5
Since α = 0.5  ω0 = 2, we have the underdamped response
s1,2 = −α ±

α2 − ω2
0 = −0.5 ± j1.936
The total response is therefore
v(t) = 24 + (A1 cos 1.936t + A2 sin 1.936t)e−0.5t
(8.7.13)
We now determine A1 and A2. We write
v(0) = 12 = 24 + A1 ⇒ A1 = −12 (8.7.14)
Since i(0) = C dv(0)/dt = 12,
dv(0)
dt
=
12
C
= 48 (8.7.15)
But
dv
dt
= e−0.5t
(−1.936A1 sin 1.936t + 1.936A2 cos 1.936t)
− 0.5e−0.5t
(A1 cos 1.936t + A2 sin 1.936t)
(8.7.16)
At t = 0,
dv(0)
dt
= 48 = (−0 + 1.936A2) − 0.5(A1 + 0)
Substituting A1 = −12 gives A2 = 21.694, and Eq. (8.7.13) becomes
v(t) = 24 + (21.694 sin 1.936t − 12 cos 1.936t)e−0.5t
V (8.7.17)
The inductor current is
i(t) = C
dv
dt
Multiplying Eq. (8.7.16) by C = 0.25 and substituting the values of A1
and A2 gives
i(t) = (3.1 sin 1.936t + 12 cos 1.936t)e−0.5t
A (8.7.18)
Note that i(0) = 12 A, as expected.
Figure 8.20 plots the responses for the three cases. From this figure,
we observe that the critically damped response approaches the step input
of 24 V the fastest.
CHAPTER 8 Second-Order Circuits 319
t (s)
v(t) V
0 1
8
16
24
32
40
2 3 4 5 6 7
Overdamped
Critically damped
Underdamped
Figure 8.20 For Example 8.7: response for three degrees of
damping.
P R A C T I C E P R O B L E M 8 . 7
Having been in position a for a long time, the switch in Fig. 8.21 is moved
to position b at t = 0. Find v(t) and vR(t) for t  0.
t = 0
a b
12 V
1 Ω
+
− 10 V +
−
10 Ω
2 Ω
2.5 H
− +
vR
v
+
−
F
1
40
Figure8.21 For Practice Prob. 8.7.
Answer: 10 − (1.1547 sin 3.464t + 2 cos 3.464t)e−2t
V,
2.31e−2t
sin 3.464t V.
8.6 STEP RESPONSE OF A PARALLEL RLC CIRCUIT
Is C
R L
t = 0
i
v
+
−
Figure 8.22 Parallel RLC circuit with an
applied current.
Consider the parallel RLC circuit shown in Fig. 8.22. We want to find
i due to a sudden application of a dc current. Applying KCL at the top
node for t  0,
v
R
+ i + C
dv
dt
= Is (8.46)
But
v = L
di
dt
Substituting for v in Eq. (8.46) and dividing by LC, we get
320 PART 1 DC Circuits
d2
i
dt2
+
1
RC
di
dt
+
i
LC
=
Is
LC
(8.47)
which has the same characteristic equation as Eq. (8.29).
The complete solution to Eq. (8.47) consists of the natural response
in(t) and the forced response if ; that is,
i(t) = in(t) + if (t) (8.48)
The natural response is the same as what we had in Section 8.3. The
forced response is the steady state or final value of i. In the circuit in Fig.
8.22, the final value of the current through the inductor is the same as the
source current Is. Thus,
i(t) = Is + A1es1t
+ A2es2t
(Overdamped)
i(t) = Is + (A1 + A2t)e−αt
(Critically damped) (8.49)
i(t) = Is + (A1 cos ωdt + A2 sin ωdt)e−αt
(Underdamped)
The constants A1 and A2 in each case can be determined from the initial
conditions for i and di/dt. Again, we should keep in mind that Eq. (8.49)
only applies for finding the inductor current i. But once the inductor cur-
rent iL = i is known, we can find v = L di/dt, which is the same voltage
across inductor, capacitor, and resistor. Hence, the current through the
resistor is iR = v/R, while the capacitor current is iC = C dv/dt. Al-
ternatively, the complete response for any variable x(t) may be found
directly, using
x(t) = xf (t) + xn(t) (8.50)
where xf and xn are its final value and natural response, respectively.
E X A M P L E 8 . 8
In the circuit in Fig. 8.23, find i(t) and iR(t) for t  0.
4 A 20 Ω
20 H
iR
i
+
− 30u(–t) V
t = 0
8 mF
20 Ω
v
+
−
Figure8.23 For Example 8.8.
Solution:
For t  0, the switch is open, and the circuit is partitioned into two
independent subcircuits. The 4-A current flows through the inductor, so
that
i(0) = 4 A
CHAPTER 8 Second-Order Circuits 321
Since 30u(−t) = 30 when t  0 and 0 when t  0, the voltage source
is operative for t  0 under consideration. The capacitor acts like an
open circuit and the voltage across it is the same as the voltage across
the 20- resistor connected in parallel with it. By voltage division, the
initial capacitor voltage is
v(0) =
20
20 + 20
(30) = 15 V
For t  0, the switch is closed, and we have a parallel RLC circuit
with a current source. The voltage source is off or short-circuited. The
two 20- resistors are now in parallel. They are combined to give R =
20 20 = 10 . The characteristic roots are determined as follows:
α =
1
2RC
=
1
2 × 10 × 8 × 10−3
= 6.25
ω0 =
1
√
LC
=
1
√
20 × 8 × 10−3
= 2.5
s1,2 = −α ±

α2 − ω2
0 = −6.25 ±
√
39.0625 − 6.25
= −6.25 ± 5.7282
or
s1 = −11.978, s2 = −0.5218
Since α  ω0, we have the overdamped case. Hence,
i(t) = Is + A1e−11.978t
+ A2e−0.5218t
(8.8.1)
where Is = 4 is the final value of i(t). We now use the initial conditions
to determine A1 and A2. At t = 0,
i(0) = 4 = 4 + A1 + A2 ⇒ A2 = −A1 (8.8.2)
Taking the derivative of i(t) in Eq. (8.8.1),
di
dt
= −11.978A1e−11.978t
− 0.5218A2e−0.5218t
so that at t = 0,
di(0)
dt
= −11.978A1 − 0.5218A2 (8.8.3)
But
L
di(0)
dt
= v(0) = 15 ⇒
di(0)
dt
=
15
L
=
15
20
= 0.75
Substituting this into Eq. (8.8.3) and incorporating Eq. (8.8.2), we get
0.75 = (11.978 − 0.5218)A2 ⇒ A2 = 0.0655
Thus, A1 = −0.0655 and A2 = 0.0655. Inserting A1 and A2 in Eq.
(8.8.1) gives the complete solution as
322 PART 1 DC Circuits
i(t) = 4 + 0.0655(e−0.5218t
− e−11.978t
) A
From i(t), we obtain v(t) = L di/dt and
iR(t) =
v(t)
20
=
L
20
di
dt
= 0.785e−11.978t
− 0.0342e−0.5218t
A
P R A C T I C E P R O B L E M 8 . 8
Find i(t) and v(t) for t  0 in the circuit in Fig. 8.24.
20u(t) A 5 H
i
0.2 F
v
+
−
Figure8.24 For Practice Prob. 8.8.
Answer: 20(1 − cos t) A, 100 sin t V.
8.7 GENERAL SECOND-ORDER CIRCUITS
Now that we have mastered series and parallel RLC circuits, we are
prepared to apply the ideas to any second-order circuit. Although the
series and parallel RLC circuits are the second-order circuits of greatest
interest, other second-order circuits including op amps are also useful.
Given a second-order circuit, we determine its step response x(t) (which
may be voltage or current) by taking the following four steps:
Acircuitmaylookcomplicatedatfirst. Butonce
the sources are turned off in an attempt to find
the natural response, it may be reducible to a
first-order circuit, when the storage elements
can be combined, or to a parallel/series RLC cir-
cuit. If it is reducible to a first-order circuit, the
solution becomes simply what we had in Chap-
ter 7. If it is reducible to a parallel or series
RLC circuit, we apply the techniques of previous
sections in this chapter.
1. We first determine the initial conditions x(0) and dx(0)/dt
and the final value x(∞), as discussed in Section 8.2.
2. We find the natural response xn(t) by turning off independent
sources and applying KCL and KVL. Once a second-order
differential equation is obtained, we determine its characteristic
roots. Depending on whether the response is overdamped,
critically damped, or underdamped, we obtain xn(t) with two
unknown constants as we did in the previous sections.
3. We obtain the forced response as
xf (t) = x(∞) (8.51)
where x(∞) is the final value of x, obtained in step 1.
4. The total response is now found as the sum of the natural
response and forced response
x(t) = xn(t) + xf (t) (8.52)
We finally determine the constants associated with the natural
response by imposing the initial conditions x(0) and dx(0)/dt,
determined in step 1.
We can apply this general procedure to find the step response of
any second-order circuit, including those with op amps. The following
examples illustrate the four steps.
CHAPTER 8 Second-Order Circuits 323
E X A M P L E 8 . 9
Find the complete response v and then i for t  0 in the circuit of Fig.
8.25.
12 V +
−
4 Ω
2 Ω
t = 0
1 H
i
v
+
−
F
1
2
Figure8.25 For Example 8.9.
Solution:
We first find the initial and final values. At t = 0−
, the circuit is at steady
state. The switch is open, the equivalent circuit is shown in Fig. 8.26(a).
It is evident from the figure that
v(0−
) = 12 V, i(0−
) = 0
At t = 0+
, the switch is closed; the equivalent circuit is in Fig. 8.26(b).
By the continuity of capacitor voltage and inductor current, we know that
v(0+
) = v(0−
) = 12 V, i(0+
) = i(0−
) = 0 (8.9.1)
To get dv(0+
)/dt, we use C dv/dt = iC or dv/dt = iC/C. Applying
KCL at node a in Fig. 8.26(b),
i(0+
) = iC(0+
) +
v(0+
)
2
0 = iC(0+
) +
12
2
⇒ iC(0+
) = −6 A
12 V +
−
4 Ω
2 Ω
1 H i
0.5 F
v
+
−
iC
(b)
12 V +
−
4 Ω i
v
+
−
(a)
a
Figure8.26 Equivalent circuit of the circuit
in Fig. 8.25 for: (a) t = 0, (b) t  0.
Hence,
dv(0+
)
dt
=
−6
0.5
= −12 V/s (8.9.2)
The final values are obtained when the inductor is replaced by a short
circuit and the capacitor by an open circuit in Fig. 8.26(b), giving
i(∞) =
12
4 + 2
= 2 A, v(∞) = 2i(∞) = 4 V (8.9.3)
4 Ω
2 Ω
1 H
i
a
v
v
+
−
F
1
2
Figure 8.27 Obtaining the natural
response for Example 8.9.
Next, we obtain the natural response for t  0. By turning off the
12-V voltage source, we have the circuit in Fig. 8.27. Applying KCL at
node a in Fig. 8.27 gives
i =
v
2
+
1
2
dv
dt
(8.9.4)
Applying KVL to the left mesh results in
4i + 1
di
dt
+ v = 0 (8.9.5)
Since we are interested in v for the moment, we substitute i from Eq.
(8.9.4) into Eq. (8.9.5). We obtain
2v + 2
dv
dt
+
1
2
dv
dt
+
1
2
d2
v
dt2
+ v = 0
or
d2
v
dt2
+ 5
dv
dt
+ 6v = 0
From this, we obtain the characteristic equation as
324 PART 1 DC Circuits
s2
+ 5s + 6 = 0
with roots s = −2 and s = −3. Thus, the natural response is
vn(t) = Ae−2t
+ Be−3t
(8.9.6)
where A and B are unknown constants to be determined later. The forced
response is
vf (t) = v(∞) = 4 (8.9.7)
The complete response is
v(t) = vn + vf = 4 + Ae−2t
+ Be−3t
(8.9.8)
We now determine A and B using the initial values. From Eq. (8.9.1),
v(0) = 12. Substituting this into Eq. (8.9.8) at t = 0 gives
12 = 4 + A + B ⇒ A + B = 8 (8.9.9)
Taking the derivative of v in Eq. (8.9.8),
dv
dt
= −2Ae−2t
− 3Be−3t
(8.9.10)
Substituting Eq. (8.9.2) into Eq. (8.9.10) at t = 0 gives
−12 = −2A − 3B ⇒ 2A + 3B = 12 (8.9.11)
From Eqs. (8.9.9) and (8.9.11), we obtain
A = 12, B = −4
so that Eq. (8.9.8) becomes
v(t) = 4 + 12e−2t
− 4e−3t
V, t  0 (8.9.12)
From v, we can obtain other quantities of interest by referring to Fig.
8.26(b). To obtain i, for example,
i =
v
2
+
1
2
dv
dt
= 2 + 6e−2t
− 2e−3t
− 12e−2t
+ 6e−3t
= 2 − 6e−2t
+ 4e−3t
A, t  0
(8.9.13)
Notice that i(0) = 0, in agreement with Eq. (8.9.1).
P R A C T I C E P R O B L E M 8 . 9
Determine v and i for t  0 in the circuit of Fig. 8.28.
t = 0
2 A
10 Ω 4 Ω
2 H
i
v
+
−
F
1
20
Figure8.28 For Practice Prob. 8.9.
Answer: 8(1 − e−5t
) V, 2(1 − e−5t
) A.
CHAPTER 8 Second-Order Circuits 325
E X A M P L E 8 . 1 0
Find vo(t) for t  0 in the circuit of Fig. 8.29.
7u(t) V +
−
3 Ω
1 Ω vo
+
−
i1
i2
H
1
2
H
1
5
Figure8.29 For Example 8.10.
Solution:
This is an example of a second-order circuit with two inductors. We
first obtain the mesh currents i1 and i2, which happen to be the currents
through the inductors. We need to obtain the initial and final values of
these currents.
For t  0, 7u(t) = 0, so that i1(0−
) = 0 = i2(0−
). For t  0,
7u(t) = 7, so that the equivalent circuit is as shown in Fig. 8.30(a). Due
to the continuity of inductor current,
i1(0+
) = i1(0−
) = 0, i2(0+
) = i2(0−
) = 0 (8.10.1)
vL2(0+
) = vo(0+
) = 1[(i1(0+
) − i2(0+
)] = 0 (8.10.2)
Applying KVL to the left loop in Fig. 8.30(a) at t = 0+
,
7 = 3i1(0+
) + vL1(0+) + vo(0+
)
or
vL1(0+
) = 7 V
Since L1 di1/dt = vL1,
di1(0+
)
dt
=
vL1
L1
=
7
1
2
= 14 V/s (8.10.3)
Similarly, since L2 di2/dt = vL2,
di2(0+
)
dt
=
vL2
L2
= 0 (8.10.4)
As t → ∞, the circuit reaches steady state, and the inductors can be
replaced by short circuits, as shown in Fig. 8.30(b). From this figure,
i1(∞) = i2(∞) =
7
3
A (8.10.5)
7 V +
−
3 Ω
1 Ω vo vL2
+
−
+ −
vL1
i1
+
−
i2
(a)
7 V +
−
3 Ω
1 Ω
i1
i2
(b)
L1 = 1
2 H
L2 = 1
5 H
Figure8.30 Equivalent circuit of that in Fig. 8.29 for: (a) t  0, (b) t → ∞.
Next, we obtain the natural responses by removing the voltage
source, as shown in Fig. 8.31. Applying KVL to the two meshes yields
4i1 − i2 +
1
2
di1
dt
= 0 (8.10.6)
326 PART 1 DC Circuits
and
i2 +
1
5
di2
dt
− i1 = 0 (8.10.7)
From Eq. (8.10.6),
i2 = 4i1 +
1
2
di1
dt
(8.10.8)
Substituting Eq. (12.8.8) into Eq. (8.10.7) gives
4i1 +
1
2
di1
dt
+
4
5
di1
dt
+
1
10
d2
i1
dt2
− i1 = 0
d2
i1
dt2
+ 13
di1
dt
+ 30i1 = 0
From this we obtain the characteristic equation as
s2
+ 13s + 30 = 0
which has roots s = −3 and s = −10. Hence, the natural response is
i1n = Ae−3t
+ Be−10t
(8.10.9)
where A and B are constants. The forced response is
i1f = i1(∞) =
7
3
A (8.10.10)
From Eqs. (8.10.9) and (8.10.10), we obtain the complete response as
i1(t) =
7
3
+ Ae−3t
+ Be−10t
(8.10.11)
We finally obtain A and B from the initial values. From Eqs. (8.10.1)
and (8.10.11),
0 =
7
3
+ A + B (8.10.12)
Taking the derivative of Eq. (8.10.11), setting t = 0 in the derivative, and
enforcing Eq. (8.10.3), we obtain
14 = −3A − 10B (8.10.13)
From Eqs. (8.10.12) and (8.10.13), A = −4/3 and B = −1. Thus,
i1(t) =
7
3
−
4
3
e−3t
− e−10t
(8.10.14)
3 Ω
1 Ω
i1 i2
H
1
2
H
1
5
Figure 8.31 Obtaining the natural
response for Example 8.10.
We now obtain i2 from i1. Applying KVL to the left loop in Fig.
8.30(a) gives
7 = 4i1 − i2 +
1
2
di1
dt
⇒ i2 = −7 + 4i1 +
1
2
di1
dt
Substituting for i1 in Eq. (8.10.14) gives
i2(t) = −7 +
28
3
−
16
3
e−3t
− 4e−10t
+ 2e−3t
+ 5e−10t
=
7
3
−
10
3
e−3t
+ e−10t
(8.10.15)
CHAPTER 8 Second-Order Circuits 327
From Fig. 8.29,
vo(t) = 1[i1(t) − i2(t)] (8.10.16)
Substituting Eqs. (8.10.14) and (8.10.15) into Eq. (8.10.16) yields
vo(t) = 2(e−3t
− e−10t
) (8.10.17)
Note that vo(0) = 0, as expected from Eq. (8.10.2).
P R A C T I C E P R O B L E M 8 . 1 0
For t  0, obtain vo(t) in the circuit of Fig. 8.32.
(Hint: First find v1 and v2.)
5u(t) V +
−
1 Ω 1 Ω
+ −
vo
v1 v2
F
1
2 F
1
3
Figure8.32 For Practice Prob. 8.10.
Answer: 2(e−t
− e−6t
) V, t  0.
8.8 SECOND-ORDER OP AMP CIRCUITS
An op amp circuit with two storage elements that cannot be combined
into a single equivalent element is second-order. Because inductors are
bulky and heavy, they are rarely used in practical op amp circuits. For
this reason, we will only consider RC second-order op amp circuits here.
Such circuits find a wide range of applications in devices such as filters
and oscillators.
The use of op amps in second-order circuits
avoidstheuseofinductors, whicharesomewhat
undesirable in some applications.
The analysis of a second-order op amp circuit follows the same four
steps given and demonstrated in the previous section.
E X A M P L E 8 . 1 1
In the op amp circuit of Fig. 8.33, find vo(t) for t  0 when vs =
10u(t) mV. Let R1 = R2 = 10 k, C1 = 20 µF, and C2 = 100 µF.
vs
R1 v1
+
− C1
vo
R2
–
+
C2
v2
+ −
1
2
vo
+
−
Figure8.33 For Example 8.11.
328 PART 1 DC Circuits
Solution:
Although we could follow the same four steps given in the previous
section to solve this problem, we will solve it a little differently. Due to
the voltage follower configuration, the voltage across C1 is vo. Applying
KCL at node 1,
vs − v1
R1
= C2
dv2
dt
+
v1 − vo
R2
(8.11.1)
At node 2, KCL gives
v1 − vo
R2
= C1
dvo
dt
(8.11.2)
But
v2 = v1 − vo (8.11.3)
We now try to eliminate v1 and v2 in Eqs. (8.11.1) to (8.11.3). Substituting
Eqs. (8.11.2) and (8.11.3) into Eq. (8.11.1) yields
vs − v1
R1
= C2
dv1
dt
− C2
dvo
dt
+ C1
dvo
dt
(8.11.4)
From Eq. (8.11.2),
v1 = vo + R2C1
dvo
dt
(8.11.5)
Substituting Eq. (8.11.5) into Eq. (8.11.4), we obtain
vs
R1
=
vo
R1
+
R2C1
R1
dvo
dt
+ C2
dvo
dt
+ R2C1C2
d2
vo
dt2
− C2
dvo
dt
+ C1
dvo
dt
or
d2
vo
dt2
+

1
R1C2
+
1
R2C2

dvo
dt
+
vo
R1R2C1C2
=
vs
R1R2C1C2
(8.11.6)
With the given values of R1, R2, C1, and C2, Eq. (8.11.6) becomes
d2
vo
dt2
+ 2
dvo
dt
+ 5vo = 5vs (8.11.7)
To obtain the natural response, set vs = 0 in Eq. (8.11.7), which is the
same as turning off the source. The characteristic equation is
s2
+ 2s + 5 = 0
which has complex roots s1,2 = −1 ± j2. Hence, the natural response is
von = e−t
(A cos 2t + B sin 2t) (8.11.8)
where A and B are unknown constants to be determined.
As t → ∞, the circuit reaches the steady-state condition, and
the capacitors can be replaced by open circuits. Since no current flows
through C1 and C2 under steady-state conditions and no current can enter
the input terminals of the ideal op amp, current does not flow through R1
and R2. Thus,
vo(∞) = v1(∞) = vs
CHAPTER 8 Second-Order Circuits 329
The forced response is then
vof = vo(∞) = vs = 10 mV, t  0 (8.11.9)
The complete response is
vo(t) = von + vof = 10 + e−t
(A cos 2t + B sin 2t) mV (8.11.10)
To determine A and B, we need the initial conditions. For t  0, vs = 0,
so that
vo(0−
) = v2(0−
) = 0
For t  0, the source is operative. However, due to capacitor voltage
continuity,
vo(0+
) = v2(0+
) = 0 (8.11.11)
From Eq. (8.11.3),
v1(0+
) = v2(0+
) + vo(0+
) = 0
and hence, from Eq. (8.11.2),
dvo(0+
)
dt
=
v1 − vo
R2C1
= 0 (8.11.12)
We now impose Eq. (8.11.11) on the complete response in Eq. (8.11.10)
at t = 0, for
0 = 10 + A ⇒ A = −10 (8.11.13)
Taking the derivative of Eq. (8.11.10),
dvo
dt
= e−t
(−A cos 2t − B sin 2t − 2A sin 2t + 2B cos 2t)
Setting t = 0 and incorporating Eq. (8.11.12), we obtain
0 = −A + 2B (8.11.14)
From Eqs. (8.11.13) and (8.11.14), A = −10 and B = −5. Thus the
step response becomes
vo(t) = 10 − e−t
(10 cos 2t + 5 sin 2t) mV, t  0
P R A C T I C E P R O B L E M 8 . 1 1
In the op amp circuit shown in Fig. 8.34, vs = 4u(t) V, find vo(t) for t  0.
Assume that R1 = R2 = 10 k, C1 = 20 µF, and C2 = 100 µF.
vs
R1
+
− C2 vo
+
−
R2
C1
–
+
Figure8.34 For Practice Prob. 8.11.
Answer: 4 − 5e−t
+ e−5t
V, t  0.
330 PART 1 DC Circuits
8.9 PSPICE ANALYSIS OF RLC CIRCUITS
RLC circuits can be analyzed with great ease using PSpice, just like
the RC or RL circuits of Chapter 7. The following two examples will
illustrate this. The reader may review Section D.4 in Appendix D on
PSpice for transient analysis.
E X A M P L E 8 . 1 2
The input voltage in Fig. 8.35(a) is applied to the circuit in Fig. 8.35(b).
Use PSpice to plot v(t) for 0  t  4s.
2
0 t (s)
12
vs
(a)
(b)
vs
3 H
60 Ω
60 Ω
+
− v
+
−
F
1
27
Figure8.35 For Example 8.12.
Solution:
The given circuit is drawn using Schematics as in Fig. 8.36. The pulse
is specified using VPWL voltage source, but VPULSE could be used
instead. Using the piecewise linear function, we set the attributes of
VPWL as T1 = 0, V1 = 0, T2 = 0.001, V2 = 12, and so forth, as
shown in Fig. 8.36. Two voltage markers are inserted to plot the input
and output voltages. Once the circuit is drawn and the attributes are set,
we select Analysis/Setup/Transient to open up the Transient Analysis
dialog box. As a parallel RLC circuit, the roots of the characteristic
equation are −1 and −9. Thus, we may set Final Time as 4 s (four times
the magnitude of the lower root). When the schematic is saved, we select
Analysis/Simulate and obtain the plots for the input and output voltages
under the Probe window as shown in Fig. 8.37.
T1=0
T2=0.001
T3=2
T4=2.001
V1=0
V2=12
V3=12
V4=0
V1
R1
60
R2 60 0.037 C1
0
3H
L1
+
−
V V
Figure8.36 Schematic for the circuit in Fig. 8.35(b).
12 V
4 V
8 V
0 V
0 s 1.0 s 2.0 s 3.0 s 4.0 s
V(L1:2)
Time
V(V1:+)
Figure8.37 For Example 8.12: the input and output
voltages.
P R A C T I C E P R O B L E M 8 . 1 2
Find i(t) using PSpice for 0  t  4 s if the pulse voltage in Fig. 8.35(a)
is applied to the circuit in Fig. 8.38.
Answer: See Fig. 8.39.
CHAPTER 8 Second-Order Circuits 331
vs
5 Ω
1 mF 2 H
+
−
i
Figure8.38 For Practice Prob. 8.12.
3.0 A
1.0 A
2.0 A
0 A
0 s 1.0 s 2.0 s 3.0 s 4.0 s
I(L1)
Time
Figure8.39 Plot of i(t) for Practice Prob. 8.12.
E X A M P L E 8 . 1 3
For the circuit in Fig. 8.40, use PSpice to obtain i(t) for 0  t  3 s.
4 A 7 H
5 Ω 6 Ω
i(t)
t = 0
a
b
F
1
42
Figure8.40 For Example 8.13.
Solution:
When the switch is in position a, the 6- resistor is redundant. The
schematic for this case is shown in Fig. 8.41(a). To ensure that current
i(t) enters pin 1, the inductor is rotated three times before it is placed in the
circuit. The same applies for the capacitor. We insert pseudocomponents
IDC
4 A R1 5 7 H L1
0
23.81m C1
(a)
R2 6 7 H L1
0
23.81m C1
IC=0
IC=4A
(b)
I
0.0000 4.000E+00
Figure8.41 For Example 8.13: (a) for dc analysis, (b) for transient analysis.
332 PART 1 DC Circuits
VIEWPOINT and IPROBE to determine the initial capacitor voltage and
initial inductor current. We carry out a dc PSpice analysis by selecting
Analysis/Simulate. As shown in Fig. 8.41(a), we obtain the initial ca-
pacitor voltage as 0 V and the initial inductor current i(0) as 4 A from the
dc analysis. These initial values will be used in the transient analysis.
When the switch is moved to position b, the circuit becomes a
source-free parallel RLC circuit with the schematic in Fig. 8.41(b). We
set the initial condition IC = 0 for the capacitor and IC = 4 A for
the inductor. A current marker is inserted at pin 1 of the inductor. We
select Analysis/Setup/Transient to open up the Transient Analysis dialog
box and set Final Time to 3 s. After saving the schematic, we select
Analysis/Transient. Figure 8.42 shows the plot of i(t). The plot agrees
withi(t) = 4.8e−t
− 0.8e−6t
A,whichisthesolutionbyhandcalculation.
4.00 A
3.92 A
3.96 A
3.88 A
0 s 1.0 s 2.0 s 3.0 s
I(L1)
Time
Figure8.42 Plot of i(t) for Example 8.13.
P R A C T I C E P R O B L E M 8 . 1 3
Refer to the circuit in Fig. 8.21 (see Practice Prob. 8.7). Use PSpice to
obtain v(t) for 0  t  2.
Answer: See Fig. 8.43.
11 V
9 V
10 V
8 V
0 s 0.5 s 1.0 s 1.5 s 2.0 s
V(C1:1)
Time
Figure8.43 Plot of v(t) for Practice Prob. 8.13.
†8.10 DUALITY
The concept of duality is a time-saving, effort-effective measure of solv-
ing circuit problems. Consider the similarity between Eq. (8.4) and Eq.
(8.29). The two equations are the same, except that we must interchange
the following quantities: (1) voltage and current, (2) resistance and con-
ductance, (3) capacitance and inductance. Thus, it sometimes occurs in
circuit analysis that two different circuits have the same equations and
solutions, except that the roles of certain complementary elements are in-
terchanged. This interchangeability is known as the principle of duality.
CHAPTER 8 Second-Order Circuits 333
The duality principle asserts a parallelism between pairs of characterizing
equations and theorems of electric circuits.
Dual pairs are shown in Table 8.1. Note that power does not appear in
Table 8.1, because power has no dual. The reason for this is the principle
of linearity; since power is not linear, duality does not apply. Also notice
from Table 8.1 that the principle of duality extends to circuit elements,
configurations, and theorems.
TABLE 8.1 Dual pairs.
Resistance R Conductance G
Inductance L Capacitance C
Voltage v Current i
Voltage source Current source
Node Mesh
Series path Parallel path
Open circuit Short circuit
KVL KCL
Thevenin Norton
Even when the principle of linearity applies, a
circuit element or variable may not have a dual.
For example, mutual inductance (to be covered
in Chapter 13) has no dual.
Two circuits that are described by equations of the same form, but
in which the variables are interchanged, are said to be dual to each other.
Two circuits are said to be duals of one another if they are described by the same
characterizing equations with dual quantities interchanged.
The usefulness of the duality principle is self-evident. Once we
know the solution to one circuit, we automatically have the solution for
the dual circuit. It is obvious that the circuits in Figs. 8.8 and 8.13 are
dual. Consequently, the result in Eq. (8.32) is the dual of that in Eq.
(8.11). We must keep in mind that the principle of duality is limited
to planar circuits. Nonplanar circuits have no duals, as they cannot be
described by a system of mesh equations.
To find the dual of a given circuit, we do not need to write down
the mesh or node equations. We can use a graphical technique. Given a
planar circuit, we construct the dual circuit by taking the following three
steps:
1. Place a node at the center of each mesh of the given circuit.
Place the reference node (the ground) of the dual circuit
outside the given circuit.
2. Draw lines between the nodes such that each line crosses an
element. Replace that element by its dual (see Table 8.1).
3. To determine the polarity of voltage sources and direction of
current sources, follow this rule: A voltage source that pro-
duces a positive (clockwise) mesh current has as its dual a cur-
rent source whose reference direction is from the ground to the
nonreference node.
In case of doubt, one may verify the dual circuit by writing the nodal or
mesh equations. The mesh (or nodal) equations of the original circuit are
similar to the nodal (or mesh) equations of the dual circuit. The duality
principle is illustrated with the following two examples.
E X A M P L E 8 . 1 4
Construct the dual of the circuit in Fig. 8.44.
334 PART 1 DC Circuits
Solution:
As shown in Fig. 8.45(a), we first locate nodes 1 and 2 in the two meshes
and also the ground node 0 for the dual circuit. We draw a line between
one node and another crossing an element. We replace the line joining
the nodes by the duals of the elements which it crosses. For example, a
line between nodes 1 and 2 crosses a 2-H inductor, and we place a 2-F
capacitor (an inductor’s dual) on the line. A line between nodes 1 and
0 crossing the 6-V voltage source will contain a 6-A current source. By
drawing lines crossing all the elements, we construct the dual circuit on
the given circuit as in Fig. 8.45(a). The dual circuit is redrawn in Fig.
8.45(b) for clarity.
6 V
2 Ω
10 mF
2 H
+
−
t = 0
Figure8.44 For Example 8.14.
6 V
6 A
10 mF
10 mH
2 H
2 F
+
−
2 F
t = 0
2
0
1
1 2
2 Ω
0.5 Ω
t = 0
6 A 10 mH
0.5 Ω
t = 0
0
(a) (b)
Figure8.45 (a) Construction of the dual circuit of Fig. 8.44, (b) dual circuit redrawn.
P R A C T I C E P R O B L E M 8 . 1 4
Draw the dual circuit of the one in Fig. 8.46.
Answer: See Fig. 8.47.
50 mA 4 H
3 F
10 Ω
Figure8.46 For Practice Prob. 8.14.
50 mV 4 F
+
− 0.1 Ω
3 H
Figure8.47 Dual of the circuit in Fig. 8.46.
E X A M P L E 8 . 1 5
Obtain the dual of the circuit in Fig. 8.48.
Solution:
The dual circuit is constructed on the original circuit as in Fig. 8.49(a).
We first locate nodes 1 to 3 and the reference node 0. Joining nodes 1
and 2, we cross the 2-F capacitor, which is replaced by a 2-H inductor.
CHAPTER 8 Second-Order Circuits 335
10 V +
− 20 Ω
5 H
3 A
i2 i3
i1 2 F
Figure8.48 For Example 8.15.
Joining nodes 2 and 3, we cross the 20- resistor, which is replaced by
a 1/20- resistor. We keep doing this until all the elements are crossed.
The result is in Fig. 8.49(a). The dual circuit is redrawn in Fig. 8.49(b).
1 2 3
0
10 A 3 V
5 F
0
2 H
+
−
1 2 3
(b)
(a)
10 V
10 A
+
− 20 Ω
5 H
3 A
3 V
2 F
2 H
5 F
+
−
Ω
1
20
Ω
1
20
Figure8.49 For Example 8.15: (a) construction of the dual circuit of Fig. 8.48, (b) dual circuit redrawn.
To verify the polarity of the voltage source and the direction of
the current source, we may apply mesh currents i1, i2, and i3 (all in the
clockwise direction) in the original circuit in Fig. 8.48. The 10-V voltage
source produces positive mesh current i1, so that its dual is a 10-A current
source directed from 0 to 1. Also, i3 = −3 A in Fig. 8.48 has as its dual
v3 = −3 V in Fig. 8.49(b).
P R A C T I C E P R O B L E M 8 . 1 5
For the circuit in Fig. 8.50, obtain the dual circuit.
Answer: See Fig. 8.51.
2 A 20 V
3 Ω
0.2 F 4 H
+
−
5 Ω
Figure8.50 For Practice Prob. 8.15.
2 V 20 A
4 F
0.2 H
+
−
Ω
1
3
Ω
1
5
Figure8.51 Dual of the circuit in Fig. 8.50.
336 PART 1 DC Circuits
†8.11 APPLICATIONS
Practical applications of RLC circuits are found in control and com-
munications circuits such as ringing circuits, peaking circuits, resonant
circuits, smoothing circuits, and filters. Most of the circuits cannot be
covered until we treat ac sources. For now, we will limit ourselves to two
simple applications: automobile ignition and smoothing circuits.
8.11.1 Automobile Ignition System
In Section 7.9.4, we considered the automobile ignition system as a charg-
ing system. That was only a part of the system. Here, we consider another
part—thevoltagegeneratingsystem. Thesystemismodeledbythecircuit
shown in Fig. 8.52. The 12-V source is due to the battery and alternator.
The 4- resistor represents the resistance of the wiring. The ignition
coil is modeled by the 8-mH inductor. The 1-µF capacitor (known as
the condenser to automechanics) is in parallel with the switch (known as
the breaking points or electronic ignition). In the following example, we
determine how the RLC circuit in Fig. 8.52 is used in generating high
voltage.
12 V
4 Ω
8 mH
i
vL
+
−
t = 0
1 mF
vC
+ −
Ignition coil
Spark plug
Figure8.52 Automobile ignition circuit.
E X A M P L E 8 . 1 6
Assuming that the switch in Fig. 8.52 is closed prior to t = 0−
, find the
inductor voltage vL for t  0.
Solution:
If the switch is closed prior to t = 0−
and the circuit is in steady state,
then
i(0−
) =
12
4
= 3 A, vC(0−
) = 0
At t = 0+
, the switch is opened. The continuity conditions require that
i(0+
) = 3 A, vC(0+
) = 0 (8.16.1)
We obtain di(0+
)/dt from vL(0+
). Applying KVL to the mesh at t = 0+
yields
−12 + 4i(0+
) + vL(0+
) + vC(0+
) = 0
−12 + 4 × 3 + vL(0+
) + 0 = 0 ⇒ vL(0+
) = 0
CHAPTER 8 Second-Order Circuits 337
Hence,
di(0+
)
dt
=
vL(0+
)
L
= 0 (8.16.2)
As t → ∞, the system reaches steady state, so that the capacitor acts like
an open circuit. Then
i(∞) = 0 (8.16.3)
If we apply KVL to the mesh for t  0, we obtain
12 = Ri + L
di
dt
+
1
C
 t
0
i dt + vC(0)
Taking the derivative of each term yields
d2
i
dt2
+
R
L
di
dt
+
i
LC
= 0 (8.16.4)
We obtain the natural response by following the procedure in Section 8.3.
Substituting R = 4 , L = 8 mH, and C = 1 µF, we get
α =
R
2L
= 250, ω0 =
1
√
LC
= 1.118 × 104
Since α  ω0, the response is underdamped. The damped natural fre-
quency is
ωd =

ω2
0 − α2 ω0 = 1.118 × 104
The natural response is
in(t) = e−α
(A cos ωdt + B sin ωdt) (8.16.5)
where A and B are constants. The forced response is
if (t) = i(∞) = 0 (8.16.6)
so that the complete response is
i(t) = in(t) + if (t) = e−250t
(A cos 11,180t + B sin 11,180t) (8.16.7)
We now determine A and B.
i(0) = 3 = A + 0 ⇒ A = 3
Taking the derivative of Eq. (8.16.7),
di
dt
= −250e−250t
(A cos 11,180t + B sin 11,180t)
+ e−250t
(−11,180A sin 11,180t + 11,180B cos 11,180t)
Setting t = 0 and incorporating Eq. (8.16.2),
0 = −250A + 11,180B ⇒ B = 0.0671
Thus
i(t) = e−250t
(3 cos 11,180t + 0.0671 sin 11,180t) (8.16.8)
338 PART 1 DC Circuits
The voltage across the inductor is then
vL(t) = L
di
dt
= −268e−250t
sin 11,180t (8.16.9)
This has a maximum value when sine is unity, that is, at 11,180t0 = π/2
or t0 = 140.5 µs. At time = t0, the inductor voltage reaches its peak,
which is
vL(t0) = −268e−250t0
= −259 V (8.16.10)
Although this is far less than the voltage range of 6000 to 10,000 V
required to fire the spark plug in a typical automobile, a device known
as a transformer (to be discussed in Chapter 13) is used to step up the
inductor voltage to the required level.
P R A C T I C E P R O B L E M 8 . 1 6
In Fig. 8.52, find the capacitor voltage vC for t  0.
Answer: 12 − 12e−250t
cos 11,180t + 267.7e−250t
sin 11,180t V.
8.11.2 Smoothing Circuits
In a typical digital communication system, the signal to be transmitted
is first sampled. Sampling refers to the procedure of selecting samples
of a signal for processing, as opposed to processing the entire signal.
Each sample is converted into a binary number represented by a series
of pulses. The pulses are transmitted by a transmission line such as a
coaxial cable, twisted pair, or optical fiber. At the receiving end, the
signal is applied to a digital-to-analog (D/A) converter whose output is
a “staircase” function, that is, constant at each time interval. In order to
recover the transmitted analog signal, the output is smoothed by letting it
pass through a “smoothing” circuit, as illustrated in Fig. 8.53. An RLC
circuit may be used as the smoothing circuit.
vs(t)
Smoothing
circuit
p(t)
D/A
v0(t)
Figure8.53 A series of pulses is applied to
the digital-to-analog (D/A) converter, whose
output is applied to the smoothing circuit.
E X A M P L E 8 . 1 7
The output of a D/A converter is shown in Fig. 8.54(a). If the RLC cir-
cuit in Fig. 8.54(b) is used as the smoothing circuit, determine the output
voltage vo(t).
vs
1 Ω 1 H
1 F
+
−
1 3
0 0
2
(b)
(a)
t (s)
–2
0
4
10
v0
+
−
vs
Figure 8.54 For Example 8.17: (a) output of a D/A converter, (b) an RLC
smoothing circuit.
CHAPTER 8 Second-Order Circuits 339
Solution:
This problem is best solved using PSpice. The schematic is shown in Fig.
8.55(a). The pulse in Fig. 8.54(a) is specified using the piecewise linear
function. The attributes of V1 are set as T1 = 0, V1 = 0, T2 = 0.001,
V2 = 4, T3 = 1, V3 = 4, and so on. To be able to plot both input
and output voltages, we insert two voltage markers as shown. We select
Analysis/Setup/Transient to open up the Transient Analysis dialog box
and set Final Time as 6 s. Once the schematic is saved, we select Anal-
ysis/Simulate to run Probe and obtain the plots shown in Fig. 8.55(b).
T1=0
T2=0.001
T3=1
T4=1.001
T5=2
T6=2.001
T7=3
T8=3.001
V1=0
V2=4
V3=4
V4=10
V5=10
V6=-2
V7=-2
V8=0
V1
R1
1
1 C1
0
1H
L1
+
−
V V
10 V
0 V
5 V
-5 V
0 s 2.0 s 4.0 s 6.0 s
V(V1:+)
Time
V(C1:1)
(a) (b)
Figure8.55 For Example 8.17: (a) schematic, (b) input and output voltages.
P R A C T I C E P R O B L E M 8 . 1 7
Rework Example 8.17 if the output of the D/A converter is as shown in
Fig. 8.56.
Answer: See Fig. 8.57.
t (s)
–3
–1
0
8
7
1 2 3 4
vs
Figure8.56 For Practice
Prob. 8.17.
8.0 V
0 V
4.0 V
-4.0 V
0 s 2.0 s 4.0 s 6.0 s
V(V1:+)
Time
V(C1:1)
Figure8.57 Result of Practice Prob. 8.17.
340 PART 1 DC Circuits
8.12 SUMMARY
1. The determination of the initial values x(0) and dx(0)/dt and final
value x(∞) is crucial to analyzing second-order circuits.
2. The RLC circuit is second-order because it is described by a
second-order differential equation. Its characteristic equation is
s2
+ 2αs + ω2
0 = 0, where α is the damping factor and ω0 is the
undamped natural frequency. For a series circuit, α = R/2L, for a
parallel circuit α = 1/2RC, and for both cases ω0 = 1/
√
LC.
3. If there are no independent sources in the circuit after switching (or
sudden change), we regard the circuit as source-free. The complete
solution is the natural response.
4. The natural response of an RLC circuit is overdamped, under-
damped, or critically damped, depending on the roots of the char-
acteristic equation. The response is critically damped when the roots
are equal (s1 = s2 or α = ω0), overdamped when the roots are real
and unequal (s1 = s2 or α  ω0), or underdamped when the roots are
complex conjugate (s1 = s∗
2 or α  ω0).
5. If independent sources are present in the circuit after switching, the
complete response is the sum of the natural response and the forced
or steady-state response.
6. PSpice is used to analyze RLC circuits in the same way as for RC or
RL circuits.
7. Two circuits are dual if the mesh equations that describe one circuit
have the same form as the nodal equations that describe the other.
The analysis of one circuit gives the analysis of its dual circuit.
8. The automobile ignition circuit and the smoothing circuit are typical
applications of the material covered in this chapter.
REVIEW QUESTIONS
8.1 For the circuit in Fig. 8.58, the capacitor voltage at
t = 0−
(just before the switch is closed) is:
(a) 0 V (b) 4 V (c) 8 V (d) 12 V
4 Ω
2 F
1 H
12 V +
−
t = 0
2 Ω
Figure 8.58 For Review Questions 8.1 and 8.2.
8.2 For the circuit in Fig. 8.58, the initial inductor
current (at t = 0) is:
(a) 0 A (b) 2 A (c) 6 A (d) 12 A
8.3 When a step input is applied to a second-order
circuit, the final values of the circuit variables are
found by:
(a) Replacing capacitors with closed circuits and
inductors with open circuits.
(b) Replacing capacitors with open circuits and
inductors with closed circuits.
(c) Doing neither of the above.
8.4 If the roots of the characteristic equation of an RLC
circuit are −2 and −3, the response is:
(a) (A cos 2t + B sin 2t)e−3t
(b) (A + 2Bt)e−3t
(c) Ae−2t
+ Bte−3t
(d) Ae−2t
+ Be−3t
where A and B are constants.
8.5 In a series RLC circuit, setting R = 0 will produce:
(a) an overdamped response
CHAPTER 8 Second-Order Circuits 341
(b) a critically damped response
(c) an underdamped response
(d) an undamped response
(e) none of the above
8.6 A parallel RLC circuit has L = 2 H and
C = 0.25 F. The value of R that will produce unity
damping factor is:
(a) 0.5  (b) 1  (c) 2  (d) 4 
8.7 Refer to the series RLC circuit in Fig. 8.59. What
kind of response will it produce?
(a) overdamped
(b) underdamped
(c) critically damped
(d) none of the above
1 H
1 F
1 Ω
Figure 8.59 For Review Question 8.7.
8.8 Consider the parallel RLC circuit in Fig. 8.60.
What type of response will it produce?
(a) overdamped
(b) underdamped
(c) critically damped
(d) none of the above
1 F
1 H
1 Ω
Figure 8.60 For Review Question 8.8.
8.9 Match the circuits in Fig. 8.61 with the following
items:
(i) first-order circuit
(ii) second-order series circuit
(iii) second-order parallel circuit
(iv) none of the above
vs
R
C1
(c)
is C2
C1
L
R1
L
(d)
(e)
is
C
(f)
R1
C2
R2
vs
R L
+
−
(a)
is C
(b)
R
C
+
−
vs
R1 R2
+
− L
L
C
R2
Figure 8.61 For Review Question 8.9.
8.10 In an electric circuit, the dual of resistance is:
(a) conductance (b) inductance
(c) capacitance (d) open circuit
(e) short circuit
Answers: 8.1a, 8.2c, 8.3b, 8.4d, 8.5d, 8.6c, 8.7b, 8.8b, 8.9 (i)-c,
(ii)-b,e, (iii)-a, (iv)-d,f, 8.10a.
PROBLEMS
Section 8.2 Finding Initial and Final Values
8.1 For the circuit in Fig. 8.62, find:
(a) i(0+
) and v(0+
),
(b) di(0+
)/dt and dv(0+
)/dt,
(c) i(∞) and v(∞).
12 V
0.4 F
6 Ω
+
−
2 H
4 Ω
i
t = 0
v
+
−
Figure 8.62 For Prob. 8.1.
342 PART 1 DC Circuits
8.2 In the circuit of Fig. 8.63, determine:
(a) iR(0+
), iL(0+
), and iC(0+
),
(b) diR(0+
)/dt, diL(0+
)/dt, and diC(0+
)/dt,
(c) iR(∞), iL(∞), and iC(∞).
80 V
20 kΩ
2 mH
1 mF
60 kΩ
+
−
iL
iC
25 kΩ
iR
t = 0
Figure 8.63 For Prob. 8.2.
8.3 Refer to the circuit shown in Fig. 8.64. Calculate:
(a) iL(0+
), vC(0+
), and vR(0+
),
(b) diL(0+
)/dt, dvC(0+
)/dt, and dvR(0+
)/dt,
(c) iL(∞), vC(∞), and vR(∞).
2u(t) A
40 Ω
10 V
vR
+
−
10 Ω
+
−
IL
vC
+
−
F
1
4
H
1
8
Figure 8.64 For Prob. 8.3.
8.4 In the circuit of Fig. 8.65, find:
(a) v(0+
) and i(0+
),
(b) dv(0+
)/dt and di(0+
)/dt,
(c) v(∞) and i(∞).
4u(–t) V 4u(t) A
3 Ω 0.25 H
0.1 F 5 Ω
+
−
i
v
+
−
Figure 8.65 For Prob. 8.4.
8.5 Refer to the circuit in Fig. 8.66. Determine:
(a) i(0+
) and v(0+
),
(b) di(0+
)/dt and dv(0+
)/dt,
(c) i(∞) and v(∞).
4u(t) A
1 H
4 Ω v
+
−
6 Ω
i
F
1
4
Figure 8.66 For Prob. 8.5.
8.6 In the circuit of Fig. 8.67, find:
(a) vR(0+
) and vL(0+
),
(b) dvR(0+
)/dt and dvL(0+
)/dt,
(c) vR(∞) and vL(∞).
Vsu(t)
Rs R
+
− L
C
+ −
vR +
−
vL
Figure 8.67 For Prob. 8.6.
Section 8.3 Source-Free Series RLC Circuit
8.7 The voltage in an RLC network is described by the
differential equation
d2
v
dt2
+ 4
dv
dt
+ 4v = 0
subject to the initial conditions v(0) = 1 and
dv(0)/dt = −1. Determine the characteristic
equation. Find v(t) for t  0.
8.8 The branch current in an RLC circuit is described
by the differential equation
d2
i
dt2
+ 6
di
dt
+ 9i = 0
and the initial conditions are i(0) = 0,
di(0)/dt = 4. Obtain the characteristic equation
and determine i(t) for t  0.
8.9 The current in an RLC circuit is described by
d2
i
dt2
+ 10
di
dt
+ 25i = 0
If i(0) = 10 and di(0)/dt = 0, find i(t) for t  0.
8.10 The differential equation that describes the voltage
in an RLC network is
d2
v
dt2
+ 5
dv
dt
+ 4v = 0
Given that v(0) = 0, dv(0)/dt = 10, obtain v(t).
8.11 The natural response of an RLC circuit is described
by the differential equation
d2
v
dt2
+ 2
dv
dt
+ v = 0
CHAPTER 8 Second-Order Circuits 343
for which the initial conditions are v(0) = 10 and
dv(0)/dt = 0. Solve for v(t).
8.12 If R = 20 , L = 0.6 H, what value of C will make
an RLC series circuit:
(a) overdamped, (b) critically damped,
(c) underdamped?
8.13 For the circuit in Fig. 8.68, calculate the value of R
needed to have a critically damped response.
R 4 H
0.01 F
60 Ω
Figure 8.68 For Prob. 8.13.
8.14 Find v(t) for t  0 if v(0) = 6 V and i(0) = 2 A in
the circuit shown in Fig. 8.69.
2 H
30 Ω
60 Ω v(t)
+
−
0.02 F
i(t)
Figure 8.69 For Prob. 8.14.
8.15 The responses of a series RLC circuit are
vC(t) = 30 − 10e−20t
+ 30e−10t
V
iL(t) = 40e−20t
− 60e−10t
mA
where vC and iL are the capacitor voltage and
inductor current, respectively. Determine the values
of R, L, and C.
8.16 Find i(t) for t  0 in the circuit of Fig. 8.70.
t = 0
30 V
10 Ω
2.5 H
1 mF
40 Ω
+
−
60 Ω
i(t)
Figure 8.70 For Prob. 8.16.
8.17 Obtain v(t) for t  0 in the circuit of Fig. 8.71.
t = 0
120 V
10 Ω
4 H
1 F
v
+
−
+
−
Figure 8.71 For Prob. 8.17.
8.18 The switch in the circuit of Fig. 8.72 has been closed
for a long time but is opened at t = 0. Determine
i(t) for t  0.
2 Ω
12 V
i(t)
+ −
t = 0
H
1
2
F
1
4
Figure 8.72 For Prob. 8.18.
8.19
∗
Calculate v(t) for t  0 in the circuit of Fig. 8.73.
t = 0
24 V
12 Ω
60 Ω
+
−
3 H
6 Ω
15 Ω
25 Ω
v
+
−
F
1
27
Figure 8.73 For Prob. 8.19.
Section 8.4 Source-Free Parallel RLC Circuit
8.20 For a parallel RLC circuit, the responses are
vL(t) = 4e−20t
cos 50t − 10e−20t
sin 50t V
iC(t) = −6.5e−20t
cos 50t mA
where iC and vL are the capacitor current and
inductor voltage, respectively. Determine the values
of R, L, and C.
8.21 For the network in Fig. 8.74, what value of C is
needed to make the response underdamped with
unity damping factor (α = 1)?
∗An asterisk indicates a challenging problem.
344 PART 1 DC Circuits
0.5 H 10 mF
C
10 Ω
Figure 8.74 For Prob. 8.21.
8.22 Find v(t) for t  0 in the circuit in Fig. 8.75.
25u(–t)
5 Ω
+
− 1 mF
i
0.1 H
v
+
−
Figure 8.75 For Prob. 8.22.
8.23 In the circuit in Fig. 8.76, calculate io(t) and vo(t)
for t  0.
t = 0
30 V
2 Ω
vo(t)
8 Ω
+
−
io(t)
1 H
+
−
F
1
4
Figure 8.76 For Prob. 8.23.
Section 8.5 Step Response of a Series RLC
Circuit
8.24 The step response of an RLC circuit is given by
d2
i
dt2
+ 2
di
dt
+ 5i = 10
Given that i(0) = 2 and di(0)/dt = 4, solve for i(t).
8.25 A branch voltage in an RLC circuit is described by
d2
v
dt2
+ 4
dv
dt
+ 8v = 24
If the initial conditions are v(0) = 0 = dv(0)/dt,
find v(t).
8.26 The current in an RLC network is governed by the
differential equation
d2
i
dt2
+ 3
di
dt
+ 2i = 4
subject to i(0) = 1, di(0)/dt = −1. Solve for i(t).
8.27 Solve the following differential equations subject to
the specified initial conditions
(a) d2
v/dt2
+ 4v = 12, v(0) = 0, dv(0)/dt = 2
(b) d2
i/dt2
+ 5 di/dt + 4i = 8, i(0) = −1,
di(0)/dt = 0
(c) d2
v/dt2
+ 2 dv/dt + v = 3, v(0) = 5,
dv(0)/dt = 1
(d) d2
i/dt2
+ 2 di/dt + 5i = 10, i(0) = 4,
di(0)/dt = −2
8.28 Consider the circuit in Fig. 8.77. Find vL(0) and
vC(0).
2u(t)
40 Ω
50 V
1 F
vL
+
−
0.5 H +
−
10 Ω
vC
+
−
Figure 8.77 For Prob. 8.28.
8.29 For the circuit in Fig. 8.78, find v(t) for t  0.
1 H
4 Ω
50u(t) V
2u(–t) A
0.04 F
+ −
2 Ω
v
+ −
Figure 8.78 For Prob. 8.29.
8.30 Find v(t) for t  0 in the circuit in Fig. 8.79.
3 A
1 H
10 Ω 5 Ω
4 F
t = 0
4u(t) A
v
+
−
Figure 8.79 For Prob. 8.30.
8.31 Calculate i(t) for t  0 in the circuit in Fig. 8.80.
CHAPTER 8 Second-Order Circuits 345
20 V
5 Ω
+
−
t = 0
i
v
+ −
F
1
16
H
1
4
Figure 8.80 For Prob. 8.31.
8.32 Determine v(t) for t  0 in the circuit in Fig. 8.81.
t = 0
8 V +
− 12 V
+
−
1 H
2 Ω
v
+
−
F
1
5
Figure 8.81 For Prob. 8.32.
8.33 Obtain v(t) and i(t) for t  0 in the circuit in Fig.
8.82.
3u(t) A
5 H
0.2 F
2 Ω
1 Ω
20 V
5 Ω
+ −
i(t)
v(t)
+
−
Figure 8.82 For Prob. 8.33.
8.34
∗
For the network in Fig. 8.83, solve for i(t) for t  0.
30 V
6 Ω
+
− 10 V +
−
6 Ω
t = 0
6 Ω
i(t)
H
1
2
F
1
8
Figure 8.83 For Prob. 8.34.
8.35 Refer to the circuit in Fig. 8.84. Calculate i(t) for
t  0.
10 Ω
2 A
t = 0
10 Ω
5 Ω
i(t)
F
1
3
H
3
4
Figure 8.84 For Prob. 8.35.
8.36 Determine v(t) for t  0 in the circuit in Fig. 8.85.
60u(t) V +
− 30u(t) V
+
−
20 Ω
0.25 H
30 Ω 0.5 F
v
+ −
Figure 8.85 For Prob. 8.36.
8.37 The switch in the circuit of Fig. 8.86 is moved from
position a to b at t = 0. Determine i(t) for t  0.
12 V
2 H
+
−
2 Ω
14 Ω
6 Ω
4 A
i(t)
a
b
0.02 F
t = 0
Figure 8.86 For Prob. 8.37.
346 PART 1 DC Circuits
8.38
∗
For the network in Fig. 8.87, find i(t) for t  0.
5 Ω
1 H
100 V 5 Ω
+
−
t = 0
20 Ω
i
F
1
25
Figure 8.87 For Prob. 8.38.
8.39
∗
Given the network in Fig. 8.88, find v(t) for t  0.
4 A 1 Ω t = 0
2 A
6 Ω
1 H
v
+
−
F
1
25
Figure 8.88 For Prob. 8.39.
Section 8.6 Step Response of a Parallel RLC
Circuit
8.40 In the circuit of Fig. 8.89, find v(t) and i(t) for
t  0. Assume v(0) = 0 V and i(0) = 1 A.
4u(t) A 0.5 F 1 H
2 Ω
i
v
+
−
Figure 8.89 For Prob. 8.40.
8.41 Find i(t) for t  0 in the circuit in Fig. 8.90.
12u(t) V +
−
8 mH
2 kΩ
i(t)
5 mF
Figure 8.90 For Prob. 8.41.
8.42 Find the output voltage vo(t) in the circuit of Fig.
8.91.
3 A 10 mF
5 Ω 1 H
10 Ω
t = 0
vo
+
−
Figure 8.91 For Prob. 8.42.
8.43 Given the circuit in Fig. 8.92, find i(t) and v(t) for
t  0.
1 Ω
6 V +
−
2 Ω
t = 0
1 H
i(t)
v(t)
+
−
F
1
4
Figure 8.92 For Prob. 8.43.
8.44 Determine i(t) for t  0 in the circuit of Fig. 8.93.
3 A
5 Ω
5 H
i(t)
12 V
t = 0
4 Ω
F
1
20
+
−
Figure 8.93 For Prob. 8.44.
8.45 For the circuit in Fig. 8.94, find i(t) for t  0.
6u(t) A 40 Ω
10 mF 4 H
i(t)
30 V +
−
10 Ω
Figure 8.94 For Prob. 8.45.
CHAPTER 8 Second-Order Circuits 347
8.46 Find v(t) for t  0 in the circuit in Fig. 8.95.
io C
L
R
t = 0
v
+
−
Figure 8.95 For Prob. 8.46.
Section 8.7 General Second-Order Circuits
8.47 Derive the second-order differential equation for vo
in the circuit of Fig. 8.96.
R2
vs C2
+
− R1
C1
vo
+
−
Figure 8.96 For Prob. 8.47.
8.48 Obtain the differential equation for vo in the circuit
in Fig. 8.97.
R1
vs C
+
− R2
L
vo
+
−
Figure 8.97 For Prob. 8.48.
8.49 For the circuit in Fig. 8.98, find v(t) for t  0.
Assume that v(0+
) = 4 V and i(0+
) = 2 A.
2 Ω
0.5 F
0.1 F
i
4
v
+
−
i
Figure 8.98 For Prob. 8.49.
8.50 In the circuit of Fig. 8.99, find i(t) for t  0.
20 V
6 Ω
4 Ω
t = 0
+
−
i
F
1
25
H
1
4
Figure 8.99 For Prob. 8.50.
8.51 If the switch in Fig. 8.100 has been closed for a long
time before t = 0 but is opened at t = 0, determine:
(a) the characteristic equation of the circuit,
(b) ix and vR for t  0.
t = 0
16 V
1 H
+
−
8 Ω
12 Ω
vR
+
−
ix
F
1
36
Figure 8.100 For Prob. 8.51.
8.52 Obtain i1 and i2 for t  0 in the circuit of Fig. 8.101.
4u(t) A 1 H
2 Ω
i2
i1
1 H
3 Ω
Figure 8.101 For Prob. 8.52.
8.53 For the circuit in Prob. 8.5, find i and v for t  0.
8.54 Find the response vR(t) for t  0 in the circuit in
Fig. 8.102. Let R = 3 , L = 2 H, and C = 1/18 F.
10u(t) V
R
+
− L
C
+ −
vR
Figure 8.102 For Prob. 8.54.
348 PART 1 DC Circuits
Section 8.8 Second-Order Op Amp Circuits
8.55 Derive the differential equation relating vo to vs in
the op amp circuit of Fig. 8.103.
R2
C2
R1
C1
vs
vo
+
−
Figure 8.103 For Prob. 8.55.
8.56 Obtain the differential equation for vo(t) in the
network of Fig. 8.104.
R2
C2
R1
C1
vs
vo
+
−
Figure 8.104 For Prob. 8.56.
8.57 Determine the differential equation for the op amp
circuit in Fig. 8.105. If v1(0+
) = 2 V and
v2(0+
) = 0 V, find vo for t  0. Let R = 100 k
and C = 1 µF.
R
vo
+
−
−
C
v2
+ −
C
v1
+ −
R
+
+
−
Figure 8.105 For Prob. 8.57.
8.58 Given that vs = 2u(t) V in the op amp circuit of Fig.
8.106, find vo(t) for t  0. Let R1 = R2 = 10 k,
R3 = 20 k, R4 = 40 k, C1 = C2 = 100 µF.
vs
R1
C1 R4
vo
R2
–
+
C2
R3
Figure 8.106 For Prob. 8.58.
8.59
∗
In the op amp circuit of Fig. 8.107, determine vo(t)
for t  0. Let vin = u(t) V, R1 = R2 = 10 k,
C1 = C2 = 100 µF.
R2
C1
R1
C2
vin
vo
+
−
Figure 8.107 For Prob. 8.59.
Section 8.9 PSpice Analysis of RLC Circuit
8.60 For the step function vs = u(t), use PSpice to find
the response v(t) for 0  t  6 s in the circuit of
Fig. 8.108.
2 Ω
vs
+
−
1 H
1 F v(t)
+
−
Figure 8.108 For Prob. 8.60.
8.61 Given the source-free circuit in Fig. 8.109, use
PSpice to get i(t) for 0  t  20 s. Take
v(0) = 30 V and i(0) = 2 A.
CHAPTER 8 Second-Order Circuits 349
1 Ω 10 H 2.5 F
i
v
+
−
Figure 8.109 For Prob. 8.61.
8.62 Obtain v(t) for 0  t  4 s in the circuit of Fig.
8.110 using PSpice.
13u(t) A 39u(t) V
6 Ω
6 Ω
+
−
1 H
v(t)
+
−
20 Ω
0.4 F
Figure 8.110 For Prob. 8.62.
8.63 Rework Prob. 8.23 using PSpice. Plot vo(t) for
0  t  4 s.
Section 8.10 Duality
8.64 Draw the dual of the network in Fig. 8.111.
4 A 5 mH 10 mH
20 Ω
2 mF
Figure 8.111 For Prob. 8.64.
8.65 Obtain the dual of the circuit in Fig. 8.112.
12 V +
−
24 V
+
−
4 Ω
10 Ω
2 H
0.5 F
Figure 8.112 For Prob. 8.65.
8.66 Find the dual of the circuit in Fig. 8.113.
20 Ω
10 Ω 30 Ω
4 H
60 V
1 F 2 A
+ −
120 V
− +
Figure 8.113 For Prob. 8.66.
8.67 Draw the dual of the circuit in Fig. 8.114.
+
−
2 Ω 3 Ω
12 V
5 A
1 Ω
0.25 H
1 F
Figure 8.114 For Prob. 8.67.
Section 8.11 Applications
8.68 An automobile airbag igniter is modeled by the
circuit in Fig. 8.115. Determine the time it takes the
voltage across the igniter to reach its first peak after
switching from A to B. Let R = 3 , C = 1/30 F,
and L = 60 mH.
t = 0
A B
12 V +
− L R
C
Airbag igniter
Figure 8.115 For Prob. 8.68.
8.69 A passive interface is to be designed to connect an
electric motor to an ideal voltage source. If the
motor is modeled as a 40-mH inductor in parallel
with a 16- resistor, design the interface circuit so
that the overall circuit is critically damped at the
natural frequency of 60 Hz.
350 PART 1 DC Circuits
COMPREHENSIVE PROBLEMS
8.70 A mechanical system is modeled by a series RLC
circuit. It is desired to produce an overdamped
response with time constants 0.1 ms and 0.5 ms. If a
series 50-k resistor is used, find the values of L
and C.
8.71 An oscillogram can be adequately modeled by a
second-order system in the form of a parallel RLC
circuit. It is desired to give an underdamped voltage
across a 200- resistor. If the damping frequency is
4 kHz and the time constant of the envelope is
0.25 s, find the necessary values of L and C.
8.72 The circuit in Fig. 8.116 is the electrical analog of
body functions used in medical schools to study
convulsions. The analog is as follows:
C1 = Volume of fluid in a drug
C2 = Volume of blood stream in a specified
region
R1 = Resistance in the passage of the drug from
the input to the blood stream
R2 = Resistance of the excretion mechanism,
such as kidney, etc.
v0 = Initial concentration of the drug dosage
v(t) = Percentage of the drug in the blood stream
Find v(t) for t  0 given that C1 = 0.5 µF,
C2 = 5 µF, R1 = 5 M, R2 = 2.5 M, and
v0 = 60u(t) V.
R1
t = 0
C2
C1
vo
+
−
R2
v(t)
+
−
Figure 8.116 For Prob. 8.72.
8.73 Figure 8.117 shows a typical tunnel-diode oscillator
circuit. The diode is modeled as a nonlinear resistor
with iD = f (vD), i.e., the diode current is a
nonlinear function of the voltage across the diode.
Derive the differential equation for the circuit in
terms of v and iD.
R L i
C
v
+
−
+
−
vs
ID
vD
+
−
Figure 8.117 For Prob. 8.73.
351
AC CIRCUITS
P A R T 2
C h a p t e r 9 Sinusoids and Phasors
C h a p t e r 1 0 Sinusoidal Steady-State Analysis
C h a p t e r 1 1 AC Power Analysis
C h a p t e r 1 2 Three-Phase Circuits
C h a p t e r 1 3 Magnetically Coupled Circuits
C h a p t e r 1 4 Frequency Response
352
353
C H A P T E R
SINUSOIDS AND PHASORS
9
The desire to understand the world and the desire to reform it are the two
great engines of progress.
— Bertrand Russell
Historical Profiles
Heinrich Rudorf Hertz (1857–1894), a German experimental physicist, demonstrated
thatelectromagneticwavesobeythesamefundamentallawsaslight. Hisworkconfirmed
James Clerk Maxwell’s celebrated 1864 theory and prediction that such waves existed.
Hertz was born into a prosperous family in Hamburg, Germany. He attended
the University of Berlin and did his doctorate under the prominent physicist Hermann
von Helmholtz. He became a professor at Karlsruhe, where he began his quest for
electromagnetic waves. Hertz successfully generated and detected electromagnetic
waves; he was the first to show that light is electromagnetic energy. In 1887, Hertz
noted for the first time the photoelectric effect of electrons in a molecular structure.
Although Hertz only lived to the age of 37, his discovery of electromagnetic waves
paved the way for the practical use of such waves in radio, television, and other
communication systems. The unit of frequency, the hertz, bears his name.
Charles Proteus Steinmetz (1865–1923), a German-Austrian mathematician and en-
gineer, introduced the phasor method (covered in this chapter) in ac circuit analysis. He
is also noted for his work on the theory of hysteresis.
Steinmetz was born in Breslau, Germany, and lost his mother at the age of one.
As a youth, he was forced to leave Germany because of his political activities just as
he was about to complete his doctoral dissertation in mathematics at the University
of Breslau. He migrated to Switzerland and later to the United States, where he
was employed by General Electric in 1893. That same year, he published a paper in
which complex numbers were used to analyze ac circuits for the first time. This led to
one of his many textbooks, Theory and Calculation of ac Phenomena, published by
McGraw-Hill in 1897. In 1901, he became the president of the American Institute of
Electrical Engineers, which later became the IEEE.
354 PART 2 AC Circuits
9.1 INTRODUCTION
Thus far our analysis has been limited for the most part to dc circuits:
those circuits excited by constant or time-invariant sources. We have
restricted the forcing function to dc sources for the sake of simplicity, for
pedagogic reasons, and also for historic reasons. Historically, dc sources
were the main means of providing electric power up until the late 1800s.
At the end of that century, the battle of direct current versus alternating
current began. Both had their advocates among the electrical engineers
of the time. Because ac is more efficient and economical to transmit over
long distances, ac systems ended up the winner. Thus, it is in keeping
with the historical sequence of events that we considered dc sources first.
We now begin the analysis of circuits in which the source voltage or
current is time-varying. In this chapter, we are particularly interested in
sinusoidally time-varying excitation, or simply, excitation by a sinusoid.
A sinusoid is a signal that has the form of the sine or cosine function.
A sinusoidal current is usually referred to as alternating current (ac).
Such a current reverses at regular time intervals and has alternately posi-
tive and negative values. Circuits driven by sinusoidal current or voltage
sources are called ac circuits.
We are interested in sinusoids for a number of reasons. First, nature
itself is characteristically sinusoidal. We experience sinusoidal variation
in the motion of a pendulum, the vibration of a string, the ripples on the
ocean surface, the political events of a nation, the economic fluctuations
of the stock market, and the natural response of underdamped second-
order systems, to mention but a few. Second, a sinusoidal signal is easy
to generate and transmit. It is the form of voltage generated throughout
the world and supplied to homes, factories, laboratories, and so on. It is
the dominant form of signal in the communications and electric power
industries. Third, through Fourier analysis, any practical periodic signal
can be represented by a sum of sinusoids. Sinusoids, therefore, play an
important role in the analysis of periodic signals. Lastly, a sinusoid is
easy to handle mathematically. The derivative and integral of a sinusoid
are themselves sinusoids. For these and other reasons, the sinusoid is an
extremely important function in circuit analysis.
A sinusoidal forcing function produces both a natural (or transient)
response and a forced (or steady-state) response, much like the step func-
tion, which we studied in Chapters 7 and 8. The natural response of a
circuit is dictated by the nature of the circuit, while the steady-state re-
sponse always has a form similar to the forcing function. However, the
natural response dies out with time so that only the steady-state response
remains after a long time. When the natural response has become negligi-
bly small compared with the steady-state response, we say that the circuit
is operating at sinusoidal steady state. It is this sinusoidal steady-state
response that is of main interest to us in this chapter.
CHAPTER 9 Sinusoids and Phasors 355
We begin with a basic discussion of sinusoids and phasors. We
then introduce the concepts of impedance and admittance. The basic
circuit laws, Kirchhoff’s and Ohm’s, introduced for dc circuits, will be
applied to ac circuits. Finally, we consider applications of ac circuits in
phase-shifters and bridges.
9.2 SINUSOIDS
Consider the sinusoidal voltage
v(t) = Vm sin ωt (9.1)
where
Vm = the amplitude of the sinusoid
ω = the angular frequency in radians/s
ωt = the argument of the sinusoid
The sinusoid is shown in Fig. 9.1(a) as a function of its argument and in
Fig. 9.1(b) as a function of time. It is evident that the sinusoid repeats
itself every T seconds; thus, T is called the period of the sinusoid. From
the two plots in Fig. 9.1, we observe that ωT = 2π,
T =
2π
ω
(9.2)
The fact that v(t) repeats itself every T seconds is shown by replacing t
by t + T in Eq. (9.1). We get
v(t + T ) = Vm sin ω(t + T ) = Vm sin ω

t +
2π
ω

= Vm sin(ωt + 2π) = Vm sin ωt = v(t)
(9.3)
Hence,
v(t + T ) = v(t) (9.4)
that is, v has the same value at t + T as it does at t and v(t) is said to be
periodic. In general,
A periodic function is one that satisfies f(t) = f(t + nT), for all t and for all
integers n.
0
Vm
–Vm
π 2π 4π vt
(a)
v(t)
0
Vm
–Vm
(b)
v(t)
T
2
T 2T t
3π 3T
2
Figure9.1 A sketch of Vm sin ωt: (a) as a function of ωt, (b) as a function of t.
356 PART 2 AC Circuits
As mentioned, the period T of the periodic function is the time of one
complete cycle or the number of seconds per cycle. The reciprocal of
this quantity is the number of cycles per second, known as the cyclic
frequency f of the sinusoid. Thus,
f =
1
T
(9.5)
From Eqs. (9.2) and (9.5), it is clear that
ω = 2πf (9.6)
While ω is in radians per second (rad/s), f is in hertz (Hz).
Theunitoff isnamedaftertheGermanphysicist
Heinrich R. Hertz (1857–1894). Let us now consider a more general expression for the sinusoid,
v(t) = Vm sin(ωt + φ) (9.7)
where (ωt + φ) is the argument and φ is the phase. Both argument and
phase can be in radians or degrees.
Let us examine the two sinusoids
v1(t) = Vm sin ωt and v2(t) = Vm sin(ωt + φ) (9.8)
shown in Fig. 9.2. The starting point of v2 in Fig. 9.2 occurs first in time.
Therefore, we say that v2 leads v1 by φ or that v1 lags v2 by φ. If φ = 0,
we also say that v1 and v2 are out of phase. If φ = 0, then v1 and v2 are
said to be in phase; they reach their minima and maxima at exactly the
same time. We can compare v1 and v2 in this manner because they operate
at the same frequency; they do not need to have the same amplitude.
Vm
–Vm
vt
f
v2 = Vm sin(vt + f)
v1 = Vm sin vt
π 2π
Figure9.2 Two sinusoids with different phases.
A sinusoid can be expressed in either sine or cosine form. When
comparing two sinusoids, it is expedient to express both as either sine or
cosine with positive amplitudes. This is achieved by using the following
trigonometric identities:
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B
(9.9)
CHAPTER 9 Sinusoids and Phasors 357
With these identities, it is easy to show that
sin(ωt ± 180◦
) = − sin ωt
cos(ωt ± 180◦
) = − cos ωt
sin(ωt ± 90◦
) = ± cos ωt
cos(ωt ± 90◦
) = ∓ sin ωt
(9.10)
Using these relationships, we can transform a sinusoid from sine form to
cosine form or vice versa.
A graphical approach may be used to relate or compare sinusoids
as an alternative to using the trigonometric identities in Eqs. (9.9) and
(9.10). Consider the set of axes shown in Fig. 9.3(a). The horizontal
axis represents the magnitude of cosine, while the vertical axis (pointing
down) denotes the magnitude of sine. Angles are measured positively
counterclockwise from the horizontal, as usual in polar coordinates. This
graphical technique can be used to relate two sinusoids. For example, we
see in Fig. 9.3(a) that subtracting 90◦
from the argument of cos ωt gives
sin ωt, or cos(ωt −90◦
) = sin ωt. Similarly, adding 180◦
to the argument
of sin ωt gives − sin ωt, or sin(ωt − 180◦
) = − sin ωt, as shown in Fig.
9.3(b).
–90°
180°
+ sin vt
+ sin vt
+ cos vt
+ cos vt
(a)
(b)
Figure9.3 A graphical means
of relating cosine and sine:
(a) cos(ωt − 90◦) = sin ωt,
(b) sin(ωt + 180◦) = − sin ωt.
The graphical technique can also be used to add two sinusoids of
the same frequency when one is in sine form and the other is in cosine
form. To add A cos ωt and B sin ωt, we note that A is the magnitude
of cos ωt while B is the magnitude of sin ωt, as shown in Fig. 9.4(a).
The magnitude and argument of the resultant sinusoid in cosine form is
readily obtained from the triangle. Thus,
A cos ωt + B sin ωt = C cos(ωt − θ) (9.11)
where
C =

A2 + B2, θ = tan−1 B
A
(9.12)
For example, we may add 3 cos ωt and −4 sin ωt as shown in Fig. 9.4(b)
and obtain
3 cos ωt − 4 sin ωt = 5 cos(ωt + 53.1◦
) (9.13)
A
C
B
–u
sin vt
cos vt
(a)
sin vt
cos vt
0
53.1°
+3
–4
5
(b)
Figure 9.4 (a) Adding A cos ωt and B sin ωt, (b) adding 3 cos ωt and −4 sin ωt.
358 PART 2 AC Circuits
Compared with the trigonometric identities in Eqs. (9.9) and (9.10),
the graphical approach eliminates memorization. However, we must not
confuse the sine and cosine axes with the axes for complex numbers to
be discussed in the next section. Something else to note in Figs. 9.3 and
9.4 is that although the natural tendency is to have the vertical axis point
up, the positive direction of the sine function is down in the present case.
E X A M P L E 9 . 1
Find the amplitude, phase, period, and frequency of the sinusoid
v(t) = 12 cos(50t + 10◦
)
Solution:
The amplitude is Vm = 12 V.
The phase is φ = 10◦
.
The angular frequency is ω = 50 rad/s.
The period T =
2π
ω
=
2π
50
= 0.1257 s.
The frequency is f =
1
T
= 7.958 Hz.
P R A C T I C E P R O B L E M 9 . 1
Given the sinusoid 5 sin(4πt − 60◦
), calculate its amplitude, phase, an-
gular frequency, period, and frequency.
Answer: 5, −60◦
, 12.57 rad/s, 0.5 s, 2 Hz.
E X A M P L E 9 . 2
Calculate the phase angle between v1 = −10 cos(ωt + 50◦
) and v2 =
12 sin(ωt − 10◦
). State which sinusoid is leading.
Solution:
Let us calculate the phase in three ways. The first two methods use trigo-
nometric identities, while the third method uses the graphical approach.
METHOD 1 In order to compare v1 and v2, we must express them in the
same form. If we express them in cosine form with positive amplitudes,
v1 = −10 cos(ωt + 50◦
) = 10 cos(ωt + 50◦
− 180◦
)
v1 = 10 cos(ωt − 130◦
) or v1 = 10 cos(ωt + 230◦
) (9.2.1)
and
v2 = 12 sin(ωt − 10◦
) = 12 cos(ωt − 10◦
− 90◦
)
v2 = 12 cos(ωt − 100◦
) (9.2.2)
It can be deduced from Eqs. (9.2.1) and (9.2.2) that the phase difference
between v1 and v2 is 30◦
. We can write v2 as
CHAPTER 9 Sinusoids and Phasors 359
v2 = 12 cos(ωt − 130◦
+ 30◦
) or v2 = 12 cos(ωt + 260◦
) (9.2.3)
Comparing Eqs. (9.2.1) and (9.2.3) shows clearly that v2 leads v1 by 30◦
.
METHOD 2 Alternatively, we may express v1 in sine form:
v1 = −10 cos(ωt + 50◦
) = 10 sin(ωt + 50◦
− 90◦
)
= 10 sin(ωt − 40◦
) = 10 sin(ωt − 10◦
− 30◦
)
But v2 = 12 sin(ωt − 10◦
). Comparing the two shows that v1 lags v2 by
30◦
. This is the same as saying that v2 leads v1 by 30◦
.
50°
10°
v1
v2
sin vt
cos vt
Figure9.5 For Example 9.2.
METHOD 3 We may regard v1 as simply −10 cos ωt with a phase shift
of +50◦
. Hence, v1 is as shown in Fig. 9.5. Similarly, v2 is 12 sin ωt with
a phase shift of −10◦
, as shown in Fig. 9.5. It is easy to see from Fig. 9.5
that v2 leads v1 by 30◦
, that is, 90◦
− 50◦
− 10◦
.
P R A C T I C E P R O B L E M 9 . 2
Find the phase angle between
i1 = −4 sin(377t + 25◦
) and i2 = 5 cos(377t − 40◦
)
Does i1 lead or lag i2?
Answer: 155◦
, i1 leads i2.
9.3 PHASORS
Sinusoids are easily expressed in terms of phasors, which are more con-
venient to work with than sine and cosine functions.
A phasor is a complex number that represents the amplitude and phase
of a sinusoid.
Phasors provide a simple means of analyzing linear circuits excited by
sinusoidal sources; solutions of such circuits would be intractable other-
wise. The notion of solving ac circuits using phasors was first introduced
by Charles Steinmetz in 1893. Before we completely define phasors and
apply them to circuit analysis, we need to be thoroughly familiar with
complex numbers.
Charles Proteus Steinmetz (1865–1923) was a
German-Austrian mathematician and electrical
engineer.
AppendixBpresentsashorttutorialoncomplex
numbers.
A complex number z can be written in rectangular form as
z = x + jy (9.14a)
where j =
√
−1; x is the real part of z; y is the imaginary part of z.
In this context, the variables x and y do not represent a location as in
two-dimensional vector analysis but rather the real and imaginary parts
of z in the complex plane. Nevertheless, we note that there are some
360 PART 2 AC Circuits
resemblances between manipulating complex numbers and manipulating
two-dimensional vectors.
The complex number z can also be written in polar or exponential
form as
z = r φ = rejφ
(9.14b)
where r is the magnitude of z, and φ is the phase of z. We notice that z
can be represented in three ways:
z = x + jy Rectangular form
z = r φ Polar form
z = rejφ
Exponential form
(9.15)
The relationship between the rectangular form and the polar form
is shown in Fig. 9.6, where the x axis represents the real part and the y
axis represents the imaginary part of a complex number. Given x and y,
we can get r and φ as
r =

x2 + y2, φ = tan−1 y
x
(9.16a)
On the other hand, if we know r and φ, we can obtain x and y as
x = r cos φ, y = r sin φ (9.16b)
Thus, z may be written as
z = x + jy = r φ = r(cos φ + j sin φ) (9.17)
0
2j
j
–2j
–j
z
y
r
x
Real axis
Imaginary axis
f
Figure9.6 Representation of a
complex number z = x + jy = r φ.
Addition and subtraction of complex numbers are better performed
in rectangular form; multiplication and division are better done in polar
form. Given the complex numbers
z = x + jy = r φ, z1 = x1 + jy1 = r1 φ1
z2 = x2 + jy2 = r2 φ2
the following operations are important.
Addition:
z1 + z2 = (x1 + x2) + j(y1 + y2) (9.18a)
Subtraction:
z1 − z2 = (x1 − x2) + j(y1 − y2) (9.18b)
Multiplication:
z1z2 = r1r2 φ1 + φ2 (9.18c)
Division:
z1
z2
=
r1
r2
φ1 − φ2 (9.18d)
Reciprocal:
1
z
=
1
r
− φ (9.18e)
Square Root:
√
z =
√
r φ/2 (9.18f)
CHAPTER 9 Sinusoids and Phasors 361
Complex Conjugate:
z∗
= x − jy = r − φ = re−jφ
(9.18g)
Note that from Eq. (9.18e),
1
j
= −j (9.18h)
These are the basic properties of complex numbers we need. Other prop-
erties of complex numbers can be found in Appendix B.
The idea of phasor representation is based on Euler’s identity. In
general,
e±jφ
= cos φ ± j sin φ (9.19)
which shows that we may regard cos φ and sin φ as the real and imaginary
parts of ejφ
; we may write
cos φ = Re(ejφ
) (9.20a)
sin φ = Im(ejφ
) (9.20b)
where Re and Im stand for the real part of and the imaginary part of.
Given a sinusoid v(t) = Vm cos(ωt + φ), we use Eq. (9.20a) to express
v(t) as
v(t) = Vm cos(ωt + φ) = Re(Vmej(ωt+φ)
) (9.21)
or
v(t) = Re(Vmejφ
ejωt
) (9.22)
Thus,
v(t) = Re(Vejωt
) (9.23)
where
V = Vmejφ
= Vm φ (9.24)
V is thus the phasor representation of the sinusoid v(t), as we said earlier.
In other words, a phasor is a complex representation of the magnitude
and phase of a sinusoid. Either Eq. (9.20a) or Eq. (9.20b) can be used to
develop the phasor, but the standard convention is to use Eq. (9.20a).
A phasor may be regarded as a mathematical
equivalent of a sinusoid with the time depen-
dence dropped.
If we use sine for the phasor instead of cosine,
then v(t) = Vm sin (ωt + φ) = Im (Vmej(ωt + φ)
)
andthecorrespondingphasoristhesameasthat
in Eq. (9.24).
One way of looking at Eqs. (9.23) and (9.24) is to consider the plot
of the sinor Vejωt
= Vmej(ωt+φ)
on the complex plane. As time increases,
the sinor rotates on a circle of radius Vm at an angular velocity ω in the
counterclockwise direction, as shown in Fig. 9.7(a). In other words, the
entire complex plane is rotating at an angular velocity of ω. We may
regard v(t) as the projection of the sinor Vejωt
on the real axis, as shown
in Fig. 9.7(b). The value of the sinor at time t = 0 is the phasor V of
the sinusoid v(t). The sinor may be regarded as a rotating phasor. Thus,
whenever a sinusoid is expressed as a phasor, the term ejωt
is implicitly
present. It is therefore important, when dealing with phasors, to keep
in mind the frequency ω of the phasor; otherwise we can make serious
mistakes.
362 PART 2 AC Circuits
Rotation at v rad ⁄s
at t = to
at t = 0
f
Vm
Im
Re 0 to t
Vm
v(t) = Re(Ve jvt
)
(a) (b)
Figure 9.7 Representation of Vejωt : (a) sinor rotating counterclockwise, (b) its
projection on the real axis, as a function of time.
Equation (9.23) states that to obtain the sinusoid corresponding to
a given phasor V, multiply the phasor by the time factor ejωt
and take
the real part. As a complex quantity, a phasor may be expressed in
rectangular form, polar form, or exponential form. Since a phasor has
magnitude and phase (“direction”), it behaves as a vector and is printed
in boldface. For example, phasors V = Vm φ and I = Im − θ are
graphically represented in Fig. 9.8. Such a graphical representation of
phasors is known as a phasor diagram.
We use lightface italic letters such as z to repre-
sent complex numbers but boldface letters such
as V to represent phasors, because phasors are
vectorlike quantities.
Lagging direction
Leading direction
Real axis
Imaginary axis
Vm
Im
v
v
V
I
–u
f
Figure 9.8 A phasor diagram showing V = Vm φ and I = Im − θ .
Equations (9.21) through (9.23) reveal that to get the phasor corre-
sponding to a sinusoid, we first express the sinusoid in the cosine form
so that the sinusoid can be written as the real part of a complex number.
Then we take out the time factor ejωt
, andwhatever is left is the pha-
CHAPTER 9 Sinusoids and Phasors 363
sor corresponding to the sinusoid. By suppressing the time factor, we
transform the sinusoid from the time domain to the phasor domain. This
transformation is summarized as follows:
v(t) = Vm cos(ωt + φ)
(Time-domain
representation)
⇐⇒ V = Vm φ
(Phasor-domain
representation)
(9.25)
Givenasinusoidv(t) = Vm cos(ωt + φ),weobtainthecorrespond-
ing phasor as V = Vm φ. Equation (9.25) is also demonstrated in Table
9.1, where the sine function is considered in addition to the cosine func-
tion. From Eq. (9.25), we see that to get the phasor representation of a
sinusoid, we express it in cosine form and take the magnitude and phase.
Given a phasor, we obtain the time-domain representation as the cosine
function with the same magnitude as the phasor and the argument as
ωt plus the phase of the phasor. The idea of expressing information in
alternate domains is fundamental to all areas of engineering.
TABLE 9.1 Sinusoid-phasor transformation.
Time-domain representation Phasor-domain representation
Vm cos(ωt + φ) Vm φ
Vm sin(ωt + φ) Vm φ − 90◦
Im cos(ωt + θ) Im θ
Im sin(ωt + θ) Im θ − 90◦
Note that in Eq. (9.25) the frequency (or time) factor ejωt
is sup-
pressed, and the frequency is not explicitly shown in the phasor-domain
representation because ω is constant. However, the response depends on
ω. For this reason, the phasor domain is also known as the frequency
domain.
From Eqs. (9.23) and (9.24), v(t) = Re(Vejωt
) = Vm cos (ωt +φ),
so that
dv
dt
= −ωVm sin(ωt + φ) = ωVm cos(ωt + φ + 90◦
)
= Re(ωVmejωt
ejφ
ej90◦
) = Re(jωVejωt
)
(9.26)
This shows that the derivative v(t) is transformed to the phasor domain
as jωV
dv
dt
(Time domain)
⇐⇒ jωV
(Phasor domain)
(9.27)
Similarly, the integral of v(t) is transformed to the phasor domain as
V/jω

v dt
(Time domain)
⇐⇒
V
jω
(Phasor domain)
(9.28)
Differentiating a sinusoid is equivalent to multi-
plying its corresponding phasor by jω.
Integrating a sinusoid is equivalent to dividing its
corresponding phasor by jω.
364 PART 2 AC Circuits
Equation (9.27) allows the replacement of a derivative with respect
to time with multiplication of jω in the phasor domain, whereas Eq.
(9.28) allows the replacement of an integral with respect to time with
division by jω in the phasor domain. Equations (9.27) and (9.28) are
usefulinfindingthesteady-statesolution, whichdoesnotrequireknowing
the initial values of the variable involved. This is one of the important
applications of phasors.
Besides time differentiation and integration, another important use
of phasors is found in summing sinusoids of the same frequency. This is
best illustrated with an example, and Example 9.6 provides one.
Adding sinusoids of the same frequency is equiv-
alent to adding their corresponding phasors.
The differences between v(t) and V should be emphasized:
1. v(t) is the instantaneous or time-domain representation, while
V is the frequency or phasor-domain representation.
2. v(t) is time dependent, while V is not. (This fact is often
forgotten by students.)
3. v(t) is always real with no complex term, while V is generally
complex.
Finally, we should bear in mind that phasor analysis applies only when
frequency is constant; it applies in manipulating two or more sinusoidal
signals only if they are of the same frequency.
E X A M P L E 9 . 3
Evaluate these complex numbers:
(a) (40 50◦
+ 20 − 30◦
)1/2
(b)
10 − 30◦
+ (3 − j4)
(2 + j4)(3 − j5)∗
Solution:
(a) Using polar to rectangular transformation,
40 50◦
= 40(cos 50◦
+ j sin 50◦
) = 25.71 + j30.64
20 − 30◦
= 20[cos(−30◦
) + j sin(−30◦
)] = 17.32 − j10
Adding them up gives
40 50◦
+ 20 − 30◦
= 43.03 + j20.64 = 47.72 25.63◦
Taking the square root of this,
(40 50◦
+ 20 − 30◦
)1/2
= 6.91 12.81◦
(b) Using polar-rectangular transformation, addition, multiplication, and
division,
10 − 30◦
+ (3 − j4)
(2 + j4)(3 − j5)∗
=
8.66 − j5 + (3 − j4)
(2 + j4)(3 + j5)
=
11.66 − j9
−14 + j22
=
14.73 − 37.66◦
26.08 122.47◦
= 0.565 − 160.31◦
CHAPTER 9 Sinusoids and Phasors 365
P R A C T I C E P R O B L E M 9 . 3
Evaluate the following complex numbers:
(a) [(5 + j2)(−1 + j4) − 5 60◦
]∗
(b)
10 + j5 + 3 40◦
−3 + j4
+ 10 30◦
Answer: (a) −15.5 − j13.67, (b) 8.293 + j2.2.
E X A M P L E 9 . 4
Transform these sinusoids to phasors:
(a) v = −4 sin(30t + 50◦
)
(b) i = 6 cos(50t − 40◦
)
Solution:
(a) Since − sin A = cos(A + 90◦
),
v = −4 sin(30t + 50◦
) = 4 cos(30t + 50◦
+ 90◦
)
= 4 cos(30t + 140◦
)
The phasor form of v is
V = 4 140◦
(b) i = 6 cos(50t − 40◦
) has the phasor
I = 6 − 40◦
P R A C T I C E P R O B L E M 9 . 4
Express these sinusoids as phasors:
(a) v = −7 cos(2t + 40◦
)
(b) i = 4 sin(10t + 10◦
)
Answer: (a) V = 7 220◦
, (b) I = 4 − 80◦
.
E X A M P L E 9 . 5
Find the sinusoids represented by these phasors:
(a) V = j8e−j20◦
(b) I = −3 + j4
Solution:
(a) Since j = 1 90◦
,
V = j8 − 20◦
= (1 90◦
)(8 − 20◦
)
= 8 90◦
− 20◦
= 8 70◦
V
Converting this to the time domain gives
v(t) = 8 cos(ωt + 70◦
) V
(b) I = −3 + j4 = 5 126.87◦
. Transforming this to the time domain
gives
i(t) = 5 cos(ωt + 126.87◦
) A
366 PART 2 AC Circuits
P R A C T I C E P R O B L E M 9 . 5
Find the sinusoids corresponding to these phasors:
(a) V = −10 30◦
(b) I = j(5 − j12)
Answer: (a) v(t) = 10 cos(ωt +210◦
), (b) i(t) = 13 cos(ωt +22.62◦
).
E X A M P L E 9 . 6
Given i1(t) = 4 cos(ωt + 30◦
) and i2(t) = 5 sin(ωt − 20◦
), find their
sum.
Solution:
Here is an important use of phasors—for summing sinusoids of the same
frequency. Current i1(t) is in the standard form. Its phasor is
I1 = 4 30◦
We need to express i2(t) in cosine form. The rule for converting sine to
cosine is to subtract 90◦
. Hence,
i2 = 5 cos(ωt − 20◦
− 90◦
) = 5 cos(ωt − 110◦
)
and its phasor is
I2 = 5 − 110◦
If we let i = i1 + i2, then
I = I1 + I2 = 4 30◦
+ 5 − 110◦
= 3.464 + j2 − 1.71 − j4.698 = 1.754 − j2.698
= 3.218 − 56.97◦
A
Transforming this to the time domain, we get
i(t) = 3.218 cos(ωt − 56.97◦
) A
Of course, we can find i1 + i2 using Eqs. (9.9), but that is the hard way.
P R A C T I C E P R O B L E M 9 . 6
If v1 = −10 sin(ωt +30◦
) and v2 = 20 cos(ωt −45◦
), find V = v1 +v2.
Answer: v(t) = 10.66 cos(ωt − 30.95◦
).
E X A M P L E 9 . 7
Using the phasor approach, determine the current i(t) in a circuit de-
scribed by the integrodifferential equation
4i + 8

i dt − 3
di
dt
= 50 cos(2t + 75◦
)
CHAPTER 9 Sinusoids and Phasors 367
Solution:
We transform each term in the equation from time domain to phasor
domain. Keeping Eqs. (9.27) and (9.28) in mind, we obtain the phasor
form of the given equation as
4I +
8I
jω
− 3jωI = 50 75◦
But ω = 2, so
I(4 − j4 − j6) = 50 75◦
I =
50 75◦
4 − j10
=
50 75◦
10.77 − 68.2◦
= 4.642 143.2◦
A
Converting this to the time domain,
i(t) = 4.642 cos(2t + 143.2◦
) A
Keep in mind that this is only the steady-state solution, and it does not
require knowing the initial values.
P R A C T I C E P R O B L E M 9 . 7
Find the voltage v(t) in a circuit described by the integrodifferential equa-
tion
2
dv
dt
+ 5v + 10

v dt = 20 cos(5t − 30◦
)
using the phasor approach.
Answer: v(t) = 2.12 cos(5t − 88◦
).
9.4 PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS
Now that we know how to represent a voltage or current in the phasor or
frequency domain, one may legitimately ask how we apply this to circuits
involving the passive elements R, L, and C. What we need to do is to
transform the voltage-current relationship from the time domain to the
frequency domain for each element. Again, we will assume the passive
sign convention.
We begin with the resistor. If the current through a resistor R is
i = Im cos(ωt + φ), the voltage across it is given by Ohm’s law as
v = iR = RIm cos(ωt + φ) (9.29)
The phasor form of this voltage is
V = RIm φ (9.30)
But the phasor representation of the current is I = Im φ. Hence,
V = RI (9.31)
368 PART 2 AC Circuits
showing that the voltage-current relation for the resistor in the phasor
domain continues to be Ohm’s law, as in the time domain. Figure 9.9
illustrates the voltage-current relations of a resistor. We should note from
Eq. (9.31) that voltage and current are in phase, as illustrated in the phasor
diagram in Fig. 9.10.
(a)
i
v
+
−
R
v = iR
(b)
I
V
+
−
R
V = IR
Figure9.9 Voltage-current
relations for a resistor in the:
(a) time domain, (b) frequency
domain.
I
f
V
0 Re
Im
Figure 9.10 Phasor diagram for the
resistor.
For the inductor L, assume the current through it is i =
Im cos(ωt + φ). The voltage across the inductor is
v = L
di
dt
= −ωLIm sin(ωt + φ) (9.32)
Recall from Eq. (9.10) that − sin A = cos(A + 90◦
). We can write the
voltage as
v = ωLIm cos(ωt + φ + 90◦
) (9.33)
which transforms to the phasor
V = ωLImej(φ+90◦
)
= ωLImejφ
ej90◦
= ωLIm φej90◦
(9.34)
But Im φ = I, and from Eq. (9.19), ej90◦
= j. Thus,
V = jωLI (9.35)
showing that the voltage has a magnitude of ωLIm and a phase of φ+90◦
.
The voltage and current are 90◦
out of phase. Specifically, the current
lags the voltage by 90◦
. Figure 9.11 shows the voltage-current relations
for the inductor. Figure 9.12 shows the phasor diagram.
Although it is equally correct to say that the in-
ductor voltage leads the current by 90◦
, con-
vention gives the current phase relative to the
voltage.
i
v
+
−
L
v = L di
dt
(a)
I
V
+
−
L
V = jvLI
(b)
Figure9.11 Voltage-current
relations for an inductor in the:
(a) time domain, (b) frequency
domain.
For the capacitor C, assume the voltage across it is v =
Vm cos(ωt + φ). The current through the capacitor is
i = C
dv
dt
(9.36)
By following the same steps as we took for the inductor or by applying
Eq. (9.27) on Eq. (9.36), we obtain
I = jωCV ⇒ V =
I
jωC
(9.37)
showing that the current and voltage are 90◦
out of phase. To be specific,
thecurrentleadsthevoltageby90◦
. Figure9.13showsthevoltage-current
v
Re
Im
V
I
0
f
Figure9.12 Phasor diagram for the
inductor; I lags V.
i
v
+
−
C
(a)
i = C
dv
dt
I
V
+
−
C
(b)
I = jvCV
Figure9.13 Voltage-current
relations for a capacitor in the:
(a) time domain, (b) frequency
domain.
CHAPTER 9 Sinusoids and Phasors 369
relations for the capacitor; Fig. 9.14 gives the phasor diagram. Table 9.2
summarizes the time-domain and phasor-domain representations of the
circuit elements. v
Re
Im
I
V
0
f
Figure 9.14 Phasor diagram for the capa-
citor; I leads V.
TABLE 9.2 Summary of voltage-current
relationships.
Element Time domain Frequency domain
R v = Ri V = RI
L v = L
di
dt
V = jωLI
C i = C
dv
dt
V =
I
jωC
E X A M P L E 9 . 8
The voltage v = 12 cos(60t + 45◦
) is applied to a 0.1-H inductor. Find
the steady-state current through the inductor.
Solution:
For the inductor, V = jωLI, where ω = 60 rad/s and V = 12 45◦
V.
Hence
I =
V
jωL
=
12 45◦
j60 × 0.1
=
12 45◦
6 90◦
= 2 − 45◦
A
Converting this to the time domain,
i(t) = 2 cos(60t − 45◦
) A
P R A C T I C E P R O B L E M 9 . 8
If voltage v = 6 cos(100t −30◦
) is applied to a 50 µF capacitor, calculate
the current through the capacitor.
Answer: 30 cos(100t + 60◦
) mA.
9.5 IMPEDANCE AND ADMITTANCE
In the preceding section, we obtained the voltage-current relations for the
three passive elements as
V = RI, V = jωLI, V =
I
jωC
(9.38)
These equations may be written in terms of the ratio of the phasor voltage
to the phasor current as
V
I
= R,
V
I
= jωL,
V
I
=
1
jωC
(9.39)
From these three expressions, we obtain Ohm’s law in phasor form for
any type of element as
370 PART 2 AC Circuits
Z =
V
I
or V = ZI (9.40)
where Z is a frequency-dependent quantity known as impedance, mea-
sured in ohms.
The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor
current I, measured in ohms ().
The impedance represents the opposition which the circuit exhibits to the
flow of sinusoidal current. Although the impedance is the ratio of two
phasors, it is not a phasor, because it does not correspond to a sinusoidally
varying quantity.
The impedances of resistors, inductors, and capacitors can be read-
ily obtained from Eq. (9.39). Table 9.3 summarizes their impedances and
admittance. From the table we notice that ZL = jωL and ZC = −j/ωC.
Consider two extreme cases of angular frequency. When ω = 0 (i.e., for
dc sources), ZL = 0 and ZC → ∞, confirming what we already know—
that the inductor acts like a short circuit, while the capacitor acts like an
open circuit. When ω → ∞ (i.e., for high frequencies), ZL → ∞ and
ZC = 0, indicating that the inductor is an open circuit to high frequencies,
while the capacitor is a short circuit. Figure 9.15 illustrates this.
TABLE 9.3 Impedances and
admittances of passive elements.
Element Impedance Admittance
R Z = R Y =
1
R
L Z = jωL Y =
1
jωL
C Z =
1
jωC
Y = jωC
Short circuit at dc
Open circuit at
high frequencies
(a)
Open circuit at dc
Short circuit at
high frequencies
(b)
L
C
Figure9.15 Equivalent circuits at dc and
high frequencies: (a) inductor, (b) capacitor.
As a complex quantity, the impedance may be expressed in rectan-
gular form as
Z = R + jX (9.41)
where R = Re Z is the resistance and X = Im Z is the reactance. The
reactance X may be positive or negative. We say that the impedance is
inductive when X is positive or capacitive when X is negative. Thus,
impedance Z = R + jX is said to be inductive or lagging since current
lags voltage, while impedance Z = R − jX is capacitive or leading
because current leads voltage. The impedance, resistance, and reactance
are all measured in ohms. The impedance may also be expressed in polar
form as
Z = |Z| θ (9.42)
Comparing Eqs. (9.41) and (9.42), we infer that
Z = R + jX = |Z| θ (9.43)
where
|Z| =

R2 + X2, θ = tan−1 X
R
(9.44)
and
R = |Z| cos θ, X = |Z| sin θ (9.45)
Itissometimesconvenienttoworkwiththereciprocalofimpedance,
known as admittance.
CHAPTER 9 Sinusoids and Phasors 371
The admittance Y is the reciprocal of impedance, measured in siemens (S).
The admittance Y of an element (or a circuit) is the ratio of the phasor
current through it to the phasor voltage across it, or
Y =
1
Z
=
I
V
(9.46)
The admittances of resistors, inductors, and capacitors can be obtained
from Eq. (9.39). They are also summarized in Table 9.3.
As a complex quantity, we may write Y as
Y = G + jB (9.47)
where G = Re Y is called the conductance and B = Im Y is called the sus-
ceptance. Admittance, conductance, and susceptance are all expressed
in the unit of siemens (or mhos). From Eqs. (9.41) and (9.47),
G + jB =
1
R + jX
(9.48)
By rationalization,
G + jB =
1
R + jX
·
R − jX
R − jX
=
R − jX
R2 + X2
(9.49)
Equating the real and imaginary parts gives
G =
R
R2 + X2
, B = −
X
R2 + X2
(9.50)
showing that G = 1/R as it is in resistive circuits. Of course, if X = 0,
then G = 1/R.
E X A M P L E 9 . 9
Find v(t) and i(t) in the circuit shown in Fig. 9.16.
+
−
i
+
−
5 Ω
v
0.1 F
vs = 10 cos 4t
Figure9.16 For Example 9.9.
Solution:
From the voltage source 10 cos 4t, ω = 4,
Vs = 10 0◦
V
The impedance is
Z = 5 +
1
jωC
= 5 +
1
j4 × 0.1
= 5 − j2.5 
Hence the current
I =
Vs
Z
=
10 0◦
5 − j2.5
=
10(5 + j2.5)
52 + 2.52
= 1.6 + j0.8 = 1.789 26.57◦
A
(9.9.1)
The voltage across the capacitor is
372 PART 2 AC Circuits
V = IZC =
I
jωC
=
1.789 26.57◦
j4 × 0.1
=
1.789 26.57◦
0.4 90◦
= 4.47 − 63.43◦
V
(9.9.2)
Converting I and V in Eqs. (9.9.1) and (9.9.2) to the time domain, we get
i(t) = 1.789 cos(4t + 26.57◦
) A
v(t) = 4.47 cos(4t − 63.43◦
) V
Notice that i(t) leads v(t) by 90◦
as expected.
P R A C T I C E P R O B L E M 9 . 9
Refer to Fig. 9.17. Determine v(t) and i(t).
+
−
i 4 Ω
v
0.2 H
vs = 5 sin 10t
+
−
Figure9.17 For Practice Prob. 9.9.
Answer: 2.236 sin(10t + 63.43◦
) V, 1.118 sin(10t − 26.57◦
) A.
†9.6 KIRCHHOFF’S LAWS IN THE FREQUENCY DOMAIN
We cannot do circuit analysis in the frequency domain without Kirch-
hoff’s current and voltage laws. Therefore, we need to express them in
the frequency domain.
For KVL, let v1, v2, . . . , vn be the voltages around a closed loop.
Then
v1 + v2 + · · · + vn = 0 (9.51)
In the sinusoidal steady state, each voltage may be written in cosine form,
so that Eq. (9.51) becomes
Vm1 cos(ωt + θ1) + Vm2 cos(ωt + θ2)
+ · · · + Vmn cos(ωt + θn) = 0
(9.52)
This can be written as
Re(Vm1ejθ1
ejωt
) + Re(Vm2ejθ2
ejωt
) + · · · + Re(Vmnejθn
ejωt
) = 0
or
Re[(Vm1ejθ1
+ Vm2ejθ2
+ · · · + Vmnejθn
)ejωt
] = 0 (9.53)
If we let Vk = Vmkejθk
, then
Re[(V1 + V2 + · · · + Vn)ejωt
] = 0 (9.54)
Since ejωt
= 0,
V1 + V2 + · · · + Vn = 0 (9.55)
indicating that Kirchhoff’s voltage law holds for phasors.
CHAPTER 9 Sinusoids and Phasors 373
By following a similar procedure, we can show that Kirchhoff’s
current law holds for phasors. If we let i1, i2, . . . , in be the current
leaving or entering a closed surface in a network at time t, then
i1 + i2 + · · · + in = 0 (9.56)
If I1, I2, . . . , In are the phasor forms of the sinusoids i1, i2, . . . , in, then
I1 + I2 + · · · + In = 0 (9.57)
which is Kirchhoff’s current law in the frequency domain.
Once we have shown that both KVL and KCL hold in the frequency
domain, it is easy to do many things, such as impedance combination,
nodal and mesh analyses, superposition, and source transformation.
9.7 IMPEDANCE COMBINATIONS
Consider the N series-connected impedances shown in Fig. 9.18. The
same current I flows through the impedances. Applying KVL around the
loop gives
V = V1 + V2 + · · · + VN = I(Z1 + Z2 + · · · + ZN ) (9.58)
The equivalent impedance at the input terminals is
Zeq =
V
I
= Z1 + Z2 + · · · + ZN
or
Zeq = Z1 + Z2 + · · · + ZN (9.59)
showingthatthetotalorequivalentimpedanceofseries-connectedimped-
ances is the sum of the individual impedances. This is similar to the series
connection of resistances.
+ − + − + −
+
−
I Z1
Zeq
Z2 ZN
V1
V
V2 V
N
Figure9.18 N impedances in series.
+
−
+ −
I
+
−
Z1
V
1
Z2
V2
V
Figure9.19 Voltage division.
If N = 2, as shown in Fig. 9.19, the current through the impedances
is
I =
V
Z1 + Z2
(9.60)
Since V1 = Z1I and V2 = Z2I, then
V1 =
Z1
Z1 + Z2
V, V2 =
Z2
Z1 + Z2
V (9.61)
which is the voltage-division relationship.
374 PART 2 AC Circuits
In the same manner, we can obtain the equivalent impedance or
admittance of the N parallel-connected impedances shown in Fig. 9.20.
The voltage across each impedance is the same. Applying KCL at the
top node,
I = I1 + I2 + · · · + IN = V

1
Z1
+
1
Z2
+ · · · +
1
ZN

(9.62)
The equivalent impedance is
1
Zeq
=
I
V
=
1
Z1
+
1
Z2
+ · · · +
1
ZN
(9.63)
and the equivalent admittance is
Yeq = Y1 + Y2 + · · · + YN (9.64)
This indicates that the equivalent admittance of a parallel connection of
admittances is the sum of the individual admittances.
I
+
−
I1 I2 IN
V
I Z1 Z2 ZN
Zeq
Figure9.20 N impedances in parallel.
When N = 2, as shown in Fig. 9.21, the equivalent impedance
becomes
Zeq =
1
Yeq
=
1
Y1 + Y2
=
1
1/Z1 + 1/Z2
=
Z1Z2
Z1 + Z2
(9.65)
Also, since
V = IZeq = I1Z1 = I2Z2
the currents in the impedances are
I1 =
Z2
Z1 + Z2
I, I2 =
Z1
Z1 + Z2
I (9.66)
which is the current-division principle.
I1 I2
+
−
I Z1 Z2
V
Figure9.21 Current division.
The delta-to-wye and wye-to-delta transformations that we applied
to resistive circuits are also valid for impedances. With reference to Fig.
9.22, the conversion formulas are as follows.
CHAPTER 9 Sinusoids and Phasors 375
a b
c
n
Z1
Zb
Zc
Za
Z2
Z3
Figure9.22 Superimposed Y and ' networks.
Y-' Conversion:
Za =
Z1Z2 + Z2Z3 + Z3Z1
Z1
Zb =
Z1Z2 + Z2Z3 + Z3Z1
Z2
Zc =
Z1Z2 + Z2Z3 + Z3Z1
Z3
(9.67)
'-Y Conversion:
Z1 =
ZbZc
Za + Zb + Zc
Z2 =
ZcZa
Za + Zb + Zc
Z3 =
ZaZb
Za + Zb + Zc
(9.68)
A delta or wye circuit is said to be balanced if it has equal impedances in all
three branches.
When a '-Y circuit is balanced, Eqs. (9.67) and (9.68) become
Z' = 3ZY or ZY =
1
3
Z' (9.69)
where ZY = Z1 = Z2 = Z3 and Z' = Za = Zb = Zc.
As you see in this section, the principles of voltage division, cur-
rent division, circuit reduction, impedance equivalence, and Y-' trans-
formation all apply to ac circuits. Chapter 10 will show that other circuit
techniques—such as superposition, nodal analysis, mesh analysis, source
transformation, the Thevenin theorem, and the Norton theorem—are all
applied to ac circuits in a manner similar to their application in dc circuits.
376 PART 2 AC Circuits
E X A M P L E 9 . 1 0
Find the input impedance of the circuit in Fig. 9.23. Assume that the cir-
cuit operates at ω = 50 rad/s.
3 Ω
10 mF
Zin
8 Ω
2 mF 0.2 H
Figure9.23 For Example 9.10.
Solution:
Let
Z1 = Impedance of the 2-mF capacitor
Z2 = Impedance of the 3- resistor in series with the 10-mF
capacitor
Z3 = Impedance of the 0.2-H inductor in series with the 8-
resistor
Then
Z1 =
1
jωC
=
1
j50 × 2 × 10−3
= −j10 
Z2 = 3 +
1
jωC
= 3 +
1
j50 × 10 × 10−3
= (3 − j2) 
Z3 = 8 + jωL = 8 + j50 × 0.2 = (8 + j10) 
The input impedance is
Zin = Z1 + Z2  Z3 = −j10 +
(3 − j2)(8 + j10)
11 + j8
= −j10 +
(44 + j14)(11 − j8)
112 + 82
= −j10 + 3.22 − j1.07 
Thus,
Zin = 3.22 − j11.07 
P R A C T I C E P R O B L E M 9 . 1 0
Determine the input impedance of the circuit in Fig. 9.24 at ω =
10 rad/s.
20 Ω
4 mF
2 mF
Zin
50 Ω
2 H
Figure9.24 For Practice Prob. 9.10.
Answer: 32.38 − j73.76 .
E X A M P L E 9 . 1 1
Determine vo(t) in the circuit in Fig. 9.25.
+
−
+
−
60 Ω
10 mF vo
20 cos(4t − 15°) 5 H
Figure9.25 For Example 9.11.
Solution:
To do the analysis in the frequency domain, we must first transform the
time-domain circuit in Fig. 9.25 to the phasor-domain equivalent in Fig.
9.26. The transformation produces
CHAPTER 9 Sinusoids and Phasors 377
vs = 20 cos(4t − 15◦
) ⇒ Vs = 20 − 15◦
V, ω = 4
10 mF ⇒
1
jωC
=
1
j4 × 10 × 10−3
= −j25 
5 H ⇒ jωL = j4 × 5 = j20 
Let
Z1 = Impedance of the 60- resistor
Z2 = Impedance of the parallel combination of the 10-mF
capacitor and the 5-H inductor
Then Z1 = 60  and
Z2 = −j25  j20 =
−j25 × j20
−j25 + j20
= j100 
By the voltage-division principle,
Vo =
Z2
Z1 + Z2
Vs =
j100
60 + j100
(20 − 15◦
)
= (0.8575 30.96◦
)(20 − 15◦
) = 17.15 15.96◦
V.
We convert this to the time domain and obtain
vo(t) = 17.15 cos(4t + 15.96◦
)V
+
−
+
−
−j25 Ω j20 Ω
60 Ω
20 −15° V
o
Figure9.26 The frequency-domain
equivalent of the circuit in Fig. 9.25.
P R A C T I C E P R O B L E M 9 . 1 1
Calculate vo in the circuit in Fig. 9.27.
+
−
+
−
10 Ω vo
0.5 H
F
10 cos (10t + 75°)
1
20
Figure9.27 For Practice Prob. 9.11.
Answer: vo(t) = 7.071 cos(10t − 60◦
) V.
E X A M P L E 9 . 1 2
Find current I in the circuit in Fig. 9.28.
+
−
12 Ω 8 Ω
8 Ω
j4 Ω
j6 Ω
−j4 Ω
−j3 Ω
2 Ω
50 0°
I
a
b
c
Figure9.28 For Example 9.12.
378 PART 2 AC Circuits
Solution:
The delta network connected to nodes a, b, and c can be converted to the
Y network of Fig. 9.29. We obtain the Y impedances as follows using
Eq. (9.68):
Zan =
j4(2 − j4)
j4 + 2 − j4 + 8
=
4(4 + j2)
10
= (1.6 + j0.8) 
Zbn =
j4(8)
10
= j3.2 , Zcn =
8(2 − j4)
10
= (1.6 − j3.2) 
The total impedance at the source terminals is
Z = 12 + Zan + (Zbn − j3)  (Zcn + j6 + 8)
= 12 + 1.6 + j0.8 + (j0.2)  (9.6 + j2.8)
= 13.6 + j0.8 +
j0.2(9.6 + j2.8)
9.6 + j3
= 13.6 + j1 = 13.64 4.204◦

The desired current is
I =
V
Z
=
50 0◦
13.64 4.204◦
= 3.666 − 4.204◦
A
Zcn
+
−
I
Zan Zcn
50 0°
12 Ω
8 Ω
j6 Ω
−j3 Ω
c
b
a
n
Zbn
Figure 9.29 The circuit in Fig. 9.28 after delta-to-wye transformation.
P R A C T I C E P R O B L E M 9 . 1 2
Find I in the circuit in Fig. 9.30.
+
−
I
−j2 Ω
−j3 Ω
j5 Ω
j4 Ω
5 Ω
10 Ω
8 Ω
30 0° V
Figure9.30 For Practice Prob. 9.12.
Answer: 6.364 3.802◦
A.
CHAPTER 9 Sinusoids and Phasors 379
†9.8 APPLICATIONS
In Chapters 7 and 8, we saw certain uses of RC, RL, and RLC circuits
in dc applications. These circuits also have ac applications; among them
are coupling circuits, phase-shifting circuits, filters, resonant circuits, ac
bridge circuits, and transformers. This list of applications is inexhaustive.
We will consider some of them later. It will suffice here to observe two
simple ones: RC phase-shifting circuits, and ac bridge circuits.
9.8.1 Phase-Shifters
A phase-shifting circuit is often employed to correct an undesirable phase
shift already present in a circuit or to produce special desired effects. An
RC circuit is suitable for this purpose because its capacitor causes the
circuit current to lead the applied voltage. Two commonly used RC
circuits are shown in Fig. 9.31. (RL circuits or any reactive circuits
could also serve the same purpose.)
(a)
I
+
−
+
−
V
o
Vi R
C
(b)
I
+
−
V
o
Vi
+
−
R
C
Figure9.31 Series RC
shift circuits: (a) leading
output, (b) lagging output.
In Fig. 9.31(a), the circuit current I leads the applied voltage Vi by
some phase angle θ, where 0  θ  90◦
, depending on the values of R
and C. If XC = −1/ωC, then the total impedance is Z = R + jXC, and
the phase shift is given by
θ = tan−1 XC
R
(9.70)
This shows that the amount of phase shift depends on the values of R,
C, and the operating frequency. Since the output voltage Vo across the
resistor is in phase with the current, Vo leads (positive phase shift) Vi as
shown in Fig. 9.32(a).
vo
t
vi
(a)
t
vi
vo
(b)
u
Phase shift
u
Phase shift
Figure9.32 Phase shift in RC circuits: (a) leading output, (b) lagging output.
In Fig. 9.31(b), the output is taken across the capacitor. The current
I leads the input voltage Vi by θ, but the output voltage vo(t) across the
capacitor lags (negative phase shift) the input voltage vi(t) as illustrated
in Fig. 9.32(b).
We should keep in mind that the simple RC circuits in Fig. 9.31
also act as voltage dividers. Therefore, as the phase shift θ approaches
90◦
, the output voltage Vo approaches zero. For this reason, these simple
RC circuits are used only when small amounts of phase shift are required.
380 PART 2 AC Circuits
If it is desired to have phase shifts greater than 60◦
, simple RC networks
are cascaded, thereby providing a total phase shift equal to the sum of
the individual phase shifts. In practice, the phase shifts due to the stages
are not equal, because the succeeding stages load down the earlier stages
unless op amps are used to separate the stages.
E X A M P L E 9 . 1 3
Design an RC circuit to provide a phase of 90◦
leading.
+
−
+
−
20 Ω 20 Ω
V
i
−j20 Ω −j20 Ω
V
o
Z
V1
Figure9.33 An RC phase shift circuit with
90◦ leading phase shift; for Example 9.13.
Solution:
If we select circuit components of equal ohmic value, say R = |XC| =
20 , at a particular frequency, according to Eq. (9.70), the phase shift
is exactly 45◦
. By cascading two similar RC circuits in Fig. 9.31(a), we
obtain the circuit in Fig. 9.33, providing a positive or leading phase shift
of 90◦
, as we shall soon show. Using the series-parallel combination
technique, Z in Fig. 9.33 is obtained as
Z = 20  (20 − j20) =
20(20 − j20)
40 − j20
= 12 − j4  (9.13.1)
Using voltage division,
V1 =
Z
Z − j20
Vi =
12 − j4
12 − j24
Vi =
√
2
3
45◦
Vi (9.13.2)
and
Vo =
20
20 − j20
V1 =
√
2
2
45◦
V1 (9.13.3)
Substituting Eq. (9.13.2) into Eq. (9.13.3) yields
Vo =
√
2
2
45◦
 √
2
3
45◦
Vi

=
1
3
90◦
Vi
Thus, the output leads the input by 90◦
but its magnitude is only about
33 percent of the input.
P R A C T I C E P R O B L E M 9 . 1 3
Design an RC circuit to provide a 90◦
lagging phase shift. If a voltage
of 10 V is applied, what is the output voltage?
Answer: Figure 9.34 shows a typical design; 3.33 V.
+
−
+
−
10 Ω 10 Ω
−j10 Ω −j10 Ω V
o
V
i
Figure9.34 For Practice Prob. 9.13.
E X A M P L E 9 . 1 4
For the RL circuit shown in Fig. 9.35(a), calculate the amount of phase
shift produced at 2 kHz.
CHAPTER 9 Sinusoids and Phasors 381
Solution:
At 2 kHz, we transform the 10-mH and 5-mH inductances to the corre-
sponding impedances.
10 mH ⇒ XL = ωL = 2π × 2 × 103
× 10 × 10−3
= 40π = 125.7 
5 mH ⇒ XL = ωL = 2π × 2 × 103
× 5 × 10−3
= 20π = 62.83 
Consider the circuit in Fig. 9.35(b). The impedance Z is the parallel
combination of j125.7  and 100 + j62.83 . Hence,
Z = j125.7  (100 + j62.83)
=
j125.7(100 + j62.83)
100 + j188.5
= 69.56 60.1◦

(9.14.1)
Using voltage division,
V1 =
Z
Z + 150
Vi =
69.56 60.1◦
184.7 + j60.3
Vi
= 0.3582 42.02◦
Vi
(9.14.2)
and
Vo =
j62.832
100 + j62.832
V1 = 0.532 57.86◦
V1 (9.14.3)
Combining Eqs. (9.14.2) and (9.14.3),
Vo = (0.532 57.86◦
)(0.3582 42.02◦
) Vi = 0.1906 100◦
Vi
showing that the output is about 19 percent of the input in magnitude but
leading the input by 100◦
. If the circuit is terminated by a load, the load
will affect the phase shift.
150 Ω 100 Ω
10 mH 5 mH
(a)
150 Ω 100 Ω
(b)
+
−
+
−
Z
Vi
V1
V
o
j125.7 Ω j62.83 Ω
Figure9.35 For Example 9.14.
P R A C T I C E P R O B L E M 9 . 1 4
Refer to the RL circuit in Fig. 9.36. If 1 V is applied, find the magnitude
and the phase shift produced at 5 kHz. Specify whether the phase shift
is leading or lagging.
10 Ω 50 Ω
+
−
+
−
Vi V
o
1 mH 2 mH
Figure9.36 For Practice Prob. 9.14.
Answer: 0.172, 120.4◦
, lagging.
9.8.2 AC Bridges
An ac bridge circuit is used in measuring the inductance L of an inductor
or the capacitance C of a capacitor. It is similar in form to the Wheatstone
bridge for measuring an unknown resistance (discussed in Section 4.10)
and follows the same principle. To measure L and C, however, an ac
source is needed as well as an ac meter instead of the galvanometer. The
ac meter may be a sensitive ac ammeter or voltmeter.
382 PART 2 AC Circuits
Consider the general ac bridge circuit displayed in Fig. 9.37. The
bridge is balanced when no current flows through the meter. This means
that V1 = V2. Applying the voltage division principle,
V1 =
Z2
Z1 + Z2
Vs = V2 =
Zx
Z3 + Zx
Vs (9.71)
Thus,
Z2
Z1 + Z2
=
Zx
Z3 + Zx
⇒ Z2Z3 = Z1Zx (9.72)
or
Zx =
Z3
Z1
Z2 (9.73)
This is the balanced equation for the ac bridge and is similar to Eq. (4.30)
for the resistance bridge except that the R’s are replaced by Z’s.
AC
meter
+
−
+
−
≈
Vs
Z1 Z3
Z2 V1 V2
Zx
Figure9.37 A general ac bridge.
Specific ac bridges for measuring L and C are shown in Fig. 9.38,
where Lx and Cx are the unknown inductance and capacitance to be
measured while Ls and Cs are a standard inductance and capacitance (the
values of which are known to great precision). In each case, two resistors,
R1 and R2, are varied until the ac meter reads zero. Then the bridge is
balanced. From Eq. (9.73), we obtain
Lx =
R2
R1
Ls (9.74)
and
Cx =
R1
R2
Cs (9.75)
Notice that the balancing of the ac bridges in Fig. 9.38 does not depend on
the frequency f of the ac source, since f does not appear in the relation-
ships in Eqs. (9.74) and (9.75).
AC
meter
≈
R1 R2
Ls Lx
(a)
AC
meter
≈
R1 R2
Cs Cx
(b)
Figure 9.38 Specific ac bridges: (a) for measuring L, (b) for measuring C.
E X A M P L E 9 . 1 5
The ac bridge circuit of Fig. 9.37 balances when Z1 is a 1-k resistor,
Z2 is a 4.2-k resistor, Z3 is a parallel combination of a 1.5-M resistor
CHAPTER 9 Sinusoids and Phasors 383
and a 12-pF capacitor, and f = 2 kHz. Find: (a) the series components
that make up Zx, and (b) the parallel components that make up Zx.
Solution:
From Eq. (9.73),
Zx =
Z3
Z1
Z2 (9.15.1)
where Zx = Rx + jXx,
Z1 = 1000 , Z2 = 4200  (9.15.2)
and
Z3 = R3 
1
jωC3
=
R3
jωC3
R3 + 1/jωC3
=
R3
1 + jωR3C3
Since R3 = 1.5 M and C3 = 12 pF,
Z3 =
1.5 × 106
1 + j2π × 2 × 103 × 1.5 × 106 × 12 × 10−12
=
1.5 × 106
1 + j0.2262
or
Z3 = 1.427 − j0.3228 M (9.15.3)
(a) Assuming that Zx is made up of series components, we substitute Eqs.
(9.15.2) and (9.15.3) in Eq. (9.15.1) and obtain
Rx + jXx =
4200
1000
(1.427 − j0.3228) × 106
= (5.993 − j1.356) M
Equating the real and imaginary parts yields Rx = 5.993 M and a
capacitive reactance
Xx =
1
ωC
= 1.356 × 106
or
C =
1
ωXx
=
1
2π × 2 × 103 × 1.356 × 106
= 58.69 pF
(b) If Zx is made up of parallel components, we notice that Z3 is also a
parallel combination. Hence, Eq. (9.15.1) becomes
Zx =
4200
1000
R3




1
jωC3
= 4.2R3




1
jωC3
= 4.2Z3 (9.15.4)
This simply means that the unknown impedance Zx is 4.2 times Z3.
Since Z3 consists of R3 and X3 = 1/ωC3, there are many ways we can
get 4.2Z3. Therefore, there is no unique answer to the problem. If we
suppose that 4.2 = 3 × 1.4 and we decide to multiply R3 by 1.4 while
multiplying X3 by 3, then the answer is
Rx = 1.4R3 = 2.1 M
and
Xx =
1
ωCx
= 3X3 =
3
ωC3
⇒ Cx =
1
3
C3 = 4 pF
384 PART 2 AC Circuits
Alternatively, we may decide to multiply R3 by 3 while multiplying Xx
by 1.4 and obtain Rx = 4.5 M and Cx = C3/1.4 = 8.571 pF. Of
course, there are several other possibilities. In a situation like this when
there is no unique solution, care must be taken to select reasonably sized
component values whenever possible.
P R A C T I C E P R O B L E M 9 . 1 5
In the ac bridge circuit of Fig. 9.37, suppose that balance is achieved
when Z1 is a 4.8-k resistor, Z2 is a 10- resistor in series with a
0.25-µH inductor, Z3 is a 12-k resistor, and f = 6 MHz. Determine
the series components that make up Zx.
Answer: A 25- resistor in series with a 0.625-µH inductor.
9.9 SUMMARY
1. A sinusoid is a signal in the form of the sine or cosine function. It
has the general form
v(t) = Vm cos(ωt + φ)
where Vm is the amplitude, ω = 2πf is the angular frequency,
(ωt + φ) is the argument, and φ is the phase.
2. A phasor is a complex quantity that represents both the magnitude
and the phase of a sinusoid. Given the sinusoid
v(t) = Vm cos(ωt + φ), its phasor V is
V = Vm φ
3. In ac circuits, voltage and current phasors always have a fixed
relation to one another at any moment of time. If v(t) =
Vm cos(ωt + φv) represents the voltage through an element and
i(t) = Im cos(ωt + φi) represents the current through the element,
then φi = φv if the element is a resistor, φi leads φv by 90◦
if the
element is a capacitor, and φi lags φv by 90◦
if the element is an
inductor.
4. The impedance Z of a circuit is the ratio of the phasor voltage across
it to the phasor current through it:
Z =
V
I
= R(ω) + jX(ω)
The admittance Y is the reciprocal of impedance:
Z =
1
Y
= G(ω) + jB(ω)
Impedances are combined in series or in parallel the same way as
resistances in series or parallel; that is, impedances in series add
while admittances in parallel add.
5. For a resistor Z = R, for an inductor Z = jX = jωL, and for a
capacitor Z = −jX = 1/jωC.
6. Basic circuit laws (Ohm’s and Kirchhoff’s) apply to ac circuits in the
same manner as they do for dc circuits; that is,
CHAPTER 9 Sinusoids and Phasors 385
V = ZI

Ik = 0 (KCL)

Vk = 0 (KVL)
7. The techniques of voltage/current division, series/parallel combina-
tion of impedance/admittance, circuit reduction, and Y-' trans-
formation all apply to ac circuit analysis.
8. AC circuits are applied in phase-shifters and bridges.
REVIEW QUESTIONS
9.1 Which of the following is not a right way to express
the sinusoid A cos ωt?
(a) A cos 2πf t (b) A cos(2πt/T )
(c) A cos ω(t − T ) (d) A sin(ωt − 90◦
)
9.2 A function that repeats itself after fixed intervals is
said to be:
(a) a phasor (b) harmonic
(c) periodic (d) reactive
9.3 Which of these frequencies has the shorter period?
(a) 1 krad/s (b) 1 kHz
9.4 If v1 = 30 sin(ωt + 10◦
) and v2 = 20 sin(ωt + 50◦
),
which of these statements are true?
(a) v1 leads v2 (b) v2 leads v1
(c) v2 lags v1 (d) v1 lags v2
(e) v1 and v2 are in phase
9.5 The voltage across an inductor leads the current
through it by 90◦
.
(a) True (b) False
9.6 The imaginary part of impedance is called:
(a) resistance (b) admittance
(c) susceptance (d) conductance
(e) reactance
9.7 The impedance of a capacitor increases with
increasing frequency.
(a) True (b) False
9.8 At what frequency will the output voltage vo(t) in
Fig. 9.39 be equal to the input voltage v(t)?
(a) 0 rad/s (b) 1 rad/s (c) 4 rad/s
(d) ∞ rad/s (e) none of the above
+
−
+
−
1 Ω
H
v(t) vo(t)
1
4
Figure 9.39 For Review Question 9.8.
9.9 A series RC circuit has VR = 12 V and VC = 5 V.
The supply voltage is:
(a) −7 V (b) 7 V (c) 13 V (d) 17 V
9.10 A series RCL circuit has R = 30 , XC = −50 ,
and XL = 90 . The impedance of the circuit is:
(a) 30 + j140  (b) 30 + j40 
(c) 30 − j40  (d) −30 − j40 
(e) −30 + j40 
Answers: 9.1d, 9.2c, 9.3b, 9.4b,d, 9.5a, 9.6e, 9.7b, 9.8d, 9.9c, 9.10b.
PROBLEMS
Section 9.2 Sinusoids
9.1 In a linear circuit, the voltage source is
vs = 12 sin(103
t + 24◦
) V
(a) What is the angular frequency of the voltage?
(b) What is the frequency of the source?
(c) Find the period of the voltage.
(d) Express vs in cosine form.
(e) Determine vs at t = 2.5 ms.
9.2 A current source in a linear circuit has
is = 8 cos(500πt − 25◦
) A
(a) What is the amplitude of the current?
(b) What is the angular frequency?
386 PART 2 AC Circuits
(c) Find the frequency of the current.
(d) Calculate is at t = 2 ms.
9.3 Express the following functions in cosine form:
(a) 4 sin(ωt − 30◦
) (b) −2 sin 6t
(c) −10 sin(ωt + 20◦
)
9.4 (a) Express v = 8 cos(7t + 15◦
) in sine form.
(b) Convert i = −10 sin(3t − 85◦
) to cosine form.
9.5 Given v1 = 20 sin(ωt + 60◦
) and v2 =
60 cos(ωt − 10◦
), determine the phase angle
between the two sinusoids and which one lags the
other.
9.6 For the following pairs of sinusoids, determine
which one leads and by how much.
(a) v(t) = 10 cos(4t − 60◦
) and
i(t) = 4 sin(4t + 50◦
)
(b) v1(t) = 4 cos(377t + 10◦
) and
v2(t) = −20 cos 377t
(c) x(t) = 13 cos 2t + 5 sin 2t and
y(t) = 15 cos(2t − 11.8◦
)
Section 9.3 Phasors
9.7 If f (φ) = cos φ + j sin φ, show that f (φ) = ejφ
.
9.8 Calculate these complex numbers and express your
results in rectangular form:
(a)
15 45◦
3 − j4
+ j2
(b)
8 − 20◦
(2 + j)(3 − j4)
+
10
−5 + j12
(c) 10 + (8 50◦
)(5 − j12)
9.9 Evaluate the following complex numbers and
express your results in rectangular form:
(a) 2 +
3 + j4
5 − j8
(b) 4 − 10◦
+
1 − j2
3 6◦
(c)
8 10◦
+ 6 − 20◦
9 80◦ − 4 50◦
9.10 Given the complex numbers z1 = −3 + j4 and
z2 = 12 + j5, find:
(a) z1z2 (b)
z1
z∗
2
(c)
z1 + z2
z1 − z2
9.11 Let X = 8 40◦
and Y = 10 − 30◦
. Evaluate the
following quantities and express your results in
polar form.
(a) (X + Y)X∗
(b) (X − Y)∗
(c) (X + Y)/X
9.12 Evaluate these determinants:
(a)
10 + j6
−5
2 − j3
−1 + j
(b)
20 − 30◦
16 0◦
−4 − 10◦
3 45◦
(c)
1 − j
j
1
−j
1
j
0
−j
1 + j
9.13 Transform the following sinusoids to phasors:
(a) −10 cos(4t + 75◦
) (b) 5 sin(20t − 10◦
)
(c) 4 cos 2t + 3 sin 2t
9.14 Express the sum of the following sinusoidal signals
in the form of A cos(ωt + θ) with A  0 and
0  θ  360◦
.
(a) 8 cos(5t − 30◦
) + 6 cos 5t
(b) 20 cos(120πt + 45◦
) − 30 sin(120πt + 20◦
)
(c) 4 sin 8t + 3 sin(8t − 10◦
)
9.15 Obtain the sinusoids corresponding to each of the
following phasors:
(a) V1 = 60 15◦
, ω = 1
(b) V2 = 6 + j8, ω = 40
(c) I1 = 2.8e−jπ/3
, ω = 377
(d) I2 = −0.5 − j1.2, ω = 103
9.16 Using phasors, find:
(a) 3 cos(20t + 10◦
) − 5 cos(20t − 30◦
)
(b) 40 sin 50t + 30 cos(50t − 45◦
)
(c) 20 sin 400t + 10 cos(400t + 60◦
)
− 5 sin(400t − 20◦
)
9.17 Find a single sinusoid corresponding to each of
these phasors:
(a) V = 40 − 60◦
(b) V = −30 10◦
+ 50 60◦
(c) I = j6e−j10◦
(d) I =
2
j
+ 10 − 45◦
9.18 Find v(t) in the following integrodifferential
equations using the phasor approach:
(a) v(t) +

v dt = 10 cos t
(b)
dv
dt
+ 5v(t) + 4

v dt = 20 sin(4t + 10◦
)
9.19 Using phasors, determine i(t) in the following
equations:
(a) 2
di
dt
+ 3i(t) = 4 cos(2t − 45◦
)
(b) 10

i dt +
di
dt
+ 6i(t) = 5 cos(5t + 22◦
)
9.20 The loop equation for a series RLC circuit gives
di
dt
+ 2i +
 t
−∞
i dt = cos 2t
Assuming that the value of the integral at t = −∞ is
zero, find i(t) using the phasor method.
CHAPTER 9 Sinusoids and Phasors 387
9.21 A parallel RLC circuit has the node equation
dv
dt
+ 50v + 100

v dt = 110 cos(377t − 10◦
)
Determine v(t) using the phasor method. You may
assume that the value of the integral at t = −∞ is
zero.
Section 9.4 Phasor Relationships for Circuit
Elements
9.22 Determine the current that flows through an 8-
resistor connected to a voltage source
vs = 110 cos 377t V.
9.23 What is the instantaneous voltage across a 2-µF
capacitor when the current through it is
i = 4 sin(106
t + 25◦
) A?
9.24 The voltage across a 4-mH inductor is
v = 60 cos(500t − 65◦
) V. Find the instantaneous
current through it.
9.25 A current source of i(t) = 10 sin(377t + 30◦
) A is
applied to a single-element load. The resulting
voltage across the element is v(t) =
−65 cos(377t + 120◦
) V. What type of element is
this? Calculate its value.
9.26 Two elements are connected in series as shown in
Fig. 9.40. If i = 12 cos(2t − 30◦
) A, find the
element values.
+
−
i
180 cos(2t + 10°) V
Figure 9.40 For Prob. 9.26.
9.27 A series RL circuit is connected to a 110-V ac
source. If the voltage across the resistor is 85 V, find
the voltage across the inductor.
9.28 What value of ω will cause the forced response vo in
Fig. 9.41 to be zero?
+
−
2 Ω
+
−
5 mF
vo
50 cos vt V
20 mH
Figure 9.41 For Prob. 9.28.
Section 9.5 Impedance and Admittance
9.29 If vs = 5 cos 2t V in the circuit of Fig. 9.42,
find vo.
+
−
+
−
2 Ω
vo
0.25 F
vs 1 H
Figure 9.42 For Prob. 9.29.
9.30 Find ix when is = 2 sin 5t A is supplied to the
circuit in Fig. 9.43.
2 Ω 1 H
is
ix
0.2 F
Figure 9.43 For Prob. 9.30.
9.31 Find i(t) and v(t) in each of the circuits of Fig. 9.44.
+
−
v
i
4 Ω F
10 cos(3t + 45°) A
(a)
i
4 Ω
8 Ω
F
50 cos 4t V +
− +
−
3 H
(b)
v
1
12
1
6
Figure 9.44 For Prob. 9.31.
9.32 Calculate i1(t) and i2(t) in the circuit of Fig. 9.45 if
the source frequency is 60 Hz.
+
−
8 Ω
40 0° V j5 Ω −j10 Ω
i1 i2
Figure 9.45 For Prob. 9.32.
388 PART 2 AC Circuits
9.33 In the circuit of Fig. 9.46, find io when:
(a) ω = 1 rad/s (b) ω = 5 rad/s
(c) ω = 10 rad/s
+
− 2 Ω
4 cos vt V 0.05 F
io 1 H
Figure 9.46 For Prob. 9.33.
9.34 Find v(t) in the RLC circuit of Fig. 9.47.
+
−
+
−
1 Ω
1 Ω
1 H
1 F v
10 cos t V
Figure 9.47 For Prob. 9.34.
9.35 Calculate vo(t) in the circuit in Fig. 9.48.
+
−
+
−
30 Ω
vo(t)
50 Ω
0.1 H
60 sin 200t V
50 mF
Figure 9.48 For Prob. 9.35.
9.36 Determine io(t) in the RLC circuit of Fig. 9.49.
io
1 Ω
4 cos 2t A
1 F
1 H
Figure 9.49 For Prob. 9.36.
9.37 Calculate i(t) in the circuit of Fig. 9.50.
+
− 3 Ω
10 mH
5 mF
6 cos 200t V 4 Ω
5 Ω
i
Figure 9.50 For Prob. 9.37.
9.38 Find current Io in the network of Fig. 9.51.
2 Ω
2 Ω
Io
−j2 Ω
j4 Ω
−j2 Ω
5 0° A
Figure 9.51 For Prob. 9.38.
9.39 If is = 5 cos(10t + 40◦
) A in the circuit in Fig. 9.52,
find io.
0.2 H 0.1 F
4 Ω 3 Ω
io
is
Figure 9.52 For Prob. 9.39.
9.40 Find vs(t) in the circuit of Fig. 9.53 if the current ix
through the 1- resistor is 0.5 sin 200t A.
+
−
1 Ω
2 Ω
vs j2 Ω −j1 Ω
ix
Figure 9.53 For Prob. 9.40.
9.41 If the voltage vo across the 2- resistor in the circuit
of Fig. 9.54 is 10 cos 2t V, obtain is.
+
−
vo
0.1 F
1 Ω 2 Ω
is
0.5 H
Figure 9.54 For Prob. 9.41.
CHAPTER 9 Sinusoids and Phasors 389
9.42 If Vo = 8 30◦
V in the circuit of Fig. 9.55,
find Is.
+
−
5 Ω
10 Ω V
o
−j5 Ω
j5 Ω
Is
Figure 9.55 For Prob. 9.42.
9.43 In the circuit of Fig. 9.56, find Vs if Io = 2 0◦
A.
+ −
Io
1 Ω
2 Ω
V
s
j2 Ω
j4 Ω
−j2 Ω −j1 Ω
Figure 9.56 For Prob. 9.43.
9.44 Find Z in the network of Fig. 9.57, given that
Vo = 4 0◦
V.
+
−
+
−
Z
12 Ω
V
o
20 −90° V j8 Ω
−j4 Ω
Figure 9.57 For Prob. 9.44.
Section 9.7 Impedance Combinations
9.45 At ω = 50 rad/s, determine Zin for each of the
circuits in Fig. 9.58.
10 Ω
20 Ω
0.4 H
0.2 H
Zin
1 mF
(b)
1 Ω 1 Ω
10 mH 10 mF
Zin
(a)
Figure 9.58 For Prob. 9.45.
9.46 Calculate Zeq for the circuit in Fig. 9.59.
1 Ω
2 Ω
6 Ω
Zeq
j4 Ω
−j2 Ω
Figure 9.59 For Prob. 9.46.
9.47 Find Zeq in the circuit of Fig. 9.60.
Zeq 1 − j Ω
1 + j2 Ω
j5 Ω
1 + j3 Ω
Figure 9.60 For Prob. 9.47.
9.48 For the circuit in Fig. 9.61, find the input impedance
Zin at 10 krad/s.
+
−
+ −
v
2v
50 Ω 2 mH
Zin
1 mF
Figure 9.61 For Prob. 9.48.
9.49 Determine I and ZT for the circuit in Fig. 9.62.
+
−
2 Ω
3 Ω
4 Ω
ZT
120 10° V
j4 Ω
−j6 Ω
I
Figure 9.62 For Prob. 9.49.
390 PART 2 AC Circuits
9.50 For the circuit in Fig. 9.63, calculate ZT and Vab.
+
−
20 Ω
+ −
ZT
V
ab
60 90° V
j10 Ω
−j5 Ω 40 Ω
a b
Figure 9.63 For Prob. 9.50.
9.51 At ω = 103
rad/s, find the input admittance of each
of the circuits in Fig. 9.64.
Yin
(a)
20 mH 12.5 mF
60 Ω 60 Ω
Yin
(b)
30 Ω 10 mH
20 mF
60 Ω
40 Ω
Figure 9.64 For Prob. 9.51.
9.52 Determine Yeq for the circuit in Fig. 9.65.
Yeq
3 Ω
5 Ω
j1 Ω
−j2 Ω
−j4 Ω
Figure 9.65 For Prob. 9.52.
9.53 Find the equivalent admittance Yeq of the circuit in
Fig. 9.66.
2 S
4 S
1 S
j5 S j1 S
−j3 S −j2 S
Figure 9.66 For Prob. 9.53.
9.54 Find the equivalent impedance of the circuit in Fig.
9.67.
10 Ω
Zeq
j15 Ω
−j5 Ω
−j10 Ω
2 Ω
5 Ω
8 Ω
Figure 9.67 For Prob. 9.54.
9.55 Obtain the equivalent impedance of the circuit in
Fig. 9.68.
Zeq
1 Ω j2 Ω
j4 Ω
−j2 Ω
−j Ω 2 Ω
Figure 9.68 For Prob. 9.55.
9.56 Calculate the value of Zab in the network of Fig.
9.69.
20 Ω
20 Ω
j6 Ω −j9 Ω
10 Ω
−j9 Ω
−j9 Ω
j6 Ω
j6 Ω
a
b
Figure 9.69 For Prob. 9.56.
9.57 Determine the equivalent impedance of the circuit in
Fig. 9.70.
2 Ω 4 Ω
j6 Ω j8 Ω j8 Ω j12 Ω
−j4 Ω
−j6 Ω
a
b
Figure 9.70 For Prob. 9.57.
CHAPTER 9 Sinusoids and Phasors 391
Section 9.8 Applications
9.58 Design an RL circuit to provide a 90◦
leading phase
shift.
9.59 Design a circuit that will transform a sinusoidal
input to a cosinusoidal output.
9.60 Refer to the RC circuit in Fig. 9.71.
(a) Calculate the phase shift at 2 MHz.
(b) Find the frequency where the phase shift is 45◦
.
+
−
+
−
5 Ω
20 nF V
o
V
i
Figure 9.71 For Prob. 9.60.
9.61 (a) Calculate the phase shift of the circuit in Fig.
9.72.
(b) State whether the phase shift is leading or
lagging (output with respect to input).
(c) Determine the magnitude of the output when the
input is 120 V.
+
−
+
−
20 Ω 40 Ω 30 Ω
V
o
j10 Ω j30 Ω j60 Ω
V
i
Figure 9.72 For Prob. 9.61.
9.62 Consider the phase-shifting circuit in Fig. 9.73. Let
Vi = 120 V operating at 60 Hz. Find:
(a) Vo when R is maximum
(b) Vo when R is minimum
(c) the value of R that will produce a phase shift
of 45◦
+
−
+
−
50 Ω
200 mH vo
vi
0  R  100 Ω
Figure 9.73 For Prob. 9.62.
9.63 The ac bridge in Fig. 9.37 is balanced when
R1 = 400 , R2 = 600 , R3 = 1.2 k, and
C2 = 0.3 µF. Find Rx and Cx.
9.64 A capacitance bridge balances when R1 = 100 ,
R2 = 2 k, and Cs = 40 µF. What is Cx, the
capacitance of the capacitor under test?
9.65 An inductive bridge balances when R1 = 1.2 k,
R2 = 500 , and Ls = 250 mH. What is the value
of Lx, the inductance of the inductor under test?
9.66 The ac bridge shown in Fig. 9.74 is known as a
Maxwell bridge and is used for accurate
measurement of inductance and resistance of a coil
in terms of a standard capacitance Cs. Show that
when the bridge is balanced,
Lx = R2R3Cs and Rx =
R2
R1
R3
Find Lx and Rx for R1 = 40 k, R2 = 1.6 k,
R3 = 4 k, and Cs = 0.45 µF.
AC
meter
R3
Lx
Rx
R2
R1
Cs
Figure 9.74 Maxwell bridge; for Prob. 9.66.
9.67 The ac bridge circuit of Fig. 9.75 is called a Wien
bridge. It is used for measuring the frequency of a
source. Show that when the bridge is balanced,
f =
1
2π
√
R2R4C2C4
AC
meter
R3
R2
R1
C4
C2
R4
Figure 9.75 Wein bridge; for Prob. 9.67.
392 PART 2 AC Circuits
COMPREHENSIVE PROBLEMS
9.68 The circuit shown in Fig. 9.76 is used in a television
receiver. What is the total impedance of this circuit?
240 Ω j95 Ω −j84 Ω
Figure 9.76 For Prob. 9.68.
9.69 The network in Fig. 9.77 is part of the schematic
describing an industrial electronic sensing device.
What is the total impedance of the circuit at 2 kHz?
50 Ω 10 mH
2 mF 80 Ω
100 Ω
Figure 9.77 For Prob. 9.69.
9.70 A series audio circuit is shown in Fig. 9.78.
(a) What is the impedance of the circuit?
(b) If the frequency were halved, what would be the
impedance of the circuit?
250 Hz ≈
j30 Ω 120 Ω
−j20 Ω
−j20 Ω
Figure 9.78 For Prob. 9.70.
9.71 An industrial load is modeled as a series
combination of a capacitance and a resistance as
shown in Fig. 9.79. Calculate the value of an
inductance L across the series combination so that
the net impedance is resistive at a frequency of
5 MHz.
200 Ω
50 nF
L
Figure 9.79 For Prob. 9.71.
9.72 An industrial coil is modeled as a series
combination of an inductance L and resistance R, as
shown in Fig. 9.80. Since an ac voltmeter measures
only the magnitude of a sinusoid, the following
measurements are taken at 60 Hz when the circuit
operates in the steady state:
|Vs| = 145 V, |V1| = 50 V, |Vo| = 110 V
Use these measurements to determine the values of
L and R.
80 Ω
+
−
+ −
V
1
V
s
+
− V
o
R
L
Coil
Figure 9.80 For Prob. 9.72.
9.73 Figure 9.81 shows a parallel combination of an
inductance and a resistance. If it is desired to
connect a capacitor in series with the parallel
combination such that the net impedance is resistive
at 10 MHz, what is the required value of C?
300 Ω 20 mH
C
Figure 9.81 For Prob. 9.73.
9.74 A power transmission system is modeled as shown
in Fig. 9.82. Given the source voltage
Vs = 115 0◦
V, source impedance
Zs = 1 + j0.5 , line impedance
Z. = 0.4 + j0.3 , and load impedance
ZL = 23.2 + j18.9 , find the load current IL.
+
−
vs
Zᐉ
Zs
Zᐉ
ZL
IL
Source Transmission line Load
Figure 9.82 For Prob. 9.74.
393
C H A P T E R
SINUSOIDAL STEADY-STATE ANALYSIS
1 0
An expert problem solver must be endowed with two incompatible quan-
tities, a restless imagination and a patient pertinacity.
—Howard W. Eves
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394 PART 2 AC Circuits
10.1 INTRODUCTION
In Chapter 9, we learned that the forced or steady-state response of cir-
cuits to sinusoidal inputs can be obtained by using phasors. We also know
that Ohm’s and Kirchhoff’s laws are applicable to ac circuits. In this
chapter, we want to see how nodal analysis, mesh analysis, Thevenin’s
theorem, Norton’s theorem, superposition, and source transformations
are applied in analyzing ac circuits. Since these techniques were already
introduced for dc circuits, our major effort here will be to illustrate with
examples.
Analyzing ac circuits usually requires three steps.
Steps to Analyze ac Circuits:
1. Transform the circuit to the phasor or frequency domain.
2. Solve the problem using circuit techniques (nodal analysis, mesh
analysis, superposition, etc.).
3. Transform the resulting phasor to the time domain.
Step 1 is not necessary if the problem is specified in the frequency domain.
In step 2, the analysis is performed in the same manner as dc circuit
analysis except that complex numbers are involved. Having read Chapter
9, we are adept at handling step 3.
Frequency-domain analysis of an ac circuit via
phasors is much easier than analysis of the cir-
cuit in the time domain.
Toward the end of the chapter, we learn how to apply PSpice in
solving ac circuit problems. We finally apply ac circuit analysis to two
practical ac circuits: oscillators and ac transistor circuits.
10.2 NODAL ANALYSIS
The basis of nodal analysis is Kirchhoff’s current law. Since KCL is valid
for phasors, as demonstrated in Section 9.6, we can analyze ac circuits
by nodal analysis. The following examples illustrate this.
E X A M P L E 1 0 . 1
Find ix in the circuit of Fig. 10.1 using nodal analysis.
0.5 H
0.1 F
1 H
10 Ω
2ix
ix
+
−
20 cos 4t V
Figure10.1 For Example 10.1.
Solution:
We first convert the circuit to the frequency domain:
CHAPTER 10 Sinusoidal Steady-State Analysis 395
20 cos 4t ⇒ 20 0◦
, ω = 4 rad/s
1 H ⇒ jωL = j4
0.5 H ⇒ jωL = j2
0.1 F ⇒
1
jωC
= −j2.5
Thus, the frequency-domain equivalent circuit is as shown in Fig. 10.2.
–j2.5 Ω j2 Ω
j4 Ω
10 Ω
2Ix
Ix
+
−
V
1 V2
20 0° V
Figure10.2 Frequency-domain equivalent of the circuit in Fig. 10.1.
Applying KCL at node 1,
20 − V1
10
=
V1
−j2.5
+
V1 − V2
j4
or
(1 + j1.5)V1 + j2.5V2 = 20 (10.1.1)
At node 2,
2Ix +
V1 − V2
j4
=
V2
j2
But Ix = V1/−j2.5. Substituting this gives
2V1
−j2.5
+
V1 − V2
j4
=
V2
j2
By simplifying, we get
11V1 + 15V2 = 0 (10.1.2)
Equations (10.1.1) and (10.1.2) can be put in matrix form as

1 + j1.5 j2.5
11 15
 
V1
V2

=

20
0

We obtain the determinants as
=




1 + j1.5 j2.5
11 15



 = 15 − j5
1 =




20 j2.5
0 15



 = 300, 2 =




1 + j1.5 20
11 0



 = −220
V1 =
1
=
300
15 − j5
= 18.97 18.43◦
V
V2 =
2
=
−220
15 − j5
= 13.91 198.3◦
V
396 PART 2 AC Circuits
The current Ix is given by
Ix =
V1
−j2.5
=
18.97 18.43◦
2.5 − 90◦
= 7.59 108.4◦
A
Transforming this to the time domain,
ix = 7.59 cos(4t + 108.4◦
) A
P R A C T I C E P R O B L E M 1 0 . 1
Using nodal analysis, find v1 and v2 in the circuit of Fig. 10.3.
4 Ω
2 Ω 3vx
vx 2 H
0.2 F
v1 v2
+
−
+
−
10 sin 2t A
Figure10.3 For Practice Prob. 10.1.
Answer: v1(t) = 20.96 sin(2t + 58◦
) V,
v2(t) = 44.11 sin(2t + 41◦
) V.
E X A M P L E 1 0 . 2
Compute V1 and V2 in the circuit of Fig. 10.4.
4 Ω
12 Ω
1 2
V1 V2
–j3 Ω j6 Ω
+ −
10 45° V
3 0° A
Figure10.4 For Example 10.2.
Solution:
Nodes 1 and 2 form a supernode as shown in Fig. 10.5. Applying KCL
at the supernode gives
3 =
V1
−j3
+
V2
j6
+
V2
12
or
36 = j4V1 + (1 − j2)V2 (10.2.1)
But a voltage source is connected between nodes 1 and 2, so that
CHAPTER 10 Sinusoidal Steady-State Analysis 397
–j3 Ω j6 Ω 12 Ω
3 A
Supernode
V1 V2
Figure10.5 A supernode in the circuit of Fig. 10.4.
V1 = V2 + 10 45◦
(10.2.2)
Substituting Eq. (10.2.2) in Eq. (10.2.1) results in
36 − 40 135◦
= (1 + j2)V2 ⇒ V2 = 31.41 − 87.18◦
V
From Eq. (10.2.2),
V1 = V2 + 10 45◦
= 25.78 − 70.48◦
V
P R A C T I C E P R O B L E M 1 0 . 2
Calculate V1 and V2 in the circuit shown in Fig. 10.6.
4 Ω
2 Ω
j4 Ω –j1 Ω
+ −
+
−
V1 V2
15 0° V
20 60° V
Figure10.6 For Practice Prob. 10.2.
Answer: V1 = 19.36 69.67◦
V, V2 = 3.376 165.7◦
V.
10.3 MESH ANALYSIS
Kirchhoff’s voltage law (KVL) forms the basis of mesh analysis. The
validity of KVL for ac circuits was shown in Section 9.6 and is illustrated
in the following examples.
E X A M P L E 1 0 . 3
Determine current Io in the circuit of Fig. 10.7 using mesh analysis.
Solution:
Applying KVL to mesh 1, we obtain
(8 + j10 − j2)I1 − (−j2)I2 − j10I3 = 0 (10.3.1)
398 PART 2 AC Circuits
4 Ω
8 Ω –j2 Ω
–j2 Ω
j10 Ω
+
−
Io
I2
I3
I1
5 0° A
20 90° V
Figure10.7 For Example 10.3.
For mesh 2,
(4 − j2 − j2)I2 − (−j2)I1 − (−j2)I3 + 20 90◦
= 0 (10.3.2)
For mesh 3, I3 = 5. Substituting this in Eqs. (10.3.1) and (10.3.2), we
get
(8 + j8)I1 + j2I2 = j50 (10.3.3)
j2I1 + (4 − j4)I2 = −j20 − j10 (10.3.4)
Equations (10.3.3) and (10.3.4) can be put in matrix form as

8 + j8 j2
j2 4 − j4
 
I1
I2

=

j50
−j30

from which we obtain the determinants
=




8 + j8 j2
j2 4 − j4



 = 32(1 + j)(1 − j) + 4 = 68
2 =




8 + j8 j50
j2 −j30



 = 340 − j240 = 416.17 − 35.22◦
I2 =
2
=
416.17 − 35.22◦
68
= 6.12 − 35.22◦
A
The desired current is
Io = −I2 = 6.12 144.78◦
A
P R A C T I C E P R O B L E M 1 0 . 3
Find Io in Fig. 10.8 using mesh analysis.
8 Ω
j4 Ω
–j2 Ω 6 Ω
+
−
Io
2 0° A
10 30° V
Figure10.8 For Practice Prob. 10.3.
Answer: 1.194 65.45◦
A.
CHAPTER 10 Sinusoidal Steady-State Analysis 399
E X A M P L E 1 0 . 4
Solve for Vo in the circuit in Fig. 10.9 using mesh analysis.
8 Ω
6 Ω
–j2 Ω
–j4 Ω
j5 Ω
4 0˚ A
+
− Vo
+
−
3 0° A
10 0° V
Figure10.9 For Example 10.4.
Solution:
As shown in Fig. 10.10, meshes 3 and 4 form a supermesh due to the
current source between the meshes. For mesh 1, KVL gives
−10 + (8 − j2)I1 − (−j2)I2 − 8I3 = 0
or
(8 − j2)I1 + j2I2 − 8I3 = 10 (10.4.1)
For mesh 2,
I2 = −3 (10.4.2)
For the supermesh,
(8 − j4)I3 − 8I1 + (6 + j5)I4 − j5I2 = 0 (10.4.3)
Due to the current source between meshes 3 and 4, at node A,
I4 = I3 + 4 (10.4.4)
Combining Eqs. (10.4.1) and (10.4.2),
(8 − j2)I1 − 8I3 = 10 + j6 (10.4.5)
Combining Eqs. (10.4.2) to (10.4.4),
−8I1 + (14 + j)I3 = −24 − j35 (10.4.6)
8 Ω
6 Ω
–j2 Ω
–j4 Ω
j5 Ω
10 V 3 A
4 A
A
+
−
+
−
I2
I3
I3 I4
I4
I1
Supermesh
Vo
Figure10.10 Analysis of the circuit in Fig. 10.9.
400 PART 2 AC Circuits
From Eqs. (10.4.5) and (10.4.6), we obtain the matrix equation

8 − j2 −8
−8 14 + j
 
I1
I3

=

10 + j6
−24 − j35

We obtain the following determinants
=




8 − j2 −8
−8 14 + j



 = 112 + j8 − j28 + 2 − 64 = 50 − j20
1 =




10 + j6 −8
−24 − j35 14 + j



 = 140 + j10 + j84 − 6 − 192 − j280
= −58 − j186
Current I1 is obtained as
I1 =
1
=
−58 − j186
50 − j20
= 3.618 274.5◦
A
The required voltage Vo is
Vo = −j2(I1 − I2) = −j2(3.618 274.5◦
+ 3)
= −7.2134 − j6.568 = 9.756 222.32◦
V
P R A C T I C E P R O B L E M 1 0 . 4
Calculate current Io in the circuit of Fig. 10.11.
j8 Ω
–j6 Ω
–j4 Ω
5 Ω
10 Ω
Io
+
−
50 0° V
2 0° A
Figure10.11 For Practice Prob. 10.4.
Answer: 5.075 5.943◦
A.
10.4 SUPERPOSITION THEOREM
Since ac circuits are linear, the superposition theorem applies to ac circuits
the same way it applies to dc circuits. The theorem becomes important
if the circuit has sources operating at different frequencies. In this case,
since the impedances depend on frequency, we must have a different
frequency-domain circuit for each frequency. The total response must
be obtained by adding the individual responses in the time domain. It is
incorrect to try to add the responses in the phasor or frequency domain.
Why? Because the exponential factor ejωt
is implicit in sinusoidal analy-
sis, and that factor would change for every angular frequency ω. It would
therefore not make sense to add responses at different frequencies in the
phasor domain. Thus, when a circuit has sources operating at different
CHAPTER 10 Sinusoidal Steady-State Analysis 401
frequencies, one must add the responses due to the individual frequencies
in the time domain.
E X A M P L E 1 0 . 5
Use the superposition theorem to find Io in the circuit in Fig. 10.7.
Solution:
Let
Io = I
o + I
o (10.5.1)
where I
o and I
o are due to the voltage and current sources, respectively.
To find I
o, consider the circuit in Fig. 10.12(a). If we let Z be the parallel
combination of −j2 and 8 + j10, then
Z =
−j2(8 + j10)
−2j + 8 + j10
= 0.25 − j2.25
and current I
o is
I
o =
j20
4 − j2 + Z
=
j20
4.25 − j4.25
or
I
o = −2.353 + j2.353 (10.5.2)
4 Ω
8 Ω –j2 Ω
–j2 Ω
j10 Ω
j20 V
+
−
I'o
(a)
(b)
4 Ω
8 Ω –j2 Ω
–j2 Ω
j10 Ω
5 A
I''
o
I2
I3
I1
Figure10.12 Solution of Example 10.5.
To get I
o, consider the circuit in Fig. 10.12(b). For mesh 1,
(8 + j8)I1 − j10I3 + j2I2 = 0 (10.5.3)
For mesh 2,
(4 − j4)I2 + j2I1 + j2I3 = 0 (10.5.4)
For mesh 3,
I3 = 5 (10.5.5)
From Eqs. (10.5.4) and (10.5.5),
(4 − j4)I2 + j2I1 + j10 = 0
Expressing I1 in terms of I2 gives
I1 = (2 + j2)I2 − 5 (10.5.6)
Substituting Eqs. (10.5.5) and (10.5.6) into Eq. (10.5.3), we get
(8 + j8)[(2 + j2)I2 − 5] − j50 + j2I2 = 0
or
I2 =
90 − j40
34
= 2.647 − j1.176
Current I
o is obtained as
I
o = −I2 = −2.647 + j1.176 (10.5.7)
From Eqs. (10.5.2) and (10.5.7), we write
Io = I
o + I
o = −5 + j3.529 = 6.12 144.78◦
A
402 PART 2 AC Circuits
which agrees with what we got in Example 10.3. It should be noted
that applying the superposition theorem is not the best way to solve this
problem. It seems that we have made the problem twice as hard as
the original one by using superposition. However, in Example 10.6,
superposition is clearly the easiest approach.
P R A C T I C E P R O B L E M 1 0 . 5
Find current Io in the circuit of Fig. 10.8 using the superposition theorem.
Answer: 1.194 65.45◦
A.
E X A M P L E 1 0 . 6
Find vo in the circuit in Fig. 10.13 using the superposition theorem.
2 H 1 Ω 4 Ω
0.1 F 5 V
+
−
+
−
10 cos 2t V 2 sin 5t A
−
+ vo
Figure10.13 For Example 10.6.
Solution:
Since the circuit operates at three different frequencies (ω = 0 for the
dc voltage source), one way to obtain a solution is to use superposition,
which breaks the problem into single-frequency problems. So we let
vo = v1 + v2 + v3 (10.6.1)
where v1 is due to the 5-V dc voltage source, v2 is due to the 10 cos 2t V
voltage source, and v3 is due to the 2 sin 5t A current source.
To find v1, we set to zero all sources except the 5-V dc source. We
recall that at steady state, a capacitor is an open circuit to dc while an
inductor is a short circuit to dc. There is an alternative way of looking at
this. Since ω = 0, jωL = 0, 1/jωC = ∞. Either way, the equivalent
circuit is as shown in Fig. 10.14(a). By voltage division,
−v1 =
1
1 + 4
(5) = 1 V (10.6.2)
To find v2, we set to zero both the 5-V source and the 2 sin 5t current
source and transform the circuit to the frequency domain.
10 cos 2t ⇒ 10 0◦
, ω = 2 rad/s
2 H ⇒ jωL = j4 
0.1 F ⇒
1
jωC
= −j5
CHAPTER 10 Sinusoidal Steady-State Analysis 403
The equivalent circuit is now as shown in Fig. 10.14(b). Let
Z = −j5 4 =
−j5 × 4
4 − j5
= 2.439 − j1.951
By voltage division,
V2 =
1
1 + j4 + Z
(10 0◦
) =
10
3.439 + j2.049
= 2.498 − 30.79◦
In the time domain,
v2 = 2.498 cos(2t − 30.79◦
) (10.6.3)
1 Ω 4 Ω
5 V
+
−
−
+ v1
(a) (b) (c)
1 Ω
j4 Ω
–j5 Ω
4 Ω
+
−
1 Ω
4 Ω
–j2 Ω
j10 Ω
I1
10 0° V 2 –90° A
+ −
V2
+ −
V3
Figure 10.14 Solution of Example 10.6: (a) setting all sources to zero except the 5-V dc source, (b) setting all sources to zero except the ac
voltage source, (c) setting all sources to zero except the ac current source.
To obtain v3, we set the voltage sources to zero and transform what
is left to the frequency domain.
2 sin 5t ⇒ 2 − 90◦
, ω = 5 rad/s
2 H ⇒ jωL = j10 
0.1 F ⇒
1
jωC
= −j2 
The equivalent circuit is in Fig. 10.14(c). Let
Z1 = −j2 4 =
−j2 × 4
4 − j2
= 0.8 − j1.6 
By current division,
I1 =
j10
j10 + 1 + Z1
(2 − 90◦
) A
V3 = I1 × 1 =
j10
1.8 + j8.4
(−j2) = 2.328 − 77.91◦
V
In the time domain,
v3 = 2.33 cos(5t − 80◦
) = 2.33 sin(5t + 10◦
) V (10.6.4)
Substituting Eqs. (10.6.2) to (10.6.4) into Eq. (10.6.1), we have
vo(t) = −1 + 2.498 cos(2t − 30.79◦
) + 2.33 sin(5t + 10◦
) V
P R A C T I C E P R O B L E M 1 0 . 6
Calculate vo in the circuit of Fig. 10.15 using the superposition theorem.
404 PART 2 AC Circuits
8 Ω
0.2 F 1 H
+
−
30 sin 5t V 2 cos 10t A
+
−
vo
Figure10.15 For Practice Prob. 10.6.
Answer: 4.631 sin(5t − 81.12◦
) + 1.051 cos(10t − 86.24◦
) V.
10.5 SOURCE TRANSFORMATION
As Fig. 10.16 shows, source transformation in the frequency domain
involves transforming a voltage source in series with an impedance to a
current source in parallel with an impedance, or vice versa. As we go
from one source type to another, we must keep the following relationship
in mind:
Vs = ZsIs ⇐⇒ Is =
Vs
Zs
(10.1)
a
b
Vs
Vs = ZsIs
Zs
Zs
+
−
a
b
Is
Is =
Zs
Vs
Figure10.16 Source transformation.
E X A M P L E 1 0 . 7
Calculate Vx in the circuit of Fig. 10.17 using the method of source trans-
formation.
5 Ω
j4 Ω
–j13 Ω
3 Ω
10 Ω
4 Ω
+
−
+
−
Vx
20 –90° V
Figure10.17 For Example 10.7.
CHAPTER 10 Sinusoidal Steady-State Analysis 405
Solution:
We transform the voltage source to a current source and obtain the circuit
in Fig. 10.18(a), where
Is =
20 − 90◦
5
= 4 − 90◦
= −j4 A
The parallel combination of 5- resistance and (3+j4) impedance gives
Z1 =
5(3 + j4)
8 + j4
= 2.5 + j1.25 
Converting the current source to a voltage source yields the circuit in Fig.
10.18(b), where
Vs = IsZ1 = −j4(2.5 + j1.25) = 5 − j10 V
By voltage division,
Vx =
10
10 + 2.5 + j1.25 + 4 − j13
(5 − j10) = 5.519 − 28◦
V
5 Ω
j4 Ω
–j13 Ω
3 Ω
10 Ω
4 Ω
+
−
+
−
Vx
Is = –j4 Α
–j13 Ω
10 Ω
4 Ω
2.5 Ω j1.25 Ω
Vx
Vs = 5 – j10 V +
−
(a) (b)
Figure10.18 Solution of the circuit in Fig. 10.17.
P R A C T I C E P R O B L E M 1 0 . 7
Find Io in the circuit of Fig. 10.19 using the concept of source transfor-
mation.
–j3 Ω
j5 Ω
j1 Ω
2 Ω
Io
–j2 Ω
4 90° Α
4 Ω
1 Ω
Figure10.19 For Practice Prob. 10.7.
Answer: 3.288 99.46◦
A.
406 PART 2 AC Circuits
10.6 THEVENIN AND NORTON EQUIVALENT CIRCUITS
Thevenin’s and Norton’s theorems are applied to ac circuits in the same
way as they are to dc circuits. The only additional effort arises from the
need to manipulate complex numbers. The frequency-domain version of
a Thevenin equivalent circuit is depicted in Fig. 10.20, where a linear
circuit is replaced by a voltage source in series with an impedance. The
Norton equivalent circuit is illustrated in Fig. 10.21, where a linear circuit
is replaced by a current source in parallel with an impedance. Keep in
mind that the two equivalent circuits are related as
VTh = ZN IN , ZTh = ZN (10.2)
just as in source transformation. VTh is the open-circuit voltage while IN
is the short-circuit current.
a
b
ZTh
a
b
VTh
Linear
circuit
+
−
Figure10.20 Thevenin equivalent.
a
b
ZN
a
b
IN
Linear
circuit
Figure10.21 Norton equivalent.
If the circuit has sources operating at different frequencies (see
Example 10.6, for example), the Thevenin or Norton equivalent circuit
must be determined at each frequency. This leads to entirely different
equivalent circuits, one for each frequency, not one equivalent circuit
with equivalent sources and equivalent impedances.
E X A M P L E 1 0 . 8
ObtaintheTheveninequivalentatterminalsa-b ofthecircuitinFig.10.22.
4 Ω
d
a b
f
c
e
–j6 Ω
j12 Ω
8 Ω
+
−
a b
120 75° V
Figure10.22 For Example 10.8.
Solution:
We find ZTh by setting the voltage source to zero. As shown in Fig.
10.23(a), the 8- resistance is now in parallel with the −j6 reactance, so
that their combination gives
Z1 = −j6 8 =
−j6 × 8
8 − j6
= 2.88 − j3.84 
Similarly, the 4- resistance is in parallel with the j12 reactance, and
their combination gives
Z2 = 4 j12 =
j12 × 4
4 + j12
= 3.6 + j1.2
CHAPTER 10 Sinusoidal Steady-State Analysis 407
4 Ω
8 Ω –j6 Ω j12 Ω
ZTh
VTh
a
e c
f,d f,d
b
(a)
(b)
8 Ω
4 Ω
j12 Ω
–j6 Ω
+
−
I2
I1
d
e
a b
c
f
−
+
120 75° V
Figure10.23 Solution of the circuit in Fig. 10.22: (a) finding ZTh, (b) finding VTh.
The Thevenin impedance is the series combination of Z1 and Z2; that is,
ZTh = Z1 + Z2 = 6.48 − j2.64 
To find VTh, consider the circuit in Fig. 10.23(b). Currents I1 and
I2 are obtained as
I1 =
120 75◦
8 − j6
A, I2 =
120 75◦
4 + j12
A
Applying KVL around loop bcdeab in Fig. 10.23(b) gives
VTh − 4I2 + (−j6)I1 = 0
or
VTh = 4I2 + j6I1 =
480 75◦
4 + j12
+
720 75◦
+ 90◦
8 − j6
= 37.95 3.43◦
+ 72 201.87◦
= −28.936 − j24.55 = 37.95 220.31◦
V
P R A C T I C E P R O B L E M 1 0 . 8
Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 10.24.
–j4 Ω
j2 Ω
6 Ω
10 Ω
+
−
a b
30 20° V
Figure10.24 For Practice Prob. 10.8.
Answer: ZTh = 12.4 − j3.2 , VTh = 18.97 − 51.57◦
V.
408 PART 2 AC Circuits
E X A M P L E 1 0 . 9
Find the Thevenin equivalent of the circuit in Fig. 10.25 as seen from ter-
minals a-b.
–j4 Ω
j3 Ω
4 Ω
2 Ω
a
b
Io
0.5 Io
15 0° A
Figure10.25 For Example 10.9.
Solution:
To find VTh, we apply KCL at node 1 in Fig. 10.26(a).
15 = Io + 0.5Io ⇒ Io = 10 A
Applying KVL to the loop on the right-hand side in Fig. 10.26(a), we
obtain
−Io(2 − j4) + 0.5Io(4 + j3) + VTh = 0
or
VTh = 10(2 − j4) − 5(4 + j3) = −j55
Thus, the Thevenin voltage is
VTh = 55 − 90◦
V
4 + j3 Ω
2 – j4 Ω
a
b
Io
0.5Io
0.5Io VTh
15 A
+
−
2
1
4 + j3 Ω
2 – j4 Ω
a
b
Vs Is
0.5Io
Io
Is = 3 0° A
(a) (b)
+
−
V
s
Figure10.26 Solution of the problem in Fig. 10.25: (a) finding VTh, (b) finding ZTh.
To obtain ZTh, we remove the independent source. Due to the
presence of the dependent current source, we connect a 3-A current source
(3 is an arbitrary value chosen for convenience here, a number divisible
by the sum of currents leaving the node) to terminals a-b as shown in Fig.
10.26(b). At the node, KCL gives
3 = Io + 0.5Io ⇒ Io = 2 A
Applying KVL to the outer loop in Fig. 10.26(b) gives
Vs = Io(4 + j3 + 2 − j4) = 2(6 − j)
CHAPTER 10 Sinusoidal Steady-State Analysis 409
The Thevenin impedance is
ZTh =
Vs
Is
=
2(6 − j)
3
= 4 − j0.6667 
P R A C T I C E P R O B L E M 1 0 . 9
DeterminetheTheveninequivalentofthecircuitinFig.10.27asseenfrom
the terminals a-b.
–j2 Ω
j4 Ω
8 Ω
4 Ω
a
b
0.2Vo
5 0° A
−
+ V
o
Figure10.27 For Practice Prob. 10.9.
Answer: ZTh = 12.166 136.3◦
, VTh = 7.35 72.9◦
V.
E X A M P L E 1 0 . 1 0
Obtain current Io in Fig. 10.28 using Norton’s theorem.
3 0° A
40 90° V
8 Ω
5 Ω
20 Ω
10 Ω
–j2 Ω
j4 Ω
j15 Ω
+
−
Io
a
b
Figure10.28 For Example 10.10.
Solution:
Our first objective is to find the Norton equivalent at terminals a-b. ZN
is found in the same way as ZTh. We set the sources to zero as shown
in Fig. 10.29(a). As evident from the figure, the (8 − j2) and (10 + j4)
impedances are short-circuited, so that
ZN = 5 
To get IN , we short-circuit terminals a-b as in Fig. 10.29(b) and
apply mesh analysis. Notice that meshes 2 and 3 form a supermesh
because of the current source linking them. For mesh 1,
−j40 + (18 + j2)I1 − (8 − j2)I2 − (10 + j4)I3 = 0 (10.10.1)
For the supermesh,
(13 − j2)I2 + (10 + j4)I3 − (18 + j2)I1 = 0 (10.10.2)
410 PART 2 AC Circuits
3
8
5
10
–j2
j4
j40 +
−
IN
I3
I2
I3
I2
I1
a
b
(b)
5
20
j15
3 + j8
Io
(c)
8
5
10
–j2
j4
ZN
(a)
Figure10.29 Solution of the circuit in Fig. 10.28: (a) finding ZN , (b) finding VN , (c) calculating Io.
At node a, due to the current source between meshes 2 and 3,
I3 = I2 + 3 (10.10.3)
Adding Eqs. (10.10.1) and (10.10.2) gives
−j40 + 5I2 = 0 ⇒ I2 = j8
From Eq. (10.10.3),
I3 = I2 + 3 = 3 + j8
The Norton current is
IN = I3 = (3 + j8) A
Figure 10.29(c) shows the Norton equivalent circuit along with the imped-
ance at terminals a-b. By current division,
Io =
5
5 + 20 + j15
IN =
3 + j8
5 + j3
= 1.465 38.48◦
A
P R A C T I C E P R O B L E M 1 0 . 1 0
Determine the Norton equivalent of the circuit in Fig. 10.30 as seen from
terminals a-b. Use the equivalent to find Io.
j2 Ω
a
b
Io
–j3 Ω
–j5 Ω
+
−
8 Ω
4 Ω
1 Ω
10 Ω
20 0° V 4 –90° A
Figure10.30 For Practice Prob. 10.10.
Answer: ZN = 3.176 + j0.706 , IN = 8.396 − 32.68◦
A,
Io = 1.971 − 2.101◦
A.
CHAPTER 10 Sinusoidal Steady-State Analysis 411
10.7 OP AMP AC CIRCUITS
The three steps stated in Section 10.1 also apply to op amp circuits, as
long as the op amp is operating in the linear region. As usual, we will
assume ideal op amps. (See Section 5.2.) As discussed in Chapter 5, the
key to analyzing op amp circuits is to keep two important properties of
an ideal op amp in mind:
1. No current enters either of its input terminals.
2. The voltage across its input terminals is zero.
The following examples will illustrate these ideas.
E X A M P L E 1 0 . 1 1
Determine vo(t) for the op amp circuit in Fig. 10.31(a) if vs =
3 cos 1000t V.
+
−
+
−
Vo
Vo
V1
–j5 kΩ
–j10 kΩ
10 kΩ 10 kΩ
20 kΩ
3 0° V
+
−
+
−
vs
vo
10 kΩ 10 kΩ
0.1 mF
0.2 mF
20 kΩ
1 2
0 V
(a) (b)
Figure10.31 For Example 10.11: (a) the original circuit in the time domain, (b) its frequency-domain equivalent.
Solution:
We first transform the circuit to the frequency domain, as shown in Fig.
10.31(b), where Vs = 3 0◦
, ω = 1000 rad/s. Applying KCL at node 1,
we obtain
3 0◦
− V1
10
=
V1
−j5
+
V1 − 0
10
+
V1 − Vo
20
or
6 = (5 + j4)V1 − Vo (10.11.1)
At node 2, KCL gives
V1 − 0
10
=
0 − Vo
−j10
which leads to
V1 = −jVo (10.11.2)
Substituting Eq. (10.11.2) into Eq. (10.11.1) yields
6 = −j(5 + j4)Vo − Vo = (3 − j5)Vo
Vo =
6
3 − j5
= 1.029 59.04◦
Hence,
vo(t) = 1.029 cos(1000t + 59.04◦
) V
412 PART 2 AC Circuits
P R A C T I C E P R O B L E M 1 0 . 1 1
Find vo and io in the op amp circuit of Fig. 10.32. Let vs =
2 cos 5000t V.
+
−
+
−
vs
vo
10 kΩ
20 kΩ
20 nF
10 nF
io
Figure10.32 For Practice Prob. 10.11.
Answer: 0.667 sin 5000t V, 66.67 sin 5000t µA.
E X A M P L E 1 0 . 1 2
Compute the closed-loop gain and phase shift for the circuit in Fig. 10.33.
Assume that R1 = R2 = 10 k, C1 = 2 µF, C2 = 1 µF, and ω =
200 rad/s.
+
−
+
−
vs vo
R1
R2
C2
C1
+
−
Figure10.33 For Example 10.12.
Solution:
The feedback and input impedances are calculated as
Zf = R2




1
jωC2
=
R2
1 + jωR2C2
Zi = R1 +
1
jωC1
=
1 + jωR1C1
jωC1
Since the circuit in Fig. 10.33 is an inverting amplifier, the closed-loop
gain is given by
G =
Vo
Vs
= −
Zf
Zi
=
jωC1R2
(1 + jωR1C1)(1 + jωR2C2)
Substituting the given values of R1, R2, C1, C2, and ω, we obtain
G =
j4
(1 + j4)(1 + j2)
= 0.434 − 49.4◦
Thus the closed-loop gain is 0.434 and the phase shift is −49.4◦
.
P R A C T I C E P R O B L E M 1 0 . 1 2
Obtain the closed-loop gain and phase shift for the circuit in Fig. 10.34.
Let R = 10 k, C = 1 µF, and ω = 1000 rad/s.
+
−
+
−
vs
vo
R
R
C
Figure10.34 For Practice Prob. 10.12.
Answer: 1.015, −5.599◦
.
CHAPTER 10 Sinusoidal Steady-State Analysis 413
10.8 AC ANALYSIS USING PSPICE
PSpice affords a big relief from the tedious task of manipulating complex
numbers in ac circuit analysis. The procedure for using PSpice for ac
analysis is quite similar to that required for dc analysis. The reader should
read Section D.5 in Appendix D for a review of PSpice concepts for ac
analysis. AC circuit analysis is done in the phasor or frequency domain,
and all sources must have the same frequency. Although AC analysis with
PSpice involves using AC Sweep, our analysis in this chapter requires a
single frequency f = ω/2π. The output file of PSpice contains voltage
and current phasors. If necessary, the impedances can be calculated using
the voltages and currents in the output file.
E X A M P L E 1 0 . 1 3
Obtain vo and io in the circuit of Fig. 10.35 using PSpice.
2 mF
50 mH
4 kΩ
2 kΩ
io
0.5io
+
−
8 sin(1000t + 50°) V vo
+
−
Figure10.35 For Example 10.13.
Solution:
We first convert the sine function to cosine.
8 sin(1000t + 50◦
) = 8 cos(1000t + 50◦
− 90◦
) = 8 cos(1000t − 40◦
)
The frequency f is obtained from ω as
f =
ω
2π
=
1000
2π
= 159.155 Hz
The schematic for the circuit is shown in Fig. 10.36. Notice the current-
controlled current source F1 is connected such that its current flows from
ACMAG=8
ACPHASE=-40
AC=ok
MAG=ok
PHASE=ok
AC=yes
MAG=yes
PHASE=ok
V
R1
C1 2u
L1
F1
4k
IPRINT
50mH
GAIN=0.5 2k
R2
0
2 3
+
−
Figure10.36 The schematic of the circuit in Fig. 10.35.
414 PART 2 AC Circuits
node0tonode3inconformitywiththeoriginalcircuitinFig.10.35. Since
we only want the magnitude and phase of vo and io, we set the attributes
of IPRINT AND VPRINT1 each to AC = yes, MAG = yes, PHASE = yes.
As a single-frequency analysis, we select Analysis/Setup/AC Sweep and
enter Total Pts = 1, Start Freq = 159.155, and Final Freq = 159.155. Af-
ter saving the schematic, we simulate it by selecting Analysis/Simulate.
The output file includes the source frequency in addition to the attributes
checked for the pseudocomponents IPRINT and VPRINT1,
FREQ IM(V_PRINT3) IP(V_PRINT3)
1.592E+02 3.264E-03 -3.743E+01
FREQ VM(3) VP(3)
1.592E+02 1.550E+00 -9.518E+01
From this output file, we obtain
Vo = 1.55 − 95.18◦
V, Io = 3.264 − 37.43◦
mA
which are the phasors for
vo = 1.55 cos(1000t − 95.18◦
) = 1.55 sin(1000t − 5.18◦
) V
and
io = 3.264 cos(1000t − 37.43◦
) mA
P R A C T I C E P R O B L E M 1 0 . 1 3
Use PSpice to obtain vo and io in the circuit of Fig. 10.37.
1 mF
2 H
2 kΩ
3 kΩ
1 kΩ
io
2vo
+
−
10 cos 3000t A vo
+
−
+
−
Figure10.37 For Practice Prob. 10.13.
Answer: 0.2682 cos(3000t−154.6◦
) V, 0.544 cos(3000t−55.12◦
) mA.
E X A M P L E 1 0 . 1 4
Find V1 and V2 in the circuit of Fig. 10.38.
Solution:
The circuit in Fig. 10.35 is in the time domain, whereas the one in Fig.
10.38 is in the frequency domain. Since we are not given a particular
CHAPTER 10 Sinusoidal Steady-State Analysis 415
2 Ω 2 Ω
V1 V2
–j1 Ω
–j2
j2 Ω
–j1 Ω
1 Ω
3 0° A 18 30° V
+
−
j2 Ω
0.2Vx
−
+
Vx
Figure10.38 For Example 10.14.
frequency and PSpice requires one, we select any frequency consistent
with the given impedances. For example, if we select ω = 1 rad/s, the
corresponding frequency is f = ω/2π = 0.159155 Hz. We obtain the
values of the capacitance (C = 1/ωXC) and inductances (L = XL/ω).
Making these changes results in the schematic in Fig. 10.39. To ease
wiring, we have exchanged the positions of the voltage-controlled cur-
rent source G1 and the 2 + j2  impedance. Notice that the current of
G1 flows from node 1 to node 3, while the controlling voltage is across
the capacitor c2, as required in Fig. 10.38. The attributes of pseudocom-
ponents VPRINT1 are set as shown. As a single-frequency analysis, we
select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq =
0.159155, and Final Freq = 0.159155. After saving the schematic, we
select Analysis/Simulate to simulate the circuit. When this is done, the
output file includes
FREQ VM(1) VP(1)
1.592E-01 2.708E+00 -5.673E+01
FREQ VM(3) VP(3)
1.592E-01 4.468E+00 -1.026E+02
AC=3
AC
R1
I1
R2 L1
L2 R3
C1
C3
G1
V1
C2
1
2
3 4
5
0.5
2
2H
2H 2
1
1
1
AC=ok
MAG=ok
PHASE=ok
AC=ok
MAG=ok
PHASE=yes
ACMAG=18
ACPHASE=30
0
GAIN=0.2
+ −
+
−
−
−
Figure10.39 Schematic for the circuit in Fig. 10.38.
416 PART 2 AC Circuits
from which we obtain
V1 = 2.708 − 56.73◦
V, V2 = 4.468 − 102.6◦
V
P R A C T I C E P R O B L E M 1 0 . 1 4
Obtain Vx and Ix in the circuit depicted in Fig. 10.40.
1 Ω 1 Ω
Vx
Ix
4Ix
–j1 Ω
j2 Ω
2 Ω
4 60° A
12 0° V
j2 Ω
–j0.25
+ −
+
−
Figure10.40 For Practice Prob. 10.14.
Answer: 13.02 − 76.08◦
V, 8.234 − 4.516◦
A.
†10.9 APPLICATIONS
The concepts learned in this chapter will be applied in later chapters to
calculate electric power and determine frequency response. The concepts
are also used in analyzing coupled circuits, three-phase circuits, ac tran-
sistor circuits, filters, oscillators, and other ac circuits. In this section, we
apply the concepts to develop two practical ac circuits: the capacitance
multiplier and the sine wave oscillators.
10.9.1 Capacitance Multiplier
The op amp circuit in Fig. 10.41 is known as a capacitance multiplier,
for reasons that will become obvious. Such a circuit is used in integrated-
circuit technology to produce a multiple of a small physical capacitance
C when a large capacitance is needed. The circuit in Fig. 10.41 can be
used to multiply capacitance values by a factor up to 1000. For example,
a 10-pF capacitor can be made to behave like a 100-nF capacitor.
In Fig. 10.41, the first op amp operates as a voltage follower, while
the second one is an inverting amplifier. The voltage follower isolates
the capacitance formed by the circuit from the loading imposed by the
inverting amplifier. Since no current enters the input terminals of the op
amp, the input current Ii flows through the feedback capacitor. Hence, at
node 1,
Ii =
Vi − Vo
1/jωC
= jωC(Vi − Vo) (10.3)
CHAPTER 10 Sinusoidal Steady-State Analysis 417
+
−
R1
A1
R2
+
−
Vo
Ii
Zi
Vi
A2
C
0 V
2
1
+
−
Vi
Figure10.41 Capacitance multiplier.
Applying KCL at node 2 gives
Vi − 0
R1
=
0 − Vo
R2
or
Vo = −
R2
R1
Vi (10.4)
Substituting Eq. (10.4) into (10.3) gives
Ii = jωC

1 +
R2
R1

Vi
or
Ii
Vi
= jω

1 +
R2
R1

C (10.5)
The input impedance is
Zi =
Vi
Ii
=
1
jωCeq
(10.6)
where
Ceq =

1 +
R2
R1

C (10.7)
Thus, by a proper selection of the values of R1 and R2, the op amp
circuit in Fig. 10.41 can be made to produce an effective capacitance
between the input terminal and ground, which is a multiple of the physical
capacitance C. The size of the effective capacitance is practically limited
by the inverted output voltage limitation. Thus, the larger the capacitance
multiplication, the smaller is the allowable input voltage to prevent the
op amps from reaching saturation.
A similar op amp circuit can be designed to simulate inductance.
(See Prob. 10.69.) There is also an op amp circuit configuration to create
a resistance multiplier.
E X A M P L E 1 0 . 1 5
Calculate Ceq in Fig. 10.41 when R1 = 10 k, R2 = 1 M, and C = 1 nF.
418 PART 2 AC Circuits
Solution:
From Eq. (10.7)
Ceq =

1 +
R2
R1

C =

1 +
1 × 106
10 × 103

1 nF = 101 nF
P R A C T I C E P R O B L E M 1 0 . 1 5
Determine the equivalent capacitance of the op amp circuit in Fig. 10.41
if R1 = 10 k, R2 = 10 M, and C = 10 nF.
Answer: 10 µF.
10.9.2 Oscillators
We know that dc is produced by batteries. But how do we produce ac?
One way is using oscillators, which are circuits that convert dc to ac.
An oscillator is a circuit that produces an ac waveform as output
when powered by a dc input.
The only external source an oscillator needs is the dc power supply.
Ironically, the dc power supply is usually obtained by converting the ac
supplied by the electric utility company to dc. Having gone through the
trouble of conversion, one may wonder why we need to use the oscillator
to convert the dc to ac again. The problem is that the ac supplied by the
utility company operates at a preset frequency of 60 Hz in the United
States (50 Hz in some other nations), whereas many applications such
as electronic circuits, communication systems, and microwave devices
require internally generated frequencies that range from 0 to 10 GHz or
higher. Oscillators are used for generating these frequencies.
This corresponds to ω = 2πf = 377 rad/s.
In order for sine wave oscillators to sustain oscillations, they must
meet the Barkhausen criteria:
1. The overall gain of the oscillator must be unity or greater.
Therefore, losses must be compensated for by an amplifying
device.
2. The overall phase shift (from input to output and back to the
input) must be zero.
Three common types of sine wave oscillators are phase-shift, twin T ,
and Wien-bridge oscillators. Here we consider only the Wien-bridge
oscillator.
+
−
Rf
Rg
R1
R2
C1
C2
+
−
v2
+
−
vo
Positive feedback path
to create oscillations
Negative feedback
path to control gain
Figure10.42 Wien-bridge oscillator.
The Wien-bridge oscillator is widely used for generating sinusoids
in the frequency range below 1 MHz. It is an RC op amp circuit with only
a few components, easily tunable and easy to design. As shown in Fig.
10.42, the oscillator essentially consists of a noninverting amplifier with
two feedback paths: the positive feedback path to the noninverting input
creates oscillations, while the negative feedback path to the inverting
CHAPTER 10 Sinusoidal Steady-State Analysis 419
input controls the gain. If we define the impedances of the RC series and
parallel combinations as Zs and Zp, then
Zs = R1 +
1
jωC1
= R1 −
j
ωC1
(10.8)
Zp = R2
1
jωC2
=
R2
1 + jωR2C2
(10.9)
The feedback ratio is
V2
Vo
=
Zp
Zs + Zp
(10.10)
Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives
V2
Vo
=
R2
R2 +

R1 −
j
ωC1

(1 + jωR2C2)
=
ωR2C1
ω(R2C1 + R1C1 + R2C2) + j(ω2R1C1R2C2 − 1)
(10.11)
To satisfy the second Barkhausen criterion, V2 must be in phase with Vo,
which implies that the ratio in Eq. (10.11) must be purely real. Hence,
the imaginary part must be zero. Setting the imaginary part equal to zero
gives the oscillation frequency ωo as
ω2
oR1C1R2C2 − 1 = 0
or
ωo =
1
√
R1R2C1C2
(10.12)
In most practical applications, R1 = R2 = R and C1 = C2 = C, so that
ωo =
1
RC
= 2πfo (10.13)
or
fo =
1
2πRC
(10.14)
Substituting Eq. (10.13) and R1 = R2 = R, C1 = C2 = C into Eq.
(10.11) yields
V2
Vo
=
1
3
(10.15)
Thus in order to satisfy the first Barkhausen criterion, the op amp must
compensate by providing a gain of 3 or greater so that the overall gain is
at least 1 or unity. We recall that for a noninverting amplifier,
Vo
V2
= 1 +
Rf
Rg
= 3 (10.16)
or
Rf = 2Rg (10.17)
420 PART 2 AC Circuits
Due to the inherent delay caused by the op amp, Wien-bridge oscil-
lators are limited to operating in the frequency range of 1 MHz or less.
E X A M P L E 1 0 . 1 6
Design a Wien-bridge circuit to oscillate at 100 kHz.
Solution:
Using Eq. (10.14), we obtain the time constant of the circuit as
RC =
1
2πfo
=
1
2π × 100 × 103
= 1.59 × 10−6
(10.16.1)
If we select R = 10 k, then we can select C = 159 pF to satisfy Eq.
(10.16.1). Since the gain must be 3, Rf /Rg = 2. We could select Rf =
20 k while Rg = 10 k.
P R A C T I C E P R O B L E M 1 0 . 1 6
In the Wien-bridge oscillator circuit in Fig. 10.42, let R1 = R2 = 2.5 k,
C1 = C2 = 1 nF. Determine the frequency fo of the oscillator.
Answer: 63.66 kHz.
10.10 SUMMARY
1. We apply nodal and mesh analysis to ac circuits by applying KCL
and KVL to the phasor form of the circuits.
2. In solving for the steady-state response of a circuit that has indepen-
dent sources with different frequencies, each independent source
must be considered separately. The most natural approach to analyz-
ing such circuits is to apply the superposition theorem. A separate
phasor circuit for each frequency must be solved independently, and
the corresponding response should be obtained in the time domain.
The overall response is the sum of the time-domain responses of all
the individual phasor circuits.
3. The concept of source transformation is also applicable in the fre-
quency domain.
4. The Thevenin equivalent of an ac circuit consists of a voltage source
VTh in series with the Thevenin impedance ZTh.
5. The Norton equivalent of an ac circuit consists of a current source IN
in parallel with the Norton impedance ZN (= ZTh).
6. PSpice is a simple and powerful tool for solving ac circuit problems.
It relieves us of the tedious task of working with the complex num-
bers involved in steady-state analysis.
7. The capacitance multiplier and the ac oscillator provide two typical
applications for the concepts presented in this chapter. A capacitance
multiplier is an op amp circuit used in producing a multiple of a
physical capacitance. An oscillator is a device that uses a dc input to
generate an ac output.
CHAPTER 10 Sinusoidal Steady-State Analysis 421
REVIEW QUESTIONS
10.1 The voltage Vo across the capacitor in Fig. 10.43 is:
(a) 5 0◦
V (b) 7.071 45◦
V
(c) 7.071 − 45◦
V (d) 5 − 45◦
V
1 Ω
+
−
Vo
+
−
–j1 Ω
10 0° V
Figure 10.43 For Review Question 10.1.
10.2 The value of the current Io in the circuit in Fig.
10.44 is:
(a) 4 0◦
A (b) 2.4 − 90◦
A
(c) 0.6 0◦
A (d) −1 A
j8 Ω –j2 Ω
3 0° A
Io
Figure 10.44 For Review Question 10.2.
10.3 Using nodal analysis, the value of Vo in the circuit
of Fig. 10.45 is:
(a) −24 V (b) −8 V
(c) 8 V (d) 24 V
–j3 Ω
j6 Ω 4 90° A
V
o
Figure 10.45 For Review Question 10.3.
10.4 In the circuit of Fig. 10.46, current i(t) is:
(a) 10 cos t A (b) 10 sin t A (c) 5 cos t A
(d) 5 sin t A (e) 4.472 cos(t − 63.43◦
) A
1 H 1 F
+
− 1 Ω
10 cos t V i(t)
Figure 10.46 For Review Question 10.4.
10.5 Refer to the circuit in Fig. 10.47 and observe that the
two sources do not have the same frequency. The
current ix(t) can be obtained by:
(a) source transformation
(b) the superposition theorem
(c) PSpice
1 F
+
−
+
−
sin 2t V sin 10t V
1 H 1 Ω
ix
Figure 10.47 For Review Question 10.5.
10.6 For the circuit in Fig. 10.48, the Thevenin
impedance at terminals a-b is:
(a) 1  (b) 0.5 − j0.5 
(c) 0.5 + j0.5  (d) 1 + j2 
(e) 1 − j2 
1 Ω 1 H
+
− 1 F
a
b
5 cos t V
Figure 10.48 For Review Questions 10.6 and 10.7.
10.7 In the circuit of Fig. 10.48, the Thevenin voltage at
terminals a-b is:
(a) 3.535 − 45◦
V (b) 3.535 45◦
V
(c) 7.071 − 45◦
V (d) 7.071 45◦
V
10.8 Refer to the circuit in Fig. 10.49. The Norton
equivalent impedance at terminals a-b is:
(a) −j4  (b) −j2 
(c) j2  (d) j4 
–j2 Ω
j4 Ω
+
−
a
b
6 0° V
Figure 10.49 For Review Questions 10.8 and 10.9.
422 PART 2 AC Circuits
10.9 The Norton current at terminals a-b in the circuit of
Fig. 10.49 is:
(a) 1 0◦
A (b) 1.5 − 90◦
A
(c) 1.5 90◦
A (d) 3 90◦
A
10.10 PSpice can handle a circuit with two independent
sources of different frequencies.
(a) True (b) False
Answers: 10.1c, 10.2a, 10.3d, 10.4a, 10.5b, 10.6c, 10.7a, 10.8a,
10.9d, 10.10b.
PROBLEMS
Section 10.2 Nodal Analysis
10.1 Find vo in the circuit in Fig. 10.50.
1 F
+
−
+
−
1 H
3 Ω
vo
10 cos(t – 45°) V 5 sin(t + 30°) V
+
−
Figure 10.50 For Prob. 10.1.
10.2 For the circuit depicted in Fig. 10.51 below,
determine io.
10.3 Determine vo in the circuit of Fig. 10.52.
+
−
2 H
4 Ω
vo
16 sin 4t V 2 cos 4t A
+
−
1 Ω 6 Ω
F
1
12
Figure 10.52 For Prob. 10.3.
10.4 Compute vo(t) in the circuit of Fig. 10.53.
+
−
1 H 0.25 F
1 Ω
0.5ix vo
+
−
16 sin (4t – 10°) V
ix
Figure 10.53 For Prob. 10.4.
10.5 Use nodal analysis to find vo in the circuit of Fig.
10.54.
+
−
10 mH
50 mF
20 Ω
20 Ω 30 Ω
10 cos 103
t V
io
4io vo
+
−
Figure 10.54 For Prob. 10.5.
10.6 Using nodal analysis, find io(t) in the circuit in Fig.
10.55.
0.02 F
+
− 1 H
10 Ω
20 sin (10t – 4)V 4 cos (10t – 3)A
io
Figure10.51 For Prob. 10.2.
CHAPTER 10 Sinusoidal Steady-State Analysis 423
0.5 F
+
−
1 H
2 H
2 Ω
0.25 F
8 sin (2t + 30°) V cos 2t A
io
Figure 10.55 For Prob. 10.6.
10.7 By nodal analysis, find io in the circuit in Fig. 10.56.
10 Ω
20 Ω 50 mF 10 mH
20 sin1000t A
2io
io
Figure 10.56 For Prob. 10.7.
10.8 Calculate the voltage at nodes 1 and 2 in the circuit
of Fig. 10.57 using nodal analysis.
10 Ω
1 2
–j2 Ω –j5 Ω
j2 Ω
j4 Ω
20 30° A
Figure 10.57 For Prob. 10.8.
10.9 Solve for the current I in the circuit of Fig. 10.58
using nodal analysis.
2 Ω
4 Ω
–j2 Ω
j1 Ω
2I
5 0° A
20 –90° V +
−
I
Figure 10.58 For Prob. 10.9.
10.10 Using nodal analysis, find V1 and V2 in the circuit
of Fig. 10.59.
20 Ω
10 Ω
j2 A 1 + j A
–j5 Ω
j10 Ω
V1 V2
Figure 10.59 For Prob. 10.10.
10.11 By nodal analysis, obtain current Io in the circuit in
Fig. 10.60.
3 Ω
2 Ω
1 Ω
j4 Ω
–j2 Ω
+
−
100 20° V
Io
Figure 10.60 For Prob. 10.11.
10.12 Use nodal analysis to obtain Vo in the circuit of Fig.
10.61 below.
8 Ω
2 Ω –j1 Ω –j2 Ω
j6 Ω 4 Ω j5 Ω
2Vx Vo
4 45° A
+
−
Vx
+
−
Figure10.61 For Prob. 10.12.
424 PART 2 AC Circuits
10.13 Obtain Vo in Fig. 10.62 using nodal analysis.
4 Ω
2 Ω –j4 Ω
j2 Ω
Vo 0.2Vo
+
−
+ −
12 0° V
Figure 10.62 For Prob. 10.13.
10.14 Refer to Fig. 10.63. If vs(t) = Vm sin ωt and
vo(t) = A sin(ωt + φ), derive the expressions for A
and φ.
+
− vo(t)
vs(t)
+
−
L
R
C
Figure 10.63 For Prob. 10.14.
10.15 For each of the circuits in Fig. 10.64, find Vo/Vi for
ω = 0, ω → ∞, and ω2
= 1/LC.
Vo
+
−
Vo
+
−
Vi
+
−
Vi
+
−
C
R R C
L
L
(b)
(a)
Figure 10.64 For Prob. 10.15.
10.16 For the circuit in Fig. 10.65, determine Vo/Vs.
+
−
Vs Vo
+
−
L
R1
R2
C
Figure 10.65 For Prob. 10.16.
Section 10.3 Mesh Analysis
10.17 Obtain the mesh currents I1 and I2 in the circuit of
Fig. 10.66.
+
−
Vs L
R
C2
C1
I2
I1
Figure 10.66 For Prob. 10.17.
10.18 Solve for io in Fig. 10.67 using mesh analysis.
+
−
+
−
2 H
0.25 F
4 Ω
10 cos 2t V 6 sin 2t V
io
Figure 10.67 For Prob. 10.18.
10.19 Rework Prob. 10.5 using mesh analysis.
10.20 Using mesh analysis, find I1 and I2 in the circuit of
Fig. 10.68.
+
−
+
−
I2
I1
j10 Ω
–j20 Ω
40 Ω
50 0° V
40 30° V
Figure 10.68 For Prob. 10.20.
10.21 By using mesh analysis, find I1 and I2 in the circuit
depicted in Fig. 10.69.
I2
I1
j4 Ω
j2 Ω
j1 Ω
–j6 Ω
3 Ω
2 Ω
30 20° V
3 Ω
+ −
Figure 10.69 For Prob. 10.21.
CHAPTER 10 Sinusoidal Steady-State Analysis 425
10.22 Repeat Prob. 10.11 using mesh analysis.
10.23 Use mesh analysis to determine current Io in the
circuit of Fig. 10.70 below.
10.24 Determine Vo and Io in the circuit of Fig. 10.71
using mesh analysis.
j4 Ω
Io
3Vo
–j2 Ω
4 –30° A 2 Ω +
−
Vo
+
−
Figure 10.71 For Prob. 10.24.
10.25 Compute I in Prob. 10.9 using mesh analysis.
10.26 Use mesh analysis to find Io in Fig. 10.28 (for
Example 10.10).
10.27 Calculate Io in Fig. 10.30 (for Practice Prob. 10.10)
using mesh analysis.
10.28 Compute Vo in the circuit of Fig. 10.72 using mesh
analysis.
–j3 Ω
2 Ω
j4 Ω
+
−
2 Ω
2 Ω
12 0° V
2 0° A
4 90° A V
o
+
−
Figure 10.72 For Prob. 10.28.
10.29 Using mesh analysis, obtain Io in the circuit shown
in Fig. 10.73.
–j4 Ω
j2 Ω
2 Ω
1 Ω 1 Ω
Io
+
− 10 90° V
4 0° A
2 0° A
Figure 10.73 For Prob. 10.29.
Section 10.4 Superposition Theorem
10.30 Find io in the circuit shown in Fig. 10.74 using
superposition.
4 Ω
+
−
+
−
2 Ω
8 V
1 H
10 cos 4t V
io
Figure 10.74 For Prob. 10.30.
10.31 Using the superposition principle, find ix in the
circuit of Fig. 10.75.
+
−
3 Ω
4 H 10 cos(2t – 60°) V
5 cos(2t + 10°) A
ix
F
1
8
Figure 10.75 For Prob. 10.31.
10.32 Rework Prob. 10.2 using the superposition theorem.
10.33 Solve for vo(t) in the circuit of Fig. 10.76 using the
superposition principle.
–j40 Ω –j40 Ω
j60 Ω
80 Ω 20 Ω
Io
+
−
+
−
100 120° V 60 –30° V
Figure10.70 For Prob. 10.23.
426 PART 2 AC Circuits
+
−
+
−
6 Ω 2 H
10 V
12 cos 3t V 4 sin 2t A
+
−
vo
F
1
12
Figure 10.76 For Prob. 10.33.
10.34 Determine io in the circuit of Fig. 10.77.
+
−
1 Ω 2 H
24 V
2 cos 3t
2 Ω 4 Ω
+
−
io
10 sin(3t – 30°) V
F
1
6
Figure 10.77 For Prob. 10.34.
10.35 Find io in the circuit in Fig. 10.78 using
superposition.
80 Ω
60 Ω
40 mH
20 mF
24 V
100 Ω
+
−
+
−
50 cos 2000t V
2 sin 4000t A
io
Figure 10.78 For Prob. 10.35.
Section 10.5 Source Transformation
10.36 Using source transformation, find i in the circuit of
Fig. 10.79.
3 Ω
5 Ω
5 mH
1 mF
8 sin(200t + 30°) A
i
Figure 10.79 For Prob. 10.36.
10.37 Use source transformation to find vo in the circuit in
Fig. 10.80.
20 Ω
80 Ω
0.4 mH
0.2 mF
+
−
5 cos 105
t V vo
+
−
Figure 10.80 For Prob. 10.37.
10.38 Solve Prob. 10.20 using source transformation.
10.39 Use the method of source transformation to find Ix
in the circuit of Fig. 10.81.
+
−
2 Ω j4 Ω –j2 Ω
–j3 Ω
6 Ω
4 Ω
Ix
60 0° V 5 90° A
Figure 10.81 For Prob. 10.39.
10.40 Use the concept of source transformation to find Vo
in the circuit of Fig. 10.82.
+
−
4 Ω j4 Ω
–j3 Ω
–j2 Ω
j2 Ω 2 Ω
20 0° V Vo
+
−
Figure 10.82 For Prob. 10.40.
Section 10.6 Thevenin and Norton Equivalent
Circuits
10.41 Find the Thevenin and Norton equivalent circuits at
terminals a-b for each of the circuits in Fig. 10.83.
CHAPTER 10 Sinusoidal Steady-State Analysis 427
–j10 Ω
j20 Ω 10 Ω
a
b
50 30° V +
−
a
b
4 0° A
–j5 Ω
j10 Ω
8 Ω
(b)
(a)
Figure 10.83 For Prob. 10.41.
10.42 For each of the circuits in Fig. 10.84, obtain
Thevenin and Norton equivalent circuits at terminals
a-b.
–j5 Ω
j4 Ω
6 Ω
30 Ω
a
b
2 0° A
a
b
120 45° V
–j2 Ω
j10 Ω
60 Ω
(b)
(a)
+
−
Figure 10.84 For Prob. 10.42.
10.43 Find the Thevenin and Norton equivalent circuits for
the circuit shown in Fig. 10.85.
j20 Ω
5 Ω 2 Ω
60 120° V +
−
–j10 Ω
Figure 10.85 For Prob. 10.43.
10.44 For the circuit depicted in Fig. 10.86, find the
Thevenin equivalent circuit at terminals a-b.
a
b
5 45° A j10 Ω
8 Ω
–j6 Ω
Figure 10.86 For Prob. 10.44.
10.45 Repeat Prob. 10.1 using Thevenin’s theorem.
10.46 Find the Thevenin equivalent of the circuit in Fig.
10.87 as seen from:
(a) terminals a-b (b) terminals c-d
10 Ω
a
b
4 0° A
20 0° V
–j4 Ω
j5 Ω 4 Ω
+
−
c d
Figure 10.87 For Prob. 10.46.
10.47 Solve Prob. 10.3 using Thevenin’s theorem.
10.48 Using Thevenin’s theorem, find vo in the circuit in
Fig. 10.88.
2 H
4 Ω
2 Ω vo
io
3io
+
−
12 cos t V
+
−
F
1
4 F
1
8
Figure 10.88 For Prob. 10.48.
10.49 Obtain the Norton equivalent of the circuit depicted
in Fig. 10.89 at terminals a-b.
a
b
5 mF
10 H 2 kΩ
4 cos(200t + 30°) V
Figure 10.89 For Prob. 10.49.
428 PART 2 AC Circuits
10.50 For the circuit shown in Fig. 10.90, find the Norton
equivalent circuit at terminals a-b.
60 Ω 40 Ω
–j30 Ω
j80 Ω
a b
3 60° A
Figure 10.90 For Prob. 10.50.
10.51 Compute io in Fig. 10.91 using Norton’s theorem.
2 Ω
4 H
5 cos 2t V
+ −
io
F
1
4 F
1
2
Figure 10.91 For Prob. 10.51.
10.52 At terminals a-b, obtain Thevenin and Norton
equivalent circuits for the network depicted in Fig.
10.92. Take ω = 10 rad/s.
a
b
10 mF
10 Ω
2 sin vt V
12 cos vt
+
−
vo 2vo
+
−
H
1
2
Figure 10.92 For Prob. 10.52.
Section 10.7 Op Amp AC Circuits
10.53 For the differentiator shown in Fig. 10.93, obtain
Vo/Vs. Find vo(t) when vs(t) = Vm sin ωt and
ω = 1/RC.
+
−
vs vo
R
C
+
−
+
−
Figure 10.93 For Prob. 10.53.
10.54 The circuit in Fig. 10.94 is an integrator with a
feedback resistor. Calculate vo(t) if
vs = 2 cos 4 × 104
t V.
+
−
vs vo
+
−
10 nF
100 kΩ
50 kΩ
+
−
Figure 10.94 For Prob. 10.54.
10.55 Compute io(t) in the op amp circuit in Fig. 10.95 if
vs = 4 cos 104
t V.
+
−
vs
50 kΩ
1 nF
100 kΩ
io
+
−
Figure 10.95 For Prob. 10.55.
10.56 If the input impedance is defined as Zin = Vs/Is,
find the input impedance of the op amp circuit in
Fig. 10.96 when R1 = 10 k, R2 = 20 k,
C1 = 10 nF, C2 = 20 nF, and ω = 5000 rad/s.
Vs C2
C1
R1 R2
Is
Zin
Vo
+
−
+
−
Figure 10.96 For Prob. 10.56.
10.57 Evaluate the voltage gain Av = Vo/Vs in the op
amp circuit of Fig. 10.97. Find Av at ω = 0,
ω → ∞, ω = 1/R1C1, and ω = 1/R2C2.
+
−
Vs Vo
+
−
C1
R1
C2
R2
+
−
Figure 10.97 For Prob. 10.57.
CHAPTER 10 Sinusoidal Steady-State Analysis 429
10.58 In the op amp circuit of Fig. 10.98, find the
closed-loop gain and phase shift if C1 = C2 = 1 nF,
R1 = R2 = 100 k, R3 = 20 k, R4 = 40 k, and
ω = 2000 rad/s.
vs vo
C1
R1
R2
+
−
C2
R4
R3
+
−
+
−
Figure 10.98 For Prob. 10.58.
10.59 Compute the closed-loop gain Vo/Vs for the op amp
circuit of Fig. 10.99.
+
−
+
−
vs
vo
+
−
C1
R1
R3 C2 R2
Figure 10.99 For Prob. 10.59.
10.60 Determine vo(t) in the op amp circuit in Fig. 10.100
below.
10.61 For the op amp circuit in Fig. 10.101, obtain vo(t).
vo
+
−
10 kΩ
20 kΩ
40 kΩ
0.1 mF
0.2 mF
+
−
+
−
+
−
5 cos 103
t V
Figure 10.101 For Prob. 10.61.
10.62 Obtain vo(t) for the op amp circuit in Fig. 10.102 if
vs = 4 cos(1000t − 60◦
) V.
vo
vs +
−
10 kΩ
50 kΩ
20 kΩ 0.2 mF
0.1 mF
+
−
+
−
+
−
Figure 10.102 For Prob. 10.62.
Section 10.8 AC Analysis Using PSpice
10.63 Use PSpice to solve Example 10.10.
10.64 Solve Prob. 10.13 using PSpice.
vo
+
−
10 kΩ
20 kΩ
20 kΩ
40 kΩ
10 kΩ
0.25 mF
0.5 mF
+
−
2 sin 400t V
Figure10.100 For Prob. 10.60.
430 PART 2 AC Circuits
10.65 Obtain Vo in the circuit of Fig. 10.103 using PSpice.
1 Ω
j4 Ω
–j2 Ω
2 Ω
+
−
Vx
2Vx
+
−
Vo
3 0° A
Figure 10.103 For Prob. 10.65.
10.66 Use PSpice to find V1, V2, and V3 in the network of
Fig. 10.104.
+
−
8 Ω
j10 Ω j10 Ω
–j4 Ω –j4 Ω
V1 V3
V2
60 30° V 4 0° A
Figure 10.104 For Prob. 10.66.
10.67 Determine V1, V2, and V3 in the circuit of Fig.
10.105 using PSpice.
8 Ω
j10 Ω
1 Ω
2 Ω
j6 Ω –j2 Ω
–j4 Ω
V1 V3
V2
4 0° A 2 0° A
Figure 10.105 For Prob. 10.67.
10.68 Use PSpice to find vo and io in the circuit of Fig.
10.106 below.
Section 10.9 Applications
10.69 The op amp circuit in Fig. 10.107 is called an
inductance simulator. Show that the input
impedance is given by
Zin =
Vin
Iin
= jωLeq
where
Leq =
R1R3R4
R2
C
Vin
Iin
+
−
+
−
R1 R2 R3
C R4
+
−
Figure 10.107 For Prob. 10.69.
10.70 Figure 10.108 shows a Wien-bridge network. Show
that the frequency at which the phase shift between
the input and output signals is zero is f = 1
2
πRC,
and that the necessary gain is Av = Vo/Vi = 3 at
that frequency.
Vi
+
−
R
R1
R2
R
C
C
+ −
Vo
Figure 10.108 For Prob. 10.70.
10.71 Consider the oscillator in Fig. 10.109.
(a) Determine the oscillation frequency.
20 mF
25 mF
2 H
4 Ω
10 Ω vo
0.5vo
io
4io
+
−
6 cos 4t V
+
−
+
−
Figure10.106 For Prob. 10.68.
CHAPTER 10 Sinusoidal Steady-State Analysis 431
(b) Obtain the minimum value of R for which
oscillation takes place.
+
−
R
10 kΩ
20 kΩ
80 kΩ
0.4 mH 2 nF
Figure 10.109 For Prob. 10.71.
10.72 The oscillator circuit in Fig. 10.110 uses an ideal op
amp.
(a) Calculate the minimum value of Ro that will
cause oscillation to occur.
(b) Find the frequency of oscillation.
+
−
10 kΩ
100 kΩ
1 MΩ
10 mH 2 nF
Ro
Figure 10.110 For Prob. 10.72.
10.73 Figure 10.111 shows a Colpitts oscillator. Show
that the oscillation frequency is
fo =
1
2π
√
LCT
where CT = C1C2/(C1 + C2). Assume Ri  XC2
.
+
−
Rf
Ri
C2 C1
L
V
o
Figure 10.111 A Colpitts oscillator; for Prob. 10.73.
(Hint: Set the imaginary part of the impedance in
the feedback circuit equal to zero.)
10.74 Design a Colpitts oscillator that will operate at
50 kHz.
10.75 Figure 10.112 shows a Hartley oscillator. Show that
the frequency of oscillation is
fo =
1
2π
√
C(L1 + L2)
+
−
Rf
Ri
L2 L1
C
Vo
Figure 10.112 A Hartley oscillator; for Prob. 10.75.
10.76 Refer to the oscillator in Fig. 10.113.
(a) Show that
V2
Vo
=
1
3 + j(ωL/R − R/ωL)
(b) Determine the oscillation frequency fo.
(c) Obtain the relationship between R1 and R2 in
order for oscillation to occur.
+
−
R L
R
L
R1
R2
V
o
V2
Figure 10.113 For Prob. 10.76.
433
C H A P T E R
AC POWER ANALYSIS
1 1
An engineer is an unordinary person who can do for one dollar what any
ordinary person can do for two dollars.
—Anonymous
Enhancing Your Career
Career in Power Systems The discovery of the principle
of an ac generator by Michael Faraday in 1831 was a major
breakthrough in engineering; it provided a convenient way
of generating the electric power that is needed in every elec-
tronic, electrical, or electromechanical device we use now.
Electric power is obtained by converting energy from
sources such as fossil fuels (gas, oil, and coal), nuclear
fuel (uranium), hydro energy (water falling through a head),
geothermal energy (hot water, steam), wind energy, tidal en-
ergy, and biomass energy (wastes). These various ways of
generating electric power are studied in detail in the field of
power engineering, which has become an indispensable sub-
discipline of electrical engineering. An electrical engineer
should be familiar with the analysis, generation, transmis-
sion, distribution, and cost of electric power.
The electric power industry is a very large employer
of electrical engineers. The industry includes thousands of
electric utility systems ranging from large, interconnected
systems serving large regional areas to small power
companies serving individual communities or factories.
Due to the complexity of the power industry, there are
numerous electrical engineering jobs in different areas of
the industry: power plant (generation), transmission and
distribution, maintenance, research, data acquisition and
flow control, and management. Since electric power is used
everywhere, electric utility companies are everywhere, of-
fering exciting training and steady employment for men and
women in thousands of communities throughout the world.
A pole-type transformer with a low-voltage, three-wire distribution
system. Source: W. N. Alerich, Electricity, 3rd ed. Albany, NY:
Delmar Publishers, 1981, p. 152. (Courtesy of General Electric.)
434 PART 2 AC Circuits
11.1 INTRODUCTION
Our effort in ac circuit analysis so far has been focused mainly on cal-
culating voltage and current. Our major concern in this chapter is power
analysis.
Power analysis is of paramount importance. Power is the most
important quantity in electric utilities, electronic, and communication
systems, because such systems involve transmission of power from one
point to another. Also, every industrial and household electrical device—
every fan, motor, lamp, pressing iron, TV, personal computer—has a
power rating that indicates how much power the equipment requires;
exceeding the power rating can do permanent damage to an appliance.
The most common form of electric power is 50- or 60-Hz ac power. The
choice of ac over dc allowed high-voltage power transmission from the
power generating plant to the consumer.
We will begin by defining and deriving instantaneous power and
average power. We will then introduce other power concepts. As practi-
cal applications of these concepts, we will discuss how power is measured
and reconsider how electric utility companies charge their customers.
11.2 INSTANTANEOUS AND AVERAGE POWER
As mentioned in Chapter 2, the instantaneous power p(t) absorbed by an
element is the product of the instantaneous voltage v(t) across the element
and the instantaneous current i(t) through it. Assuming the passive sign
convention,
p(t) = v(t)i(t) (11.1)
The instantaneous power is the power at any instant of time. It is the rate
at which an element absorbs energy.
We can also think of the instantaneous power
as the power absorbed by the element at a spe-
cificinstantoftime. Instantaneousquantitiesare
denoted by lowercase letters.
Sinusoidal
source
Passive
linear
network
i(t)
+
−
v(t)
Figure 11.1 Sinusoidal source and passive
linear circuit.
Consider the general case of instantaneous power absorbed by an
arbitrary combination of circuit elements under sinusoidal excitation, as
shown in Fig. 11.1. Let the voltage and current at the terminals of the
circuit be
v(t) = Vm cos(ωt + θv) (11.2a)
i(t) = Im cos(ωt + θi) (11.2b)
where Vm and Im are the amplitudes (or peak values), and θv and θi are the
phase angles of the voltage and current, respectively. The instantaneous
power absorbed by the circuit is
p(t) = v(t)i(t) = VmIm cos(ωt + θv) cos(ωt + θi) (11.3)
We apply the trigonometric identity
cos A cos B =
1
2
[cos(A − B) + cos(A + B)] (11.4)
and express Eq. (11.3) as
p(t) =
1
2
VmIm cos(θv − θi) +
1
2
VmIm cos(2ωt + θv + θi) (11.5)
CHAPTER 11 AC Power Analysis 435
This shows us that the instantaneous power has two parts. The first part is
constant or time independent. Its value depends on the phase difference
between the voltage and the current. The second part is a sinusoidal
function whose frequency is 2ω, which is twice the angular frequency of
the voltage or current.
A sketch of p(t) in Eq. (11.5) is shown in Fig. 11.2, where T =
2π/ω is the period of voltage or current. We observe that p(t) is periodic,
p(t) = p(t + T0), and has a period of T0 = T/2, since its frequency
is twice that of voltage or current. We also observe that p(t) is positive
for some part of each cycle and negative for the rest of the cycle. When
p(t) is positive, power is absorbed by the circuit. When p(t) is negative,
power is absorbed by the source; that is, power is transferred from the
circuit to the source. This is possible because of the storage elements
(capacitors and inductors) in the circuit.
0
VmIm cos(uv − ui)
VmIm
p(t)
T
2
T t
1
2
1
2
Figure11.2 The instantaneous power p(t) entering a circuit.
Theinstantaneouspowerchangeswithtimeandisthereforedifficult
to measure. The average power is more convenient to measure. In fact,
the wattmeter, the instrument for measuring power, responds to average
power.
The average power is the average of the instantaneous power over one period.
Thus, the average power is given by
P =
1
T
 T
0
p(t) dt (11.6)
Although Eq. (11.6) shows the averaging done over T , we would get the
same result if we performed the integration over the actual period of p(t)
which is T0 = T/2.
Substituting p(t) in Eq. (11.5) into Eq. (11.6) gives
P =
1
T
 T
0
1
2
VmIm cos(θv − θi) dt
+
1
T
 T
0
1
2
VmIm cos(2ωt + θv + θi) dt
436 PART 2 AC Circuits
=
1
2
VmIm cos(θv − θi)
1
T
 T
0
dt
+
1
2
VmIm
1
T
 T
0
cos(2ωt + θv + θi) dt (11.7)
The first integrand is constant, and the average of a constant is the same
constant. The second integrand is a sinusoid. We know that the average of
asinusoidoveritsperiodiszerobecausetheareaunderthesinusoidduring
a positive half-cycle is canceled by the area under it during the following
negative half-cycle. Thus, the second term in Eq. (11.7) vanishes and the
average power becomes
P =
1
2
VmIm cos(θv − θi) (11.8)
Since cos(θv − θi) = cos(θi − θv), what is important is the difference in
the phases of the voltage and current.
Note that p(t) is time-varying while P does not depend on time.
To find the instantaneous power, we must necessarily have v(t) and i(t)
in the time domain. But we can find the average power when voltage
and current are expressed in the time domain, as in Eq. (11.2), or when
they are expressed in the frequency domain. The phasor forms of v(t)
and i(t) in Eq. (11.2) are V = Vm θv and I = Im θi, respectively. P is
calculated using Eq. (11.8) or using phasors V and I. To use phasors, we
notice that
1
2
VI∗
=
1
2
VmIm θv − θi
=
1
2
VmIm [cos(θv − θi) + j sin(θv − θi)]
(11.9)
We recognize the real part of this expression as the average power P
according to Eq. (11.8). Thus,
P =
1
2
Re

VI∗

=
1
2
VmIm cos(θv − θi) (11.10)
Consider two special cases of Eq. (11.10). When θv = θi, the
voltage and current are in phase. This implies a purely resistive circuit
or resistive load R, and
P =
1
2
VmIm =
1
2
I2
mR =
1
2
|I|2
R (11.11)
where |I|2
= I × I∗
. Equation (11.11) shows that a purely resistive
circuit absorbs power at all times. When θv − θi = ±90◦
, we have a
purely reactive circuit, and
P =
1
2
VmIm cos 90◦
= 0 (11.12)
CHAPTER 11 AC Power Analysis 437
showing that a purely reactive circuit absorbs no average power. In sum-
mary,
A resistive load (R) absorbs power at all times, while a reactive load (L or C)
absorbs zero average power.
E X A M P L E 1 1 . 1
Given that
v(t) = 120 cos(377t +45◦
) V and i(t) = 10 cos(377t −10◦
) A
find the instantaneous power and the average power absorbed by the
passive linear network of Fig. 11.1.
Solution:
The instantaneous power is given by
p = vi = 1200 cos(377t + 45◦
) cos(377t − 10◦
)
Applying the trigonometric identity
cos A cos B =
1
2
[cos(A + B) + cos(A − B)]
gives
p = 600[cos(754t + 35◦
) + cos 55◦
]
or
p(t) = 344.2 + 600 cos(754t + 35◦
) W
The average power is
P =
1
2
VmIm cos(θv − θi) =
1
2
120(10) cos[45◦
− (−10◦
)]
= 600 cos 55◦
= 344.2 W
which is the constant part of p(t) above.
P R A C T I C E P R O B L E M 1 1 . 1
Calculate the instantaneous power and average power absorbed by the
passive linear network of Fig. 11.1 if
v(t) = 80 cos(10t + 20◦
) V and i(t) = 15 sin(10t + 60◦
) A
Answer: 385.7 + 600 cos(20t − 10◦
) W, 385.7 W.
E X A M P L E 1 1 . 2
Calculate the average power absorbed by an impedance Z = 30 − j70 
when a voltage V = 120 0◦
is applied across it.
438 PART 2 AC Circuits
Solution:
The current through the impedance is
I =
V
Z
=
120 0◦
30 − j70
=
120 0◦
76.16 − 66.8◦
= 1.576 66.8◦
A
The average power is
P =
1
2
VmIm cos(θv − θi) =
1
2
(120)(1.576) cos(0 − 66.8◦
) = 37.24 W
P R A C T I C E P R O B L E M 1 1 . 2
A current I = 10 30◦
flows through an impedance Z = 20 − 22◦
.
Find the average power delivered to the impedance.
Answer: 927.2 W.
E X A M P L E 1 1 . 3
For the circuit shown in Fig. 11.3, find the average power supplied by the
source and the average power absorbed by the resistor.
4 Ω
+
−
I
−j2 Ω
5 30° V
Figure11.3 For Example 11.3.
Solution:
The current I is given by
I =
5 30◦
4 − j2
=
5 30◦
4.472 − 26.57◦
= 1.118 56.57◦
A
The average power supplied by the voltage source is
P =
1
2
(5)(1.118) cos(30◦
− 56.57◦
) = 2.5 W
The current through the resistor is
I = IR = 1.118 56.57◦
A
and the voltage across it is
VR = 4IR = 4.472 56.57◦
V
The average power absorbed by the resistor is
P =
1
2
(4.472)(1.118) = 2.5 W
which is the same as the average power supplied. Zero average power is
absorbed by the capacitor.
P R A C T I C E P R O B L E M 1 1 . 3
In the circuit of Fig. 11.4, calculate the average power absorbed by the
resistor and inductor. Find the average power supplied by the voltage
source.
3 Ω
+
− j1 Ω
8 45° V
Figure11.4 For Practice Prob. 11.3.
Answer: 9.6 W, 0 W, 9.6 W.
CHAPTER 11 AC Power Analysis 439
E X A M P L E 1 1 . 4
Determine the power generated by each source and the average power ab-
sorbed by each passive element in the circuit of Fig. 11.5(a).
20 Ω
+
−
j10 Ω
−j5 Ω
4 0° Α 60 30° V
1 3 5
4
2
(a)
20 Ω
+
−
j10 Ω
−j5 Ω
4 0° Α 60 30° V
(b)
+
−
+ −
V2
V1
I1 I2
Figure11.5 For Example 11.4.
Solution:
We apply mesh analysis as shown in Fig. 11.5(b). For mesh 1,
I1 = 4 A
For mesh 2,
(j10 − j5)I2 − j10I1 + 60 30◦
= 0, I1 = 4 A
or
j5I2 = −60 30◦
+ j40 ⇒ I2 = −12 − 60◦
+ 8
= 10.58 79.1◦
A
For the voltage source, the current flowing from it is I2 = 10.58 79.1◦
A
and the voltage across it is 60 30◦
V, so that the average power is
P5 =
1
2
(60)(10.58) cos(30◦
− 79.1◦
) = 207.8 W
Following the passive sign convention (see Fig. 1.8), this average power
is absorbed by the source, in view of the direction of I2 and the polarity
of the voltage source. That is, the circuit is delivering average power to
the voltage source.
For the current source, the current through it is I1 = 4 0◦
and the
voltage across it is
V1 = 20I1 + j10(I1 − I2) = 80 + j10(4 − 2 − j10.39)
= 183.9 + j20 = 184.984 6.21◦
V
The average power supplied by the current source is
P1 = −
1
2
(184.984)(4) cos(6.21◦
− 0) = −367.8 W
It is negative according to the passive sign convention, meaning that the
current source is supplying power to the circuit.
For the resistor, the current through it is I1 = 4 0◦
and the voltage
across it is 20I1 = 80 0◦
, so that the power absorbed by the resistor is
P2 =
1
2
(80)(4) = 160 W
440 PART 2 AC Circuits
For the capacitor, the current through it is I2 = 10.58 79.1◦
and the
voltage across it is −j5I2 = (5 − 90◦
)(10.58 79.1◦
) =
52.9 79.1◦
− 90◦
. The average power absorbed by the capacitor is
P4 =
1
2
(52.9)(10.58) cos(−90◦
) = 0
For the inductor, the current through it is I1 − I2 = 2 − j10.39 =
10.58 − 79.1◦
. The voltage across it is j10(I1 − I2) =
105.8 − 79.1◦
+ 90◦
. Hence, the average power absorbed by the in-
ductor is
P3 =
1
2
(105.8)(10.58) cos 90◦
= 0
Notice that the inductor and the capacitor absorb zero average power
and that the total power supplied by the current source equals the power
absorbed by the resistor and the voltage source, or
P1 + P2 + P3 + P4 + P5 = −367.8 + 160 + 0 + 0 + 207.8 = 0
indicating that power is conserved.
P R A C T I C E P R O B L E M 1 1 . 4
Calculate the average power absorbed by each of the five elements in the
circuit of Fig. 11.6.
8 Ω
+
−
+
− −j2 Ω
j4 Ω
40 0° V 20 90° V
Figure11.6 For Practice Prob. 11.4.
Answer: 40-V Voltage source: −100 W; resistor: 100 W; others: 0 W.
11.3 MAXIMUM AVERAGE POWER TRANSFER
In Section 4.8 we solved the problem of maximizing the power deliv-
ered by a power-supplying resistive network to a load RL. Represent-
ing the circuit by its Thevenin equivalent, we proved that the maximum
power would be delivered to the load if the load resistance is equal to the
Thevenin resistance RL = RTh. We now extend that result to ac circuits.
Consider the circuit in Fig. 11.7, where an ac circuit is connected
to a load ZL and is represented by its Thevenin equivalent. The load
is usually represented by an impedance, which may model an electric
motor, an antenna, a TV, and so forth. In rectangular form, the Thevenin
impedance ZTh and the load impedance ZL are
ZTh = RTh + jXTh (11.13a)
ZL = RL + jXL (11.13b)
CHAPTER 11 AC Power Analysis 441
The current through the load is
I =
VTh
ZTh + ZL
=
VTh
(RTh + jXTh) + (RL + jXL)
(11.14)
From Eq. (11.11), the average power delivered to the load is
P =
1
2
|I|2
RL =
|VTh|2
RL/2
(RTh + RL)2 + (XTh + XL)2
(11.15)
Our objective is to adjust the load parameters RL and XL so that P is
maximum. To do this we set ∂P/∂RL and ∂P/∂XL equal to zero. From
Eq. (11.15), we obtain
∂P
∂XL
= −
|VTh|2
RL(XTh + XL)
[(RTh + RL)2 + (XTh + XL)2]2
(11.16a)
∂P
∂RL
=
|VTh|2
[(RTh + RL)2
+ (XTh + XL)2
− 2RL(RTh + RL)]
2[(RTh + RL)2 + (XTh + XL)2]2
(11.16b)
Setting ∂P/∂XL to zero gives
XL = −XTh (11.17)
and setting ∂P/∂RL to zero results in
RL =

R2
Th + (XTh + XL)2 (11.18)
Combining Eqs. (11.17) and (11.18) leads to the conclusion that for max-
imum average power transfer, ZL must be selected so that XL = −XTh
and RL = RTh, i.e.,
ZL = RL + jXL = RTh − jXTh = Z∗
Th (11.19)
I
ZL
(a)
VTh
ZTh
(b)
ZL
+
−
Linear
circuit
Figure11.7 Finding the
maximum average power transfer:
(a) circuit with a load, (b) the
Thevenin equivalent.
When ZL =Z*
Th,wesaythattheloadismatched
to the source.
For maximum average power transfer, the load impedance ZL must be equal to the
complex conjugate of the Thevenin impedance ZTh.
This result is known as the maximum average power transfer theorem for
the sinusoidal steady state. Setting RL = RTh and XL = −XTh in Eq.
(11.15) gives us the maximum average power as
Pmax =
|VTh|2
8RTh
(11.20)
In a situation in which the load is purely real, the condition for
maximum power transfer is obtained from Eq. (11.18) by setting XL = 0;
that is,
RL =

R2
Th + X2
Th = |ZTh| (11.21)
This means that for maximum average power transfer to a purely resistive
load, the load impedance (or resistance) is equal to the magnitude of the
Thevenin impedance.
442 PART 2 AC Circuits
E X A M P L E 1 1 . 5
Determine the load impedance ZL that maximizes the average power
drawn from the circuit of Fig. 11.8. What is the maximum average power?
4 Ω
8 Ω
+
−
−j6 Ω
j5 Ω
10 0° V ZL
Figure11.8 For Example 11.5.
Solution:
First we obtain the Thevenin equivalent at the load terminals. To get ZTh,
consider the circuit shown in Fig. 11.9(a). We find
ZTh = j5 + 4 (8 − j6) = j5 +
4(8 − j6)
4 + 8 − j6
= 2.933 + j4.467 
To find VTh, consider the circuit in Fig. 11.8(b). By voltage division,
VTh =
8 − j6
4 + 8 − j6
(10) = 7.454 − 10.3◦
V
The load impedance draws the maximum power from the circuit when
ZL = Z∗
Th = 2.933 − j4.467 
According to Eq. (11.20), the maximum average power is
Pmax =
|VTh|2
8RTh
=
(7.454)2
8(2.933)
= 2.368 W
4 Ω
8 Ω
−j6 Ω
j5 Ω
10 V
ZTh
(a)
4 Ω
8 Ω
−j6 Ω
j5 Ω
VTh
(b)
+
−
+
−
Figure11.9 Finding the Thevenin equivalent of the circuit in Fig. 11.8.
P R A C T I C E P R O B L E M 1 1 . 5
For the circuit shown in Fig. 11.10, find the load impedance ZL that ab-
sorbs the maximum average power. Calculate that maximum average
power.
5 Ω
8 Ω
−j4 Ω j10 Ω
ZL
2 A
Figure11.10 For Practice Prob. 11.5.
Answer: 3.415 − j0.7317 , 1.429 W.
E X A M P L E 1 1 . 6
In the circuit in Fig. 11.11, find the value of RL that will absorb the max-
imum average power. Calculate that power.
CHAPTER 11 AC Power Analysis 443
Solution:
We first find the Thevenin equivalent at the terminals of RL.
ZTh = (40 − j30) j20 =
j20(40 − j30)
j20 + 40 − j30
= 9.412 + j22.35 
By voltage division,
VTh =
j20
j20 + 40 − j30
(150 30◦
) = 72.76 134◦
V
The value of RL that will absorb the maximum average power is
RL = |ZTh| =

9.4122 + 22.352 = 24.25 
The current through the load is
I =
VTh
ZTh + RL
=
72.76 134◦
33.39 + j22.35
= 1.8 100.2◦
A
The maximum average power absorbed by RL is
Pmax =
1
2
|I|2
RL =
1
2
(1.8)2
(24.25) = 39.29 W
40 Ω
+
− j20 Ω
−j30 Ω
150 30° V RL
Figure11.11 For Example 11.6.
P R A C T I C E P R O B L E M 1 1 . 6
In Fig. 11.12, the resistor RL is adjusted until it absorbs the maximum
average power. Calculate RL and the maximum average power absorbed
by it.
80 Ω
+
− 90 Ω
j60 Ω
120 60° V RL
−j30 Ω
Figure11.12 For Practice Prob. 11.6.
Answer: 30 , 9.883 W.
11.4 EFFECTIVE OR RMS VALUE
The idea of effective value arises from the need to measure the effec-
tiveness of a voltage or current source in delivering power to a resistive
load.
The effective value of a periodic current is the dc current that delivers the same
average power to a resistor as the periodic current.
444 PART 2 AC Circuits
In Fig. 11.13, the circuit in (a) is ac while that of (b) is dc. Our objective
is to find Ieff that will transfer the same power to resistor R as the sinusoid
i. The average power absorbed by the resistor in the ac circuit is
P =
1
T
 T
0
i2
R dt =
R
T
 T
0
i2
dt (11.22)
while the power absorbed by the resistor in the dc circuit is
P = I2
effR (11.23)
Equating the expressions in Eqs. (11.22) and (11.23) and solving for Ieff,
we obtain
Ieff =

1
T
 T
0
i2 dt (11.24)
The effective value of the voltage is found in the same way as current;
that is,
Veff =

1
T
 T
0
v2 dt (11.25)
This indicates that the effective value is the (square) root of the mean (or
average) of the square of the periodic signal. Thus, the effective value is
often known as the root-mean-square value, or rms value for short; and
we write
Ieff = Irms, Veff = Vrms (11.26)
For any periodic function x(t) in general, the rms value is given by
Xrms =

1
T
 T
0
x2 dt (11.27)
R
+
−
i(t)
v(t)
(a)
R
Ieff
Veff
(b)
+
−
Figure11.13 Finding the
effective current: (a) ac circuit,
(b) dc circuit.
The effective value of a periodic signal is its root mean square (rms) value.
Equation 11.27 states that to find the rms value of x(t), we first find
its square x2
and then find the mean of that, or
1
T
 T
0
x2
dt
and then the square root (
√
) of that mean. The rms value of a
constant is the constant itself. For the sinusoid i(t) = Im cos ωt, the
effective or rms value is
Irms =

1
T
 T
0
I2
m cos2 ωt dt
=

I2
m
T
 T
0
1
2
(1 + cos 2ωt) dt =
Im
√
2
(11.28)
CHAPTER 11 AC Power Analysis 445
Similarly, for v(t) = Vm cos ωt,
Vrms =
Vm
√
2
(11.29)
Keep in mind that Eqs. (11.28) and (11.29) are only valid for sinusoidal
signals.
The average power in Eq. (11.8) can be written in terms of the rms
values.
P =
1
2
VmIm cos(θv − θi) =
Vm
√
2
Im
√
2
cos(θv − θi)
= VrmsIrms cos(θv − θi)
(11.30)
Similarly, the average power absorbed by a resistor R in Eq. (11.11) can
be written as
P = I2
rmsR =
V 2
rms
R
(11.31)
When a sinusoidal voltage or current is specified, it is often in terms
of its maximum (or peak) value or its rms value, since its average value
is zero. The power industries specify phasor magnitudes in terms of their
rms values rather than peak values. For instance, the 110 V available at
every household is the rms value of the voltage from the power company.
It is convenient in power analysis to express voltage and current in their
rms values. Also, analog voltmeters and ammeters are designed to read
directly the rms value of voltage and current, respectively.
E X A M P L E 1 1 . 7
Determine the rms value of the current waveform in Fig. 11.14. If the
current is passed through a 2- resistor, find the average power absorbed
by the resistor.
0
t
10
−10
i(t)
2 4 6 8 10
Figure11.14 For Example 11.7.
Solution:
The period of the waveform is T = 4. Over a period, we can write the
current waveform as
i(t) =

5t, 0  t  2
−10, 2  t  4
The rms value is
Irms =

1
T
 T
0
i2 dt =

1
4
 2
0
(5t)2 dt +
 4
2
(−10)2 dt
=
1
4
25
t3
3




2
0
+ 100t




4
2

=

1
4

200
3
+ 200

= 8.165 A
The power absorbed by a 2- resistor is
P = I2
rmsR = (8.165)2
(2) = 133.3 W
446 PART 2 AC Circuits
P R A C T I C E P R O B L E M 1 1 . 7
Find the rms value of the current waveform of Fig. 11.15. If the current
flows through a 9- resistor, calculate the average power absorbed by the
resistor.
2 3
1
0 4 5 6 t
4
i(t)
Figure11.15 For Practice Prob. 11.7.
Answer: 2.309 A, 48 W.
E X A M P L E 1 1 . 8
The waveform shown in Fig. 11.16 is a half-wave rectified sine wave.
Find the rms value and the amount of average power dissipated in a 10-
resistor.
0 t
10
v(t)
p 2p 3p
Figure11.16 For Example 11.8.
Solution:
The period of the voltage waveform is T = 2π, and
v(t) =

10 sin t, 0  t  π
0, π  t  2π
The rms value is obtained as
V 2
rms =
1
T
 T
0
v2
(t) dt =
1
2π
 π
0
(10 sin t)2
dt +
 2π
π
02
dt
But sin2
t = 1
2
(1 − cos 2t). Hence
V 2
rms =
1
2π
 π
0
100
2
(1 − cos 2t) dt =
50
2π

t −
sin 2t
2




π
0
=
50
2π

π −
1
2
sin 2π − 0

= 25, Vrms = 5 V
The average power absorbed is
P =
V 2
rms
R
=
52
10
= 2.5 W
P R A C T I C E P R O B L E M 1 1 . 8
Find the rms value of the full-wave rectified sine wave in Fig. 11.17. Cal-
culate the average power dissipated in a 6- resistor.
0 t
8
v(t)
p 2p 3p
Figure11.17 For Practice Prob. 11.8.
Answer: 5.657 V, 5.334 W.
CHAPTER 11 AC Power Analysis 447
11.5 APPARENT POWER AND POWER FACTOR
In Section 11.2 we see that if the voltage and current at the terminals of
a circuit are
v(t) = Vm cos(ωt + θv) and i(t) = Im cos(ωt + θi) (11.32)
or, in phasor form, V = Vm θv and I = Im θi, the average power is
P =
1
2
VmIm cos(θv − θi) (11.33)
In Section 11.4, we saw that
P = VrmsIrms cos(θv − θi) = S cos(θv − θi) (11.34)
We have added a new term to the equation:
S = VrmsIrms (11.35)
The average power is a product of two terms. The product VrmsIrms is
known as the apparent power S. The factor cos(θv − θi) is called the
power factor (pf).
The apparent power (in VA) is the product of the rms values of voltage and current.
The apparent power is so called because it seems apparent that the power
should be the voltage-current product, by analogy with dc resistive cir-
cuits. It is measured in volt-amperes or VA to distinguish it from the
average or real power, which is measured in watts. The power factor is
dimensionless, since it is the ratio of the average power to the apparent
power,
pf =
P
S
= cos(θv − θi) (11.36)
The angle θv − θi is called the power factor angle, since it is the
angle whose cosine is the power factor. The power factor angle is equal
to the angle of the load impedance if V is the voltage across the load and
I is the current through it. This is evident from the fact that
Z =
V
I
=
Vm θv
Im θi
=
Vm
Im
θv − θi (11.37)
Alternatively, since
Vrms =
V
√
2
= Vrms θv (11.38a)
and
Irms =
I
√
2
= Irms θi (11.38b)
the impedance is
Z =
V
I
=
Vrms
Irms
=
Vrms
Irms
θv − θi (11.39)
448 PART 2 AC Circuits
The power factor is the cosine of the phase difference between voltage and current.
It is also the cosine of the angle of the load impedance.
From Eq. (11.36), the power factor may also be
regardedastheratiooftherealpowerdissipated
in the load to the apparent power of the load.
From Eq. (11.36), the power factor may be seen as that factor by which the
apparent power must be multiplied to obtain the real or average power.
The value of pf ranges between zero and unity. For a purely resistive
load, the voltage and current are in phase, so that θv − θi = 0 and pf
= 1. This implies that the apparent power is equal to the average power.
For a purely reactive load, θv − θi = ±90◦
and pf = 0. In this case the
average power is zero. In between these two extreme cases, pf is said
to be leading or lagging. Leading power factor means that current leads
voltage, which implies a capacitive load. Lagging power factor means
that current lags voltage, implying an inductive load. Power factor affects
the electric bills consumers pay the electric utility companies, as we will
see in Section 11.9.2.
E X A M P L E 1 1 . 9
A series-connected load draws a current i(t) = 4 cos(100πt + 10◦
) A
when the applied voltage is v(t) = 120 cos(100πt − 20◦
) V. Find the
apparent power and the power factor of the load. Determine the element
values that form the series-connected load.
Solution:
The apparent power is
S = VrmsIrms =
120
√
2
4
√
2
= 240 VA
The power factor is
pf = cos(θv − θi) = cos(−20◦
− 10◦
) = 0.866 (leading)
The pf is leading because the current leads the voltage. The pf may also
be obtained from the load impedance.
Z =
V
I
=
120 − 20◦
4 10◦
= 30 − 30◦
= 25.98 − j15 
pf = cos(−30◦
) = 0.866 (leading)
The load impedance Z can be modeled by a 25.98- resistor in series
with a capacitor with
XC = −15 = −
1
ωC
or
C =
1
15ω
=
1
15 × 100π
= 212.2 µF
P R A C T I C E P R O B L E M 1 1 . 9
Obtain the power factor and the apparent power of a load whose imped-
ance is Z = 60 + j40  when the applied voltage is v(t) =
150 cos(377t + 10◦
) V.
Answer: 0.832 lagging, 156 VA.
CHAPTER 11 AC Power Analysis 449
E X A M P L E 1 1 . 1 0
Determine the power factor of the entire circuit of Fig. 11.18 as seen by
the source. Calculate the average power delivered by the source.
6 Ω
4 Ω
+
−
30 0° V rms −j2 Ω
Figure11.18 For Example 11.10.
Solution:
The total impedance is
Z = 6 + 4 (−j2) = 6 +
−j2 × 4
4 − j2
= 6.8 − j1.6 = 7 − 13.24 
The power factor is
pf = cos(−13.24) = 0.9734 (leading)
since the impedance is capacitive. The rms value of the current is
Irms =
Vrms
Z
=
30 0◦
7 − 13.24◦
= 4.286 13.24◦
A
The average power supplied by the source is
P = VrmsIrms pf = (30)(4.286)0.9734 = 125 W
or
P = I2
rmsR = (4.286)2
(6.8) = 125 W
where R is the resistive part of Z.
P R A C T I C E P R O B L E M 1 1 . 1 0
Calculate the power factor of the entire circuit of Fig. 11.19 as seen by
the source. What is the average power supplied by the source?
10 Ω
+
−
8 Ω
j4 Ω −j6 Ω
40 0° V rms
Figure11.19 For Practice Prob. 11.10.
Answer: 0.936 lagging, 118 W.
11.6 COMPLEX POWER
Considerable effort has been expended over the years to express power
relations as simply as possible. Power engineers have coined the term
complex power, which they use to find the total effect of parallel loads.
Complex power is important in power analysis because it contains all the
information pertaining to the power absorbed by a given load.
V
I
+
−
Load
Z
Figure11.20 The
voltage and current
phasors associated
with a load.
Consider the ac load in Fig. 11.20. Given the phasor form V =
Vm θv and I = Im θi of voltage v(t) and current i(t), the complex
power S absorbed by the ac load is the product of the voltage and the
complex conjugate of the current, or
S =
1
2
VI∗
(11.40)
assuming the passive sign convention (see Fig. 11.20). In terms of the
rms values,
S = VrmsI∗
rms (11.41)
450 PART 2 AC Circuits
where
Vrms =
V
√
2
= Vrms θv (11.42)
and
Irms =
I
√
2
= Irms θi (11.43)
Thus we may write Eq. (11.41) as
S = VrmsIrms θv − θi
= VrmsIrms cos(θv − θi) + jVrmsIrms sin(θv − θi)
(11.44)
This equation can also be obtained from Eq. (11.9). We notice from Eq.
(11.44) that the magnitude of the complex power is the apparent power;
hence, the complex power is measured in volt-amperes (VA). Also, we
notice that the angle of the complex power is the power factor angle.
When working with the rms values of currents
or voltages, we may drop the subscript rms if no
confusion will be caused by doing so.
Thecomplexpowermaybeexpressedintermsoftheloadimpedance
Z. From Eq. (11.37), the load impedance Z may be written as
Z =
V
I
=
Vrms
Irms
=
Vrms
Irms
θv − θi (11.45)
Thus, Vrms = ZIrms. Substituting this into Eq. (11.41) gives
S = I2
rmsZ =
V 2
rms
Z∗
(11.46)
Since Z = R + jX, Eq. (11.46) becomes
S = I2
rms(R + jX) = P + jQ (11.47)
where P and Q are the real and imaginary parts of the complex power;
that is,
P = Re(S) = I2
rmsR (11.48)
Q = Im(S) = I2
rmsX (11.49)
P is the average or real power and it depends on the load’s resistance
R. Q depends on the load’s reactance X and is called the reactive (or
quadrature) power.
Comparing Eq. (11.44) with Eq. (11.47), we notice that
P = VrmsIrms cos(θv − θi), Q = VrmsIrms sin(θv − θi) (11.50)
The real power P is the average power in watts delivered to a load; it
is the only useful power. It is the actual power dissipated by the load.
The reactive power Q is a measure of the energy exchange between the
source and the reactive part of the load. The unit of Q is the volt-ampere
reactive(VAR)todistinguishitfromtherealpower, whoseunitisthewatt.
We know from Chapter 6 that energy storage elements neither dissipate
nor supply power, but exchange power back and forth with the rest of
the network. In the same way, the reactive power is being transferred
back and forth between the load and the source. It represents a lossless
interchange between the load and the source. Notice that:
CHAPTER 11 AC Power Analysis 451
1. Q = 0 for resistive loads (unity pf).
2. Q  0 for capacitive loads (leading pf).
3. Q  0 for inductive loads (lagging pf).
Thus,
Complex power (in VA) is the product of the rms voltage phasor and the
complex conjugate of the rms current phasor. As a complex quantity, its
real part is real power P and its imaginary part is reactive power Q.
Introducing the complex power enables us to obtain the real and reactive
powers directly from voltage and current phasors.
Complex Power = S = P + jQ =
1
2
VI∗
= VrmsIrms θv − θi
Apparent Power = S = |S| = VrmsIrms =

P 2 + Q2
Real Power = P = Re(S) = S cos(θv − θi)
Reactive Power = Q = Im(S) = S sin(θv − θi)
Power Factor =
P
S
= cos(θv − θi)
(11.51)
This shows how the complex power contains all the relevant power in-
formation in a given load. S contains all power information of a load. The
real part of S is the real power P; its imaginary
part is the reactive power Q; its magnitude is the
apparent power S; and the cosine of its phase
angle is the power factor pf.
It is a standard practice to represent S, P, and Q in the form of
a triangle, known as the power triangle, shown in Fig. 11.21(a). This
is similar to the impedance triangle showing the relationship between
Z, R, and X, illustrated in Fig. 11.21(b). The power triangle has four
items—the apparent/complex power, real power, reactive power, and the
power factor angle. Given two of these items, the other two can easily
be obtained from the triangle. As shown in Fig. 11.22, when S lies in the
first quadrant, we have an inductive load and a lagging pf. When S lies
in the fourth quadrant, the load is capacitive and the pf is leading. It is
also possible for the complex power to lie in the second or third quadrant.
This requires that the load impedance have a negative resistance, which
is possible with active circuits.
P Re
Im
S
S
+Q (lagging pf)
−Q (leading pf)
uv − ui
uv − ui
Figure11.22 Power triangle.
S Q
P
u
(a)
|Z| X
R
u
(b)
Figure 11.21 (a) Power triangle,
(b) impedance triangle.
452 PART 2 AC Circuits
E X A M P L E 1 1 . 1 1
The voltage across a load is v(t) = 60 cos(ωt − 10◦
) V and the cur-
rent through the element in the direction of the voltage drop is i(t) =
1.5 cos(ωt + 50◦
) A. Find: (a) the complex and apparent powers, (b) the
real and reactive powers, and (c) the power factor and the load impedance.
Solution:
(a) For the rms values of the voltage and current, we write
Vrms =
60
√
2
− 10◦
, Irms =
1.5
√
2
+ 50◦
The complex power is
S = VrmsI∗
rms =

60
√
2
− 10◦
 
1.5
√
2
− 50◦

= 45 − 60◦
VA
The apparent power is
S = |S| = 45 VA
(b) We can express the complex power in rectangular form as
S = 45 − 60◦
= 45[cos(−60◦
) + j sin(−60◦
)] = 22.5 − j38.97
Since S = P + jQ, the real power is
P = 22.5 W
while the reactive power is
Q = −38.97 VAR
(c) The power factor is
pf = cos(−60◦
) = 0.5 (leading)
It is leading, because the reactive power is negative. The load impedance
is
Z =
V
I
=
60 − 10◦
1.5 + 50◦
= 40 − 60◦

which is a capacitive impedance.
P R A C T I C E P R O B L E M 1 1 . 1 1
For a load, Vrms = 110 85◦
V, Irms = 0.4 15◦
A. Determine: (a) the
complex and apparent powers, (b) the real and reactive powers, and (c)
the power factor and the load impedance.
Answer: (a) 44 70◦
VA, 44 VA, (b) 15.05 W, 41.35 VAR,
(c) 0.342 lagging, 94.06 + j258.4 .
E X A M P L E 1 1 . 1 2
A load Z draws 12 kVA at a power factor of 0.856 lagging from a 120-V
rms sinusoidal source. Calculate: (a) the average and reactive powers
delivered to the load, (b) the peak current, and (c) the load impedance.
CHAPTER 11 AC Power Analysis 453
Solution:
(a) Given that pf = cos θ = 0.856, we obtain the power angle as θ =
cos−1
0.856 = 31.13◦
. If the apparent power is S = 12,000 VA, then the
average or real power is
P = S cos θ = 12,000 × 0.856 = 10.272 kW
while the reactive power is
Q = S sin θ = 12,000 × 0.517 = 6.204 kVA
(b) Since the pf is lagging, the complex power is
S = P + jQ = 10.272 + j6.204 kVA
From S = VrmsI∗
rms, we obtain
I∗
rms =
S
Vrms
=
10,272 + j6204
120 0◦
= 85.6 + j51.7 A = 100 31.13◦
A
Thus Irms = 100 − 31.13◦
and the peak current is
Im =
√
2Irms =
√
2(100) = 141.4 A
(c) The load impedance
Z =
Vrms
Irms
=
120 0◦
100 − 31.13◦
= 1.2 31.13◦

which is an inductive impedance.
P R A C T I C E P R O B L E M 1 1 . 1 2
A sinusoidal source supplies 10 kVA reactive power to load Z =
250 − 75◦
. Determine: (a) the power factor, (b) the apparent power
delivered to the load, and (c) the peak voltage.
Answer: (a) 0.2588 leading, (b) −10.35 kVAR, (c) 2.275 kV.
†11.7 CONSERVATION OF AC POWER
Infact, wealreadysawinExamples11.3and11.4
that average power is conserved in ac circuits.
The principle of conservation of power applies to ac circuits as well as to
dc circuits (see Section 1.5).
To see this, consider the circuit in Fig. 11.23(a), where two load
impedances Z1 and Z2 are connected in parallel across an ac source V.
KCL gives
I = I1 + I2 (11.52)
The complex power supplied by the source is
S =
1
2
VI∗
=
1
2
V(I∗
1 + I∗
2) =
1
2
VI∗
1 +
1
2
VI∗
2 = S1 + S2 (11.53)
where S1 and S2 denote the complex powers delivered to loads Z1 and
Z2, respectively.
454 PART 2 AC Circuits
(a) (b)
I
V Z2
Z1
Z2
Z1
+
−
I1 I2
I
V +
−
+ −
V1
+ −
V2
Figure 11.23 An ac voltage source supplied loads connected in:
(a) parallel, (b) series.
If the loads are connected in series with the voltage source, as shown
in Fig. 11.23(b), KVL yields
V = V1 + V2 (11.54)
The complex power supplied by the source is
S =
1
2
VI∗
=
1
2
(V1 + V2)I∗
=
1
2
V1I∗
+
1
2
V2I∗
= S1 + S2 (11.55)
where S1 and S2 denote the complex powers delivered to loads Z1 and
Z2, respectively.
We conclude from Eqs. (11.53) and (11.55) that whether the loads
are connected in series or in parallel (or in general), the total power
supplied by the source equals the total power delivered to the load. Thus,
in general, for a source connected to N loads,
S = S1 + S1 + · · · + SN (11.56)
This means that the total complex power in a network is the sum of the
complex powers of the individual components. (This is also true of real
power and reactive power, but not true of apparent power.) This expresses
the principle of conservation of ac power:
In fact, all forms of ac power are conserved: in-
stantaneous, real, reactive, and complex.
The complex, real, and reactive powers of the sources equal the respective sums
of the complex, real, and reactive powers of the individual loads.
From this we imply that the real (or reactive) power flow from sources in
a network equals the real (or reactive) power flow into the other elements
in the network.
E X A M P L E 1 1 . 1 3
Figure 11.24 shows a load being fed by a voltage source through a trans-
mission line. The impedance of the line is represented by the (4 + j2) 
impedance and a return path. Find the real power and reactive power
absorbed by: (a) the source, (b) the line, and (c) the load.
CHAPTER 11 AC Power Analysis 455
4 Ω
+
−
j2 Ω
220 0° V rms
−j10 Ω
15 Ω
I
Source Line Load
Figure11.24 For Example 11.13.
Solution:
The total impedance is
Z = (4 + j2) + (15 − j10) = 19 − j8 = 20.62 − 22.83◦

The current through the circuit is
I =
Vs
Z
=
220 0◦
20.62 − 22.83◦
= 10.67 22.83◦
A rms
(a) For the source, the complex power is
Ss = VsI∗
= (220 0◦
)(10.67 − 22.83◦
)
= 2347.4 − 22.83◦
= (2163.5 − j910.8) VA
From this, we obtain the real power as 2163.5 W and the reactive power
as 910.8 VAR (leading).
(b) For the line, the voltage is
Vline = (4 + j2)I = (4.472 26.57◦
)(10.67 22.83◦
)
= 47.72 49.4◦
V rms
The complex power absorbed by the line is
Sline = VlineI∗
= (47.72 49.4◦
)(10.67 − 22.83◦
)
= 509.2 26.57◦
= 455.4 + j227.7 VA
or
Sline = |I|2
Zline = (10.67)2
(4 + j2) = 455.4 + j227.7 VA
That is, the real power is 455.4 W and the reactive power is 227.76 VAR
(lagging).
(c) For the load, the voltage is
VL = (15 − j10)I = (18.03 − 33.7◦
)(10.67 22.83◦
)
= 192.38 − 10.87◦
V rms
The complex power absorbed by the load is
SL = VLI∗
= (192.38 − 10.87◦
)(10.67 − 22.83◦
)
= 2053 − 33.7◦
= (1708 − j1139) VA
456 PART 2 AC Circuits
The real power is 1708 W and the reactive power is 1139 VAR (leading).
Note that Ss = Sline + SL, as expected. We have used the rms values of
voltages and currents.
P R A C T I C E P R O B L E M 1 1 . 1 3
In the circuit in Fig. 11.25, the 60- resistor absorbs an average power
of 240 W. Find V and the complex power of each branch of the circuit.
What is the overall complex power of the circuit?
20 Ω
30 Ω
+
−
−j10 Ω
j20 Ω
V
60 Ω
Figure11.25 For Practice Prob. 11.13.
Answer: 240.67 21.45◦
V (rms); the 20- resistor: 656 VA; the
(30 − j10)  impedance: 480 − j160 VA; the (60 + j20)  impedance:
240 + j80 VA; overall: 1376 − j80 VA.
E X A M P L E 1 1 . 1 4
In the circuit of Fig. 11.26, Z1 = 60 − 30◦
 and Z2 = 40 45◦
.
Calculate the total: (a) apparent power, (b) real power, (c) reactive power,
and (d) pf.
It
Z2
Z1
+
−
I1 I2
120 10° V rms
Figure11.26 For Example 11.14.
Solution:
The current through Z1 is
I1 =
V
Z1
=
120 10◦
60 − 30◦
= 2 40◦
A rms
while the current through Z2 is
I2 =
V
Z2
=
120 10◦
40 45◦
= 3 − 35◦
A rms
The complex powers absorbed by the impedances are
S1 =
V 2
rms
Z∗
1
=
(120)2
60 30◦
= 240 − 30◦
= 207.85 − j120 VA
S2 =
V 2
rms
Z∗
2
=
(120)2
40 − 45◦
= 360 45◦
= 254.6 + j254.6 VA
The total complex power is
St = S1 + S2 = 462.4 + j134.6 VA
(a) The total apparent power is
|St | =
√
462.42 + 134.62 = 481.6 VA.
(b) The total real power is
Pt = Re(St ) = 462.4 W or Pt = P1 + P2.
(c) The total reactive power is
Qt = Im(St ) = 134.6 VAR or Qt = Q1 + Q2.
(d) The pf = Pt /|St | = 462.4/481.6 = 0.96 (lagging).
CHAPTER 11 AC Power Analysis 457
We may cross check the result by finding the complex power Ss supplied
by the source.
It = I1 + I2 = (1.532 + j1.286) + (2.457 − j1.721)
= 4 − j0.435 = 4.024 − 6.21◦
A rms
Ss = VI∗
t = (120 10◦
)(4.024 6.21◦
)
= 482.88 16.21◦
= 463 + j135 VA
which is the same as before.
P R A C T I C E P R O B L E M 1 1 . 1 4
Two loads connected in parallel are respectively 2 kW at a pf of 0.75 lead-
ing and 4 kW at a pf of 0.95 lagging. Calculate the pf of the two loads.
Find the complex power supplied by the source.
Answer: 0.9972 (leading), 6 − j0.4495 kVA.
11.8 POWER FACTOR CORRECTION
Most domestic loads (such as washing machines, air conditioners, and
refrigerators) and industrial loads (such as induction motors) are inductive
and operate at a low lagging power factor. Although the inductive nature
of the load cannot be changed, we can increase its power factor.
The process of increasing the power factor without altering the voltage or current
to the original load is known as power factor correction.
Alternatively, power factor correction may be
viewedastheadditionofareactiveelement(usu-
ally a capacitor) in parallel with the load in order
to make the power factor closer to unity.
An inductive load is modeled as a series combi-
nation of an inductor and a resistor.
Since most loads are inductive, as shown in Fig. 11.27(a), a load’s
powerfactorisimprovedorcorrectedbydeliberatelyinstallingacapacitor
in parallel with the load, as shown in Fig. 11.27(b). The effect of adding
the capacitor can be illustrated using either the power triangle or the
phasor diagram of the currents involved. Figure 11.28 shows the latter,
where it is assumed that the circuit in Fig. 11.27(a) has a power factor of
cos θ1, while the one in Fig. 11.27(b) has a power factor of cos θ2. It is
V
+
−
(a)
IL
Inductive
load
V
+
−
(b)
IL IC
Inductive
load
C
I
Figure11.27 Power factor correction: (a) original inductive load,
(b) inductive load with improved power factor.
V
IC
IC
IL
I
u1
u2
Figure11.28 Phasor diagram showing the
effect of adding a capacitor in parallel with
the inductive load.
458 PART 2 AC Circuits
evident from Fig. 11.28 that adding the capacitor has caused the phase
angle between the supplied voltage and current to reduce from θ1 to θ2,
thereby increasing the power factor. We also notice from the magnitudes
of the vectors in Fig. 11.28 that with the same supplied voltage, the circuit
in Fig. 11.27(a) draws larger current IL than the current I drawn by the
circuit in Fig. 11.27(b). Power companies charge more for larger currents,
because they result in increased power losses (by a squared factor, since
P = I2
LR). Therefore, it is beneficial to both the power company and the
consumer that every effort is made to minimize current level or keep the
power factor as close to unity as possible. By choosing a suitable size for
the capacitor, the current can be made to be completely in phase with the
voltage, implying unity power factor.
S1
S2
QC
Q2
Q1
u1
u2
P
Figure 11.29 Powertriangleillustratingpower
factor correction.
Wecanlookatthepowerfactorcorrectionfromanotherperspective.
Consider the power triangle in Fig. 11.29. If the original inductive load
has apparent power S1, then
P = S1 cos θ1, Q1 = S1 sin θ1 = P tan θ1 (11.57)
If we desire to increase the power factor from cos θ1 to cos θ2 without
altering the real power (i.e., P = S2 cos θ2), then the new reactive power
is
Q2 = P tan θ2 (11.58)
The reduction in the reactive power is caused by the shunt capacitor, that
is,
QC = Q1 − Q2 = P(tan θ1 − tan θ2) (11.59)
But from Eq. (11.49), QC = V 2
rms/XC = ωCV 2
rms. The value of the
required shunt capacitance C is determined as
C =
QC
ωV 2
rms
=
P(tan θ1 − tan θ2)
ωV 2
rms
(11.60)
Note that the real power P dissipated by the load is not affected by the
power factor correction because the average power due to the capacitance
is zero.
Although the most common situation in practice is that of an in-
ductive load, it is also possible that the load is capacitive, that is, the load
is operating at a leading power factor. In this case, an inductor should
be connected across the load for power factor correction. The required
shunt inductance L can be calculated from
QL =
V 2
rms
XL
=
V 2
rms
ωL
⇒ L =
V 2
rms
ωQL
(11.61)
where QL = Q1 − Q2, the difference between the new and old reactive
powers.
E X A M P L E 1 1 . 1 5
When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW
at a lagging power factor of 0.8. Find the value of capacitance necessary
to raise the pf to 0.95.
CHAPTER 11 AC Power Analysis 459
Solution:
If the pf = 0.8, then
cos θ1 = 0.8 ⇒ θ1 = 36.87◦
where θ1 is the phase difference between voltage and current. We obtain
the apparent power from the real power and the pf as
S1 =
P
cos θ1
=
4000
0.8
= 5000 VA
The reactive power is
Q1 = S1 sin θ = 5000 sin 36.87 = 3000 VAR
When the pf is raised to 0.95,
cos θ2 = 0.95 ⇒ θ2 = 18.19◦
The real power P has not changed. But the apparent power has changed;
its new value is
S2 =
P
cos θ2
=
4000
0.95
= 4210.5 VA
The new reactive power is
Q2 = S2 sin θ2 = 1314.4 VAR
The difference between the new and old reactive powers is due to the
parallel addition of the capacitor to the load. The reactive power due to
the capacitor is
QC = Q1 − Q2 = 3000 − 1314.4 = 1685.6 VAR
and
C =
QC
ωV 2
rms
=
1685.6
2π × 60 × 1202
= 310.5 µF
P R A C T I C E P R O B L E M 1 1 . 1 5
Find the value of parallel capacitance needed to correct a load of
140 kVAR at 0.85 lagging pf to unity pf. Assume that the load is supplied
by a 110-V (rms), 60-Hz line.
Answer: 30.69 mF.
Reactive power is measured by an instrument
called the varmeter. The varmeter is often con-
nected to the load in the same way as the
wattmeter.
†11.9 APPLICATIONS
In this section, we consider two important application areas: how power
is measured and how electric utility companies determine the cost of
electricity consumption.
11.9.1 Power Measurement
The average power absorbed by a load is measured by an instrument
called the wattmeter.
460 PART 2 AC Circuits
The wattmeter is the instrument for measuring the average power.
Figure 11.30 shows a wattmeter that consists essentially of two
coils: the current coil and the voltage coil. A current coil with very
low impedance (ideally zero) is connected in series with the load (Fig.
11.31) and responds to the load current. The voltage coil with very high
impedance(ideallyinfinite)isconnectedinparallelwiththeloadasshown
in Fig. 11.31 and responds to the load voltage. The current coil acts like
a short circuit because of its low impedance; the voltage coil behaves like
an open circuit because of its high impedance. As a result, the presence
of the wattmeter does not disturb the circuit or have an effect on the power
measurement.
Some wattmeters do not have coils; the watt-
meter considered here is the electromagnetic
type.
i
+
−
v
R
±
±
Figure11.30 A wattmeter.
i
i
+
−
v
Current coil
Voltage coil
±
±
ZL
Figure11.31 The wattmeter connected to the load.
When the two coils are energized, the mechanical inertia of the
moving system produces a deflection angle that is proportional to the
average value of the product v(t)i(t). If the current and voltage of the
load are v(t) = Vm cos(ωt + θv) and i(t) = Im cos(ωt + θi), their corre-
sponding rms phasors are
Vrms =
Vm
√
2
θv and Irms =
Im
√
2
θi (11.62)
and the wattmeter measures the average power given by
P = |Vrms||Irms| cos(θv − θi) =
1
2
VmIm cos(θv − θi) (11.63)
As shown in Fig. 11.31, each wattmeter coil has two terminals
with one marked ±. To ensure upscale deflection, the ± terminal of the
current coil is toward the source, while the ± terminal of the voltage coil
is connected to the same line as the current coil. Reversing both coil
connections still results in upscale deflection. However, reversing one
coil and not the other results in downscale deflection and no wattmeter
reading.
CHAPTER 11 AC Power Analysis 461
E X A M P L E 1 1 . 1 6
Find the wattmeter reading of the circuit in Fig. 11.32.
±
±
+
−
12 Ω
150 0° V rms
j10 Ω
−j6 Ω
8 Ω
Figure11.32 For Example 11.16.
Solution:
In Fig. 11.32, the wattmeter reads the average power absorbed by the
(8 − j6)  impedance because the current coil is in series with the
impedance while the voltage coil is in parallel with it. The current through
the circuit is
I =
150 0◦
(12 + j10) + (8 − j6)
=
150
20 + j4
A rms
The voltage across the (8 − j6)  impedance is
V = I(8 − j6) =
150(8 − j6)
20 + j4
V rms
The complex power is
S = VI∗
=
150(8 − j6)
20 + j4
·
150
20 − j4
=
1502
(8 − j6)
202 + 42
= 423.7 − j324.6 VA
The wattmeter reads
P = Re(S) = 432.7 W
P R A C T I C E P R O B L E M 1 1 . 1 6
For the circuit in Fig. 11.33, find the wattmeter reading.
±
±
+
−
4 Ω
120 30° V rms j9 Ω
−j2 Ω
12 Ω
Figure11.33 For Practice Prob. 11.16.
Answer: 1437 W.
462 PART 2 AC Circuits
11.9.2 Electricity Consumption Cost
In Section 1.7, we considered a simplified model of the way the cost of
electricity consumption is determined. But the concept of power factor
was not included in the calculations. Now we consider the importance of
power factor in electricity consumption cost.
Loads with low power factors are costly to serve because they re-
quire large currents, as explained in Section 11.8. The ideal situation
would be to draw minimum current from a supply so that S = P, Q = 0,
andpf = 1. AloadwithnonzeroQmeansthatenergyflowsforthandback
between the load and the source, giving rise to additional power losses.
In view of this, power companies often encourage their customers to have
power factors as close to unity as possible and penalize some customers
who do not improve their load power factors.
Utility companies divide their customers into categories: as resi-
dential(domestic), commercial, andindustrial, orassmallpower, medium
power, and large power. They have different rate structures for each
category. The amount of energy consumed in units of kilowatt-hours
(kWh) is measured using a kilowatt-hour meter installed at the customer’s
premises.
Although utility companies use different methods for charging cus-
tomers, the tariff or charge to a consumer is often two-part. The first part
is fixed and corresponds to the cost of generation, transmission, and dis-
tribution of electricity to meet the load requirements of the consumers.
This part of the tariff is generally expressed as a certain price per kW of
maximum demand. Or it may instead be based on kVA of maximum de-
mand, to account for the power factor (pf) of the consumer. A pf penalty
charge may be imposed on the consumer whereby a certain percentage of
kW or kVA maximum demand is charged for every 0.01 fall in pf below
a prescribed value, say 0.85 or 0.9. On the other hand, a pf credit may be
given for every 0.01 that the pf exceeds the prescribed value.
The second part is proportional to the energy consumed in kWh; it
may be in graded form, for example, the first 100 kWh at 16 cents/kWh,
the next 200 kWh at 10 cents/kWh and so forth. Thus, the bill is deter-
mined based on the following equation:
Total Cost = Fixed Cost + Cost of Energy (11.64)
E X A M P L E 1 1 . 1 7
A manufacturing industry consumes 200 MWh in one month. If the
maximum demand is 1600 kW, calculate the electricity bill based on the
following two-part rate:
Demand charge: $5.00 per month per kW of billing demand.
Energy charge: 8 cents per kWh for the first 50,000 kWh, 5 cents
per kWh for the remaining energy.
Solution:
The demand charge is
$5.00 × 1600 = $8000 (11.17.1)
The energy charge for the first 50,000 kWh is
CHAPTER 11 AC Power Analysis 463
$0.08 × 50,000 = $4000 (11.17.2)
The remaining energy is 200,000 kWh − 50,000 kWh = 150,000 kWh,
and the corresponding energy charge is
$0.05 × 150,000 = $7500 (11.17.3)
Adding Eqs. (11.17.1) to (11.17.3) gives
Total bill for the month = $8000 + $4000 + $7500 = $19,500
It may appear that the cost of electricity is too high. But this is often a
small fraction of the overall cost of production of the goods manufactured
or the selling price of the finished product.
P R A C T I C E P R O B L E M 1 1 . 1 7
The monthly reading of a paper mill’s meter is as follows:
Maximum demand: 32,000 kW
Energy consumed: 500 MWh
Using the two-part rate in Example 11.17, calculate the monthly bill for
the paper mill.
Answer: $186,500.
E X A M P L E 1 1 . 1 8
A 300-kW load supplied at 13 kV (rms) operates 520 hours a month at
80 percent power factor. Calculate the average cost per month based on
this simplified tariff:
Energy charge: 6 cents per kWh
Power-factor penalty: 0.1 percent of energy charge for every 0.01
that pf falls below 0.85.
Power-factor credit: 0.1 percent of energy charge for every 0.01
that pf exceeds 0.85.
Solution:
The energy consumed is
W = 300 kW × 520 h = 156,000 kWh
The operating power factor pf = 80% = 0.8 is 5 × 0.01 below the
prescribed power factor of 0.85. Since there is 0.1 percent energy charge
for every 0.01, there is a power-factor penalty charge of 0.5 percent. This
amounts to an energy charge of
%W = 156,000 ×
5 × 0.1
100
= 780 kWh
The total energy is
464 PART 2 AC Circuits
Wt = W + %W = 156,000 + 780 = 156,780 kWh
The cost per month is given by
Cost = 6 cents × Wt = $0.06 × 156,780 = $9406.80
P R A C T I C E P R O B L E M 1 1 . 1 8
An 800-kW induction furnace at 0.88 power factor operates 20 hours per
day for 26 days in a month. Determine the electricity bill per month based
on the tariff in Example 11.16.
Answer: $24,885.12.
11.10 SUMMARY
1. The instantaneous power absorbed by an element is the product of
the element’s terminal voltage and the current through the element:
p = vi.
2. Average or real power P (in watts) is the average of instantaneous
power p:
P =
1
T
 T
0
p dt
If v(t) = Vm cos(ωt + θv) and i(t) = Im cos(ωt + θi), then Vrms =
Vm/
√
2, Irms = Im/
√
2, and
P =
1
2
VmIm cos(θv − θi) = VrmsIrms cos(θv − θi)
Inductors and capacitors absorb no average power, while the aver-
age power absorbed by a resistor is 1/2 I2
mR = I2
rmsR.
3. Maximum average power is transferred to a load when the load
impedance is the complex conjugate of the Thevenin impedance as
seen from the load terminals, ZL = Z∗
Th.
4. The effective value of a periodic signal x(t) is its root-mean-square
(rms) value.
Xeff = Xrms =

1
T
 T
0
x2 dt
For a sinusoid, the effective or rms value is its amplitude divided by
√
2.
5. The power factor is the cosine of the phase difference between volt-
age and current:
pf = cos(θv − θi)
It is also the cosine of the angle of the load impedance or the ratio
of real power to apparent power. The pf is lagging if the current
lags voltage (inductive load) and is leading when the current leads
voltage (capacitive load).
CHAPTER 11 AC Power Analysis 465
6. Apparent power S (in VA) is the product of the rms values of volt-
age and current:
S = VrmsIrms
It is also given by S = |S| =

P 2 + Q2, where Q is reactive
power.
7. Reactive power (in VAR) is:
Q =
1
2
VmIm sin(θv − θi) = VrmsIrms sin(θv − θi)
8. Complex power S (in VA) is the product of the rms voltage phasor
and the complex conjugate of the rms current phasor. It is also the
complex sum of real power P and reactive power Q.
S = VrmsI∗
rms = VrmsIrms θv − θi = P + jQ
Also,
S = I2
rmsZ =
V 2
rms
Z∗
9. The total complex power in a network is the sum of the complex
powers of the individual components. Total real power and reactive
power are also, respectively, the sums of the individual real powers
and the reactive powers, but the total apparent power is not calcu-
lated by the process.
10. Power factor correction is necessary for economic reasons; it is the
process of improving the power factor of a load by reducing the
overall reactive power.
11. The wattmeter is the instrument for measuring the average power.
Energy consumed is measured with a kilowatt-hour meter.
REVIEW QUESTIONS
11.1 The average power absorbed by an inductor is zero.
(a) True (b) False
11.2 The Thevenin impedance of a network seen from the
load terminals is 80 + j55 . For maximum power
transfer, the load impedance must be:
(a) −80 + j55  (b) −80 − j55 
(c) 80 − j55  (d) 80 + j55 
11.3 The amplitude of the voltage available in the 60-Hz,
120-V power outlet in your home is:
(a) 110 V (b) 120 V
(c) 170 V (d) 210 V
11.4 If the load impedance is 20 − j20, the power factor
is
(a) − 45◦
(b) 0 (c) 1
(d) 0.7071 (e) noneofthese
11.5 A quantity that contains all the power information in
a given load is the
(a) power factor (b) apparent power
(c) average power (d) reactive power
(e) complex power
11.6 Reactive power is measured in:
(a) watts (b) VA
(c) VAR (d) none of these
11.7 In the power triangle shown in Fig. 11.34(a), the
reactive power is:
(a) 1000 VAR leading (b) 1000 VAR lagging
(c) 866 VAR leading (d) 866 VAR lagging
466 PART 2 AC Circuits
(a) (b)
60°
500 W
30°
1000 VAR
Figure 11.34 For Review Questions 11.7 and 11.8.
11.8 For the power triangle in Fig. 11.34(b), the apparent
power is:
(a) 2000 VA (b) 1000 VAR
(c) 866 VAR (d) 500 VAR
11.9 A source is connected to three loads Z1, Z2, and Z3
in parallel. Which of these is not true?
(a) P = P1 + P2 + P3 (b) Q = Q1 + Q2 + Q3
(c) S = S1 + S2 + S3 (d) S = S1 + S2 + S3
11.10 The instrument for measuring average power is the:
(a) voltmeter (b) ammeter
(c) wattmeter (d) varmeter
(e) kilowatt-hour meter
Answers: 11.1a, 11.2c, 11.3c, 11.4d, 11.5e, 11.6c, 11.7d, 11.8a,
11.9c, 11.10c.
PROBLEMS
Section 11.2 Instantaneous and Average Power
11.1 If v(t) = 160 cos 50tV and i(t) =
−20 sin(50t − 30◦
) A, calculate the instantaneous
power and the average power.
11.2 At t = 2 s, find the instantaneous power on each of
the elements in the circuit of Fig. 11.35.
+
−
30 cos 500t A 0.3 H
20 mF
200 Ω
Figure 11.35 For Prob. 11.2.
11.3 Refer to the circuit depicted in Fig. 11.36. Find the
average power absorbed by each element.
4 Ω
+
−
2 Ω
1 H 0.25 F
10 cos(2t + 30°) V
Figure 11.36 For Prob. 11.3.
11.4 Given the circuit in Fig. 11.37, find the average
power absorbed by each of the elements.
20 Ω
+
−
10 Ω
j5 Ω
−j10 Ω
50 0° V
Figure 11.37 For Prob. 11.4.
11.5 Compute the average power absorbed by the 4-
resistor in the circuit of Fig. 11.38.
4 Ω
j2 Ω
−j1 Ω
2 Ω
+ −
+ −
Vo
4 60° Α
4Vo
Figure 11.38 For Prob. 11.5.
11.6 Given the circuit of Fig. 11.39, find the average
power absorbed by the 10- resistor.
+
− 10 Ω
+
−
Io
8 20° V 0.1V
o
4 Ω
j5 Ω
−j5 Ω
8Io
+
−
Vo
Figure 11.39 For Prob. 11.6.
CHAPTER 11 AC Power Analysis 467
11.7 In the circuit of Fig. 11.40, determine the average
power absorbed by the 40- resistor.
Io
6 0° A j10 Ω 0.5Io 40 Ω
−j20 Ω
Figure 11.40 For Prob. 11.7.
11.8 Calculate the average power absorbed by each
resistor in the op amp circuit of Fig. 11.41 if the rms
value of vs is 2 V.
+
−
6 kΩ
2 kΩ
+
−
vs 10 kΩ
Figure 11.41 For Prob. 11.8.
11.9 In the op amp circuit in Fig. 11.42, find the total
average power absorbed by the resistors.
+
−
R
R
R
Vo cos vt V
+
−
+
−
Figure 11.42 For Prob. 11.9.
11.10 For the network in Fig. 11.43, assume that the port
impedance is
Zab =
R
√
1 + ω2R2C2
− tan−1
ωRC
Find the average power consumed by the network
when R = 10 k, C = 200 nF, and i =
2 sin(377t + 22◦
) mA.
v
i
+
−
Linear
network
a
b
Figure 11.43 For Prob. 11.10.
Section 11.3 Maximum Average Power
Transfer
11.11 For each of the circuits in Fig. 11.44, determine the
value of load Z for maximum power transfer and the
maximum average power transferred.
−j2 Ω
8 Ω 4 0° Α
Z
(a)
+
−
−j3 Ω
j2 Ω
4 Ω
5 Ω
10 30° V
Z
(b)
Figure 11.44 For Prob. 11.11.
11.12 For the circuit in Fig. 11.45, find:
(a) the value of the load impedance that absorbs the
maximum average power
(b) the value of the maximum average power
absorbed
−j40 Ω
j100 Ω
80 Ω
3 20° Α Load
Figure 11.45 For Prob. 11.12.
11.13 In the circuit of Fig. 11.46, find the value of ZL that
will absorb the maximum power and the value of the
maximum power.
j1 Ω
1 Ω −j1 Ω
12 0° V ZL
+
−
+
−
Vo 2Vo
Figure 11.46 For Prob. 11.13.
468 PART 2 AC Circuits
11.14 Calculate the value of ZL in the circuit of Fig. 11.47
in order for ZL to receive maximum average power.
What is the maximum average power received by Z?
−j10 Ω
j20 Ω
30 Ω
40 Ω
5 90° A
ZL
Figure 11.47 For Prob. 11.14.
11.15 Find the value of ZL in the circuit of Fig. 11.48 for
maximum power transfer.
40 Ω
40 Ω −j10 Ω
j20 Ω
+
−
80 Ω
60 0° V
5 0° A ZL
Figure 11.48 For Prob. 11.15.
11.16 The variable resistor R in the circuit of Fig. 11.49 is
adjusted until it absorbs the maximum average
power. Find R and the maximum average power
absorbed.
j1 Ω
−j2 Ω
3 Ω
6 Ω
4 0° A R
Figure 11.49 For Prob. 11.16.
11.17 The load resistance RL in Fig. 11.50 is adjusted until
it absorbs the maximum average power. Calculate
the value of RL and the maximum average power.
−j10 Ω
−j10 Ω
+
−
Io
120 0° V
40 Ω
j20 Ω
4Io
RL
+ −
Figure 11.50 For Prob. 11.17.
11.18 Assuming that the load impedance is to be purely
resistive, what load should be connected to terminals
a-b of the circuits in Fig. 11.51 so that the maximum
power is transferred to the load?
100 Ω
40 Ω
−j10 Ω
j30 Ω
50 Ω
+
−
120 60° V 2 90° A
a
b
Figure 11.51 For Prob. 11.18.
Section 11.4 Effective or RMS Value
11.19 Find the rms value of the periodic signal in Fig.
11.52.
2
0 4 6 8 t
15
5
v(t)
Figure 11.52 For Prob. 11.19.
11.20 Determine the rms value of the waveform in Fig.
11.53.
1
0
2 3 4 t
5
−5
v(t)
Figure 11.53 For Prob. 11.20.
11.21 Find the effective value of the voltage waveform in
Fig. 11.54.
2
0 4 6 8 10 t
5
10
v(t)
Figure 11.54 For Prob. 11.21.
11.22 Calculate the rms value of the current waveform of
Fig. 11.55.
CHAPTER 11 AC Power Analysis 469
5
0 10 15 20 25 t
5
i(t)
Figure 11.55 For Prob. 11.22.
11.23 Find the rms value of the voltage waveform of Fig.
11.56 as well as the average power absorbed by a
2- resistor when the voltage is applied across the
resistor.
2
0 5 7 10 12 t
8
v(t)
Figure 11.56 For Prob. 11.23.
11.24 Calculate the effective value of the current
waveform in Fig. 11.57 and the average power
delivered to a 12- resistor when the current runs
through the resistor.
5
0
15 25 t
10
−10
i(t)
10 20 30
Figure 11.57 For Prob. 11.24.
11.25 Compute the rms value of the waveform depicted in
Fig. 11.58.
2
−1
0
4 6 8 10 t
2
v(t)
Figure 11.58 For Prob. 11.25.
11.26 Obtain the rms value of the current waveform shown
in Fig. 11.59.
1
0 2 3 4 5 t
10
i(t)
10t2
Figure 11.59 For Prob. 11.26.
11.27 Determine the effective value of the periodic
waveform in Fig. 11.60.
1
0 2 3 4 5 t
10
i(t)
Figure 11.60 For Prob. 11.27.
11.28 One cycle of a periodic voltage waveform is depicted
in Fig. 11.61. Find the effective value of the voltage.
1
0 2 3 4 6
5 t
10
20
30
v(t)
Figure 11.61 For Prob. 11.28.
Section 11.5 Apparent Power and Power
Factor
11.29 A relay coil is connected to a 210-V, 50-Hz supply.
If it has a resistance of 30  and an inductance of
0.5 H, calculate the apparent power and the power
factor.
11.30 A certain load comprises 12 − j8  in parallel with
j4 . Determine the overall power factor.
11.31 Obtain the power factor for each of the circuits in
Fig. 11.62. Specify each power factor as leading or
lagging.
470 PART 2 AC Circuits
−j2 Ω
j2 Ω j1 Ω
−j1 Ω
1 Ω
−j2 Ω
(a)
4 Ω
(b)
j5 Ω
4 Ω
Figure 11.62 For Prob. 11.31.
Section 11.6 Complex Power
11.32 A load draws 5 kVAR at a power factor of 0.86
(leading) from a 220-V rms source. Calculate the
peak current and the apparent power supplied to the
load.
11.33 For the following voltage and current phasors,
calculate the complex power, apparent power, real
power, and reactive power. Specify whether the pf is
leading or lagging.
(a) V = 220 30◦
V rms, I = 0.5 60◦
A rms
(b) V = 250 − 10◦
V rms,
I = 6.2 − 25◦
A rms
(c) V = 120 0◦
V rms, I = 2.4 − 15◦
A rms
(d) V = 160 45◦
V rms, I = 8.5 90◦
A rms
11.34 For each of the following cases, find the complex
power, the average power, and the reactive power:
(a) v(t) = 112 cos(ωt + 10◦
) V,
i(t) = 4 cos(ωt − 50◦
) A
(b) v(t) = 160 cos 377t V,
i(t) = 4 cos(377t + 45◦
) A
(c) V = 80 60◦
V rms, Z = 50 30◦

(d) I = 10 60◦
V rms, Z = 100 45◦

11.35 Determine the complex power for the following
cases:
(a) P = 269 W, Q = 150 VAR (capacitive)
(b) Q = 2000 VAR, pf = 0.9 (leading)
(c) S = 600 VA, Q = 450 VAR (inductive)
(d) Vrms = 220 V, P = 1 kW,
|Z| = 40  (inductive)
11.36 Find the complex power for the following cases:
(a) P = 4 kW, pf = 0.86 (lagging)
(b) S = 2 kVA, P = 1.6 kW (capacitive)
(c) Vrms = 208 20◦
V, Irms = 6.5 − 50◦
A
(d) Vrms = 120 30◦
V, Z = 40 + j60 
11.37 Obtain the overall impedance for the following
cases:
(a) P = 1000 W, pf = 0.8 (leading),
Vrms = 220 V
(b) P = 1500 W, Q = 2000 VAR (inductive),
Irms = 12 A
(c) S = 4500 60◦
VA, V = 120 45◦
V
11.38 For the entire circuit in Fig. 11.63, calculate:
(a) the power factor
(b) the average power delivered by the source
(c) the reactive power
(d) the apparent power
(e) the complex power
2 Ω
10 Ω
+
−
−j5 Ω j6 Ω
8 Ω
16 45° V
Figure 11.63 For Prob. 11.38.
Section 11.7 Conservation of AC Power
11.39 For the network in Fig. 11.64, find the complex
power absorbed by each element.
4 Ω
+
−
−j3 Ω
j5 Ω
8 −20° V
Figure 11.64 For Prob. 11.39.
11.40 Find the complex power absorbed by each of the
five elements in the circuit of Fig. 11.65.
+
−
+
−
20 Ω
40 0° V rms
j10 Ω
−j20 Ω
50 90° V rms
Figure 11.65 For Prob. 11.40.
11.41 Obtain the complex power delivered by the source in
the circuit of Fig. 11.66.
CHAPTER 11 AC Power Analysis 471
−j2 Ω
j4 Ω
5 Ω
3 Ω
2 30° Α 6 Ω
Figure 11.66 For Prob. 11.41.
11.42 For the circuit in Fig. 11.67, find the average,
reactive, and complex power delivered by the
dependent voltage source.
4 Ω
1 Ω
−j1 Ω
j2 Ω
24 0° V
2 Ω
+
− 2Vo
+
−
Vo
Figure 11.67 For Prob. 11.42.
11.43 Obtain the complex power delivered to the 10-k
resistor in Fig. 11.68 below.
11.44 Calculate the reactive power in the inductor and
capacitor in the circuit of Fig. 11.69.
−j20 Ω
j30 Ω
50 Ω
240 0° V 4 0° A 40 Ω
+
−
Figure 11.69 For Prob. 11.44.
11.45 For the circuit in Fig. 11.70, find Vo and the input
power factor.
6 0° A rms
+
−
V
o
20 kW
0.8 pf lagging
16 kW
0.9 pf lagging
Figure 11.70 For Prob. 11.45.
11.46 Given the circuit in Fig. 11.71, find Io and the
overall complex power supplied.
100 90° V 2 kVA
0.707 pf leading
1.2 kW
0.8 kVAR (cap)
4 kW
0.9 pf lagging
Io
+
−
Figure 11.71 For Prob. 11.46.
11.47 For the circuit in Fig. 11.72, find Vs.
V
s
+
−
120 V rms
10 W
0.9 pf lagging
15 W
0.8 pf leading
0.2 Ω 0.3 Ω
j0.04 Ω j0.15 Ω
+
−
Figure 11.72 For Prob. 11.47.
11.48 Find Io in the circuit of Fig. 11.73 on the bottom of
the next page.
11.49 In the op amp circuit of Fig. 11.74, vs =
4 cos 104
t V. Find the average power delivered to the
50-k resistor.
+
−
100 kΩ
+
−
vs 50 kΩ
1 nF
Figure 11.74 For Prob. 11.49.
11.50 Obtain the average power absorbed by the 6-k
resistor in the op amp circuit in Fig. 11.75.
+
− 10 kΩ
Io
0.2 0° V rms
500 Ω −j3 kΩ j1 kΩ
20Io 4 kΩ
Figure11.68 For Prob. 11.43.
472 PART 2 AC Circuits
+
−
4 kΩ j3 kΩ
−j2 kΩ
j4 kΩ
+
−
6 kΩ
2 kΩ
4 45° V
Figure 11.75 For Prob. 11.50.
11.51 Calculate the complex power delivered to each
resistor and capacitor in the op amp circuit of Fig.
11.76. Let vs = 2 cos 103
t V.
+
−
10 kΩ
+
− 40 kΩ
20 kΩ
vs 0.1 mF
0.2 mF
Figure 11.76 For Prob. 11.51.
11.52 Compute the complex power supplied by the current
source in the series RLC circuit in Fig. 11.77.
L
R
Io cos vt C
Figure 11.77 For Prob. 11.52.
Section 11.8 Power Factor Correction
11.53 Refer to the circuit shown in Fig. 11.78.
(a) What is the power factor?
(b) What is the average power dissipated?
(c) What is the value of the capacitance that will
give a unity power factor when connected to the
load?
Z = 10 + j12 Ω
+
− C
120 V
60 Hz
Figure 11.78 For Prob. 11.53.
11.54 An 880-VA, 220-V, 50-Hz load has a power factor of
0.8 lagging. What value of parallel capacitance will
correct the load power factor to unity?
11.55 An 40-kW induction motor, with a lagging power
factor of 0.76, is supplied by a 120-V rms 60-Hz
sinusoidal voltage source. Find the capacitance
needed in parallel with the motor to raise the power
factor to:
(a) 0.9 lagging (b) 1.0.
11.56 A 240-V rms 60-Hz supply serves a load that is
10 kW (resistive), 15 kVAR (capacitive), and
22 kVAR (inductive). Find:
(a) the apparent power
(b) the current drawn from the supply
(c) the kVAR rating and capacitance required to
improve the power factor to 0.96 lagging
(d) the current drawn from the supply under the new
power-factor conditions
11.57 A 120-V rms 60-Hz source supplies two loads
connected in parallel, as shown in Fig. 11.79.
(a) Find the power factor of the parallel
combination.
(b) Calculate the value of the capacitance connected
in parallel that will raise the power factor to
unity.
220 0° V
12 kW
0.866 pf leading
20 kVAR
0.6 pf lagging
16 kW
0.85 pf lagging
Io
+
−
Figure11.73 For Prob. 11.48.
CHAPTER 11 AC Power Analysis 473
Load 1
24 kW
pf = 0.8
lagging
Load 2
40 kW
pf = 0.95
lagging
Figure 11.79 For Prob. 11.57.
11.58 Consider the power system shown in Fig. 11.80.
Calculate:
(a) the total complex power
(b) the power factor
(c) the capacitance necessary to establish a unity
power factor
+
−
240 V rms, 50 Hz
80 − j50 Ω
120 + j70 Ω
60 + j0
Figure 11.80 For Prob. 11.58.
Section 11.9 Applications
11.59 Obtain the wattmeter reading of the circuit in Fig.
11.81 below.
11.60 What is the reading of the wattmeter in the network
of Fig. 11.82?
6 Ω 4 H
15 Ω
0.1 F
+
−
120 cos 2t V
±
±
Figure 11.82 For Prob. 11.60.
11.61 Find the wattmeter reading of the circuit shown in
Fig. 11.83 below.
11.62 The circuit of Fig. 11.84 portrays a wattmeter
connected into an ac network.
(a) Find the load current.
(b) Calculate the wattmeter reading.
ZL = 6.4 Ω
pf = 0.825
+
−
110 V
WM
Figure 11.84 For Prob. 11.62.
11.63 The kilowatthour-meter of a home is read once a
month. For a particular month, the previous and
present readings are as follows:
Previous reading: 3246 kWh
Present reading: 4017 kWh
Calculate the electricity bill for that month
based on the following residential rate schedule:
4 Ω −j3 Ω
j2 Ω 8 Ω
+
−
12 0° V 3 30° A
±
±
Figure11.81 For Prob. 11.59.
10 Ω
4 Ω
+
−
5 Ω 1 H
20 cos 4t V
±
±
F
1
12
Figure11.83 For Prob. 11.61.
474 PART 2 AC Circuits
Minimum monthly charge—$12.00
First 100 kWh per month at 16 cents/kWh
Next 200 kWh per month at 10 cents/kWh
Over 300 kWh per month at 6 cents/kWh
11.64 A consumer has an annual consumption of
1200 MWh with a maximum demand of 2.4 MVA.
The maximum demand charge is $30 per kVA per
annum, and the energy charge per kWh is 4 cents.
(a) Determine the annual cost of energy.
(b) Calculate the charge per kWh with a flat-rate
tariff if the revenue to the utility company is to
remain the same as for the two-part tariff.
COMPREHENSIVE PROBLEMS
11.65 A transmitter delivers maximum power to an
antenna when the antenna is adjusted to represent a
load of 75- resistance in series with an inductance
of 4 µH. If the transmitter operates at 4.12 MHz,
find its internal impedance.
11.66 In a TV transmitter, a series circuit has an
impedance of 3 k and a total current of 50 mA. If
the voltage across the resistor is 80 V, what is the
power factor of the circuit?
11.67 A certain electronic circuit is connected to a 110-V
ac line. The root-mean-square value of the current
drawn is 2 A, with a phase angle of 55◦
.
(a) Find the true power drawn by the circuit.
(b) Calculate the apparent power.
11.68 An industrial heater has a nameplate which reads:
210 V 60 Hz 12 kVA 0.78 pf lagging
Determine:
(a) the apparent and the complex power
(b) the impedance of the heater
11.69
∗
A 2000-kW turbine-generator of 0.85 power factor
operates at the rated load. An additional load of
300 kW at 0.8 power factor is added. What kVAR of
capacitors is required to operate the turbine
-generator but keep it from being overloaded?
11.70 The nameplate of an electric motor has the
following information:
Line voltage: 220 V rms
Line current: 15 A rms
Line frequency: 60 Hz
Power: 2700 W
Determine the power factor (lagging) of the motor.
Find the value of the capacitance C that must be
connected across the motor to raise the pf to unity.
11.71 As shown in Fig. 11.85, a 550-V feeder line supplies
an industrial plant consisting of a motor drawing
60 kW at 0.75 pf (inductive), a capacitor with a
rating of 20 kVAR, and lighting drawing 20 kW.
(a) Calculate the total reactive power and apparent
power absorbed by the plant.
*An asterisk indicates a challenging problem.
(b) Determine the overall pf.
(c) Find the current in the feeder line.
60 kW
pf = 0.75
550 V 20 kVAR 10 kW
+
−
Figure 11.85 For Prob. 11.71.
11.72 A factory has the following four major loads:
• A motor rated at 5 hp, 0.8 pf lagging
(1 hp = 0.7457 kW).
• A heater rated at 1.2 kW, 1.0 pf.
• Ten 120-W lightbulbs.
• A synchronous motor rated at 1.6 kVA, 0.6 pf
leading.
(a) Calculate the total real and reactive power.
(b) Find the overall power factor.
11.73 A 1-MVA substation operates at full load at 0.7
power factor. It is desired to improve the power
factor to 0.95 by installing capacitors. Assume that
new substation and distribution facilities cost $120
per kVA installed, and capacitors cost $30 per kVA
installed.
(a) Calculate the cost of capacitors needed.
(b) Find the savings in substation capacity released.
(c) Are capacitors economical for releasing the
amount of substation capacity?
11.74 A coupling capacitor is used to block dc current
from an amplifier as shown in Fig. 11.86(a). The
amplifier and the capacitor act as the source, while
the speaker is the load as in Fig. 11.86(b).
(a) At what frequency is maximum power
transferred to the speaker?
(b) If Vs = 4.6 V rms, how much power is delivered
to the speaker at that frequency?
CHAPTER 11 AC Power Analysis 475
10 Ω 40 nF
80 mH
4 Ω
vs
Amplifier
Vin Speaker
Coupling capacitor
Amplifier Speaker
(a)
(b)
Figure 11.86 For Prob. 11.74.
11.75 A power amplifier has an output impedance of
40 + j8 . It produces a no-load output voltage of
146 V at 300 Hz.
(a) Determine the impedance of the load that
achieves maximum power transfer.
(b) Calculate the load power under this matching
condition.
11.76 A power transmission system is modeled as shown
in Fig. 11.87. If Vs = 240 0◦
rms, find the average
power absorbed by the load.
0.1 Ω
0.1 Ω
j1 Ω
j1 Ω
+
−
j1 Ω
100 Ω
Vs
Source Line Load
Figure 11.87 For Prob. 11.76.
477
C H A P T E R
THREE-PHASE CIRCUITS
1 2
Society is never prepared to receive any invention. Every new thing is
resisted, and it takes years for the inventor to get people to listen to him
and years more before it can be introduced.
—Thomas Alva Edison
Historical Profiles
Thomas Alva Edison (1847–1931) was perhaps the greatest American inventor. He
patented 1093 inventions, including such history-making inventions as the incandescent
electric bulb, the phonograph, and the first commercial motion pictures.
Born in Milan, Ohio, the youngest of seven children, Edison received only three
months of formal education because he hated school. He was home-schooled by his
mother and quickly began to read on his own. In 1868, Edison read one of Faraday’s
books and found his calling. He moved to Menlo Park, New Jersey, in 1876, where he
managed a well-staffed research laboratory. Most of his inventions came out of this lab-
oratory. His laboratory served as a model for modern research organizations. Because
of his diverse interests and the overwhelming number of his inventions and patents,
Edison began to establish manufacturing companies for making the devices he invented.
He designed the first electric power station to supply electric light. Formal electri-
cal engineering education began in the mid-1880s with Edison as a role model and leader.
Nikola Tesla (1856–1943) was a Croatian-American engineer whose inventions—
among them the induction motor and the first polyphase ac power system—greatly
influenced the settlement of the ac versus dc debate in favor of ac. He was also re-
sponsible for the adoption of 60 Hz as the standard for ac power systems in the United
States.
Born in Austria-Hungary (now Croatia), to a clergyman, Tesla had an incredible
memory and a keen affinity for mathematics. He moved to the United States in 1884
and first worked for Thomas Edison. At that time, the country was in the “battle of the
currents” with George Westinghouse (1846–1914) promoting ac and Thomas Edison
rigidly leading the dc forces. Tesla left Edison and joined Westinghouse because of
his interest in ac. Through Westinghouse, Tesla gained the reputation and acceptance
of his polyphase ac generation, transmission, and distribution system. He held 700
patents in his lifetime. His other inventions include high-voltage apparatus (the tesla
coil) and a wireless transmission system. The unit of magnetic flux density, the tesla,
was named in honor of him.
478 PART 2 AC Circuits
12.1 INTRODUCTION
So far in this text, we have dealt with single-phase circuits. A single-
phase ac power system consists of a generator connected through a pair
of wires (a transmission line) to a load. Figure 12.1(a) depicts a single-
phase two-wire system, where Vp is the magnitude of the source voltage
and φ is the phase. What is more common in practice is a single-phase
three-wiresystem, showninFig.12.1(b). Itcontainstwoidenticalsources
(equal magnitude and the same phase) which are connected to two loads
by two outer wires and the neutral. For example, the normal household
system is a single-phase three-wire system because the terminal voltages
have the same magnitude and the same phase. Such a system allows the
connection of both 120-V and 240-V appliances.
Historical note: Thomas Edison invented a three-
wire system, using three wires instead of four.
ZL
Vp
+
−
(a)
f
ZL1
Vp
a A
n N
b B
+
−
(b)
f
ZL2
Vp
+
−
f
Figure12.1 Single-phase systems: (a) two-wire type, (b) three-wire type.
Circuits or systems in which the ac sources operate at the same
frequencybutdifferentphasesareknownaspolyphase. Figure12.2shows
a two-phase three-wire system, and Fig. 12.3 shows a three-phase four-
wire system. As distinct from a single-phase system, a two-phase system
is produced by a generator consisting of two coils placed perpendicular
to each other so that the voltage generated by one lags the other by 90◦
.
By the same token, a three-phase system is produced by a generator
consisting of three sources having the same amplitude and frequency but
out of phase with each other by 120◦
. Since the three-phase system is by
far the most prevalent and most economical polyphase system, discussion
in this chapter is mainly on three-phase systems.
ZL1
Vp
a A
n N
b B
+
−
ZL2
+
−
−90°
Vp 0°
Figure12.2 Two-phase three-wire system.
ZL1
a A
b B
c C
n N
Vp 0°
− +
ZL2
Vp −120°
Vp +120°
− +
ZL3
− +
Figure12.3 Three-phase four-wire system.
Three-phase systems are important for at least three reasons. First,
nearly all electric power is generated and distributed in three-phase, at
the operating frequency of 60 Hz (or ω = 377 rad/s) in the United States
or 50 Hz (or ω = 314 rad/s) in some other parts of the world. When one-
phase or two-phase inputs are required, they are taken from the three-
phase system rather than generated independently. Even when more
than three phases are needed—such as in the aluminum industry, where
48 phases are required for melting purposes—they can be provided by
manipulating the three phases supplied. Second, the instantaneous power
in a three-phase system can be constant (not pulsating), as we will see
in Section 12.7. This results in uniform power transmission and less
vibration of three-phase machines. Third, for the same amount of power,
the three-phase system is more economical than the single-phase. The
CHAPTER 12 Three-Phase Circuits 479
amount of wire required for a three-phase system is less than that required
for an equivalent single-phase system.
We begin with a discussion of balanced three-phase voltages. Then
we analyze each of the four possible configurations of balanced three-
phase systems. We also discuss the analysis of unbalanced three-phase
systems. We learn how to use PSpice for Windows to analyze a balanced
or unbalanced three-phase system. Finally, we apply the concepts devel-
oped in this chapter to three-phase power measurement and residential
electrical wiring.
12.2 BALANCED THREE-PHASE VOLTAGES
Three-phase voltages are often produced with a three-phase ac generator
(or alternator) whose cross-sectional view is shown in Fig. 12.4. The gen-
erator basically consists of a rotating magnet (called the rotor) surrounded
by a stationary winding (called the stator). Three separate windings or
coils with terminals a-a
, b-b
, and c-c
are physically placed 120◦
apart
around the stator. Terminals a and a
, for example, stand for one of the
ends of coils going into and the other end coming out of the page. As the
rotor rotates, its magnetic field “cuts” the flux from the three coils and
induces voltages in the coils. Because the coils are placed 120◦
apart,
the induced voltages in the coils are equal in magnitude but out of phase
by 120◦
(Fig. 12.5). Since each coil can be regarded as a single-phase
generator by itself, the three-phase generator can supply power to both
single-phase and three-phase loads.
Stator
Three-
phase
output
a
b
c
n
c
N
S
b′
c′
b
a
a′
Rotor
Figure12.4 A three-phase generator.
0
120°
V
an
vt
V
bn V
cn
240°
Figure12.5 The generated voltages are 120◦
apart from each other.
A typical three-phase system consists of three voltage sources con-
nected to loads by three or four wires (or transmission lines). (Three-
phase current sources are very scarce.) A three-phase system is equiv-
alent to three single-phase circuits. The voltage sources can be either
wye-connected as shown in Fig. 12.6(a) or delta-connected as in Fig.
12.6(b).
Let us consider the wye-connected voltages in Fig. 12.6(a) for now.
The voltages Van, Vbn, and Vcn are respectively between lines a, b, and
480 PART 2 AC Circuits
+
−
+
−
+
−
(a)
a
V
an
V
bn
V
cn
V
ca V
ab
V
bc
n
b
c
+
−
+
−
(b)
a
b
c
− +
Figure 12.6 Three-phase voltage sources: (a) Y-connected source,
(b) -connected source.
c, and the neutral line n. These voltages are called phase voltages. If the
voltage sources have the same amplitude and frequency ω and are out of
phase with each other by 120◦
, the voltages are said to be balanced. This
implies that
Van + Vbn + Vcn = 0 (12.1)
|Van| = |Vbn| = |Vcn| (12.2)
Thus,
Balanced phase voltages are equal in magnitude and are out
of phase with each other by 120◦
.
120°
Vcn
Van
Vbn
120°
−120°
(a)
120°
Vbn
Van
Vcn
120°
−120°
(b)
v
v
Figure12.7 Phase sequences:
(a) abc or positive sequence,
(b) acb or negative sequence.
As a common tradition in power systems, volt-
age and current in this chapter are in rms values
unless otherwise stated.
Since the three-phase voltages are 120◦
out of phase with each other,
there are two possible combinations. One possibility is shown in Fig.
12.7(a) and expressed mathematically as
Van = Vp 0◦
Vbn = Vp − 120◦
Vcn = Vp − 240◦
= Vp + 120◦
(12.3)
where Vp is the effective or rms value. This is known as the abc sequence
or positive sequence. In this phase sequence, Van leads Vbn, which in
turn leads Vcn. This sequence is produced when the rotor in Fig. 12.4
rotates counterclockwise. The other possibility is shown in Fig. 12.7(b)
and is given by
Van = Vp 0◦
Vcn = Vp − 120◦
Vbn = Vp − 240◦
= Vp + 120◦
(12.4)
This is called the acb sequence or negative sequence. For this phase
sequence, Van leads Vcn, which in turn leads Vbn. The acb sequence is
produced when the rotor in Fig. 12.4 rotates in the clockwise direction.
CHAPTER 12 Three-Phase Circuits 481
It is easy to show that the voltages in Eqs. (12.3) or (12.4) satisfy Eqs.
(12.1) and (12.2). For example, from Eq. (12.3),
Van + Vbn + Vcn = Vp 0◦
+ Vp − 120◦
+ Vp + 120◦
= Vp(1.0 − 0.5 − j0.866 − 0.5 + j0.866)
= 0
(12.5)
The phase sequence is the time order in which the voltages pass through
their respective maximum values.
The phase sequence may also be regarded as the
order in which the phase voltages reach their
peak (or maximum) values with respect to time.
The phase sequence is determined by the order in which the phasors pass
through a fixed point in the phase diagram. Reminder: As time increases, each phasor (or
sinor) rotates at an angular velocity ω.
In Fig. 12.7(a), as the phasors rotate in the counterclockwise direc-
tion with frequency ω, they pass through the horizontal axis in a sequence
abcabca . . . . Thus, the sequence is abc or bca or cab. Similarly, for the
phasors in Fig. 12.7(b), as they rotate in the counterclockwise direction,
they pass the horizontal axis in a sequence acbacba . . . . This describes
the acb sequence. The phase sequence is important in three-phase power
distribution. It determines the direction of the rotation of a motor con-
nected to the power source, for example.
Like the generator connections, a three-phase load can be either
wye-connected or delta-connected, depending on the end application.
Figure 12.8(a) shows a wye-connected load, and Fig. 12.8(b) shows a
delta-connected load. The neutral line in Fig. 12.8(a) may or may not
be there, depending on whether the system is four- or three-wire. (And,
of course, a neutral connection is topologically impossible for a delta
connection.) A wye- or delta-connected load is said to be unbalanced if
the phase impedances are not equal in magnitude or phase.
a
b
n
c
(a)
Z2
Z1
Z3
a
b
c
(b)
Zb
Zc
Zb
Figure12.8 Two possible three-
phase load configurations:
(a) a Y-connected load,
(b) a -connected load
A balanced load is one in which the phase impedances
are equal in magnitude and in phase.
Reminder: A Y-connected load consists of three
impedancesconnectedtoaneutralnode,whilea
-connected load consists of three impedances
connected around a loop. The load is balanced
when the three impedances are equal in either
case.
For a balanced wye-connected load,
Z1 = Z2 = Z3 = ZY (12.6)
whereZY istheloadimpedanceperphase. Forabalanced delta-connected
load,
Za = Zb = Zc = Z (12.7)
where Z is the load impedance per phase in this case. We recall from
Eq. (9.69) that
Z = 3ZY or ZY =
1
3
Z (12.8)
so we know that a wye-connected load can be transformed into a delta-
connected load, or vice versa, using Eq. (12.8).
Since both the three-phase source and the three-phase load can be
either wye- or delta-connected, we have four possible connections:
482 PART 2 AC Circuits
• Y-Y connection (i.e., Y-connected source with a Y-connected
load).
• Y- connection.
• - connection.
• -Y connection.
In subsequent sections, we will consider each of these possible configu-
rations.
It is appropriate to mention here that a balanced delta-connected
load is more common than a balanced wye-connected load. This is due
to the ease with which loads may be added or removed from each phase
of a delta-connected load. This is very difficult with a wye-connected
load because the neutral may not be accessible. On the other hand, delta-
connected sources are not common in practice because of the circulating
current that will result in the delta-mesh if the three-phase voltages are
slightly unbalanced.
E X A M P L E 1 2 . 1
Determine the phase sequence of the set of voltages
van = 200 cos(ωt + 10◦
)
vbn = 200 cos(ωt − 230◦
), vcn = 200 cos(ωt − 110◦
)
Solution:
The voltages can be expressed in phasor form as
Van = 200 10◦
, Vbn = 200 − 230◦
, Vcn = 200 − 110◦
We notice that Van leads Vcn by 120◦
and Vcn in turn leads Vbn by 120◦
.
Hence, we have an acb sequence.
P R A C T I C E P R O B L E M 1 2 . 1
Given that Vbn = 110 30◦
, find Van and Vcn, assuming a positive (abc)
sequence.
Answer: 110 150◦
, 110 − 90◦
.
12.3 BALANCED WYE-WYE CONNECTION
We begin with the Y-Y system, because any balanced three-phase system
can be reduced to an equivalent Y-Y system. Therefore, analysis of this
system should be regarded as the key to solving all balanced three-phase
systems.
A balanced Y-Y system is a three-phase system with a balanced Y-connected
source and a balanced Y-connected load.
Consider the balanced four-wire Y-Y system of Fig. 12.9, where
a Y-connected load is connected to a Y-connected source. We assume a
CHAPTER 12 Three-Phase Circuits 483
balanced load so that load impedances are equal. Although the impedance
ZY is the total load impedance per phase, it may also be regarded as the
sum of the source impedance Zs, line impedance Z, and load impedance
ZL for each phase, since these impedances are in series. As illustrated in
Fig. 12.9, Zs denotes the internal impedance of the phase winding of the
generator; Z is the impedance of the line joining a phase of the source
with a phase of the load; ZL is the impedance of each phase of the load;
and Zn is the impedance of the neutral line. Thus, in general
ZY = Zs + Z + ZL (12.9)
Zs and Z are often very small compared with ZL, so one can assume
that ZY = ZL if no source or line impedance is given. In any event,
by lumping the impedances together, the Y-Y system in Fig. 12.9 can be
simplified to that shown in Fig. 12.10.
+
−
Zl
Zl
Zl
Zn
Zs
Zs Zs
ZL
ZL
ZL
V
an
V
cn V
bn
A
N
B
C
b
n
c
a
+
−
+
−
Figure12.9 A balanced Y-Y system, showing the
source, line, and load impedances.
+
− ZY
ZY
ZY
V
an
V
cn V
bn
A
N
B
C
b
n
c
a
In
Ib
Ia
Ic
+
−
+
−
Figure12.10 Balanced Y-Y connection.
Assuming the positive sequence, the phase voltages (or line-to-
neutral voltages) are
Van = Vp 0◦
Vbn = Vp − 120◦
, Vcn = Vp + 120◦
(12.10)
The line-to-line voltages or simply line voltages Vab, Vbc, and Vca are
related to the phase voltages. For example,
Vab = Van + Vnb = Van − Vbn = Vp 0◦
− Vp − 120◦
= Vp

1 +
1
2
+ j
√
3
2

=
√
3Vp 30◦
(12.11a)
Similarly, we can obtain
Vbc = Vbn − Vcn =
√
3Vp − 90◦
(12.11b)
Vca = Vcn − Van =
√
3Vp − 210◦
(12.11c)
484 PART 2 AC Circuits
Thus, the magnitude of the line voltages VL is
√
3 times the magnitude
of the phase voltages Vp, or
VL =
√
3Vp (12.12)
where
Vp = |Van| = |Vbn| = |Vcn| (12.13)
and
VL = |Vab| = |Vbc| = |Vca| (12.14)
Also the line voltages lead their corresponding phase voltages by 30◦
.
Figure 12.11(a) illustrates this. Figure 12.11(a) also shows how to deter-
mine Vab from the phase voltages, while Fig. 12.11(b) shows the same
for the three line voltages. Notice that Vab leads Vbc by 120◦
, and Vbc
leads Vca by 120◦
, so that the line voltages sum up to zero as do the phase
voltages.
(a)
30°
Vcn
Vnb Vab = Van + Vnb
Van
Vbn
(b)
Vca Vcn Vab
Van
Vbc
Vbn
Figure12.11 Phasor diagrams illustrating
the relationship between line voltages and
phase voltages.
Applying KVL to each phase in Fig. 12.10, we obtain the line cur-
rents as
Ia =
Van
ZY
, Ib =
Vbn
ZY
=
Van − 120◦
ZY
= Ia − 120◦
Ic =
Vcn
ZY
=
Van − 240◦
ZY
= Ia − 240◦
(12.15)
We can readily infer that the line currents add up to zero,
Ia + Ib + Ic = 0 (12.16)
so that
In = −(Ia + Ib + Ic) = 0 (12.17a)
or
VnN = ZnIn = 0 (12.17b)
that is, the voltage across the neutral wire is zero. The neutral line can
thus be removed without affecting the system. In fact, in long distance
power transmission, conductors in multiples of three are used with the
earth itself acting as the neutral conductor. Power systems designed in
this way are well grounded at all critical points to ensure safety.
While the line current is the current in each line, the phase current
is the current in each phase of the source or load. In the Y-Y system, the
line current is the same as the phase current. We will use single subscripts
for line currents because it is natural and conventional to assume that line
currents flow from the source to the load.
ZY
Van
+
−
a A
n N
Ia
Figure 12.12 A single-phase
equivalent circuit.
An alternative way of analyzing a balanced Y-Y system is to do so
on a “per phase” basis. We look at one phase, say phase a, and analyze the
single-phase equivalent circuit in Fig. 12.12. The single-phase analysis
yields the line current Ia as
Ia =
Van
ZY
(12.18)
CHAPTER 12 Three-Phase Circuits 485
From Ia, we use the phase sequence to obtain other line currents. Thus,
as long as the system is balanced, we need only analyze one phase. We
may do this even if the neutral line is absent, as in the three-wire system.
E X A M P L E 1 2 . 2
Calculate the line currents in the three-wire Y-Y system of Fig. 12.13.
+
−
5 – j2Ω
10 + j8Ω
10 + j8Ω
A
B
c
b
a
5 – j2Ω 10 + j8Ω
C
5 – j2Ω
110 −120° V
110 −240° V
110 0° V
+
−
+
−
Figure12.13 Three-wire Y-Y system; for Example 12.2.
Solution:
The three-phase circuit in Fig. 12.13 is balanced; we may replace it with
its single-phase equivalent circuit such as in Fig. 12.12. We obtain Ia
from the single-phase analysis as
Ia =
Van
ZY
where ZY = (5 − j2) + (10 + j8) = 15 + j6 = 16.155 21.8◦
. Hence,
Ia =
110 0◦
16.155 21.8◦
= 6.81 − 21.8◦
A
Since the source voltages in Fig. 12.13 are in positive sequence and the
line currents are also in positive sequence,
Ib = Ia − 120◦
= 6.81 − 141.8◦
A
Ic = Ia − 240◦
= 6.81 − 261.8◦
A = 6.81 98.2◦
A
P R A C T I C E P R O B L E M 1 2 . 2
A Y-connected balanced three-phase generator with an impedance of
0.4+j0.3  per phase is connected to a Y-connected balanced load with
an impedance of 24 + j19  per phase. The line joining the generator
and the load has an impedance of 0.6 + j0.7  per phase. Assuming
a positive sequence for the source voltages and that Van = 120 30◦
V,
find: (a) the line voltages, (b) the line currents.
486 PART 2 AC Circuits
Answer: (a) 207.85 60◦
V, 207.85 − 60◦
V, 207.85 − 180◦
V,
(b) 3.75 − 8.66◦
A, 3.75 − 128.66◦
A, 3.75 − 248.66◦
A.
12.4 BALANCED WYE-DELTA CONNECTION
A balanced Y-∆ system consists of a balanced Y-connected source
feeding a balanced -connected load.
This is perhaps the most practical three-phase
system, asthethree-phasesourcesareusuallyY-
connected while the three-phase loads are usu-
ally -connected.
The balanced Y-delta system is shown in Fig. 12.14, where the
source is wye-connected and the load is -connected. There is, of course,
no neutral connection from source to load for this case. Assuming the
positive sequence, the phase voltages are again
Van = Vp 0◦
Vbn = Vp − 120◦
, Vcn = Vp + 120◦
(12.19)
As shown in Section 12.3, the line voltages are
Vab =
√
3Vp 30◦
= VAB, Vbc =
√
3Vp − 90◦
= VBC
Vca =
√
3Vp − 210◦
= VCA
(12.20)
showing that the line voltages are equal to the voltages across the load
impedances for this system configuration. From these voltages, we can
obtain the phase currents as
IAB =
VAB
Z
, IBC =
VBC
Z
, ICA =
VCA
Z
(12.21)
These currents have the same magnitude but are out of phase with each
other by 120◦
.
+
−
Z∆
Z∆
Z∆
V
an
V
cn V
bn
IAB
ICA
A
C
b
c
B
n
a
+
−
+
−
Ia
Ib
IBC
Ic
Figure12.14 Balanced Y- connection.
CHAPTER 12 Three-Phase Circuits 487
Another way to get these phase currents is to apply KVL. For ex-
ample, applying KVL around loop aABbna gives
−Van + Z IAB + Vbn = 0
or
IAB =
Van − Vbn
Z
=
Vab
Z
=
VAB
Z
(12.22)
which is the same as Eq. (12.21). This is the more general way of finding
the phase currents.
The line currents are obtained from the phase currents by applying
KCL at nodes A, B, and C. Thus,
Ia = IAB − ICA, Ib = IBC − IAB, Ic = ICA − IBC (12.23)
Since ICA = IAB − 240◦
,
Ia = IAB − ICA = IAB(1 − 1 − 240◦
)
= IAB(1 + 0.5 − j0.866) = IAB
√
3 − 30◦
(12.24)
showing that the magnitude IL of the line current is
√
3 times the magni-
tude Ip of the phase current, or
IL =
√
3Ip (12.25)
where
IL = |Ia| = |Ib| = |Ic| (12.26)
and
Ip = |IAB| = |IBC| = |ICA| (12.27)
Also, the line currents lag the corresponding phase currents by 30◦
, as-
suming the positive sequence. Figure 12.15 is a phasor diagram illustrat-
ing the relationship between the phase and line currents.
30°
30°
30°
ICA
IAB
Ib IBC
Ia
Ic
Figure12.15 Phasor diagram
illustrating the relationship between
phase and line currents.
An alternative way of analyzing the Y- circuit is to transform the
-connected load to an equivalent Y-connected load. Using the -Y
transformation formula in Eq. (9.69),
ZY =
Z
3
(12.28)
After this transformation, we now have a Y-Y system as in Fig. 12.10.
The three-phase Y- system in Fig. 12.14 can be replaced by the single-
phase equivalent circuit in Fig. 12.16. This allows us to calculate only
the line currents. The phase currents are obtained using Eq. (12.25) and
utilizing the fact that each of the phase currents leads the corresponding
line current by 30◦
.
V
an
+
−
Ia
Z∆
3
Figure12.16 A single-phase equivalent
circuit of a balanced Y- circuit.
E X A M P L E 1 2 . 3
A balanced abc-sequence Y-connected source with Van = 100 10◦
V
is connected to a -connected balanced load (8 + j4)  per phase. Cal-
culate the phase and line currents.
488 PART 2 AC Circuits
Solution:
This can be solved in two ways.
METHOD 1 The load impedance is
Z = 8 + j4 = 8.944 26.57◦

If the phase voltage Van = 100 10◦
, then the line voltage is
Vab = Van
√
3 30◦
= 100
√
3 10◦
+ 30◦
= VAB
or
VAB = 173.2 40◦
V
The phase currents are
IAB =
VAB
Z
=
173.2 40◦
8.944 26.57◦
= 19.36 13.43◦
A
IBC = IAB − 120◦
= 19.36 − 106.57◦
A
ICA = IAB + 120◦
= 19.36 133.43◦
A
The line currents are
Ia = IAB
√
3 − 30◦
=
√
3(19.36) 13.43◦
− 30◦
= 33.53 − 16.57◦
A
Ib = Ia − 120◦
= 33.53 − 136.57◦
A
Ic = Ia + 120◦
= 33.53 103.43◦
A
METHOD 2 Alternatively, using single-phase analysis,
Ia =
Van
Z /3
=
100 10◦
2.981 26.57◦
= 33.54 − 16.57◦
A
as above. Other line currents are obtained using the abc phase sequence.
P R A C T I C E P R O B L E M 1 2 . 3
One line voltage of a balanced Y-connected source is VAB =
180 − 20◦
V. If the source is connected to a -connected load of
20 40◦
, find the phase and line currents. Assume the abc sequence.
Answer: 9 − 60◦
, 9 − 180◦
, 9 60◦
, 15.59 − 90◦
,
15.59 − 210◦
, 15.59 30◦
A.
12.5 BALANCED DELTA-DELTA CONNECTION
A balanced ∆-∆ system is one in which both the balanced source
and balanced load are -connected.
CHAPTER 12 Three-Phase Circuits 489
The source as well as the load may be delta-connected as shown
in Fig. 12.17. Our goal is to obtain the phase and line currents as usual.
Assuming a positive sequence, the phase voltages for a delta-connected
source are
Vab = Vp 0◦
Vbc = Vp − 120◦
, Vca = Vp + 120◦
(12.29)
The line voltages are the same as the phase voltages. From Fig. 12.17,
assuming there is no line impedances, the phase voltages of the delta-
connected source are equal to the voltages across the impedances; that
is,
Vab = VAB, Vbc = VBC, Vca = VCA (12.30)
Hence, the phase currents are
IAB =
VAB
Z
=
Vab
Z
, IBC =
VBC
Z
=
Vbc
Z
ICA =
VCA
Z
=
Vca
Z
(12.31)
Since the load is delta-connected just as in the previous section, some
of the formulas derived there apply here. The line currents are obtained
from the phase currents by applying KCL at nodes A, B, and C, as we
did in the previous section:
Ia = IAB − ICA, Ib = IBC − IAB, Ic = ICA − IBC (12.32)
Also, as shown in the last section, each line current lags the corresponding
phase current by 30◦
; the magnitude IL of the line current is
√
3 times
the magnitude Ip of the phase current,
IL =
√
3Ip (12.33)
Z∆
Z∆
Z∆
V
ca
V
bc
V
ab
IAB
ICA
A
C
b
c
B
a
+
−
Ia
Ib
IBC
Ic
+
−
− +
Figure12.17 A balanced - connection.
An alternative way of analyzing the - circuit is to convert both
the source and the load to their Y equivalents. We already know that
ZY = Z /3. To convert a -connected source to a Y-connected source,
see the next section.
490 PART 2 AC Circuits
E X A M P L E 1 2 . 4
A balanced -connected load having an impedance 20 − j15  is con-
nected to a -connected, positive-sequence generator having Vab =
330 0◦
V. Calculate the phase currents of the load and the line currents.
Solution:
The load impedance per phase is
Z = 20 − j15 = 25 − 36.87◦

The phase currents are
IAB =
VAB
Z
=
330 0◦
25 − 36.87
= 13.2 36.87◦
A
IBC = IAB − 120◦
= 13.2 − 83.13◦
A
ICA = IAB + 120◦
= 13.2 156.87◦
A
For a delta load, the line current always lags the corresponding phase
current by 30◦
and has a magnitude
√
3 times that of the phase current.
Hence, the line currents are
Ia = IAB
√
3 − 30◦
= (13.2 36.87◦
)(
√
3 − 30◦
)
= 22.86 6.87◦
A
Ib = Ia − 120◦
= 22.86 − 113.13◦
A
Ic = Ia + 120◦
= 22.86 126.87◦
A
P R A C T I C E P R O B L E M 1 2 . 4
A positive-sequence, balanced -connected source supplies a balanced
-connected load. If the impedance per phase of the load is 18 + j12 
and Ia = 22.5 35◦
A, find IAB and VAB.
Answer: 13 65◦
A, 281.2 98.69◦
V.
12.6 BALANCED DELTA-WYE CONNECTION
A balanced ∆-Y system consists of a balanced -connected
source feeding a balanced Y-connected load.
Consider the -Y circuit in Fig. 12.18. Again, assuming the abc
sequence, the phase voltages of a delta-connected source are
Vab = Vp 0◦
, Vbc = Vp − 120◦
Vca = Vp + 120◦
(12.34)
These are also the line voltages as well as the phase voltages.
CHAPTER 12 Three-Phase Circuits 491
ZY
ZY
ZY
V
ca
V
bc
V
ab
A
C
b
c
B
N
a
+
−
Ia
Ib
Ic
+
−
− +
Figure12.18 A balanced -Y connection.
We can obtain the line currents in many ways. One way is to apply
KVL to loop aANBba in Fig. 12.18, writing
−Vab + ZY Ia − ZY Ib = 0
or
ZY (Ia − Ib) = Vab = Vp 0◦
Thus,
Ia − Ib =
Vp 0◦
ZY
(12.35)
But Ib lags Ia by 120◦
, since we assumed the abc sequence; that is,
Ib = Ia − 120◦
. Hence,
Ia − Ib = Ia(1 − 1 − 120◦
)
= Ia

1 +
1
2
+ j
√
3
2

= Ia
√
3 30◦
(12.36)
Substituting Eq. (12.36) into Eq. (12.35) gives
Ia =
Vp/
√
3 − 30◦
ZY
(12.37)
From this, we obtain the other line currents Ib and Ic using the positive
phase sequence, i.e., Ib = Ia − 120◦
, Ic = Ia + 120◦
. The phase
currents are equal to the line currents.
V
an
V
ab
V
ca
V
bn
V
bc
V
cn
a
c b
n
+
−
+ −
+ −
− +
+
−
+
−
Figure12.19 Transforming a -connected
source to an equivalent Y-connected source.
Another way to obtain the line currents is to replace the delta-
connected source with its equivalent wye-connected source, as shown in
Fig. 12.19. In Section 12.3, we found that the line-to-line voltages of
a wye-connected source lead their corresponding phase voltages by 30◦
.
Therefore, we obtain each phase voltage of the equivalent wye-connected
source by dividing the corresponding line voltage of the delta-connected
source by
√
3 and shifting its phase by −30◦
. Thus, the equivalent wye-
connected source has the phase voltages
Van =
Vp
√
3
− 30◦
Vbn =
Vp
√
3
− 150◦
, Vcn =
Vp
√
3
+ 90◦
(12.38)
492 PART 2 AC Circuits
If the delta-connected source has source impedance Zs per phase, the
equivalent wye-connected source will have a source impedance of Zs/3
per phase, according to Eq. (9.69).
Once the source is transformed to wye, the circuit becomes a wye-
wye system. Therefore, we can use the equivalent single-phase circuit
shown in Fig. 12.20, from which the line current for phase a is
Ia =
Vp/
√
3 − 30◦
ZY
(12.39)
which is the same as Eq. (12.37).
ZY
+
−
Ia
Vp −30°
√3
Figure 12.20 The single-phase equivalent
circuit.
Alternatively, we may transform the wye-connected load to an
equivalent delta-connected load. This results in a delta-delta system,
which can be analyzed as in Section 12.5. Note that
VAN = IaZY =
Vp
√
3
− 30◦
(12.40)
VBN = VAN − 120◦
, VCN = VAN + 120◦
As stated earlier, the delta-connected load is more desirable than
the wye-connected load. It is easier to alter the loads in any one phase of
the delta-connected loads, as the individual loads are connected directly
across the lines. However, the delta-connected source is hardly used in
practice, because any slight imbalance in the phase voltages will result in
unwanted circulating currents.
Table 12.1 presents a summary of the formulas for phase currents
and voltages and line currents and voltages for the four connections.
Students are advised not to memorize the formulas but to understand
how they are derived. The formulas can always be obtained by directly
applying KCL and KVL to the appropriate three-phase circuits.
TABLE 12.1 Summary of phase and line voltages/currents for
balanced three-phase systems1
.
Connection Phase voltages/currents Line voltages/currents
Y-Y Van = Vp 0◦
Vab =
√
3Vp 30◦
Vbn = Vp − 120◦
Vbc = Vab − 120◦
Vcn = Vp + 120◦
Vca = Vab + 120◦
Same as line currents Ia = Van/ZY
Ib = Ia − 120◦
Ic = Ia + 120◦
Y- Van = Vp 0◦
Vab = VAB =
√
3Vp 30◦
Vbn = Vp − 120◦
Vbc = VBC = Vab − 120◦
Vcn = Vp + 120◦
Vca = VCA = Vab + 120◦
IAB = VAB /Z Ia = IAB
√
3 − 30◦
IBC = VBC/Z Ib = Ia − 120◦
ICA = VCA/Z Ic = Ia + 120◦
1Positive or abc sequence is assumed.
CHAPTER 12 Three-Phase Circuits 493
TABLE 12.1 (continued)
Connection Phase voltages/currents Line voltages/currents
- Vab = Vp 0◦
Same as phase voltages
Vbc = Vp − 120◦
Vca = Vp + 120◦
IAB = Vab/Z Ia = IAB
√
3 − 30◦
IBC = Vbc/Z Ib = Ia − 120◦
ICA = Vca/Z Ic = Ia + 120◦
-Y Vab = Vp 0◦
Same as phase voltages
Vbc = Vp − 120◦
Vca = Vp + 120◦
Same as line currents Ia =
Vp − 30◦
√
3ZY
Ib = Ia − 120◦
Ic = Ia + 120◦
E X A M P L E 1 2 . 5
A balanced Y-connected load with a phase resistance of 40  and a reac-
tance of 25  is supplied by a balanced, positive sequence -connected
source with a line voltage of 210 V. Calculate the phase currents. Use
Vab as reference.
Solution:
The load impedance is
ZY = 40 + j25 = 47.17 32◦

and the source voltage is
Vab = 210 0◦
V
When the -connected source is transformed to a Y-connected source,
Van =
Vab
√
3
− 30◦
= 121.2 − 30◦
V
The line currents are
Ia =
Van
ZY
=
121.2 − 30◦
47.12 32◦
= 2.57 − 62◦
A
Ib = Ia − 120◦
= 2.57 − 182◦
A
Ic = Ia 120◦
= 2.57 58◦
A
which are the same as the phase currents.
494 PART 2 AC Circuits
P R A C T I C E P R O B L E M 1 2 . 5
In a balanced -Y circuit, Vab = 240 15◦
and ZY = (12 + j15) .
Calculate the line currents.
Answer: 7.21 − 66.34◦
, 7.21 − 186.34◦
, 7.21 53.66◦
A.
12.7 POWER IN A BALANCED SYSTEM
Let us now consider the power in a balanced three-phase system. We
begin by examining the instantaneous power absorbed by the load. This
requires that the analysis be done in the time domain. For a Y-connected
load, the phase voltages are
vAN =
√
2Vp cos ωt, vBN =
√
2Vp cos(ωt − 120◦
)
vCN =
√
2Vp cos(ωt + 120◦
)
(12.41)
where the factor
√
2 is necessary because Vp has been defined as the rms
value of the phase voltage. If ZY = Z θ, the phase currents lag behind
their corresponding phase voltages by θ. Thus,
ia =
√
2Ip cos(ωt − θ), ib =
√
2Ip cos(ωt − θ − 120◦
)
ic =
√
2Ip cos(ωt − θ + 120◦
)
(12.42)
where Ip is the rms value of the phase current. The total instantaneous
power in the load is the sum of the instantaneous powers in the three
phases; that is,
p = pa + pb + pc = vAN ia + vBN ib + vCN ic
= 2VpIp[cos ωt cos(ωt − θ)
+ cos(ωt − 120◦
) cos(ωt − θ − 120◦
)
+ cos(ωt + 120◦
) cos(ωt − θ + 120◦
)]
(12.43)
Applying the trigonometric identity
cos A cos B =
1
2
[cos(A + B) + cos(A − B)] (12.44)
gives
p = VpIp[3 cos θ + cos(2ωt − θ) + cos(2ωt − θ − 240◦
)
+ cos(2ωt − θ + 240◦
)]
= VpIp[3 cos θ + cos α + cos α cos 240◦
+ sin α sin 240◦
+ cos α cos 240◦
− sin α sin 240◦
]
where α = 2ωt − θ
= VpIp

3 cos θ + cos α + 2

−
1
2

cos α

= 3VpIp cos θ
(12.45)
Thus the total instantaneous power in a balanced three-phase system is
constant—it does not change with time as the instantaneous power of each
phase does. This result is true whether the load is Y- or -connected.
CHAPTER 12 Three-Phase Circuits 495
This is one important reason for using a three-phase system to generate
and distribute power. We will look into another reason a little later.
Since the total instantaneous power is independent of time, the
average power per phase Pp for either the -connected load or the Y-
connected load is p/3, or
Pp = VpIp cos θ (12.46)
and the reactive power per phase is
Qp = VpIp sin θ (12.47)
The apparent power per phase is
Sp = VpIp (12.48)
The complex power per phase is
Sp = Pp + jQp = VpI∗
p (12.49)
where Vp and Ip are the phase voltage and phase current with magnitudes
Vp and Ip, respectively. The total average power is the sum of the average
powers in the phases:
P = Pa + Pb + Pc = 3Pp = 3VpIp cos θ =
√
3VLIL cos θ (12.50)
For a Y-connected load, IL = Ip but VL =
√
3Vp, whereas for a -
connected load, IL =
√
3Ip but VL = Vp. Thus, Eq. (12.50) applies for
both Y-connected and -connected loads. Similarly, the total reactive
power is
Q = 3VpIp sin θ = 3Qp =
√
3VLIL sin θ (12.51)
and the total complex power is
S = 3Sp = 3VpI∗
p = 3I2
pZp =
3V 2
p
Z∗
p
(12.52)
where Zp = Zp θ is the load impedance per phase. (Zp could be ZY or
Z .) Alternatively, we may write Eq. (12.52) as
S = P + jQ =
√
3VLIL θ (12.53)
Remember that Vp, Ip, VL, and IL are all rms values and that θ is the
angle of the load impedance or the angle between the phase voltage and
the phase current.
A second major advantage of three-phase systems for power dis-
tribution is that the three-phase system uses a lesser amount of wire than
the single-phase system for the same line voltage VL and the same ab-
sorbed power PL. We will compare these cases and assume in both that
the wires are of the same material (e.g., copper with resistivity ρ), of the
same length , and that the loads are resistive (i.e., unity power factor).
For the two-wire single-phase system in Fig. 12.21(a), IL = PL/VL, so
the power loss in the two wires is
Ploss = 2I2
LR = 2R
P 2
L
V 2
L
(12.54)
496 PART 2 AC Circuits
Single-
phase
source
R
R
Transmission lines
(a)
PL
IL
Load
−
+
VL
Three-
phase
balanced
load
Three-
phase
balanced
source
R′
R′
R′
Transmission lines
(b)
Ia
Ib
Ic −
+
−
+
VL
VL −120°
0°
Figure12.21 Comparing the power loss in (a) a single-phase system, and (b) a three-phase system.
For the three-wire three-phase system in Fig. 12.21(b), I
L = |Ia| = |Ib| =
|Ic| = PL/
√
3VL from Eq. (12.50). The power loss in the three wires is
P 
loss = 3(I
L)2
R
= 3R P 2
L
3V 2
L
= R P 2
L
V 2
L
(12.55)
Equations(12.54)and(12.55)showthatforthesametotalpowerdelivered
PL and same line voltage VL,
Ploss
P 
loss
=
2R
R
(12.56)
But from Chapter 2, R = ρ/πr2
and R
= ρ/πr2
, where r and r
are
the radii of the wires. Thus,
Ploss
P 
loss
=
2r2
r2
(12.57)
If the same power loss is tolerated in both systems, then r2
= 2r2
. The
ratio of material required is determined by the number of wires and their
volumes, so
Material for single-phase
Material for three-phase
=
2(πr2
)
3(πr2)
=
2r2
3r2
=
2
3
(2) = 1.333
(12.58)
sincer2
= 2r2
. Equation(12.58)showsthatthesingle-phasesystemuses
33 percent more material than the three-phase system or that the three-
phase system uses only 75 percent of the material used in the equivalent
single-phase system. In other words, considerably less material is needed
to deliver the same power with a three-phase system than is required for
a single-phase system.
E X A M P L E 1 2 . 6
Refer to the circuit in Fig. 12.13 (in Example 12.2). Determine the total
average power, reactive power, and complex power at the source and at
the load.
CHAPTER 12 Three-Phase Circuits 497
Solution:
It is sufficient to consider one phase, as the system is balanced. For phase
a,
Vp = 110 0◦
V and Ip = 6.81 − 21.8◦
A
Thus, at the source, the complex power supplied is
Ss = −3VpI∗
p = 3(110 0◦
)(6.81 21.8◦
)
= −2247 21.8◦
= −(2087 + j834.6) VA
The real or average power supplied is −2087 W and the reactive power
is −834.6 VAR.
At the load, the complex power absorbed is
SL = 3|Ip|2
Zp
where Zp = 10 + j8 = 12.81 38.66◦
and Ip = Ia = 6.81 − 21.8◦
.
Hence
SL = 3(6.81)2
12.81 38.66◦
= 1782 38.66
= (1392 + j1113) VA
The real power absorbed is 1391.7 W and the reactive power absorbed is
1113.3 VAR. The difference between the two complex powers is absorbed
by the line impedance (5 − j2) . To show that this is the case, we find
the complex power absorbed by the line as
S = 3|Ip|2
Z = 3(6.81)2
(5 − j2) = 695.6 − j278.3 VA
which is the difference between Ss and SL, that is, Ss + S + SL = 0, as
expected.
P R A C T I C E P R O B L E M 1 2 . 6
For the Y-Y circuit in Practice Prob. 12.2, calculate the complex power
at the source and at the load.
Answer: (1054 + j843.3) VA, (1012 + j801.6) VA.
E X A M P L E 1 2 . 7
A three-phase motor can be regarded as a balanced Y-load. A three-phase
motor draws 5.6 kW when the line voltage is 220 V and the line current
is 18.2 A. Determine the power factor of the motor.
Solution:
The apparent power is
S =
√
3VLIL =
√
3(220)(18.2) = 6935.13 VA
Since the real power is
P = S cos θ = 5600 W
498 PART 2 AC Circuits
the power factor is
pf = cos θ =
P
S
=
5600
6935.13
= 0.8075
P R A C T I C E P R O B L E M 1 2 . 7
Calculate the line current required for a 30-kW three-phase motor having
a power factor of 0.85 lagging if it is connected to a balanced source with
a line voltage of 440 V.
Answer: 50.94 A.
E X A M P L E 1 2 . 8
Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown
in Fig. 12.22(a). Load 1 draws 30 kW at a power factor of 0.6 lagging,
while load 2 draws 45 kVAR at a power factor of 0.8 lagging. Assuming
the abc sequence, determine: (a) the complex, real, and reactive powers
absorbed by the combined load, (b) the line currents, and (c) the kVAR
rating of the three capacitors -connected in parallel with the load that
will raise the power factor to 0.9 lagging and the capacitance of each
capacitor.
(a)
C
C
C
Balanced
load 1
Balanced
load 2
(b)
Combined
load
Figure12.22 For Example 12.8: (a) The
original balanced loads, (b) the combined load
with improved power factor.
Solution:
(a) For load 1, given that P1 = 30 kW and cos θ1 = 0.6, then sin θ1 = 0.8.
Hence,
S1 =
P1
cos θ1
=
30 kW
0.6
= 50 kVA
and Q1 = S1 sin θ1 = 50(0.8) = 40 kVAR. Thus, the complex power
due to load 1 is
S1 = P1 + jQ1 = 30 + j40 kVA (12.8.1)
For load 2, if Q2 = 45 kVAR and cos θ2 = 0.8, then sin θ2 = 0.6. We
find
S2 =
Q2
sin θ2
=
45 kVA
0.6
= 75 kVA
and P2 = S2 cos θ2 = 75(0.8) = 60 kW. Therefore the complex power
due to load 2 is
S2 = P2 + jQ2 = 60 + j45 kVA (12.8.2)
From Eqs. (12.8.1) and (12.8.2), the total complex power absorbed by
the load is
S = S1 + S2 = 90 + j85 kVA = 123.8 43.36◦
kVA (12.8.3)
which has a power factor of cos 43.36◦
= 0.727 lagging. The real power
is then 90 kW, while the reactive power is 85 kVAR.
(b) Since S =
√
3VLIL, the line current is
IL =
S
√
3VL
(12.8.4)
CHAPTER 12 Three-Phase Circuits 499
We apply this to each load, keeping in mind that for both loads, VL = 240
kV. For load 1,
IL1 =
50,000
√
3 240,000
= 120.28 mA
Since the power factor is lagging, the line current lags the line voltage by
θ1 = cos−1
0.6 = 53.13◦
. Thus,
Ia1 = 120.28 − 53.13◦
For load 2,
IL2 =
75,000
√
3 240,000
= 180.42 mA
and the line current lags the line voltage by θ2 = cos−1
0.8 = 36.87◦
.
Hence,
Ia2 = 180.42 − 36.87◦
The total line current is
Ia = Ia1 + Ia2 = 120.28 − 53.13◦
+ 180.42 − 36.87◦
= (72.168 − j96.224) + (144.336 − j108.252)
= 216.5 − j204.472 = 297.8 − 43.36◦
mA
Alternatively, we could obtain the current from the total complex
power using Eq. (12.8.4) as
IL =
123,800
√
3 240,000
= 297.82 mA
and
Ia = 297.82 − 43.36◦
mA
which is the same as before. The other line currents, Ib2 and Ica, can be
obtainedaccordingtotheabc sequence(i.e., Ib = 297.82 −163.36◦
mA
and Ic = 297.82 76.64◦
mA).
(c) We can find the reactive power needed to bring the power factor to 0.9
lagging using Eq. (11.59),
QC = P(tan θold − tan θnew)
where P = 90 kW, θold = 43.36◦
, and θnew = cos−1
0.9 = 25.84◦
.
Hence,
QC = 90,000(tan 43.36◦
− tan 25.04◦
) = 41.4 kVAR
This reactive power is for the three capacitors. For each capacitor, the
rating Q
C = 13.8 kVAR. From Eq. (11.60), the required capacitance is
C =
Q
C
ωV 2
rms
Since the capacitors are -connected as shown in Fig. 12.22(b), Vrms in
the above formula is the line-to-line or line voltage, which is 240 kV.
Thus,
C =
13,800
(2π60)(240,000)2
= 635.5 pF
500 PART 2 AC Circuits
P R A C T I C E P R O B L E M 1 2 . 8
Assume that the two balanced loads in Fig. 12.22(a) are supplied by an
840-V rms 60-Hz line. Load 1 is Y-connected with 30+j40  per phase,
while load 2 is a balanced three-phase motor drawing 48 kW at a power
factor of 0.8 lagging. Assuming the abc sequence, calculate: (a) the
complex power absorbed by the combined load, (b) the kVAR rating of
each of the three capacitors -connected in parallel with the load to raise
the power factor to unity, and (c) the current drawn from the supply at
unity power factor condition.
Answer: (a) 56.47 + j47.29 kVA, (b) 15.7 kVAR, (c) 38.813 A.
†12.8 UNBALANCED THREE-PHASE SYSTEMS
This chapter would be incomplete without mentioning unbalanced three-
phase systems. An unbalanced system is caused by two possible situa-
tions: (1) the source voltages are not equal in magnitude and/or differ
in phase by angles that are unequal, or (2) load impedances are unequal.
Thus,
An unbalanced system is due to unbalanced voltage sources or an unbalanced load.
To simplify analysis, we will assume balanced source voltages, but an
unbalanced load.
A special technique for handling unbalanced
three-phase systems is the method of symmet-
rical components, which is beyond the scope of
this text.
Unbalanced three-phase systems are solved by direct application
of mesh and nodal analysis. Figure 12.23 shows an example of an unbal-
anced three-phase system that consists of balanced source voltages (not
shown in the figure) and an unbalanced Y-connected load (shown in the
figure). Since the load is unbalanced, ZA, ZB, and ZC are not equal. The
line currents are determined by Ohm’s law as
Ia =
VAN
ZA
, Ib =
VBN
ZB
, Ic =
VCN
ZC
(12.59)
This set of unbalanced line currents produces current in the neutral line,
which is not zero as in a balanced system. Applying KCL at node N
gives the neutral line current as
In = −(Ia + Ib + Ic) (12.60)
ZA
ZC
ZB
A
N
C
B
Ia
In
Ib
Ic
V
AN
V
BN
V
CN
Figure 12.23 Unbalanced three-phase
Y-connected load.
In a three-wire system where the neutral line is absent, we can still
find the line currents Ia, Ib, and Ic using mesh analysis. At node N,
KCL must be satisfied so that Ia + Ib + Ic = 0 in this case. The same
could be done for a -Y, Y- , or - three-wire system. As mentioned
earlier, in long distance power transmission, conductors in multiples of
three (multiple three-wire systems) are used, with the earth itself acting
as the neutral conductor.
CHAPTER 12 Three-Phase Circuits 501
To calculate power in an unbalanced three-phase system requires
that we find the power in each phase using Eqs. (12.46) to (12.49). The
total power is not simply three times the power in one phase but the sum
of the powers in the three phases.
E X A M P L E 1 2 . 9
The unbalanced Y-load of Fig. 12.23 has balanced voltages of 100 V and
the acb sequence. Calculate the line currents and the neutral current.
Take ZA = 15 , ZB = 10 + j5 , ZC = 6 − j8 .
Solution:
Using Eq. (12.59), the line currents are
Ia =
100 0◦
15
= 6.67 0◦
A
Ib =
100 120◦
10 + j5
=
100 120◦
11.18 26.56◦
= 8.94 93.44◦
A
Ic =
100 − 120◦
6 − j8
=
100 − 120◦
10 − 53.13◦
= 10 − 66.87◦
A
Using Eq. (12.60), the current in the neutral line is
In = −(Ia + Ib + Ic) = −(6.67 − 0.54 + j8.92 + 3.93 − j9.2)
= −10.06 + j0.28 = 10.06 178.4◦
A
P R A C T I C E P R O B L E M 1 2 . 9
The unbalanced -load of Fig. 12.24 is supplied by balanced voltages
of 200 V in the positive sequence. Find the line currents. Take Vab as
reference.
16 Ω
8 Ω
j6 Ω
10 Ω
–j5 Ω
A
C
B
Ia
Ib
Ic
Figure 12.24 Unbalanced -load; for
Practice Prob. 12.9.
Answer: 18.05 − 41.06◦
, 29.15 220.2◦
, 31.87 74.27◦
A.
E X A M P L E 1 2 . 1 0
For the unbalanced circuit in Fig. 12.25, find: (a) the line currents,
(b)thetotalcomplexpowerabsorbedbytheload, and(c)thetotalcomplex
power supplied by the source.
Solution:
(a) We use mesh analysis to find the required currents. For mesh 1,
120 − 120◦
− 120 0◦
+ (10 + j5)I1 − 10I2 = 0
502 PART 2 AC Circuits
+
− j5 Ω
A
N
10 Ω
–j10 Ω
C
b B
n
c
a
Ib
Ic
Ia
120 0° rms
120 120° rms 120 −120° rms
+ − − +
I2
I1
Figure12.25 For Example 12.10.
or
(10 + j5)I1 − 10I2 = 120
√
3 30◦
(12.10.1)
For mesh 2,
120 120◦
− 120 − 120◦
+ (10 − j10)I2 − 10I1 = 0
or
−10I1 + (10 − j10)I2 = 120
√
3 − 90◦
(12.10.2)
Equations (12.10.1) and (12.10.2) form a matrix equation:

10 + j5 −10
−10 10 − j10
 
I1
I2

=

120
√
3 30◦
120
√
3 − 90◦

The determinants are
=
10 + j5 −10
−10 10 − j10
= 50 − j50 = 70.71 − 45◦
1 =
120
√
3 30◦
−10
120
√
3 − 90◦
10 − j10
= 207.85(13.66 − j13.66)
= 4015 − 45◦
2 =
10 + j5 120
√
3 30◦
−10 120
√
3 − 90◦
= 207.85(13.66 − j5)
= 3023 − 20.1◦
CHAPTER 12 Three-Phase Circuits 503
The mesh currents are
I1 =
1
=
4015.23 − 45◦
70.71 − 45◦
= 56.78 A
I2 =
2
=
3023.4 − 20.1◦
70.71 − 45◦
= 42.75 24.9◦
A
The line currents are
Ia = I1 = 56.78 A, Ic = −I2 = 42.75 − 155.1◦
A
Ib = I2 − I1 = 38.78 + j18 − 56.78 = 25.46 135◦
A
(b) We can now calculate the complex power absorbed by the load. For
phase A,
SA = |Ia|2
ZA = (56.78)2
(j5) = j16,120 VA
For phase B,
SB = |Ib|2
ZB = (25.46)2
(10) = 6480 VA
For phase C,
SC = |Ic|2
ZC = (42.75)2
(−j10) = −j18,276 VA
The total complex power absorbed by the load is
SL = SA + SB + SC = 6480 − j2156 VA
(c) We check the result above by finding the power supplied by the source.
For the voltage source in phase a,
Sa = −VanI∗
a = −(120 0◦
)(56.78) = −6813.6 VA
For the source in phase b,
Sb = −VbnI∗
b = −(120 − 120◦
)(25.46 − 135◦
)
= −3055.2 105◦
= 790 − j2951.1 VA
For the source in phase c,
Sc = −VbnI∗
c = −(120 120◦
)(42.75 155.1◦
)
= −5130 275.1◦
= −456.03 + j5109.7 VA
The total complex power supplied by the three-phase source is
Ss = Sa + Sb + Sc = −6480 + j2156 VA
showing that Ss + SL = 0 and confirming the conservation principle of
ac power.
P R A C T I C E P R O B L E M 1 2 . 1 0
Find the line currents in the unbalanced three-phase circuit of Fig. 12.26
and the real power absorbed by the load.
504 PART 2 AC Circuits
10 Ω
A
j10 Ω
−j5 Ω
C
b B
c
a
220 −0° rms V
220 120° rms V
+
− + −
− +
220 −120° rms V
Figure12.26 For Practice Prob. 12.10.
Answer: 64 80.1◦
, 38.1 − 60◦
, 42.5 225◦
A, 4.84 kW.
12.9 PSPICE FOR THREE-PHASE CIRCUITS
PSpice can be used to analyze three-phase balanced or unbalanced circuits
in the same way it is used to analyze single-phase ac circuits. However,
a delta-connected source presents two major problems to PSpice. First, a
delta-connected source is a loop of voltage sources—which PSpice does
not like. To avoid this problem, we insert a resistor of negligible resistance
(say, 1 µ per phase) into each phase of the delta-connected source.
Second, the delta-connected source does not provide a convenient node
for the ground node, which is necessary to run PSpice. This problem can
be eliminated by inserting balanced wye-connected large resistors (say,
1 M per phase) in the delta-connected source so that the neutral node of
the wye-connected resistors serves as the ground node 0. Example 12.12
will illustrate this.
E X A M P L E 1 2 . 1 1
For the balanced Y- circuit in Fig. 12.27, use PSpice to find the line cur-
rent IaA, the phase voltage VAB, and the phase current IAC. Assume that
the source frequency is 60 Hz.
100 0° V
a
n
A
C
B
100 Ω
100 Ω
1 Ω
0.2 H
0.2 H
100 Ω
0.2 H
− +
100 −120° V
b 1 Ω
− +
100 120° V
c 1 Ω
− +
Figure12.27 For Example 12.10.
CHAPTER 12 Three-Phase Circuits 505
Solution:
The schematic is shown in Fig. 12.28. The pseudocomponents IPRINT
are inserted in the appropriate lines to obtain IaA and IAC, while VPRINT2
is inserted between nodes A and B to print differential voltage VAB. We
set the attributes of IPRINT and VPRINT2 each to AC = yes, MAG = yes,
PHASE = yes, to print only the magnitude and phase of the currents and
voltages. As a single-frequency analysis, we select Analysis/Setup/AC
Sweep and enter Total Pts = 1, Start Freq = 60, and Final Freq = 60.
Once the circuit is saved, it is simulated by selecting Analysis/Simulate.
The output file includes the following:
FREQ V(A,B) VP(A,B)
6.000E+01 1.699E+02 3.081E+01
FREQ IM(V_PRINT2) IP(V_PRINT2)
6.000E+01 2.350E+00 -3.620E+01
FREQ IM(V_PRINT3) IP(V_PRINT3)
6.000E+01 1.357E+00 -6.620E+01
From this, we obtain
IaA = 2.35 − 36.2◦
A
VAB = 169.9 30.81◦
V, IAC = 1.357 − 66.2◦
A
A
B
C
R4
R6
100
0.2H L1
AC = yes
MAG = yes
PHASE = yes
AC = yes
MAG = yes
PHASE = yes
AC = yes
MAG = yes
PHASE = yes
ACMAG = 100 V
ACPHASE = 0
1
IPRINT
IPRINT
ACMAG = 100
ACPHASE = −120
1
R2
R1
V2
V1
ACMAG = 100 V
ACPHASE = 120
1
R3
V3
R5 100
0.2H
0.2H
L3
L2
100
−
+
−
+
−
+
0
Figure12.28 Schematic for the circuit in Fig. 12.27.
P R A C T I C E P R O B L E M 1 2 . 1 1
Refer to the balanced Y-Y circuit of Fig. 12.29. Use PSpice to find the
line current IbB and the phase voltage VAN . Take f = 100 Hz.
506 PART 2 AC Circuits
120 60° V
a
n
A
C
N
10 Ω
2 Ω
10 mH
− +
120 −60° V
b 2 Ω
− +
120 180° V
c 2 Ω
1.6 mH
1.6 mH
1.6 mH
− +
10 Ω 10 mH
10 Ω
10 mH
B
Figure12.29 For Practice Prob. 12.11.
Answer: 100.9 60.87◦
V, 8.547 − 91.27◦
A.
E X A M P L E 1 2 . 1 2
Consider the unbalanced - circuit in Fig. 12.30. Use PSpice to find
the generator current Iab, the line current IbB, and the phase current IBC.
208 130° V
208 −110° V
208 10° V
A
B
b
C
c
a
50 Ω
2 Ω
j30 Ω
j5 Ω
−j40 Ω
2 Ω j5 Ω
2 Ω j5 Ω
+
−
+
−
+
−
Figure12.30 For Example 12.12.
Solution:
As mentioned above, we avoid the loop of voltage sources by inserting a
1-µ series resistor in the delta-connected source. To provide a ground
node 0, we insert balanced wye-connected resistors (1 M per phase)
in the delta-connected source, as shown in the schematic in Fig. 12.31.
Three IPRINT pseudocomponents with their attributes are inserted to be
able to get the required currents Iab, IbB, and IBC. Since the operating
frequency is not given and the inductances and capacitances should be
specified instead of impedances, we assume ω = 1 rad/s so that f =
1/2π = 0.159155 Hz. Thus,
CHAPTER 12 Three-Phase Circuits 507
L =
XL
ω
and C =
1
ωXC
We select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start
Freq = 0.159155, and Final Freq = 0.159155. Once the schematic
is saved, we select Analysis/Simulate to simulate the circuit. The output
file includes:
FREQ IM(V_PRINT1) IP(V_PRINT1)
1.592E-01 9.106E+00 1.685E+02
FREQ IM(V_PRINT2) IP(V_PRINT2)
1.592E-01 5.959E+00 2.821E+00
FREQ IM(V_PRINT3) IP(V_PRINT3)
1.592E-01 5.500E+00 -7.532E+00
from which we get
Iab = 5.96 2.82◦
A
IbB = 9.106 168.5◦
A, IBC = 5.5 − 7.53◦
A
IPRINT
ACMAG = 208V
ACPHASE = 130
ACMAG = 208V
ACPHASE = -110
V1
V2
30H L4
25m C1
R1
2
1Meg
R8
L1
5H
R2
2
L2
5H
R3
2
L3
5H
R5 1Meg
R4 1u
R6 1Meg
R7 1u
R9 1u
−
+
−
+
ACMAG = 208V
ACPHASE = 110
V3
−
+
IPRINT
R10 50
AC = yes
MAG = yes
PHASE = yes
AC = yes
MAG = yes
PHASE = yes
AC = yes
MAG = yes
PHASE = yes
IPRINT
0
Figure12.31 Schematic for the circuit in Fig. 12.30.
P R A C T I C E P R O B L E M 1 2 . 1 2
For the unbalanced circuit in Fig. 12.32, use PSpice to find the generator
current Ica, the line current IcC, and the phase current IAB.
508 PART 2 AC Circuits
220 90° V
220 −150° V
220 −30° V
A
B
b
C
c
a
10 Ω
j10 Ω
10 Ω
+
−
+
−
+
−
10 Ω
−j10 Ω
Figure12.32 For Practice Prob. 12.12.
Answer: 24.68 − 90◦
A, 15.56 105◦
A, 37.24 83.79◦
A.
†12.10 APPLICATIONS
Both wye and delta source connections have important practical applica-
tions. The wye source connection is used for long distance transmission
of electric power, where resistive losses (I2
R) should be minimal. This
is due to the fact that the wye connection gives a line voltage that is
√
3
greater than the delta connection; hence, for the same power, the line
current is
√
3 smaller. The delta source connection is used when three
single-phase circuits are desired from a three-phase source. This conver-
sion from three-phase to single-phase is required in residential wiring, be-
cause household lighting and appliances use single-phase power. Three-
phase power is used in industrial wiring where a large power is required.
In some applications, it is immaterial whether the load is wye- or delta-
connected. For example, both connections are satisfactory with induction
motors. In fact, some manufacturers connect a motor in delta for 220 V
and in wye for 440 V so that one line of motors can be readily adapted to
two different voltages.
Here we consider two practical applications of those concepts cov-
ered in this chapter: power measurement in three-phase circuits and res-
idential wiring.
Three-phase
load (wye
or delta,
balanced or
unbalanced)
W1
a
b
c
W3
±
±
W2
±
±
±
±
o
Figure 12.33 Three-wattmeter method for
measuring three-phase power.
12.10.1 Three-Phase Power Measurement
Section 11.9 presented the wattmeter as the instrument for measuring the
average (or real) power in single-phase circuits. A single wattmeter can
also measure the average power in a three-phase system that is balanced,
so that P1 = P2 = P3; the total power is three times the reading of that
one wattmeter. However, two or three single-phase wattmeters are neces-
sary to measure power if the system is unbalanced. The three-wattmeter
method of power measurement, shown in Fig. 12.33, will work regardless
of whether the load is balanced or unbalanced, wye- or delta-connected.
CHAPTER 12 Three-Phase Circuits 509
The three-wattmeter method is well suited for power measurement in a
three-phase system where the power factor is constantly changing. The
total average power is the algebraic sum of the three wattmeter readings,
PT = P1 + P2 + P3 (12.61)
where P1, P2, and P3 correspond to the readings of wattmeters W1, W2,
and W3, respectively. Notice that the common or reference point o in Fig.
12.33 is selected arbitrarily. If the load is wye-connected, point o can be
connected to the neutral point n. For a delta-connected load, point o can
be connected to any point. If point o is connected to point b, for example,
the voltage coil in wattmeter W2 reads zero and P2 = 0, indicating that
wattmeter W2 is not necessary. Thus, two wattmeters are sufficient to
measure the total power.
The two-wattmeter method is the most commonly used method for
three-phase power measurement. The two wattmeters must be properly
connected to any two phases, as shown typically in Fig. 12.34. Notice
that the current coil of each wattmeter measures the line current, while the
respective voltage coil is connected between the line and the third line and
measures the line voltage. Also notice that the ± terminal of the voltage
coil is connected to the line to which the corresponding current coil is
connected. Although the individual wattmeters no longer read the power
taken by any particular phase, the algebraic sum of the two wattmeter
readingsequalsthetotalaveragepowerabsorbedbytheload, regardlessof
whether it is wye- or delta-connected, balanced or unbalanced. The total
real power is equal to the algebraic sum of the two wattmeter readings,
PT = P1 + P2 (12.62)
We will show here that the method works for a balanced three-phase
system.
Three-phase
load (wye
or delta,
balanced or
unbalanced)
W1
a
b
c
W2
±
±
±
±
Figure 12.34 Two-wattmeter method for
measuring three-phase power.
Consider the balanced, wye-connected load in Fig. 12.35. Our
objective is to apply the two-wattmeter method to find the average power
absorbed by the load. Assume the source is in the abc sequence and the
load impedance ZY = ZY θ. Due to the load impedance, each voltage
coil leads its current coil by θ, so that the power factor is cos θ. We recall
that each line voltage leads the corresponding phase voltage by 30◦
. Thus,
the total phase difference between the phase current Ia and line voltage
W1
a
b
c
W2
± ±
±
±
Ib
Ic
Ia
ZY ZY
ZY
+
−
V
ab
−
+
V
cb
Figure12.35 Two-wattmeter method applied to a balanced wye load.
510 PART 2 AC Circuits
Vab is θ + 30◦
, and the average power read by wattmeter W1 is
P1 = Re[VabI∗
a] = VabIa cos(θ + 30◦
) = VLIL cos(θ + 30◦
) (12.63)
Similarly, we can show that the average power read by wattmeter 2 is
P2 = Re[VcbI∗
c] = VcbIc cos(θ − 30◦
) = VLIL cos(θ − 30◦
) (12.64)
We now use the trigonometric identities
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
(12.65)
to find the sum and the difference of the two wattmeter readings in Eqs.
(12.63) and (12.64):
P1 + P2 = VLIL[cos(θ + 30◦
) + cos(θ − 30◦
)]
= VlIL(cos θ cos 30◦
− sin θ sin 30◦
+ cos θ cos 30◦
+ sin θ sin 30◦
)
= VLIL2 cos 30◦
cos θ =
√
3VLIL cos θ
(12.66)
since 2 cos 30◦
=
√
3. Comparing Eq. (12.66) with Eq. (12.50) shows
that the sum of the wattmeter readings gives the total average power,
PT = P1 + P2 (12.67)
Similarly,
P1 − P2 = VLIL[cos(θ + 30◦
) − cos(θ − 30◦
)]
= VlIL(cos θ cos 30◦
− sin θ sin 30◦
− cos θ cos 30◦
− sin θ sin 30◦
)
= −VLIL2 sin 30◦
sin θ
P2 − P1 = VLIL sin θ
(12.68)
since 2 sin 30◦
= 1. Comparing Eq. (12.68) with Eq. (12.51) shows that
thedifferenceofthewattmeterreadingsisproportionaltothetotalreactive
power, or
QT =
√
3(P2 − P1) (12.69)
From Eqs. (12.67) and (12.69), the total apparent power can be obtained
as
ST = P 2
T + Q2
T (12.70)
Dividing Eq. (12.69) by Eq. (12.67) gives the tangent of the power factor
angle as
tan θ =
QT
PT
=
√
3
P2 − P1
P2 + P1
(12.71)
from which we can obtain the power factor as pf = cos θ. Thus, the two-
wattmeter method not only provides the total real and reactive powers, it
can also be used to compute the power factor. From Eqs. (12.67), (12.69),
and (12.71), we conclude that:
CHAPTER 12 Three-Phase Circuits 511
1. If P2 = P1, the load is resistive.
2. If P2  P1, the load is inductive.
3. If P2  P1, the load is capacitive.
Although these results are derived from a balanced wye-connected load,
they are equally valid for a balanced delta-connected load. However,
the two-wattmeter method cannot be used for power measurement in a
three-phase four-wire system unless the current through the neutral line
is zero. We use the three-wattmeter method to measure the real power in
a three-phase four-wire system.
E X A M P L E 1 2 . 1 3
Three wattmeters W1, W2, and W3 are connected, respectively, to phases
a, b, and c to measure the total power absorbed by the unbalanced wye-
connectedloadinExample12.9(seeFig.12.23). (a)Predictthewattmeter
readings. (b) Find the total power absorbed.
Solution:
Part of the problem is already solved in Example 12.9. Assume that the
wattmeters are properly connected as in Fig. 12.36.
−
+
V
CN
−
+
V
AN
−
+
V
BN
W3
Ia
Ib
Ic
In
W1
A
N
C
B
W2
j5 Ω
−j8 Ω
10 Ω 6 Ω
15 Ω
Figure12.36 For Example 12.13.
(a) From Example 12.9,
VAN = 100 0◦
, VBN = 100 120◦
, VCN = 100 − 120◦
V
while
Ia = 6.67 0◦
, Ib = 8.94 93.44◦
, Ic = 10 − 66.87◦
A
We calculate the wattmeter readings as follows:
P1 = Re(VAN I∗
a) = VAN Ia cos(θVAN
− θIa
)
= 100 × 6.67 × cos(0◦
− 0◦
) = 667 W
P2 = Re(VBN I∗
b) = VBN Ib cos(θVBN
− θIb
)
= 100 × 8.94 × cos(120◦
− 93.44◦
) = 800 W
P3 = Re(VCN I∗
c) = VCN Ic cos(θVCN
− θIc
)
= 100 × 10 × cos(−120◦
+ 66.87◦
) = 600 W
512 PART 2 AC Circuits
(b) The total power absorbed is
PT = P1 + P2 + P3 = 667 + 800 + 600 = 2067 W
We can find the power absorbed by the resistors in Fig. 12.36 and use that
to check or confirm this result.
PT = |Ia|2
(15) + |Ib|2
(10) + |Ic|2
(6)
= 6.672
(15) + 8.942
(10) + 102
(6)
= 667 + 800 + 600 = 2067 W
which is exactly the same thing.
P R A C T I C E P R O B L E M 1 2 . 1 3
Repeat Example 12.13 for the network in Fig. 12.24 (see Practice Prob.
12.9). Hint: Connect the reference point o in Fig. 12.33 to point B.
Answer: (a) 2961 W, 0 W, 4339 W, (b) 7300 W.
E X A M P L E 1 2 . 1 4
The two-wattmeter method produces wattmeter readings P1 = 1560 W
and P2 = 2100 W when connected to a delta-connected load. If the line
voltage is 220 V, calculate: (a) the per-phase average power, (b) the per-
phase reactive power, (c) the power factor, and (d) the phase impedance.
Solution:
We can apply the given results to the delta-connected load.
(a) The total real or average power is
PT = P1 + P2 = 1560 + 2100 = 3660 W
The per-phase average power is then
Pp =
1
3
PT = 1220 W
(b) The total reactive power is
QT =
√
3(P2 − P1) =
√
3(2100 − 1560) = 935.3 VAR
so that the per-phase reactive power is
Qp =
1
3
QT = 311.77 VAR
(c) The power angle is
θ = tan−1 QT
PT
= tan−1 935.3
3660
= 14.33◦
Hence, the power factor is
cos θ = 0.9689 (leading)
It is a leading pf because QT is positive or P2  P1.
CHAPTER 12 Three-Phase Circuits 513
(c) The phase impedance is Zp = Zp θ. We know that θ is the same as
the pf angle; that is, θ = 14.57◦
.
Zp =
Vp
Ip
We recall that for a delta-connected load, Vp = VL = 220 V. From Eq.
(12.46),
Pp = VpIp cos θ ⇒ Ip =
1220
220 × 0.9689
= 5.723 A
Hence,
Zp =
Vp
Ip
=
220
5.723
= 38.44 
and
Zp = 38.44 14.33◦

P R A C T I C E P R O B L E M 1 2 . 1 4
Let the line voltage VL = 208 V and the wattmeter readings of the bal-
anced system in Fig. 12.35 be P1 = −560 W and P2 = 800 W. Deter-
mine:
(a) the total average power
(b) the total reactive power
(c) the power factor
(d) the phase impedance
Is the impedance inductive or capacitive?
Answer: (a) 240 W, (b) 2355.6 VAR, (c) 0.1014, (d) 18.25 84.18◦
,
inductive.
E X A M P L E 1 2 . 1 5
The three-phase balanced load in Fig. 12.35 has impedance per phase of
ZY = 8 + j6 . If the load is connected to 208-V lines, predict the read-
ings of the wattmeters W1 and W2. Find PT and QT .
Solution:
The impedance per phase is
ZY = 8 + j6 = 10 36.87◦

so that the pf angle is 36.87◦
. Since the line voltage VL = 208 V, the line
current is
IL =
Vp
|ZY |
=
208/
√
3
10
= 12 A
514 PART 2 AC Circuits
Then
P1 = VLIL cos(θ + 30◦
) = 208 × 12 × cos(36.87◦
+ 30◦
)
= 980.48 W
P2 = VLIL cos(θ − 30◦
) = 208 × 12 × cos(36.87◦
− 30◦
)
= 2478.1 W
Thus, wattmeter 1 reads 980.48 W, while wattmeter 2 reads 2478.1 W.
Since P2  P1, the load is inductive. This is evident from the load ZY
itself. Next,
PT = P1 + P2 = 3.4586 kW
and
QT =
√
3(P2 − P1) =
√
3(1497.6) VAR = 2.594 kVAR
P R A C T I C E P R O B L E M 1 2 . 1 5
If the load in Fig. 12.35 is delta-connected with impedance per phase of
Zp = 30−j40  and VL = 440 V, predict the readings of the wattmeters
W1 and W2. Calculate PT and QT .
Answer: 6.166 kW, 0.8021 kW, 6.968 kW, −9.291 kVAR.
12.10.2 Residential Wiring
In the United States, most household lighting and appliances operate
on 120-V, 60-Hz, single-phase alternating current. (The electricity may
also be supplied at 110, 115, or 117 V, depending on the location.) The
local power company supplies the house with a three-wire ac system.
Typically, as in Fig. 12.37, the line voltage of, say, 12,000 V is stepped
down to 120/240 V with a transformer (more details on transformers
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
Light
pole
Grounded metal
stake
Ground
Wall of
house
Circuit
# 1
120 V
Circuit
# 2
120 V
Circuit
# 3
120 V
Fuse
Fuse Fuses
Switch
Watt-hour meter
Step-down
transformer
Figure 12.37 A 120/240 household power system.
(Source: A. Marcus and C. M. Thomson, Electricity for Technicians,
2nd ed. [Englewood Cliffs, NJ: Prentice Hall, 1975], p. 324.)
CHAPTER 12 Three-Phase Circuits 515
in the next chapter). The three wires coming from the transformer are
typically colored red (hot), black (hot), and white (neutral). As shown in
Fig. 12.38, the two 120-V voltages are opposite in phase and hence add up
to zero. That is, VW = 0 0◦
, VB = 120 0◦
, VR = 120 180◦
= −VB.
VBR = VB − VR = VB − (−VB) = 2VB = 240 0◦
(12.72)
Since most appliances are designed to operate with 120 V, the lighting
and appliances are connected to the 120-V lines, as illustrated in Fig.
12.39 for a room. Notice in Fig. 12.37 that all appliances are connected
in parallel. Heavy appliances that consume large currents, such as air
conditioners, dishwashers, ovens, and laundry machines, are connected
to the 240-V power line.
120 V
lights
120 V
appliance
120 V
lights
120 V
appliance
120 V
120 V
−
+
+
− 240 V
appliance
Black
(hot) B
White
(neutral)
Red (hot)
W
R
Ground
To other houses
Transformer
House
Figure12.38 Single-phase three-wire residential wiring.
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
Lamp sockets
Base outlets
120 volts
Ungrounded conductor
Switch
Neutral
Figure 12.39 A typical wiring diagram of
a room.
(Source: A. Marcus and C. M.
Thomson, ElectricityforTech-
nicians, 2nd ed. [Englewood
Cliffs, NJ:PrenticeHall, 1975],
p. 325.)
Because of the dangers of electricity, house wiring is carefully reg-
ulated by a code drawn by local ordinances and by the National Electrical
Code (NEC). To avoid trouble, insulation, grounding, fuses, and circuit
breakers are used. Modern wiring codes require a third wire for a sep-
arate ground. The ground wire does not carry power like the neutral
wire but enables appliances to have a separate ground connection. Figure
12.40 shows the connection of the receptacle to a 120-V rms line and to
the ground. As shown in the figure, the neutral line is connected to the
ground (the earth) at many critical locations. Although the ground line
+
−
Fuse or circuit breaker
120 V rms
Hot wire
Receptacle
To other appliances
Neutral wire
Power system
ground
Service
panel ground
Ground wire
Figure12.40 Connection of a receptacle to the hot line and to the ground.
516 PART 2 AC Circuits
seems redundant, grounding is important for many reasons. First, it is re-
quired by NEC. Second, grounding provides a convenient path to ground
for lightning that strikes the power line. Third, grounds minimize the risk
of electric shock. What causes shock is the passage of current from one
part of the body to another. The human body is like a big resistor R. If V
is the potential difference between the body and the ground, the current
through the body is determined by Ohm’s law as
I =
V
R
(12.73)
The value of R varies from person to person and depends on whether the
body is wet or dry. How great or how deadly the shock is depends on
the amount of current, the pathway of the current through the body, and
the length of time the body is exposed to the current. Currents less than
1 mA may not be harmful to the body, but currents greater than 10 mA can
cause severe shock. A modern safety device is the ground-fault circuit
interrupter (GFCI), used in outdoor circuits and in bathrooms, where the
risk of electric shock is greatest. It is essentially a circuit breaker that
opens when the sum of the currents iR, iW , and iB through the red, white,
and the black lines is not equal to zero, or iR + iW + iB = 0.
The best way to avoid electric shock is to follow safety guidelines
concerning electrical systems and appliances. Here are some of them:
• Never assume that an electrical circuit is dead. Always check to
be sure.
• Use safety devices when necessary, and wear suitable clothing
(insulated shoes, gloves, etc.).
• Never use two hands when testing high-voltage circuits, since
the current through one hand to the other hand has a direct path
through your chest and heart.
• Do not touch an electrical appliance when you are wet.
Remember that water conducts electricity.
• Be extremely careful when working with electronic appliances
such as radio and TV because these appliances have large
capacitors in them. The capacitors take time to discharge after
the power is disconnected.
• Always have another person present when working on a wiring
system, just in case of an accident.
12.11 SUMMARY
1. The phase sequence is the order in which the phase voltages of a
three-phase generator occur with respect to time. In an abc
sequence of balanced source voltages, Van leads Vbn by 120◦
,
which in turn leads Vcn by 120◦
. In an acb sequence of balanced
voltages, Van leads Vcn by 120◦
, which in turn leads Vbn by 120◦
.
2. A balanced wye- or delta-connected load is one in which the three-
phase impedances are equal.
3. The easiest way to analyze a balanced three-phase circuit is to
transform both the source and the load to a Y-Y system and then
CHAPTER 12 Three-Phase Circuits 517
analyze the single-phase equivalent circuit. Table 12.1 presents a
summary of the formulas for phase currents and voltages and line
currents and voltages for the four possible configurations.
4. The line current IL is the current flowing from the generator to the
load in each transmission line in a three-phase system. The line
voltage VL is the voltage between each pair of lines, excluding the
neutral line if it exists. The phase current Ip is the current flowing
through each phase in a three-phase load. The phase voltage Vp is
the voltage of each phase. For a wye-connected load,
VL =
√
3Vp and IL = Ip
For a delta-connected load,
VL = Vp and IL =
√
3Ip
5. The total instantaneous power in a balanced three-phase system is
constant and equal to the average power.
6. The total complex power absorbed by a balanced three-phase
Y-connected or -connected load is
S = P + jQ =
√
3VLIL θ
where θ is the angle of the load impedances.
7. An unbalanced three-phase system can be analyzed using nodal or
mesh analysis.
8. PSpice is used to analyze three-phase circuits in the same way as it
is used for analyzing single-phase circuits.
9. The total real power is measured in three-phase systems using
either the three-wattmeter method or the two-wattmeter method.
10. Residential wiring uses a 120/240-V, single-phase, three-wire
system.
REVIEW QUESTIONS
12.1 What is the phase sequence of a three-phase motor
for which VAN = 220 − 100◦
V and
VBN = 220 140◦
V?
(a) abc (b) acb
12.2 If in an acb phase sequence, Van = 100 − 20◦
,
then Vcn is:
(a) 100 − 140◦
(b) 100 100◦
(c) 100 − 50◦
(d) 100 10◦
12.3 Which of these is not a required condition for a
balanced system:
(a) |Van| = |Vbn| = |Vcn|
(b) Ia + Ib + Ic = 0
(c) Van + Vbn + Vcn = 0
(d) Source voltages are 120◦
out of phase with each
other.
(e) Load impedances for the three phases are equal.
12.4 In a Y-connected load, the line current and phase
current are equal.
(a) True (b) False
12.5 In a -connected load, the line current and phase
current are equal.
(a) True (b) False
12.6 In a Y-Y system, a line voltage of 220 V produces a
phase voltage of:
(a) 381 V (b) 311 V (c) 220 V
(d) 156 V (e) 127 V
518 PART 2 AC Circuits
12.7 In a - system, a phase voltage of 100 V produces
a line voltage of:
(a) 58 V (b) 71 V (c) 100 V
(d) 173 V (e) 141 V
12.8 When a Y-connected load is supplied by voltages in
abc phase sequence, the line voltages lag the
corresponding phase voltages by 30◦
.
(a) True (b) False
12.9 In a balanced three-phase circuit, the total
instantaneous power is equal to the average power.
(a) True (b) False
12.10 The total power supplied to a balanced -load is
found in the same way as for a balanced Y-load.
(a) True (b) False
Answers: 12.1a, 12.2a, 12.3c, 12.4a, 12.5b, 12.6e, 12.7c, 12.8b,
12.9a, 12.10a.
PROBLEMS1
Section 12.2 Balanced Three-Phase Voltages
12.1 If Vab = 400 V in a balanced Y-connected
three-phase generator, find the phase voltages,
assuming the phase sequence is:
(a) abc (b) acb
12.2 What is the phase sequence of a balanced
three-phase circuit for which Van = 160 30◦
V and
Vcn = 160 − 90◦
V? Find Vbn.
12.3 Determine the phase sequence of a balanced
three-phase circuit in which Vbn = 208 130◦
V
and Vcn = 208 10◦
V. Obtain Van.
12.4 Assuming the abc sequence, if Vca = 208 20◦
V
in a balanced three-phase circuit, find Vab, Vbc, Van,
and Vbn.
12.5 Given that the line voltages of a three-phase circuit
are
Vab = 420 0◦
, Vbc = 420 − 120◦
Vac = 420 120◦
V
find the phase voltages Van, Vbn, and Vcn.
Section 12.3 Balanced Wye-Wye Connection
12.6 For the Y-Y circuit of Fig. 12.41, find the line
currents, the line voltages, and the load voltages.
a A
b B
c C
n N
− +
− +
− +
220 0° V
220 −120° V
220 120° V
10 Ω j5 Ω
10 Ω j5 Ω
10 Ω j5 Ω
Figure 12.41 For Prob. 12.6.
12.7 Obtain the line currents in the three-phase circuit of
Fig. 12.42 below.
+
−
A
N
n
a
440 0° V
440 120° V 440 −120° V
+
− − +
6 − j8 Ω 6 − j8 Ω
6 − j8 Ω
Ia
Ib
Ic
Figure12.42 For Prob. 12.7.
1Remember that unless stated otherwise, all given voltages and currents are rms values.
CHAPTER 12 Three-Phase Circuits 519
12.8 A balanced Y-connected load with a phase
impedance of 16 + j9  is connected to a balanced
three-phase source with a line voltage of 220 V.
Calculate the line current IL.
12.9 A balanced Y-Y four-wire system has phase voltages
Van = 120 0◦
, Vbn = 120 − 120◦
Vcn = 120 120◦
V
The load impedance per phase is 19 + j13 , and
the line impedance per phase is 1 + j2 . Solve for
the line currents and neutral current.
12.10 For the circuit in Fig. 12.43, determine the current in
the neutral line.
+
−
+
−
− +
2 Ω
2 Ω
20 Ω
2 Ω
220 −120° V
220 120° V
10 + j5 Ω
25 − j10 Ω
220 0° V
Figure 12.43 For Prob. 12.10.
Section 12.4 Balanced Wye-Delta Connection
12.11 For the three-phase circuit of Fig. 12.44,
IbB = 30 60◦
A and VBC = 220 10◦
V. Find Van,
VAB , IAC, and Z.
+
−
+
−
− +
C
B
A
Z
Z
Z
Vcn
V
an
V
bn
n
b
a
c
Figure 12.44 For Prob. 12.11.
12.12 Solve for the line currents in the Y- circuit of Fig.
12.45. Take Z = 60 45◦
.
Z∆
A
C
c
B
a
+
−
+
−
Ia
Ib
Ic
+
−
n
b
Z∆ Z∆
110 0° V
110 −120° V
110 120° V
Figure 12.45 For Prob. 12.12.
12.13 The circuit in Fig. 12.46 is excited by a balanced
three-phase source with a line voltage of 210 V. If
Z = 1 + j1 , Z = 24 − j30 , and
ZY = 12 + j5 , determine the magnitude of the
line current of the combined loads.
a
b
c
Zl
Zl
Zl
Z∆
Z∆
Z∆
ZY
ZY
ZY
Figure 12.46 For Prob. 12.13.
12.14 A balanced delta-connected load has a phase current
IAC = 10 − 30◦
A.
(a) Determine the three line currents assuming that
the circuit operates in the positive phase
sequence.
(b) Calculate the load impedance if the line voltage
is VAB = 110 0◦
V.
12.15 In a wye-delta three-phase circuit, the source is a
balanced, positive phase sequence with
Van = 120 0◦
V. It feeds a balanced load with
Z = 9 + j12  per phase through a balanced line
with Z = 1 + j0.5  per phase. Calculate the
phase voltages and currents in the load.
520 PART 2 AC Circuits
12.16 If Van = 440 60◦
V in the network of Fig. 12.47,
find the load phase currents IAB , IBC, and ICA.
Three-phase,
Y-connected
generator
(+) phase
sequence
12 Ω
j9 Ω
j9 Ω
12 Ω
12 Ω j9 Ω
a
b
c
A
B
C
Figure 12.47 For Prob. 12.16.
Section 12.5 Balanced Delta-Delta Connection
12.17 For the - circuit of Fig. 12.48, calculate the
phase and line currents.
+
−
+
−
+
−
30 Ω
30 Ω
173 −120° V
173 0° V
j10 Ω
j10 Ω 30 Ω
j10 Ω
A
B
a
b
c C
173 120° V
Figure 12.48 For Prob. 12.17.
12.18 Refer to the - circuit in Fig. 12.49. Find the line
and phase currents. Assume that the load impedance
is 12 + j9  per phase.
A
C
B
Ia
Ib
Ic
+
−
+
−
− +
210 0° V
210 −120° V
210 120° V
IBC
IAB
ICA
ZL
ZL
ZL
Figure 12.49 For Prob. 12.18.
12.19 Find the line currents Ia, Ib, and Ic in the three-phase
network of Fig. 12.50 below. Take
Z = 12 − j15 , ZY = 4 + j6 ,
and Z = 2 .
12.20 A balanced delta-connected source has phase
voltage Vab = 416 30◦
V and a positive phase
sequence. If this is connected to a balanced
delta-connected load, find the line and phase
currents. Take the load impedance per phase as
60 30◦
 and line impedance per phase as
1 + j1 .
A
C
B
Ia
Ib
Ic
+
−
+
−
− +
208 0° V
208 −120° V
208 120° V
Zl
Zl
Zl
ZY
ZY
Z∆ Z∆
Z∆
ZY
Figure12.50 For Prob. 12.19.
CHAPTER 12 Three-Phase Circuits 521
Section 12.6 Balanced Delta-Wye Connection
12.21 In the circuit of Fig. 12.51, if Vab = 440 10◦
,
Vbc = 440 250◦
, Vca = 440 130◦
V, find the line
currents.
b
c
a
3 + j2 Ω
3 + j2 Ω 10 − j8 Ω
10 − j8 Ω
10 − j8 Ω
3 + j2 Ω
+
−
+
−
+
−
Vca
Vab
Vbc
Ib
Ic
Ia
Figure 12.51 For Prob. 12.21.
12.22 For the balanced circuit in Fig. 12.52,
Vab = 125 0◦
V. Find the line currents IaA, IbB ,
and IcC.
N
IbB
IcC
IaA
24 Ω
24 Ω
24 Ω
−j15 Ω
−j15 Ω
−j15 Ω
a
b
c
A
C
B
Three-phase,
∆-connected
generator
(+) phase
sequence
Figure 12.52 For Prob. 12.22.
12.23 In a balanced three-phase -Y circuit, the source is
connected in the positive sequence, with
Vab = 220 20◦
V and ZY = 20 + j15 . Find the
line currents.
12.24 A delta-connected generator supplies a balanced
wye-connected load with an impedance of
30 − 60◦
. If the line voltages of the generator
have a magnitude of 400 V and are in the positive
phase sequence, find the line current IL and phase
voltage Vp at the load.
Section 12.7 Power in a Balanced System
12.25 A balanced wye-connected load absorbs a total
power of 5 kW at a leading power factor of 0.6 when
connected to a line voltage of 240 V. Find the
impedance of each phase and the total complex
power of the load.
12.26 A balanced wye-connected load absorbs 50 kVA at a
0.6 lagging power factor when the line voltage is
440 V. Find the line current and the phase
impedance.
12.27 A three-phase source delivers 4800 VA to a
wye-connected load with a phase voltage of 208 V
and a power factor of 0.9 lagging. Calculate the
source line current and the source line voltage.
12.28 A balanced wye-connected load with a phase
impedance of 10 − j16  is connected to a balanced
three-phase generator with a line voltage of 220 V.
Determine the line current and the complex power
absorbed by the load.
12.29 The total power measured in a three-phase system
feeding a balanced wye-connected load is 12 kW at
a power factor of 0.6 leading. If the line voltage is
208 V, calculate the line current IL and the load
impedance ZY .
12.30 Given the circuit in Fig. 12.53 below, find the total
complex power absorbed by the load.
+
−
+
−
− +
1 Ω 9 Ω
9 Ω
9 Ω
110 120° V
j2 Ω
1 Ω j2 Ω
1 Ω j2 Ω
1 Ω j2 Ω
j12 Ω j12 Ω
j12 Ω
110 240° V
110 0° V
Figure12.53 For Prob. 12.30.
522 PART 2 AC Circuits
12.31 Find the real power absorbed by the load in Fig.
12.54.
A
−j6 Ω
j3 Ω
C
b B
c
a
100 −120° V
− +
+
−
+
−
100 120° V 100 0° V
5 Ω
5 Ω
5 Ω
8 Ω
4 Ω
10 Ω
Figure 12.54 For Prob. 12.31.
12.32 For the three-phase circuit in Fig. 12.55, find the
average power absorbed by the delta-connected load
with Z = 21 + j24 .
1 Ω
− +
1 Ω
1 Ω
j0.5 Ω
j0.5 Ω
j0.5 Ω
100 0° V rms
100 −120° V rms
100 120° V rms
− +
− +
Z∆
Z∆
Z∆
Figure 12.55 For Prob. 12.32.
12.33 A balanced delta-connected load draws 5 kW at a
power factor of 0.8 lagging. If the three-phase
system has an effective line voltage of 400 V, find
the line current.
12.34 A balanced three-phase generator delivers 7.2 kW to
a wye-connected load with impedance 30 − j40 
per phase. Find the line current IL and the line
voltage VL.
12.35 Refer to Fig. 12.46. Obtain the complex power
absorbed by the combined loads.
12.36 A three-phase line has an impedance of 1 + j3 
per phase. The line feeds a balanced
delta-connected load, which absorbs a total complex
power of 12 + j5 kVA. If the line voltage at the load
end has a magnitude of 240 V, calculate the
magnitude of the line voltage at the source end and
the source power factor.
12.37 A balanced wye-connected load is connected to the
generator by a balanced transmission line with an
impedance of 0.5 + j2  per phase. If the load is
rated at 450 kW, 0.708 power factor lagging, 440-V
line voltage, find the line voltage at the generator.
12.38 A three-phase load consists of three 100- resistors
that can be wye- or delta-connected. Determine
which connection will absorb the most average
power from a three-phase source with a line voltage
of 110 V. Assume zero line impedance.
12.39 The following three parallel-connected three-phase
loads are fed by a balanced three-phase source.
Load 1: 250 kVA, 0.8 pf lagging
Load 2: 300 kVA, 0.95 pf leading
Load 3: 450 kVA, unity pf
If the line voltage is 13.8 kV, calculate the line
current and the power factor of the source. Assume
that the line impedance is zero.
Section 12.8 Unbalanced Three-Phase Systems
12.40 For the circuit in Fig. 12.56, Za = 6 − j8 ,
Zb = 12 + j9 , and Zc = 15 . Find the line
currents Ia, Ib, and Ic.
Ib
Ic
Ia
150 120° V
150 0° V
150 −120° V
+
−
+
−
+
−
Zb
Za
Zc
Figure 12.56 For Prob. 12.40.
12.41 A four-wire wye-wye circuit has
Van = 120 120◦
, Vbn = 120 0◦
Vcn = 120 − 120◦
V
If the impedances are
ZAN = 20 60◦
, ZBN = 30 0◦
Zcn = 40 30◦

find the current in the neutral line.
12.42 For the wye-connected load of Fig. 12.57, the line
voltages all have a magnitude of 250 V and are in a
positive phase sequence. Calculate the line currents
and the neutral current.
40 60° Ω
Ia
60 −45° Ω
Ib
20 0° Ω
Ic
In
Figure 12.57 For Prob. 12.42.
CHAPTER 12 Three-Phase Circuits 523
12.43 A delta-connected load whose phase impedances are
ZAB = 50 , ZBC = −j50 , and ZCA = j50  is
fed by a balanced wye-connected three-phase source
with Vp = 100 V. Find the phase currents.
12.44 A balanced three-phase wye-connected generator
with Vp = 220 V supplies an unbalanced
wye-connected load with ZAN = 60 + j80 ,
ZBN = 100 − j120 , and ZCN = 30 + j40 .
Find the total complex power absorbed by the load.
12.45 Refer to the unbalanced circuit of Fig. 12.58.
Calculate:
(a) the line currents
(b) the real power absorbed by the load
(c) the total complex power supplied by the source
440 0° V
440 120° V 440 −120° V
+
−
+
−
− +
a
b B
c
A
C
20 Ω
j10 Ω
−j5 Ω
Figure 12.58 For Prob. 12.45.
Section 12.9 PSpice for Three-Phase Circuits
12.46 Solve Prob. 12.10 using PSpice.
12.47 The source in Fig. 12.59 is balanced and exhibits a
positive phase sequence. If f = 60 Hz, use PSpice
to find VAN , VBN , and VCN .
100 0° V
+
−
+
−
− +
a
b B
c
A
C
40 Ω
n N
0.2 mF
10 mF
Figure 12.59 For Prob. 12.47.
12.48 Use PSpice to determine Io in the single-phase,
three-wire circuit of Fig. 12.60. Let
Z1 = 15 − j10 , Z2 = 30 + j20 , and
Z3 = 12 + j5 .
Z1
Z2
Z3
4 Ω
4 Ω
4 Ω
+
−
+
−
220 0° V
220 0° V
Io
Figure 12.60 For Prob. 12.48.
12.49 Given the circuit in Fig. 12.61, use PSpice to
determine currents IaA and voltage VBN .
a
n
A
C
N
4 Ω
− +
b 4 Ω
c 4 Ω
10 Ω
B
j3 Ω j15 Ω
j15 Ω
j3 Ω
j3 Ω
−j36 Ω −j36 Ω
240 0° V
240 −120° V
240 120° V
−j36 Ω
− +
10 Ω j15 Ω
10 Ω
− +
Figure 12.61 For Prob. 12.49.
12.50 The circuit in Fig. 12.62 operates at 60 Hz. Use
PSpice to find the source current Iab and the line
current IbB .
A
N
b
c
a 1 Ω 16 Ω
2 mH
2 mH 27 mH
2 mH 133 mF
1 Ω
1 Ω
+
−
+
−
+
−
B
C
110 120° V
110 −120° V
110 0° V
Figure 12.62 For Prob. 12.50.
12.51 For the circuit in Fig. 12.54, use PSpice to find the
line currents and the phase currents.
12.52 A balanced three-phase circuit is shown in Fig.
12.63 on the next page. Use PSpice to find the line
currents IaA, IbB , and IcC.
Section 12.10 Applications
12.53 A three-phase, four-wire system operating with a
208-V line voltage is shown in Fig. 12.64. The
source voltages are balanced. The power absorbed
by the resistive wye-connected load is measured by
the three-wattmeter method. Calculate:
(a) the voltage to neutral
(b) the currents I1, I2, I3, and In
(c) the readings of the wattmeters
(d) the total power absorbed by the load
524 PART 2 AC Circuits
A
b
c
a 0.6 Ω
0.2 Ω
0.2 Ω
30 Ω
j0.5 Ω
j1 Ω
j1 Ω
−j20 Ω
j0.5 Ω
j0.5 Ω
0.2 Ω
j1 Ω
0.6 Ω
0.6 Ω
30 Ω
−j20 Ω
+
−
+
−
+
−
B
C
240 130° V
240 −110° V
240 10° V
30 Ω
−j20 Ω
Figure12.63 For Prob. 12.52.
n
I2
In
I3
I1
40 Ω
48 Ω
60 Ω
W1
W2
W3
Figure 12.64 For Prob. 12.53.
12.54
∗
As shown in Fig. 12.65, a three-phase four-wire line
with a phase voltage of 120 V supplies a balanced
motor load at 260 kVA at 0.85 pf lagging. The
motor load is connected to the three main lines
marked a, b, and c. In addition, incandescent lamps
(unity pf) are connected as follows: 24 kW from
line a to the neutral, 15 kW from line b to the
neutral, and 9 kW from line a to the neutral.
(a) If three wattmeters are arranged to measure the
power in each line, calculate the reading of each
meter.
(b) Find the current in the neutral line.
*An asterisk indicates a challenging problem.
a
b
c
d
24 kW 15 kW 9 kW
Motor load
260 kVA,
0.85 pf, lagging
Lighting loads
Figure 12.65 For Prob. 12.54.
12.55 Meter readings for a three-phase wye-connected
alternator supplying power to a motor indicate that
the line voltages are 330 V, the line currents are
8.4 A, and the total line power is 4.5 kW. Find:
(a) the load in VA
(b) the load pf
(c) the phase current
(d) the phase voltage
12.56 The two-wattmeter method gives P1 = 1200 W and
P2 = −400 W for a three-phase motor running on a
240-V line. Assume that the motor load is wye-
connected and that it draws a line current of 6 A.
Calculate the pf of the motor and its phase
impedance.
CHAPTER 12 Three-Phase Circuits 525
12.57 In Fig. 12.66, two wattmeters are properly
connected to the unbalanced load supplied by a
balanced source such that Vab = 208 0◦
V with
positive phase sequence.
(a) Determine the reading of each wattmeter.
(b) Calculate the total apparent power absorbed by
the load.
a
b
0 B
c
A
C
10 Ω
12 Ω
20 Ω
j5 Ω
−j10 Ω
W1
W2
Figure 12.66 For Prob. 12.57.
12.58 If wattmeters W1 and W2 are properly connected
respectively between lines a and b and lines b and c
to measure the power absorbed by the
delta-connected load in Fig. 12.44, predict their
readings.
12.59 For the circuit displayed in Fig. 12.67, find the
wattmeter readings.
Z
Z
Z = 10 + j30 Ω
+
−
+
−
240 −60° V
240 −120° V
±
±
±
±
W1
W2
Figure 12.67 For Prob. 12.59.
12.60 Predict the wattmeter readings for the circuit in Fig.
12.68.
Z
Z
Z = 60 − j30 Ω
+
−
+
−
208 0° V
208 −60° V
±
±
±
±
W1
W2
Figure 12.68 For Prob. 12.60.
12.61 A man has a body resistance of 600 . How much
current flows through his ungrounded body:
(a) when he touches the terminals of a 12-V
autobattery?
(b) when he sticks his finger into a 120-V light
socket?
12.62 Show that the I2
R losses will be higher for a 120-V
appliance than for a 240-V appliance if both have
the same power rating.
COMPREHENSIVE PROBLEMS
12.63 A three-phase generator supplied 3.6 kVA at a
power factor of 0.85 lagging. If 2500 W are
delivered to the load and line losses are 80 W per
phase, what are the losses in the generator?
12.64 A three-phase 440-V, 51-kW, 60-kVA inductive load
operates at 60 Hz and is wye-connected. It is
desired to correct the power factor to 0.95 lagging.
What value of capacitor should be placed in parallel
with each load impedance?
12.65 A balanced three-phase generator has an abc phase
sequence with phase voltage Van = 255 0◦
V. The
generator feeds an induction motor which may be
represented by a balanced Y-connected load with an
impedance of 12 + j5  per phase. Find the line
currents and the load voltages. Assume a line
impedance of 2  per phase.
526 PART 2 AC Circuits
12.66 Three balanced loads are connected to a distribution
line as depicted in Fig. 12.69. The loads are
Transformer: 12 kVA at 0.6 pf lagging
Motor: 16 kVA at 0.8 pf lagging
Unknown load: − − −−
If the line voltage is 220 V, the line current is 120 A,
and the power factor of the combined load is 0.95
lagging, determine the unknown load.
Transformer Motor Unknown Load
Figure 12.69 For Prob. 12.66.
12.67 A professional center is supplied by a balanced
three-phase source. The center has four plants, each
a balanced three-phase load as follows:
Load 1: 150 kVA at 0.8 pf leading
Load 2: 100 kW at unity pf
Load 3: 200 kVA at 0.6 pf lagging
Load 4: 80 kW and 95 kVAR (inductive)
If the line impedance is 0.02 + j0.05  per phase
and the line voltage at the loads is 480 V, find the
magnitude of the line voltage at the source.
12.68
∗
Figure 12.70 displays a three-phase delta-connected
motor load which is connected to a line voltage of
440 V and draws 4 kVA at a power factor of 72
percent lagging. In addition, a single 1.8 kVAR
capacitor is connected between lines a and b, while
a 800-W lighting load is connected between line c
and neutral. Assuming the abc sequence and taking
Van = Vp 0◦
, find the magnitude and phase angle
of currents Ia, Ib, Ic, and In.
a
b
c
d
Motor load
4 kVA,
pf = 72%, lagging
800 W lighting load
Ia
Ib
Ic
In
1.8 kVAR
Figure 12.70 For Prob. 12.68.
12.69 Design a three-phase heater with suitable symmetric
loads using wye-connected pure resistance. Assume
that the heater is supplied by a 240-V line voltage
and is to give 27 kW of heat.
12.70 For the single-phase three-wire system in Fig. 12.71,
find currents IaA, IbB , and InN .
24 − j2 Ω
15 + j4 Ω
1 Ω
1 Ω
1 Ω
+
−
+
−
120 0° V rms
120 0° V rms
a A
n
b B
N
Figure 12.71 For Prob. 12.70.
12.71 Consider the single-phase three-wire system shown
in Fig. 12.72. Find the current in the neutral wire
and the complex power supplied by each source.
Take Vs as a 115 0◦
-V, 60-Hz source.
1 Ω
2 Ω
20 Ω
15 Ω
30 Ω 50 mH
1 Ω
+
−
+
−
V
s
Vs
Figure 12.72 For Prob. 12.71.
527
C H A P T E R
MAGNETICALLY COUPLED CIRCUITS
1 3
People want success but keep running away from problems, and yet it is
only in tackling problems that success is achieved.
— Josiah J. Bonire
Enhancing Your Career
Career in Electromagnetics Electromagnetics is the
branch of electrical engineering (or physics) that deals with
the analysis and application of electric and magnetic fields.
In electromagnetics, electric circuit analysis is applied at low
frequencies.
The principles of electromagnetics (EM) are applied
in various allied disciplines, such as electric machines,
electromechanical energy conversion, radar meteorology,
remote sensing, satellite communications, bioelectromag-
netics, electromagnetic interference and compatibility, plas-
mas, and fiber optics. EM devices include electric motors
and generators, transformers, electromagnets, magnetic lev-
itation, antennas, radars, microwave ovens, microwave
dishes, superconductors, and electrocardiograms. The de-
sign of these devices requires a thorough knowledge of the
laws and principles of EM.
EM is regarded as one of the more difficult disci-
plines in electrical engineering. One reason is that EM
phenomena are rather abstract. But if one enjoys working
with mathematics and can visualize the invisible, one should
consider being a specialist in EM, since few electrical
engineers specialize in this area. Electrical engineers who
specialize in EM are needed in microwave industries,
radio/TV broadcasting stations, electromagnetic research
laboratories, and several communications industries.
Telemetry receiving station for space satellites. Source: T. J. Mal-
oney, Modern Industrial Electronics, 3rd ed. Englewood Cliffs, NJ:
Prentice Hall, 1996, p. 718.
528 PART 2 AC Circuits
13.1 INTRODUCTION
The circuits we have considered so far may be regarded as conductively
coupled, because one loop affects the neighboring loop through current
conduction. When two loops with or without contacts between them
affect each other through the magnetic field generated by one of them,
they are said to be magnetically coupled.
The transformer is an electrical device designed on the basis of
the concept of magnetic coupling. It uses magnetically coupled coils to
transfer energy from one circuit to another. Transformers are key circuit
elements. They are used in power systems for stepping up or stepping
down ac voltages or currents. They are used in electronic circuits such as
radio and television receivers for such purposes as impedance matching,
isolating one part of a circuit from another, and again for stepping up or
down ac voltages and currents.
We will begin with the concept of mutual inductance and introduce
the dot convention used for determining the voltage polarities of induc-
tively coupled components. Based on the notion of mutual inductance,
we then introduce the circuit element known as the transformer. We will
consider the linear transformer, the ideal transformer, the ideal autotrans-
former, and the three-phase transformer. Finally, among their important
applications, we look at transformers as isolating and matching devices
and their use in power distribution.
13.2 MUTUAL INDUCTANCE
When two inductors (or coils) are in a close proximity to each other,
the magnetic flux caused by current in one coil links with the other coil,
thereby inducing voltage in the latter. This phenomenon is known as
mutual inductance.
i(t) v
+
−
f
Figure13.1 Magnetic flux produced
by a single coil with N turns.
Let us first consider a single inductor, a coil with N turns. When
current i flows through the coil, a magnetic flux φ is produced around it
(Fig. 13.1). According to Faraday’s law, the voltage v induced in the coil
is proportional to the number of turns N and the time rate of change of
the magnetic flux φ; that is,
v = N
dφ
dt
(13.1)
But the flux φ is produced by current i so that any change in φ is caused
by a change in the current. Hence, Eq. (13.1) can be written as
v = N
dφ
di
di
dt
(13.2)
or
v = L
di
dt
(13.3)
which is the voltage-current relationship for the inductor. From Eqs.
(13.2) and (13.3), the inductance L of the inductor is thus given by
L = N
dφ
di
(13.4)
CHAPTER 13 Magnetically Coupled Circuits 529
This inductance is commonly called self-inductance, because it relates
the voltage induced in a coil by a time-varying current in the same coil.
Now consider two coils with self-inductances L1 and L2 that are in
close proximity with each other (Fig. 13.2). Coil 1 has N1 turns, while
coil 2 has N2 turns. For the sake of simplicity, assume that the second
inductor carries no current. The magnetic flux φ1 emanating from coil 1
has two components: one component φ11 links only coil 1, and another
component φ12 links both coils. Hence,
φ1 = φ11 + φ12 (13.5)
Although the two coils are physically separated, they are said to be mag-
netically coupled. Since the entire flux φ1 links coil 1, the voltage induced
in coil 1 is
v1 = N1
dφ1
dt
(13.6)
Only flux φ12 links coil 2, so the voltage induced in coil 2 is
v2 = N2
dφ12
dt
(13.7)
Again, as the fluxes are caused by the current i1 flowing in coil 1, Eq.
(13.6) can be written as
v1 = N1
dφ1
di1
di1
dt
= L1
di1
dt
(13.8)
where L1 = N1 dφ1/di1 is the self-inductance of coil 1. Similarly, Eq.
(13.7) can be written as
v2 = N2
dφ12
di1
di1
dt
= M21
di1
dt
(13.9)
where
M21 = N2
dφ12
di1
(13.10)
M21 is known as the mutual inductance of coil 2 with respect to coil 1.
Subscript 21 indicates that the inductance M21 relates the voltage induced
in coil 2 to the current in coil 1. Thus, the open-circuit mutual voltage
(or induced voltage) across coil 2 is
v2 = M21
di1
dt
(13.11)
i1(t) v1
+
−
v2
+
−
f11
f12
L1 L2
N1 turns N2 turns
Figure 13.2 Mutual inductance M21 of
coil 2 with respect to coil 1.
v1
+
−
v2
+
−
i2(t)
f22
f21
L1 L2
N1 turns N2 turns
Figure 13.3 Mutual inductance M12 of
coil 1 with respect to coil 2.
Suppose we now let current i2 flow in coil 2, while coil 1 carries no
current(Fig.13.3). Themagneticfluxφ2 emanatingfromcoil2comprises
flux φ22 that links only coil 2 and flux φ21 that links both coils. Hence,
φ2 = φ21 + φ22 (13.12)
The entire flux φ2 links coil 2, so the voltage induced in coil 2 is
v2 = N2
dφ2
dt
= N2
dφ2
di2
di2
dt
= L2
di2
dt
(13.13)
530 PART 2 AC Circuits
where L2 = N2 dφ2/di2 is the self-inductance of coil 2. Since only flux
φ21 links coil 1, the voltage induced in coil 1 is
v1 = N1
dφ21
dt
= N1
dφ21
di2
di2
dt
= M12
di2
dt
(13.14)
where
M12 = N1
dφ21
di2
(13.15)
which is the mutual inductance of coil 1 with respect to coil 2. Thus, the
open-circuit mutual voltage across coil 1 is
v1 = M12
di2
dt
(13.16)
We will see in the next section that M12 and M21 are equal, that is,
M12 = M21 = M (13.17)
and we refer to M as the mutual inductance between the two coils. Like
self-inductance L, mutual inductance M is measured in henrys (H). Keep
in mind that mutual coupling only exists when the inductors or coils are
in close proximity, and the circuits are driven by time-varying sources.
We recall that inductors act like short circuits to dc.
From the two cases in Figs. 13.2 and 13.3, we conclude that mutual
inductance results if a voltage is induced by a time-varying current in
another circuit. It is the property of an inductor to produce a voltage in
reaction to a time-varying current in another inductor near it. Thus,
Mutual inductance is the ability of one inductor to induce a voltage
across a neighboring inductor, measured in henrys (H).
Although mutual inductance M is always a positive quantity, the
mutual voltage M di/dt may be negative or positive, just like the self-
induced voltage L di/dt. However, unlike the self-induced L di/dt,
whose polarity is determined by the reference direction of the current and
the reference polarity of the voltage (according to the passive sign con-
vention), the polarity of mutual voltage M di/dt is not easy to determine,
because four terminals are involved. The choice of the correct polarity for
M di/dt is made by examining the orientation or particular way in which
both coils are physically wound and applying Lenz’s law in conjunction
with the right-hand rule. Since it is inconvenient to show the construction
details of coils on a circuit schematic, we apply the dot convention in cir-
cuit analysis. By this convention, a dot is placed in the circuit at one end
of each of the two magnetically coupled coils to indicate the direction of
the magnetic flux if current enters that dotted terminal of the coil. This is
illustrated in Fig. 13.4. Given a circuit, the dots are already placed beside
the coils so that we need not bother about how to place them. The dots
are used along with the dot convention to determine the polarity of the
mutual voltage. The dot convention is stated as follows:
CHAPTER 13 Magnetically Coupled Circuits 531
i1
f21
f11 f22
f12
v1
+
−
i2
Coil 1 Coil 2
v2
+
−
Figure13.4 Illustration of the dot convention.
If a current enters the dotted terminal of one coil, the reference
polarity of the mutual voltage in the second coil is positive
at the dotted terminal of the second coil.
Alternatively,
If a current leaves the dotted terminal of one coil, the reference
polarity of the mutual voltage in the second coil is negative
at the dotted terminal of the second coil.
+
−
M
i1
v2 = M
di1
dt
(a)
+
−
M
i1
v2 = –M
di1
dt
v1 = –M
di2
dt
(b)
+
−
M
(c)
(d)
i2
v1 = M
di2
dt
+
−
M
i2
Figure13.5 Examples
illustrating how to apply the
dot convention.
Thus, the reference polarity of the mutual voltage depends on the refer-
ence direction of the inducing current and the dots on the coupled coils.
Application of the dot convention is illustrated in the four pairs of mu-
tually coupled coils in Fig. 13.5. For the coupled coils in Fig. 13.5(a),
the sign of the mutual voltage v2 is determined by the reference polarity
for v2 and the direction of i1. Since i1 enters the dotted terminal of coil
1 and v2 is positive at the dotted terminal of coil 2, the mutual voltage is
+M di1/dt. For the coils in Fig. 13.5(b), the current i1 enters the dot-
ted terminal of coil 1 and v2 is negative at the dotted terminal of coil 2.
Hence, the mutual voltage is −M di1/dt. The same reasoning applies to
the coils in Fig. 13.5(c) and 13.5(d). Figure 13.6 shows the dot conven-
tion for coupled coils in series. For the coils in Fig. 13.6(a), the total
inductance is
L = L1 + L2 + 2M (Series-aiding connection) (13.18)
For the coil in Fig. 13.6(b),
L = L1 + L2 − 2M (Series-opposing connection) (13.19)
Now that we know how to determine the polarity of the mutual
voltage, we are prepared to analyze circuits involving mutual inductance.
532 PART 2 AC Circuits
i i
L1 L2
M
(+)
(a)
i i
L1 L2
M
(−)
(b)
Figure 13.6 Dot convention for coils in series; the sign indicates the polarity of the mutual
voltage: (a) series-aiding connection, (b) series-opposing connection.
As the first example, consider the circuit in Fig. 13.7. Applying KVL to
coil 1 gives
v1 = i1R1 + L1
di1
dt
+ M
di2
dt
(13.20a)
For coil 2, KVL gives
v2 = i2R2 + L2
di2
dt
+ M
di1
dt
(13.20b)
We can write Eq. (13.20) in the frequency domain as
V1 = (R1 + jωL1)I1 + jωMI2 (13.21a)
V2 = jωMI1 + (R2 + jωL2)I2 (13.21b)
As a second example, consider the circuit in Fig. 13.8. We analyze this
in the frequency domain. Applying KVL to coil 1, we get
V = (Z1 + jωL1)I1 − jωMI2 (13.22a)
For coil 2, KVL yields
0 = −jωMI1 + (ZL + jωL2)I2 (13.22b)
Equations (13.21) and (13.22) are solved in the usual manner to determine
the currents.
v1 v2
R1 R2
+
− L1 L2
i1 i2
M
+
−
Figure 13.7 Time-domain analysis of a circuit containing
coupled coils.
V ZL
Z1
+
− jvL1 jvL2
I1 I2
jvM
Figure 13.8 Frequency-domain analysis of a circuit
containing coupled coils.
At this introductory level we are not concerned with the determi-
nation of the mutual inductances of the coils and their dot placements.
Like R, L, and C, calculation of M would involve applying the theory
of electromagnetics to the actual physical properties of the coils. In this
text, we assume that the mutual inductance and the dots placement are the
“givens” of the circuit problem, like the circuit components R, L, and C.
CHAPTER 13 Magnetically Coupled Circuits 533
E X A M P L E 1 3 . 1
Calculate the phasor currents I1 and I2 in the circuit of Fig. 13.9.
V 12 Ω
−j4 Ω
+
− j5 Ω j6 Ω
I1 I2
j3 Ω
12 0°
Figure13.9 For Example 13.1.
Solution:
For coil 1, KVL gives
−12 + (−j4 + j5)I1 − j3I2 = 0
or
jI1 − j3I2 = 12 (13.1.1)
For coil 2, KVL gives
−j3I1 + (12 + j6)I2 = 0
or
I1 =
(12 + j6)I2
j3
= (2 − j4)I2 (13.1.2)
Substituting this in Eq. (13.1.1), we get
(j2 + 4 − j3)I2 = (4 − j)I2 = 12
or
I2 =
12
4 − j
= 2.91 14.04◦
A (13.1.3)
From Eqs. (13.1.2) and (13.1.3),
I1 = (2 − j4)I2 = (4.472 − 63.43◦
)(2.91 14.04◦
)
= 13.01 − 49.39◦
A
P R A C T I C E P R O B L E M 1 3 . 1
Determine the voltage Vo in the circuit of Fig. 13.10.
V 10 Ω
4 Ω
+
− j8 Ω j5 Ω
I1 I2
j1 Ω
Vo
+
−
6 90°
Figure13.10 For Practice Prob. 13.1.
Answer: 0.6 − 90◦
V.
534 PART 2 AC Circuits
E X A M P L E 1 3 . 2
Calculate the mesh currents in the circuit of Fig. 13.11.
V 5 Ω
4 Ω j8 Ω
+
− j6 Ω
j2 Ω
I1
I2
−j3 Ω
100 0°
Figure13.11 For Example 13.2.
Solution:
The key to analyzing a magnetically coupled circuit is knowing the po-
larity of the mutual voltage. We need to apply the dot rule. In Fig. 13.11,
suppose coil 1 is the one whose reactance is 6 , and coil 2 is the one
whose reactance is 8 . To figure out the polarity of the mutual voltage
in coil 1 due to current I2, we observe that I2 leaves the dotted terminal of
coil 2. Since we are applying KVL in the clockwise direction, it implies
that the mutual voltage is negative, that is, −j2I2.
+
−
j2
I2
V1
(a) V1 = –2jI2
I1 j6 Ω j8 Ω
Coil 1 Coil 2
−
+
j2 Ω
I1
V2
(b) V2 = –2jI1
I2
j6 Ω j8 Ω
Coil 1 Coil 2
Figure13.12 For Example 13.2;
redrawing the relevant portion of the
circuit in Fig. 13.11 to find mutual
voltages by the dot convention.
Alternatively, it might be best to figure out the mutual voltage by
redrawing the relevant portion of the circuit, as shown in Fig. 13.12(a),
where it becomes clear that the mutual voltage is V1 = −2jI2.
Thus, for mesh 1 in Fig. 13.11, KVL gives
−100 + I1(4 − j3 + j6) − j6I2 − j2I2 = 0
or
100 = (4 + j3)I1 − j8I2 (13.2.1)
Similarly, to figure out the mutual voltage in coil 2 due to current I1,
consider the relevant portion of the circuit, as shown in Fig. 13.12(b).
Applying the dot convention gives the mutual voltage as V2 = −2jI1.
Also, current I2 sees the two coupled coils in series in Fig. 13.11; since it
leaves the dotted terminals in both coils, Eq. (13.18) applies. Therefore,
for mesh 2, KVL gives
0 = −2jI1 − j6I1 + (j6 + j8 + j2 × 2 + 5)I2
or
0 = −j8I1 + (5 + j18)I2 (13.2.2)
Putting Eqs. (13.2.1) and (13.2.2) in matrix form, we get

100
0

=

4 + j3
−j8
−j8
5 + j18
 
I1
I2
CHAPTER 13 Magnetically Coupled Circuits 535
The determinants are
 =




4 + j3 −j8
−j8 5 + j18



 = 30 + j87
1 =




100 −j8
0 5 + j18



 = 100(5 + j18)
2 =




4 + j3 100
−j8 0



 = j800
Thus, we obtain the mesh currents as
I1 =
1

=
100(5 + j18)
30 + j87
=
1868.2 74.5◦
92.03 71◦
= 20.3 3.5◦
A
I2 =
2

=
j800
30 + j87
=
800 90◦
92.03 71◦
= 8.693 19◦
A
P R A C T I C E P R O B L E M 1 3 . 2
Determine the phasor currents I1 and I2 in the circuit of Fig. 13.13.
V
5 Ω j2 Ω
+
− j6 Ω
j3 Ω
I1 I2 −j4 Ω
12 60°
Figure13.13 For Practice Prob. 13.2.
Answer: 2.15 86.56◦
, 3.23 86.56◦
A.
13.3 ENERGY IN A COUPLED CIRCUIT
In Chapter 6, we saw that the energy stored in an inductor is given by
w =
1
2
Li2
(13.23)
We now want to determine the energy stored in magnetically coupled
coils.
+
−
M
i1
v1
+
−
v2
i2
L1 L2
Figure13.14 The circuit
for deriving energy stored in
a coupled circuit.
Consider the circuit in Fig. 13.14. We assume that currents i1 and
i2 are zero initially, so that the energy stored in the coils is zero. If we let
i1 increase from zero to I1 while maintaining i2 = 0, the power in coil 1
is
p1(t) = v1i1 = i1L1
di1
dt
(13.24)
and the energy stored in the circuit is
w1 =

p1 dt = L1
 I1
0
i1 di1 =
1
2
L1I2
1 (13.25)
536 PART 2 AC Circuits
If we now maintain i1 = I1 and increase i2 from zero to I2, the mutual
voltage induced in coil 1 is M12 di2/dt, while the mutual voltage induced
in coil 2 is zero, since i1 does not change. The power in the coils is now
p2(t) = i1M12
di2
dt
+ i2v2 = I1M12
di2
dt
+ i2L2
di2
dt
(13.26)
and the energy stored in the circuit is
w2 =

p2 dt = M12I1
 I2
0
di2 + L2
 I2
0
i2 di2
= M12I1I2 +
1
2
L2I2
2
(13.27)
The total energy stored in the coils when both i1 and i2 have reached
constant values is
w = w1 + w2 =
1
2
L1I2
1 +
1
2
L2I2
2 + M12I1I2 (13.28)
If we reverse the order by which the currents reach their final values, that
is, if we first increase i2 from zero to I2 and later increase i1 from zero to
I1, the total energy stored in the coils is
w =
1
2
L1I2
1 +
1
2
L2I2
2 + M21I1I2 (13.29)
Since the total energy stored should be the same regardless of how we
reach the final conditions, comparing Eqs. (13.28) and (13.29) leads us
to conclude that
M12 = M21 = M (13.30a)
and
w =
1
2
L1I2
1 +
1
2
L2I2
2 + MI1I2 (13.30b)
This equation was derived based on the assumption that the coil currents
bothenteredthedottedterminals. Ifonecurrententersonedottedterminal
while the other current leaves the other dotted terminal, the mutual voltage
is negative, so that the mutual energy MI1I2 is also negative. In that case,
w =
1
2
L1I2
1 +
1
2
L2I2
2 − MI1I2 (13.31)
Also, since I1 and I2 are arbitrary values, they may be replaced by i1 and
i2, which gives the instantaneous energy stored in the circuit the general
expression
w =
1
2
L1i2
1 +
1
2
L2i2
2 ± Mi1i2 (13.32)
The positive sign is selected for the mutual term if both currents enter
or leave the dotted terminals of the coils; the negative sign is selected
otherwise.
We will now establish an upper limit for the mutual inductance M.
The energy stored in the circuit cannot be negative because the circuit is
CHAPTER 13 Magnetically Coupled Circuits 537
passive. This means that the quantity 1/2L1i2
1 + 1/2L2i2
2 − Mi1i2 must
be greater than or equal to zero,
1
2
L1i2
1 +
1
2
L2i2
2 − Mi1i2 ≥ 0 (13.33)
To complete the square, we both add and subtract the term i1i2
√
L1L2 on
the right-hand side of Eq. (13.33) and obtain
1
2
(i1

L1 − i2

L2)2
+ i1i2(

L1L2 − M) ≥ 0 (13.34)
The squared term is never negative; at its least it is zero. Therefore, the
second term on the right-hand side of Eq. (13.34) must be greater than
zero; that is,

L1L2 − M ≥ 0
or
M ≤

L1L2 (13.35)
Thus, the mutual inductance cannot be greater than the geometric mean
of the self-inductances of the coils. The extent to which the mutual
inductance M approaches the upper limit is specified by the coefficient
of coupling k, given by
k =
M
√
L1L2
(13.36)
or
M = k

L1L2 (13.37)
where 0 ≤ k ≤ 1 or equivalently 0 ≤ M ≤
√
L1L2. The coupling
coefficient is the fraction of the total flux emanating from one coil that
links the other coil. For example, in Fig. 13.2,
k =
φ12
φ1
=
φ12
φ11 + φ12
(13.38)
and in Fig. 13.3,
k =
φ21
φ2
=
φ21
φ21 + φ22
(13.39)
If the entire flux produced by one coil links another coil, then k = 1
and we have 100 percent coupling, or the coils are said to be perfectly
coupled. Thus,
The coupling coefficient k is a measure of the magnetic
coupling between two coils; 0 ≤ k ≤ 1.
For k  0.5, coils are said to be loosely coupled; and for k  0.5, they
are said to be tightly coupled.
(a) (b)
Air or ferrite core
Figure13.15 Windings: (a) loosely coupled,
(b) tightly coupled; cutaway view demonstrates
both windings.
We expect k to depend on the closeness of the two coils, their core,
their orientation, and their windings. Figure 13.15 shows loosely coupled
538 PART 2 AC Circuits
windings and tightly coupled windings. The air-core transformers used
in radio frequency circuits are loosely coupled, whereas iron-core trans-
formers used in power systems are tightly coupled. The linear transform-
ers discussed in Section 3.4 are mostly air-core; the ideal transformers
discussed in Sections 13.5 and 13.6 are principally iron-core.
E X A M P L E 1 3 . 3
Consider the circuit in Fig. 13.16. Determine the coupling coefficient.
Calculate the energy stored in the coupled inductors at time t = 1 s if
v = 60 cos(4t + 30◦
) V.
v
10 Ω
+
− 5 H 4 H
2.5 H
F
1
16
Figure13.16 For Example 13.3.
Solution:
The coupling coefficient is
k =
M
√
L1L2
=
2.5
√
20
= 0.56
indicating that the inductors are tightly coupled. To find the energy stored,
we need to obtain the frequency-domain equivalent of the circuit.
60 cos(4t + 30◦
) ⇒ 60 30◦
, ω = 4 rad/s
5 H ⇒ jωL1 = j20 
2.5 H ⇒ jωM = j10 
4 H ⇒ jωL2 = j16 
1
16
F ⇒
1
jωC
= −j4 
The frequency-domain equivalent is shown in Fig. 13.17. We now apply
mesh analysis. For mesh 1,
(10 + j20)I1 + j10I2 = 60 30◦
(13.3.1)
For mesh 2,
j10I1 + (j16 − j4)I2 = 0
or
I1 = −1.2I2 (13.3.2)
Substituting this into Eq. (13.3.1) yields
I2(−12 − j14) = 60 30◦
⇒ I2 = 3.254 − 160.6◦
A
and
I1 = −1.2I2 = 3.905 − 19.4◦
A
In the time-domain,
i1 = 3.905 cos(4t − 19.4◦
), i2 = 3.254 cos(4t − 199.4◦
)
At time t = 1 s, 4t = 4 rad = 229.2◦
, and
i1 = 3.905 cos(229.2◦
− 19.4◦
) = −3.389 A
i2 = 3.254 cos(229.2◦
+ 160.6◦
) = 2.824 A
CHAPTER 13 Magnetically Coupled Circuits 539
The total energy stored in the coupled inductors is
w =
1
2
L1i2
1 +
1
2
L2i2
2 + Mi1i2
=
1
2
(5)(−3.389)2
+
1
2
(4)(2.824)2
+ 2.5(−3.389)(2.824) = 20.73 J
V
10 Ω
+
− j20 Ω j16 Ω
I1 I2
j10
−j4 Ω
60 30°
Figure13.17 Frequency-domain equivalent of the circuit in Fig. 13.16.
P R A C T I C E P R O B L E M 1 3 . 3
For the circuit in Fig. 13.18, determine the coupling coefficient and the
energy stored in the coupled inductors at t = 1.5 s.
20 cos 2t V
4 Ω
+
− 2 H 1 H
1 H
2 Ω
F
1
8
Figure13.18 For Practice Prob. 13.3.
Answer: 0.7071, 9.85 J.
13.4 LINEAR TRANSFORMERS
Hereweintroducethetransformerasanewcircuitelement. Atransformer
is a magnetic device that takes advantage of the phenomenon of mutual
inductance.
A transformer is generally a four-terminal device comprising
two (or more) magnetically coupled coils.
As shown in Fig. 13.19, the coil that is directly connected to the voltage
source is called the primary winding. The coil connected to the load is
called the secondary winding. The resistances R1 and R2 are included
to account for the losses (power dissipation) in the coils. The trans-
former is said to be linear if the coils are wound on a magnetically linear
540 PART 2 AC Circuits
material—a material for which the magnetic permeability is constant.
Such materials include air, plastic, Bakelite, and wood. In fact, most ma-
terials are magnetically linear. Linear transformers are sometimes called
air-core transformers, although not all of them are necessarily air-core.
They are used in radio and TV sets. Figure 13.20 portrays different types
of transformers.
Alineartransformermayalsoberegardedasone
whose flux is proportional to the currents in its
windings.
V ZL
+
− L1 L2
I1 I2
M
R1 R2
Primary coil Secondary coil
Figure13.19 A linear transformer.
(b)
(a)
Figure 13.20 Different types of transformers: (a) copper wound dry power transformer, (b) audio transformers.
(Courtesy of: (a) Electric Service Co., (b) Jensen Transformers.)
We would like to obtain the input impedance Zin as seen from the
source, because Zin governs the behavior of the primary circuit. Applying
KVL to the two meshes in Fig. 13.19 gives
V = (R1 + jωL1)I1 − jωMI2 (13.40a)
0 = −jωMI1 + (R2 + jωL2 + ZL)I2 (13.40b)
CHAPTER 13 Magnetically Coupled Circuits 541
In Eq. (13.40b), we express I2 in terms of I1 and substitute it into Eq.
(13.40a). We get the input impedance as
Zin =
V
I1
= R1 + jωL1 +
ω2
M2
R2 + jωL2 + ZL
(13.41)
Notice that the input impedance comprises two terms. The first term,
(R1 + jωL1), is the primary impedance. The second term is due to the
coupling between the primary and secondary windings. It is as though this
impedance is reflected to the primary. Thus, it is known as the reflected
impedance ZR, and
ZR =
ω2
M2
R2 + jωL2 + ZL
(13.42)
It should be noted that the result in Eq. (13.41) or (13.42) is not affected
by the location of the dots on the transformer, because the same result is
produced when M is replaced by −M.
Some authors call this the coupled impedance.
The little bit of experience gained in Sections 13.2 and 13.3 in
analyzingmagneticallycoupledcircuitsisenoughtoconvinceanyonethat
analyzing these circuits is not as easy as circuits in previous chapters. For
this reason, it is sometimes convenient to replace a magnetically coupled
circuit by an equivalent circuit with no magnetic coupling. We want to
replace the linear transformer in Fig. 13.19 by an equivalent T or  circuit,
a circuit that would have no mutual inductance. Ignore the resistances of
the coils and assume that the coils have a common ground as shown in
Fig. 13.21. The assumption of a common ground for the two coils is a
major restriction of the equivalent circuits. A common ground is imposed
on the linear transformer in Fig. 13.21 in view of the necessity of having
a common ground in the equivalent T or  circuit; see Figs. 13.22 and
13.23.
+
−
M
I1
V1
+
−
V2
I2
L1 L2
Figure13.21 Determining
the equivalent circuit of a
linear transformer.
+
−
I1
V1
+
−
V2
I2
Lc
La Lb
Figure 13.22 An equivalent T circuit.
+
−
I1
V1
+
−
V2
I2
LB
LA
LC
Figure 13.23 An equivalent  circuit.
The voltage-current relationships for the primary and secondary
coils give the matrix equation

V1
V2

=

jωL1 jωM
jωM jωL2
 
I1
I2

(13.43)
By matrix inversion, this can be written as

I1
I2

=






L2
jω(L1L2 − M2)
−M
jω(L1L2 − M2)
−M
jω(L1L2 − M2)
L1
jω(L1L2 − M2)







V1
V2

(13.44)
Our goal is to match Eqs. (13.43) and (13.44) with the corresponding
equations for the T and  networks.
For the T (or Y) network of Fig. 13.22, mesh analysis provides the
terminal equations as

V1
V2

=

jω(La + Lc) jωLc
jωLc jω(Lb + Lc)
 
I1
I2

(13.45)
If the circuits in Figs. 13.21 and 13.22 are equivalents, Eqs. (13.43) and
(13.45) must be identical. Equating terms in the impedance matrices of
542 PART 2 AC Circuits
Eqs. (13.43) and (13.45) leads to
La = L1 − M, Lb = L2 − M, Lc = M (13.46)
For the  (or ) network in Fig. 13.23, nodal analysis gives the
terminal equations as

I1
I2

=






1
jωLA
+
1
jωLC
−
1
jωLC
−
1
jωLC
1
jωLB
+
1
jωLC







V1
V2

(13.47)
Equating terms in admittance matrices of Eqs. (13.44) and (13.47), we
obtain
LA =
L1L2 − M2
L2 − M
, LB =
L1L2 − M2
L1 − M
LC =
L1L2 − M2
M
(13.48)
Note that in Figs. 13.23 and 13.24, the inductors are not magnetically
coupled. Also note that changing the locations of the dots in Fig. 13.21
can cause M to become −M. As Example 13.6 illustrates, a negative
value of M is physically unrealizable but the equivalent model is still
mathematically valid.
ZL
+
− j20 Ω j40 Ω
I1 I2
j5 Ω
Z1 Z2
V
50 60°
Figure13.24 For Example 13.4.
E X A M P L E 1 3 . 4
In the circuit of Fig. 13.24, calculate the input impedance and current I1.
Take Z1 = 60 − j100 , Z2 = 30 + j40 , and ZL = 80 + j60 .
Solution:
From Eq. (13.41),
Zin = Z1 + j20 +
(5)2
j40 + Z2 + ZL
= 60 − j100 + j20 +
25
110 + j140
= 60 − j80 + 0.14 − 51.84◦
= 60.09 − j80.11 = 100.14 − 53.1◦
CHAPTER 13 Magnetically Coupled Circuits 543
Thus,
I1 =
V
Zin
=
50 60◦
100.14 − 53.1◦
= 0.5 113.1◦
A
P R A C T I C E P R O B L E M 1 3 . 4
Find the input impedance of the circuit of Fig. 13.25 and the current from
the voltage source.
4 Ω
+
− j8 Ω j10 Ω
j3 Ω
j4 Ω
6 Ω
−j6 Ω
V
10 0°
Figure13.25 For Practice Prob. 13.4.
Answer: 8.58 58.05◦
, 1.165 − 58.05◦
A.
E X A M P L E 1 3 . 5
DeterminetheT-equivalentcircuitofthelineartransformerinFig.13.26(a).
2 H
(a)
10 H 4 H
a
b
c
d
(b)
a
b
c
d
2 H
8 H 2 H
Figure 13.26 For Example 13.5: (a) a linear transformer,
(b) its T-equivalent circuit.
Solution:
Given that L1 = 10, L2 = 4, and M = 2, the T equivalent network has
the following parameters:
La = L1 − M = 10 − 2 = 8 H
Lb = L2 − M = 4 − 2 = 2 H, Lc = M = 2 H
The T-equivalent circuit is shown in Fig. 13.26(b). We have assumed that
reference directions for currents and voltage polarities in the primary and
secondary windings conform to those in Fig. 13.21. Otherwise, we may
need to replace M with −M, as Example 13.6 illustrates.
544 PART 2 AC Circuits
P R A C T I C E P R O B L E M 1 3 . 5
For the linear transformer in Fig. 13.26 (a), find the  equivalent network.
Answer: LA = 18 H, LB = 4.5 H, LC = 18 H.
E X A M P L E 1 3 . 6
Solve for I1, I2, and Vo in Fig. 13.27 (the same circuit as for Practice Prob.
13.1) using the T-equivalent circuit for the linear transformer.
+
− j8 Ω j5 Ω
I1 I2
j1 Ω
4 Ω
V
60 90° 10 Ω
+
−
Vo
Figure13.27 For Example 13.6.
Solution:
Notice that the circuit in Fig. 13.27 is the same as that in Fig. 13.10 except
that the reference direction for current I2 has been reversed, just to make
the reference directions for the currents for the magnetically coupled coils
conform with those in Fig. 13.21.
+
−
j1 Ω
(a)
(b)
V1
+
−
V2
j8 Ω j5 Ω
j9 Ω j6 Ω
−j1 Ω
I1 I2
Figure13.28 For Example 13.6:
(a) circuit for coupled coils of Fig.
13.27, (b) T-equivalent circuit.
We need to replace the magnetically coupled coils with the T-
equivalent circuit. The relevant portion of the circuit in Fig. 13.27 is
shown in Fig. 13.28(a). Comparing Fig. 13.28(a) with Fig. 13.21 shows
that there are two differences. First, due to the current reference direc-
tions and voltage polarities, we need to replace M by −M to make Fig.
13.28(a) conform with Fig. 13.21. Second, the circuit in Fig. 13.21 is in
the time-domain, whereas the circuit in Fig. 13.28(a) is in the frequency-
domain. The difference is the factor jω; that is, L in Fig. 13.21 has been
replaced with jωL and M with jωM. Since ω is not specified, we can
assume ω = 1 or any other value; it really does not matter. With these
two differences in mind,
La = L1 − (−M) = 8 + 1 = 9 H
Lb = L2 − (−M) = 5 + 1 = 6 H, Lc = −M = −1 H
Thus, the T-equivalent circuit for the coupled coils is as shown in Fig.
13.28(b).
Inserting the T-equivalent circuit in Fig. 13.28(b) to replace the two
coils in Fig. 13.27 gives the equivalent circuit in Fig. 13.29, which can be
solved using nodal or mesh analysis. Applying mesh analysis, we obtain
j6 = I1(4 + j9 − j1) + I2(−j1) (13.6.1)
and
0 = I1(−j1) + I2(10 + j6 − j1) (13.6.2)
From Eq. (13.6.2),
I1 =
(10 + j5)
j
I2 = (5 − j10)I2 (13.6.3)
CHAPTER 13 Magnetically Coupled Circuits 545
Substituting Eq. (13.6.3) into Eq. (13.6.1) gives
j6 = (4 + j8)(5 − j10)I2 − jI2 = (100 − j)I2 100I2
Since 100 is very large compared to 1, the imaginary part of (100 − j)
can be ignored so that 100 − j 100. Hence,
I2 =
j6
100
= j0.06 = 0.06 90◦
A
From Eq. (13.6.3),
I1 = (5 − j10)j0.06 = 0.6 + j0.3 A
and
Vo = −10I2 = −j0.6 = 0.6 − 90◦
V
This agrees with the answer to Practice Prob. 13.1. Of course, the direc-
tion of I2 in Fig. 13.10 is opposite to that in Fig. 13.27. This will not
affect Vo, but the value of I2 in this example is the negative of that of I2
in Practice Prob. 13.1. The advantage of using the T-equivalent model
for the magnetically coupled coils is that in Fig. 13.29 we do not need to
bother with the dot on the coupled coils.
j6 V
4 Ω j9 Ω
+
− −j1 Ω
I1 I2
j6 Ω
10 Ω
I1 I2
+
−
V
o
Figure13.29 For Example 13.6.
P R A C T I C E P R O B L E M 1 3 . 6
Solve the problem in Example 13.1 (see Fig. 13.9) using the T-equivalent
model for the magnetically coupled coils.
Answer: 13 − 49.4◦
A, 2.91 14.04◦
A.
13.5 IDEAL TRANSFORMERS
An ideal transformer is one with perfect coupling (k = 1). It consists of
two (or more) coils with a large number of turns wound on a common
core of high permeability. Because of this high permeability of the core,
the flux links all the turns of both coils, thereby resulting in a perfect
coupling.
To see how an ideal transformer is the limiting case of two cou-
pled inductors where the inductances approach infinity and the coupling
is perfect, let us reexamine the circuit in Fig. 13.14. In the frequency
domain,
V1 = jωL1I1 + jωMI2 (13.49a)
V2 = jωMI1 + jωL2I2 (13.49b)
546 PART 2 AC Circuits
From Eq. (13.49a), I1 = (V1 − jωMI2)/jωL1. Substituting this in Eq.
(13.49b) gives
V2 = jωL2I2 +
MV1
L1
−
jωM2
I2
L1
But M =
√
L1L2 for perfect coupling (k = 1). Hence,
V2 = jωL2I2 +
√
L1L2V1
L1
−
jωL1L2I2
L1
=
L2
L1
V1 = nV1
where n =
√
L2/L1 and is called the turns ratio. As L1, L2, M → ∞
such that n remains the same, the coupled coils become an ideal trans-
former. A transformer is said to be ideal if it has the following properties:
1. Coils have very large reactances (L1, L2, M → ∞).
2. Coupling coefficient is equal to unity (k = 1).
3. Primary and secondary coils are lossless (R1 = 0 = R2).
An ideal transformer is a unity-coupled, lossless transformer in which the
primary and secondary coils have infinite self-inductances.
Iron-core transformers are close approximations to ideal transformers.
These are used in power systems and electronics.
Figure 13.30(a) shows a typical ideal transformer; the circuit sym-
bol is in Fig. 13.30(b). The vertical lines between the coils indicate an
iron core as distinct from the air core used in linear transformers. The
primary winding has N1 turns; the secondary winding has N2 turns.
N1 N2
(a)
(b)
N1 N2
Figure13.30 (a) Ideal transformer,
(b) circuit symbol for ideal transformers.
ZL
+
− V1 V2
1:n
V
+
−
+
−
I1 I2
Figure13.31 Relating primary and
secondary quantities in an ideal transformer.
When a sinusoidal voltage is applied to the primary winding as
shown in Fig. 13.31, the same magnetic flux φ goes through both wind-
ings. According to Faraday’s law, the voltage across the primary winding
is
v1 = N1
dφ
dt
(13.50a)
while that across the secondary winding is
v2 = N2
dφ
dt
(13.50b)
Dividing Eq. (13.50b) by Eq. (13.50a), we get
v2
v1
=
N2
N1
= n (13.51)
where n is, again, the turns ratio or transformation ratio. We can use the
phasor voltages V1 and V2 rather than the instantaneous values v1 and v2.
Thus, Eq. (13.51) may be written as
V2
V1
=
N2
N1
= n (13.52)
CHAPTER 13 Magnetically Coupled Circuits 547
For the reason of power conservation, the energy supplied to the primary
must equal the energy absorbed by the secondary, since there are no losses
in an ideal transformer. This implies that
v1i1 = v2i2 (13.53)
In phasor form, Eq. (13.53) in conjunction with Eq. (13.52) becomes
I1
I2
=
V2
V1
= n (13.54)
showing that the primary and secondary currents are related to the turns
ratio in the inverse manner as the voltages. Thus,
I2
I1
=
N1
N2
=
1
n
(13.55)
When n = 1, we generally call the transformer an isolation transformer.
The reason will become obvious in Section 13.9.1. If n  1, we have
a step-up transformer, as the voltage is increased from primary to sec-
ondary (V2  V1). On the other hand, if n  1, the transformer is a
step-down transformer, since the voltage is decreased from primary to
secondary (V2  V1).
A step-down transformer is one whose secondary voltage
is less than its primary voltage.
A step-up transformer is one whose secondary voltage
is greater than its primary voltage.
The ratings of transformers are usually specified as V1/V2. A transformer
with rating 2400/120 V should have 2400 V on the primary and 120 in the
secondary (i.e., a step-down transformer). Keep in mind that the voltage
ratings are in rms.
Power companies often generate at some convenient voltage and
use a step-up transformer to increase the voltage so that the power can be
transmitted at very high voltage and low current over transmission lines,
resulting in significant cost savings. Near residential consumer premises,
step-down transformers are used to bring the voltage down to 120 V.
Section 13.9.3 will elaborate on this.
It is important that we know how to get the proper polarity of the
voltages and the direction of the currents for the transformer in Fig. 13.31.
If the polarity of V1 or V2 or the direction of I1 or I2 is changed, n in Eqs.
(13.51) to (13.55) may need to be replaced by −n. The two simple rules
to follow are:
1. If V1 and V2 are both positive or both negative at the dotted
terminals, use +n in Eq. (13.52). Otherwise, use −n.
2. If I1 and I2 both enter into or both leave the dotted terminals,
use −n in Eq. (13.55). Otherwise, use +n.
The rules are demonstrated with the four circuits in Fig. 13.32.
V1 V2
N1:N2
+
−
+
−
I1 I2
V2
V1
N2
N1
=
I2
I1
N1
N2
=
(a)
V1 V2
N1:N2
+
−
+
−
I1 I2
V2
V1
N2
N1
=
I2
I1
N1
N2
= −
(b)
V1 V2
N1:N2
+
−
+
−
I1
I2
V2
V1
N2
N1
= −
V2
V1
N2
N1
= −
I2
I1
N1
N2
=
(c)
V1 V2
N1:N2
+
−
+
−
I1
I2
I2
I1
N1
N2
= −
(d)
Figure13.32 Typical
circuits illustrating proper
voltage polarities and
current directions in an
ideal transformer.
548 PART 2 AC Circuits
Using Eqs. (13.52) and (13.55), we can always express V1 in terms
of V2 and I1 in terms of I2, or vice versa:
V1 =
V2
n
or V2 = nV1 (13.56)
I1 = nI2 or I2 =
I1
n
(13.57)
The complex power in the primary winding is
S1 = V1I∗
1 =
V2
n
(nI2)∗
= V2I∗
2 = S2 (13.58)
showing that the complex power supplied to the primary is delivered to the
secondary without loss. The transformer absorbs no power. Of course,
we should expect this, since the ideal transformer is lossless. The input
impedance as seen by the source in Fig. 13.31 is found from Eqs. (13.56)
and (13.57) as
Zin =
V1
I1
=
1
n2
V2
I2
(13.59)
It is evident from Fig. 13.31 that V2/I2 = ZL, so that
Zin =
ZL
n2
(13.60)
The input impedance is also called the reflected impedance, since it ap-
pears as if the load impedance is reflected to the primary side. This ability
of the transformer to transform a given impedance into another impedance
provides us a means of impedance matching to ensure maximum power
transfer. The idea of impedance matching is very useful in practice and
will be discussed more in Section 13.9.2.
Notice that an ideal transformer reflects an im-
pedance as the square of the turns ratio.
In analyzing a circuit containing an ideal transformer, it is common
practicetoeliminatethetransformerbyreflectingimpedancesandsources
from one side of the transformer to the other. In the circuit of Fig. 13.33,
suppose we want to reflect the secondary side of the circuit to the primary
side. We find the Thevenin equivalent of the circuit to the right of the
terminals a-b. We obtain VTh as the open-circuit voltage at terminals a-b,
as shown in Fig. 13.34(a). Since terminals a-b are open, I1 = 0 = I2 so
that V2 = Vs2. Hence, from Eq. (13.56),
VTh = V1 =
V2
n
=
Vs2
n
(13.61)
+
−
Z1 Z2
Vs1
+
− Vs2
I1 I2
a
b
c
d
V1 V2
+
−
+
−
1:n
Figure 13.33 Ideal transformer circuit whose equivalent circuits are
to be found.
CHAPTER 13 Magnetically Coupled Circuits 549
Z2
+
− Vs2
I1 I2
a
b
a
b
V1
VTh V2
+
−
+
−
1:n
(a)
+
−
I1 I2
V1 V2
+
−
+
−
1:n
(b)
+
−
Z2
V
1 0°
Figure13.34 (a) Obtaining VTh for the circuit in Fig. 13.33, (b) obtaining ZTh for the circuit in Fig. 13.33.
To get ZTh, we remove the voltage source in the secondary winding and
insert a unit source at terminals a-b, as in Fig. 13.34(b). From Eqs. (13.56)
and (13.57), I1 = nI2 and V1 = V2/n, so that
ZTh =
V1
I1
=
V2/n
nI2
=
Z2
n2
, V2 = Z2I2 (13.62)
which is what we should have expected from Eq. (13.60). Once we have
VTh and ZTh, we add the Thevenin equivalent to the part of the circuit in
Fig. 13.33 to the left of terminals a-b. Figure 13.35 shows the result.
+
−
Z1
Vs1
+
−
Vs2
n
a
b
V1
+
−
n2
Z2
Figure13.35 Equivalent circuit for Fig. 13.33
obtained by reflecting the secondary circuit to
the primary side.
The general rule for eliminating the transformer and reflecting the secondary circuit
to the primary side is: divide the secondary impedance by n2
, divide the secondary
voltage by n, and multiply the secondary current by n.
We can also reflect the primary side of the circuit in Fig. 13.33 to
the secondary side. Figure 13.36 shows the equivalent circuit.
The rule for eliminating the transformer and reflecting the primary circuit to the
secondary side is: multiply the primary impedance by n2
, multiply the primary
voltage by n, and divide the primary current by n.
According to Eq. (13.58), the power remains the same, whether calculated
on the primary or the secondary side. But realize that this reflection
approach only applies if there are no external connections between the
primary and secondary windings. When we have external connections
between the primary and secondary windings, we simply use regular
mesh and nodal analysis. Examples of circuits where there are external
connections between the primary and secondary windings are in Figs.
13.39 and 13.40. Also note that if the locations of the dots in Fig. 13.33
are changed, we might have to replace n by −n in order to obey the dot
rule, illustrated in Fig. 13.32.
+
−
n2
Z1 Z2
nVs1 Vs2
+
−
c
d
V2
+
−
Figure13.36 Equivalent circuit for Fig. 13.33
obtained by reflecting the primary circuit to the
secondary side.
E X A M P L E 1 3 . 7
An ideal transformer is rated at 2400/120 V, 9.6 kVA, and has 50 turns
on the secondary side. Calculate: (a) the turns ratio, (b) the number of
550 PART 2 AC Circuits
turns on the primary side, and (c) the current ratings for the primary and
secondary windings.
Solution:
(a) This is a step-down transformer, since V1 = 2400 V  V2 = 120 V.
n =
V2
V1
=
120
2400
= 0.05
(b)
n =
N2
N1
⇒ 0.05 =
50
N1
or
N1 =
50
0.05
= 1000 turns
(c) S = V1I1 = V2I2 = 9.6 kVA. Hence,
I1 =
9600
V1
=
9600
2400
= 4 A
I2 =
9600
V2
=
9600
120
= 80 A or I2 =
I1
n
=
4
0.05
= 80 A
P R A C T I C E P R O B L E M 1 3 . 7
The primary current to an ideal transformer rated at 3300/110 V is 3 A.
Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current.
Answer: (a) 1/30, (b) 9.9 kVA, (c) 90 A.
E X A M P L E 1 3 . 8
For the ideal transformer circuit of Fig. 13.37, find: (a) the source current
I1, (b) the output voltage Vo, and (c) the complex power supplied by the
source.
4 Ω
+
− 20 Ω
−j6 Ω
V1 V2 Vo
1:2
+
−
+
−
I1 I2
V rms
120 0°
+
−
Figure13.37 For Example 13.8.
Solution:
(a) The 20- impedance can be reflected to the primary side and we get
ZR =
20
n2
=
20
4
= 5 
Thus,
Zin = 4 − j6 + ZR = 9 − j6 = 10.82 − 33.69◦

I1 =
120 0◦
Zin
=
120 0◦
10.82 − 33.69◦
= 11.09 33.69◦
A
CHAPTER 13 Magnetically Coupled Circuits 551
(b) Since both I1 and I2 leave the dotted terminals,
I2 = −
1
n
I1 = −5.545 33.69◦
A
Vo = 20I2 = 110.9 213.69◦
V
(c) The complex power supplied is
S = VsI∗
1 = (120 0◦
)(11.09 − 33.69◦
) = 1330.8 − 33.69◦
VA
P R A C T I C E P R O B L E M 1 3 . 8
In the ideal transformer circuit of Fig. 13.38, find Vo and the complex
power supplied by the source.
2 Ω
+
−
16 Ω
V1 V2 Vo
1:4
+
−
+
−
I1 I2
V rms
100 0°
+
−
−j24 Ω
Figure13.38 For Practice Prob. 13.8.
Answer: 178.9 116.56◦
V, 2981.5 − 26.56◦
VA.
E X A M P L E 1 3 . 9
Calculate the power supplied to the 10- resistor in the ideal transformer
circuit of Fig. 13.39.
20 Ω
10 Ω
V1 V2
2:1
+
−
+
−
V rms
120 0°
30 Ω
+
− I1 I2
Figure13.39 For Example 13.9.
Solution:
Reflection to the secondary or primary side cannot be done with this
circuit: there is direct connection between the primary and secondary
sides due to the 30- resistor. We apply mesh analysis. For mesh 1,
−120 + (20 + 30)I1 − 30I2 + V1 = 0
or
50I1 − 30I2 + V1 = 120 (13.9.1)
552 PART 2 AC Circuits
For mesh 2,
−V2 + (10 + 30)I2 − 30I1 = 0
or
−30I1 + 40I2 − V2 = 0 (13.9.2)
At the transformer terminals,
V2 = −
1
2
V1 (13.9.3)
I2 = −2I1 (13.9.4)
(Note that n = 1/2.) We now have four equations and four unknowns,
but our goal is to get I2. So we substitute for V1 and I1 in terms of V2
and I2 in Eqs. (13.9.1) and (13.9.2). Equation (13.9.1) becomes
−55I2 − 2V2 = 120 (13.9.5)
and Eq. (13.9.2) becomes
15I2 + 40I2 − V2 = 0 ⇒ V2 = 55I2 (13.9.6)
Substituting Eq. (13.9.6) in Eq. (13.9.5),
−165I2 = 120 ⇒ I2 = −
120
165
= −0.7272 A
The power absorbed by the 10- resistor is
P = (−0.7272)2
(10) = 5.3 W
P R A C T I C E P R O B L E M 1 3 . 9
Find Vo in the circuit in Fig. 13.40.
4 Ω
8 Ω
1:2
V
60 0° +
−
2 Ω
8 Ω
+ −
Vo
Figure13.40 For Practice Prob. 13.9.
Answer: 24 V.
13.6 IDEAL AUTOTRANSFORMERS
Unlike the conventional two-winding transformer we have considered so
far, an autotransformer has a single continuous winding with a connection
point called a tap between the primary and secondary sides. The tap is
CHAPTER 13 Magnetically Coupled Circuits 553
often adjustable so as to provide the desired turns ratio for stepping up
or stepping down the voltage. This way, a variable voltage is provided to
the load connected to the autotransformer.
An autotransformer is a transformer in which both the primary
and the secondary are in a single winding.
Figure 13.41 A typical autotransformer.
(Courtesy of Todd Systems, Inc.)
Figure 13.41 shows a typical autotransformer. As shown in Fig.
13.42, the autotransformer can operate in the step-down or step-up mode.
The autotransformer is a type of power transformer. Its major advantage
over the two-winding transformer is its ability to transfer larger apparent
power. Example 13.10 will demonstrate this. Another advantage is that
an autotransformer is smaller and lighter than an equivalent two-winding
transformer. However, since both the primary and secondary windings
are one winding, electrical isolation (no direct electrical connection) is
lost. (We will see how the property of electrical isolation in the conven-
tional transformer is practically employed in Section 13.9.1.) The lack
of electrical isolation between the primary and secondary windings is a
major disadvantage of the autotransformer.
Some of the formulas we derived for ideal transformers apply to
ideal autotransformers as well. For the step-down autotransformer circuit
of Fig. 13.42(a), Eq. (13.52) gives
V1
V2
=
N1 + N2
N2
= 1 +
N1
N2
(13.63)
As an ideal autotransformer, there are no losses, so the complex power
remains the same in the primary and secondary windings:
S1 = V1I∗
1 = S2 = V2I∗
2 (13.64)
Equation (13.64) can also be expressed with rms values as
V1I1 = V2I2
or
V2
V1
=
I1
I2
(13.65)
Thus, the current relationship is
I1
I2
=
N2
N1 + N2
(13.66)
+
−
I1
I2
V1
N1
N2
+
−
V2
+
−
(a)
V
+
−
I1
I2
V1
N1
N2
+
−
V2
+
−
(b)
ZL
ZL
V
Figure 13.42 (a) Step-down autotransformer,
(b) step-up autotransformer.
For the step-up autotransformer circuit of Fig. 13.42(b),
V1
N1
=
V2
N1 + N2
or
V1
V2
=
N1
N1 + N2
(13.67)
554 PART 2 AC Circuits
The complex power given by Eq. (13.64) also applies to the step-up auto-
transformer so that Eq. (13.65) again applies. Hence, the current relation-
ship is
I1
I2
=
N1 + N2
N1
= 1 +
N2
N1
(13.68)
A major difference between conventional transformers and auto-
transformers is that the primary and secondary sides of the autotrans-
former are not only coupled magnetically but also coupled conductively.
The autotransformer can be used in place of a conventional transformer
when electrical isolation is not required.
E X A M P L E 1 3 . 1 0
Compare the power ratings of the two-winding transformer in Fig.
13.43(a) and the autotransformer in Fig. 13.43(b).
+
−
(a) (b)
240 V
+
−
12 V
Vs
Vp
0.2 A 4 A
4 A
+
−
Vs = 12 V
+
−
Vp = 240 V
+
−
+
−
+
−
240 V
+
−
252 V
4 A
0.2 A
Figure13.43 For Example 13.10.
Solution:
Although the primary and secondary windings of the autotransformer
are together as a continuous winding, they are separated in Fig. 13.43(b)
for clarity. We note that the current and voltage of each winding of the
autotransformer in Fig. 13.43(b) are the same as those for the two-winding
transformer in Fig. 13.43(a). This is the basis of comparing their power
ratings.
For the two-winding transformer, the power rating is
S1 = 0.2(240) = 48 VA or S2 = 4(12) = 48 VA
For the autotransformer, the power rating is
S1 = 4.2(240) = 1008 VA or S2 = 4(252) = 1008 VA
which is 21 times the power rating of the two-winding transformer.
P R A C T I C E P R O B L E M 1 3 . 1 0
Refer to Fig. 13.43. If the two-winding transformer is a 60-VA,
120 V/10 V transformer, what is the power rating of the autotransformer?
Answer: 780 VA.
CHAPTER 13 Magnetically Coupled Circuits 555
E X A M P L E 1 3 . 1 1
Refer to the autotransformer circuit in Fig. 13.44. Calculate: (a) I1, I2,
and Io if ZL = 8+j6 , and (b) the complex power supplied to the load.
+
−
I1
I2
V1
120 turns
80 turns
+
−
V2
+
−
ZL
Io
120 30° V rms
Figure13.44 For Example 13.11.
Solution:
(a) This is a step-up autotransformer with N1 = 80, N2 = 120, V1 =
120 30◦
, so Eq. (13.67) can be used to find V2 by
V1
V2
=
N1
N1 + N2
=
80
200
or
V2 =
200
80
V1 =
200
80
(120 30◦
) = 300 30◦
V
I2 =
V2
ZL
=
300 30◦
8 + j6
=
300 30◦
10 36.87◦
= 30 − 6.87◦
A
But
I1
I2
=
N1 + N2
N1
=
200
80
or
I1 =
200
80
I2 =
200
80
(30 − 6.87◦
) = 75 − 6.87◦
A
At the tap, KCL gives
I1 + Io = I2
or
Io = I2 − I1 = 30 − 6.87◦
− 75 − 6.87◦
= 45 173.13◦
A
(b) The complex power supplied to the load is
S2 = V2I∗
2 = |I2|2
ZL = (30)2
(10 36.87◦
) = 9 36.87◦
kVA
556 PART 2 AC Circuits
P R A C T I C E P R O B L E M 1 3 . 1 1
In the autotransformer circuit in Fig. 13.45, find currents I1, I2, and Io.
Take V1 = 1250 V, V2 = 800 V.
I1
I2
V2
+
−
V1
+
− Io
16 kVA load
Figure13.45 For Practice Prob. 13.11.
Answer: 12.8 A, 20 A, 7.2 A.
†13.7 THREE-PHASE TRANSFORMERS
To meet the demand for three-phase power transmission, transformer
connections compatible with three-phase operations are needed. We can
achieve the transformer connections in two ways: by connecting three
single-phase transformers, thereby forming a so-called transformer bank,
or by using a special three-phase transformer. For the same kVA rat-
ing, a three-phase transformer is always smaller and cheaper than three
single-phase transformers. When single-phase transformers are used, one
must ensure that they have the same turns ratio n to achieve a balanced
three-phase system. There are four standard ways of connecting three
single-phase transformers or a three-phase transformer for three-phase
operations: Y-Y, -, Y-, and -Y.
For any of the four connections, the total apparent power ST , real
power PT , and reactive power QT are obtained as
ST =
√
3VLIL (13.69a)
PT = ST cos θ =
√
3VLIL cos θ (13.69b)
QT = ST sin θ =
√
3VLIL sin θ (13.69c)
where VL and IL are, respectively, equal to the line voltage VLp and the
line current ILp for the primary side, or the line voltage VLs and the line
current ILs for the secondary side. Notice from Eq. (13.69) that for each of
the four connections, VLsILs = VLpILp, since power must be conserved
in an ideal transformer.
For the Y-Y connection (Fig. 13.46), the line voltage VLp at the
primary side, the line voltage VLs on the secondary side, the line current
ILp on the primary side, and the line current ILs on the secondary side
are related to the transformer per phase turns ratio n according to Eqs.
(13.52) and (13.55) as
VLs = nVLp (13.70a)
ILs =
ILp
n
(13.70b)
CHAPTER 13 Magnetically Coupled Circuits 557
For the - connection (Fig. 13.47), Eq. (13.70) also applies for
the line voltages and line currents. This connection is unique in the sense
that if one of the transformers is removed for repair or maintenance, the
other two form an open delta, which can provide three-phase voltages at
a reduced level of the original three-phase transformer.
+
−
VLp
+
−
VLs = nVLp
ILp
ILs =
ILp
n
1:n
Figure13.46 Y-Y three-phase transformer connection.
+
−
VLp
+
−
VLs = nVLp
ILp
ILs =
ILp
n
1:n
Figure13.47 - three-phase transformer connection.
For the Y- connection (Fig. 13.48), there is a factor of
√
3 arising
from the line-phase values in addition to the transformer per phase turns
ratio n. Thus,
VLs =
nVLp
√
3
(13.71a)
ILs =
√
3ILp
n
(13.71b)
Similarly, for the -Y connection (Fig. 13.49),
VLs = n
√
3VLp (13.72a)
ILs =
ILp
n
√
3
(13.72b)
+
−
VLp
+
−
ILp
1:n
ILs =
3ILp
n
VLs =
nVLp
3
Figure13.48 Y- three-phase transformer connection.
+
−
VLs = n VLp
ILp
1:n
ILs =
ILp
3
n
3
+
−
VLp
Figure13.49 -Y three-phase transformer connection.
558 PART 2 AC Circuits
E X A M P L E 1 3 . 1 2
The 42-kVA balanced load depicted in Fig. 13.50 is supplied by a three-
phase transformer. (a) Determine the type of transformer connections.
(b) Find the line voltage and current on the primary side. (c) Determine
the kVA rating of each transformer used in the transformer bank. Assume
that the transformers are ideal.
a
b
c
A
B
C
42 kVA
Three-phase
load
240 V
1:5
Figure13.50 For Example 13.12.
Solution:
(a) A careful observation of Fig. 13.50 shows that the primary side is
Y-connected, while the secondary side is -connected. Thus, the three-
phase transformer is Y-, similar to the one shown in Fig. 13.48.
(b) Given a load with total apparent power ST = 42 kVA, the turns ra-
tio n = 5, and the secondary line voltage VLs = 240 V, we can find the
secondary line current using Eq. (13.69a), by
ILs =
ST
√
3VLs
=
42,000
√
3(240)
= 101 A
From Eq. (13.71),
ILp =
n
√
3
ILs =
5 × 101
√
3
= 292 A
VLp =
√
3
n
VLs =
√
3 × 240
5
= 83.14 V
(c) Because the load is balanced, each transformer equally shares the
total load and since there are no losses (assuming ideal transformers), the
kVA rating of each transformer is S = ST /3 = 14 kVA. Alternatively,
the transformer rating can be determined by the product of the phase
current and phase voltage of the primary or secondary side. For the
primary side, for example, we have a delta connection, so that the phase
voltage is the same as the line voltage of 240 V, while the phase current
is ILp/
√
3 = 58.34 A. Hence, S = 240 × 58.34 = 14 kVA.
P R A C T I C E P R O B L E M 1 3 . 1 2
A three-phase - transformer is used to step down a line voltage of
625 kV, to supply a plant operating at a line voltage of 12.5 kV. The plant
CHAPTER 13 Magnetically Coupled Circuits 559
draws 40 MW with a lagging power factor of 85 percent. Find: (a) the
currentdrawnbytheplant, (b)theturnsratio, (c)thecurrentontheprimary
side of the transformer, and (d) the load carried by each transformer.
Answer: (a) 2.1736 kA, (b) 0.02, (c) 43.47 A, (d) 15.69 MVA.
13.8 PSPICE ANALYSIS OF MAGNETICALLY COUPLED
CIRCUITS
PSpice analyzes magnetically coupled circuits just like inductor circuits
except that the dot convention must be followed. In PSpice Schematic, the
dot (not shown) is always next to pin 1, which is the left-hand terminal of
the inductor when the inductor with part name L is placed (horizontally)
without rotation on a schematic. Thus, the dot or pin 1 will be at the
bottom after one 90◦
counterclockwise rotation, since rotation is always
about pin 1. Once the magnetically coupled inductors are arranged with
the dot convention in mind and their value attributes are set in henries, we
use the coupling symbol K−LINEAR to define the coupling. For each
pair of coupled inductors, take the following steps:
1. Select Draw/Get New Part and type K−LINEAR.
2. Hit enter or click OK and place the K−LINEAR symbol on
the schematic, as shown in Fig. 13.51. (Notice that
K−LINEAR is not a component and therefore has no pins.)
3. DCLICKL on COUPLING and set the value of the coupling
coefficient k.
4. DCLICKL on the boxed K (the coupling symbol) and enter
the reference designator names for the coupled inductors as
values of Li, i = 1, 2, . . . , 6. For example, if inductors L20
and L23 are coupled, we set L1 = L20 and L2 = L23. L1 and
at least one other Li must be assigned values; other Li’s may be
left blank.
In step 4, up to six coupled inductors with equal coupling can be specified.
For the air-core transformer, the partname is XFRM−LINEAR. It
can be inserted in a circuit by selecting Draw/Get Part Name and then
typing in the part name or by selecting the part name from the analog.slb
library. As shown typically in Fig. 13.51, the main attributes of the linear
transformer are the coupling coefficient k and the inductance values L1
and L2 in henries. If the mutual inductance M is specified, its value must
be used along with L1 and L2 to calculate k. Keep in mind that the value
of k should lie between 0 and 1.
TX2
COUPLING=0.5
L1_VALUE=1mH
L2_VALUE=25mH
Figure13.51 Linear trans-
former XFRM LINEAR.
TX4
COUPLING=0.5
L1_TURNS=500
L2_TURNS=1000
kbreak
Figure13.52 Ideal trans-
former XFRM NONLINEAR.
For the ideal transformer, the part name is XFRM−NONLINEAR
and is located in the breakout.slb library. Select it by clicking Draw/Get
Part Name and then typing in the part name. Its attributes are the cou-
pling coefficient and the numbers of turns associated with L1 and L2, as
illustrated typically in Fig. 13.52. The value of the coefficient of mutual
coupling must lie between 0 and 1.
560 PART 2 AC Circuits
PSpice has some additional transformer configurations that we will
not discuss here.
E X A M P L E 1 3 . 1 3
Use PSpice to find i1, i2, and i3 in the circuit displayed in Fig. 13.53.
40 cos 12pt V
60 cos (12pt – 10°) V 4 H
2 H
1.5 H
270 mF
3 H
3 H
1 H
2 H
+
−
+
−
70 Ω
100 Ω
i2
i1 i3
Figure13.53 For Example 13.13.
Solution:
The coupling coefficients of the three coupled inductors are determined
as follows.
k12 =
M12
√
L1L2
=
1
√
3 × 3
= 0.3333
k13 =
M13
√
L1L3
=
1.5
√
3 × 4
= 0.433
k23 =
M23
√
L2L3
=
2
√
3 × 4
= 0.5774
The operating frequency f is obtained from Fig. 13.53 as ω = 12π =
2πf → f = 6 Hz.
The right-hand values are the reference designa-
tors of the inductors on the schematic.
The schematic of the circuit is portrayed in Fig. 13.54. Notice how
the dot convention is adhered to. For L2, the dot (not shown) is on pin
1 (the left-hand terminal) and is therefore placed without rotation. For
L1, in order for the dot to be on the right-hand side of the inductor, the
inductor must be rotated through 180◦
. For L3, the inductor must be
rotated through 90◦
so that the dot will be at the bottom. Note that the
2-H inductor (L4) is not coupled. To handle the three coupled inductors,
we use three K−LINEAR parts provided in the analog library and set the
following attributes (by double-clicking on the symbol K in the box):
K1 - K_LINEAR
L1 = L1
L2 = L2
COUPLING = 0.3333
K2 - K_LINEAR
L1 = L2
L2 = L3
COUPLING = 0.433
CHAPTER 13 Magnetically Coupled Circuits 561
K3 - K_LINEAR
L1 = L1
L2 = L3
COUPLING = 0.5774
ThreeIPRINTpseudocomponentsareinsertedintheappropriatebranches
to obtain the required currents i1, i2, and i3. As an AC single-frequency
analysis, we select Analysis/Setup/AC Sweep and enter Total Pts = 1,
Start Freq = 6, and Final Freq = 6. After saving the schematic, we select
Analysis/Simulate to simulate it. The output file includes:
FREQ IM(V_PRINT2) IP(V_PRINT2)
6.000E+00 2.114E-01 -7.575E+01
FREQ IM(V_PRINT1) IP(V_PRINT1)
6.000E+00 4.654E-01 -7.025E+01
FREQ IM(V_PRINT3) IP(V_PRINT3)
6.000E+00 1.095E-01 1.715E+01
From this we obtain
I1 = 0.4654 − 70.25◦
I2 = 0.2114 − 75.75◦
, I3 = 0.1095 17.15◦
Thus,
i1 = 0.4654 cos(12πt − 70.25◦
) A
i2 = 0.2114 cos(12πt − 75.75◦
) A
i3 = 0.1095 cos(12πt + 17.15◦
) A
V2
4H
L3
270u
0
L2
3H
3H
L1
L4
R1
2H
70
R2
100
−
+ ACMAG=40V
ACPHASE=0
V1
−
+
ACMAG=60V
ACPHASE=-10
MAG=ok
AC=ok
PHASE=ok
K K1
K_Linear
COUPLING=0.3333
L1=L1
L2=L2
K K2
K_Linear
COUPLING=0.433
L1=L2
L2=L3
K K3
K_Linear
COUPLING=0.5774
L1=L1
L2=L3
IPRINT
IPRINT
IPRINT
C1
Figure13.54 Schematic of the circuit of Fig. 13.53.
P R A C T I C E P R O B L E M 1 3 . 1 3
Find io in the circuit of Fig. 13.55.
562 PART 2 AC Circuits
20 Ω
6 H
5 H 4 H
8 cos (4t + 50°) V +
−
10 Ω
8 Ω
12 Ω
k = 0.4
25 mF
io
Figure13.55 For Practice Prob. 13.13.
Answer: 0.1006 cos(4t + 68.52◦
) A.
E X A M P L E 1 3 . 1 4
Find V1 and V2 in the ideal transformer circuit of Fig. 13.56 using PSpice.
80 Ω
6 Ω
V1 V2
4:1
+
−
+
−
V
20 Ω
+
−
120 30°
−j40 Ω
j10 Ω
Figure13.56 For Example 13.14.
Solution:
As usual, we assume ω = 1 and find the corresponding values of capac-
itance and inductance of the elements:
j10 = jωL ⇒ L = 10 H
−j40 =
1
jωC
⇒ C = 25 mF
Figure 13.57 shows the schematic. For the ideal transformer, we set
the coupling factor to 0.999 and the numbers of turns to 400,000 and
100,000. The two VPRINT2 pseudocomponents are connected across
the transformer terminals to obtain V1 and V2. As a single-frequency
analysis, we select Analysis/Setup/AC Sweep and enter Total Pts =
1, Start Freq = 0.1592, and Final Freq = 0.1592. After saving the
schematic, we select Analysis/Simulate to simulate it. The output file
includes:
Reminder: For an ideal transformer, the induc-
tances of both the primary and secondary wind-
ings are infinitely large.
FREQ VM(C,A) VP(C,A)
1.592E-01 1.212E+02 -1.435E+02
CHAPTER 13 Magnetically Coupled Circuits 563
FREQ VM(B,C) VP(B,C)
1.592E-01 2.775E+02 2.789E+01
From this we obtain
V1 = −V(C,A) = 121.1 36.5◦
V
V2 = V(B,C) = 27.75 27.89◦
V
R1
V1
80
6
R3
10H
L1
TX1
ACMAG=120V
ACPHASE=30
AC=ok
MAG=ok
PHASE=ok
AC=ok
MAG=ok
PHASE=ok
L1_TURNS=400000
L2_TURNS=100000
COUPLING=0.999
20
R2
C1
A B
C
25m
−
+
0
kbreak
Figure13.57 The schematic for the circuit in Fig. 13.56.
P R A C T I C E P R O B L E M 1 3 . 1 4
Obtain V1 and V2 in the circuit of Fig. 13.58 using PSpice.
10 Ω
2:3
30 Ω
20 Ω
V1 V2
+
−
+
−
V +
−
100 20° −j16 Ω
j15 Ω
Figure13.58 For Practice Prob. 13.14.
Answer: 63.1 28.65◦
V, 94.64 − 151.4◦
V.
†13.9 APPLICATIONS
Transformers are the largest, the heaviest, and often the costliest of circuit
components. Nevertheless, they are indispensable passive devices in
564 PART 2 AC Circuits
electric circuits. They are among the most efficient machines, 95 percent
efficiency being common and 99 percent being achievable. They have
numerous applications. For example, transformers are used:
• To step up or step down voltage and current, making them
useful for power transmission and distribution.
• To isolate one portion of a circuit from another (i.e., to transfer
power without any electrical connection).
• As an impedance-matching device for maximum power transfer.
• In frequency-selective circuits whose operation depends on the
response of inductances.
Because of these diverse uses, there are many special designs for
transformers (only some of which are discussed in this chapter): voltage
transformers, currenttransformers, powertransformers, distributiontrans-
formers, impedance-matching transformers, audio transformers, single-
phase transformers, three-phase transformers, rectifier transformers,
inverter transformers, and more. In this section, we consider three im-
portant applications: transformer as an isolation device, transformer as a
matching device, and power distribution system.
Formoreinformationonthemanykindsoftrans-
formers, a good text is W. M. Flanagan, Hand-
book of Transformer Design and Applications, 2nd
ed. (New York: McGraw-Hill, 1993).
13.9.1 Transformer as an Isolation Device
Electrical isolation is said to exist between two devices when there is no
physical connection between them. In a transformer, energy is transferred
by magnetic coupling, without electrical connection between the primary
circuit and secondary circuit. We now consider three simple practical
examples of how we take advantage of this property.
First, consider the circuit in Fig. 13.59. A rectifier is an electronic
circuit that converts an ac supply to a dc supply. A transformer is often
used to couple the ac supply to the rectifier. The transformer serves two
purposes. First, it steps up or steps down the voltage. Second, it provides
electrical isolation between the ac power supply and the rectifier, thereby
reducing the risk of shock hazard in handling the electronic device.
va
+
−
1:n
Fuse
Rectifier
Isolation transformer
Figure 13.59 A transformer used to isolate an
ac supply from a rectifier.
As a second example, a transformer is often used to couple two
stages of an amplifier, to prevent any dc voltage in one stage from affecting
the dc bias of the next stage. Biasing is the application of a dc voltage to
a transistor amplifier or any other electronic device in order to produce
a desired mode of operation. Each amplifier stage is biased separately
to operate in a particular mode; the desired mode of operation will be
compromised without a transformer providing dc isolation. As shown in
Fig. 13.60, only the ac signal is coupled through the transformer from one
stage to the next. We recall that magnetic coupling does not exist with
a dc voltage source. Transformers are used in radio and TV receivers to
couple stages of high-frequency amplifiers. When the sole purpose of a
transformer is to provide isolation, its turns ratio n is made unity. Thus,
an isolation transformer has n = 1.
As a third example, consider measuring the voltage across 13.2-kV
lines. It is obviously not safe to connect a voltmeter directly to such
high-voltage lines. A transformer can be used both to electrically isolate
the line power from the voltmeter and to step down the voltage to a safe
level, as shown in Fig. 13.61. Once the voltmeter is used to measure the
CHAPTER 13 Magnetically Coupled Circuits 565
secondary voltage, the turns ratio is used to determine the line voltage on
the primary side.
1:1
Amplifier
stage 2
Amplifier
stage 1
Isolation transformer
ac only
ac + dc
Figure13.60 A transformer providing dc isolation between
two amplifier stages.
1:n
V
120 V
–
+
13,200 V
–
+
Power lines
Voltmeter
Figure13.61 A transformer providing isolation between
the power lines and the voltmeter.
E X A M P L E 1 3 . 1 5
Determine the voltage across the load in Fig. 13.62.
Solution:
We can apply the superposition principle to find the load voltage. Let
vL = vL1 + vL2, where vL1 is due to the dc source and vL2 is due to the
ac source. We consider the dc and ac sources separately, as shown in
Fig. 13.63. The load voltage due to the dc source is zero, because a time-
varying voltage is necessary in the primary circuit to induce a voltage in
the secondary circuit. Thus, vL1 = 0. For the ac source,
V2
V1
=
V2
120
=
1
3
or V2 =
120
3
= 40 V
Hence, VL2 = 40 V ac or vL2 = 40 cos ωt; that is, only the ac voltage
is passed to the load by the transformer. This example shows how the
transformer provides dc isolation.
3:1
+
−
120 V
ac
12 V
dc
RL = 5 kΩ
Figure13.62 For Example 13.15.
3:1
6 V
dc RL
V2 = 0
3:1
+
−
120 V
ac RL
+
−
+
−
V2
V1
+
−
(a) (b)
Figure13.63 For Example 13.15: (a) dc source, (b) ac source.
P R A C T I C E P R O B L E M 1 3 . 1 5
Refer to Fig. 13.61. Calculate the turns ratio required to step down the
13.2-kV line voltage to a safe level of 120 V.
Answer: 1/110.
566 PART 2 AC Circuits
13.9.2 Transformer as a Matching Device
Werecallthatformaximumpowertransfer, theloadresistanceRL mustbe
matched with the source resistance Rs. In most cases, the two resistances
are not matched; both are fixed and cannot be altered. However, an iron-
core transformer can be used to match the load resistance to the source
resistance. This is called impedance matching. For example, to connect a
loudspeaker to an audio power amplifier requires a transformer, because
the speaker’s resistance is only a few ohms while the internal resistance
of the amplifier is several thousand ohms.
ConsiderthecircuitshowninFig.13.64. WerecallfromEq.(13.60)
that the ideal transformer reflects its load back to the primary with a
scaling factor of n2
. To match this reflected load RL/n2
with the source
resistance Rs, we set them equal,
Rs =
RL
n2
(13.73)
Equation (13.73) can be satisfied by proper selection of the turns ratio
n. From Eq. (13.73), we notice that a step-down transformer (n  1) is
needed as the matching device when Rs  RL, and a step-up (n  1) is
required when Rs  RL.
vs
+
−
1:n
Matching transformer
Load
Source
Rs
RL
Figure 13.64 Transformer used as a matching
device.
E X A M P L E 1 3 . 1 6
The ideal transformer in Fig. 13.65 is used to match the amplifier circuit
to the loudspeaker to achieve maximum power transfer. The Thevenin (or
output) impedance of the amplifier is 192 , and the internal impedance
of the speaker is 12 . Determine the required turns ratio.
Amplifier
circuit
1:n
Speaker
Figure13.65 Using an ideal transformer to
match the speaker to the amplifier; for
Example 13.16.
Solution:
We replace the amplifier circuit with the Thevenin equivalent and reflect
the impedance ZL = 12  of the speaker to the primary side of the
ideal transformer. Figure 13.66 shows the result. For maximum power
transfer,
ZTh =
ZL
n2
or n2
=
ZL
ZTh
=
12
192
=
1
16
Thus, the turns ratio is n = 1/4 = 0.25.
VTh
ZL
n2
ZTh
+
−
Figure13.66 Equivalent circuit
of the circuit in Fig. 13.65, for
Example 13.16.
Using P = I2
R, we can show that indeed the power delivered to
the speaker is much larger than without the ideal transformer. Without
the ideal transformer, the amplifier is directly connected to the speaker.
The power delivered to the speaker is
PL =
VTh
ZTh + ZL
2
ZL = 288 V2
Th µW
With the transformer in place, the primary and secondary currents are
Ip =
VTh
ZTh + ZL/n2
, Is =
Ip
n
CHAPTER 13 Magnetically Coupled Circuits 567
Hence,
PL = I2
s ZL =
VTh/n
ZTh + ZL/n2
2
ZL
=
nVTh
n2ZTh + ZL
2
ZL = 1302V2
Th µW
confirming what was said earlier.
P R A C T I C E P R O B L E M 1 3 . 1 6
Calculate the turns ratio of an ideal transformer required to match a 100-
load to a source with internal impedance of 2.5 k. Find the load voltage
when the source voltage is 30 V.
Answer: 0.2, 3 V.
One may ask, How would increasing the voltage
not increase the current, thereby increasing I2
R
losses? Keep in mind that I = V-/R, where V-
is the potential difference between the sending
and receiving ends of the line. The voltage that
is stepped up is the sending end voltage V, not
V-. If the receiving end is VR, then V- = V −
VR. Since V and VR are close to each other, V- is
small even when V is stepped up.
13.9.3 Power Distribution
A power system basically consists of three components:generation, trans-
mission, and distribution. The local electric company operates a plant
that generates several hundreds of megavolt-amperes (MVA), typically at
about 18 kV. As Fig. 13.67 illustrates, three-phase step-up transformers
are used to feed the generated power to the transmission line. Why do
we need the transformer? Suppose we need to transmit 100,000 VA over
a distance of 50 km. Since S = V I, using a line voltage of 1000 V
implies that the transmission line must carry 100 A and this requires a
transmission line of a large diameter. If, on the other hand, we use a
line voltage of 10,000 V, the current is only 10 A. The smaller current
reduces the required conductor size, producing considerable savings as
well as minimizing transmission line I2
R losses. To minimize losses
requires a step-up transformer. Without the transformer, the majority of
the power generated would be lost on the transmission line. The ability
3f
345,000 V
3f 60 Hz ac
18,000 V
Generator
3f
Step-up
transformer
3f
Step-down
transformer
Neutral
3f 60 Hz ac
208 V
345,000 V 345,000 V
Neutral
Neutral
Neutral
Tower
Tower
Insulators
Figure 13.67 A typical power distribution system.
(Source: A. Marcus and C. M. Thomson, Electricity for Technicians,
2nd ed. [Englewood Cliffs, NJ: Prentice Hall, 1975], p. 337.)
568 PART 2 AC Circuits
of the transformer to step up or step down voltage and distribute power
economically is one of the major reasons for generating ac rather than dc.
Thus, for a given power, the larger the voltage, the better. Today, 1 MV
is the largest voltage in use; the level may increase as a result of research
and experiments.
Beyond the generation plant, the power is transmitted for hundreds
of miles through an electric network called the power grid. The three-
phase power in the power grid is conveyed by transmission lines hung
overhead from steel towers which come in a variety of sizes and shapes.
The (aluminum-conductor, steel-reinforced) lines typically have overall
diameters up to about 40 mm and can carry current of up to 1380 A.
At the substations, distribution transformers are used to step down
the voltage. The step-down process is usually carried out in stages. Power
may be distributed throughout a locality by means of either overhead or
underground cables. The substations distribute the power to residential,
commercial, and industrial customers. At the receiving end, a residen-
tial customer is eventually supplied with 120/240 V, while industrial or
commercial customers are fed with higher voltages such as 460/208 V.
Residential customers are usually supplied by distribution transformers
often mounted on the poles of the electric utility company. When direct
current is needed, the alternating current is converted to dc electronically.
E X A M P L E 1 3 . 1 7
A distribution transformer is used to supply a household as in Fig. 13.68.
The load consists of eight 100-W bulbs, a 350-W TV, and a 15-kW kitchen
range. If the secondary side of the transformer has 72 turns, calculate:
(a) the number of turns of the primary winding, and (b) the current Ip in
the primary winding.
Ip
2400 V
120 V
+
−
+
−
120 V
–
+
TV
Kitchen
range
8 bulbs
Figure13.68 For Example 13.17.
Solution:
(a) The dot locations on the winding are not important, since we are only
interested in the magnitudes of the variables involved. Since
Np
Ns
=
Vp
Vs
CHAPTER 13 Magnetically Coupled Circuits 569
we get
Np = Ns
Vp
Vs
= 72
2400
240
= 720 turns
(b) The total power absorbed by the load is
S = 8 × 100 + 350 + 15,000 = 16.15 kW
But S = VpIp = VsIs, so that
Ip =
S
Vp
=
16,150
2400
= 6.729 A
P R A C T I C E P R O B L E M 1 3 . 1 7
In Example 13.17, if the eight 100-W bulbs are replaced by twelve 60-W
bulbs and the kitchen range is replaced by a 4.5-kW air-conditioner, find:
(a) the total power supplied, (b) the current Ip in the primary winding.
Answer: (a) 5.57 kW, (b) 2.321 A.
13.10 SUMMARY
1. Two coils are said to be mutually coupled if the magnetic flux φ
emanating from one passes through the other. The mutual induc-
tance between the two coils is given by
M = k

L1L2
where k is the coupling coefficient, 0  k  1.
2. If v1 and i1 are the voltage and current in coil 1, while v2 and i2 are
the voltage and current in coil 2, then
v1 = L1
di1
dt
+ M
di2
dt
and v2 = L2
di2
dt
+ M
di1
dt
Thus, the voltage induced in a coupled coil consists of self-induced
voltage and mutual voltage.
3. The polarity of the mutually induced voltage is expressed in the
schematic by the dot convention.
4. The energy stored in two coupled coils is
1
2
L1i2
1 +
1
2
L2i2
2 ± Mi1i2
5. A transformer is a four-terminal device containing two or more
magnetically coupled coils. It is used in changing the current, volt-
age, or impedance level in a circuit.
6. A linear (or loosely coupled) transformer has its coils wound on a
magnetically linear material. It can be replaced by an equivalent T
or  network for the purposes of analysis.
7. An ideal (or iron-core) transformer is a lossless (R1 = R2 = 0)
transformer with unity coupling coefficient (k = 1) and infinite
inductances (L1, L2, M → ∞).
570 PART 2 AC Circuits
8. For an ideal transformer,
V2 = nV1, I2 =
I1
n
, S1 = S2, ZR =
ZL
n2
where n = N2/N1 is the turns ratio. N1 is the number of turns of
the primary winding and N2 is the number of turns of the secondary
winding. The transformer steps up the primary voltage when
n  1, steps it down when n  1, or serves as a matching device
when n = 1.
9. An autotransformer is a transformer with a single winding common
to both the primary and the secondary circuits.
10. PSpice is a useful tool for analyzing magnetically coupled circuits.
11. Transformers are necessary in all stages of power distribution
systems. Three-phase voltages may be stepped up or down by
three-phase transformers.
12. Important uses of transformers in electronics applications are as
electrical isolation devices and impedance-matching devices.
REVIEW QUESTIONS
13.1 Refer to the two magnetically coupled coils of Fig.
13.69(a). The polarity of the mutual voltage is:
(a) Positive (b) Negative
M
i1
(b)
i2
M
i1
(a)
i2
Figure 13.69 For Review Questions 13.1 and 13.2.
13.2 For the two magnetically coupled coils of Fig.
13.69(b), the polarity of the mutual voltage is:
(a) Positive (b) Negative
13.3 The coefficient of coupling for two coils having
L1 = 2 H, L2 = 8 H, M = 3 H is:
(a) 0.1875 (b) 0.75
(c) 1.333 (d) 5.333
13.4 A transformer is used in stepping down or stepping
up:
(a) dc voltages (b) ac voltages
(c) both dc and ac voltages
13.5 The ideal transformer in Fig. 13.70(a) has
N2/N1 = 10. The ratio V2/V1 is:
(a) 10 (b) 0.1 (c) −0.1 (d) −10
N1:N2
I1
(a)
I2
+
−
+
−
N1:N2
I1
(b)
I2
V1 V2
Figure 13.70 For Review Questions 13.5 and 13.6.
13.6 For the ideal transformer in Fig. 13.70(b),
N2/N1 = 10. The ratio I2/I1 is:
(a) 10 (b) 0.1 (c) −0.1 (d) −10
13.7 A three-winding transformer is connected as
portrayed in Fig. 13.71(a). The value of the output
voltage Vo is:
(a) 10 (b) 6 (c) −6 (d) −10
50 V
+
−
Vo
2 V
8 V
+
−
(a)
50 V
+
−
Vo
2 V
8 V
+
−
(b)
Figure 13.71 For Review Questions 13.7 and 13.8.
13.8 If the three-winding transformer is connected as in
Fig. 13.71(b), the value of the output voltage Vo is:
(a) 10 (b) 6 (c) −6 (d) −10
CHAPTER 13 Magnetically Coupled Circuits 571
13.9 In order to match a source with internal impedance
of 500  to a 15- load, what is needed is:
(a) step-up linear transformer
(b) step-down linear transformer
(c) step-up ideal transformer
(d) step-down ideal transformer
(e) autotransformer
13.10 Which of these transformers can be used as an
isolation device?
(a) linear transformer (b) ideal transformer
(c) autotransformer (d) all of the above
Answers: 13.1b, 13.2a, 13.3b, 13.4b, 13.5d, 13.6b, 13.7c, 13.8a,
13.9d, 13.10b.
PROBLEMS
Section 13.2 Mutual Inductance
13.1 For the three coupled coils in Fig. 13.72, calculate
the total inductance.
6 H 10 H
2 H
8 H
4 H 5 H
Figure 13.72 For Prob. 13.1.
13.2 Determine the inductance of the three series-
connected inductors of Fig. 13.73.
10 H 8 H
4 H
12 H
6 H 6 H
Figure 13.73 For Prob. 13.2.
13.3 Two coils connected in series-aiding fashion have a
total inductance of 250 mH. When connected in a
series-opposing configuration, the coils have a total
inductance of 150 mH. If the inductance of one coil
(L1) is three times the other, find L1, L2, and M.
What is the coupling coefficient?
13.4 (a) For the coupled coils in Fig. 13.74(a), show that
Leq = L1 + L2 + 2M
(b) For the coupled coils in Fig. 13.74(b), show that
Leq =
L1L2 − M2
L1L2 − 2M2
M
L2
L1
L1
Leq (b)
M L2
Leq (a)
Figure 13.74 For Prob. 13.4.
13.5 Determine V1 and V2 in terms of I1 and I2 in the
circuit in Fig. 13.75.
+
−
jvM
I1
V1
+
−
V2
I2
jvL1 jvL2
R1 R2
Figure 13.75 For Prob. 13.5.
13.6 Find Vo in the circuit of Fig. 13.76.
V
–j6 Ω j8 Ω
+
−
j12 Ω
10 Ω
j4 Ω
20 30°
+
−
Vo
Figure 13.76 For Prob. 13.6.
572 PART 2 AC Circuits
13.7 Obtain Vo in the circuit of Fig. 13.77.
1 Ω
+
− j6 Ω j4 Ω −j3 Ω
4 Ω
j2 Ω
Vo
+
−
V
10 0°
Figure 13.77 For Prob. 13.7.
13.8 Find Vx in the network shown in Fig. 13.78.
2 Ω
+
− j4 Ω j4 Ω −j1 Ω
2 Ω
j1 Ω
V
8 30°
+ −
Vx
A
2 0°
Figure 13.78 For Prob. 13.8.
13.9 Find Io in the circuit of Fig. 13.79.
Im cos vt
L
C
L
R
io
k = 1
Figure 13.79 For Prob. 13.9.
13.10 Obtain the mesh equations for the circuit in Fig.
13.80.
V1
V2
R1
I1
I2
I3
R2
jvL1
jvL2
jvM
+
−
+ −
jvC
1
Figure 13.80 For Prob. 13.10.
13.11 Obtain the Thevenin equivalent circuit for the circuit
in Fig. 13.81 at terminals a-b.
j6 Ω
b
a
j8 Ω
2 Ω
j2 Ω
+
−
V
10 90° A
4 0°
5 Ω −j3 Ω
Figure 13.81 For Prob. 13.11.
13.12 Find the Norton equivalent for the circuit in Fig.
13.82 at terminals a-b.
b
a
j20 Ω
j5 Ω
j10 Ω
+
−
V
60 30°
20 Ω
Figure 13.82 For Prob. 13.12.
Section 13.3 Energy in a Coupled Circuit
13.13 Determine currents I1, I2, and I3 in the circuit of
Fig. 13.83. Find the energy stored in the coupled
coils at t = 2 ms. Take ω = 1000 rad/s.
j10 Ω j10 Ω
4 Ω
k = 0.5
+
−
A
3 90° V
20 0°
8 Ω
−j5 Ω
I2
I3
I1
Figure 13.83 For Prob. 13.13.
CHAPTER 13 Magnetically Coupled Circuits 573
13.14 Find I1 and I2 in the circuit of Fig. 13.84. Calculate
the power absorbed by the 4- resistor.
5 Ω
4 Ω
j6 Ω j3 Ω
2 Ω
+
− I1 I2
V
36 30°
j1 Ω
–j4 Ω
Figure 13.84 For Prob. 13.14.
13.15
∗
Find current Io in the circuit of Fig. 13.85.
j80 Ω
j30 Ω
j10 Ω
j60 Ω
j20 Ω
+
−
Io
V
50 0° 100 Ω
j40 Ω
–j50 Ω
Figure 13.85 For Prob. 13.15.
13.16 If M = 0.2 H and vs = 12 cos 10t V in the circuit of
Fig. 13.86, find i1 and i2. Calculate the energy
stored in the coupled coils at t = 15 ms.
0.5 H
25 mF
1 H
5 Ω
M
i1 i2
+
−
vs
Figure 13.86 For Prob. 13.16.
13.17 In the circuit of Fig. 13.87,
(a) find the coupling coefficient,
(b) calculate vo,
(c) determine the energy stored in the coupled
inductors at t = 2 s.
2 Ω
+
− 4 H 2 H 1 Ω vo
1 H
12 cos 4t V
+
−
F
1
4
Figure 13.87 For Prob. 13.17.
13.18 For the network in Fig. 13.88, find Zab and Io.
4 Ω
0.5 F
1 Ω
2 Ω
io
+
−
12 sin 2t V 1 H 1 H 2 H
k = 0.5
3 Ω
a
b
Figure 13.88 For Prob. 13.18.
13.19 Find Io in the circuit of Fig. 13.89. Switch the dot
on the winding on the right and calculate Io again.
−j30 Ω
j20 Ω j40 Ω 10 Ω
50 Ω
k = 0.601
Io
A
4 60°
Figure 13.89 For Prob. 13.19.
13.20 Rework Example 13.1 using the concept of reflected
impedance.
Section 13.4 Linear Transformers
13.21 In the circuit of Fig. 13.90, find the value of the
coupling coefficient k that will make the 10-
resistor dissipate 320 W. For this value of k, find the
energy stored in the coupled coils at t = 1.5 s.
∗An asterisk indicates a challenging problem.
574 PART 2 AC Circuits
10 Ω
+
− 30 mH 50 mH 20 Ω
k
165 cos 103
t V
Figure 13.90 For Prob. 13.21.
13.22 (a) Find the input impedance of the circuit in Fig.
13.91 using the concept of reflected impedance.
(b) Obtain the input impedance by replacing the
linear transformer by its T equivalent.
j30 Ω j20 Ω −j6 Ω
j10 Ω
8 Ω
25 Ω
j40 Ω
Zin
Figure 13.91 For Prob. 13.22.
13.23 For the circuit in Fig. 13.92, find:
(a) the T -equivalent circuit,
(b) the -equivalent circuit.
15 H 20 H
5 H
Figure 13.92 For Prob. 13.23.
13.24
∗
Two linear transformers are cascaded as shown in
Fig. 13.93. Show that
Zin =
ω2
R(L2
a + LaLb − M2
a )
+jω3
(L2
aLb + LaL2
b − LaM2
b − LbM2
a )
ω2(LaLb + L2
b − M2
b ) − jωR(La + Lb)
La La
Ma
Lb Lb
Mb
R
Zin
Figure 13.93 For Prob. 13.24.
13.25 Determine the input impedance of the air-core
transformer circuit of Fig. 13.94.
j12 Ω j40 Ω −j5 Ω
j15 Ω
20 Ω
10 Ω
Zin
Figure 13.94 For Prob. 13.25.
Section 13.5 Ideal Transformers
13.26 As done in Fig. 13.32, obtain the relationships
between terminal voltages and currents for each of
the ideal transformers in Fig. 13.95.
V1 V2
1:n
+
−
+
−
I1 I2
(a)
V1 V2
1:n
+
−
+
−
I1 I2
(b)
V1 V2
1:n
+
−
+
−
I1 I2
(d)
V1 V2
1:n
+
−
+
−
I1 I2
(c)
Figure 13.95 For Prob. 13.26.
13.27 A 4-kVA, 2300/230-V rms transformer has an
equivalent impedance of 2 10◦
 on the primary
side. If the transformer is connected to a load with
0.6 power factor leading, calculate the input
impedance.
13.28 A 1200/240-V rms transformer has impedance
60 − 30◦
 on the high-voltage side. If the
transformer is connected to a 0.8 10◦
- load on
the low-voltage side, determine the primary and
secondary currents.
13.29 Determine I1 and I2 in the circuit of Fig. 13.96.
2 Ω
10 Ω
3:1
I1 I2
+
−
V
14 0°
Figure 13.96 For Prob. 13.29.
CHAPTER 13 Magnetically Coupled Circuits 575
13.30 Obtain V1 and V2 in the ideal transformer circuit of
Fig. 13.97.
10 Ω
A 12 Ω
2 0° A
1 0°
V1 V2
+
−
+
−
1:4
Figure 13.97 For Prob. 13.30.
13.31 In the ideal transformer circuit of Fig. 13.98, find
i1(t) and i2(t).
R
1:n
i1(t) i2(t)
+
− Vm cos vt
Vo
dc
Figure 13.98 For Prob. 13.31.
13.32 (a) Find I1 and I2 in the circuit of Fig. 13.99 below.
(b) Switch the dot on one of the windings. Find I1
and I2 again.
13.33 For the circuit in Fig. 13.100, find Vo. Switch the
dot on the secondary side and find Vo again.
10 Ω
2 Ω
3:1
20 mF
10 cos 5t V +
−
+
−
Vo
Figure 13.100 For Prob. 13.33.
13.34 Calculate the input impedance for the network in
Fig. 13.101 below.
13.35 Use the concept of reflected impedance to find the
input impedance and current I1 in Fig. 13.102 below.
12 Ω
10 Ω
j16 Ω
+
−
V
16 60° 10 30°
I1 I2
1:2
+
−
–j8 Ω
Figure13.99 For Prob. 13.32.
Zin
1:5 4:1
a
b
8 Ω 24 Ω 6 Ω
j12 Ω
−j10 Ω
Figure13.101 For Prob. 13.34.
1:2 1:3
5 Ω 8 Ω 36 Ω
j18 Ω
+
−
I1
–j2 Ω
V
24 0°
Figure13.102 For Prob. 13.35.
576 PART 2 AC Circuits
13.36 For the circuit in Fig. 13.103, determine the turns
ratio n that will cause maximum average power
transfer to the load. Calculate that maximum
average power.
40 Ω
1:n
+
− 10 Ω
V rms
120 0°
Figure 13.103 For Prob. 13.36.
13.37 Refer to the network in Fig. 13.104.
(a) Find n for maximum power supplied to the
200- load.
(b) Determine the power in the 200- load if
n = 10.
3 Ω
1:n
200 Ω
5 Ω
A rms
4 0°
Figure 13.104 For Prob. 13.37.
13.38 A transformer is used to match an amplifier with an
8- load as shown in Fig. 13.105. The Thevenin
equivalent of the amplifier is: VTh = 10 V,
ZTh = 128 .
(a) Find the required turns ratio for maximum
energy power transfer.
(b) Determine the primary and secondary currents.
(c) Calculate the primary and secondary voltages.
1:n
8 Ω
Amplifier
circuit
Figure 13.105 For Prob. 13.38.
13.39 In Fig. 13.106 below, determine the average power
delivered to Zs.
13.40 Find the power absorbed by the 10- resistor in the
ideal transformer circuit of Fig. 13.107.
2 Ω
10 Ω
1:2
V
46 0°
5 Ω
+
−
Figure 13.107 For Prob. 13.40.
13.41 For the ideal transformer circuit of Fig. 13.108
below, find:
(a) I1 and I2,
(b) V1, V2, and Vo,
(c) the complex power supplied by the source.
1:10
+
−
V rms
120 0° Zs = 500 – j200 Ω
Zp = 3 + j4 Ω
Figure13.106 For Prob. 13.39.
12 Ω
j3 Ω
V1 V2 Vo
− −
1:2
+
−
V rms
60 90°
I1 I2
2 Ω
+ + +
−
−j6 Ω
Figure13.108 For Prob. 13.41.
CHAPTER 13 Magnetically Coupled Circuits 577
13.42 Determine the average power absorbed by each
resistor in the circuit of Fig. 13.109.
20 Ω
100 Ω
1:5
80 cos 4t V +
−
20 Ω
Figure 13.109 For Prob. 13.42.
13.43 Find the average power delivered to each resistor in
the circuit of Fig. 13.110.
8 Ω
4 Ω
2:1
V
20 0° +
−
2 Ω
Figure 13.110 For Prob. 13.43.
13.44 Refer to the circuit in Fig. 13.111 below.
(a) Find currents I1, I2, and I3.
(b) Find the power dissipated in the 40- resistor.
13.45
∗
For the circuit in Fig. 13.112 below, find I1, I2, and
Vo.
13.46 For the network in Fig. 13.113 below, find
(a) the complex power supplied by the source,
(b) the average power delivered to the 18- resistor.
1:4 1:2
4 Ω 5 Ω
+
−
I1 I2 I3
V
120 0° 40 Ω
10 Ω
Figure13.111 For Prob. 13.44.
1:5 3:4
2 Ω 14 Ω
+
−
I1 I2
V
24 0° 160 Ω
60 Ω
Vo
+
−
Figure13.112 For Prob. 13.45.
2:5 1:3
6 Ω 8 Ω
+
−
V
40 0°
18 Ω
–j20 Ω
j4 Ω
j45 Ω
Figure13.113 For Prob. 13.46.
578 PART 2 AC Circuits
13.47 Find the mesh currents in the circuit of Fig. 13.114
below.
Section 13.6 Ideal Autotransformers
13.48 An ideal autotransformer with a 1:4 step-up turns
ratio has its secondary connected to a 120- load
and the primary to a 420-V source. Determine the
primary current.
13.49 In the ideal autotransformer of Fig. 13.115, calculate
I1, I2, and Io. Find the average power delivered to
the load.
+
−
I1
I2
10 + j40 Ω
2 – j6 Ω
Io
20 30° V rms
80 turns
200 turns
Figure 13.115 For Prob. 13.49.
13.50
∗
In the circuit of Fig. 13.116, ZL is adjusted until
maximum average power is delivered to ZL. Find
ZL and the maximum average power transferred to
it. Take N1 = 600 turns and N2 = 200 turns.
+
−
V rms
N1
N2
75 Ω j125 Ω
ZL
120 0°
Figure 13.116 For Prob. 13.50.
13.51 In the ideal transformer circuit shown in Fig. 13.117,
determine the average power delivered to the load.
+
−
V rms
20 – j40 Ω
120 0°
30 + j12 Ω
1000 turns
200 turns
Figure 13.117 For Prob. 13.51.
13.52 In the autotransformer circuit in Fig. 13.118, show
that
Zin = 1 +
N1
N2
2
ZL
ZL
Zin
Figure 13.118 For Prob. 13.52.
Section 13.7 Three-Phase Transformers
13.53 In order to meet an emergency, three single-phase
transformers with 12,470/7200 V rms are connected
in -Y to form a three-phase transformer which is
fed by a 12,470-V transmission line. If the
transformer supplies 60 MVA to a load, find:
(a) the turns ratio for each transformer,
(b) the currents in the primary and secondary
windings of the transformer,
(c) the incoming and outgoing transmission line
currents.
1:2 1:3
1 Ω 9 Ω
7 Ω
+
−
V
12 0°
–j6 Ω
j18 Ω
I1 I1
I2
Figure13.114 For Prob. 13.47.
CHAPTER 13 Magnetically Coupled Circuits 579
13.54 Figure 13.119 below shows a three-phase
transformer that supplies a Y-connected load.
(a) Identify the transformer connection.
(b) Calculate currents I2 and Ic.
(c) Find the average power absorbed by the load.
13.55 Consider the three-phase transformer shown in Fig.
13.120. The primary is fed by a three-phase source
with line voltage of 2.4 kV rms, while the secondary
supplies a three-phase 120-kW balanced load at pf
of 0.8. Determine:
(a) the type of transformer connections,
(b) the values of ILS and IP S,
(c) the values of ILP and IP P ,
(d) the kVA rating of each phase of the transformer.
13.56 A balanced three-phase transformer bank with the
-Y connection depicted in Fig. 13.121 below is
used to step down line voltages from 4500 V rms to
900 V rms. If the transformer feeds a 120-kVA load,
find:
(a) the turns ratio for the transformer,
(b) the line currents at the primary and secondary
sides.
Load
120 kW pf = 0.8
4:1
2.4 kV
ILP
ILS
IPS
IPP
Figure 13.120 For Prob. 13.55.
3:1
V
450 0°
I1
I2
I3
Ic
Ib
Ia
450 120° V
450 –120° V
8 Ω
−j6 Ω
8 Ω
−j6 Ω
8 Ω
−j6 Ω
Figure13.119 For Prob. 13.54.
1:n
4500 V 900 V 42 kVA
Three-phase
load
Figure13.121 For Prob. 13.56.
580 PART 2 AC Circuits
13.57 A Y- three-phase transformer is connected to a
60-kVA load with 0.85 power factor (leading)
through a feeder whose impedance is 0.05 + j0.1 
per phase, as shown in Fig. 13.122 below. Find the
magnitude of:
(a) the line current at the load,
(b) the line voltage at the secondary side of the
transformer,
(c) the line current at the primary side of the
transformer.
13.58 The three-phase system of a town distributes power
with a line voltage of 13.2 kV. A pole transformer
connected to single wire and ground steps down the
high-voltage wire to 120 V rms and serves a house
as shown in Fig. 13.123.
(a) Calculate the turns ratio of the pole transformer
to get 120 V.
(b) Determine how much current a 100-W lamp
connected to the 120-V hot line draws from the
high-voltage line.
13.2 kV 120 V
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
Figure 13.123 For Prob. 13.58.
Section 13.8 PSpice Analysis of Magnetically
Coupled Circuits
13.59 Rework Prob. 13.14 using PSpice.
13.60 Use PSpice to find I1, I2, and I3 in the circuit of Fig.
13.124.
j15 Ω
j80 Ω
j0 Ω
j100 Ω
j10 Ω
+
−
V
60 0°
j50 Ω –j20 Ω
20 90°
+
− V
I1 I3
I2
40 Ω
80 Ω
Figure 13.124 For Prob. 13.60.
13.61 Rework Prob. 13.15 using PSpice.
13.62 Use PSpice to find I1, I2, and I3 in the circuit of Fig.
13.125.
I1 I2
I3
100 Ω
8 H
2 H
1 H
4 H 3 H
2 H 60 mF
200 Ω
70 Ω 50 mF
+
−
V
120 0°
f = 100
Figure 13.125 For Prob. 13.62.
1:n
0.05 Ω j0.1 Ω
0.05 Ω j0.1 Ω
0.05 Ω j0.1 Ω
Balanced
load
60 kVA
0.85 pf
leading
2640 V
240 V
Figure13.122 For Prob. 13.57.
CHAPTER 13 Magnetically Coupled Circuits 581
13.63 Use PSpice to find V1, V2, and Io in the circuit of
Fig. 13.126.
2 Ω
1:2
+
−
20 Ω
16 Ω
V1 V2
+
−
+
−
V
40 60° +
− V
30 0°
–j12 Ω
–j4 Ω
j8 Ω
Io
Figure 13.126 For Prob. 13.63.
13.64 Find Ix and Vx in the circuit of Fig. 13.127 below
using PSpice.
13.65 Determine I1, I2, and I3 in the ideal transformer
circuit of Fig. 13.128 using PSpice.
j80 Ω
−j30 Ω
50 Ω
+
−
I1 I2
1:2
1:3
V
440 0°
40 Ω
j50 Ω
I3 60 Ω
Figure 13.128 For Prob. 13.65.
Section 13.9 Applications
13.66 A stereo amplifier circuit with an output impedance
of 7.2 k is to be matched to a speaker with an input
impedance of 8  by a transformer whose primary
side has 3000 turns. Calculate the number of turns
required on the secondary side.
13.67 A transformer having 2400 turns on the primary and
48 turns on the secondary is used as an
impedance-matching device. What is the reflected
value of a 3- load connected to the secondary?
13.68 A radio receiver has an input resistance of 300 .
When it is connected directly to an antenna system
with a characteristic impedance of 75 , an
impedance mismatch occurs. By inserting an
impedance-matching transformer ahead of the
receiver, maximum power can be realized. Calculate
the required turns ratio.
13.69 A step-down power transformer with a turns ratio of
n = 0.1 supplies 12.6 V rms to a resistive load. If
the primary current is 2.5 A rms, how much power is
delivered to the load?
13.70 A 240/120-V rms power transformer is rated at
10 kVA. Determine the turns ratio, the primary
current, and the secondary current.
13.71 A 4-kVA, 2400/240-V rms transformer has 250
turns on the primary side. Calculate:
(a) the turns ratio,
(b) the number of turns on the secondary side,
(c) the primary and secondary currents.
13.72 A 25,000/240-V rms distribution transformer has a
primary current rating of 75 A.
(a) Find the transformer kVA rating.
(b) Calculate the secondary current.
13.73 A 4800-V rms transmission line feeds a distribution
transformer with 1200 turns on the primary and 28
turns on the secondary. When a 10- load is
connected across the secondary, find:
(a) the secondary voltage,
(b) the primary and secondary currents,
(c) the power supplied to the load.
1:2 2:1
1 Ω 6 Ω
2Vx
8 Ω
j2 Ω
+
−
Ix
–j10 Ω
V
6 0°
+
−
Vx
4 Ω
+
−
Vo
+ −
Figure13.127 For Prob. 13.64.
582 PART 2 AC Circuits
COMPREHENSIVE PROBLEMS
13.74 A four-winding transformer (Fig. 13.129) is often
used in equipment (e.g., PCs, VCRs) that may be
operated from either 110 V or 220 V. This makes the
equipment suitable for both domestic and foreign
use. Show which connections are necessary to
provide:
(a) an output of 12 V with an input of 110 V,
(b) an output of 50 V with an input of 220 V.
a
b
c
d
e
f
g
h
32 V
18 V
110 V
110 V
Figure 13.129 For Prob. 13.74.
13.75
∗
A 440/110-V ideal transformer can be connected to
become a 550/440-V ideal autotransformer. There
are four possible connections, two of which are
wrong. Find the output voltage of:
(a) a wrong connection,
(b) the right connection.
13.76 Ten bulbs in parallel are supplied by a 7200/120-V
transformer as shown in Fig. 13.130, where the
bulbs are modeled by the 144- resistors. Find:
(a) the turns ratio n,
(b) the current through the primary winding.
1:n
144 Ω
7200 V 120 V 144 Ω
Figure 13.130 For Prob. 13.76.
583
C H A P T E R
FREQUENCY RESPONSE
1 4
One machine can do the work of fifty ordinary men. No machine can do
the work of one extraordinary man.
— Elbert G. Hubbard
Enhancing Your Career
Career in Control Systems Control systems are another
area of electrical engineering where circuit analysis is used.
A control system is designed to regulate the behavior of
one or more variables in some desired manner. Control
systems play major roles in our everyday life. Household
appliances such as heating and air-conditioning systems,
switch-controlled thermostats, washers and dryers, cruise
controllers in automobiles, elevators, traffic lights, manu-
facturing plants, navigation systems—all utilize control sys-
tems. In the aerospace field, precision guidance of space
probes, the wide range of operational modes of the space
shuttle, and the ability to maneuver space vehicles remotely
from earth all require knowledge of control systems. In
the manufacturing sector, repetitive production line opera-
tions are increasingly performed by robots, which are pro-
grammable control systems designed to operate for many
hours without fatigue.
Control engineering integrates circuit theory and
communication theory. It is not limited to any specific engi-
neering discipline but may involve environmental, chemical,
aeronautical, mechanical, civil, and electrical engineering.
For example, a typical task for a control system engineer
might be to design a speed regulator for a disk drive head.
A thorough understanding of control systems tech-
niques is essential to the electrical engineer and is of great
value for designing control systems to perform the desired
task. A welding robot.
(Courtesy of Shela Terry/Science Photo Library.)
584 PART 2 AC Circuits
14.1 INTRODUCTION
Inoursinusoidalcircuitanalysis, wehavelearnedhowtofindvoltagesand
currents in a circuit with a constant frequency source. If we let the ampli-
tude of the sinusoidal source remain constant and vary the frequency, we
obtain the circuit’s frequency response. The frequency response may be
regarded as a complete description of the sinusoidal steady-state behavior
of a circuit as a function of frequency.
The frequency response of a circuit is the variation in its
behavior with change in signal frequency.
The frequency response of a circuit may also be
considered as the variation of the gain and phase
with frequency.
The sinusoidal steady-state frequency responses of circuits are of
significance in many applications, especially in communications and con-
trol systems. A specific application is in electric filters that block out or
eliminate signals with unwanted frequencies and pass signals of the de-
sired frequencies. Filters are used in radio, TV, and telephone systems to
separate one broadcast frequency from another.
We begin this chapter by considering the frequency response of sim-
ple circuits using their transfer functions. We then consider Bode plots,
which are the industry-standard way of presenting frequency response.
We also consider series and parallel resonant circuits and encounter im-
portant concepts such as resonance, quality factor, cutoff frequency, and
bandwidth. We discuss different kinds of filters and network scaling. In
the last section, we consider one practical application of resonant circuits
and two applications of filters.
14.2 TRANSFER FUNCTION
The transfer function H(ω) (also called the network function) is a useful
analytical tool for finding the frequency response of a circuit. In fact, the
frequency response of a circuit is the plot of the circuit’s transfer function
H(ω) versus ω, with ω varying from ω = 0 to ω = ∞.
A transfer function is the frequency-dependent ratio of a forced
function to a forcing function (or of an output to an input). The idea of a
transfer function was implicit when we used the concepts of impedance
and admittance to relate voltage and current. In general, a linear network
can be represented by the block diagram shown in Fig. 14.1.
Input Output
Linear network
H(v)
Y(v)
X(v)
Figure 14.1 A block diagram representation
of a linear network. The transfer function H(ω) of a circuit is the frequency-dependent ratio of a
phasor output Y(ω) (an element voltage or current) to a phasor input
X(ω) (source voltage or current).
In this context, X(ω) and Y(ω) denote the input
andoutputphasorsofanetwork;theyshouldnot
be confused with the same symbolism used for
reactance and admittance. The multiple usage
of symbols is conventionally permissible due to
lack of enough letters in the English language to
express all circuit variables distinctly.
Thus,
H(ω) =
Y(ω)
X(ω)
(14.1)
CHAPTER 14 Frequency Response 585
assuming zero initial conditions. Since the input and output can be either
voltageorcurrentatanyplaceinthecircuit, therearefourpossibletransfer
functions:
H(ω) = Voltage gain =
Vo(ω)
Vi(ω)
(14.2a)
H(ω) = Current gain =
Io(ω)
Ii(ω)
(14.2b)
H(ω) = Transfer Impedance =
Vo(ω)
Ii(ω)
(14.2c)
H(ω) = Transfer Admittance =
Io(ω)
Vi(ω)
(14.2d)
where subscripts i and o denote input and output values. Being a complex
quantity, H(ω) has a magnitude H(ω) and a phase φ; that is, H(ω) =
H(ω) φ.
To obtain the transfer function using Eq. (14.2), we first obtain
the frequency-domain equivalent of the circuit by replacing resistors,
inductors, and capacitors with their impedances R, jωL, and 1/jωC.
We then use any circuit technique(s) to obtain the appropriate quantity in
Eq. (14.2). We can obtain the frequency response of the circuit by plotting
the magnitude and phase of the transfer function as the frequency varies.
A computer is a real time-saver for plotting the transfer function.
Some authors use H(jω) for transfer instead of
H(ω), since ω and j are an inseparable pair.
The transfer function H(ω) can be expressed in terms of its numer-
ator polynomial N(ω) and denominator polynomial D(ω) as
H(ω) =
N(ω)
D(ω)
(14.3)
where N(ω) and D(ω) are not necessarily the same expressions for the
input and output functions, respectively. The representation of H(ω) in
Eq. (14.3) assumes that common numerator and denominator factors in
H(ω) have canceled, reducing the ratio to lowest terms. The roots of
N(ω) = 0 are called the zeros of H(ω) and are usually represented as
jω = z1, z2, . . . . Similarly, the roots of D(ω) = 0 are the poles of H(ω)
and are represented as jω = p1, p2, . . . .
A zero, as a root of the numerator polynomial, is a value that results in a zero
value of the function. A pole, as a root of the denominator polynomial,
is a value for which the function is infinite.
A zero may also be regarded as the value of s =
jω that makes H(s) zero, and a pole as the value
of s = jω that makes H(s) infinite.
To avoid complex algebra, it is expedient to replace jω temporarily
with s when working with H(ω) and replace s with jω at the end.
E X A M P L E 1 4 . 1
For the RC circuit in Fig. 14.2(a), obtain the transfer function Vo/Vs and
its frequency response. Let vs = Vm cos ωt.
586 PART 2 AC Circuits
Solution:
The frequency-domain equivalent of the circuit is in Fig. 14.2(b). By
voltage division, the transfer function is given by
H(ω) =
Vo
Vs
=
1/jωC
R + 1/jωC
=
1
1 + jωRC
Comparing this with Eq. (9.18e), we obtain the magnitude and phase of
H(ω) as
H =
1

1 + (ω/ω0)2
, φ = − tan−1 ω
ω0
where ω0 = 1/RC. To plot H and φ for 0  ω  ∞, we obtain their
values at some critical points and then sketch.
vs(t) vo(t)
R
(a) (b)
C
+
− Vs Vo
R
jvC
1
+
−
+
−
+
−
Figure 14.2 For Example 14.1: (a) time-domain RC circuit,
(b) frequency-domain RC circuit.
At ω = 0, H = 1 and φ = 0. At ω = ∞, H = 0 and φ = −90◦
.
Also, at ω = ω0, H = 1/
√
2 and φ = −45◦
. With these and a few more
points as shown in Table 14.1, we find that the frequency response is as
shown in Fig. 14.3. Additional features of the frequency response in Fig.
14.3 will be explained in Section 14.6.1 on lowpass filters.
TABLE 14.1 For Example 14.1.
ω/ω0 H φ ω/ω0 H φ
0 1 0 10 0.1 −84◦
1 0.71 −45◦
20 0.05 −87◦
2 0.45 −63◦
100 0.01 −89◦
3 0.32 −72◦
∞ 0 −90◦
0
0.707
H
1
v0 = 1
RC
v0 = 1
RC
v
0 v
−90°
−45°
(a)
(b)
f
Figure14.3 Frequency response of the
RC circuit: (a) amplitude response,
(b) phase response.
P R A C T I C E P R O B L E M 1 4 . 1
Obtain the transfer function Vo/Vs of the RL circuit in Fig. 14.4, assuming
vs = Vm cos ωt. Sketch its frequency response.
Answer: jωL/(R + jωL); see Fig. 14.5 for the response.
vs
vo
R
L
+
−
+
−
Figure14.4 RL circuit
for Practice Prob. 14.1.
CHAPTER 14 Frequency Response 587
1
H
0.707
0 v0 = R
L
v0 = R
L
v
(a) (b)
90°
45°
f
0 v
Figure14.5 Frequency response of the RL circuit in Fig. 14.4.
E X A M P L E 1 4 . 2
For the circuit in Fig. 14.6, calculate the gain Io(ω)/Ii(ω) and its poles
and zeros.
ii (t)
io(t)
0.5 F
2 H
4 Ω
Figure14.6 For Example 14.2.
Solution:
By current division,
Io(ω) =
4 + j2ω
4 + j2ω + 1/j0.5ω
Ii(ω)
or
Io(ω)
Ii(ω)
=
j0.5ω(4 + j2ω)
1 + j2ω + (jω)2
=
s(s + 2)
s2 + 2s + 1
, s = jω
The zeros are at
s(s + 2) = 0 ⇒ z1 = 0, z2 = −2
The poles are at
s2
+ 2s + 1 = (s + 1)2
= 0
Thus, there is a repeated pole (or double pole) at p = −1.
P R A C T I C E P R O B L E M 1 4 . 2
Find the transfer function Vo(ω)/Ii(ω) for the circuit of Fig. 14.7. Obtain
its poles and zeros.
vo(t)
ii(t)
0.1 F 2 H
3 Ω
5 Ω
+
−
Figure14.7 For Practice Prob. 14.2.
Answer:
5(s + 2)(s + 1.5)
s2 + 4s + 5
, s = jω; poles: −2, −1.5; zeros: −2±j.
588 PART 2 AC Circuits
†14.3 THE DECIBEL SCALE
It is not always easy to get a quick plot of the magnitude and phase
of the transfer function as we did above. A more systematic way of
obtaining the frequency response is to use Bode plots. Before we begin
to construct Bode plots, we should take care of two important issues: the
use of logarithms and decibels in expressing gain.
Since Bode plots are based on logarithms, it is important that we
keep the following properties of logarithms in mind:
1. log P1P2 = log P1 + log P2
2. log P1/P2 = log P1 − log P2
3. log Pn
= n log P
4. log 1 = 0
In communications systems, gain is measured in bels. Historically,
the bel is used to measure the ratio of two levels of power or power gain
G; that is,
G = Number of bels = log10
P2
P1
(14.4)
The decibel (dB) provides us with a unit of less magnitude. It is 1/10th
of a bel and is given by
GdB = 10 log10
P2
P1
(14.5)
When P1 = P2, there is no change in power and the gain is 0 dB. If
P2 = 2P1, the gain is
GdB = 10 log10 2 = 3 dB (14.6)
and when P2 = 0.5P1, the gain is
GdB = 10 log10 0.5 = −3 dB (14.7)
Equations (14.6) and (14.7) show another reason why logarithms are
greatly used: The logarithm of the reciprocal of a quantity is simply
negative the logarithm of that quantity.
Historical note: The bel is named after Alexander
Graham Bell, the inventor of the telephone.
V2
−
+
V1 R2
Network
I1 I2
P1 P2
R1
+
−
Figure 14.8 Voltage-current relationships
for a four-terminal network.
Alternatively, the gain G can be expressed in terms of voltage or
current ratio. To do so, consider the network shown in Fig. 14.8. If P1 is
the input power, P2 is the output (load) power, R1 is the input resistance,
and R2 is the load resistance, then P1 = 0.5V 2
1 /R1 and P2 = 0.5V 2
2 /R2,
and Eq. (14.5) becomes
GdB = 10 log10
P2
P1
= 10 log10
V 2
2 /R2
V 2
1 /R1
= 10 log10

V2
V1
2
+ 10 log10
R1
R2
(14.8)
GdB = 20 log10
V2
V1
− 10 log10
R2
R1
(14.9)
CHAPTER 14 Frequency Response 589
For the case when R2 = R1, a condition that is often assumed when
comparing voltage levels, Eq. (14.9) becomes
GdB = 20 log10
V2
V1
(14.10)
Instead, if P1 = I2
1 R1 and P2 = I2
2 R2, for R1 = R2, we obtain
GdB = 20 log10
I2
I1
(14.11)
Two things are important to note from Eqs. (14.5), (14.10), and (14.11):
1. That 10 log is used for power, while 20 log is used for voltage
or current, because of the square relationship between them
(P = V 2
/R = I2
R).
2. That the dB value is a logarithmic measurement of the ratio of
one variable to another of the same type. Therefore, it applies
in expressing the transfer function H in Eqs. (14.2a) and
(14.2b), which are dimensionless quantities, but not in
expressing H in Eqs. (14.2c) and (14.2d).
With this in mind, we now apply the concepts of logarithms and decibels
to construct Bode plots.
14.4 BODE PLOTS
Obtaining the frequency response from the transfer function as we did in
Section 14.2 is an uphill task. The frequency range required in frequency
response is often so wide that it is inconvenient to use a linear scale for
the frequency axis. Also, there is a more systematic way of locating
the important features of the magnitude and phase plots of the transfer
function. For these reasons, it has become standard practice to use a
logarithmic scale for the frequency axis and a linear scale in each of the
separate plots of magnitude and phase. Such semilogarithmic plots of
the transfer function—known as Bode plots—have become the industry
standard.
Historical note: Named after Hendrik W. Bode
(1905–1982),anengineerwiththeBellTelephone
Laboratories,forhispioneeringworkinthe1930s
and 1940s.
Bode plots are semilog plots of the magnitude (in decibels) and phase (in degrees)
of a transfer function versus frequency.
Bode plots contain the same information as the nonlogarithmic plots dis-
cussed in the previous section, but they are much easier to construct, as
we shall see shortly.
The transfer function can be written as
H = H φ = Hejφ
(14.12)
Taking the natural logarithm of both sides,
ln H = ln H + ln ejφ
= ln H + jφ (14.13)
590 PART 2 AC Circuits
Thus, the real part of ln H is a function of the magnitude while the imag-
inary part is the phase. In a Bode magnitude plot, the gain
HdB = 20 log10 H (14.14)
is plotted in decibels (dB) versus frequency. Table 14.2 provides a few
values of H with the corresponding values in decibels. In a Bode phase
plot, φ is plotted in degrees versus frequency. Both magnitude and phase
plots are made on semilog graph paper.
TABLE 14.2 Specific gains
and their decibel values.
Magnitude H 20 log10 H (dB)
0.001 −60
0.01 −40
0.1 −20
0.5 −6
1/
√
2 −3
1 0
√
2 3
2 6
10 20
20 26
100 40
1000 60
A transfer function in the form of Eq. (14.3) may be written in terms
of factors that have real and imaginary parts. One such representation
might be
H(ω) =
K(jω)±1
(1 + jω/z1)[1 + j2ζ1ω/ωk + (jω/ωk)2
] · · ·
(1 + jω/p1)[1 + j2ζ2ω/ωn + (jω/ωn)2] · · ·
(14.15)
which is obtained by dividing out the poles and zeros in H(ω). The
representation of H(ω) as in Eq. (14.15) is called the standard form. In
this particular case, H(ω) has seven different factors that can appear in
various combinations in a transfer function. These are:
The origin is where ω = 1 or log ω = 0 and the
gain is zero.
1. A gain K
2. A pole (jω)−1
or zero (jω) at the origin
3. A simple pole 1/(1 + jω/p1) or zero (1 + jω/z1)
4. A quadratic pole 1/[1 + j2ζ2ω/ωn + (jω/ωn)2
] or zero
[1 + j2ζ1ω/ωk + (jω/ωk)2
]
In constructing a Bode plot, we plot each factor separately and then com-
bine them graphically. The factors can be considered one at a time and
then combined additively because of the logarithms involved. It is this
mathematical convenience of the logarithm that makes Bode plots a pow-
erful engineering tool.
We will now make straight-line plots of the factors listed above. We
shall find that these straight-line plots known as Bode plots approximate
the actual plots to a surprising degree of accuracy.
Constant term: For the gain K, the magnitude is 20 log10 K and the
phase is 0◦
; both are constant with frequency. Thus the magnitude and
phase plots of the gain are shown in Fig. 14.9. If K is negative, the
magnitude remains 20 log10 |K| but the phase is ±180◦
.
(a)
0.1 1 10 100 v
20 log10K
H
(b)
0.1 1 10 100 v
0
f
Figure14.9 Bode plots for gain K: (a) magnitude plot, (b) phase plot.
CHAPTER 14 Frequency Response 591
Pole/zero at the origin: For the zero (jω) at the origin, the magnitude
is 20 log10 ω and the phase is 90◦
. These are plotted in Fig. 14.10, where
we notice that the slope of the magnitude plot is 20 dB/decade, while the
phase is constant with frequency.
A decade is an interval between two frequen-
cies with a ratio of 10; e.g., between ω0
and 10ω0, or between 10 and 100 Hz. Thus,
20 dB/decade means that the magnitude changes
20 dB whenever the frequency changes tenfold
or one decade.
The special case of dc (ω = 0) does not appear
on Bode plots because log 0 = −∞, implying
that zero frequency is infinitely far to the left of
the origin of Bode plots.
The Bode plots for the pole (jω)−1
are similar except that the slope
of the magnitude plot is −20 dB/decade while the phase is −90◦
. In
general, for (jω)N
, where N is an integer, the magnitude plot will have
a slope of 20N dB/decade, while the phase is 90N degrees.
Simple pole/zero: For the simple zero (1 + jω/z1), the magnitude is
20 log10 |1 + jω/z1| and the phase is tan−1
ω/z1. We notice that
HdB = 20 log10



1 +
jω
z1



 ⇒ 20 log10 1 = 0
as ω → 0
(14.16)
HdB = 20 log10



1 +
jω
z1



 ⇒ 20 log10
ω
z1
as ω → ∞
(14.17)
showing that we can approximate the magnitude as zero (a straight line
with zero slope) for small values of ω and by a straight line with slope
20 dB/decade for large values of ω. The frequency ω = z1 where the two
asymptotic lines meet is called the corner frequency or break frequency.
Thus the approximate magnitude plot is shown in Fig. 14.11(a), where
the actual plot is also shown. Notice that the approximate plot is close
to the actual plot except at the break frequency, where ω = z1 and the
deviation is 20 log10 |(1 + j1)| = 20 log10
√
2 = 3 dB.
(a)
(b)
0.1 1.0
Slope = 20 dB/decade
10 v
0
20
–20
H
0.1 1.0 10 v
90°
0°
f
Figure14.10 Bode plot for a zero (jω) at
the origin: (a) magnitude plot, (b) phase plot.
The phase tan−1
(ω/z1) can be expressed as
φ = tan−1

ω
z1

=



0, ω = 0
45◦
, ω = z1
90◦
, ω → ∞
(14.18)
As a straight-line approximation, we let φ  0 for ω ≤ z1/10, φ  45◦
for ω = z1, and φ  90◦
for ω ≥ 10z1. As shown in Fig. 14.11(b) along
with the actual plot, the straight-line plot has a slope of 45◦
per decade.
The Bode plots for the pole 1/(1 + jω/p1) are similar to those in
Fig. 14.11 except that the corner frequency is at ω = p1, the magnitude
(a)
Approximate
Exact
3 dB
0.1z1 10z1
z1 v
20
H
(b)
Approximate
Exact
45°/decade
0.1z1 10z1
z1 v
45°
0°
90°
f
Figure14.11 Bode plots of zero (1 + jω/z1): (a) magnitude plot, (b) phase plot.
592 PART 2 AC Circuits
has a slope of −20 dB/decade, and the phase has a slope of −45◦
per
decade.
Quadratic pole/zero: The magnitude of the quadratic pole 1/[1 +
j2ζ2ω/ωn + (jω/ωn)2
] is −20 log10 |1 + j2ζ2ω/ωn + (jω/ωn)2
| and
the phase is − tan−1
(2ζ2ω/ωn)/(1 − ω/ω2
n). But
HdB = −20 log10





1 +
j2ζ2ω
ωn
+

jω
ωn
2





⇒ 0
as ω → 0
(14.19)
and
HdB = −20 log10





1 +
j2ζ2ω
ωn
+

jω
ωn
2





⇒ −40 log10
ω
ωn
as ω → ∞
(14.20)
Thus, the amplitude plot consists of two straight asymptotic lines: one
with zero slope for ω  ωn and the other with slope −40 dB/decade
for ω  ωn, with ωn as the corner frequency. Figure 14.12(a) shows
the approximate and actual amplitude plots. Note that the actual plot
depends on the damping factor ζ2 as well as the corner frequency ωn. The
significant peaking in the neighborhood of the corner frequency should
be added to the straight-line approximation if a high level of accuracy
is desired. However, we will use the straight-line approximation for the
sake of simplicity.
(a)
0.01vn 100vn
10vn
0.1vn
z2 = 0.05
z2 = 0.2
z2 = 0.4
z2 = 0.707
z2 = 1.5
vn v
20
0
–20
–40
H
–40 dB/dec
(b)
0.01vn 100vn
10vn
0.1vn
z2 = 0.4
z2 = 1.5
z2 = 0.2
z2 = 0.05
vn v
0°
–90°
–180°
f
–90°/dec
z2 = 0.707
Figure14.12 Bode plots of quadratic pole [1 + j2ζω/ωn − ω2/ω2
n]−1: (a) magnitude plot, (b) phase plot.
The phase can be expressed as
φ = − tan−1 2ζ2ω/ωn
1 − ω2/ω2
n
=



0, ω = 0
−90◦
, ω = ωn
−180◦
, ω → ∞
(14.21)
The phase plot is a straight line with a slope of 90◦
per decade starting
at ωn/10 and ending at 10ωn, as shown in Fig. 14.12(b). We see again
that the difference between the actual plot and the straight-line plot is
due to the damping factor. Notice that the straight-line approximations
for both magnitude and phase plots for the quadratic pole are the same
CHAPTER 14 Frequency Response 593
as those for a double pole, i.e. (1 + jω/ωn)−2
. We should expect this
because the double pole (1 + jω/ωn)−2
equals the quadratic pole 1/[1 +
j2ζ2ω/ωn + (jω/ωn)2
] when ζ2 = 1. Thus, the quadratic pole can be
treated as a double pole as far as straight-line approximation is concerned.
For the quadratic zero [1+j2ζ1ω/ωk +(jω/ωk)2
], the plots in Fig.
14.12 are inverted because the magnitude plot has a slope of 40 dB/decade
while the phase plot has a slope of 90◦
per decade.
Table 14.3 presents a summary of Bode plots for the seven factors.
To sketch the Bode plots for a function H(ω) in the form of Eq. (14.15), for
example, we first record the corner frequencies on the semilog graph pa-
per, sketch the factors one at a time as discussed above, and then combine
additively the graphs of the factors. The combined graph is often drawn
from left to right, changing slopes appropriately each time a corner fre-
quency is encountered. The following examples illustrate this procedure.
There is another procedure for obtaining Bode
plots that is faster and perhaps more efficient
than the one we have just discussed. It consists
in realizing that zeros cause an increase in slope,
while poles cause a decrease. By starting with
the low-frequency asymptote of the Bode plot,
moving along the frequency axis, and increasing
ordecreasingtheslopeateachcornerfrequency,
one can sketch the Bode plot immediately from
thetransferfunctionwithouttheeffortofmaking
individualplotsandaddingthem. Thisprocedure
can be used once you become proficient in the
one discussed here.
Digital computers have rendered the pro-
cedure discussed here almost obsolete. Several
software packages such as PSpice, Matlab, Math-
cad, and Micro-Cap can be used to generate fre-
quency response plots. We will discuss PSpice
later in the chapter.
TABLE 14.3 Summary of Bode straight-line magnitude and phase plots.
Factor Magnitude Phase
K
v
20 log10 K
v
0°
(jω)N
v
20N dB⁄decade
1 v
90N°
1
(jω)N
v
1
−20N dB⁄decade
v
−90N°

1 +
jω
z
N
v
z
20N dB⁄decade
v
90N°
0°
z
10
z 10z
1
(1 + jω/p)N
v
p
−20N dB⁄decade
v
−90N°
0°
p
10 p 10p
594 PART 2 AC Circuits
TABLE 14.3 (continued)
Factor Magnitude Phase

1 +
2jωζ
ωn
+

jω
ωn
2 N
v
vn
40N dB⁄decade
v
180N°
0°
vn vn 10vn
10
1
[1 + 2jωζ/ωk + (jω/ωk)2]N
v
vk
−40N dB⁄decade
v
−180N°
0°
vk 10vk
vk
10
E X A M P L E 1 4 . 3
Construct the Bode plots for the transfer function
H(ω) =
200jω
(jω + 2)(jω + 10)
Solution:
We first put H(ω) in the standard form by dividing out the poles and zeros.
Thus,
H(ω) =
10jω
(1 + jω/2)(1 + jω/10)
=
10|jω|
|1 + jω/2| |1 + jω/10|
90◦
− tan−1
ω/2 − tan−1
ω/10
Hence the magnitude and phase are
HdB = 20 log10 10 + 20 log10 |jω| − 20 log10



1 +
jω
2




− 20 log10



1 +
jω
10




φ = 90◦
− tan−1 ω
2
− tan−1 ω
10
We notice that there are two corner frequencies at ω = 2, 10. For both the
magnitude and phase plots, we sketch each term as shown by the dotted
lines in Fig. 14.13. We add them up graphically to obtain the overall plots
shown by the solid curves.
CHAPTER 14 Frequency Response 595
(a)
1

1 + jv/2 
1 2 10 100
20 log1010
20 log10
20 log10
20 log10 jv
200 v
0
20
H (dB)
0.1 20
1

1 + jv/10 
(b)
0.2
0.2 100 200 v
90°
90°
0°
–90°
f
0.1 20
1 2 10
–tan–1 v
2 –tan–1 v
10
Figure14.13 For Example 14.3: (a) magnitude plot, (b) phase plot.
P R A C T I C E P R O B L E M 1 4 . 3
Draw the Bode plots for the transfer function
H(ω) =
5(jω + 2)
jω(jω + 10)
Answer: See Fig. 14.14.
(a)
20 log10 1 +
20 log10
20 log101
v
20
0
–20
H (dB)
100
1
1
2 10
jv 
20 log10
1

1+ jv/10 
(b)
90°
−90°
0°
–90°
f
v
100
1
0.2 2 10 20
0.1
tan–1
v
2
–tan–1
v
10
0.1
jv
2
Figure 14.14 For Practice Prob. 14.3: (a) magnitude plot, (b) phase plot.
596 PART 2 AC Circuits
E X A M P L E 1 4 . 4
Obtain the Bode plots for
H(ω) =
jω + 10
jω(jω + 5)2
Solution:
Putting H(ω) in the standard form, we get
H(ω) =
0.4 (1 + jω/10)
jω (1 + jω/5)2
From this, we obtain the magnitude and phase as
HdB = 20 log10 0.4 + 20 log10



1 +
jω
10



 − 20 log10 |jω|
− 40 log10



1 +
jω
5




φ = 0◦
+ tan−1 ω
10
− 90◦
− 2 tan−1 ω
5
There are two corner frequencies at ω = 5, 10 rad/s. For the pole with cor-
ner frequency at ω = 5, the slope of the magnitude plot is −40 dB/decade
and that of the phase plot is −90◦
per decade due to the power of 2. The
magnitude and the phase plots for the individual terms (in dotted lines)
and the entire H(jω) (in solid lines) are in Fig. 14.15.
(a)
20
0
–20
–8
–40
H (dB)
v
100
50
1
0.5 10
–20 dB/decade
–60 dB/decade
–40 dB/decade
5
20 log10
20 log100.4
0.1
1
jv 
40 log10
1

1 + jv/5 
20 log10 1 +
jv
10
(b)
90°
0°
–90°
–180°
f
v
100
50
–90°
1
0.5 10
–90°/decade
–45°/decade
45°/decade
5
0.1
tan–1
10
v
–2 tan–1
v
5
Figure 14.15 Bode plots for Example 14.4: (a) magnitude plot, (b) phase plot.
CHAPTER 14 Frequency Response 597
P R A C T I C E P R O B L E M 1 4 . 4
Sketch the Bode plots for
H(ω) =
50jω
(jω + 4)(jω + 10)2
Answer: See Fig. 14.16.
20
–20
–40
H (dB)
100
40
1 10
4
(a)
20 log10 jv 
0.1
–20 log108
20 log10
1

1 + jv/4 
40 log10
1

1 + jv/10 
v
0
90°
–90°
–180°
f
v
100
90°
40
1
0.4 10
4
(b)
0.1
– tan–1
4
v
–2 tan–1
v
10
0°
Figure 14.16 For Practice Prob. 14.4: (a) magnitude plot, (b) phase plot.
E X A M P L E 1 4 . 5
Draw the Bode plots for
H(s) =
s + 1
s2 + 60s + 100
Solution:
We express H(s) as
H(ω) =
1/100(1 + jω)
1 + jω6/10 + (jω/10)2
For the quadratic pole, ωn = 10 rad/s, which serves as the corner fre-
598 PART 2 AC Circuits
quency. The magnitude and phase are
HdB = −20 log10 100 + 20 log10 |1 + jω|
− 20 log10



1 +
jω6
10
−
ω2
100




φ = 0◦
+ tan−1
ω − tan−1 ω6/10
1 − ω2/100
Figure 14.17 shows the Bode plots. Notice that the quadratic pole is
treated as a repeated pole at ωk, that is, (1 + jω/ωk)2
, which is an ap-
proximation.
20
0
–20
–40
H (dB)
v
100
1 10
(a)
20 log10 
1 + jv 
0.1
20 log10
–20 log10 100
1

1 + j6v/10 – v2
/100 
90°
0°
–90°
–180°
f
v
100
1
6v/10
1 – v2
/100
10
(b)
0.1
–tan–1
tan–1
v
Figure 14.17 Bode plots for Example 14.5: (a) magnitude plot, (b) phase plot.
P R A C T I C E P R O B L E M 1 4 . 5
Construct the Bode plots for
H(s) =
10
s(s2 + 80s + 400)
Answer: See Fig. 14.18.
CHAPTER 14 Frequency Response 599
20
0
–20
–40
–32
H (dB)
v
100 200
1 20
10
(a)
0.1
20 log10
–20 log10 40
–20 dB/decade
–60 dB/decade
1

1 + jv0.2 – v2
/400 
20 log10
1
jv 
–90°
–90°
0°
–180°
–270°
f
v
1 2 20
10
(b)
0.1
–tan–1 v

1 – v2
/400
2
100 200
Figure 14.18 For Practice Prob. 14.5: (a) magnitude plot, (b) phase plot.
E X A M P L E 1 4 . 6
Given the Bode plot in Fig. 14.19, obtain the transfer function H(ω).
0.1 1 5 10 20 100
–20 dB/decade
v
40 dB
0
H
+20 dB/decade
–40 dB/decade
Figure14.19 For Example 14.6.
Solution:
To obtain H(ω) from the Bode plot, we keep in mind that a zero always
causes an upward turn at a corner frequency, while a pole causes a down-
ward turn. We notice from Fig. 14.19 that there is a zero jω at the origin
which should have intersected the frequency axis at ω = 1. This is indi-
cated by the straight line with slope +20 dB/decade. The fact that this
straight line is shifted by 40 dB indicates that there is a 40-dB gain; that
is,
40 = 20 log10 K ⇒ log10 K = 2
600 PART 2 AC Circuits
or
K = 102
= 100
In addition to the zero jω at the origin, we notice that there are three
factors with corner frequencies at ω = 1, 5, and 20 rad/s. Thus, we have:
1. A pole at p = 1 with slope −20 dB/decade to cause a down-
ward turn and counteract the pole at the origin. The pole at
z = 1 is determined as 1/(1 + jω/1).
2. Another pole at p = 5 with slope −20 dB/decade causing a
downward turn. The pole is 1/(1 + jω/5).
3. A third pole at p = 20 with slope −20 dB/decade causing a
further downward turn. The pole is 1/(1 + jω/20).
Putting all these together gives the corresponding transfer function
as
H(ω) =
100jω
(1 + jω/1)(1 + jω/5)(1 + jω/20)
=
jω104
(jω + 1)(jω + 5)(jω + 20)
or
H(s) =
104
s
(s + 1)(s + 5)(s + 20)
, s = jω
P R A C T I C E P R O B L E M 1 4 . 6
Obtain the transfer function H(ω) corresponding to the Bode plot in Fig.
14.20.
0.1 1
0.5 10 100
–40 dB/decade
v
0 dB
0
H
+20 dB/decade
Figure14.20 For Practice Prob. 14.6.
Answer: H(ω) =
200(s + 0.5)
(s + 1)(s + 10)2
.
14.5 SERIES RESONANCE
The most prominent feature of the frequency response of a circuit may be
the sharp peak (or resonant peak) exhibited in its amplitude characteristic.
The concept of resonance applies in several areas of science and engi-
neering. Resonance occurs in any system that has a complex conjugate
pair of poles; it is the cause of oscillations of stored energy from one form
to another. It is the phenomenon that allows frequency discrimination in
communications networks. Resonance occurs in any circuit that has at
least one inductor and one capacitor.
CHAPTER 14 Frequency Response 601
Resonance is a condition in an RLC circuit in which the capacitive and inductive
reactances are equal in magnitude, thereby resulting in a purely resistive impedance.
Resonant circuits (series or parallel) are useful for constructing filters, as
their transfer functions can be highly frequency selective. They are used
in many applications such as selecting the desired stations in radio and
TV receivers.
R jvL
jvC
1
I
+
−
Vs = Vm u
Figure14.21 The series resonant circuit.
Consider the series RLC circuit shown in Fig. 14.21 in the fre-
quency domain. The input impedance is
Z = H(ω) =
Vs
I
= R + jωL +
1
jωC
(14.22)
or
Z = R + j

ωL −
1
ωC

(14.23)
Resonance results when the imaginary part of the transfer function is
zero, or
Im(Z) = ωL −
1
ωC
= 0 (14.24)
The value of ω that satisfies this condition is called the resonant frequency
ω0. Thus, the resonance condition is
ω0L =
1
ω0C
(14.25)
or
ω0 =
1
√
LC
rad/s (14.26)
Since ω0 = 2πf0,
f0 =
1
2π
√
LC
Hz (14.27)
Note that at resonance:
Note No. 4 becomes evident from the fact that
|VL| =
Vm
R
ω0L = QVm
|VC| =
Vm
R
1
ω0C
= QVm
whereQisthequalityfactor,definedinEq.(14.38).
1. The impedance is purely resistive, thus, Z = R. In other
words, the LC series combination acts like a short circuit, and
the entire voltage is across R.
2. The voltage Vs and the current I are in phase, so that the power
factor is unity.
3. The magnitude of the transfer function H(ω) = Z(ω) is
minimum.
4. The inductor voltage and capacitor voltage can be much more
than the source voltage.
602 PART 2 AC Circuits
The frequency response of the circuit’s current magnitude
I = |I| =
Vm

R2 + (ωL − 1/ωC)2
(14.28)
is shown in Fig. 14.22; the plot only shows the symmetry illustrated in
this graph when the frequency axis is a logarithm. The average power
dissipated by the RLC circuit is
P(ω) =
1
2
I2
R (14.29)
The highest power dissipated occurs at resonance, when I = Vm/R, so
that
P(ω0) =
1
2
V 2
m
R
(14.30)
At certain frequencies ω = ω1, ω2, the dissipated power is half the
maximum value; that is,
P(ω1) = P(ω2) =
(Vm/
√
2)2
2R
=
V 2
m
4R
(14.31)
Hence, ω1 and ω2 are called the half-power frequencies.
0
Bandwidth B
v
v1 v0 v2
I
Vm/R
0.707Vm/R
Figure14.22 The current amplitude versus
frequency for the series resonant circuit of
Fig. 14.21.
The half-power frequencies are obtained by setting Z equal to
√
2R,
and writing
R2 +

ωL −
1
ωC
2
=
√
2R (14.32)
Solving for ω, we obtain
ω1 = −
R
2L
+

R
2L
2
+
1
LC
ω2 =
R
2L
+

R
2L
2
+
1
LC
(14.33)
We can relate the half-power frequencies with the resonant frequency.
From Eqs. (14.26) and (14.33),
ω0 =
√
ω1ω2 (14.34)
showing that the resonant frequency is the geometric mean of the half-
power frequencies. Notice that ω1 and ω2 are in general not symmetrical
around the resonant frequency ω0, because the frequency response is not
generally symmetrical. However, as will be explained shortly, symmetry
of the half-power frequencies around the resonant frequency is often a
reasonable approximation.
Although the height of the curve in Fig. 14.22 is determined by
R, the width of the curve depends on other factors. The width of the
response curve depends on the bandwidth B, which is defined as the
difference between the two half-power frequencies,
B = ω2 − ω1 (14.35)
CHAPTER 14 Frequency Response 603
This definition of bandwidth is just one of several that are commonly used.
Strictly speaking, B in Eq. (14.35) is a half-power bandwidth, because it
is the width of the frequency band between the half-power frequencies.
The “sharpness” of the resonance in a resonant circuit is measured
quantitatively by the quality factor Q. At resonance, the reactive energy
inthecircuitoscillatesbetweentheinductorandthecapacitor. Thequality
factor relates the maximum or peak energy stored to the energy dissipated
in the circuit per cycle of oscillation:
Q = 2π
Peak energy stored in the circuit
Energy dissipated by the circuit
in one period at resonance
(14.36)
It is also regarded as a measure of the energy storage property of a circuit
in relation to its energy dissipation property. In the series RLC circuit,
the peak energy stored is 1
2
LI2
, while the energy dissipated in one period
is 1
2
(I2
R)(1/f ). Hence,
Q = 2π
1
2
LI2
1
2
I2R(1/f )
=
2πf L
R
(14.37)
or
Q =
ω0L
R
=
1
ω0CR
(14.38)
Notice that the quality factor is dimensionless. The relationship between
the bandwidth B and the quality factor Q is obtained by substituting Eq.
(14.33) into Eq. (14.35) and utilizing Eq. (14.38).
B =
R
L
=
ω0
Q
(14.39)
or B = ω2
0CR. Thus
AlthoughthesamesymbolQisusedforthereac-
tivepower, thetwoarenotequalandshouldnot
be confused. Q here is dimensionless, whereas
reactive power Q is in VAR. This may help distin-
guish between the two.
The quality factor of a resonant circuit is the ratio of its
resonant frequency to its bandwidth.
Keep in mind that Eqs. (14.26), (14.33), (14.38), and (14.39) only apply
to a series RLC circuit.
As illustrated in Fig. 14.23, the higher the value of Q, the more
selective the circuit is but the smaller the bandwidth. The selectivity of
an RLC circuit is the ability of the circuit to respond to a certain frequency
and discriminate against all other frequencies. If the band of frequencies
to be selected or rejected is narrow, the quality factor of the resonant
circuit must be high. If the band of frequencies is wide, the quality factor
must be low.
The quality factor is a measure of the selectivity
(or “sharpness” of resonance) of the circuit.
B3
Q3 (greatest selectivity)
Q2 (medium selectivity)
Q1 (least selectivity)
B2
B1
v
Amplitude
Figure 14.23 The higher the circuit Q, the
smaller the bandwidth.
A resonant circuit is designed to operate at or near its resonant
frequency. It is said to be a high-Q circuit when its quality factor is
604 PART 2 AC Circuits
equal to or greater than 10. For high-Q circuits (Q ≥ 10), the half-
power frequencies are, for all practical purposes, symmetrical around the
resonant frequency and can be approximated as
ω1  ω0 −
B
2
, ω2  ω0 +
B
2
(14.40)
High-Q circuits are used often in communications networks.
We see that a resonant circuit is characterized by five related param-
eters: the two half-power frequencies ω1 and ω2, the resonant frequency
ω0, the bandwidth B, and the quality factor Q.
E X A M P L E 1 4 . 7
In the circuit in Fig. 14.24, R = 2 , L = 1 mH, and C = 0.4 µF.
(a) Find the resonant frequency and the half-power frequencies. (b) Cal-
culate the quality factor and bandwidth. (c) Determine the amplitude of
the current at ω0, ω1, and ω2.
20 sin vt
R L
C
+
−
Figure14.24 For Example 14.7.
Solution:
(a) The resonant frequency is
ω0 =
1
√
LC
=
1
√
10−3 × 0.4 × 10−6
= 50 krad/s
METHOD 1 The lower half-power frequency is
ω1 = −
R
2L
+

R
2L
2
+
1
LC
= −
2
2 × 10−3
+

(103)2 + (50 × 103)2
= −1 +
√
1 + 2500 krad/s = 49 krad/s
Similarly, the upper half-power frequency is
ω2 = 1 +
√
1 + 2500 krad/s = 51 krad/s
(b) The bandwidth is
B = ω2 − ω1 = 2 krad/s
or
B =
R
L
=
2
10−3
= 2 krad/s
The quality factor is
Q =
ω0
B
=
50
2
= 25
METHOD 2 Alternatively, we could find
Q =
ω0L
R
=
50 × 103
× 10−3
2
= 25
From Q, we find
B =
ω0
Q
=
50 × 103
25
= 2 krad/s
CHAPTER 14 Frequency Response 605
Since Q  10, this is a high-Q circuit and we can obtain the half-power
frequencies as
ω1 = ω0 −
B
2
= 50 − 1 = 49 krad/s
ω2 = ω0 +
B
2
= 50 + 1 = 51 krad/s
as obtained earlier.
(c) At ω = ω0,
I =
Vm
R
=
20
2
= 10 A
At ω = ω1, ω2,
I =
Vm
√
2R
=
10
√
2
= 7.071 A
P R A C T I C E P R O B L E M 1 4 . 7
A series-connected circuit has R = 4  and L = 25 mH. (a) Calculate
the value of C that will produce a quality factor of 50. (b) Find ω1, ω2,
and B. (c) Determine the average power dissipated at ω = ω0, ω1, ω2.
Take Vm = 100 V.
Answer: (a) 0.625 µF, (b) 7920 rad/s, 8080 rad/s, 160 rad/s,
(c) 1.25 kW, 0.625 kW, 0.625 kW.
14.6 PARALLEL RESONANCE
1
jvC
jvL
R
V
+
−
I = Im u
Figure14.25 The parallel resonant circuit.
0
Bandwidth B
v
v1 v0 v2
V 
ImR
0.707 ImR
Figure14.26 The current amplitude versus
frequency for the series resonant circuit of
Fig. 14.25.
The parallel RLC circuit in Fig. 14.25 is the dual of the series RLC
circuit. So we will avoid needless repetition. The admittance is
Y = H(ω) =
I
V
=
1
R
+ jωC +
1
jωL
(14.41)
or
Y =
1
R
+ j

ωC −
1
ωL

(14.42)
Resonance occurs when the imaginary part of Y is zero,
ωC −
1
ωL
= 0 (14.43)
or
ω0 =
1
√
LC
rad/s (14.44)
which is the same as Eq. (14.26) for the series resonant circuit. The
voltage |V| is sketched in Fig. 14.26 as a function of frequency. Notice
that at resonance, the parallel LC combination acts like an open circuit, so
606 PART 2 AC Circuits
that the entire currents flows through R. Also, the inductor and capacitor
current can be much more than the source current at resonance.
We can see this from the fact that
|IL| =
ImR
ω0L
= QIm
|IC| = ω0CImR = QIm
whereQisthequalityfactor,definedinEq.(14.47).
We exploit the duality between Figs. 14.21 and 14.25 by comparing
Eq. (14.42) with Eq. (14.23). By replacing R, L, and C in the expressions
for the series circuit with 1/R, 1/C, and 1/L respectively, we obtain for
the parallel circuit
ω1 = −
1
2RC
+

1
2RC
2
+
1
LC
ω2 =
1
2RC
+

1
2RC
2
+
1
LC
(14.45)
B = ω2 − ω1 =
1
RC
(14.46)
Q =
ω0
B
= ω0RC =
R
ω0L
(14.47)
Using Eqs. (14.45) and (14.47), we can express the half-power frequen-
cies in terms of the quality factor. The result is
ω1 = ω0 1 +

1
2Q
2
−
ω0
2Q
, ω2 = ω0 1 +

1
2Q
2
+
ω0
2Q
(14.48)
Again, for high-Q circuits (Q ≥ 10)
ω1  ω0 −
B
2
, ω2  ω0 +
B
2
(14.49)
Table 14.4 presents a summary of the characteristics of the series and
parallel resonant circuits. Besides the series and parallel RLC considered
here, other resonant circuits exist. Example 14.9 treats a typical example.
TABLE 14.4 Summary of the characteristics of resonant RLC circuits.
Characteristic Series circuit Parallel circuit
Resonant frequency, ω0
1
√
LC
1
√
LC
Quality factor, Q
ω0L
R
or
1
ω0RC
R
ω0L
or ω0RC
Bandwidth, B
ω0
Q
ω0
Q
Half-power frequencies, ω1, ω2 ω0 1 +

1
2Q
2
±
ω0
2Q
ω0 1 +

1
2Q
2
±
ω0
2Q
For Q ≥ 10, ω1, ω2 ω0 ±
B
2
ω0 ±
B
2
CHAPTER 14 Frequency Response 607
E X A M P L E 1 4 . 8
In the parallel RLC circuit in Fig. 14.27, let R = 8 k, L = 0.2 mH, and
C = 8 µF. (a) Calculate ω0, Q, and B. (b) Find ω1 and ω2. (c) Deter-
mine the power dissipated at ω0, ω1, and ω2.
10 sin vt C
L
R
io
+
−
Figure14.27 For Example 14.8.
Solution:
(a)
ω0 =
1
√
LC
=
1
√
0.2 × 10−3 × 8 × 10−6
=
105
4
= 25 krad/s
Q =
R
ω0L
=
8 × 103
25 × 103 × 0.2 × 10−3
= 1600
B =
ω0
Q
= 15.625 rad/s
(b) Due to the high value of Q, we can regard this as a high-Q circuit.
Hence,
ω1 = ω0 −
B
2
= 25,000 − 7.812 = 24,992 rad/s
ω2 = ω0 +
B
2
= 25,000 + 7.8125 = 25,008 rad/s
(c) At ω = ω0, Y = 1/R or Z = R = 8 k. Then
Io =
V
Z
=
10 − 90◦
8000
= 1.25 − 90◦
mA
Since the entire current flows through R at resonance, the average power
dissipated at ω = ω0 is
P =
1
2
|Io|2
R =
1
2
(1.25 × 10−3
)2
(8 × 103
) = 6.25 mW
or
P =
V 2
m
2R
=
100
2 × 8 × 103
= 6.25 mW
At ω = ω1, ω2,
P =
V 2
m
4R
= 3.125 mW
P R A C T I C E P R O B L E M 1 4 . 8
A parallel resonant circuit has R = 100 k, L = 20 mH, and C = 5 nF.
Calculate ω0, ω1, ω2, Q, and B.
Answer: 100 krad/s, 99 krad/s, 101 krad/s, 50, 2 krad/s.
E X A M P L E 1 4 . 9
Determine the resonant frequency of the circuit in Fig. 14.28.
608 PART 2 AC Circuits
Solution:
The input admittance is
Y = jω0.1 +
1
10
+
1
2 + jω2
= 0.1 + jω0.1 +
2 − jω2
4 + 4ω2
At resonance, Im(Y) = 0 and
ω00.1 −
2ω0
4 + 4ω2
0
= 0 ⇒ ω0 = 2 rad/s
Im cos vt 0.1 F 10 Ω
2 H
2 Ω
Figure14.28 For Example 14.9.
P R A C T I C E P R O B L E M 1 4 . 9
Calculate the resonant frequency of the circuit in Fig. 14.29.
Vm cos vt 10 Ω
0.2 F
1 H
+
−
Figure14.29 For Practice Prob. 14.9.
Answer: 2.179 rad/s.
14.7 PASSIVE FILTERS
The concept of filters has been an integral part of the evolution of electri-
cal engineering from the beginning. Several technological achievements
would not have been possible without electrical filters. Because of this
prominent role of filters, much effort has been expended on the theory,
design, and construction of filters and many articles and books have been
written on them. Our discussion in this chapter should be considered
introductory.
A filter is a circuit that is designed to pass signals with desired frequencies
and reject or attenuate others.
As a frequency-selective device, a filter can be used to limit the frequency
spectrum of a signal to some specified band of frequencies. Filters are the
circuits used in radio and TV receivers to allow us to select one desired
signal out of a multitude of broadcast signals in the environment.
A filter is a passive filter if it consists of only passive elements R,
L, and C. It is said to be an active filter if it consists of active elements
(such as transistors and op amps) in addition to passive elements R, L,
and C. We consider passive filters in this section and active filters in
the next section. Besides the filters we study in these sections, there are
other kinds of filters—such as digital filters, electromechanical filters,
and microwave filters—which are beyond the level of the text.
As shown in Fig. 14.30, there are four types of filters whether
passive or active:
CHAPTER 14 Frequency Response 609
1. A lowpass filter passes low frequencies and stops high
frequencies, as shown ideally in Fig. 14.30(a).
2. A highpass filter passes high frequencies and rejects low
frequencies, as shown ideally in Fig. 14.30(b).
3. A bandpass filter passes frequencies within a frequency band
and blocks or attenuates frequencies outside the band, as
shown ideally in Fig. 14.30(c).
4. A bandstop filter passes frequencies outside a frequency band
and blocks or attenuates frequencies within the band, as shown
ideally in Fig. 14.30(d).
0
(b)
v
vc
H(v)
1
0
(a)
v
vc
H(v)
1
0
(c)
v
v1 v2
H(v)
1
0
(d)
v
v1 v2
H(v)
1
Figure14.30 Ideal frequency response
of four types of filter: (a) lowpass filter,
(b) highpass filter, (c) bandpass filter,
(d) bandstop filter.
Table 14.5 presents a summary of the characteristics of these filters. Be
aware that the characteristics in Table 14.5 are only valid for first- or
second-order filters—but one should not have the impression that only
these kinds of filter exist. We now consider typical circuits for realizing
the filters shown in Table 14.5.
TABLE 14.5 Summary of the characteristics of filters.
Type of Filter H(0) H(∞) H(ωc) or H(ω0)
Lowpass 1 0 1/
√
2
Highpass 0 1 1/
√
2
Bandpass 0 0 1
Bandstop 1 1 0
ωc is the cutoff frequency for lowpass and highpass filters; ω0 is
the center frequency for bandpass and bandstop filters.
vi(t)
R
C
+
− vo(t)
+
−
Figure14.31 A lowpass filter.
14.7.1 Lowpass Filter
A typical lowpass filter is formed when the output of an RC circuit is
taken off the capacitor as shown in Fig. 14.31. The transfer function (see
also Example 14.1) is
H(ω) =
Vo
Vi
=
1/jωC
R + 1/jωC
H(ω) =
1
1 + jωRC
(14.50)
Note that H(0) = 1, H(∞) = 0. Figure 14.32 shows the plot of |H(ω)|,
along with the ideal characteristic. The half-power frequency, which is
equivalent to the corner frequency on the Bode plots but in the context of
filters is usually known as the cutoff frequency ωc, is obtained by setting
the magnitude of H(ω) equal to 1/
√
2, thus
H(ωc) =
1

1 + ω2
c R2C2
=
1
√
2
or
ωc =
1
RC
(14.51)
610 PART 2 AC Circuits
vc v
0.707
Ideal
Actual
1
0
H(v)
Figure14.32 Ideal and actual fre-
quency response of a lowpass filter.
The cutoff frequency is also called the rolloff frequency.
The cutoff frequency is the frequency at which
the transfer function H drops in magnitude to
70.71% of its maximum value. It is also regarded
as the frequency at which the power dissipated
in a circuit is half of its maximum value.
A lowpass filter is designed to pass only frequencies from dc up
to the cutoff frequency ωc.
A lowpass filter can also be formed when the output of an RL
circuit is taken off the resistor. Of course, there are many other circuits
for lowpass filters.
vi(t) R
C
+
−
vo(t)
+
−
Figure14.33 A highpass filter.
14.7.2 Highpass Filter
A highpass filter is formed when the output of an RC circuit is taken off
the resistor as shown in Fig. 14.33. The transfer function is
H(ω) =
Vo
Vi
=
R
R + 1/jωC
H(ω) =
jωRC
1 + jωRC
(14.52)
Note that H(0) = 0, H(∞) = 1. Figure 14.34 shows the plot of |H(ω)|.
Again, the corner or cutoff frequency is
ωc =
1
RC
(14.53)
vc v
0.707
Ideal
Actual
1
0
H(v)
Figure14.34 Ideal and actual fre-
quency response of a highpass filter.
A highpass filter is designed to pass all frequencies above its cutoff frequency ωc.
vi(t) R
C
+
− vo(t)
L
+
−
Figure14.35 A bandpass filter.
A highpass filter can also be formed when the output of an RL
circuit is taken off the inductor.
14.7.3 Bandpass Filter
The RLC series resonant circuit provides a bandpass filter when the out-
put is taken off the resistor as shown in Fig. 14.35. The transfer function
is
H(ω) =
Vo
Vi
=
R
R + j(ωL − 1/ωC)
(14.54)
CHAPTER 14 Frequency Response 611
We observe that H(0) = 0, H(∞) = 0. Figure 14.36 shows the plot of
|H(ω)|. The bandpass filter passes a band of frequencies (ω1  ω  ω2)
centered on ω0, the center frequency, which is given by
ω0 =
1
√
LC
(14.55)
A bandpass filter is designed to pass all frequencies within a band
of frequencies, ω1  ω  ω2.
Since the bandpass filter in Fig. 14.35 is a series resonant circuit, the half-
power frequencies, the bandwidth, and the quality factor are determined
as in Section 14.5. A bandpass filter can also be formed by cascading
the lowpass filter (where ω2 = ωc) in Fig. 14.31 with the highpass filter
(where ω1 = ωc) in Fig. 14.33.
v0
v1 v2 v
0.707
Ideal
Actual
1
0
 H(v)
Figure 14.36 Ideal and actual frequency
response of a bandpass filter.
14.7.4 Bandstop Filter
A filter that prevents a band of frequencies between two designated values
(ω1 and ω2) from passing is variably known as a bandstop, bandreject,
or notch filter. A bandstop filter is formed when the output RLC series
resonant circuit is taken off the LC series combination as shown in Fig.
14.37. The transfer function is
H(ω) =
Vo
Vi
=
j(ωL − 1/ωC)
R + j(ωL − 1/ωC)
(14.56)
Notice that H(0) = 1, H(∞) = 1. Figure 14.38 shows the plot of |H(ω)|.
Again, the center frequency is given by
ω0 =
1
√
LC
(14.57)
whilethehalf-powerfrequencies, thebandwidth, andthequalityfactorare
calculated using the formulas in Section 14.5 for a series resonant circuit.
Here, ω0 is called the frequency of rejection, while the corresponding
bandwidth (B = ω2 − ω1) is known as the bandwidth of rejection. Thus,
vi(t)
R
C
+
−
–
+
vo(t)
L
Figure14.37 A bandstop filter.
v0
v1 v2 v
0.707
Ideal
Actual
1
0
 H(v)
Figure 14.38 Ideal and actual frequency
response of a bandstop filter.
A bandstop filter is designed to stop or eliminate all frequencies within
a band of frequencies, ω1  ω  ω2.
Notice that adding the transfer functions of the bandpass and the
bandstop gives unity at any frequency for the same values of R, L, and
C. Of course, this is not true in general but true for the circuits treated
here. This is due to the fact that the characteristic of one is the inverse of
the other.
In concluding this section, we should note that:
1. From Eqs. (14.50), (14.52), (14.54), and (14.56), the maximum
gain of a passive filter is unity. To generate a gain greater than
unity, one should use an active filter as the next section shows.
612 PART 2 AC Circuits
2. There are other ways to get the types of filters treated in this
section.
3. The filters treated here are the simple types. Many other filters
have sharper and complex frequency responses.
E X A M P L E 1 4 . 1 0
Determine what type of filter is shown in Fig. 14.39. Calculate the corner
or cutoff frequency. Take R = 2 k, L = 2 H, and C = 2 µF.
vi(t) C
R
+
−
vo(t)
L
+
−
Figure14.39 For Example 14.10.
Solution:
The transfer function is
H(s) =
Vo
Vi
=
R  1/sC
sL + R  1/sC
, s = jω (14.10.1)
But
R
1
sC
=
R/sC
R + 1/sC
=
R
1 + sRC
Substituting this into Eq. (14.10.1) gives
H(s) =
R/(1 + sRC)
sL + R/(1 + sRC)
=
R
s2RLC + sL + R
, s = jω
or
H(ω) =
R
−ω2RLC + jωL + R
(14.10.2)
Since H(0) = 1 and H(∞) = 0, we conclude from Table 14.5 that the
circuit in Fig. 14.39 is a second-order lowpass filter. The magnitude of
H is
H =
R

(R − ω2RLC)2 + ω2L2
(14.10.3)
The corner frequency is the same as the half-power frequency, i.e., where
H is reduced by a factor of 1
√
2. Since the dc value of H(ω) is 1, at the
corner frequency, Eq. (14.10.3) becomes after squaring
H2
=
1
2
=
R2
(R − ω2
c RLC)2 + ω2
c L2
or
2 = (1 − ω2
c LC)2
+

ωcL
R
2
Substituting the values of R, L, and C, we obtain
2 =

1 − ω2
c 4 × 10−6
2
+ (ωc 10−3
)2
Assuming that ωc is in krad/s,
2 = (1 − 4ωc)2
+ ω2
c or 16ω4
c − 7ω2
c − 1 = 0
Solving the quadratic equation in ω2
c , we get ω2
c = 0.5509, or
ωc = 0.742 krad/s = 742 rad/s
CHAPTER 14 Frequency Response 613
P R A C T I C E P R O B L E M 1 4 . 1 0
For the circuit in Fig. 14.40, obtain the transfer function Vo(ω)/Vi(ω).
Identify the type of filter the circuit represents and determine the corner
frequency. Take R1 = 100  = R2, L = 2 mH.
vi(t)
R1
R2
+
−
vo(t)
L
+
−
Figure14.40 For Practice Prob. 14.10.
Answer: Highpass filter,
R2
R1 + R2

jω
jω + ωc

,
ωc =
R1R2
(R1 + R2)L
= 25 krad/s.
E X A M P L E 1 4 . 1 1
IfthebandstopfilterinFig.14.37istorejecta200-Hzsinusoidwhilepass-
ing other frequencies, calculate the values of L and C. Take R = 150 
and the bandwidth as 100 Hz.
Solution:
We use the formulas for a series resonant circuit in Section 14.5.
B = 2π(100) = 200π rad/s
But
B =
R
L
⇒ L =
R
B
=
150
200π
= 0.2387 H
Rejection of the 200-Hz sinusoid means that f0 is 200 Hz, so that ω0 in
Fig. 14.38 is
ω0 = 2πf0 = 2π(200) = 400π
Since ω0 = 1/
√
LC,
C =
1
ω2
0L
=
1
(400π)2(0.2387)
= 2.66 µF
P R A C T I C E P R O B L E M 1 4 . 1 1
Design a bandpass filter of the form in Fig. 14.35 with a lower cutoff fre-
quency of 20.1 kHz and an upper cutoff frequency of 20.3 kHz. Take
R = 20 k. Calculate L, C, and Q.
Answer: 7.96 H, 3.9 pF, 101.
14.8 ACTIVE FILTERS
Therearethreemajorlimitstothepassivefiltersconsideredintheprevious
section. First, they cannot generate gain greater than 1; passive elements
cannot add energy to the network. Second, they may require bulky and
expensive inductors. Third, they perform poorly at frequencies below the
audio frequency range (300 Hz  f  3000 Hz). Nevertheless, passive
filters are useful at high frequencies.
614 PART 2 AC Circuits
Active filters consist of combinations of resistors, capacitors, and
op amps. They offer some advantages over passive RLC filters. First,
they are often smaller and less expensive, because they do not require
inductors. This makes feasible the integrated circuit realizations of fil-
ters. Second, they can provide amplifier gain in addition to providing
the same frequency response as RLC filters. Third, active filters can be
combined with buffer amplifiers (voltage followers) to isolate each stage
of the filter from source and load impedance effects. This isolation allows
designing the stages independently and then cascading them to realize the
desired transfer function. (Bode plots, being logarithmic, may be added
when transfer functions are cascaded.) However, active filters are less
reliable and less stable. The practical limit of most active filters is about
100 kHz—most active filters operate well below that frequency.
Filters are often classified according to their order (or number of
poles) or their specific design type.
14.8.1 First-Order Lowpass Filter
One type of first-order filter is shown in Fig. 14.41. The components
selected for Zi and Zf determine whether the filter is lowpass or highpass,
but one of the components must be reactive.
+
−
−
+
V
o
+
–
V
i
Zi
Zf
Figure 14.41 A general first-
order active filter.
Figure 14.42 shows a typical active low-pass filter. For this filter,
the transfer function is
H(ω) =
Vo
Vi
= −
Zf
Zi
(14.58)
where Zi = Ri and
Zf = Rf
1
jωCf
=
Rf /jωCf
Rf + 1/jωCf
=
Rf
1 + jωCf Rf
(14.59)
Therefore,
H(ω) = −
Rf
Ri
1
1 + jωCf Rf
(14.60)
We notice that Eq. (14.60) is similar to Eq. (14.50), except that there is
a low frequency (ω → 0) gain or dc gain of −Rf /Ri. Also, the corner
frequency is
ωc =
1
Rf Cf
(14.61)
which does not depend on Ri. This means that several inputs with dif-
ferent Ri could be summed if required, and the corner frequency would
remain the same for each input.
+
−
+
–
V
o
+
–
V
i
Ri
Rf
Cf
Figure 14.42 Active first-order
lowpass filter.
14.8.2 First-Order Highpass Filter
+
−
+
–
V
o
+
–
V
i
Ri
Ci
Rf
Figure 14.43 Active first-order
highpass filter.
Figure 14.43 shows a typical highpass filter. As before,
H(ω) =
Vo
Vi
= −
Zf
Zi
(14.62)
where Zi = Ri + 1/jωCi and Zf = Rf so that
H(ω) = −
Rf
Ri + 1/jωCi
= −
jωCiRf
1 + jωCiRi
(14.63)
CHAPTER 14 Frequency Response 615
This is similar to Eq. (14.52), except that at very high frequencies (ω →
∞), the gain tends to −Rf /Ri. The corner frequency is
ωc =
1
RiCi
(14.64)
14.8.3 Bandpass Filter
The circuit in Fig. 14.42 may be combined with that in Fig. 14.43 to form
a bandpass filter that will have a gain K over the required range of fre-
quencies. By cascading a unity-gain lowpass filter, a unity-gain highpass
filter, and an inverter with gain −Rf /Ri, as shown in the block diagram
of Fig. 14.44(a), we can construct a bandpass filter whose frequency re-
sponse is that in Fig. 14.44(b). The actual construction of the bandpass
filter is shown in Fig. 14.45.
This way of creating a bandpass filter, not neces-
sarily the best, is perhaps the easiest to under-
stand.
v0
v1 v2 v
0.707 K
K
B
0
(a) (b)
Low-pass
filter
vi vo
H
High-pass
filter
Inverter
Figure14.44 Active bandpass filter: (a) block diagram, (b) frequency response.
+
−
+
–
vi
R
R
C1
C2
+
−
R
Stage 1
Low-pass filter
sets v2 value
Stage 2
High-pass filter
sets v1 value
Stage 3
An inverter
provides gain
R
+
−
+
–
vo
Ri
Rf
Figure14.45 Active bandpass filter.
The analysis of the bandpass filter is relatively simple. Its transfer
function is obtained by multiplying Eqs. (14.60) and (14.63) with the gain
of the inverter; that is
H(ω) =
Vo
Vi
=

−
1
1 + jωC1R
 
−
jωC2R
1 + jωC2R
 
−
Rf
Ri

= −
Rf
Ri
1
1 + jωC1R
jωC2R
1 + jωC2R
(14.65)
616 PART 2 AC Circuits
The lowpass section sets the upper corner frequency as
ω2 =
1
RC1
(14.66)
while the highpass section sets the lower corner frequency as
ω1 =
1
RC2
(14.67)
With these values of ω1 and ω2, the center frequency, bandwidth, and
quality factor are found as follows:
ω0 =
√
ω1ω2 (14.68)
B = ω2 − ω1 (14.69)
Q =
ω0
B
(14.70)
To find the passband gain K, we write Eq. (14.65) in the standard
form of Eq. (14.15),
H(ω) =
−Kjω/ω1
(1 + jω/ω1)(1 + jω/ω2)
=
−Kjωω2
(ω1 + jω)(ω2 + jω)
(14.71)
At the center frequency ω0 =
√
ω1ω2, the magnitude of the transfer
function is
H(ω0) =




−Kjω0ω2
(ω1 + jω0)(ω2 + jω0)



 =
Kω2
ω1 + ω2
(14.72)
We set this equal to the gain of the inverting amplifier, as
Kω2
ω1 + ω2
=
Rf
Ri
(14.73)
from which the gain K can be determined.
14.8.4 Bandreject (or Notch) Filter
A bandreject filter may be constructed by parallel combination of a low-
pass filter and a highpass filter and a summing amplifier, as shown in
the block diagram of Fig. 14.46(a). The circuit is designed such that the
v0
v1 v2 v
0.707 K
K
B
(b)
(a)
0
H
vi vo = v1 + v2
v1
v2
Low-pass
filter sets
v1
High-pass
filter sets
v2  v1
Summing
amplifier
Figure14.46 Active bandreject filter: (a) block diagram, (b) frequency response.
CHAPTER 14 Frequency Response 617
lower cutoff frequency ω1 is set by the lowpass filter while the upper cut-
off frequency ω2 is set by the highpass filter. The gap between ω1 and ω2
is the bandwidth of the filter. As shown in Fig. 14.46(b), the filter passes
frequencies below ω1 and above ω2. The block diagram in Fig. 14.46(a)
is actually constructed as shown in Fig. 14.47. The transfer function is
H(ω) =
Vo
Vi
= −
Rf
Ri

−
1
1 + jωC1R
−
jωC2R
1 + jωC2R

(14.74)
The formulas for calculating the values of ω1, ω2, the center frequency,
bandwidth, and quality factor are the same as in Eqs. (14.66) to (14.70).
+
−
+
–
vi
+
–
vo
R
R
C1
Rf
C2
+
−
+
−
R
R Ri
Ri
Figure14.47 Active bandreject filter.
To determine the passband gain K of the filter, we can write Eq.
(14.74) in terms of the upper and lower corner frequencies as
H(ω) =
Rf
Ri

1
1 + jω/ω2
+
jω/ω1
1 + jω/ω1

=
Rf
Ri
(1 + j2ω/ω1 + (jω)2
/ω1ω1)
(1 + jω/ω2)(1 + jω/ω1)
(14.75)
Comparing this with the standard form in Eq. (14.15) indicates that in the
two passbands (ω → 0 and ω → ∞) the gain is
K =
Rf
Ri
(14.76)
We can also find the gain at the center frequency by finding the magnitude
of the transfer function at ω0 =
√
ω1ω2, writing
H(ω0) =




Rf
Ri
(1 + j2ω0/ω1 + (jω0)2
/ω1ω1)
(1 + jω0/ω2)(1 + jω0/ω1)




=
Rf
Ri
2ω1
ω1 + ω2
(14.77)
Again, the filters treated in this section are only typical. There are
many other active filters that are more complex.
618 PART 2 AC Circuits
E X A M P L E 1 4 . 1 2
Design a low-pass active filter with a dc gain of 4 and a corner frequency
of 500 Hz.
Solution:
From Eq. (14.61), we find
ωc = 2πfc = 2π(500) =
1
Rf Cf
(14.12.1)
The dc gain is
H(0) = −
Rf
Ri
= −4 (14.12.2)
We have two equations and three unknowns. If we select Cf = 0.2 µF,
then
Rf =
1
2π(500)0.2 × 10−6
= 1.59 k
and
Ri =
Rf
4
= 397.5 
We use a 1.6-k resistor for Rf and a 400- resistor for Ri. Figure 14.42
shows the filter.
P R A C T I C E P R O B L E M 1 4 . 1 2
Design a highpass filter with a high-frequency gain of 5 and a corner fre-
quency of 2 kHz. Use a 0.1-µF capacitor in your design.
Answer: Ri = 800  and Rf = 4 k.
E X A M P L E 1 4 . 1 3
Design a bandpass filter in the form of Fig. 14.45 to pass frequencies be-
tween 250 Hz and 3000 Hz and with K = 10. Select R = 20 k.
Solution:
Since ω1 = 1/RC2, we obtain
C2 =
1
Rω1
=
1
2πf1R
=
1
2π × 250 × 20 × 103
= 31.83 nF
Similarly, since ω2 = 1/RC1,
C1 =
1
Rω2
=
1
2πf2R
=
1
2π × 3000 × 20 × 103
= 2.65 nF
From Eq. (14.73),
Rf
Ri
=
Kω2
ω1 + ω2
=
Kf2
f1 + f2
= 10
3000
3250
= 9.223
If we select Ri = 10 k, then Rf = 9.223Ri  92 k.
CHAPTER 14 Frequency Response 619
P R A C T I C E P R O B L E M 1 4 . 1 3
Design a notch filter based on Fig. 14.47 for ω0 = 20 krad/s, K = 5, and
Q = 10. Use R = Ri = 10 k.
Answer: C1 = 47.62 nF, C2 = 52.63 nF, and Rf = 50 k.
†14.9 SCALING
In designing and analyzing filters and resonant circuits or in circuit anal-
ysis in general, it is sometimes convenient to work with element values
of 1 , 1 H, or 1 F, and then transform the values to realistic values by
scaling. We have taken advantage of this idea by not using realistic el-
ement values in most of our examples and problems; mastering circuit
analysis is made easy by using convenient component values. We have
thus eased calculations, knowing that we could use scaling to then make
the values realistic.
There are two ways of scaling a circuit: magnitude or impedance
scaling, and frequency scaling. Both are useful in scaling responses and
circuit elements to values within the practical ranges. While magnitude
scaling leaves the frequency response of a circuit unaltered, frequency
scaling shifts the frequency response up or down the frequency spectrum.
14.9.1 Magnitude Scaling
Magnitude scaling is the process of increasing all impedance in a network by a factor,
the frequency response remaining unchanged.
Recall that impedances of individual elements R, L, and C are
given by
ZR = R, ZL = jωL, ZC =
1
jωC
(14.78)
In magnitude scaling, we multiply the impedance of each circuit element
by a factor Km and let the frequency remain constant. This gives the new
impedances as
Z
R = KmZR = KmR, Z
L = KmZL = jωKmL
Z
C = KmZC =
1
jωC/Km
(14.79)
Comparing Eq. (14.79) with Eq. (14.78), we notice the following changes
in the element values: R → KmR, L → KmL, and C → C/Km. Thus,
in magnitude scaling, the new values of the elements and frequency are
R
= KmR, L
= KmL
C
=
C
Km
, ω
= ω
(14.80)
620 PART 2 AC Circuits
The primed variables are the new values and the unprimed variables are
the old values. Consider the series or parallel RLC circuit. We now have
ω
0 =
1
√
LC
=
1
√
KmLC/Km
=
1
√
LC
= ω0 (14.81)
showing that the resonant frequency, as expected, has not changed. Sim-
ilarly, the quality factor and the bandwidth are not affected by magnitude
scaling. Also, magnitude scaling does not affect transfer functions in the
forms of Eqs. (14.2a) and (14.2b), which are dimensionless quantities.
14.9.2 Frequency Scaling
Frequency scaling is the process of shifting the frequency response of a network up
or down the frequency axis while leaving the impedance the same.
We achieve frequency scaling by multiplying the frequency by a factor
Kf while keeping the impedance the same.
Frequency scaling is equivalent to relabeling the
frequency axis of a frequency response plot. It is
needed when translating such frequencies such
as a resonant frequency, a corner frequency, a
bandwidth, etc., to a realistic level. It can be
used to bring capacitance and inductance values
into a range that is convenient to work with.
From Eq. (14.78), we see that the impedances of L and C are
frequency-dependent. If we apply frequency scaling to ZL(ω) and ZC(ω)
in Eq. (14.78), we obtain
ZL = j(ωKf )L
= jωL ⇒ L
=
L
Kf
(14.82a)
ZC =
1
j(ωKf )C
=
1
jωC
⇒ C
=
C
Kf
(14.82b)
since the impedance of the inductor and capacitor must remain the same
after frequency scaling. We notice the following changes in the element
values: L → L/Kf and C → C/Kf . The value of R is not affected,
since its impedance does not depend on frequency. Thus, in frequency
scaling, the new values of the elements and frequency are
R
= R, L
=
L
Kf
C
=
C
Kf
, ω
= Kf ω
(14.83)
Again, if we consider the series or parallel RLC circuit, for the resonant
frequency
ω
0 =
1
√
LC
=
1

(L/Kf )(C/Kf )
=
Kf
√
LC
= Kf ω0 (14.84)
and for the bandwidth
B
= Kf B (14.85)
but the quality factor remains the same (Q
= Q).
CHAPTER 14 Frequency Response 621
14.9.3 Magnitude and Frequency Scaling
If a circuit is scaled in magnitude and frequency at the same time, then
R
= KmR, L
=
Km
Kf
L
C
=
1
KmKf
C, ω
= Kf ω
(14.86)
These are more general formulas than those in Eqs. (14.80) and (14.83).
We set Km = 1 in Eq. (14.86) when there is no magnitude scaling or
Kf = 1 when there is no frequency scaling.
E X A M P L E 1 4 . 1 4
A fourth-order Butterworth lowpass filter is shown in Fig. 14.48(a). The
filter is designed such that the cutoff frequency ωc = 1 rad/s. Scale the
circuit for a cutoff frequency of 50 kHz using 10-k resistors.
1 Ω
1 Ω
(a)
+
− vo
vs
+
−
1.848 F
0.765 F
1.848 H 0.765 H 10 kΩ
10 kΩ
(b)
+
− vo
vs
+
−
588.2 pF
243.5 pF
58.82 mH 24.35 H
Figure14.48 For Example 14.14: (a) Normalized Butterworth lowpass filter, (b) scaled version of the same lowpass filter.
Solution:
If the cutoff frequency is to shift from ωc = 1 rad/s to ω
c = 2π(50)
krad/s, then the frequency scale factor is
Kf =
ω
c
ωc
=
100π × 103
1
= π × 105
Also, if each 1- resistor is to be replaced by a 10-k resistor, then the
magnitude scale factor must be
Km =
R
R
=
10 × 103
1
= 104
Using Eq. (14.86),
L
1 =
Km
Kf
L1 =
104
π × 105
(1.848) = 58.82 mH
L
2 =
Km
Kf
L2 =
104
π × 105
(0.765) = 24.35 mH
C
1 =
C1
KmKf
=
0.765
π × 109
= 243.5 pF
C
2 =
C2
KmKf
=
1.848
π × 109
= 588.2 pF
622 PART 2 AC Circuits
The scaled circuit is as shown in Fig. 14.48(b). This circuit uses practical
values and will provide the same transfer function as the prototype in Fig.
14.48(a), but shifted in frequency.
P R A C T I C E P R O B L E M 1 4 . 1 4
A third-order Butterworth filter normalized to ωc = 1 rad/s is shown in
Fig. 14.49. Scale the circuit to a cutoff frequency of 10 kHz. Use 15-nF
capacitors.
1 Ω
1 Ω
+
− vo
vs
+
−
1 F
1 F
2 H
Figure14.49 For Practice Prob. 14.14.
Answer: R
1 = R
2 = 1.061 k, C
1 = C
2 =15 nF,
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Network_theory_by_alaxander_and_sadiku.pdf

  • 1. Preface v Acknowledgments vi A Note to the Student ix 1.1 Introduction 4 1.2 Systems of Units 4 1.3 Charge and Current 6 1.4 Voltage 9 1.5 Power and Energy 10 1.6 Circuit Elements 13 †1.7 Applications 15 1.7.1 TV Picture Tube 1.7.2 Electricity Bills †1.8 Problem Solving 18 1.9 Summary 21 Review Questions 22 Problems 23 Comprehensive Problems 25 2.1 Introduction 28 2.2 Ohm’s Laws 28 †2.3 Nodes, Branches, and Loops 33 2.4 Kirchhoff’s Laws 35 2.5 Series Resistors and Voltage Division 41 2.6 Parallel Resistors and Current Division 42 †2.7 Wye-Delta Transformations 50 †2.8 Applications 54 2.8.1 Lighting Systems 2.8.2 Design of DC Meters 2.9 Summary 60 Review Questions 61 Problems 63 Comprehensive Problems 72 3.1 Introduction 76 3.2 Nodal Analysis 76 3.3 Nodal Analysis with Voltage Sources 82 3.4 Mesh Analysis 87 3.5 Mesh Analysis with Current Sources 92 †3.6 Nodal and Mesh Analyses by Inspection 95 3.7 Nodal Versus Mesh Analysis 99 3.8 Circuit Analysis with PSpice 100 †3.9 Applications: DC Transistor Circuits 102 3.10 Summary 107 Review Questions 107 Problems 109 Comprehensive Problems 117 4.1 Introduction 120 4.2 Linearity Property 120 4.3 Superposition 122 4.4 Source Transformation 127 4.5 Thevenin’s Theorem 131 4.6 Norton’s Theorem 137 †4.7 Derivations of Thevenin’s and Norton’s Theorems 140 4.8 Maximum Power Transfer 142 4.9 Verifying Circuit Theorems with PSpice 144 †4.10 Applications 147 4.10.1 Source Modeling 4.10.2 Resistance Measurement 4.11 Summary 153 Review Questions 153 Problems 154 Comprehensive Problems 162 5.1 Introduction 166 5.2 Operational Amplifiers 166 5.3 Ideal Op Amp 170 5.4 Inverting Amplifier 171 5.5 Noninverting Amplifier 174 5.6 Summing Amplifier 176 5.7 Difference Amplifier 177 5.8 Cascaded Op Amp Circuits 181 5.9 Op Amp Circuit Analysis with PSpice 183 †5.10 Applications 185 5.10.1 Digital-to Analog Converter 5.10.2 Instrumentation Amplifiers 5.11 Summary 188 Review Questions 190 Problems 191 Comprehensive Problems 200 Contents xi Chapter 2 Basic Laws 27 Chapter 3 Methods of Analysis 75 PART 1 DC CIRCUITS 1 Chapter 1 Basic Concepts 3 Chapter 4 Circuit Theorems 119 Chapter 5 Operational Amplifiers 165 f51-cont.qxd 3/16/00 4:22 PM Page xi
  • 2. 6.1 Introduction 202 6.2 Capacitors 202 6.3 Series and Parallel Capacitors 208 6.4 Inductors 211 6.5 Series and Parallel Inductors 216 †6.6 Applications 219 6.6.1 Integrator 6.6.2 Differentiator 6.6.3 Analog Computer 6.7 Summary 225 Review Questions 226 Problems 227 Comprehensive Problems 235 7.1 Introduction 238 7.2 The Source-free RC Circuit 238 7.3 The Source-free RL Circuit 243 7.4 Singularity Functions 249 7.5 Step Response of an RC Circuit 257 7.6 Step Response of an RL Circuit 263 †7.7 First-order Op Amp Circuits 268 7.8 Transient Analysis with PSpice 273 †7.9 Applications 276 7.9.1 Delay Circuits 7.9.2 Photoflash Unit 7.9.3 Relay Circuits 7.9.4 Automobile Ignition Circuit 7.10 Summary 282 Review Questions 283 Problems 284 Comprehensive Problems 293 8.1 Introduction 296 8.2 Finding Initial and Final Values 296 8.3 The Source-Free Series RLC Circuit 301 8.4 The Source-Free Parallel RLC Circuit 308 8.5 Step Response of a Series RLC Circuit 314 8.6 Step Response of a Parallel RLC Circuit 319 8.7 General Second-Order Circuits 322 8.8 Second-Order Op Amp Circuits 327 8.9 PSpice Analysis of RLC Circuits 330 †8.10 Duality 332 †8.11 Applications 336 8.11.1 Automobile Ignition System 8.11.2 Smoothing Circuits 8.12 Summary 340 Review Questions 340 Problems 341 Comprehensive Problems 350 9.1 Introduction 354 9.2 Sinusoids 355 9.3 Phasors 359 9.4 Phasor Relationships for Circuit Elements 367 9.5 Impedance and Admittance 369 9.6 Kirchhoff’s Laws in the Frequency Domain 372 9.7 Impedance Combinations 373 †9.8 Applications 379 9.8.1 Phase-Shifters 9.8.2 AC Bridges 9.9 Summary 384 Review Questions 385 Problems 385 Comprehensive Problems 392 10.1 Introduction 394 10.2 Nodal Analysis 394 10.3 Mesh Analysis 397 10.4 Superposition Theorem 400 10.5 Source Transformation 404 10.6 Thevenin and Norton Equivalent Circuits 406 10.7 Op Amp AC Circuits 411 10.8 AC Analysis Using PSpice 413 †10.9 Applications 416 10.9.1 Capacitance Multiplier 10.9.2 Oscillators 10.10 Summary 420 Review Questions 421 Problems 422 11.1 Introduction 434 11.2 Instantaneous and Average Power 434 11.3 Maximum Average Power Transfer 440 11.4 Effective or RMS Value 443 11.5 Apparent Power and Power Factor 447 11.6 Complex Power 449 †11.7 Conservation of AC Power 453 xii CONTENTS Chapter 8 Second-Order Circuits 295 Chapter 10 Sinusoidal Steady-State Analysis 393 Chapter 11 AC Power Analysis 433 Chapter 6 Capacitors and Inductors 201 Chapter 7 First-Order Circuits 237 PART 2 AC CIRCUITS 351 Chapter 9 Sinusoids and Phasors 353
  • 3. 11.8 Power Factor Correction 457 †11.9 Applications 459 11.9.1 Power Measurement 11.9.2 Electricity Consumption Cost 11.10 Summary 464 Review Questions 465 Problems 466 Comprehensive Problems 474 12.1 Introduction 478 12.2 Balanced Three-Phase Voltages 479 12.3 Balanced Wye-Wye Connection 482 12.4 Balanced Wye-Delta Connection 486 12.5 Balanced Delta-Delta Connection 488 12.6 Balanced Delta-Wye Connection 490 12.7 Power in a Balanced System 494 †12.8 Unbalanced Three-Phase Systems 500 12.9 PSpice for Three-Phase Circuits 504 †12.10 Applications 508 12.10.1 Three-Phase Power Measurement 12.10.2 Residential Wiring 12.11 Summary 516 Review Questions 517 Problems 518 Comprehensive Problems 525 13.1 Introduction 528 13.2 Mutual Inductance 528 13.3 Energy in a Coupled Circuit 535 13.4 Linear Transformers 539 13.5 Ideal Transformers 545 13.6 Ideal Autotransformers 552 †13.7 Three-Phase Transformers 556 13.8 PSpice Analysis of Magnetically Coupled Circuits 559 †13.9 Applications 563 13.9.1 Transformer as an Isolation Device 13.9.2 Transformer as a Matching Device 13.9.3 Power Distribution 13.10 Summary 569 Review Questions 570 Problems 571 Comprehensive Problems 582 14.1 Introduction 584 14.2 Transfer Function 584 †14.3 The Decibel Scale 588 14.4 Bode Plots 589 14.5 Series Resonance 600 14.6 Parallel Resonance 605 14.7 Passive Filters 608 14.7.1 Lowpass Filter 14.7.2 Highpass Filter 14.7.3 Bandpass Filter 14.7.4 Bandstop Filter 14.8 Active Filters 613 14.8.1 First-Order Lowpass Filter 14.8.2 First-Order Highpass Filter 14.8.3 Bandpass Filter 14.8.4 Bandreject (or Notch) Filter †14.9 Scaling 619 14.9.1 Magnitude Scaling 14.9.2 Frequency Scaling 14.9.3 Magnitude and Frequency Scaling 14.10 Frequency Response Using PSpice 622 †14.11 Applications 626 14.11.1 Radio Receiver 14.11.2 Touch-Tone Telephone 14.11.3 Crossover Network 14.12 Summary 631 Review Questions 633 Problems 633 Comprehensive Problems 640 15.1 Introduction 646 15.2 Definition of the Laplace Transform 646 15.3 Properties of the Laplace Transform 649 15.4 The Inverse Laplace Transform 659 15.4.1 Simple Poles 15.4.2 Repeated Poles 15.4.3 Complex Poles 15.5 Applicaton to Circuits 666 15.6 Transfer Functions 672 15.7 The Convolution Integral 677 †15.8 Application to Integrodifferential Equations 685 †15.9 Applications 687 15.9.1 Network Stability 15.9.2 Network Synthesis 15.10 Summary 694 xiii CONTENTS PART 3 ADVANCEDCIRCUITANALYSIS 643 Chapter 15 The Laplace Transform 645 Chapter 12 Three-Phase Circuits 477 Chapter 13 Magnetically Coupled Circuits 527 Chapter 14 Frequency Response 583
  • 4. Review Questions 696 Problems 696 Comprehensive Problems 705 16.1 Introduction 708 16.2 Trigonometric Fourier Series 708 16.3 Symmetry Considerations 717 16.3.1 Even Symmetry 16.3.2 Odd Symmetry 16.3.3 Half-Wave Symmetry 16.4 Circuit Applicatons 727 16.5 Average Power and RMS Values 730 16.6 Exponential Fourier Series 734 16.7 Fourier Analysis with PSpice 740 16.7.1 Discrete Fourier Transform 16.7.2 Fast Fourier Transform †16.8 Applications 746 16.8.1 Spectrum Analyzers 16.8.2 Filters 16.9 Summary 749 Review Questions 751 Problems 751 Comprehensive Problems 758 17.1 Introduction 760 17.2 Definition of the Fourier Transform 760 17.3 Properties of the Fourier Transform 766 17.4 Circuit Applications 779 17.5 Parseval’s Theorem 782 17.6 Comparing the Fourier and Laplace Transforms 784 †17.7 Applications 785 17.7.1 Amplitude Modulation 17.7.2 Sampling 17.8 Summary 789 Review Questions 790 Problems 790 Comprehensive Problems 794 18.1 Introduction 796 18.2 Impedance Parameters 796 18.3 Admittance Parameters 801 18.4 Hybrid Parameters 804 18.5 Transmission Parameters 809 †18.6 Relationships between Parameters 814 18.7 Interconnection of Networks 817 18.8 Computing Two-Port Parameters Using PSpice 823 †18.9 Applications 826 18.9.1 Transistor Circuits 18.9.2 Ladder Network Synthesis 18.10 Summary 833 Review Questions 834 Problems 835 Comprehensive Problems 844 AppendixA Solution of Simultaneous Equations Using Cramer’s Rule 845 Appendix B Complex Numbers 851 Appendix C Mathematical Formulas 859 Appendix D PSpice for Windows 865 Appendix E Answers to Odd-Numbered Problems 893 Selected Bibliography 929 Index 933 xiv CONTENTS Chapter 16 The Fourier Series 707 Chapter 17 Fourier Transform 759 Chapter 18 Two-Port Networks 795
  • 5. Features In spite of the numerous textbooks on circuit analysis available in the market, students often find the course difficult to learn. The main objective of this book is to present circuit analysis in a manner that is clearer, more interesting, and easier to understand than earlier texts. This objective is achieved in the following ways: • A course in circuit analysis is perhaps the first exposure students have to electrical engineering. We have included several features to help stu- dents feel at home with the subject. Each chapter opens with either a historical profile of some electrical engineering pioneers to be mentioned in the chapter or a career discussion on a subdisci- pline of electrical engineering. An introduction links the chapter with the previous chapters and states the chapter’s objectives. The chapter ends with a summary of the key points and formulas. • All principles are presented in a lucid, logical, step-by-step manner. We try to avoid wordiness and superfluous detail that could hide concepts and impede understanding the material. • Important formulas are boxed as a means of helping students sort what is essential from what is not; and to ensure that students clearly get the gist of the matter, key terms are defined and highlighted. • Marginal notes are used as a pedagogical aid. They serve multiple uses—hints, cross-references, more exposition, warnings, reminders, common mis- takes, and problem-solving insights. • Thoroughly worked examples are liberally given at the end of every section. The examples are regard- ed as part of the text and are explained clearly, with- out asking the reader to fill in missing steps. Thoroughly worked examples give students a good understanding of the solution and the confidence to solve problems themselves. Some of the problems are solved in two or three ways to facilitate an understanding and comparison of different approaches. • To give students practice opportunity, each illus- trative example is immediately followed by a practice problem with the answer. The students can follow the example step-by-step to solve the prac- tice problem without flipping pages or searching the end of the book for answers. The practice prob- lem is also intended to test students’ understanding of the preceding example. It will reinforce their grasp of the material before moving to the next section. • In recognition of ABET’s requirement on integrat- ing computer tools, the use of PSpice is encouraged in a student-friendly manner. Since the Windows version of PSpice is becoming popular, it is used instead of the MS-DOS version. PSpice is covered early so that students can use it throughout the text. Appendix D serves as a tutorial on PSpice for Windows. • The operational amplifier (op amp) as a basic ele- ment is introduced early in the text. • To ease the transition between the circuit course and signals/systems courses, Fourier and Laplace transforms are covered lucidly and thoroughly. • The last section in each chapter is devoted to appli- cations of the concepts covered in the chapter. Each chapter has at least one or two practical problems or devices. This helps students apply the concepts to real-life situations. • Ten multiple-choice review questions are provided at the end of each chapter, with answers. These are intended to cover the little “tricks” that the exam- ples and end-of-chapter problems may not cover. They serve as a self-test device and help students determine how well they have mastered the chapter. Organization This book was written for a two-semester or three-semes- ter course in linear circuit analysis. The book may also be used for a one-semester course by a proper selec- tion of chapters and sections. It is broadly divided into three parts. • Part 1, consisting of Chapters 1 to 8, is devoted to dc circuits. It covers the fundamental laws and the- orems, circuit techniques, passive and active ele- ments. • Part 2, consisting of Chapters 9 to 14, deals with ac circuits. It introduces phasors, sinusoidal steady- state analysis, ac power, rms values, three-phase systems, and frequency response. • Part 3, consisting of Chapters 15 to 18, is devoted to advanced techniques for network analysis. It provides a solid introduction to the Laplace transform, Fourier series, the Fourier transform, and two-port network analysis. The material in three parts is more than suffi- cient for a two-semester course, so that the instructor PREFACE v F51-pref.qxd 3/17/00 10:11 AM Page v
  • 6. must select which chapters/sections to cover. Sections marked with the dagger sign (†) may be skipped, explained briefly, or assigned as homework. They can be omitted without loss of continuity. Each chapter has plenty of problems, grouped according to the sections of the related material, and so diverse that the instruc- tor can choose some as examples and assign some as homework. More difficult problems are marked with a star (*). Comprehensive problems appear last; they are mostly applications problems that require multiple skills from that particular chapter. The book is as self-contained as possible. At the end of the book are some appendixes that review solutions of linear equations, complex numbers, math- ematical formulas, a tutorial on PSpice for Windows, and answers to odd-numbered problems. Answers to all the problems are in the solutions manual, which is available from the publisher. Prerequisites As with most introductory circuit courses, the main prerequisites are physics and calculus. Although famil- iarity with complex numbers is helpful in the later part of the book, it is not required. Supplements Solutions Manual—an Instructor’s Solutions Manual is available to instructors who adopt the text. It contains complete solutions to all the end-of-chapter problems. Transparency Masters—over 200 important figures are available as transparency masters for use as over- heads. Student CD-ROM—100 circuit files from the book are presented as Electronics Workbench (EWB) files; 15–20 of these files are accessible using the free demo of Elec- tronics Workbench. The students are able to experiment with the files. For those who wish to fully unlock all 100 circuit files, EWB’s full version may be purchased from Interactive Image Technologies for approximately $79.00. The CD-ROM also contains a selection of prob- lem-solving, analysis and design tutorials, designed to further support important concepts in the text. Problem-Solving Workbook—a paperback work- book is for sale to students who wish to practice their problem solving techniques. The workbook contains a discussion of problem solving strategies and 150 addi- tional problems with complete solutions provided. Online Learning Center (OLC)—the Web site for the book will serve as an online learning center for stu- dents as a useful resource for instructors. The OLC will provide access to: 300 test questions—for instructors only Downloadable figures for overhead presentations—for instructors only Solutions manual—for instructors only Web links to useful sites Sample pages from the Problem-Solving Workbook PageOut Lite—a service provided to adopters who want to create their own Web site. In just a few minutes, instructors can change the course syllabus into a Web site using PageOut Lite. The URL for the web site is www.mhhe.com.alexander. Although the textbook is meant to be self-explanatory and act as a tutor for the student, the personal contact involved in teaching is not to be forgotten. The book and supplements are intended to supply the instructor with all the pedagogical tools necessary to effectively present the material. We wish to take the opportunity to thank the staff of McGraw-Hill for their commitment and hard work: Lynn Cox, Senior Editor; Scott Isenberg, Senior Sponsoring Editor; Kelley Butcher, Senior Developmental Editor; Betsy Jones, Executive Editor; Catherine Fields, Sponsoring Editor; Kimberly Hooker, Project Manager; and Michelle Flomenhoft, Editorial Assistant. They got numerous reviews, kept the book on track, and helped in many ways. We really appreciate their inputs. We are greatly in debt to Richard Mickey for taking the pain ofchecking and correcting the entire manuscript. We wish to record our thanks to Steven Durbin at Florida State University and Daniel Moore at Rose Hulman Institute of Technology for serving as accuracy checkers of examples, practice problems, and end- of-chapter problems. We also wish to thank the fol- lowing reviewers for their constructive criticisms and helpful comments. Promod Vohra, Northern Illinois University Moe Wasserman, Boston University Robert J. Krueger, University of Wisconsin Milwaukee John O’Malley, University of Florida vi PREFACE ACKNOWLEDGMENTS F51-pref.qxd 3/17/00 10:11 AM Page vi
  • 7. Aniruddha Datta, Texas A&M University John Bay, Virginia Tech Wilhelm Eggimann, Worcester Polytechnic Institute A. B. Bonds, Vanderbilt University Tommy Williamson, University of Dayton Cynthia Finelli, Kettering University John A. Fleming, Texas A&M University Roger Conant, University of Illinois at Chicago Daniel J. Moore, Rose-Hulman Institute of Technology Ralph A. Kinney, Louisiana State University Cecilia Townsend, North Carolina State University Charles B. Smith, University of Mississippi H. Roland Zapp, Michigan State University Stephen M. Phillips, Case Western University Robin N. Strickland, University of Arizona David N. Cowling, Louisiana Tech University Jean-Pierre R. Bayard, California State University Jack C. Lee, University of Texas at Austin E. L. Gerber, Drexel University The first author wishes to express his apprecia- tion to his department chair, Dr. Dennis Irwin, for his outstanding support. In addition, he is extremely grate- ful to Suzanne Vazzano for her help with the solutions manual. The second author is indebted to Dr. Cynthia Hirtzel, the former dean of the college of engineering at Temple University, and Drs.. Brian Butz, Richard Klafter, and John Helferty, his departmental chairper- sons at different periods, for their encouragement while working on the manuscript. The secretarial support provided by Michelle Ayers and Carol Dahlberg is gratefully appreciated. Special thanks are due to Ann Sadiku, Mario Valenti, Raymond Garcia, Leke and Tolu Efuwape, and Ope Ola for helping in various ways. Finally, we owe the greatest debt to our wives, Paulette and Chris, without whose constant support and cooperation this project would have been impossible. Please address comments and corrections to the publisher. C. K. Alexander and M. N. O. Sadiku PREFACE vii F51-pref.qxd 3/17/00 10:11 AM Page vii
  • 9. This may be your first course in electrical engineer- ing. Although electrical engineering is an exciting and challenging discipline, the course may intimidate you. This book was written to prevent that. A good textbook and a good professor are an advantage—but you are the one who does the learning. If you keep the follow- ing ideas in mind, you will do very well in this course. • This course is the foundation on which most other courses in the electrical engineering cur- riculum rest. For this reason, put in as much effort as you can. Study the course regularly. • Problem solving is an essential part of the learn- ing process. Solve as many problems as you can. Begin by solving the practice problem following each example, and then proceed to the end-of- chapter problems. The best way to learn is to solve a lot of problems. An asterisk in front of a problem indicates a challenging problem. • Spice, a computer circuit analysis program, is used throughout the textbook. PSpice, the per- sonal computer version of Spice, is the popular standard circuit analysis program at most uni- versities. PSpice for Windows is described in Appendix D. Make an effort to learn PSpice, because you can check any circuit problem with PSpice and be sure you are handing in a correct problem solution. • Each chapter ends with a section on how the material covered in the chapter can be applied to real-life situations. The concepts in this section may be new and advanced to you. No doubt, you will learn more of the details in other courses. We are mainly interested in gaining a general familiarity with these ideas. • Attempt the review questions at the end of each chapter. They will help you discover some “tricks” not revealed in class or in the textbook. A short review on finding determinants is cov- ered in Appendix A, complex numbers in Appendix B, and mathematical formulas in Appendix C. Answers to odd-numbered problems are given in Appendix E. Have fun! C.K.A. and M.N.O.S. A NOTE TO THE STUDENT ix F51-pref.qxd 3/17/00 10:11 AM Page ix
  • 10. 1 DC CIRCUITS P A R T 1 C h a p t e r 1 Basic Concepts C h a p t e r 2 Basic Laws C h a p t e r 3 Methods of Analysis C h a p t e r 4 Circuit Theorems C h a p t e r 5 Operational Amplifier C h a p t e r 6 Capacitors and Inductors C h a p t e r 7 First-Order Circuits C h a p t e r 8 Second-Order Circuits
  • 11. 2
  • 12. 3 C H A P T E R BASIC CONCEPTS 1 It is engineering that changes the world. —Isaac Asimov Historical Profiles Alessandro Antonio Volta (1745–1827), an Italian physicist, invented the electric battery—which provided the first continuous flow of electricity—and the capacitor. Born into a noble family in Como, Italy, Volta was performing electrical experiments at age 18. His invention of the battery in 1796 revolutionized the use of electricity. The publication of his work in 1800 marked the beginning of electric circuit theory. Volta received many honors during his lifetime. The unit of voltage or potential difference, the volt, was named in his honor. Andre-Marie Ampere (1775–1836), a French mathematician and physicist, laid the foundation of electrodynamics. He defined the electric current and developed a way to measure it in the 1820s. Born in Lyons, France, Ampere at age 12 mastered Latin in a few weeks, as he was intensely interested in mathematics and many of the best mathematical works were in Latin. He was a brilliant scientist and a prolific writer. He formulated the laws of electromagnetics. He invented the electromagnet and the ammeter. The unit of electric current, the ampere, was named after him.
  • 13. 4 PART 1 DC Circuits 1.1 INTRODUCTION Electric circuit theory and electromagnetic theory are the two fundamen- tal theories upon which all branches of electrical engineering are built. Many branches of electrical engineering, such as power, electric ma- chines, control, electronics, communications, and instrumentation, are based on electric circuit theory. Therefore, the basic electric circuit the- ory course is the most important course for an electrical engineering student, and always an excellent starting point for a beginning student in electrical engineering education. Circuit theory is also valuable to students specializing in other branches of the physical sciences because circuits are a good model for the study of energy systems in general, and because of the applied mathematics, physics, and topology involved. In electrical engineering, we are often interested in communicating or transferring energy from one point to another. To do this requires an interconnection of electrical devices. Such interconnection is referred to as an electric circuit, and each component of the circuit is known as an element. An electric circuit is an interconnection of electrical elements. A simple electric circuit is shown in Fig. 1.1. It consists of three basic components: a battery, a lamp, and connecting wires. Such a simple circuit can exist by itself; it has several applications, such as a torch light, a search light, and so forth. + − Current Lamp Battery Figure1.1 A simple electric circuit. A complicated real circuit is displayed in Fig. 1.2, representing the schematic diagram for a radio receiver. Although it seems complicated, this circuit can be analyzed using the techniques we cover in this book. Our goal in this text is to learn various analytical techniques and computer software applications for describing the behavior of a circuit like this. Electric circuits are used in numerous electrical systems to accom- plish different tasks. Our objective in this book is not the study of various uses and applications of circuits. Rather our major concern is the anal- ysis of the circuits. By the analysis of a circuit, we mean a study of the behavior of the circuit: How does it respond to a given input? How do the interconnected elements and devices in the circuit interact? We commence our study by defining some basic concepts. These concepts include charge, current, voltage, circuit elements, power, and energy. Before defining these concepts, we must first establish a system of units that we will use throughout the text. 1.2 SYSTEMS OF UNITS As electrical engineers, we deal with measurable quantities. Our mea- surement, however, must be communicated in a standard language that virtually all professionals can understand, irrespective of the country where the measurement is conducted. Such an international measure- ment language is the International System of Units (SI), adopted by the General Conference on Weights and Measures in 1960. In this system,
  • 14. CHAPTER 1 Basic Concepts 5 2, 5, 6 C Oscillator E B R2 10 k R3 10 k R1 47 Y1 7 MHz C6 5 L2 22.7 mH (see text) to U1, Pin 8 R10 10 k GAIN + + C16 100 mF 16 V C11 100 mF 16 V C10 1.0 mF 16 V C9 1.0 mF 16 V C15 0.47 16 V C17 100 mF 16 V + − 12-V dc Supply Audio Output + C18 0.1 R12 10 1 4 2 3 C14 0.0022 0.1 C13 U2A 1 ⁄2 TL072 U2B 1⁄2 TL072 R9 15 k R5 100 k R8 15 k R6 100 k 5 6 R7 1 M C12 0.0033 + L3 1 mH R11 47 C8 0.1 Q1 2N2222A 7 C3 0.1 L1 0.445 mH Antenna C1 2200 pF C2 2200 pF 1 8 7 U1 SBL-1 Mixer 3, 4 C7 532 C4 910 C5 910 R4 220 U3 LM386N Audio power amp 5 4 6 3 2 + + − + − + − + 8 Figure 1.2 Electric circuit of a radio receiver. (Reproduced with permission from QST, August 1995, p. 23.) there are six principal units from which the units of all other physical quantities can be derived. Table 1.1 shows the six units, their symbols, and the physical quantities they represent. The SI units are used through- out this text. One great advantage of the SI unit is that it uses prefixes based on the power of 10 to relate larger and smaller units to the basic unit. Table 1.2 shows the SI prefixes and their symbols. For example, the following are expressions of the same distance in meters (m): 600,000,000 mm 600,000 m 600 km TABLE 1.2 The SI prefixes. Multiplier Prefix Symbol 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h 10 deka da 10−1 deci d 10−2 centi c 10−3 milli m 10−6 micro µ 10−9 nano n 10−12 pico p 10−15 femto f 10−18 atto a TABLE 1.1 The six basic SI units. Quantity Basic unit Symbol Length meter m Mass kilogram kg Time second s Electric current ampere A Thermodynamic temperature kelvin K Luminous intensity candela cd
  • 15. 6 PART 1 DC Circuits 1.3 CHARGE AND CURRENT The concept of electric charge is the underlying principle for explaining all electrical phenomena. Also, the most basic quantity in an electric circuit is the electric charge. We all experience the effect of electric charge when we try to remove our wool sweater and have it stick to our body or walk across a carpet and receive a shock. Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C). We know from elementary physics that all matter is made of fundamental building blocks known as atoms and that each atom consists of electrons, protons, and neutrons. We also know that the charge e on an electron is negative and equal in magnitude to 1.602×10−19 C, while a proton carries a positive charge of the same magnitude as the electron. The presence of equal numbers of protons and electrons leaves an atom neutrally charged. The following points should be noted about electric charge: 1. The coulomb is a large unit for charges. In 1 C of charge, there are 1/(1.602 × 10−19 ) = 6.24 × 1018 electrons. Thus realistic or laboratory values of charges are on the order of pC, nC, or µC.1 2. According to experimental observations, the only charges that occur in nature are integral multiples of the electronic charge e = −1.602 × 10−19 C. 3. The law of conservation of charge states that charge can neither be created nor destroyed, only transferred. Thus the algebraic sum of the electric charges in a system does not change. We now consider the flow of electric charges. A unique feature of electric charge or electricity is the fact that it is mobile; that is, it can be transferred from one place to another, where it can be converted to another form of energy. Battery I − − − − + − Figure1.3 Electric current due to flow of electronic charge in a conductor. A convention is a standard way of describing something so that others in the profession can understandwhatwemean. WewillbeusingIEEE conventions throughout this book. When a conducting wire (consisting of several atoms) is connected to a battery (a source of electromotive force), the charges are compelled to move; positive charges move in one direction while negative charges move in the opposite direction. This motion of charges creates electric current. It is conventional to take the current flow as the movement of positive charges, that is, opposite to the flow of negative charges, as Fig. 1.3 illustrates. This convention was introduced by Benjamin Franklin (1706–1790), the American scientist and inventor. Although we now know that current in metallic conductors is due to negatively charged electrons, we will follow the universally accepted convention that current is the net flow of positive charges. Thus, 1However, a large power supply capacitor can store up to 0.5 C of charge.
  • 16. CHAPTER 1 Basic Concepts 7 Electric current is the time rate of change of charge, measured in amperes (A). Mathematically, the relationship between current i, charge q, and time t is i = dq dt (1.1) where current is measured in amperes (A), and 1 ampere = 1 coulomb/second The charge transferred between time t0 and t is obtained by integrating both sides of Eq. (1.1). We obtain q = t t0 i dt (1.2) The way we define current as i in Eq. (1.1) suggests that current need not be a constant-valued function. As many of the examples and problems in this chapter and subsequent chapters suggest, there can be several types of current; that is, charge can vary with time in several ways that may be represented by different kinds of mathematical functions. If the current does not change with time, but remains constant, we call it a direct current (dc). A direct current (dc) is a current that remains constant with time. By convention the symbol I is used to represent such a constant current. A time-varying current is represented by the symbol i. A com- mon form of time-varying current is the sinusoidal current or alternating current (ac). An alternating current (ac) is a current that varies sinusoidally with time. Such current is used in your household, to run the air conditioner, refrig- erator, washing machine, and other electric appliances. Figure 1.4 shows direct current and alternating current; these are the two most common types of current. We will consider other types later in the book. I 0 t (a) (b) i t 0 Figure1.4 Two common types of current: (a) direct current (dc), (b) alternating current (ac). Once we define current as the movement of charge, we expect cur- rent to have an associated direction of flow. As mentioned earlier, the directionofcurrentflowisconventionallytakenasthedirectionofpositive charge movement. Based on this convention, a current of 5 A may be represented positively or negatively as shown in Fig. 1.5. In other words, a negative current of −5 A flowing in one direction as shown in Fig. 1.5(b) is the same as a current of +5 A flowing in the opposite direction. 5 A (a) −5 A (b) Figure1.5 Conventional current flow: (a) positive current flow, (b) negative current flow.
  • 17. 8 PART 1 DC Circuits E X A M P L E 1 . 1 How much charge is represented by 4,600 electrons? Solution: Each electron has −1.602 × 10−19 C. Hence 4,600 electrons will have −1.602 × 10−19 C/electron × 4,600 electrons = −7.369 × 10−16 C P R A C T I C E P R O B L E M 1 . 1 Calculate the amount of charge represented by two million protons. Answer: +3.204 × 10−13 C. E X A M P L E 1 . 2 The total charge entering a terminal is given by q = 5t sin 4πt mC. Cal- culate the current at t = 0.5 s. Solution: i = dq dt = d dt (5t sin 4πt) mC/s = (5 sin 4πt + 20πt cos 4πt) mA At t = 0.5, i = 5 sin 2π + 10π cos 2π = 0 + 10π = 31.42 mA P R A C T I C E P R O B L E M 1 . 2 If in Example 1.2, q = (10 − 10e−2t ) mC, find the current at t = 0.5 s. Answer: 7.36 mA. E X A M P L E 1 . 3 Determine the total charge entering a terminal between t = 1 s and t = 2 s if the current passing the terminal is i = (3t2 − t) A. Solution: q = 2 t=1 i dt = 2 1 (3t2 − t) dt = t3 − t2 2 2 1 = (8 − 2) − 1 − 1 2 = 5.5 C P R A C T I C E P R O B L E M 1 . 3 The current flowing through an element is i = 2 A, 0 t 1 2t2 A, t 1 Calculate the charge entering the element from t = 0 to t = 2 s. Answer: 6.667 C.
  • 18. CHAPTER 1 Basic Concepts 9 1.4 VOLTAGE As explained briefly in the previous section, to move the electron in a conductor in a particular direction requires some work or energy transfer. Thisworkisperformedbyanexternalelectromotiveforce(emf), typically represented by the battery in Fig. 1.3. This emf is also known as voltage or potential difference. The voltage vab between two points a and b in an electric circuit is the energy (or work) needed to move a unit charge from a to b; mathematically, vab = dw dq (1.3) where w is energy in joules (J) and q is charge in coulombs (C). The voltage vab or simply v is measured in volts (V), named in honor of the Italian physicist Alessandro Antonio Volta (1745–1827), who invented the first voltaic battery. From Eq. (1.3), it is evident that 1 volt = 1 joule/coulomb = 1 newton meter/coulomb Thus, Voltage (or potential difference) is the energy required to move a unit charge through an element, measured in volts (V). Figure 1.6 shows the voltage across an element (represented by a rectangular block) connected to points a and b. The plus (+) and minus (−) signs are used to define reference direction or voltage polarity. The vab can be interpreted in two ways: (1) point a is at a potential of vab volts higher than point b, or (2) the potential at point a with respect to point b is vab. It follows logically that in general vab = −vba (1.4) For example, in Fig. 1.7, we have two representations of the same vol- tage. In Fig. 1.7(a), point a is +9 V above point b; in Fig. 1.7(b), point b is −9 V above point a. We may say that in Fig. 1.7(a), there is a 9-V voltage drop from a to b or equivalently a 9-V voltage rise from b to a. In other words, a voltage drop from a to b is equivalent to a voltage rise from b to a. a b vab + − Figure1.6 Polarity of voltage vab. 9 V (a) a b + − −9 V (b) a b + − Figure1.7 Two equivalent representations of the same voltage vab: (a) point a is 9 V above point b, (b) point b is −9 V above point a. Current and voltage are the two basic variables in electric circuits. The common term signal is used for an electric quantity such as a current or a voltage (or even electromagnetic wave) when it is used for conveying information. Engineers prefer to call such variables signals rather than mathematical functions of time because of their importance in commu- nications and other disciplines. Like electric current, a constant voltage is called a dc voltage and is represented by V, whereas a sinusoidally time-varying voltage is called an ac voltage and is represented by v. A dc voltage is commonly produced by a battery; ac voltage is produced by an electric generator. Keep in mind that electric current is always through an element and that electric voltage is al- ways across the element or between two points.
  • 19. 10 PART 1 DC Circuits 1.5 POWER AND ENERGY Although current and voltage are the two basic variables in an electric circuit, they are not sufficient by themselves. For practical purposes, we need to know how much power an electric device can handle. We all know from experience that a 100-watt bulb gives more light than a 60-watt bulb. We also know that when we pay our bills to the electric utility companies, we are paying for the electric energy consumed over a certain period of time. Thus power and energy calculations are important in circuit analysis. To relate power and energy to voltage and current, we recall from physics that: Power is the time rate of expending or absorbing energy, measured in watts (W). We write this relationship as p = dw dt (1.5) where p is power in watts (W), w is energy in joules (J), and t is time in seconds (s). From Eqs. (1.1), (1.3), and (1.5), it follows that p = dw dt = dw dq · dq dt = vi (1.6) or p = vi (1.7) The power p in Eq. (1.7) is a time-varying quantity and is called the instantaneouspower. Thus, thepowerabsorbedorsuppliedbyanelement is the product of the voltage across the element and the current through it. If the power has a + sign, power is being delivered to or absorbed by the element. If, on the other hand, the power has a − sign, power is being supplied by the element. But how do we know when the power has a negative or a positive sign? Current direction and voltage polarity play a major role in deter- mining the sign of power. It is therefore important that we pay attention to the relationship between current i and voltage v in Fig. 1.8(a). The vol- tage polarity and current direction must conform with those shown in Fig. 1.8(a) in order for the power to have a positive sign. This is known as the passive sign convention. By the passive sign convention, current en- ters through the positive polarity of the voltage. In this case, p = +vi or vi 0 implies that the element is absorbing power. However, if p = −vi or vi 0, as in Fig. 1.8(b), the element is releasing or supplying power. p = +vi (a) v + − p = −vi (b) v + − i i Figure1.8 Reference polarities for power using the passive sign conven- tion: (a) absorbing power, (b) supplying power. Passive sign convention is satisfied when the current enters through the positive terminal of an element and p = +vi. If the current enters through the negative terminal, p = −vi.
  • 20. CHAPTER 1 Basic Concepts 11 When the voltage and current directions con- form to Fig. 1.8(b), we have the active sign con- vention and p = +vi. Unless otherwise stated, we will follow the passive sign convention throughout this text. For example, the element in both circuits of Fig. 1.9 has an absorbing power of +12 W because a positive current enters the positive terminal in both cases. In Fig. 1.10, however, the element is supplying power of −12 W because a positive current enters the negative terminal. Of course, an absorbing power of +12 W is equivalent to a supplying power of −12 W. In general, Power absorbed = −Power supplied (a) 4 V 3 A (a) + − 3 A 4 V 3 A (b) + − Figure1.9 Two cases of an element with an absorbing power of 12 W: (a) p = 4 × 3 = 12 W, (b) p = 4 × 3 = 12 W. 3 A (a) 4 V 3 A (a) + − 3 A 4 V 3 A (b) + − Figure1.10 Two cases of an element with a supplying power of 12 W: (a) p = 4 × (−3) = −12 W, (b) p = 4 × (−3) = −12 W. In fact, the law of conservation of energy must be obeyed in any electric circuit. For this reason, the algebraic sum of power in a circuit, at any instant of time, must be zero: p = 0 (1.8) This again confirms the fact that the total power supplied to the circuit must balance the total power absorbed. From Eq. (1.6), the energy absorbed or supplied by an element from time t0 to time t is w = t t0 p dt = t t0 vi dt (1.9) Energy is the capacity to do work, measured in joules (J). The electric power utility companies measure energy in watt-hours (Wh), where 1 Wh = 3,600 J E X A M P L E 1 . 4 An energy source forces a constant current of 2 A for 10 s to flow through a lightbulb. If 2.3 kJ is given off in the form of light and heat energy, calculate the voltage drop across the bulb.
  • 21. 12 PART 1 DC Circuits Solution: The total charge is q = it = 2 × 10 = 20 C The voltage drop is v = w q = 2.3 × 103 20 = 115 V P R A C T I C E P R O B L E M 1 . 4 To move charge q from point a to point b requires −30 J. Find the voltage drop vab if: (a) q = 2 C, (b) q = −6 C . Answer: (a) −15 V, (b) 5 V. E X A M P L E 1 . 5 Find the power delivered to an element at t = 3 ms if the current entering its positive terminal is i = 5 cos 60πt A and the voltage is: (a) v = 3i, (b) v = 3 di/dt. Solution: (a) The voltage is v = 3i = 15 cos 60πt; hence, the power is p = vi = 75 cos2 60πt W At t = 3 ms, p = 75 cos2 (60π × 3 × 10−3 ) = 75 cos2 0.18π = 53.48 W (b) We find the voltage and the power as v = 3 di dt = 3(−60π)5 sin 60πt = −900π sin 60πt V p = vi = −4500π sin 60πt cos 60πt W At t = 3 ms, p = −4500π sin 0.18π cos 0.18π W = −14137.167 sin 32.4◦ cos 32.4◦ = −6.396 kW P R A C T I C E P R O B L E M 1 . 5 Find the power delivered to the element in Example 1.5 at t = 5 ms if the current remains the same but the voltage is: (a) v = 2i V, (b) v = 10 + 5 t 0 i dt V. Answer: (a) 17.27 W, (b) 29.7 W.
  • 22. CHAPTER 1 Basic Concepts 13 E X A M P L E 1 . 6 How much energy does a 100-W electric bulb consume in two hours? Solution: w = pt = 100 (W) × 2 (h) × 60 (min/h) × 60 (s/min) = 720,000 J = 720 kJ This is the same as w = pt = 100 W × 2 h = 200 Wh P R A C T I C E P R O B L E M 1 . 6 A stove element draws 15 A when connected to a 120-V line. How long does it take to consume 30 kJ? Answer: 16.67 s. 1.6 CIRCUIT ELEMENTS As we discussed in Section 1.1, an element is the basic building block of a circuit. An electric circuit is simply an interconnection of the elements. Circuit analysis is the process of determining voltages across (or the currents through) the elements of the circuit. There are two types of elements found in electric circuits: passive elements and active elements. An active element is capable of generating energy while a passive element is not. Examples of passive elements are resistors, capacitors, and inductors. Typical active elements include generators, batteries, and operational amplifiers. Our aim in this section is to gain familiarity with some important active elements. The most important active elements are voltage or current sources that generally deliver power to the circuit connected to them. There are two kinds of sources: independent and dependent sources. An ideal independent source is an active element that provides a specified voltage or current that is completely independent of other circuit variables. V (b) + − v (a) + − Figure1.11 Symbols for independent voltage sources: (a) used for constant or time-varying voltage, (b) used for constant voltage (dc). In other words, an ideal independent voltage source delivers to the circuit whatever current is necessary to maintain its terminal voltage. Physical sources such as batteries and generators may be regarded as approxima- tions to ideal voltage sources. Figure 1.11 shows the symbols for inde- pendent voltage sources. Notice that both symbols in Fig. 1.11(a) and (b) can be used to represent a dc voltage source, but only the symbol in Fig. 1.11(a) can be used for a time-varying voltage source. Similarly, an ideal independent current source is an active element that provides a specified current completely independent of the voltage across the source. That is, the current source delivers to the circuit whatever voltage is necessary to
  • 23. 14 PART 1 DC Circuits maintain the designated current. The symbol for an independent current source is displayed in Fig. 1.12, where the arrow indicates the direction of current i. i Figure1.12 Symbol for independent current source. An ideal dependent (or controlled) source is an active element in which the source quantity is controlled by another voltage or current. Dependent sources are usually designated by diamond-shaped symbols, as shown in Fig. 1.13. Since the control of the dependent source is ac- hieved by a voltage or current of some other element in the circuit, and the source can be voltage or current, it follows that there are four possible types of dependent sources, namely: 1. A voltage-controlled voltage source (VCVS). 2. A current-controlled voltage source (CCVS). 3. A voltage-controlled current source (VCCS). 4. A current-controlled current source (CCCS). (a) (b) v + − i Figure1.13 Symbols for: (a) dependent voltage source, (b) dependent current source. Dependent sources are useful in modeling elements such as transistors, operational amplifiers and integrated circuits. An example of a current- controlled voltage source is shown on the right-hand side of Fig. 1.14, where the voltage 10i of the voltage source depends on the current i through element C. Students might be surprised that the value of the dependent voltage source is 10i V (and not 10i A) because it is a voltage source. The key idea to keep in mind is that a voltage source comes with polarities (+ −) in its symbol, while a current source comes with an arrow, irrespective of what it depends on. i A B C 10i 5 V + − + − Figure1.14 The source on the right-hand side is a current-controlled voltage source. It should be noted that an ideal voltage source (dependent or in- dependent) will produce any current required to ensure that the terminal voltage is as stated, whereas an ideal current source will produce the necessary voltage to ensure the stated current flow. Thus an ideal source could in theory supply an infinite amount of energy. It should also be noted that not only do sources supply power to a circuit, they can absorb power from a circuit too. For a voltage source, we know the voltage but not the current supplied or drawn by it. By the same token, we know the current supplied by a current source but not the voltage across it. E X A M P L E 1 . 7 Calculate the power supplied or absorbed by each element in Fig. 1.15. p2 p3 I = 5 A 20 V 6 A 8 V 0.2I 12 V + − + − + − p1 p4 Figure1.15 For Example 1.7. Solution: We apply the sign convention for power shown in Figs. 1.8 and 1.9. For p1, the 5-A current is out of the positive terminal (or into the negative terminal); hence, p1 = 20(−5) = −100 W Supplied power For p2 and p3, the current flows into the positive terminal of the element in each case.
  • 24. CHAPTER 1 Basic Concepts 15 p2 = 12(5) = 60 W Absorbed power p3 = 8(6) = 48 W Absorbed power For p4, we should note that the voltage is 8 V (positive at the top), the same as the voltage for p3, since both the passive element and the dependent source are connected to the same terminals. (Remember that voltage is always measured across an element in a circuit.) Since the current flows out of the positive terminal, p4 = 8(−0.2I) = 8(−0.2 × 5) = −8 W Supplied power We should observe that the 20-V independent voltage source and 0.2I dependent current source are supplying power to the rest of the network, while the two passive elements are absorbing power. Also, p1 + p2 + p3 + p4 = −100 + 60 + 48 − 8 = 0 In agreement with Eq. (1.8), the total power supplied equals the total power absorbed. P R A C T I C E P R O B L E M 1 . 7 Compute the power absorbed or supplied by each component of the circuit in Fig. 1.16. 8 A 5 V 3 V 2 V 3 A I = 5 A 0.6I + − + − + − + − + − p2 p1 p3 p4 Figure1.16 For Practice Prob. 1.7. Answer: p1 = −40 W, p2 = 16 W, p3 = 9 W, p4 = 15 W. †1.7 APPLICATIONS2 In this section, we will consider two practical applications of the concepts developed in this chapter. The first one deals with the TV picture tube and the other with how electric utilities determine your electric bill. 1.7.1 TV Picture Tube One important application of the motion of electrons is found in both the transmission and reception of TV signals. At the transmission end, a TV camera reduces a scene from an optical image to an electrical signal. Scanning is accomplished with a thin beam of electrons in an iconoscope camera tube. At the receiving end, the image is reconstructed by using a cath- ode-ray tube (CRT) located in the TV receiver.3 The CRT is depicted in 2The dagger sign preceding a section heading indicates a section that may be skipped, explained briefly, or assigned as homework. 3Modern TV tubes use a different technology.
  • 25. 16 PART 1 DC Circuits Fig. 1.17. Unlike the iconoscope tube, which produces an electron beam of constant intensity, the CRT beam varies in intensity according to the incoming signal. The electron gun, maintained at a high potential, fires the electron beam. The beam passes through two sets of plates for vertical and horizontal deflections so that the spot on the screen where the beam strikes can move right and left and up and down. When the electron beam strikes the fluorescent screen, it gives off light at that spot. Thus the beam can be made to “paint” a picture on the TV screen. Vertical deflection plates Horizontal deflection plates Electron trajectory Bright spot on fluorescent screen Electron gun Figure1.17 Cathode-ray tube. (Source: D. E. Tilley, Contemporary College Physics [Menlo Park, CA: Benjamin/Cummings, 1979], p. 319.) E X A M P L E 1 . 8 The electron beam in a TV picture tube carries 1015 electrons per second. As a design engineer, determine the voltage Vo needed to accelerate the electron beam to achieve 4 W. Solution: The charge on an electron is e = −1.6 × 10−19 C If the number of electrons is n, then q = ne and i = dq dt = e dn dt = (−1.6 × 10−19 )(1015 ) = −1.6 × 10−4 A The negative sign indicates that the electron flows in a direction opposite to electron flow as shown in Fig. 1.18, which is a simplified diagram of the CRT for the case when the vertical deflection plates carry no charge. The beam power is p = Voi or Vo = p i = 4 1.6 × 10−4 = 25,000 V Thus the required voltage is 25 kV. i q Vo Figure1.18 A simplified diagram of the cathode-ray tube; for Example 1.8. P R A C T I C E P R O B L E M 1 . 8 If an electron beam in a TV picture tube carries 1013 electrons/second and is passing through plates maintained at a potential difference of 30 kV, calculate the power in the beam. Answer: 48 mW.
  • 26. CHAPTER 1 Basic Concepts 17 1.7.2 Electricity Bills The second application deals with how an electric utility company charges their customers. The cost of electricity depends upon the amount of energy consumed in kilowatt-hours (kWh). (Other factors that affect the cost include demand and power factors; we will ignore these for now.) However, even if a consumer uses no energy at all, there is a minimum service charge the customer must pay because it costs money to stay connected to the power line. As energy consumption increases, the cost per kWh drops. It is interesting to note the average monthly consumption of household appliances for a family of five, shown in Table 1.3. TABLE 1.3 Typical average monthly consumption of household appliances. Appliance kWh consumed Appliance kWh consumed Water heater 500 Washing machine 120 Freezer 100 Stove 100 Lighting 100 Dryer 80 Dishwasher 35 Microwave oven 25 Electric iron 15 Personal computer 12 TV 10 Radio 8 Toaster 4 Clock 2 E X A M P L E 1 . 9 A homeowner consumes 3,300 kWh in January. Determine the electricity bill for the month using the following residential rate schedule: Base monthly charge of $12.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 200 kWh per month at 6 cents/kWh. Solution: We calculate the electricity bill as follows. Base monthly charge = $12.00 First 100 kWh @ $0.16/kWh = $16.00 Next 200 kWh @ $0.10/kWh = $20.00 Remaining 100 kWh @ $0.06/kWh = $6.00 Total Charge = $54.00 Average cost = $54 100 + 200 + 100 = 13.5 cents/kWh
  • 27. 18 PART 1 DC Circuits P R A C T I C E P R O B L E M 1 . 9 Referring to the residential rate schedule in Example 1.9, calculate the average cost per kWh if only 400 kWh are consumed in July when the family is on vacation most of the time. Answer: 13.5 cents/kWh. †1.8 PROBLEM SOLVING Although the problems to be solved during one’s career will vary in complexity and magnitude, the basic principles to be followed remain the same. The process outlined here is the one developed by the authors over many years of problem solving with students, for the solution of engineering problems in industry, and for problem solving in research. We will list the steps simply and then elaborate on them. 1. Carefully Define the problem. 2. Present everything you know about the problem. 3. Establish a set of Alternative solutions and determine the one that promises the greatest likelihood of success. 4. Attempt a problem solution. 5. Evaluate the solution and check for accuracy. 6. Has the problem been solved Satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process again. 1. Carefully Define the problem. This may be the most important part of the process, because it becomes the foundation for all the rest of the steps. In general, the presentation of engineering problems is somewhat incomplete. You must do all you can to make sure you understand the problem as thoroughly as the presenter of the problem understands it. Time spent at this point clearly identifying the problem will save you considerable time and frustration later. As a student, you can clarify a problem statement in a textbook by asking your professor to help you understand it better. A problem presented to you in industry may require that you consult several individuals. At this step, it is important to develop questionsthatneedtobeaddressedbeforecontinuingthesolutionprocess. If you have such questions, you need to consult with the appropriate individuals or resources to obtain the answers to those questions. With those answers, you can now refine the problem, and use that refinement as the problem statement for the rest of the solution process. 2. Present everything you know about the problem. You are now ready to write down everything you know about the problem and its possible solutions. This important step will save you time and frustration later.
  • 28. CHAPTER 1 Basic Concepts 19 3. Establish a set of Alternative solutions and determine the one that promises the greatest likelihood of success. Almost every problem will have a number of possible paths that can lead to a solution. It is highly desirable to identify as many of those paths as possible. At this point, you also need to determine what tools are available to you, such as Matlab and other software packages that can greatly reduce effort and increase accuracy. Again, we want to stress that time spent carefully defining the problem and investigating alternative approaches to its solution will pay big dividends later. Evaluating the alternatives and determining which promises the greatest likelihood of success may be difficult but will be well worth the effort. Document this process well since you will want to come back to it if the first approach does not work. 4. Attempt a problem solution. Now is the time to actually begin solving the problem. The process you follow must be well documented in order to present a detailed solution if successful, and to evaluate the process if you are not successful. This detailed evaluation may lead to corrections that can then lead to a successful solution. It can also lead to new alternatives to try. Many times, it is wise to fully set up a solution before putting numbers into equations. This will help in checking your results. 5. Evaluate the solution and check for accuracy. You now thor- oughly evaluate what you have accomplished. Decide if you have an acceptable solution, one that you want to present to your team, boss, or professor. 6. Has the problem been solved Satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process again. Now you need to present your solution or try another alternative. At this point, presenting your solution may bring closure to the process. Often, however, presentation of a solution leads to further refinement of the problem definition, and the process continues. Following this process will eventually lead to a satisfactory conclusion. Now let us look at this process for a student taking an electrical and computer engineering foundations course. (The basic process also applies to almost every engineering course.) Keep in mind that although the steps have been simplified to apply to academic types of problems, the process as stated always needs to be followed. We consider a simple example. Assume that we have been given the following circuit. The instruc- tor asks us to solve for the current flowing through the 8-ohm resistor. 2 Ω 4 Ω 8 Ω 5 V 3 V + − 1. Carefully Define the problem. This is only a simple example, but we can already see that we do not know the polarity on the 3-V source. We have the following options. We can ask the professor what
  • 29. 20 PART 1 DC Circuits the polarity should be. If we cannot ask, then we need to make a decision on what to do next. If we have time to work the problem both ways, we can solve for the current when the 3-V source is plus on top and then plus on the bottom. If we do not have the time to work it both ways, assume a polarity and then carefully document your decision. Let us assume that the professor tells us that the source is plus on the bottom. 2. Present everything you know about the problem. Presenting all that we know about the problem involves labeling the circuit clearly so that we define what we seek. Given the following circuit, solve for i8. 2 Ω 4 Ω 8 Ω 5 V 3 V + − + − i8Ω Wenowcheckwiththeprofessor, ifreasonable, toseeiftheproblem is properly defined. 3. Establish a set of Alternative solutions and determine the one that promises the greatest likelihood of success. There are essentially three techniques that can be used to solve this problem. Later in the text you will see that you can use circuit analysis (using Kirchoff’s laws and Ohm’s law), nodal analysis, and mesh analysis. To solve for i8 using circuit analysis will eventually lead to a solution, but it will likely take more work than either nodal or mesh analysis. To solve for i8 using mesh analysis will require writing two simultaneous equations to find the two loop currents indicated in the following circuit. Using nodal analysis requires solving for only one unknown. This is the easiest approach. 2 Ω 4 Ω 8 Ω 5 V 3 V + − + − i2 i1 i3 + − v1 + − v3 + − v2 Loop 1 Loop 2 v1 Therefore, we will solve for i8 using nodal analysis. 4. Attempt a problem solution. We first write down all of the equations we will need in order to find i8. i8 = i2, i2 = v1 8 , i8 = v1 8 v1 − 5 2 + v1 − 0 8 + v1 + 3 4 = 0
  • 30. CHAPTER 1 Basic Concepts 21 Now we can solve for v1. 8 v1 − 5 2 + v1 − 0 8 + v1 + 3 4 = 0 leads to (4v1 − 20) + (v1) + (2v1 + 6) = 0 7v1 = +14, v1 = +2 V, i8 = v1 8 = 2 8 = 0.25 A 5. Evaluate the solution and check for accuracy. We can now use Kirchoff’s voltage law to check the results. i1 = v1 − 5 2 = 2 − 5 2 = − 3 2 = −1.5 A i2 = i8 = 0.25 A i3 = v1 + 3 4 = 2 + 3 4 = 5 4 = 1.25 A i1 + i2 + i3 = −1.5 + 0.25 + 1.25 = 0 (Checks.) Applying KVL to loop 1, −5 + v1 + v2 = −5 + (−i1 × 2) + (i2 × 8) = −5 + (−(−1.5)2) + (0.25 × 8) = −5 + 3 + 2 = 0 (Checks.) Applying KVL to loop 2, −v2 + v3 − 3 = −(i2 × 8) + (i3 × 4) − 3 = −(0.25 × 8) + (1.25 × 4) − 3 = −2 + 5 − 3 = 0 (Checks.) So we now have a very high degree of confidence in the accuracy of our answer. 6. Has the problem been solved Satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process again. This problem has been solved satisfactorily. The current through the 8-ohm resistor is 0.25 amp flowing down through the 8-ohm resistor. 1.9 SUMMARY 1. An electric circuit consists of electrical elements connected together. 2. The International System of Units (SI) is the international mea- surement language, which enables engineers to communicate their results. From the six principal units, the units of other physical quantities can be derived. 3. Current is the rate of charge flow. i = dq dt
  • 31. 22 PART 1 DC Circuits 4. Voltage is the energy required to move 1 C of charge through an element. v = dw dq 5. Power is the energy supplied or absorbed per unit time. It is also the product of voltage and current. p = dw dt = vi 6. According to the passive sign convention, power assumes a positive sign when the current enters the positive polarity of the voltage across an element. 7. An ideal voltage source produces a specific potential difference across its terminals regardless of what is connected to it. An ideal current source produces a specific current through its terminals regardless of what is connected to it. 8. Voltage and current sources can be dependent or independent. A dependent source is one whose value depends on some other circuit variable. 9. Two areas of application of the concepts covered in this chapter are the TV picture tube and electricity billing procedure. REVIEW QUESTIONS 1.1 One millivolt is one millionth of a volt. (a) True (b) False 1.2 The prefix micro stands for: (a) 106 (b) 103 (c) 10−3 (d) 10−6 1.3 The voltage 2,000,000 V can be expressed in powers of 10 as: (a) 2 mV (b) 2 kV (c) 2 MV (d) 2 GV 1.4 A charge of 2 C flowing past a given point each second is a current of 2 A. (a) True (b) False 1.5 A 4-A current charging a dielectric material will accumulate a charge of 24 C after 6 s. (a) True (b) False 1.6 The unit of current is: (a) Coulomb (b) Ampere (c) Volt (d) Joule 1.7 Voltage is measured in: (a) Watts (b) Amperes (c) Volts (d) Joules per second 1.8 The voltage across a 1.1 kW toaster that produces a current of 10 A is: (a) 11 kV (b) 1100 V (c) 110 V (d) 11 V 1.9 Which of these is not an electrical quantity? (a) charge (b) time (c) voltage (d) current (e) power 1.10 The dependent source in Fig. 1.19 is: (a) voltage-controlled current source (b) voltage-controlled voltage source (c) current-controlled voltage source (d) current-controlled current source vs io 6io + − Figure 1.19 For Review Question 1.10. Answers: 1.1b, 1.2d, 1.3c, 1.4a, 1.5a, 1.6b, 1.7c, 1.8c, 1.9b, 1.10d.
  • 32. CHAPTER 1 Basic Concepts 23 PROBLEMS Section 1.3 Charge and Current 1.1 How many coulombs are represented by these amounts of electrons: (a) 6.482 × 1017 (b) 1.24 × 1018 (c) 2.46 × 1019 (d) 1.628 × 1020 1.2 Find the current flowing through an element if the charge flow is given by: (a) q(t) = (t + 2) mC (b) q(t) = (5t2 + 4t − 3) C (c) q(t) = 10e−4t pC (d) q(t) = 20 cos 50πt nC (e) q(t) = 5e−2t sin 100t µC 1.3 Find the charge q(t) flowing through a device if the current is: (a) i(t) = 3 A, q(0) = 1 C (b) i(t) = (2t + 5) mA, q(0) = 0 (c) i(t) = 20 cos(10t + π/6) µA, q(0) = 2 µC (d) i(t) = 10e−30t sin 40t A, q(0) = 0 1.4 The current flowing through a device is i(t) = 5 sin 6πt A. Calculate the total charge flow through the device from t = 0 to t = 10 ms. 1.5 Determine the total charge flowing into an element for 0 t 2 s when the current entering its positive terminal is i(t) = e−2t mA. 1.6 The charge entering a certain element is shown in Fig. 1.20. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 ms q(t) (mC) t (ms) 0 2 4 6 8 10 12 80 Figure 1.20 For Prob. 1.6. 1.7 The charge flowing in a wire is plotted in Fig. 1.21. Sketch the corresponding current. q (C) t (s) 50 −50 0 2 4 6 8 Figure 1.21 For Prob. 1.7. 1.8 The current flowing past a point in a device is shown in Fig. 1.22. Calculate the total charge through the point. i (mA) t (ms) 0 1 2 10 Figure 1.22 For Prob. 1.8. 1.9 The current through an element is shown in Fig. 1.23. Determine the total charge that passed through the element at: (a) t = 1 s (b) t = 3 s (c) t = 5 s 0 1 2 3 4 5 5 10 i (A) t (s) Figure 1.23 For Prob. 1.9. Sections 1.4 and 1.5 Voltage, Power, and Energy 1.10 A certain electrical element draws the current i(t) = 10 cos 4t A at a voltage v(t) = 120 cos 4t V. Find the energy absorbed by the element in 2 s. 1.11 The voltage v across a device and the current i through it are v(t) = 5 cos 2t V, i(t) = 10(1 − e−0.5t ) A Calculate: (a) the total charge in the device at t = 1 s (b) the power consumed by the device at t = 1 s.
  • 33. 24 PART 1 DC Circuits 1.12 The current entering the positive terminal of a device is i(t) = 3e−2t A and the voltage across the device is v(t) = 5 di/dt V. (a) Find the charge delivered to the device between t = 0 and t = 2 s. (b) Calculate the power absorbed. (c) Determine the energy absorbed in 3 s. 1.13 Figure 1.24 shows the current through and the voltage across a device. Find the total energy absorbed by the device for the period of 0 t 4 s. 0 2 4 50 i (mA) t (s) 10 0 1 3 4 v (V) t (s) Figure 1.24 For Prob. 1.13. Section 1.6 Circuit Elements 1.14 Figure 1.25 shows a circuit with five elements. If p1 = −205 W, p2 = 60 W, p4 = 45 W, p5 = 30 W, calculate the power p3 received or delivered by element 3. 3 1 2 4 5 Figure 1.25 For Prob. 1.14. 1.15 Find the power absorbed by each of the elements in Fig. 1.26. I = 10 A 10 V 30 V 8 V 14 A 20 V 12 V 4 A 0.4I + − + − + − + − + − p2 p1 p3 p4 p5 Figure 1.26 For Prob. 1.15. 1.16 Determine Io in the circuit of Fig. 1.27. 5 A 3 A 20 V 20 V 8 V 12 V 3 A Io + − + − + − + − Figure 1.27 For Prob. 1.16. 1.17 Find Vo in the circuit of Fig. 1.28. 6 A 6 A 1 A 3 A 3 A Vo 5Io Io = 2 A 28 V 12 V + − + − 28 V + − + − 30 V – + + − Figure 1.28 For Prob. 1.17. Section 1.7 Applications 1.18 It takes eight photons to strike the surface of a photodetector in order to emit one electron. If 4 × 1011 photons/second strike the surface of the photodetector, calculate the amount of current flow. 1.19 Find the power rating of the following electrical appliances in your household: (a) Lightbulb (b) Radio set (c) TV set (d) Refrigerator (e) Personal computer (f) PC printer (g) Microwave oven (h) Blender 1.20 A 1.5-kW electric heater is connected to a 120-V source. (a) How much current does the heater draw? (b) If the heater is on for 45 minutes, how much energy is consumed in kilowatt-hours (kWh)? (c) Calculate the cost of operating the heater for 45 minutes if energy costs 10 cents/kWh. 1.21 A 1.2-kW toaster takes roughly 4 minutes to heat four slices of bread. Find the cost of operating the toaster once per day for 1 month (30 days). Assume energy costs 9 cents/kWh.
  • 34. CHAPTER 1 Basic Concepts 25 1.22 A flashlight battery has a rating of 0.8 ampere-hours (Ah) and a lifetime of 10 hours. (a) How much current can it deliver? (b) How much power can it give if its terminal voltage is 6 V? (c) How much energy is stored in the battery in kWh? 1.23 A constant current of 3 A for 4 hours is required to charge an automotive battery. If the terminal voltage is 10 + t/2 V, where t is in hours, (a) how much charge is transported as a result of the charging? (b) how much energy is expended? (c) how much does the charging cost? Assume electricity costs 9 cents/kWh. 1.24 A 30-W incandescent lamp is connected to a 120-V source and is left burning continuously in an otherwise dark staircase. Determine: (a) the current through the lamp, (b) the cost of operating the light for one non-leap year if electricity costs 12 cents per kWh. 1.25 An electric stove with four burners and an oven is used in preparing a meal as follows. Burner 1: 20 minutes Burner 2: 40 minutes Burner 3: 15 minutes Burner 4: 45 minutes Oven: 30 minutes If each burner is rated at 1.2 kW and the oven at 1.8 kW, and electricity costs 12 cents per kWh, calculate the cost of electricity used in preparing the meal. 1.26 PECO (the electric power company in Philadelphia) charged a consumer $34.24 one month for using 215 kWh. If the basic service charge is $5.10, how much did PECO charge per kWh? COMPREHENSIVE PROBLEMS 1.27 A telephone wire has a current of 20 µA flowing through it. How long does it take for a charge of 15 C to pass through the wire? 1.28 A lightning bolt carried a current of 2 kA and lasted for 3 ms. How many coulombs of charge were contained in the lightning bolt? 1.29 The power consumption for a certain household for a day is shown in Fig. 1.29. Determine: (a) the total energy consumed in kWh (b) the average power per hour. 12 2 4 6 8 10 12 noon 2 4 6 8 10 12 p(t) t (hour) 400 W 1000 W 200 W 1200 W 400 W Figure 1.29 For Prob. 1.29. 1.30 The graph in Fig. 1.30 represents the power drawn by an industrial plant between 8:00 and 8:30 A.M. Calculate the total energy in MWh consumed by the plant. 8.00 8.05 8.10 8.15 8.20 8.25 8.30 5 4 3 8 p (MW) t Figure 1.30 For Prob. 1.30. 1.31 A battery may be rated in ampere-hours (Ah). An lead-acid battery is rated at 160 Ah. (a) What is the maximum current it can supply for 40 h? (b) How many days will it last if it is discharged at 1 mA? 1.32 How much work is done by a 12-V automobile battery in moving 5 × 1020 electrons from the positive terminal to the negative terminal? 1.33 How much energy does a 10-hp motor deliver in 30 minutes? Assume that 1 horsepower = 746 W. 1.34 A 2-kW electric iron is connected to a 120-V line. Calculate the current drawn by the iron.
  • 35. 27 C H A P T E R BASIC LAWS 2 The chessboard is the world, the pieces are the phenomena of the universe, the rules of the game are what we call the laws of Nature. The player on the other side is hidden from us, we know that his play is always fair, just, and patient. But also we know, to our cost, that he never overlooks a mistake, or makes the smallest allowance for ignorance. — Thomas Henry Huxley Historical Profiles Georg Simon Ohm (1787–1854), a German physicist, in 1826 experimentally deter- mined the most basic law relating voltage and current for a resistor. Ohm’s work was initially denied by critics. Born of humble beginnings in Erlangen, Bavaria, Ohm threw himself into electrical research. His efforts resulted in his famous law. He was awarded the Copley Medal in 1841 by the Royal Society of London. In 1849, he was given the Professor of Physics chair by the University of Munich. To honor him, the unit of resistance was named the ohm. Gustav Robert Kirchhoff (1824–1887), a German physicist, stated two basic laws in 1847 concerning the relationship between the currents and voltages in an electrical network. Kirchhoff’s laws, along with Ohm’s law, form the basis of circuit theory. Born the son of a lawyer in Konigsberg, East Prussia, Kirchhoff entered the University of Konigsberg at age 18 and later became a lecturer in Berlin. His collaborative work in spectroscopy with German chemist Robert Bunsen led to the discovery of cesium in 1860 and rubidium in 1861. Kirchhoff was also credited with the Kirchhoff law of radiation. Thus Kirchhoff is famous among engineers, chemists, and physicists.
  • 36. 28 PART 1 DC Circuits 2.1 INTRODUCTION Chapter 1 introduced basic concepts such as current, voltage, and power in an electric circuit. To actually determine the values of these variables in a given circuit requires that we understand some fundamental laws that govern electric circuits. These laws, known as Ohm’s law and Kirchhoff’s laws, form the foundation upon which electric circuit analysis is built. In this chapter, in addition to these laws, we shall discuss some techniques commonly applied in circuit design and analysis. These tech- niques include combining resistors in series or parallel, voltage division, current division, and delta-to-wye and wye-to-delta transformations. The application of these laws and techniques will be restricted to resistive cir- cuits in this chapter. We will finally apply the laws and techniques to real-life problems of electrical lighting and the design of dc meters. 2.2 OHM’S LAW Materials in general have a characteristic behavior of resisting the flow of electric charge. This physical property, or ability to resist current, is known as resistance and is represented by the symbol R. The resistance of any material with a uniform cross-sectional area A depends on A and its length , as shown in Fig. 2.1(a). In mathematical form, R = ρ A (2.1) where ρ is known as the resistivity of the material in ohm-meters. Good conductors, such as copper and aluminum, have low resistivities, while insulators, such as mica and paper, have high resistivities. Table 2.1 presents the values of ρ for some common materials and shows which materials are used for conductors, insulators, and semiconductors. l Cross-sectional area A (a) Material with resistivity r v R i + − (b) Figure 2.1 (a) Resistor, (b) Circuit symbol for resistance. TABLE 2.1 Resistivities of common materials. Material Resistivity (·m) Usage Silver 1.64 × 10−8 Conductor Copper 1.72 × 10−8 Conductor Aluminum 2.8 × 10−8 Conductor Gold 2.45 × 10−8 Conductor Carbon 4 × 10−5 Semiconductor Germanium 47 × 10−2 Semiconductor Silicon 6.4 × 102 Semiconductor Paper 1010 Insulator Mica 5 × 1011 Insulator Glass 1012 Insulator Teflon 3 × 1012 Insulator The circuit element used to model the current-resisting behavior of a material is the resistor. For the purpose of constructing circuits, resistors are usually made from metallic alloys and carbon compounds. The circuit
  • 37. CHAPTER 2 Basic Laws 29 symbol for the resistor is shown in Fig. 2.1(b), where R stands for the resistance of the resistor. The resistor is the simplest passive element. Georg Simon Ohm (1787–1854), a German physicist, is credited with finding the relationship between current and voltage for a resistor. This relationship is known as Ohm’s law. Ohm’s law states that the voltage v across a resistor is directly proportional to the current i flowing through the resistor. That is, v ∝ i (2.2) Ohm defined the constant of proportionality for a resistor to be the resis- tance, R. (The resistance is a material property which can change if the internal or external conditions of the element are altered, e.g., if there are changes in the temperature.) Thus, Eq. (2.2) becomes v = iR (2.3) which is the mathematical form of Ohm’s law. R in Eq. (2.3) is measured in the unit of ohms, designated . Thus, The resistance R of an element denotes its ability to resist the flow of electric current; it is measured in ohms (). We may deduce from Eq. (2.3) that R = v i (2.4) so that 1 = 1 V/A To apply Ohm’s law as stated in Eq. (2.3), we must pay careful attention to the current direction and voltage polarity. The direction of current i and the polarity of voltage v must conform with the passive sign convention, as shown in Fig. 2.1(b). This implies that current flows from a higher potential to a lower potential in order for v = iR. If current flows from a lower potential to a higher potential, v = −iR. (a) (b) R = 0 i R = ∞ i = 0 v = 0 + − v + − Figure 2.2 (a) Short circuit (R = 0), (b)Opencircuit(R = ∞). Since the value of R can range from zero to infinity, it is important that we consider the two extreme possible values of R. An element with R = 0 is called a short circuit, as shown in Fig. 2.2(a). For a short circuit, v = iR = 0 (2.5) showing that the voltage is zero but the current could be anything. In practice, a short circuit is usually a connecting wire assumed to be a perfect conductor. Thus,
  • 38. 30 PART 1 DC Circuits A short circuit is a circuit element with resistance approaching zero. Similarly, an element with R = ∞ is known as an open circuit, as shown in Fig. 2.2(b). For an open circuit, i = lim R→∞ v R = 0 (2.6) indicating that the current is zero though the voltage could be anything. Thus, An open circuit is a circuit element with resistance approaching infinity. (a) (b) Figure2.3 Fixed resistors: (a) wire- wound type, (b) carbon film type. (Courtesy of Tech America.) (a) (b) Figure2.4 Circuit symbol for: (a) a variable resistor in general, (b) a potentiometer. A resistor is either fixed or variable. Most resistors are of the fixed type, meaning their resistance remains constant. The two common types of fixed resistors (wirewound and composition) are shown in Fig. 2.3. The composition resistors are used when large resistance is needed. The circuit symbol in Fig. 2.1(b) is for a fixed resistor. Variable resistors have adjustable resistance. The symbol for a variable resistor is shown in Fig. 2.4(a). A common variable resistor is known as a potentiometer or pot for short, with the symbol shown in Fig. 2.4(b). The pot is a three-terminal element with a sliding contact or wiper. By sliding the wiper, the resistances between the wiper terminal and the fixed terminals vary. Like fixed resistors, variable resistors can either be of wirewound or composition type, as shown in Fig. 2.5. Although resistors like those in Figs.2.3and2.5areusedincircuitdesigns, todaymostcircuitcomponents including resistors are either surface mounted or integrated, as typically shown in Fig. 2.6. (a) (b) Figure2.5 Variable resistors: (a) composition type, (b) slider pot. (Courtesy of Tech America.) Figure2.6 Resistors in a thick-film circuit. (Source: G. Daryanani, Principles of Active Network Synthesis and Design [New York: John Wiley, 1976], p. 461c.) It should be pointed out that not all resistors obey Ohm’s law. A resistor that obeys Ohm’s law is known as a linear resistor. It has a con- stant resistance and thus its current-voltage characteristic is as illustrated in Fig. 2.7(a): its i-v graph is a straight line passing through the ori- gin. A nonlinear resistor does not obey Ohm’s law. Its resistance varies with current and its i-v characteristic is typically shown in Fig. 2.7(b).
  • 39. CHAPTER 2 Basic Laws 31 Examples of devices with nonlinear resistance are the lightbulb and the diode. Although all practical resistors may exhibit nonlinear behavior under certain conditions, we will assume in this book that all elements actually designated as resistors are linear. Slope = R (a) v i Slope = R (b) v i Figure 2.7 The i-v characteristic of: (a) a linear resistor, (b) a nonlinear resistor. A useful quantity in circuit analysis is the reciprocal of resistance R, known as conductance and denoted by G: G = 1 R = i v (2.7) The conductance is a measure of how well an element will conduct electric current. The unit of conductance is the mho (ohm spelled back- ward) or reciprocal ohm, with symbol , the inverted omega. Although engineers often use the mhos, in this book we prefer to use the siemens (S), the SI unit of conductance: 1 S = 1 = 1 A/V (2.8) Thus, Conductance is the ability of an element to conduct electric current; it is measured in mhos ( ) or siemens (S). The same resistance can be expressed in ohms or siemens. For example, 10 is the same as 0.1 S. From Eq. (2.7), we may write i = Gv (2.9) The power dissipated by a resistor can be expressed in terms of R. Using Eqs. (1.7) and (2.3), p = vi = i2 R = v2 R (2.10) The power dissipated by a resistor may also be expressed in terms of G as p = vi = v2 G = i2 G (2.11) We should note two things from Eqs. (2.10) and (2.11): 1. The power dissipated in a resistor is a nonlinear function of either current or voltage. 2. Since R and G are positive quantities, the power dissipated in a resistor is always positive. Thus, a resistor always absorbs power from the circuit. This confirms the idea that a resistor is a passive element, incapable of generating energy. E X A M P L E 2 . 1 An electric iron draws 2 A at 120 V. Find its resistance.
  • 40. 32 PART 1 DC Circuits Solution: From Ohm’s law, R = v i = 120 2 = 60 P R A C T I C E P R O B L E M 2 . 1 The essential component of a toaster is an electrical element (a resistor) that converts electrical energy to heat energy. How much current is drawn by a toaster with resistance 12 at 110 V? Answer: 9.167 A. E X A M P L E 2 . 2 In the circuit shown in Fig. 2.8, calculate the current i, the conductance G, and the power p. 30 V i + − 5 kΩ v + − Figure2.8 For Example 2.2. Solution: The voltage across the resistor is the same as the source voltage (30 V) because the resistor and the voltage source are connected to the same pair of terminals. Hence, the current is i = v R = 30 5 × 103 = 6 mA The conductance is G = 1 R = 1 5 × 103 = 0.2 mS We can calculate the power in various ways using either Eqs. (1.7), (2.10), or (2.11). p = vi = 30(6 × 10−3 ) = 180 mW or p = i2 R = (6 × 10−3 )2 5 × 103 = 180 mW or p = v2 G = (30)2 0.2 × 10−3 = 180 mW P R A C T I C E P R O B L E M 2 . 2 For the circuit shown in Fig. 2.9, calculate the voltage v, the conductance G, and the power p. 2 mA i 10 kΩ v + − Figure2.9 For Practice Prob. 2.2 Answer: 20 V, 100 µS, 40 mW.
  • 41. CHAPTER 2 Basic Laws 33 E X A M P L E 2 . 3 A voltage source of 20 sin πt V is connected across a 5-k resistor. Find the current through the resistor and the power dissipated. Solution: i = v R = 20 sin πt 5 × 103 = 4 sin πt mA Hence, p = vi = 80 sin2 πt mW P R A C T I C E P R O B L E M 2 . 3 A resistor absorbs an instantaneous power of 20 cos2 t mW when con- nected to a voltage source v = 10 cos t V. Find i and R. Answer: 2 cos t mA, 5 k. †2.3 NODES, BRANCHES, AND LOOPS Since the elements of an electric circuit can be interconnected in several ways, weneedtounderstandsomebasicconceptsofnetworktopology. To differentiate between a circuit and a network, we may regard a network as an interconnection of elements or devices, whereas a circuit is a network providing one or more closed paths. The convention, when addressing network topology, is to use the word network rather than circuit. We do this even though the words network and circuit mean the same thing when used in this context. In network topology, we study the properties relating to the placement of elements in the network and the geometric configuration of the network. Such elements include branches, nodes, and loops. A branch represents a single element such as a voltage source or a resistor. In other words, a branch represents any two-terminal element. The circuit in Fig. 2.10 has five branches, namely, the 10-V voltage source, the 2-A current source, and the three resistors. 10 V 2 A a b c 5 Ω + − 2 Ω 3 Ω Figure2.10 Nodes, branches, and loops. b c a 10 V 5 Ω 2 Ω 3 Ω 2 A + − Figure 2.11 The three-node circuit of Fig. 2.10 is redrawn. A node is the point of connection between two or more branches. A node is usually indicated by a dot in a circuit. If a short circuit (a connecting wire) connects two nodes, the two nodes constitute a single node. The circuit in Fig. 2.10 has three nodes a, b, and c. Notice that the three points that form node b are connected by perfectly conducting wires and therefore constitute a single point. The same is true of the four points forming node c. We demonstrate that the circuit in Fig. 2.10 has only three nodes by redrawing the circuit in Fig. 2.11. The two circuits in
  • 42. 34 PART 1 DC Circuits Figs. 2.10 and 2.11 are identical. However, for the sake of clarity, nodes b and c are spread out with perfect conductors as in Fig. 2.10. A loop is any closed path in a circuit. A loop is a closed path formed by starting at a node, passing through a set of nodes, and returning to the starting node without passing through any node more than once. A loop is said to be independent if it contains a branch which is not in any other loop. Independent loops or paths result in independent sets of equations. For example, the closed path abca containing the 2- resistor in Fig. 2.11 is a loop. Another loop is the closed path bcb containing the 3- resistor and the current source. Although one can identify six loops in Fig. 2.11, only three of them are independent. A network with b branches, n nodes, and l independent loops will satisfy the fundamental theorem of network topology: b = l + n − 1 (2.12) As the next two definitions show, circuit topology is of great value to the study of voltages and currents in an electric circuit. Two or more elements are in series if they are cascaded or connected sequentially and consequently carry the same current. Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them. Elements are in series when they are chain-connected or connected se- quentially, end to end. For example, two elements are in series if they share one common node and no other element is connected to that com- mon node. Elements in parallel are connected to the same pair of termi- nals. Elements may be connected in a way that they are neither in series nor in parallel. In the circuit shown in Fig. 2.10, the voltage source and the 5- resistor are in series because the same current will flow through them. The 2- resistor, the 3- resistor, and the current source are in parallel because they are connected to the same two nodes (b and c) and consequently have the same voltage across them. The 5- and 2- resistors are neither in series nor in parallel with each other. E X A M P L E 2 . 4 Determine the number of branches and nodes in the circuit shown in Fig. 2.12. Identify which elements are in series and which are in parallel. Solution: Since there are four elements in the circuit, the circuit has four branches: 10 V, 5 , 6 , and 2 A. The circuit has three nodes as identified in
  • 43. CHAPTER 2 Basic Laws 35 Fig. 2.13. The 5- resistor is in series with the 10-V voltage source because the same current would flow in both. The 6- resistor is in parallel with the 2-A current source because both are connected to the same nodes 2 and 3. 5 Ω 6 Ω 2 A 10 V + − Figure2.12 For Example 2.4. 1 2 5 Ω 6 Ω 2 A 10 V + − 3 Figure2.13 The three nodes in the circuit of Fig. 2.12. P R A C T I C E P R O B L E M 2 . 4 How many branches and nodes does the circuit in Fig. 2.14 have? Identify the elements that are in series and in parallel. Answer: Five branches and three nodes are identified in Fig. 2.15. The 1- and 2- resistors are in parallel. The 4- resistor and 10-V source are also in parallel. 5 Ω 1 Ω 2 Ω 4 Ω 10 V + − Figure2.14 For Practice Prob. 2.4. 3 Ω 3 1 Ω 2 Ω 4 Ω 10 V + − 1 2 Figure2.15 Answer for Practice Prob. 2.4. 2.4 KIRCHHOFF’S LAWS Ohm’s law by itself is not sufficient to analyze circuits. However, when it is coupled with Kirchhoff’s two laws, we have a sufficient, powerful set of tools for analyzing a large variety of electric circuits. Kirchhoff’s laws were first introduced in 1847 by the German physicist Gustav Robert Kirchhoff (1824–1887). These laws are formally known as Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL). Kirchhoff’s first law is based on the law of conservation of charge, which requires that the algebraic sum of charges within a system cannot change.
  • 44. 36 PART 1 DC Circuits Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero. Mathematically, KCL implies that N n=1 in = 0 (2.13) where N is the number of branches connected to the node and in is the nth current entering (or leaving) the node. By this law, currents entering a node may be regarded as positive, while currents leaving the node may be taken as negative or vice versa. To prove KCL, assume a set of currents ik(t), k = 1, 2, . . . , flow into a node. The algebraic sum of currents at the node is iT (t) = i1(t) + i2(t) + i3(t) + · · · (2.14) Integrating both sides of Eq. (2.14) gives qT (t) = q1(t) + q2(t) + q3(t) + · · · (2.15) where qk(t) = ik(t) dt and qT (t) = iT (t) dt. But the law of conser- vation of electric charge requires that the algebraic sum of electric charges at the node must not change; that is, the node stores no net charge. Thus qT (t) = 0 → iT (t) = 0, confirming the validity of KCL. i1 i5 i4 i3 i2 Figure2.16 Currents at a node illustrating KCL. Consider the node in Fig. 2.16. Applying KCL gives i1 + (−i2) + i3 + i4 + (−i5) = 0 (2.16) since currents i1, i3, and i4 are entering the node, while currents i2 and i5 are leaving it. By rearranging the terms, we get i1 + i3 + i4 = i2 + i5 (2.17) Equation (2.17) is an alternative form of KCL: The sum of the currents entering a node is equal to the sum of the currents leaving the node. Note that KCL also applies to a closed boundary. This may be regarded as a generalized case, because a node may be regarded as a closed surface shrunk to a point. In two dimensions, a closed boundary is the same as a closed path. As typically illustrated in the circuit of Fig. 2.17, the total current entering the closed surface is equal to the total current leaving the surface. Closed boundary Figure 2.17 Applying KCL to a closed boundary. Twosources(orcircuitsingeneral)aresaidtobe equivalent if they have the same i-v relationship at a pair of terminals. A simple application of KCL is combining current sources in par- allel. The combined current is the algebraic sum of the current supplied by the individual sources. For example, the current sources shown in Fig. 2.18(a) can be combined as in Fig. 2.18(b). The combined or equivalent current source can be found by applying KCL to node a. IT + I2 = I1 + I3
  • 45. CHAPTER 2 Basic Laws 37 or IT = I1 − I2 + I3 (2.18) A circuit cannot contain two different currents, I1 and I2, in series, unless I1 = I2; otherwise KCL will be violated. Kirchhoff’s second law is based on the principle of conservation of energy: a (a) (b) I1 I2 I3 b a IS = I1 – I2 + I3 b IT IT Figure2.18 Current sources in parallel: (a) original circuit, (b) equivalent circuit. Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero. Expressed mathematically, KVL states that M m=1 vm = 0 (2.19) where M is the number of voltages in the loop (or the number of branches in the loop) and vm is the mth voltage. KVLcanbeappliedintwoways: bytakingeithera clockwiseoracounterclockwisetriparoundthe loop. Either way, the algebraic sum of voltages around the loop is zero. To illustrate KVL, consider the circuit in Fig. 2.19. The sign on each voltage is the polarity of the terminal encountered first as we travel around the loop. We can start with any branch and go around the loop either clockwise or counterclockwise. Suppose we start with the voltage source and go clockwise around the loop as shown; then voltages would be −v1, +v2, +v3, −v4, and +v5, in that order. For example, as we reach branch 3, the positive terminal is met first; hence we have +v3. For branch 4, we reach the negative terminal first; hence, −v4. Thus, KVL yields −v1 + v2 + v3 − v4 + v5 = 0 (2.20) Rearranging terms gives v2 + v3 + v5 = v1 + v4 (2.21) which may be interpreted as Sum of voltage drops = Sum of voltage rises (2.22) This is an alternative form of KVL. Notice that if we had traveled coun- terclockwise, the result would have been +v1, −v5, +v4, −v3, and −v2, which is the same as before except that the signs are reversed. Hence, Eqs. (2.20) and (2.21) remain the same. v4 v1 + − + − v3 v2 v5 + − + − + − Figure 2.19 A single-loop circuit illustrating KVL. When voltage sources are connected in series, KVL can be applied to obtain the total voltage. The combined voltage is the algebraic sum of the voltages of the individual sources. For example, for the voltage sources shown in Fig. 2.20(a), the combined or equivalent voltage source in Fig. 2.20(b) is obtained by applying KVL. −Vab + V1 + V2 − V3 = 0
  • 46. 38 PART 1 DC Circuits or Vab = V1 + V2 − V3 (2.23) To avoid violating KVL, a circuit cannot contain two different voltages V1 and V2 in parallel unless V1 = V2. V1 V2 V3 a b (a) VS = V1 + V2 − V3 a b (b) + − + − + − Vab + − Vab + − + − Figure 2.20 Voltage sources in series: (a) original circuit, (b) equivalent circuit. E X A M P L E 2 . 5 For the circuit in Fig. 2.21(a), find voltages v1 and v2. (a) 20 V + − 3 Ω v2 2 Ω v1 + − + − (b) 20 V + − 3 Ω v2 2 Ω v1 + − + − i Figure2.21 For Example 2.5. Solution: To find v1 and v2, we apply Ohm’s law and Kirchhoff’s voltage law. Assume that current i flows through the loop as shown in Fig. 2.21(b). From Ohm’s law, v1 = 2i, v2 = −3i (2.5.1) Applying KVL around the loop gives −20 + v1 − v2 = 0 (2.5.2) Substituting Eq. (2.5.1) into Eq. (2.5.2), we obtain −20 + 2i + 3i = 0 or 5i = 20 ⇒ i = 4 A Substituting i in Eq. (2.5.1) finally gives v1 = 8 V, v2 = −12 V
  • 47. CHAPTER 2 Basic Laws 39 P R A C T I C E P R O B L E M 2 . 5 Find v1 and v2 in the circuit of Fig. 2.22. 10 V + − 8 V + − 4 Ω v1 2 Ω v2 + − + − Figure2.22 For Practice Prob. 2.5 Answer: 12 V, −6 V. E X A M P L E 2 . 6 Determine vo and i in the circuit shown in Fig. 2.23(a). 4 Ω (a) 12 V 2vo i 4 V i + − + − + − 4 Ω (b) 12 V 2vo 4 V + − + − + − 6 Ω vo 6 Ω vo + − + − Figure2.23 For Example 2.6. Solution: We apply KVL around the loop as shown in Fig. 2.23(b). The result is −12 + 4i + 2vo − 4 + 6i = 0 (2.6.1) Applying Ohm’s law to the 6- resistor gives vo = −6i (2.6.2) Substituting Eq. (2.6.2) into Eq. (2.6.1) yields −16 + 10i − 12i = 0 ⇒ i = −8 A and vo = 48 V. P R A C T I C E P R O B L E M 2 . 6 Find vx and vo in the circuit of Fig. 2.24. 35 V 2vx + − + − 10 Ω vx 5 Ω vo + − + − Figure2.24 For Practice Prob. 2.6. Answer: 10 V, −5 V.
  • 48. 40 PART 1 DC Circuits E X A M P L E 2 . 7 Find current io and voltage vo in the circuit shown in Fig. 2.25. a 0.5io 3 A io 4 Ω vo + − Figure2.25 For Example 2.7. Solution: Applying KCL to node a, we obtain 3 + 0.5io = io ⇒ io = 6 A For the 4- resistor, Ohm’s law gives vo = 4io = 24 V P R A C T I C E P R O B L E M 2 . 7 Find vo and io in the circuit of Fig. 2.26. io 4 6 A io 2 Ω 8 Ω vo + − Figure2.26 For Practice Prob. 2.7. Answer: 8 V, 4 A. E X A M P L E 2 . 8 Find the currents and voltages in the circuit shown in Fig. 2.27(a). 8 Ω 30 V + − (a) v1 i2 i3 i1 a 6 Ω v3 3 Ω v2 + − + − + − 8 Ω 30 V + − (b) v1 i2 i3 i1 a 6 Ω v3 3 Ω v2 + − + − + − Loop 2 Loop 1 Figure2.27 For Example 2.8. Solution: We apply Ohm’s law and Kirchhoff’s laws. By Ohm’s law, v1 = 8i1, v2 = 3i2, v3 = 6i3 (2.8.1) Since the voltage and current of each resistor are related by Ohm’s law as shown, we are really looking for three things: (v1, v2, v3) or (i1, i2, i3). At node a, KCL gives i1 − i2 − i3 = 0 (2.8.2) Applying KVL to loop 1 as in Fig. 2.27(b), −30 + v1 + v2 = 0
  • 49. CHAPTER 2 Basic Laws 41 We express this in terms of i1 and i2 as in Eq. (2.8.1) to obtain −30 + 8i1 + 3i2 = 0 or i1 = (30 − 3i2) 8 (2.8.3) Applying KVL to loop 2, −v2 + v3 = 0 ⇒ v3 = v2 (2.8.4) as expected since the two resistors are in parallel. We express v1 and v2 in terms of i1 and i2 as in Eq. (2.8.1). Equation (2.8.4) becomes 6i3 = 3i2 ⇒ i3 = i2 2 (2.8.5) Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives 30 − 3i2 8 − i2 − i2 2 = 0 or i2 = 2 A. From the value of i2, we now use Eqs. (2.8.1) to (2.8.5) to obtain i1 = 3 A, i3 = 1 A, v1 = 24 V, v2 = 6 V, v3 = 6 V P R A C T I C E P R O B L E M 2 . 8 Find the currents and voltages in the circuit shown in Fig. 2.28. 5 V 3 V + − i2 i3 i1 8 Ω v2 + − 2 Ω v1 4 Ω v3 + − + − + − Figure2.28 For Practice Prob. 2.8. Answer: v1 = 3 V, v2 = 2 V, v3 = 5 V, i1 = 1.5 A, i2 = 0.25 A, i3 =1.25 A. 2.5 SERIES RESISTORS AND VOLTAGE DIVISION v + − R1 v1 R2 v2 i + − + − a b Figure2.29 A single-loop circuit with two resistors in series. The need to combine resistors in series or in parallel occurs so frequently that it warrants special attention. The process of combining the resistors is facilitated by combining two of them at a time. With this in mind, consider the single-loop circuit of Fig. 2.29. The two resistors are in series, since the same current i flows in both of them. Applying Ohm’s law to each of the resistors, we obtain v1 = iR1, v2 = iR2 (2.24) If we apply KVL to the loop (moving in the clockwise direction), we have −v + v1 + v2 = 0 (2.25)
  • 50. 42 PART 1 DC Circuits Combining Eqs. (2.24) and (2.25), we get v = v1 + v2 = i(R1 + R2) (2.26) or i = v R1 + R2 (2.27) Notice that Eq. (2.26) can be written as v = iReq (2.28) implying that the two resistors can be replaced by an equivalent resistor Req; that is, Req = R1 + R2 (2.29) Thus, Fig. 2.29 can be replaced by the equivalent circuit in Fig. 2.30. The two circuits in Figs. 2.29 and 2.30 are equivalent because they exhibit the same voltage-current relationships at the terminals a-b. An equivalent circuit such as the one in Fig. 2.30 is useful in simplifying the analysis of a circuit. In general, v Req v + − i + − a b Figure2.30 Equivalent circuit of the Fig. 2.29 circuit. The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances. Resistors in series behave as a single resistor whose resistance is equal to the sum of the re- sistances of the individual resistors. For N resistors in series then, Req = R1 + R2 + · · · + RN = N n=1 Rn (2.30) To determine the voltage across each resistor in Fig. 2.29, we sub- stitute Eq. (2.26) into Eq. (2.24) and obtain v1 = R1 R1 + R2 v, v2 = R2 R1 + R2 v (2.31) Notice that the source voltage v is divided among the resistors in direct proportion to their resistances; the larger the resistance, the larger the voltage drop. This is called the principle of voltage division, and the circuit in Fig. 2.29 is called a voltage divider. In general, if a voltage divider has N resistors (R1, R2, . . . , RN ) in series with the source voltage v, the nth resistor (Rn) will have a voltage drop of vn = Rn R1 + R2 + · · · + RN v (2.32) 2.6 PARALLEL RESISTORS AND CURRENT DIVISION Consider the circuit in Fig. 2.31, where two resistors are connected in parallel and therefore have the same voltage across them. From Ohm’s law, v = i1R1 = i2R2
  • 51. CHAPTER 2 Basic Laws 43 or i1 = v R1 , i2 = v R2 (2.33) Applying KCL at node a gives the total current i as i = i1 + i2 (2.34) Substituting Eq. (2.33) into Eq. (2.34), we get i = v R1 + v R2 = v 1 R1 + 1 R2 = v Req (2.35) where Req is the equivalent resistance of the resistors in parallel: 1 Req = 1 R1 + 1 R2 (2.36) or 1 Req = R1 + R2 R1R2 or Req = R1R2 R1 + R2 (2.37) Thus, The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum. It must be emphasized that this applies only to two resistors in parallel. From Eq. (2.37), if R1 = R2, then Req = R1/2. Node b Node a v + − R1 R2 i1 i2 i Figure2.31 Two resistors in parallel. We can extend the result in Eq. (2.36) to the general case of a circuit with N resistors in parallel. The equivalent resistance is 1 Req = 1 R1 + 1 R2 + · · · + 1 RN (2.38) Note that Req is always smaller than the resistance of the smallest resistor in the parallel combination. If R1 = R2 = · · · = RN = R, then Req = R N (2.39) For example, if four 100- resistors are connected in parallel, their equiv- alent resistance is 25 . Conductances in parallel behave as a single con- ductance whose value is equal to the sum of the individual conductances. It is often more convenient to use conductance rather than resistance when dealing with resistors in parallel. From Eq. (2.38), the equivalent conductance for N resistors in parallel is Geq = G1 + G2 + G3 + · · · + GN (2.40) where Geq = 1/Req, G1 = 1/R1, G2 = 1/R2, G3 = 1/R3, . . . , GN = 1/RN . Equation (2.40) states:
  • 52. 44 PART 1 DC Circuits The equivalent conductance of resistors connected in parallel is the sum of their individual conductances. This means that we may replace the circuit in Fig. 2.31 with that in Fig. 2.32. Notice the similarity between Eqs. (2.30) and (2.40). The equivalent conductance of parallel resistors is obtained the same way as the equivalent resistance of series resistors. In the same manner, the equivalent conductance of resistors in series is obtained just the same way as the resistance of resistors in parallel. Thus the equivalent conductance Geq of N resistors in series (such as shown in Fig. 2.29) is 1 Geq = 1 G1 + 1 G2 + 1 G3 + · · · + 1 GN (2.41) b a v + − Req or Geq v i Figure 2.32 Equivalent circuit to Fig. 2.31. Given the total current i entering node a in Fig. 2.31, how do we obtain current i1 and i2? We know that the equivalent resistor has the same voltage, or v = iReq = iR1R2 R1 + R2 (2.42) Combining Eqs. (2.33) and (2.42) results in i1 = R2 i R1 + R2 , i2 = R1 i R1 + R2 (2.43) which shows that the total current i is shared by the resistors in inverse proportion to their resistances. This is known as the principle of current division, and the circuit in Fig. 2.31 is known as a current divider. Notice that the larger current flows through the smaller resistance. As an extreme case, suppose one of the resistors in Fig. 2.31 is zero, say R2 = 0; that is, R2 is a short circuit, as shown in Fig. 2.33(a). From Eq. (2.43), R2 = 0 implies that i1 = 0, i2 = i. This means that the entire current i bypasses R1 and flows through the short circuit R2 = 0, the path of least resistance. Thus when a circuit is short circuited, as shown in Fig. 2.33(a), two things should be kept in mind: R2 = 0 (a) R1 i i1 = 0 i2 = i R2 = ∞ (b) R1 i i1 = i i2 = 0 Figure 2.33 (a) A shorted circuit, (b) an open circuit. 1. The equivalent resistance Req = 0. [See what happens when R2 = 0 in Eq. (2.37).] 2. The entire current flows through the short circuit. As another extreme case, suppose R2 = ∞, that is, R2 is an open circuit, as shown in Fig. 2.33(b). The current still flows through the path of least resistance, R1. By taking the limit of Eq. (2.37) as R2 → ∞, we obtain Req = R1 in this case. If we divide both the numerator and denominator by R1R2, Eq. (2.43) becomes i1 = G1 G1 + G2 i (2.44a) i2 = G2 G1 + G2 i (2.44b)
  • 53. CHAPTER 2 Basic Laws 45 Thus, in general, if a current divider has N conductors (G1, G2, . . . , GN ) in parallel with the source current i, the nth conductor (Gn) will have current in = Gn G1 + G2 + · · · + GN i (2.45) In general, it is often convenient and possible to combine resistors in series and parallel and reduce a resistive network to a single equivalent resistance Req. Such an equivalent resistance is the resistance between the designated terminals of the network and must exhibit the same i-v characteristics as the original network at the terminals. E X A M P L E 2 . 9 Find Req for the circuit shown in Fig. 2.34. 2 Ω 5 Ω Req 4 Ω 8 Ω 1 Ω 6 Ω 3 Ω Figure2.34 For Example 2.9. Solution: 6 Ω Req 4 Ω (a) 8 Ω 2 Ω 2 Ω 2.4 Ω Req 4 Ω (b) 8 Ω Figure 2.35 Equivalent circuits for Example 2.9. To get Req, we combine resistors in series and in parallel. The 6- and 3- resistors are in parallel, so their equivalent resistance is 6 3 = 6 × 3 6 + 3 = 2 (The symbol is used to indicate a parallel combination.) Also, the 1- and 5- resistors are in series; hence their equivalent resistance is 1 + 5 = 6 Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig. 2.35(a), we notice that the two 2- resistors are in series, so the equivalent resistance is 2 + 2 = 4 This 4- resistor is now in parallel with the 6- resistor in Fig. 2.35(a); their equivalent resistance is 4 6 = 4 × 6 4 + 6 = 2.4 The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig. 2.35(b), the three resistors are in series. Hence, the equivalent resistance for the circuit is Req = 4 + 2.4 + 8 = 14.4 P R A C T I C E P R O B L E M 2 . 9 By combining the resistors in Fig. 2.36, find Req. 5 Ω 4 Ω 6 Ω Req 2 Ω 1 Ω 3 Ω 4 Ω 3 Ω Figure2.36 For Practice Prob. 2.9. Answer: 6 .
  • 54. 46 PART 1 DC Circuits E X A M P L E 2 . 1 0 Calculate the equivalent resistance Rab in the circuit in Fig. 2.37. a b b b c d 6 Ω 12 Ω 5 Ω 4 Ω 10 Ω 1 Ω 1 Ω Rab 3 Ω Figure2.37 For Example 2.10. Solution: The 3- and 6- resistors are in parallel because they are connected to the same two nodes c and b. Their combined resistance is 3 6 = 3 × 6 3 + 6 = 2 (2.10.1) Similarly, the 12- and 4- resistors are in parallel since they are con- nected to the same two nodes d and b. Hence 12 4 = 12 × 4 12 + 4 = 3 (2.10.2) Also the 1- and 5- resistors are in series; hence, their equivalent resistance is 1 + 5 = 6 (2.10.3) With these three combinations, we can replace the circuit in Fig. 2.37 with that in Fig. 2.38(a). In Fig. 2.38(a), 3- in parallel with 6- gives 2-, as calculated in Eq. (2.10.1). This 2- equivalent resistance is now in series with the 1- resistance to give a combined resistance of 1 +2 = 3 . Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In Fig. 2.38(b), we combine the 2- and 3- resistors in parallel to get (a) b b d b c 3 Ω 6 Ω 2 Ω 10 Ω 1 Ω a b (b) b b c 3 Ω 2 Ω 10 Ω a b Figure 2.38 Equivalent circuits for Example 2.10. 2 3 = 2 × 3 2 + 3 = 1.2 This 1.2- resistor is in series with the 10- resistor, so that Rab = 10 + 1.2 = 11.2 P R A C T I C E P R O B L E M 2 . 1 0 Find Rab for the circuit in Fig. 2.39. 1 Ω 9 Ω 18 Ω 20 Ω 20 Ω 2 Ω 5 Ω 8 Ω a b Rab Figure2.39 For Practice Prob. 2.10. Answer: 11 .
  • 55. CHAPTER 2 Basic Laws 47 E X A M P L E 2 . 1 1 Find the equivalent conductance Geq for the circuit in Fig. 2.40(a). 12 S 8 S 6 S (a) 5 S Geq 20 S 6 S (b) 5 S Geq (c) Req Ω 1 5 Ω 1 6 Ω 1 8 Ω 1 12 Figure2.40 For Example 2.11: (a) original circuit, (b) its equivalent circuit, (c) same circuit as in (a) but resistors are expressed in ohms. Solution: The 8-S and 12-S resistors are in parallel, so their conductance is 8 S + 12 S = 20 S This 20-S resistor is now in series with 5 S as shown in Fig. 2.40(b) so that the combined conductance is 20 × 5 20 + 5 = 4 S This is in parallel with the 6-S resistor. Hence Geq = 6 + 4 = 10 S We should note that the circuit in Fig. 2.40(a) is the same as that in Fig. 2.40(c). While the resistors in Fig. 2.40(a) are expressed in siemens, they are expressed in ohms in Fig. 2.40(c). To show that the circuits are the same, we find Req for the circuit in Fig. 2.40(c). Req = 1 6 1 5 + 1 8 1 12 = 1 6 1 5 + 1 20 = 1 6 1 4 = 1 6 × 1 4 1 6 + 1 4 = 1 10 Geq = 1 Req = 10 S This is the same as we obtained previously. P R A C T I C E P R O B L E M 2 . 1 1 Calculate Geq in the circuit of Fig. 2.41. 4 S 6 S 8 S 2 S 12 Ω Geq Figure2.41 For Practice Prob. 2.11. Answer: 4 S. E X A M P L E 2 . 1 2 Find io and vo in the circuit shown in Fig. 2.42(a). Calculate the power dissipated in the 3- resistor. Solution: The 6- and 3- resistors are in parallel, so their combined resistance is 6 3 = 6 × 3 6 + 3 = 2
  • 56. 48 PART 1 DC Circuits Thus our circuit reduces to that shown in Fig. 2.42(b). Notice that vo is not affected by the combination of the resistors because the resistors are in parallel and therefore have the same voltage vo. From Fig. 2.42(b), we can obtain vo in two ways. One way is to apply Ohm’s law to get i = 12 4 + 2 = 2 A and hence, vo = 2i = 2 × 2 = 4 V. Another way is to apply voltage division, since the 12 V in Fig. 2.42(b) is divided between the 4- and 2- resistors. Hence, vo = 2 2 + 4 (12 V) = 4 V a b (a) 12 V 4 Ω i io 6 Ω 3 Ω vo + − a b (b) 12 V 4 Ω i + − 2 Ω vo + − + − Figure2.42 For Example 2.12: (a) original circuit, (b) its equivalent circuit. Similarly, io can be obtained in two ways. One approach is to apply Ohm’s law to the 3- resistor in Fig. 2.42(a) now that we know vo; thus, vo = 3io = 4 ⇒ io = 4 3 A Another approach is to apply current division to the circuit in Fig. 2.42(a) now that we know i, by writing io = 6 6 + 3 i = 2 3 (2 A) = 4 3 A The power dissipated in the 3- resistor is po = voio = 4 4 3 = 5.333 W P R A C T I C E P R O B L E M 2 . 1 2 Find v1 and v2 in the circuit shown in Fig. 2.43. Also calculate i1 and i2 and the power dissipated in the 12- and 40- resistors. 15 V i1 + − 40 Ω v2 + − 10 Ω 12 Ω v1 6 Ω i2 + − Figure2.43 For Practice Prob. 2.12. Answer: v1 = 5 V, i1 = 416.7 mA, p1 = 2.083 W, v2 = 10 V, i2 = 250 mA, p2 = 2.5 W. E X A M P L E 2 . 1 3 For the circuit shown in Fig. 2.44(a), determine: (a) the voltage vo, (b) the power supplied by the current source, (c) the power absorbed by each resistor. Solution: (a) The 6-k and 12-k resistors are in series so that their combined value is 6 + 12 = 18 k. Thus the circuit in Fig. 2.44(a) reduces to that
  • 57. CHAPTER 2 Basic Laws 49 shown in Fig. 2.44(b). We now apply the current division technique to find i1 and i2. i1 = 18,000 9000 + 18,000 (30 mA) = 20 mA i2 = 9000 9000 + 18,000 (30 A) = 10 mA Notice that the voltage across the 9-k and 18-k resistors is the same, and vo = 9,000i1 = 18,000i2 = 180 V, as expected. (a) 30 mA 9 kΩ vo + − 12 kΩ 6 kΩ (b) 30 mA 9 kΩ vo + − 18 kΩ i1 io i2 Figure 2.44 For Example 2.13: (a) original circuit, (b) its equivalent circuit. (b) Power supplied by the source is po = voio = 180(30) mW = 5.4 W (c) Power absorbed by the 12-k resistor is p = iv = i2(i2R) = i2 2 R = (10 × 10−3 )2 (12,000) = 1.2 W Power absorbed by the 6-k resistor is p = i2 2 R = (10 × 10−3 )2 (6000) = 0.6 W Power absorbed by the 9-k resistor is p = v2 o R = (180)2 9000 = 3.6 W or p = voi1 = 180(20) mW = 3.6 W Notice that the power supplied (5.4 W) equals the power absorbed (1.2+ 0.6 + 3.6 = 5.4 W). This is one way of checking results. P R A C T I C E P R O B L E M 2 . 1 3 For the circuit shown in Fig. 2.45, find: (a) v1 and v2, (b) the power dis- sipated in the 3-k and 20-k resistors, and (c) the power supplied by the current source. 10 mA 3 kΩ 5 kΩ 20 kΩ 1 kΩ v1 + − v2 + − Figure2.45 For Practice Prob. 2.13. Answer: (a) 15 V, 20 V, (b) 75 mW, 20 mW, (c) 200 mW.
  • 58. 50 PART 1 DC Circuits †2.7 WYE-DELTA TRANSFORMATIONS Situations often arise in circuit analysis when the resistors are neither in parallel nor in series. For example, consider the bridge circuit in Fig. 2.46. How do we combine resistors R1 through R6 when the resistors are neither in series nor in parallel? Many circuits of the type shown in Fig. 2.46 can be simplified by using three-terminal equivalent networks. These are the wye (Y) or tee (T) network shown in Fig. 2.47 and the delta () or pi (#) network shown in Fig. 2.48. These networks occur by themselves or as part of a larger network. They are used in three-phase networks, electrical filters, and matching networks. Our main interest here is in how to identify them when they occur as part of a network and how to apply wye-delta transformation in the analysis of that network. vs + − R1 R4 R2 R5 R3 R6 Figure2.46 The bridge network. 1 3 2 4 R3 R2 R1 (a) 1 3 2 4 R3 R2 R1 (b) Figure2.47 Two forms of the same network: (a) Y, (b) T. 1 3 2 4 Rc (a) 1 3 2 4 (b) Ra Rb Rc Ra Rb Figure2.48 Two forms of the same network: (a) , (b) #. Delta to Wye Conversion Suppose it is more convenient to work with a wye network in a place where the circuit contains a delta configuration. We superimpose a wye network on the existing delta network and find the equivalent resistances in the wye network. To obtain the equivalent resistances in the wye network, we compare the two networks and make sure that the resistance between each pair of nodes in the (or #) network is the same as the resistance between the same pair of nodes in the Y (or T) network. For terminals 1 and 2 in Figs. 2.47 and 2.48, for example, R12(Y) = R1 + R3 R12() = Rb (Ra + Rc) (2.46) Setting R12(Y)= R12() gives R12 = R1 + R3 = Rb(Ra + Rc) Ra + Rb + Rc (2.47a) Similarly, R13 = R1 + R2 = Rc(Ra + Rb) Ra + Rb + Rc (2.47b) R34 = R2 + R3 = Ra(Rb + Rc) Ra + Rb + Rc (2.47c) Subtracting Eq. (2.47c) from Eq. (2.47a), we get R1 − R2 = Rc(Rb − Ra) Ra + Rb + Rc (2.48)
  • 59. CHAPTER 2 Basic Laws 51 Adding Eqs. (2.47b) and (2.48) gives R1 = RbRc Ra + Rb + Rc (2.49) and subtracting Eq. (2.48) from Eq. (2.47b) yields R2 = RcRa Ra + Rb + Rc (2.50) Subtracting Eq. (2.49) from Eq. (2.47a), we obtain R3 = RaRb Ra + Rb + Rc (2.51) We do not need to memorize Eqs. (2.49) to (2.51). To transform a network to Y, we create an extra node n as shown in Fig. 2.49 and follow this conversion rule: Each resistor in the Y network is the product of the resistors in the two adjacent branches, divided by the sum of the three resistors. R3 Ra Rb R1 R2 Rc b n a c Figure2.49 Superposition of Y and networks as an aid in transforming one to the other. Wye to Delta Conversion To obtain the conversion formulas for transforming a wye network to an equivalent delta network, we note from Eqs. (2.49) to (2.51) that R1R2 + R2R3 + R3R1 = RaRbRc(Ra + Rb + Rc) (Ra + Rb + Rc)2 = RaRbRc Ra + Rb + Rc (2.52) Dividing Eq. (2.52) by each of Eqs. (2.49) to (2.51) leads to the following equations: Ra = R1R2 + R2R3 + R3R1 R1 (2.53) Rb = R1R2 + R2R3 + R3R1 R2 (2.54) Rc = R1R2 + R2R3 + R3R1 R3 (2.55) From Eqs. (2.53) to (2.55) and Fig. 2.49, the conversion rule for Y to is as follows:
  • 60. 52 PART 1 DC Circuits Each resistor in the network is the sum of all possible products of Y resistors taken two at a time, divided by the opposite Y resistor. The Y and networks are said to be balanced when R1 = R2 = R3 = RY , Ra = Rb = Rc = R (2.56) Under these conditions, conversion formulas become RY = R 3 or R = 3RY (2.57) One may wonder why RY is less than R. Well, we notice that the Y- connection is like a “series” connection while the -connection is like a “parallel” connection. Note that in making the transformation, we do not take anything out of the circuit or put in anything new. We are merely substituting different but mathematically equivalent three-terminal network patterns to create a circuit in which resistors are either in series or in parallel, allowing us to calculate Req if necessary. E X A M P L E 2 . 1 4 Convert the network in Fig. 2.50(a) to an equivalent Y network. c b a 10 Ω 15 Ω (a) Rb Ra Rc 25 Ω c b a 5 Ω 3 Ω 7.5 Ω R2 R1 R3 (b) Figure2.50 For Example 2.14: (a) original network, (b) Y equivalent network. Solution: Using Eqs. (2.49) to (2.51), we obtain
  • 61. CHAPTER 2 Basic Laws 53 R1 = RbRc Ra + Rb + Rc = 25 × 10 25 + 10 + 15 = 250 50 = 5 R2 = RcRa Ra + Rb + Rc = 25 × 15 50 = 7.5 R3 = RaRb Ra + Rb + Rc = 15 × 10 50 = 3 The equivalent Y network is shown in Fig. 2.50(b). P R A C T I C E P R O B L E M 2 . 1 4 Transform the wye network in Fig. 2.51 to a delta network. 20 Ω R2 b a c 10 Ω R1 R3 40 Ω Figure2.51 For Practice Prob. 2.14. Answer: Ra = 140 , Rb = 70 , Rc = 35 . E X A M P L E 2 . 1 5 Obtain the equivalent resistance Rab for the circuit in Fig. 2.52 and use it to find current i. a a i b b c n 120 V 5 Ω 30 Ω 12.5 Ω 15 Ω 10 Ω 20 Ω + − Figure2.52 For Example 2.15. Solution: Inthiscircuit, therearetwoYnetworksandonenetwork. Transforming just one of these will simplify the circuit. If we convert the Y network comprising the 5-, 10-, and 20- resistors, we may select R1 = 10 , R2 = 20 , R3 = 5 Thus from Eqs. (2.53) to (2.55) we have Ra = R1R2 + R2R3 + R3R1 R1 = 10 × 20 + 20 × 5 + 5 × 10 10 = 350 10 = 35 Rb = R1R2 + R2R3 + R3R1 R2 = 350 20 = 17.5 Rc = R1R2 + R2R3 + R3R1 R3 = 350 5 = 70 With the Y converted to , the equivalent circuit (with the voltage source removed for now) is shown in Fig. 2.53(a). Combining the three pairs of resistors in parallel, we obtain
  • 62. 54 PART 1 DC Circuits 70 30 = 70 × 30 70 + 30 = 21 12.5 17.5 = 12.5 × 17.5 12.5 + 17.5 = 7.2917 15 35 = 15 × 35 15 + 35 = 10.5 so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we find Rab = (7.292 + 10.5) 21 = 17.792 × 21 17.792 + 21 = 9.632 Then i = vs Rab = 120 9.632 = 12.458 A a b 30 Ω 70 Ω 17.5 Ω 35 Ω 12.5 Ω 15 Ω (a) a b 21 Ω (b) 7.292 Ω 10.5 Ω Figure2.53 Equivalent circuits to Fig. 2.52, with the voltage removed. P R A C T I C E P R O B L E M 2 . 1 5 For the bridge network in Fig. 2.54, find Rab and i. 24 Ω 100 V i 30 Ω 10 Ω 50 Ω 13 Ω 20 Ω + − b a Figure2.54 For Practice Prob. 2.15. Answer: 40 , 2.5 A. †2.8 APPLICATIONS Resistors are often used to model devices that convert electrical energy into heat or other forms of energy. Such devices include conducting wire, lightbulbs, electric heaters, stoves, ovens, and loudspeakers. In this
  • 63. CHAPTER 2 Basic Laws 55 section, we will consider two real-life problems that apply the concepts developed in this chapter: electrical lighting systems and design of dc meters. So far, we have assumed that connecting wires are perfect conductors (i.e., conductors of zero resistance). In real physical systems, however, the resistance of the connecting wire may be ap- preciably large, and the modeling of the system must include that resistance. 2.8.1 Lighting Systems Lighting systems, such as in a house or on a Christmas tree, often consist of N lamps connected either in parallel or in series, as shown in Fig. 2.55. Each lamp is modeled as a resistor. Assuming that all the lamps are identical and Vo is the power-line voltage, the voltage across each lamp is Vo for the parallel connection and Vo/N for the series connection. The series connection is easy to manufacture but is seldom used in practice, for at least two reasons. First, it is less reliable; when a lamp fails, all the lamps go out. Second, it is harder to maintain; when a lamp is bad, one must test all the lamps one by one to detect the faulty one. Vo + − Power plug 1 2 3 N Lamp (a) Vo + − 1 2 3 N (b) Figure2.55 (a) Parallel connection of lightbulbs, (b) series connection of lightbulbs. E X A M P L E 2 . 1 6 Three lightbulbs are connected to a 9-V battery as shown in Fig. 2.56(a). Calculate: (a) the total current supplied by the battery, (b) the current through each bulb, (c) the resistance of each bulb. (a) 9 V 10 W 15 W 20 W (b) 9 V + − + − + − I1 I2 V3 V2 V1 R1 I R3 R2 Figure 2.56 (a) Lighting system with three bulbs, (b) resistive circuit equivalent model.
  • 64. 56 PART 1 DC Circuits Solution: (a) The total power supplied by the battery is equal to the total power absorbed by the bulbs, that is, p = 15 + 10 + 20 = 45 W Since p = V I, then the total current supplied by the battery is I = p V = 45 9 = 5 A (b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b). Since R1 (20-W bulb) is in parallel with the battery as well as the series com- bination of R2 and R3, V1 = V2 + V3 = 9 V The current through R1 is I1 = p1 V1 = 20 9 = 2.222 A By KCL, the current through the series combination of R2 and R3 is I2 = I − I1 = 5 − 2.222 = 2.778 A (c) Since p = I2 R, R1 = p1 I2 1 = 20 2.2222 = 4.05 R2 = p2 I2 2 = 15 2.7772 = 1.945 R3 = p3 I2 3 = 10 2.7772 = 1.297 P R A C T I C E P R O B L E M 2 . 1 6 Refer to Fig. 2.55 and assume there are 10 lightbulbs, each with a power rating of 40 W. If the voltage at the plug is 110 V for the parallel and series connections, calculate the current through each bulb for both cases. Answer: 0.364 A (parallel), 3.64 A (series). + + − − Vin Vout a b c Max Min Figure2.57 The potentiometer controlling potential levels. 2.8.2 Design of DC Meters By their nature, resistors are used to control the flow of current. We take advantage of this property in several applications, such as in a poten- tiometer (Fig. 2.57). The word potentiometer, derived from the words potential and meter, implies that potential can be metered out. The po- tentiometer (or pot for short) is a three-terminal device that operates on the principle of voltage division. It is essentially an adjustable voltage divider. As a voltage regulator, it is used as a volume or level control on radios, TVs, and other devices. In Fig. 2.57, Vout = Vbc = Rbc Rac Vin (2.58) where Rac = Rab + Rbc. Thus, Vout decreases or increases as the sliding contact of the pot moves toward c or a, respectively.
  • 65. CHAPTER 2 Basic Laws 57 Another application where resistors are used to control current flow is in the analog dc meters—the ammeter, voltmeter, and ohmmeter, which measure current, voltage, and resistance, respectively. Each of these me- ters employs the d’Arsonval meter movement, shown in Fig. 2.58. The movement consists essentially of a movable iron-core coil mounted on a pivot between the poles of a permanent magnet. When current flows through the coil, it creates a torque which causes the pointer to deflect. The amount of current through the coil determines the deflection of the pointer, which is registered on a scale attached to the meter movement. For example, if the meter movement is rated 1 mA, 50 , it would take 1 mA to cause a full-scale deflection of the meter movement. By introduc- ing additional circuitry to the d’Arsonval meter movement, an ammeter, voltmeter, or ohmmeter can be constructed. Aninstrumentcapableofmeasuringvoltage,cur- rent, and resistance is called a multimeter or a volt-ohm meter (VOM). Aloadisacomponentthatisreceivingenergy(an energysink),asopposedtoageneratorsupplying energy (an energy source). More about loading will be discussed in Section 4.9.1. Consider Fig. 2.59, where an analog voltmeter and ammeter are connected to an element. The voltmeter measures the voltage across a load and is therefore connected in parallel with the element. As shown in Fig. 2.60(a), the voltmeter consists of a d’Arsonval movement in par- allel with a resistor whose resistance Rm is deliberately made very large (theoretically, infinite), to minimize the current drawn from the circuit. To extend the range of voltage that the meter can measure, series multi- plier resistors are often connected with the voltmeters, as shown in Fig. 2.60(b). The multiple-range voltmeter in Fig. 2.60(b) can measure volt- age from 0 to 1 V, 0 to 10 V, or 0 to 100 V, depending on whether the switch is connected to R1, R2, or R3, respectively. Let us calculate the multiplier resistor Rn for the single-range volt- meter in Fig. 2.60(a), or Rn = R1, R2, or R3 for the multiple-range voltmeter in Fig. 2.60(b). We need to determine the value of Rn to be connected in series with the internal resistance Rm of the voltmeter. In any design, we consider the worst-case condition. In this case, the worst case occurs when the full-scale current Ifs = Im flows through the meter. This should also correspond to the maximum voltage reading or the full- scale voltage Vfs. Since the multiplier resistance Rn is in series with the scale pointer spring permanent magnet rotating coil stationary iron core spring N S Figure2.58 A d’Arsonval meter movement. V A V I + − Voltmeter Ammeter Element Figure2.59 Connection of a voltmeter and an ammeter to an element.
  • 66. 58 PART 1 DC Circuits Probes V + − R1 R2 R3 1 V 10 V 100 V Switch Im (b) Rn Im Multiplier Probes V + − (a) Rm Meter Rm Meter Figure2.60 Voltmeters: (a) single-range type, (b) multiple-range type. internal resistance Rm, Vfs = Ifs(Rn + Rm) (2.59) From this, we obtain Rn = Vfs Ifs − Rm (2.60) Im I Probes (a) Rn In (b) R1 R2 R3 10 mA 100 mA 1 A Switch Im I Probes Rm Meter Rm Meter Figure 2.61 Ammeters: (a) single-range type, (b) multiple-range type. Similarly, the ammeter measures the current through the load and is connected in series with it. As shown in Fig. 2.61(a), the ammeter con- sists of a d’Arsonval movement in parallel with a resistor whose resistance Rm is deliberately made very small (theoretically, zero) to minimize the voltage drop across it. To allow multiple range, shunt resistors are often connected in parallel with Rm as shown in Fig. 2.61(b). The shunt resis- tors allow the meter to measure in the range 0–10 mA, 0–100 mA, or 0–1 A, depending on whether the switch is connected to R1, R2, or R3, respectively. Now our objective is to obtain the multiplier shunt Rn for the single- range ammeter in Fig. 2.61(a), or Rn = R1, R2, or R3 for the multiple- range ammeter in Fig. 2.61(b). We notice that Rm and Rn are in parallel and that at full-scale reading I = Ifs = Im + In, where In is the current through the shunt resistor Rn. Applying the current division principle yields Im = Rn Rn + Rm Ifs
  • 67. CHAPTER 2 Basic Laws 59 or Rn = Im Ifs − Im Rm (2.61) The resistance Rx of a linear resistor can be measured in two ways. An indirect way is to measure the current I that flows through it by connecting an ammeter in series with it and the voltage V across it by connecting a voltmeter in parallel with it, as shown in Fig. 2.62(a). Then Rx = V I (2.62) The direct method of measuring resistance is to use an ohmmeter. An ohmmeterconsistsbasicallyofad’Arsonvalmovement, avariableresistor or potentiometer, and a battery, as shown in Fig. 2.62(b). Applying KVL to the circuit in Fig. 2.62(b) gives E = (R + Rm + Rx)Im or Rx = E Im − (R + Rm) (2.63) The resistor R is selected such that the meter gives a full-scale deflection, that is, Im = Ifs when Rx = 0. This implies that E = (R + Rm)Ifs (2.64) Substituting Eq. (2.64) into Eq. (2.63) leads to Rx = Ifs Im − 1 (R + Rm) (2.65) Im R E Rx Ohmmeter (b) (a) V A + − V Rx I Rm Figure2.62 Two ways of measuring resistance: (a) using an ammeter and a voltmeter, (b) using an ohmmeter. As mentioned, the types of meters we have discussed are known as analog meters and are based on the d’Arsonval meter movement. Another type of meter, called a digital meter, is based on active circuit elements such as op amps. For example, a digital multimeter displays measure- ments of dc or ac voltage, current, and resistance as discrete numbers, instead of using a pointer deflection on a continuous scale as in an ana- log multimeter. Digital meters are what you would most likely use in a modern lab. However, the design of digital meters is beyond the scope of this book. E X A M P L E 2 . 1 7 Following the voltmeter setup of Fig. 2.60, design a voltmeter for the fol- lowing multiple ranges: (a) 0–1 V (b) 0–5 V (c) 0–50 V (d) 0–100 V Assume that the internal resistance Rm = 2 k and the full-scale current Ifs = 100 µA. Solution: We apply Eq. (2.60) and assume that R1, R2, R3, and R4 correspond with ranges 0–1 V, 0–5 V, 0–50 V, and 0–100 V, respectively. (a) For range 0–1 V, R1 = 1 100 × 10−6 − 2000 = 10,000 − 2000 = 8 k
  • 68. 60 PART 1 DC Circuits (b) For range 0–5 V, R2 = 5 100 × 10−6 − 2000 = 50,000 − 2000 = 48 k (c) For range 0–50 V, R3 = 50 100 × 10−6 − 2000 = 500,000 − 2000 = 498 k (d) For range 0–100 V, R4 = 100 V 100 × 10−6 − 2000 = 1,000,000 − 2000 = 998 k Note that the ratio of the total resistance (Rn+Rm) to the full-scale voltage Vfs is constant and equal to 1/Ifs for the four ranges. This ratio (given in ohms per volt, or /V) is known as the sensitivity of the voltmeter. The larger the sensitivity, the better the voltmeter. P R A C T I C E P R O B L E M 2 . 1 7 Following the ammeter setup of Fig. 2.61, design an ammeter for the fol- lowing multiple ranges: (a) 0–1 A (b) 0–100 mA (c) 0–10 mA Take the full-scale meter current as Im = 1 mA and the internal resistance of the ammeter as Rm = 50 . Answer: Shunt resistors: 0.05 , 0.505 , 5.556 . 2.9 SUMMARY 1. A resistor is a passive element in which the voltage v across it is directly proportional to the current i through it. That is, a resistor is a device that obeys Ohm’s law, v = iR where R is the resistance of the resistor. 2. A short circuit is a resistor (a perfectly conducting wire) with zero resistance (R = 0). An open circuit is a resistor with infinite resis- tance (R = ∞). 3. The conductance G of a resistor is the reciprocal of its resistance: G = 1 R 4. A branch is a single two-terminal element in an electric circuit. A node is the point of connection between two or more branches. A loop is a closed path in a circuit. The number of branches b, the number of nodes n, and the number of independent loops l in a network are related as b = l + n − 1
  • 69. CHAPTER 2 Basic Laws 61 5. Kirchhoff’s current law (KCL) states that the currents at any node algebraically sum to zero. In other words, the sum of the currents entering a node equals the sum of currents leaving the node. 6. Kirchhoff’s voltage law (KVL) states that the voltages around a closed path algebraically sum to zero. In other words, the sum of voltage rises equals the sum of voltage drops. 7. Two elements are in series when they are connected sequentially, end to end. When elements are in series, the same current flows through them (i1 = i2). They are in parallel if they are connected to the same two nodes. Elements in parallel always have the same voltage across them (v1 = v2). 8. When two resistors R1 (= 1/G1) and R2 (= 1/G2) are in series, their equivalent resistance Req and equivalent conductance Geq are Req = R1 + R2, Geq = G1G2 G1 + G2 9. When two resistors R1 (= 1/G1) and R2 (= 1/G2) are in parallel, their equivalent resistance Req and equivalent conductance Geq are Req = R1R2 R1 + R2 , Geq = G1 + G2 10. The voltage division principle for two resistors in series is v1 = R1 R1 + R2 v, v2 = R2 R1 + R2 v 11. The current division principle for two resistors in parallel is i1 = R2 R1 + R2 i, i2 = R1 R1 + R2 i 12. The formulas for a delta-to-wye transformation are R1 = RbRc Ra + Rb + Rc , R2 = RcRa Ra + Rb + Rc R3 = RaRb Ra + Rb + Rc 13. The formulas for a wye-to-delta transformation are Ra = R1R2 + R2R3 + R3R1 R1 , Rb = R1R2 + R2R3 + R3R1 R2 Rc = R1R2 + R2R3 + R3R1 R3 14. The basic laws covered in this chapter can be applied to the prob- lems of electrical lighting and design of dc meters. REVIEW QUESTIONS 2.1 The reciprocal of resistance is: (a) voltage (b) current (c) conductance (d) coulombs 2.2 An electric heater draws 10 A from a 120-V line. The resistance of the heater is: (a) 1200 (b) 120 (c) 12 (d) 1.2
  • 70. 62 PART 1 DC Circuits 2.3 The voltage drop across a 1.5-kW toaster that draws 12 A of current is: (a) 18 kV (b) 125 V (c) 120 V (d) 10.42 V 2.4 The maximum current that a 2W, 80 k resistor can safely conduct is: (a) 160 kA (b) 40 kA (c) 5 mA (d) 25 µA 2.5 A network has 12 branches and 8 independent loops. How many nodes are there in the network? (a) 19 (b) 17 (c) 5 (d) 4 2.6 The current I in the circuit in Fig. 2.63 is: (a) −0.8 A (b) −0.2 A (c) 0.2 A (d) 0.8 A 3 V 5 V + − + − 4 Ω I 6 Ω Figure 2.63 For Review Question 2.6. 2.7 The current Io in Fig. 2.64 is: (a) −4 A (b) −2 A (c) 4 A (d) 16 A 10 A 4 A 2 A Io Figure 2.64 For Review Question 2.7. 2.8 In the circuit in Fig. 2.65, V is: (a) 30 V (b) 14 V (c) 10 V (d) 6 V + − + − + − + − 10 V 12 V 8 V V Figure 2.65 For Review Question 2.8. 2.9 Which of the circuits in Fig. 2.66 will give you Vab = 7 V? 3 V a b 5 V 1 V (a) + − + − + − 3 V a b 5 V 1 V (b) + − + − + − 3 V a 5 V 1 V (c) + − + − + − b 3 V a 5 V 1 V (d) + − + − + − b Figure 2.66 For Review Question 2.9. 2.10 The equivalent resistance of the circuit in Fig. 2.67 is: (a) 4 k (b) 5 k (c) 8 k (d) 14 k 2 kΩ 3 kΩ Req 6 kΩ 3 kΩ Figure 2.67 For Review Question 2.10. Answers: 2.1c, 2.2c, 2.3b, 2.4c, 2.5c, 2.6b, 2.7a, 2.8d, 2.9d, 2.10a.
  • 71. CHAPTER 2 Basic Laws 63 PROBLEMS Section 2.2 Ohm’s Law 2.1 The voltage across a 5-k resistor is 16 V. Find the current through the resistor. 2.2 Find the hot resistance of a lightbulb rated 60 W, 120 V. 2.3 When the voltage across a resistor is 120 V, the current through it is 2.5 mA. Calculate its conductance. 2.4 (a) Calculate current i in Fig. 2.68 when the switch is in position 1. (b) Find the current when the switch is in position 2. + − 150 Ω 100 Ω 3 V 1 2 i Figure 2.68 For Prob. 2.4. Section 2.3 Nodes, Branches, and Loops 2.5 For the network graph in Fig. 2.69, find the number of nodes, branches, and loops. Figure 2.69 For Prob. 2.5. 2.6 In the network graph shown in Fig. 2.70, determine the number of branches and nodes. Figure 2.70 For Prob. 2.6. 2.7 Determine the number of branches and nodes in the circuit in Fig. 2.71. + − 6 Ω 3 Ω 2 Ω 5 Ω 10 V 5i 4 Ω i Figure 2.71 For Prob. 2.7. Section 2.4 Kirchhoff’s Laws 2.8 Use KCL to obtain currents i1, i2, and i3 in the circuit shown in Fig. 2.72. 8 mA 9 mA 12 mA i1 i3 i2 Figure 2.72 For Prob. 2.8. 2.9 Find i1, i2, and i3 in the circuit in Fig. 2.73. i2 i3 10 A i1 3 A 2 A 1 A Figure 2.73 For Prob. 2.9.
  • 72. 64 PART 1 DC Circuits 2.10 Determine i1 and i2 in the circuit in Fig. 2.74. 3 A 4 A –2 A i2 i1 Figure 2.74 For Prob. 2.10. 2.11 Determine v1 through v4 in the circuit in Fig. 2.75. 12 V + − 10 V − + + − 8 V + − v1 + − v3 + − v2 − + v4 − + 6 V Figure 2.75 For Prob. 2.11. 2.12 In the circuit in Fig. 2.76, obtain v1, v2, and v3. + − 20 V + − v1 + − v3 25 V 10 V 15 V v2 + − + − + − + − Figure 2.76 For Prob. 2.12. 2.13 Find v1 and v2 in the circuit in Fig. 2.77. 6 V v1 + − + − v1 v2 + − + − + − + − 12 V 10 V Figure 2.77 For Prob. 2.13. 2.14 Obtain v1 through v3 in the circuit of Fig. 2.78. 24 V 12 V 10 V v3 v2 + − + − + − + − + − v1 + − Figure 2.78 For Prob. 2.14. 2.15 Find I and Vab in the circuit of Fig. 2.79. 5 Ω 3 Ω + − + − + − Vab 30 V 8 V b a + − 10 V I Figure 2.79 For Prob. 2.15. 2.16 From the circuit in Fig. 2.80, find I, the power dissipated by the resistor, and the power supplied by each source. −8 V 10 V 12 V 3 Ω + − + − + − I Figure 2.80 For Prob. 2.16.
  • 73. CHAPTER 2 Basic Laws 65 2.17 Determine io in the circuit of Fig. 2.81. 36 V + − 4 Ω + − 5io io Figure 2.81 For Prob. 2.17. 2.18 Calculate the power dissipated in the 5- resistor in the circuit of Fig. 2.82. – + 45 V 1 Ω 5 Ω 3Vo + − Vo + − Figure 2.82 For Prob. 2.18. 2.19 Find Vo in the circuit in Fig. 2.83 and the power dissipated by the controlled source. 10 A 6 Ω 2Vo + − 4 Ω Vo Figure 2.83 For Prob. 2.19. 2.20 For the circuit in Fig. 2.84, find Vo/Vs in terms of α, R1, R2, R3, and R4. If R1 = R2 = R3 = R4, what value of α will produce |Vo/Vs| = 10? Vo + − + − R4 R3 R1 R2 aIo Vs Io Figure 2.84 For Prob. 2.20. 2.21 For the network in Fig. 2.85, find the current, voltage, and power associated with the 20-k resistor. 0.01Vo Vo + − 20 kΩ 5 kΩ 10 kΩ 5 mA Figure 2.85 For Prob. 2.21. Sections 2.5 and 2.6 Series and Parallel Resistors 2.22 For the circuit in Fig. 2.86, find i1 and i2. 4 kΩ 6 kΩ 20 mA i1 i2 Figure 2.86 For Prob. 2.22. 2.23 Find v1 and v2 in the circuit in Fig. 2.87. 24 V 3 kΩ 9 kΩ v1 v2 + − + − + − Figure 2.87 For Prob. 2.23. 2.24 Find v1, v2, and v3 in the circuit in Fig. 2.88. 40 V 14 Ω 15 Ω v1 v2 + − + − + − 10 Ω v3 + − Figure 2.88 For Prob. 2.24. 2.25 Calculate v1, i1, v2, and i2 in the circuit of Fig. 2.89. 3 Ω 4 Ω 6 Ω i1 i2 v1 v2 + − + − + − 12 V Figure 2.89 For Prob. 2.25.
  • 74. 66 PART 1 DC Circuits 2.26 Find i, v, and the power dissipated in the 6- resistor in Fig. 2.90. 9 A 4 Ω 8 Ω 6 Ω i + − v Figure 2.90 For Prob. 2.26. 2.27 In the circuit in Fig. 2.91, find v, i, and the power absorbed by the 4- resistor. 20 V 6 Ω 10 Ω 5 Ω + − 4 Ω + − v i Figure 2.91 For Prob. 2.27. 2.28 Find i1 through i4 in the circuit in Fig. 2.92. 20 A 10 Ω 40 Ω i4 i3 20 Ω 30 Ω i2 i1 Figure 2.92 For Prob. 2.28. 2.29 Obtain v and i in the circuit in Fig. 2.93. 9 A 2 S 1 S 4 S 6 S 3 S + − v i Figure 2.93 For Prob. 2.29. 2.30 Determine i1, i2, v1, and v2 in the ladder network in Fig. 2.94. Calculate the power dissipated in the 2- resistor. 28 V 13 Ω 15 Ω 6 Ω + − 8 Ω 2 Ω 4 Ω − 10 Ω 12 Ω + v1 − + v2 i1 i2 Figure 2.94 For Prob. 2.30. 2.31 Calculate Vo and Io in the circuit of Fig. 2.95. 50 V 30 Ω 70 Ω + − 5 Ω 20 Ω + − Vo Io Figure 2.95 For Prob. 2.31. 2.32 Find Vo and Io in the circuit of Fig. 2.96. 4 V 6 Ω 3 Ω 1 Ω + − Vo 8 Ω 2 Ω + − Io Figure 2.96 For Prob. 2.32. 2.33 In the circuit of Fig. 2.97, find R if Vo = 4V. 20 V 6 Ω 16 Ω + − + − Vo R Figure 2.97 For Prob. 2.33.
  • 75. CHAPTER 2 Basic Laws 67 2.34 Find I and Vs in the circuit of Fig. 2.98 if the current through the 3- resistor is 2 A. 4 Ω 2 A 10 Ω + − 2 Ω 3 Ω 6 Ω V s I Figure 2.98 For Prob. 2.34. 2.35 Find the equivalent resistance at terminals a-b for each of the networks in Fig. 2.99. R (a) (b) (c) a b 3R R R R (d) (e) a b R R R R a b R R R R a b R 2R 3R a b Figure 2.99 For Prob. 2.35. 2.36 For the ladder network in Fig. 2.100, find I and Req. 10 V 6 Ω 2 Ω + − 3 Ω 1 Ω 2 Ω 4 Ω I Req Figure 2.100 For Prob. 2.36. 2.37 If Req = 50 in the circuit in Fig. 2.101, find R. Req 30 Ω 10 Ω 60 Ω R 12 Ω 12 Ω 12 Ω Figure 2.101 For Prob. 2.37. 2.38 Reduce each of the circuits in Fig. 2.102 to a single resistor at terminals a-b. 8 Ω 5 Ω 20 Ω 30 Ω a b (a) 5 Ω 4 Ω 8 Ω 5 Ω 10 Ω 4 Ω 2 Ω 3 Ω a b (b) Figure 2.102 For Prob. 2.38. 2.39 Calculate the equivalent resistance Rab at terminals a-b for each of the circuits in Fig. 2.103. 40 Ω 10 Ω 5 Ω 20 Ω (a) a b 30 Ω 80 Ω 60 Ω (b) a b 10 Ω 20 Ω Figure 2.103 For Prob. 2.39.
  • 76. 68 PART 1 DC Circuits 2.40 Obtain the equivalent resistance at the terminals a-b for each of the circuits in Fig. 2.104. 11 Ω 10 Ω 20 Ω 6 Ω 5 Ω 4 Ω 9 Ω 8 Ω 4 Ω 5 Ω 15 Ω (b) a b (a) a b 10 Ω 20 Ω 60 Ω 30 Ω Figure 2.104 For Prob. 2.40. 2.41 Find Req at terminals a-b for each of the circuits in Fig. 2.105. (a) a b 40 Ω 70 Ω 30 Ω 60 Ω 20 Ω (b) a b 6 Ω 40 Ω 60 Ω 30 Ω 20 Ω 50 Ω 80 Ω 10 Ω 70 Ω 4 Ω 8 Ω Figure 2.105 For Prob. 2.41. 2.42 Find the equivalent resistance Rab in the circuit of Fig. 2.106. a d e f b c 6 Ω 3 Ω 5 Ω 20 Ω 10 Ω 8 Ω Figure 2.106 For Prob. 2.42. Section 2.7 Wye-Delta Transformations 2.43 Convert the circuits in Fig. 2.107 from Y to . 10 Ω 10 Ω 10 Ω b a c (a) 20 Ω 30 Ω 50 Ω a (b) b c Figure 2.107 For Prob. 2.43. 2.44 Transform the circuits in Fig. 2.108 from to Y. 12 Ω 12 Ω 12 Ω (a) a b c 60 Ω 30 Ω 10 Ω (b) a b c Figure 2.108 For Prob. 2.44.
  • 77. CHAPTER 2 Basic Laws 69 2.45 What value of R in the circuit of Fig. 2.109 would cause the current source to deliver 800 mW to the resistors? 30 mA R R R R R Figure 2.109 For Prob. 2.45. 2.46 Obtain the equivalent resistance at the terminals a-b for each of the circuits in Fig. 2.110. (a) b a 30 Ω 10 Ω 10 Ω 20 Ω 20 Ω 10 Ω 20 Ω 10 Ω 30 Ω 25 Ω (b) b a 15 Ω 5 Ω Figure 2.110 For Prob. 2.46. 2.47 ∗ Find the equivalent resistance Rab in each of the circuits of Fig. 2.111. Each resistor is 100 . (a) b a (b) b a Figure 2.111 For Prob. 2.47. 2.48 ∗ Obtain the equivalent resistance Rab in each of the circuits of Fig. 2.112. In (b), all resistors have a value of 30 . (b) 40 Ω 50 Ω 10 Ω 60 Ω 30 Ω 20 Ω (a) b a 80 Ω 30 Ω a b Figure 2.112 For Prob. 2.48. 2.49 Calculate Io in the circuit of Fig. 2.113. 20 Ω 40 Ω 60 Ω 50 Ω 10 Ω 20 Ω 24 V + − Io Figure 2.113 For Prob. 2.49. ∗An asterisk indicates a challenging problem.
  • 78. 70 PART 1 DC Circuits 2.50 Determine V in the circuit of Fig. 2.114. 100 V 30 Ω 15 Ω 10 Ω 16 Ω 35 Ω 12 Ω 20 Ω + − V + − Figure 2.114 For Prob. 2.50. 2.51 ∗ Find Req and I in the circuit of Fig. 2.115. 2 Ω 4 Ω 12 Ω 6 Ω 1 Ω 8 Ω 2 Ω 3 Ω 10 Ω 5 Ω 4 Ω 20 V + − Req I Figure 2.115 For Prob. 2.51. Section 2.8 Applications 2.52 The lightbulb in Fig. 2.116 is rated 120 V, 0.75 A. Calculate Vs to make the lightbulb operate at the rated conditions. + − 40 Ω Vs 80 Ω Bulb Figure 2.116 For Prob. 2.52. 2.53 Three lightbulbs are connected in series to a 100-V battery as shown in Fig. 2.117. Find the current I through the bulbs. 30 W 40 W 50 W 100 V + − I Figure 2.117 For Prob. 2.53. 2.54 If the three bulbs of Prob. 2.53 are connected in parallel to the 100-V battery, calculate the current through each bulb. 2.55 As a design engineer, you are asked to design a lighting system consisting of a 70-W power supply and two lightbulbs as shown in Fig. 2.118. You must select the two bulbs from the following three available bulbs. R1 = 80 , cost = $0.60 (standard size) R2 = 90 , cost = $0.90 (standard size) R3 = 100 , cost = $0.75 (nonstandard size) The system should be designed for minimum cost such that I = 1.2 A ± 5 percent. I Rx Ry 70 W Power Supply + − Figure 2.118 For Prob. 2.55. 2.56 If an ammeter with an internal resistance of 100 and a current capacity of 2 mA is to measure 5 A, determine the value of the resistance needed. Calculate the power dissipated in the shunt resistor. 2.57 The potentiometer (adjustable resistor) Rx in Fig. 2.119 is to be designed to adjust current ix from 1 A to 10 A. Calculate the values of R and Rx to achieve this. + − ix R Rx ix 110 V Figure 2.119 For Prob. 2.57. 2.58 A d’Arsonval meter with an internal resistance of 1 k requires 10 mA to produce full-scale deflection. Calculate the value of a series resistance needed to measure 50 V of full scale.
  • 79. CHAPTER 2 Basic Laws 71 2.59 A 20-k/V voltmeter reads 10 V full scale. (a) What series resistance is required to make the meter read 50 V full scale? (b) What power will the series resistor dissipate when the meter reads full scale? 2.60 (a) Obtain the voltage vo in the circuit of Fig. 2.120(a). (b) Determine the voltage v o measured when a voltmeter with 6-k internal resistance is connected as shown in Fig. 2.120(b). (c) The finite resistance of the meter introduces an error into the measurement. Calculate the percent error as vo − v o vo × 100% (d) Find the percent error if the internal resistance were 36 k. + − 2 mA 1 kΩ 5 kΩ 4 kΩ vo (a) (b) 2 mA + − 1 kΩ 5 kΩ 4 kΩ Voltmeter vo Figure 2.120 For Prob. 2.60. 2.61 (a) Find the current i in the circuit of Fig. 2.121(a). (b) An ammeter with an internal resistance of 1 is inserted in the network to measure i as shown in Fig. 2.121(b). What is i ? (c) Calculate the percent error introduced by the meter as i − i i × 100% + − i 4 V 16 Ω 40 Ω 60 Ω (a) + − i' 4 V 16 Ω 40 Ω 60 Ω (b) Ammeter Figure 2.121 For Prob. 2.61. 2.62 A voltmeter is used to measure Vo in the circuit in Fig. 2.122. The voltmeter model consists of an ideal voltmeter in parallel with a 100-k resistor. Let Vs = 40 V, Rs = 10 k, and R1 = 20 k. Calculate Vo with and without the voltmeter when (a) R2 = 1 k (b) R2 = 10 k (c) R2 = 100 k + − + − V 100 kΩ Vo Vs Rs R1 R2 Figure 2.122 For Prob. 2.62. 2.63 An ammeter model consists of an ideal ammeter in series with a 20- resistor. It is connected with a current source and an unknown resistor Rx as shown in Fig. 2.123. The ammeter reading is noted. When a potentiometer R is added and adjusted until the ammeter reading drops to one half its previous reading, then R = 65 . What is the value of Rx? I A R Rx 20 Ω Ammeter model Figure 2.123 For Prob. 2.63. 2.64 The circuit in Fig. 2.124 is to control the speed of a motor such that the motor draws currents 5 A, 3 A,
  • 80. 72 PART 1 DC Circuits and 1 A when the switch is at high, medium, and low positions, respectively. The motor can be modeled as a load resistance of 20 m. Determine the series dropping resistances R1, R2, and R3. 6 V High Medium Low 10-A, 0.01-Ω fuse R1 R2 R3 Motor Figure 2.124 For Prob. 2.64. 2.65 An ohmmeter is constructed with a 2-V battery and 0.1-mA (full-scale) meter with 100- internal resistance. (a) Calculate the resistance of the (variable) resistor required to be in series with the meter and the battery. (b) Determine the unknown resistance across the terminals of the ohmmeter that will cause the meter to deflect half scale. COMPREHENSIVE PROBLEMS 2.66 An electric heater connected to a 120-V source consists of two identical 0.4- elements made of Nichrome wire. The elements provide low heat when connected in series and high heat when connected in parallel. Find the power at low and high heat settings. 2.67 Suppose your circuit laboratory has the following standard commercially available resistors in large quantities: 1.8 20 300 24 k 56 k Using series and parallel combinations and a minimum number of available resistors, how would you obtain the following resistances for an electronic circuit design? (a) 5 (b) 311.8 (c) 40 k (d) 52.32 k 2.68 In the circuit in Fig. 2.125, the wiper divides the potentiometer resistance between αR and (1 − α)R, 0 ≤ α ≤ 1. Find vo/vs. vo + − + − R R aR vs Figure 2.125 For Prob. 2.68. 2.69 An electric pencil sharpener rated 240 mW, 6 V is connected to a 9-V battery as shown in Fig. 2.126. Calculate the value of the series-dropping resistor Rx needed to power the sharpener. 9 V Switch Rx Figure 2.126 For Prob. 2.69. 2.70 A loudspeaker is connected to an amplifier as shown in Fig. 2.127. If a 10- loudspeaker draws the maximum power of 12 W from the amplifier, determine the maximum power a 4- loudspeaker will draw. Amplifier Loudspeaker Figure 2.127 For Prob. 2.70. 2.71 In a certain application, the circuit in Fig. 2.128 must be designed to meet these two criteria: (a) Vo/Vs = 0.05 (b) Req = 40 k
  • 81. CHAPTER 2 Basic Laws 73 If the load resistor 5 k is fixed, find R1 and R2 to meet the criteria. Vs + − + − 5 kΩ Vo R2 R1 Req Figure 2.128 For Prob. 2.71. 2.72 The pin diagram of a resistance array is shown in Fig. 2.129. Find the equivalent resistance between the following: (a) 1 and 2 (b) 1 and 3 (c) 1 and 4 20 Ω 20 Ω 40 Ω 10 Ω 10 Ω 1 2 3 4 80 Ω Figure 2.129 For Prob. 2.72. 2.73 Two delicate devices are rated as shown in Fig. 2.130. Find the values of the resistors R1 and R2 needed to power the devices using a 24-V battery. Device 1 Device 2 24 V R1 R2 60-mA, 2-Ω fuse 9 V, 45 mW 24 V, 480 mW Figure 2.130 For Prob. 2.73.
  • 82. 75 C H A P T E R METHODS OF ANALYSIS 3 Scientists study the world as it is, engineers create the world that never has been. —Theodore von Karman Enhancing Your Career Career in Electronics One area of application for electric circuit analysis is electronics. The term electronics was orig- inally used to distinguish circuits of very low current levels. This distinction no longer holds, as power semiconductor de- vices operate at high levels of current. Today, electronics is regarded as the science of the motion of charges in a gas, vac- uum, or semiconductor. Modern electronics involves tran- sistors and transistor circuits. The earlier electronic circuits were assembled from components. Many electronic circuits are now produced as integrated circuits, fabricated in a semi- conductor substrate or chip. Electronic circuits find applications in many areas, such as automation, broadcasting, computers, and instru- mentation. The range of devices that use electronic circuits is enormous and is limited only by our imagination. Radio, television, computers, and stereo systems are but a few. An electrical engineer usually performs diverse func- tions and is likely to use, design, or construct systems that incorporate some form of electronic circuits. Therefore, an understanding of the operation and analysis of electronics is essential to the electrical engineer. Electronics has become a specialty distinct from other disciplines within electrical engineering. Because the field of electronics is ever advancing, an electronics engineer must update his/her knowledge from time to time. The best way to do this is by being a member of a professional organization such as the Institute of Electrical and Electronics Engineers Troubleshooting an electronic circuit board. Source: T. J. Mal- oney, Modern Industrial Electronics, 3rd ed. Englewood Cliffs, NJ: Prentice Hall, 1996, p. 408. (IEEE). With a membership of over 300,000, the IEEE is the largest professional organization in the world. Members benefit immensely from the numerous magazines, journals, transactions, and conference/symposium proceedings pub- lished yearly by IEEE. You should consider becoming an IEEE member.
  • 83. 76 PART 1 DC Circuits 3.1 INTRODUCTION Having understood the fundamental laws of circuit theory (Ohm’s law and Kirchhoff’s laws), we are now prepared to apply these laws to develop two powerful techniques for circuit analysis: nodal analysis, which is based on a systematic application of Kirchhoff’s current law (KCL), and mesh analysis, which is based on a systematic application of Kirchhoff’s voltage law (KVL). The two techniques are so important that this chapter should be regarded as the most important in the book. Students are therefore encouraged to pay careful attention. With the two techniques to be developed in this chapter, we can analyze almost any circuit by obtaining a set of simultaneous equations that are then solved to obtain the required values of current or voltage. One method of solving simultaneous equations involves Cramer’s rule, whichallowsustocalculatecircuitvariablesasaquotientofdeterminants. The examples in the chapter will illustrate this method; Appendix A also briefly summarizes the essentials the reader needs to know for applying Cramer’s rule. Also in this chapter, we introduce the use of PSpice for Windows, a circuit simulation computer software program that we will use throughout thetext. Finally, weapplythetechniqueslearnedinthischaptertoanalyze transistor circuits. 3.2 NODAL ANALYSIS Nodal analysis provides a general procedure for analyzing circuits using node voltages as the circuit variables. Choosing node voltages instead of element voltages as circuit variables is convenient and reduces the number of equations one must solve simultaneously. To simplify matters, we shall assume in this section that circuits do not contain voltage sources. Circuits that contain voltage sources will be analyzed in the next section. Nodal analysis is also known as the node-voltage method. In nodal analysis, we are interested in finding the node voltages. Given a circuit with n nodes without voltage sources, the nodal analysis of the circuit involves taking the following three steps. Steps to Determine Node Voltages: 1. Select a node as the reference node. Assign voltages v1, v2, . . . , vn−1 to the remaining n − 1 nodes. The voltages are referenced with respect to the reference node. 2. Apply KCL to each of the n − 1 nonreference nodes. Use Ohm’s law to express the branch currents in terms of node voltages. 3. Solve the resulting simultaneous equations to obtain the unknown node voltages. We shall now explain and apply these three steps. The first step in nodal analysis is selecting a node as the reference or datum node. The reference node is commonly called the ground since it is assumed to have zero potential. A reference node is indicated by
  • 84. CHAPTER 3 Methods of Analysis 77 any of the three symbols in Fig. 3.1. The type of ground in Fig. 3.1(b) is called a chassis ground and is used in devices where the case, enclosure, or chassis acts as a reference point for all circuits. When the potential of the earth is used as reference, we use the earth ground in Fig. 3.1(a) or (c). We shall always use the symbol in Fig. 3.1(b). The number of nonreference nodes is equal to the number of independent equations that we will derive. Once we have selected a reference node, we assign voltage desig- nations to nonreference nodes. Consider, for example, the circuit in Fig. 3.2(a). Node 0 is the reference node (v = 0), while nodes 1 and 2 are assigned voltages v1 and v2, respectively. Keep in mind that the node voltages are defined with respect to the reference node. As illustrated in Fig. 3.2(a), each node voltage is the voltage rise from the reference node to the corresponding nonreference node or simply the voltage of that node with respect to the reference node. (a) (b) (c) Figure3.1 Common symbols for indicating a reference node. As the second step, we apply KCL to each nonreference node in the circuit. To avoid putting too much information on the same circuit, the circuit in Fig. 3.2(a) is redrawn in Fig. 3.2(b), where we now add i1, i2, and i3 as the currents through resistors R1, R2, and R3, respectively. At node 1, applying KCL gives I1 = I2 + i1 + i2 (3.1) At node 2, I2 + i2 = i3 (3.2) We now apply Ohm’s law to express the unknown currents i1, i2, and i3 in terms of node voltages. The key idea to bear in mind is that, since resistance is a passive element, by the passive sign convention, current must always flow from a higher potential to a lower potential. Current flows from a higher potential to a lower potential in a resistor. (a) (b) 1 2 v1 i1 i2 i2 i3 v2 I2 0 R3 v2 + − R3 R1 v1 + − R1 I1 I2 R2 R2 I1 Figure 3.2 Typicalcircuitfornodal analysis. We can express this principle as i = vhigher − vlower R (3.3) Note that this principle is in agreement with the way we defined resistance in Chapter 2 (see Fig. 2.1). With this in mind, we obtain from Fig. 3.2(b), i1 = v1 − 0 R1 or i1 = G1v1 i2 = v1 − v2 R2 or i2 = G2(v1 − v2) i3 = v2 − 0 R3 or i3 = G3v2 (3.4) Substituting Eq. (3.4) in Eqs. (3.1) and (3.2) results, respectively, in I1 = I2 + v1 R1 + v1 − v2 R2 (3.5) I2 + v1 − v2 R2 = v2 R3 (3.6)
  • 85. 78 PART 1 DC Circuits In terms of the conductances, Eqs. (3.5) and (3.6) become I1 = I2 + G1v1 + G2(v1 − v2) (3.7) I2 + G2(v1 − v2) = G3v2 (3.8) The third step in nodal analysis is to solve for the node voltages. If we apply KCL to n−1 nonreference nodes, we obtain n−1 simultaneous equations such as Eqs. (3.5) and (3.6) or (3.7) and (3.8). For the circuit of Fig. 3.2, we solve Eqs. (3.5) and (3.6) or (3.7) and (3.8) to obtain the node voltages v1 and v2 using any standard method, such as the substitu- tion method, the elimination method, Cramer’s rule, or matrix inversion. To use either of the last two methods, one must cast the simultaneous equations in matrix form. For example, Eqs. (3.7) and (3.8) can be cast in matrix form as G1 + G2 −G2 −G2 G2 + G3 v1 v2 = I1 − I2 I2 (3.9) which can be solved to get v1 and v2. Equation 3.9 will be generalized in Section 3.6. The simultaneous equations may also be solved using calculators such as HP48 or with software packages such as Matlab, Mathcad, Maple, and Quattro Pro. AppendixAdiscusseshowtouseCramer’srule. E X A M P L E 3 . 1 Calculate the node voltages in the circuit shown in Fig. 3.3(a). Solution: Consider Fig. 3.3(b), where the circuit in Fig. 3.3(a) has been prepared for nodal analysis. Notice how the currents are selected for the application of KCL. Except for the branches with current sources, the labeling of the currents is arbitrary but consistent. (By consistent, we mean that if, for example, we assume that i2 enters the 4 resistor from the left-hand side, i2 must leave the resistor from the right-hand side.) The reference node is selected, and the node voltages v1 and v2 are now to be determined. At node 1, applying KCL and Ohm’s law gives i1 = i2 + i3 ⇒ 5 = v1 − v2 4 + v1 − 0 2 Multiplying each term in the last equation by 4, we obtain 20 = v1 − v2 + 2v1 or 3v1 − v2 = 20 (3.1.1) 2 1 5 A 10 A 2 Ω 6 Ω 4 Ω (a) 5 A 10 A 2 Ω 6 Ω 4 Ω (b) i1 = 5 i1 = 5 i4 = 10 i2 i3 i2 i5 v2 v1 Figure 3.3 For Example 3.1: (a) original circuit, (b) circuit for analysis. At node 2, we do the same thing and get i2 + i4 = i1 + i5 ⇒ v1 − v2 4 + 10 = 5 + v2 − 0 6 Multiplying each term by 12 results in 3v1 − 3v2 + 120 = 60 + 2v2 or −3v1 + 5v2 = 60 (3.1.2)
  • 86. CHAPTER 3 Methods of Analysis 79 Now we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve the equations using any method and obtain the values of v1 and v2. METHOD 1 Using the elimination technique, we add Eqs. (3.1.1) and (3.1.2). 4v2 = 80 ⇒ v2 = 20 V Substituting v2 = 20 in Eq. (3.1.1) gives 3v1 − 20 = 20 ⇒ v1 = 40 3 = 13.33 V METHOD 2 To use Cramer’s rule, we need to put Eqs. (3.1.1) and (3.1.2) in matrix form as 3 −3 −1 5 v1 v2 = 20 60 (3.1.3) The determinant of the matrix is = 3 −3 −1 5 = 15 − 3 = 12 We now obtain v1 and v2 as v1 = 1 = 20 60 −1 5 = 100 + 60 12 = 13.33 V v2 = 2 = 3 −3 20 60 = 180 + 60 12 = 20 V giving us the same result as did the elimination method. If we need the currents, we can easily calculate them from the values of the nodal voltages. i1 = 5 A, i2 = v1 − v2 4 = −1.6667 A, i3 = v1 2 = 6.666 i4 = 10 A, i5 = v2 6 = 3.333 A The fact that i2 is negative shows that the current flows in the direction opposite to the one assumed. P R A C T I C E P R O B L E M 3 . 1 Obtain the node voltages in the circuit in Fig. 3.4. 1 A 4 A 6 Ω 2 Ω 7 Ω 1 2 Figure3.4 For Practice Prob. 3.1. Answer: v1 = −2 V, v2 = −14 V.
  • 87. 80 PART 1 DC Circuits E X A M P L E 3 . 2 Determine the voltages at the nodes in Fig. 3.5(a). Solution: The circuit in this example has three nonreference nodes, unlike the pre- vious example which has two nonreference nodes. We assign voltages to the three nodes as shown in Fig. 3.5(b) and label the currents. 4 Ω 4 Ω 2 Ω 8 Ω ix 1 3 2 0 3 A 2ix (a) ix ix i3 4 Ω 4 Ω 2 Ω 8 Ω i1 v1 v2 i2 i2 i1 v3 3 A 3 A 2ix (b) Figure3.5 For Example 3.2: (a) original circuit, (b) circuit for analysis. At node 1, 3 = i1 + ix ⇒ 3 = v1 − v3 4 + v1 − v2 2 Multiplying by 4 and rearranging terms, we get 3v1 − 2v2 − v3 = 12 (3.2.1) At node 2, ix = i2 + i3 ⇒ v1 − v2 2 = v2 − v3 8 + v2 − 0 4 Multiplying by 8 and rearranging terms, we get −4v1 + 7v2 − v3 = 0 (3.2.2) At node 3, i1 + i2 = 2ix ⇒ v1 − v3 4 + v2 − v3 8 = 2(v1 − v2) 2 Multiplying by 8, rearranging terms, and dividing by 3, we get 2v1 − 3v2 + v3 = 0 (3.2.3) We have three simultaneous equations to solve to get the node voltages v1, v2, and v3. We shall solve the equations in two ways. METHOD 1 Using the elimination technique, we add Eqs. (3.2.1) and (3.2.3). 5v1 − 5v2 = 12
  • 88. CHAPTER 3 Methods of Analysis 81 or v1 − v2 = 12 5 = 2.4 (3.2.4) Adding Eqs. (3.2.2) and (3.2.3) gives −2v1 + 4v2 = 0 ⇒ v1 = 2v2 (3.2.5) Substituting Eq. (3.2.5) into Eq. (3.2.4) yields 2v2 − v2 = 2.4 ⇒ v2 = 2.4, v1 = 2v2 = 4.8 V From Eq. (3.2.3), we get v3 = 3v2 − 2v1 = 3v2 − 4v2 = −v2 = −2.4 V Thus, v1 = 4.8 V, v2 = 2.4 V, v3 = −2.4 V METHOD 2 To use Cramer’s rule, we put Eqs. (3.2.1) to (3.2.3) in matrix form.   3 −4 2 −2 7 −3 −1 −1 1     v1 v2 v3   =   12 0 0   From this, we obtain v1 = 1 , v2 = 2 , v3 = 3 where , 1, 2, and 3 are the determinants to be calculated as follows. As explained in Appendix A, to calculate the determinant of a 3 by 3 matrix, we repeat the first two rows and cross multiply. − − − + + + = 21 − 12 + 4 + 14 − 9 − 8 = 10 3 3 7 7 −4 −2 −2 −4 = = −4 −2 −1 −1 −1 −1 −1 −1 −3 −3 1 1 2 2 3 7 Similarly, we obtain = 84 + 0 + 0 − 0 − 36 − 0 = 48 − − − + + + 7 7 −2 −2 = 0 0 0 12 12 −3 1 −1 −1 1 −1 −1
  • 89. 82 PART 1 DC Circuits = 0 + 0 − 24 − 0 − 0 + 48 = 24 − − − + + + 3 3 −4 −4 −1 −1 1 −1 −1 = 0 0 0 12 12 2 2 = 0 + 144 + 0 − 168 − 0 − 0 = −24 − − − + + + 3 3 7 7 −4 −2 −2 −4 = 0 0 0 12 12 −3 2 3 Thus, we find v1 = 1 = 48 10 = 4.8 V, v2 = 2 = 24 10 = 2.4 V v3 = 3 = −24 10 = −2.4 V as we obtained with Method 1. P R A C T I C E P R O B L E M 3 . 2 Find the voltages at the three nonreference nodes in the circuit of Fig. 3.6. 10 A 2 Ω 3 Ω 4 Ω 6 Ω ix 4ix 1 3 2 Figure3.6 For Practice Prob. 3.2. Answer: v1 = 80 V, v2 = −64 V, v3 = 156 V. 3.3 NODAL ANALYSIS WITH VOLTAGE SOURCES We now consider how voltage sources affect nodal analysis. We use the circuitinFig.3.7forillustration. Considerthefollowingtwopossibilities. CASE 1 If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source. In Fig. 3.7, for example, v1 = 10 V (3.10) Thus our analysis is somewhat simplified by this knowledge of the voltage at this node.
  • 90. CHAPTER 3 Methods of Analysis 83 10 V 5 V 4 Ω 8 Ω 6 Ω 2 Ω v1 v3 v2 i3 i1 i2 i4 Supernode + − + − Figure3.7 A circuit with a supernode. CASE 2 If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a gen- eralized node or supernode; we apply both KCL and KVL to determine the node voltages. Asupernodemayberegardedasaclosedsurface enclosing the voltage source and its two nodes. A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it. In Fig. 3.7, nodes 2 and 3 form a supernode. (We could have more than two nodes forming a single supernode. For example, see the circuit in Fig. 3.14.) We analyze a circuit with supernodes using the same three steps mentioned in the previous section except that the supernodes are treated differently. Why? Because an essential component of nodal analysis is applying KCL, which requires knowing the current through each element. There is no way of knowing the current through a voltage source in advance. However, KCL must be satisfied at a supernode like any other node. Hence, at the supernode in Fig. 3.7, i1 + i4 = i2 + i3 (3.11a) or v1 − v2 2 + v1 − v3 4 = v2 − 0 8 + v3 − 0 6 (3.11b) To apply Kirchhoff’s voltage law to the supernode in Fig. 3.7, we redraw the circuit as shown in Fig. 3.8. Going around the loop in the clockwise direction gives −v2 + 5 + v3 = 0 ⇒ v2 − v3 = 5 (3.12) From Eqs. (3.10), (3.11b), and (3.12), we obtain the node voltages. + − v2 v3 5 V + + − − Figure 3.8 Applying KVL to a supernode.
  • 91. 84 PART 1 DC Circuits Note the following properties of a supernode: 1. The voltage source inside the supernode provides a constraint equation needed to solve for the node voltages. 2. A supernode has no voltage of its own. 3. A supernode requires the application of both KCL and KVL. E X A M P L E 3 . 3 For the circuit shown in Fig. 3.9, find the node voltages. + − 2 A 2 V 7 A 4 Ω 10 Ω 2 Ω v1 v2 Figure3.9 For Example 3.3. Solution: The supernode contains the 2-V source, nodes 1 and 2, and the 10- re- sistor. Applying KCL to the supernode as shown in Fig. 3.10(a) gives 2 = i1 + i2 + 7 Expressing i1 and i2 in terms of the node voltages 2 = v1 − 0 2 + v2 − 0 4 + 7 ⇒ 8 = 2v1 + v2 + 28 or v2 = −20 − 2v1 (3.3.1) To get the relationship between v1 and v2, we apply KVL to the circuit in Fig. 3.10(b). Going around the loop, we obtain −v1 − 2 + v2 = 0 ⇒ v2 = v1 + 2 (3.3.2) From Eqs. (3.3.1) and (3.3.2), we write v2 = v1 + 2 = −20 − 2v1 or 3v1 = −22 ⇒ v1 = −7.333 V and v2 = v1 + 2 = −5.333 V. Note that the 10- resistor does not make any difference because it is connected across the supernode. 2 A 2 A 7 A 7 A 2 Ω 4 Ω v2 v1 i1 i2 1 2 (a) + − (b) 2 V 1 2 + + − − v1 v2 Figure3.10 Applying: (a) KCL to the supernode, (b) KVL to the loop.
  • 92. CHAPTER 3 Methods of Analysis 85 P R A C T I C E P R O B L E M 3 . 3 Find v and i in the circuit in Fig. 3.11. 7 V 3 V 4 Ω 3 Ω 2 Ω 6 Ω + − + − i v + − Figure3.11 For Practice Prob. 3.3. Answer: −0.2 V, 1.4 A. E X A M P L E 3 . 4 Find the node voltages in the circuit of Fig. 3.12. 20 V 2 Ω 4 Ω 6 Ω 3 Ω 1 Ω vx 3vx + − + − 10 A 1 4 3 2 + − Figure3.12 For Example 3.4. Solution: Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCL to the two supernodes as in Fig. 3.13(a). At supernode 1-2, i3 + 10 = i1 + i2 Expressing this in terms of the node voltages, v3 − v2 6 + 10 = v1 − v4 3 + v1 2 or 5v1 + v2 − v3 − 2v4 = 60 (3.4.1) At supernode 3-4, i1 = i3 + i4 + i5 ⇒ v1 − v4 3 = v3 − v2 6 + v4 1 + v3 4 or 4v1 + 2v2 − 5v3 − 16v4 = 0 (3.4.2)
  • 93. 86 PART 1 DC Circuits 10 A 3 Ω 6 Ω 2 Ω 4 Ω 1 Ω (a) i1 i2 i3 i4 i5 v1 v2 v3 v4 vx + − (b) + − + − + − 20 V 3 Ω 6 Ω i3 v1 v2 v3 v4 vx Loop 1 Loop 2 Loop 3 3vx + + + + − − − − i1 i3 Figure3.13 Applying: (a) KCL to the two supernodes, (b) KVL to the loops. We now apply KVL to the branches involving the voltage sources as shown in Fig. 3.13(b). For loop 1, −v1 + 20 + v2 = 0 ⇒ v1 − v2 = 20 (3.4.3) For loop 2, −v3 + 3vx + v4 = 0 But vx = v1 − v4 so that 3v1 − v3 − 2v4 = 0 (3.4.4) For loop 3, vx − 3vx + 6i3 − 20 = 0 But 6i3 = v3 − v2 and vx = v1 − v4. Hence −2v1 − v2 + v3 + 2v4 = 20 (3.4.5) We need four node voltages, v1, v2, v3, and v4, and it requires only four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although the fifth equation is redundant, it can be used to check results. We can eliminate one node voltage so that we solve three simultaneous equations instead of four. From Eq. (3.4.3), v2 = v1 − 20. Substituting this into Eqs. (3.4.1) and (3.4.2), respectively, gives 6v1 − v3 − 2v4 = 80 (3.4.6) and 6v1 − 5v3 − 16v4 = 40 (3.4.7) Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as   3 6 6 −1 −1 −5 −2 −2 −16     v1 v3 v4   =   0 80 40  
  • 94. CHAPTER 3 Methods of Analysis 87 Using Cramer’s rule, = 3 6 6 −1 −1 −5 −2 −2 −16 = −18, 1 = 0 80 40 −1 −1 −5 −2 −2 −16 = −480 3 = 3 6 6 0 80 40 −2 −2 −16 = −3120, 4 = 3 6 6 −1 −1 −5 0 80 40 = 840 Thus, we arrive at the node voltages as v1 = 1 = −480 −18 = 26.667 V, v3 = 3 = −3120 −18 = 173.333 V v4 = 4 = 840 −18 = −46.667 V and v2 = v1 −20 = 6.667 V. We have not used Eq. (3.4.5); it can be used to cross check results. P R A C T I C E P R O B L E M 3 . 4 Find v1, v2, and v3 in the circuit in Fig. 3.14 using nodal analysis. 2 Ω 4 Ω 3 Ω 6 Ω i v1 v2 v3 + − + − 10 V 5i Figure3.14 For Practice Prob. 3.4. Answer: v1 = 3.043 V, v2 = −6.956 V, v3 = 0.6522 V. 3.4 MESH ANALYSIS Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously. Recall that a loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it. Meshanalysisisalsoknownasloopanalysisorthe mesh-current method. Nodal analysis applies KCL to find unknown voltages in a given circuit, while mesh analysis applies KVL to find unknown currents. Mesh analysis is not quite as general as nodal analysis because it is only ap- plicable to a circuit that is planar. A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar. A circuit may have crossing branches and still be planar if it can be redrawn such that it has no crossing branches. For example, the
  • 95. 88 PART 1 DC Circuits circuit in Fig. 3.15(a) has two crossing branches, but it can be redrawn as in Fig. 3.15(b). Hence, the circuit in Fig. 3.15(a) is planar. However, the circuit in Fig. 3.16 is nonplanar, because there is no way to redraw it and avoid the branches crossing. Nonplanar circuits can be handled using nodal analysis, but they will not be considered in this text. (a) 1 A (b) 1 A 1 Ω 1 Ω 3 Ω 2 Ω 4 Ω 5 Ω 8 Ω 7 Ω 6 Ω 2 Ω 4 Ω 7 Ω 8 Ω 5 Ω 6 Ω 3 Ω Figure3.15 (a) A planar circuit with crossing branches, (b) the same circuit redrawn with no crossing branches. 5 A 1 Ω 5 Ω 4 Ω 6 Ω 10 Ω 11 Ω 12 Ω 13 Ω 9 Ω 8 Ω 3 Ω 2 Ω 7 Ω Figure3.16 A nonplanar circuit. To understand mesh analysis, we should first explain more about what we mean by a mesh. A mesh is a loop which does not contain any other loops within it. In Fig. 3.17, for example, paths abefa and bcdeb are meshes, but path abcdefa is not a mesh. The current through a mesh is known as mesh current. In mesh analysis, we are interested in applying KVL to find the mesh currents in a given circuit. Although path abcdefa is a loop and not a mesh, KVL still holds. This is the reason for loosely using the terms loop analysis and mesh analysis to mean the same thing. + − + − I1 R1 R2 R3 i1 i2 I2 I3 V1 V2 a b c d e f Figure3.17 A circuit with two meshes. In this section, we will apply mesh analysis to planar circuits that do not contain current sources. In the next sections, we will consider circuits with current sources. In the mesh analysis of a circuit with n meshes, we take the following three steps.
  • 96. CHAPTER 3 Methods of Analysis 89 Steps to Determine Mesh Currents: 1. Assign mesh currents i1, i2, . . . , in to the n meshes. 2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents. 3. Solve the resulting n simultaneous equations to get the mesh currents. To illustrate the steps, consider the circuit in Fig. 3.17. The first step requires that mesh currents i1 and i2 are assigned to meshes 1 and 2. Although a mesh current may be assigned to each mesh in an arbi- trary direction, it is conventional to assume that each mesh current flows clockwise. The direction of the mesh current is arbitrary— (clockwise or counterclockwise)—and does not affect the validity of the solution. As the second step, we apply KVL to each mesh. Applying KVL to mesh 1, we obtain −V1 + R1i1 + R3(i1 − i2) = 0 or (R1 + R3)i1 − R3i2 = V1 (3.13) For mesh 2, applying KVL gives R2i2 + V2 + R3(i2 − i1) = 0 or −R3i1 + (R2 + R3)i2 = −V2 (3.14) Note in Eq. (3.13) that the coefficient of i1 is the sum of the resistances in the first mesh, while the coefficient of i2 is the negative of the resistance common to meshes 1 and 2. Now observe that the same is true in Eq. (3.14). This can serve as a shortcut way of writing the mesh equations. We will exploit this idea in Section 3.6. The shortcut way will not apply if one mesh cur- rentisassumedclockwiseandtheotherassumed anticlockwise, although this is permissible. The third step is to solve for the mesh currents. Putting Eqs. (3.13) and (3.14) in matrix form yields R1 + R3 −R3 −R3 R2 + R3 i1 i2 = V1 −V2 (3.15) which can be solved to obtain the mesh currents i1 and i2. We are at liberty to use any technique for solving the simultaneous equations. According to Eq. (2.12), if a circuit has n nodes, b branches, and l independent loops or meshes, then l = b − n + 1. Hence, l independent simultaneous equations are required to solve the circuit using mesh analysis. Notice that the branch currents are different from the mesh currents unless the mesh is isolated. To distinguish between the two types of currents, we use i for a mesh current and I for a branch current. The current elements I1, I2, and I3 are algebraic sums of the mesh currents. It is evident from Fig. 3.17 that I1 = i1, I2 = i2, I3 = i1 − i2 (3.16)
  • 97. 90 PART 1 DC Circuits E X A M P L E 3 . 5 For the circuit in Fig. 3.18, find the branch currents I1, I2, and I3 using mesh analysis. + − + − 15 V 10 V 5 Ω 6 Ω 10 Ω 4 Ω I1 i1 I2 i2 I3 Figure3.18 For Example 3.5. Solution: We first obtain the mesh currents using KVL. For mesh 1, −15 + 5i1 + 10(i1 − i2) + 10 = 0 or 3i1 − 2i2 = 1 (3.5.1) For mesh 2, 6i2 + 4i2 + 10(i2 − i1) − 10 = 0 or i1 = 2i2 − 1 (3.5.2) METHOD 1 Using the substitution method, we substitute Eq. (3.5.2) into Eq. (3.5.1), and write 6i2 − 3 − 2i2 = 1 ⇒ i2 = 1 A From Eq. (3.5.2), i1 = 2i2 − 1 = 2 − 1 = 1 A. Thus, I1 = i1 = 1 A, I2 = i2 = 1 A, I3 = i1 − i2 = 0 METHOD 2 To use Cramer’s rule, we cast Eqs. (3.5.1) and (3.5.2) in matrix form as 3 −1 −2 2 i1 i2 = 1 1 We obtain the determinants = 3 −1 −2 2 = 6 − 2 = 4 1 = 1 1 −2 2 = 2 + 2 = 4, 2 = 3 −1 1 1 = 3 + 1 = 4 Thus, i1 = 1 = 1 A, i2 = 2 = 1 A as before. P R A C T I C E P R O B L E M 3 . 5 Calculate the mesh currents i1 and i2 in the circuit of Fig. 3.19. 12 V 8 V 2 Ω 4 Ω 3 Ω 12 Ω 9 Ω i1 i2 + − + − Figure3.19 For Practice Prob. 3.5. Answer: i1 = 2 3 A, i2 = 0 A.
  • 98. CHAPTER 3 Methods of Analysis 91 E X A M P L E 3 . 6 Use mesh analysis to find the current io in the circuit in Fig. 3.20. + − + − 24 V 12 Ω 4 Ω 10 Ω 24 Ω i1 i1 i3 i2 i2 io 4io A Figure3.20 For Example 3.6. Solution: We apply KVL to the three meshes in turn. For mesh 1, −24 + 10(i1 − i2) + 12(i1 − i3) = 0 or 11i1 − 5i2 − 6i3 = 12 (3.6.1) For mesh 2, 24i2 + 4(i2 − i3) + 10(i2 − i1) = 0 or −5i1 + 19i2 − 2i3 = 0 (3.6.2) For mesh 3, 4io + 12(i3 − i1) + 4(i3 − i2) = 0 But at node A, io = i1 − i2, so that 4(i1 − i2) + 12(i3 − i1) + 4(i3 − i2) = 0 or −i1 − i2 + 2i3 = 0 (3.6.3) In matrix form, Eqs. (3.6.1) to (3.6.3) become   11 −5 −1 −5 19 −1 −6 −2 2     i1 i2 i3   =   12 0 0   We obtain the determinants as = 418 − 30 − 10 − 114 − 22 − 50 = 192 − − − + + + 19 19 −5 −5 = −5 −1 11 −5 11 −1 −6 −2 2 −6 −2 = 456 − 24 = 432 − − − + + + 19 19 −5 −5 = 0 0 12 0 12 −1 −6 −2 2 −6 −2 1 = 24 + 120 = 144 − − − + + + 0 0 12 12 = −5 −1 11 −5 11 0 −6 −2 2 −6 −2 2
  • 99. 92 PART 1 DC Circuits = 60 + 228 = 288 − − − + + + 0 19 19 0 12 12 = −5 −5 −5 −1 −1 11 −5 11 0 3 We calculate the mesh currents using Cramer’s rule as i1 = 1 = 432 192 = 2.25 A, i2 = 2 = 144 192 = 0.75 A i3 = 3 = 288 192 = 1.5 A Thus, io = i1 − i2 = 1.5 A. P R A C T I C E P R O B L E M 3 . 6 Using mesh analysis, find io in the circuit in Fig. 3.21. + − – + 20 V 4 Ω 8 Ω 2 Ω 6 Ω i1 i2 i3 10io io Figure3.21 For Practice Prob. 3.6. Answer: −5 A. 3.5 MESH ANALYSIS WITH CURRENT SOURCES Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what we encountered in the previous section, because the presence of the current sources reduces the number of equations. Consider the following two possible cases. + − 5 A 10 V 4 Ω 3 Ω 6 Ω i1 i2 Figure3.22 A circuit with a current source. CASE 1 When a current source exists only in one mesh: Consider the circuit in Fig. 3.22, for example. We set i2 = −5 A and write a mesh equation for the other mesh in the usual way, that is, −10 + 4i1 + 6(i1 − i2) = 0 ⇒ i1 = −2 A (3.17) CASE 2 When a current source exists between two meshes: Consider the circuit in Fig. 3.23(a), for example. We create a supermesh by ex- cluding the current source and any elements connected in series with it, as shown in Fig. 3.23(b). Thus, A supermesh results when two meshes have a (dependent or independent) current source in common.
  • 100. CHAPTER 3 Methods of Analysis 93 (b) 20 V 4 Ω 6 Ω 10 Ω i1 i2 + − + − 6 A 20 V 6 Ω 10 Ω 2 Ω 4 Ω i1 i1 i2 i2 0 (a) Exclude these elements Figure 3.23 (a) Two meshes having a current source in common, (b) a supermesh, created by excluding the current source. As shown in Fig. 3.23(b), we create a supermesh as the periphery of the two meshes and treat it differently. (If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh.) Why treat the supermesh differently? Because mesh analy- sis applies KVL—which requires that we know the voltage across each branch—and we do not know the voltage across a current source in ad- vance. However, a supermesh must satisfy KVL like any other mesh. Therefore, applying KVL to the supermesh in Fig. 3.23(b) gives −20 + 6i1 + 10i2 + 4i2 = 0 or 6i1 + 14i2 = 20 (3.18) We apply KCL to a node in the branch where the two meshes intersect. Applying KCL to node 0 in Fig. 3.23(a) gives i2 = i1 + 6 (3.19) Solving Eqs. (3.18) and (3.19), we get i1 = −3.2 A, i2 = 2.8 A (3.20) Note the following properties of a supermesh: 1. The current source in the supermesh is not completely ignored; it provides the constraint equation necessary to solve for the mesh currents. 2. A supermesh has no current of its own. 3. A supermesh requires the application of both KVL and KCL. E X A M P L E 3 . 7 For the circuit in Fig. 3.24, find i1 to i4 using mesh analysis. Solution: Notethatmeshes1and2formasupermeshsincetheyhaveanindependent current source in common. Also, meshes 2 and 3 form another supermesh
  • 101. 94 PART 1 DC Circuits + − 10 V 6 Ω 8 Ω 2 Ω 4 Ω i1 i2 i3 i4 2 Ω 5 A i1 i2 i2 i3 io P Q 3io Figure3.24 For Example 3.7. because they have a dependent current source in common. The two supermeshes intersect and form a larger supermesh as shown. Applying KVL to the larger supermesh, 2i1 + 4i3 + 8(i3 − i4) + 6i2 = 0 or i1 + 3i2 + 6i3 − 4i4 = 0 (3.7.1) For the independent current source, we apply KCL to node P: i2 = i1 + 5 (3.7.2) For the dependent current source, we apply KCL to node Q: i2 = i3 + 3io But io = −i4, hence, i2 = i3 − 3i4 (3.7.3) Applying KVL in mesh 4, 2i4 + 8(i4 − i3) + 10 = 0 or 5i4 − 4i3 = −5 (3.7.4) From Eqs. (3.7.1) to (3.7.4), i1 = −7.5 A, i2 = −2.5 A, i3 = 3.93 A, i4 = 2.143 A P R A C T I C E P R O B L E M 3 . 7 Use mesh analysis to determine i1, i2, and i3 in Fig. 3.25. + − 3 A 6 V 1 Ω 2 Ω 2 Ω 8 Ω 4 Ω i1 i3 i2 Figure3.25 For Practice Prob. 3.7. Answer: i1 = 3.474 A, i2 = 0.4737 A, i3 = 1.1052 A.
  • 102. CHAPTER 3 Methods of Analysis 95 †3.6 NODAL AND MESH ANALYSES BY INSPECTION This section presents a generalized procedure for nodal or mesh analysis. It is a shortcut approach based on mere inspection of a circuit. When all sources in a circuit are independent current sources, we do not need to apply KCL to each node to obtain the node-voltage equa- tions as we did in Section 3.2. We can obtain the equations by mere inspection of the circuit. As an example, let us reexamine the circuit in Fig. 3.2, shown again in Fig. 3.26(a) for convenience. The circuit has two nonreference nodes and the node equations were derived in Section 3.2 as G1 + G2 −G2 −G2 G2 + G3 v1 v2 = I1 − I2 I2 (3.21) Observe that each of the diagonal terms is the sum of the conductances connected directly to node 1 or 2, while the off-diagonal terms are the negatives of the conductances connected between the nodes. Also, each term on the right-hand side of Eq. (3.21) is the algebraic sum of the currents entering the node. I1 v1 G1 G3 G2 I2 v2 (a) (b) i1 i3 V1 V2 + − + − R1 R2 R3 Figure 3.26 (a) The circuit in Fig. 3.2, (b) the circuit in Fig. 3.17. In general, if a circuit with independent current sources has N nonreference nodes, the node-voltage equations can be written in terms of the conductances as      G11 G21 . . . GN1 G12 G22 . . . GN2 . . . . . . . . . . . . G1N G2N . . . GNN          v1 v2 . . . vN     =     i1 i2 . . . iN     (3.22) or simply Gv = i (3.23) where Gkk = Sum of the conductances connected to node k Gkj = Gjk = Negative of the sum of the conductances directly connecting nodes k and j, k = j vk = Unknown voltage at node k ik = Sum of all independent current sources directly connected to node k, with currents entering the node treated as positive G is called the conductance matrix, v is the output vector; and i is the input vector. Equation (3.22) can be solved to obtain the unknown node voltages. Keep in mind that this is valid for circuits with only independent current sources and linear resistors. Similarly, wecanobtainmesh-currentequationsbyinspectionwhen a linear resistive circuit has only independent voltage sources. Consider the circuit in Fig. 3.17, shown again in Fig. 3.26(b) for convenience. The circuit has two nonreference nodes and the node equations were derived in Section 3.4 as R1 + R3 −R3 −R3 R2 + R3 i1 i2 = v1 −v2 (3.24)
  • 103. 96 PART 1 DC Circuits We notice that each of the diagonal terms is the sum of the resistances in the related mesh, while each of the off-diagonal terms is the negative of the resistance common to meshes 1 and 2. Each term on the right-hand side of Eq. (3.24) is the algebraic sum taken clockwise of all independent voltage sources in the related mesh. In general, if the circuit has N meshes, the mesh-current equations can be expressed in terms of the resistances as     R11 R21 . . . RN1 R12 R22 . . . RN2 . . . . . . . . . . . . R1N R2N . . . RNN         i1 i2 . . . iN     =     v1 v2 . . . vN     (3.25) or simply Ri = v (3.26) where Rkk = Sum of the resistances in mesh k Rkj = Rjk = Negative of the sum of the resistances in common with meshes k and j, k = j ik = Unknown mesh current for mesh k in the clockwise direction vk = Sum taken clockwise of all independent voltage sources in mesh k, with voltage rise treated as positive R is called the resistance matrix, i is the output vector; and v is the input vector. We can solve Eq. (3.25) to obtain the unknown mesh currents. E X A M P L E 3 . 8 Write the node-voltage matrix equations for the circuit in Fig. 3.27 by inspection. 3 A 1 A 4 A 2 A 10 Ω 5 Ω 1 Ω 8 Ω 8 Ω v1 v2 v3 v4 4 Ω 2 Ω Figure3.27 For Example 3.8.
  • 104. CHAPTER 3 Methods of Analysis 97 Solution: The circuit in Fig. 3.27 has four nonreference nodes, so we need four node equations. This implies that the size of the conductance matrix G, is 4 by 4. The diagonal terms of G, in siemens, are G11 = 1 5 + 1 10 = 0.3, G22 = 1 5 + 1 8 + 1 1 = 1.325 G33 = 1 8 + 1 8 + 1 4 = 0.5, G44 = 1 8 + 1 2 + 1 1 = 1.625 The off-diagonal terms are G12 = − 1 5 = −0.2, G13 = G14 = 0 G21 = −0.2, G23 = − 1 8 = −0.125, G24 = − 1 1 = −1 G31 = 0, G32 = −0.125, G34 = − 1 8 = −0.125 G41 = 0, G42 = −1, G43 = −0.125 The input current vector i has the following terms, in amperes: i1 = 3, i2 = −1 − 2 = −3, i3 = 0, i4 = 2 + 4 = 6 Thus the node-voltage equations are     0.3 −0.2 0 0 −0.2 1.325 −0.125 −1 0 −0.125 0.5 −0.125 0 −1 −0.125 1.625         v1 v2 v3 v4     =     3 −3 0 6     which can be solved to obtain the node voltages v1, v2, v3, and v4. P R A C T I C E P R O B L E M 3 . 8 By inspection, obtain the node-voltage equations for the circuit in Fig. 3.28. 1 A 2 A 3 A 10 Ω 1 Ω 5 Ω 4 Ω 2 Ω v1 v2 v3 v4 Figure3.28 For Practice Prob. 3.8. Answer:     1.3 −0.2 −1 0 −0.2 0.2 0 0 −1 0 1.25 −0.25 0 0 −0.25 0.75         v1 v2 v3 v4     =     0 3 −1 3     E X A M P L E 3 . 9 Byinspection, writethemesh-currentequationsforthecircuitinFig.3.29.
  • 105. 98 PART 1 DC Circuits + − + − + − + − 10 V 4 V 2 Ω 2 Ω 5 Ω 2 Ω 4 Ω 3 Ω 3 Ω 1 Ω 1 Ω 4 Ω i1 i2 i3 i4 i5 6 V 12 V Figure3.29 For Example 3.9. Solution: We have five meshes, so the resistance matrix is 5 by 5. The diagonal terms, in ohms, are: R11 = 5 + 2 + 2 = 9, R22 = 2 + 4 + 1 + 1 + 2 = 10 R33 = 2 + 3 + 4 = 9, R44 = 1 + 3 + 4 = 8, R55 = 1 + 3 = 4 The off-diagonal terms are: R12 = −2, R13 = −2, R14 = 0 = R15 R21 = −2, R23 = −4, R24 = −1, R25 = −1 R31 = −2, R32 = −4, R34 = 0 = R35 R41 = 0, R42 = −1, R43 = 0, R45 = −3 R51 = 0, R52 = −1, R53 = 0, R54 = −3 The input voltage vector v has the following terms in volts: v1 = 4, v2 = 10 − 4 = 6 v3 = −12 + 6 = −6, v4 = 0, v5 = −6 Thus the mesh-current equations are:       9 −2 −2 0 0 −2 10 −4 −1 −1 −2 −4 9 0 0 0 −1 0 8 −3 0 −1 0 −3 4             i1 i2 i3 i4 i5       =       4 6 −6 0 −6       From this, we can obtain mesh currents i1, i2, i3, i4, and i5. P R A C T I C E P R O B L E M 3 . 9 By inspection, obtain the mesh-current equations for the circuit in Fig. 3.30.
  • 106. CHAPTER 3 Methods of Analysis 99 + − + − 24 V 12 V 10 V 50 Ω 40 Ω i1 i2 i3 i4 i5 10 Ω 30 Ω 20 Ω 60 Ω 80 Ω + − Figure3.30 For Practice Prob. 3.9. Answer:       170 −40 0 −80 0 −40 80 −30 −10 0 0 −30 50 0 −20 −80 −10 0 90 0 0 0 −20 0 80             i1 i2 i3 i4 i5       =       24 0 −12 10 −10       3.7 NODAL VERSUS MESH ANALYSIS Both nodal and mesh analyses provide a systematic way of analyzing a complex network. Someone may ask: Given a network to be analyzed, how do we know which method is better or more efficient? The choice of the better method is dictated by two factors. The first factor is the nature the particular network. Networks that containmanyseries-connectedelements, voltagesources, orsupermeshes are more suitable for mesh analysis, whereas networks with parallel- connected elements, current sources, or supernodes are more suitable for nodal analysis. Also, a circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis. The key is to select the method that results in the smaller number of equations. The second factor is the information required. If node voltages are required, it may be expedient to apply nodal analysis. If branch or mesh currents are required, it may be better to use mesh analysis. It is helpful to be familiar with both methods of analysis, for at least two reasons. First, one method can be used to check the results from the other method, if possible. Second, since each method has its limitations, only one method may be suitable for a particular problem. For example,
  • 107. 100 PART 1 DC Circuits mesh analysis is the only method to use in analyzing transistor circuits, as we shall see in Section 3.9. But mesh analysis cannot easily be used to solve an op amp circuit, as we shall see in Chapter 5, because there is no direct way to obtain the voltage across the op amp itself. For nonplanar networks, nodal analysis is the only option, because mesh analysis only applies to planar networks. Also, nodal analysis is more amenable to solution by computer, as it is easy to program. This allows one to analyze complicated circuits that defy hand calculation. A computer software package based on nodal analysis is introduced next. 3.8 CIRCUIT ANALYSIS WITH PSPICE PSpice is a computer software circuit analysis program that we will grad- ually learn to use throught the course of this text. This section illustrates how to use PSpice for Windows to analyze the dc circuits we have studied so far. Appendix D provides a tutorial on using PSpice for Windows. The reader is expected to review Sections D.1 through D.3 of Ap- pendix D before proceeding in this section. It should be noted that PSpice is only helpful in determining branch voltages and currents when the nu- merical values of all the circuit components are known. E X A M P L E 3 . 1 0 Use PSpice to find the node voltages in the circuit of Fig. 3.31. + − 3 A 120 V 20 Ω 30 Ω 40 Ω 10 Ω 1 2 3 0 Figure3.31 For Example 3.10. Solution: The first step is to draw the given circuit using Schematics. If one follows the instructions given in Appendix sections D.2 and D.3, the schematic in Fig. 3.32 is produced. Since this is a dc analysis, we use voltage source VDC and current source IDC. The pseudocomponent VIEWPOINTS are added to display the required node voltages. Once the circuit is drawn and saved as exam310.sch, we run PSpice by selecting Analysis/Simulate. The circuit is simulated and the results are displayed on VIEWPOINTS and also saved in output file exam310.out. The output file includes the following: NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE (1) 120.0000 (2) 81.2900 (3) 89.0320 indicating that V1 = 120 V, V2 = 81.29 V, V3 = 89.032 V. + − R1 R3 20 10 120 V V1 R2 R4 30 40 I1 3 A IDC 0 1 2 3 120.0000 81.2900 89.0320 Figure 3.32 For Example 3.10; the schematic of the circuit in Fig. 3.31.
  • 108. CHAPTER 3 Methods of Analysis 101 P R A C T I C E P R O B L E M 3 . 1 0 For the circuit in Fig. 3.33, use PSpice to find the node voltages. + − 2 A 200 V 30 Ω 60 Ω 50 Ω 100 Ω 25 Ω 1 2 3 0 Figure3.33 For Practice Prob. 3.10. Answer: V1 = −40 V, V2 = 57.14 V, V3 = 200 V. E X A M P L E 3 . 1 1 In the circuit in Fig. 3.34, determine the currents i1, i2, and i3. Solution: The schematic is shown in Fig. 3.35. (The schematic in Fig. 3.35 includes the output results, implying that it is the schematic displayed on the screen after the simulation.) Notice that the voltage-controlled voltage source E1 in Fig. 3.35 is connected so that its input is the voltage across the 4- resistor; its gain is set equal to 3. In order to display the required currents, we insert pseudocomponent IPROBES in the appropriate branches. The schematic is saved as exam311.sch and simulated by selecting Analy- sis/Simulate. The results are displayed on IPROBES as shown in Fig. 3.35 and saved in output file exam311.out. From the output file or the IPROBES, we obtain i1 = i2 = 1.333 A and i3 = 2.667 A. + − + − 24 V 1 Ω i1 i2 i3 + − 4 Ω 2 Ω 2 Ω 8 Ω 4 Ω 3vo vo Figure3.34 For Example 3.11.
  • 109. 102 PART 1 DC Circuits + − 24 V V1 R1 4 R2 2 R3 8 R4 4 1.333E+00 1.333E+00 2.667E+00 0 R6 1 R5 2 E E1 + − − + Figure3.35 The schematic of the circuit in Fig. 3.34. P R A C T I C E P R O B L E M 3 . 1 1 Use PSpice to determine currents i1, i2, and i3 in the circuit of Fig. 3.36. + − 2 A 10 V 2 Ω i1 i1 i2 4 Ω 1 Ω 2 Ω i3 Figure3.36 For Practice Prob. 3.11. Answer: i1 = −0.4286 A, i2 = 2.286 A, i3 = 2 A. †3.9 APPLICATIONS: DC TRANSISTOR CIRCUITS Most of us deal with electronic products on a routine basis and have some experience with personal computers. A basic component for the integrated circuits found in these electronics and computers is the ac- tive, three-terminal device known as the transistor. Understanding the transistor is essential before an engineer can start an electronic circuit design. Figure 3.37 depicts various kinds of transistors commercially avail- able. There are two basic types of transistors: bipolar junction transis- tors (BJTs) and field-effect transistors (FETs). Here, we consider only the BJTs, which were the first of the two and are still used today. Our objective is to present enough detail about the BJT to enable us to apply the techniques developed in this chapter to analyze dc transistor circuits. TherearetwotypesofBJTs: npnandpnp, withtheircircuitsymbols as shown in Fig. 3.38. Each type has three terminals, designated as emit- ter (E), base (B), and collector (C). For the npn transistor, the currents and
  • 110. CHAPTER 3 Methods of Analysis 103 Figure3.37 Various types of transistors. (Courtesy of Tech America.) n n p Base Collector Emitter E B C (a) p p n Base Collector Emitter E B C (b) Figure3.38 Two types of BJTs and their circuit symbols: (a) npn, (b) pnp. B C E + + + − − − VCB VCE VBE B C E IB IC IE (a) (b) Figure3.39 The terminal variables of an npn transistor: (a) currents, (b) voltages. voltages of the transistor are specified as in Fig. 3.39. Applying KCL to Fig. 3.39(a) gives IE = IB + IC (3.27) whereIE, IC, andIB areemitter, collector, andbasecurrents, respectively. Similarly, applying KVL to Fig. 3.39(b) gives VCE + VEB + VBC = 0 (3.28) where VCE, VEB, and VBC are collector-emitter, emitter-base, and base- collector voltages. The BJT can operate in one of three modes: active, cutoff, and saturation. When transistors operate in the active mode, typ- ically VBE 0.7 V, IC = αIE (3.29) where α is called the common-base current gain. In Eq. (3.29), α denotes the fraction of electrons injected by the emitter that are collected by the collector. Also, IC = βIB (3.30) where β is known as the common-emitter current gain. The α and β are characteristic properties of a given transistor and assume constant values for that transistor. Typically, α takes values in the range of 0.98 to 0.999, while β takes values in the range 50 to 1000. From Eqs. (3.27) to (3.30), it is evident that IE = (1 + β)IB (3.31) and β = α 1 − α (3.32)
  • 111. 104 PART 1 DC Circuits These equations show that, in the active mode, the BJT can be modeled as a dependent current-controlled current source. Thus, in circuit analysis, the dc equivalent model in Fig. 3.40(b) may be used to replace the npn transistor in Fig. 3.40(a). Since β in Eq. (3.32) is large, a small base current controls large currents in the output circuit. Consequently, the bipolar transistor can serve as an amplifier, producing both current gain and voltage gain. Such amplifiers can be used to furnish a considerable amount of power to transducers such as loudspeakers or control motors. B C E IB IC VBE VCE + − + − B C E IB (a) (b) VBE VCE + + − − bIB Figure 3.40 (a) An npn transistor, (b) its dc equivalent model. In fact, transistor circuits provide motivation to study dependent sources. It should be observed in the following examples that one cannot directly analyze transistor circuits using nodal analysis because of the potential difference between the terminals of the transistor. Only when the transistor is replaced by its equivalent model can we apply nodal analysis. E X A M P L E 3 . 1 2 Find IB, IC, and vo in the transistor circuit of Fig. 3.41. Assume that the transistor operates in the active mode and that β = 50. Solution: For the input loop, KVL gives −4 + IB(20 × 103 ) + VBE = 0 IC + − + + − − + − 4 V 6 V 20 kΩ IB VBE vo Output loop Input loop 100 Ω Figure3.41 For Example 3.12.
  • 112. CHAPTER 3 Methods of Analysis 105 Since VBE = 0.7 V in the active mode, IB = 4 − 0.7 20 × 103 = 165 µA But IC = βIB = 50 × 165 µA = 8.25 mA For the output loop, KVL gives −vo − 100IC + 6 = 0 or vo = 6 − 100IC = 6 − 0.825 = 5.175 V Note that vo = VCE in this case. P R A C T I C E P R O B L E M 3 . 1 2 For the transistor circuit in Fig. 3.42, let β = 100 and VBE = 0.7 V. De- termine vo and VCE. + − + + + − − − + − 5 V 12 V 10 kΩ 500 Ω VBE VCE 200 Ω vo Figure3.42 For Practice Prob. 3.12. Answer: 2.876 V, 1.984 V. E X A M P L E 3 . 1 3 For the BJT circuit in Fig. 3.43, β = 150 and VBE = 0.7 V. Find vo. Solution: We can solve this problem in two ways. One way is by direct analysis of the circuit in Fig. 3.43. Another way is by replacing the transistor with its equivalent circuit. 2 V 100 kΩ + − + − 16 V 200 kΩ 1 kΩ + − vo Figure3.43 For Example 3.13. METHOD 1 We can solve the problem as we solved the problem in the previous example. We apply KVL to the input and output loops as shown in Fig. 3.44(a). For loop 1, 2 = 100 × 103 I1 + 200 × 103 I2 (3.13.1) For loop 2, VBE = 0.7 = 200 × 103 I2 ⇒ I2 = 3.5 µA (3.13.2) For loop 3, −vo − 1000IC + 16 = 0
  • 113. 106 PART 1 DC Circuits or vo = 16 − 1000IC (3.13.3) From Eqs. (3.13.1) and (3.13.2), I1 = 2 − 0.7 100 × 103 = 13 µA, IB = I1 − I2 = 9.5 µA IC = βIB = 150 × 9.5 µA = 1.425 mA Substituting for IC in Eq. (3.13.3), vo = 16 − 1.425 = 14.575 V + − Vo + − + − I1 IB IC I2 1 kΩ 100 kΩ 200 kΩ 2 V 16 V 16 V 2 V Loop 1 Loop 2 Loop 3 (a) (b) + − + − 0.7 V 100 kΩ 200 kΩ 1 kΩ 150IB IB I1 I2 B C E + − Vo Figure 3.44 Solution of the problem in Example 3.13: (a) method 1, (b) method 2. METHOD 2 We can modify the circuit in Fig. 3.43 by replacing the transistor by its equivalent model in Fig. 3.40(b). The result is the circuit shown in Fig. 3.44(b). Notice that the locations of the base (B), emitter (E), and collector (C) remain the same in both the original circuit in Fig. 3.43 and its equivalent circuit in Fig. 3.44(b). From the output loop, vo = 16 − 1000(150IB) But IB = I1 − I2 = 2 − 0.7 100 × 103 − 0.7 200 × 103 = (13 − 3.5) µA = 9.5 µA and so vo = 16 − 1000(150 × 9.5 × 10−6 ) = 14.575 V
  • 114. CHAPTER 3 Methods of Analysis 107 P R A C T I C E P R O B L E M 3 . 1 3 The transistor circuit in Fig. 3.45 has β = 80 and VBE = 0.7 V. Find vo and io. 1 V 10 V 30 kΩ 20 kΩ 20 kΩ + − io VBE + − vo + − + − Figure 3.45 For Practice Prob. 3.13. Answer: −3 V, −150 µA. 3.10 SUMMARY 1. Nodal analysis is the application of Kirchhoff’s current law at the nonreference nodes. (It is applicable to both planar and nonplanar circuits.) We express the result in terms of the node voltages. Solving the simultaneous equations yields the node voltages. 2. A supernode consists of two nonreference nodes connected by a (dependent or independent) voltage source. 3. Mesh analysis is the application of Kirchhoff’s voltage law around meshes in a planar circuit. We express the result in terms of mesh currents. Solving the simultaneous equations yields the mesh currents. 4. A supermesh consists of two meshes that have a (dependent or independent) current source in common. 5. Nodal analysis is normally used when a circuit has fewer node equations than mesh equations. Mesh analysis is normally used when a circuit has fewer mesh equations than node equations. 6. Circuit analysis can be carried out using PSpice. 7. DC transistor circuits can be analyzed using the techniques cover- ed in this chapter. REVIEW QUESTIONS 3.1 At node 1 in the circuit in Fig. 3.46, applying KCL gives: (a) 2 + 12 − v1 3 = v1 6 + v1 − v2 4 (b) 2 + v1 − 12 3 = v1 6 + v2 − v1 4 (c) 2 + 12 − v1 3 = 0 − v1 6 + v1 − v2 4 (d) 2 + v1 − 12 3 = 0 − v1 6 + v2 − v1 4
  • 115. 108 PART 1 DC Circuits 2 A v1 1 2 v2 12 V + − 3 Ω 4 Ω 6 Ω 6 Ω 8 Ω Figure 3.46 For Review Questions 3.1 and 3.2. 3.2 In the circuit in Fig. 3.46, applying KCL at node 2 gives: (a) v2 − v1 4 + v2 8 = v2 6 (b) v1 − v2 4 + v2 8 = v2 6 (c) v1 − v2 4 + 12 − v2 8 = v2 6 (d) v2 − v1 4 + v2 − 12 8 = v2 6 3.3 For the circuit in Fig. 3.47, v1 and v2 are related as: (a) v1 = 6i + 8 + v2 (b) v1 = 6i − 8 + v2 (c) v1 = −6i + 8 + v2 (d) v1 = −6i − 8 + v2 12 V + − 4 Ω 6 Ω 8 V v2 v1 i + − Figure 3.47 For Review Questions 3.3 and 3.4. 3.4 In the circuit in Fig. 3.47, the voltage v2 is: (a) −8 V (b) −1.6 V (c) 1.6 V (d) 8 V 3.5 The current i in the circuit in Fig. 3.48 is: (a) −2.667 A (b) −0.667 A (c) 0.667 A (d) 2.667 A 10 V + − 6 V + − 4 Ω i 2 Ω Figure 3.48 For Review Questions 3.5 and 3.6. 3.6 The loop equation for the circuit in Fig. 3.48 is: (a) −10 + 4i + 6 + 2i = 0 (b) 10 + 4i + 6 + 2i = 0 (c) 10 + 4i − 6 + 2i = 0 (d) −10 + 4i − 6 + 2i = 0 3.7 In the circuit in Fig. 3.49, current i1 is: (a) 4 A (b) 3 A (c) 2 A (d) 1 A i1 i2 2 A 20 V + − 2 Ω 1 Ω 3 Ω 4 Ω v + − Figure 3.49 For Review Questions 3.7 and 3.8. 3.8 The voltage v across the current source in the circuit of Fig. 3.49 is: (a) 20 V (b) 15 V (c) 10 V (d) 5 V 3.9 The PSpice part name for a current-controlled voltage source is: (a) EX (b) FX (c) HX (d) GX 3.10 Which of the following statements are not true of the pseudocomponent IPROBE: (a) It must be connected in series. (b) It plots the branch current. (c) It displays the current through the branch in which it is connected. (d) It can be used to display voltage by connecting it in parallel. (e) It is used only for dc analysis. (f) It does not correspond to a particular circuit element. Answers: 3.1a, 3.2c, 3.3b, 3.4d, 3.5c, 3.6a, 3.7d, 3.8b, 3.9c, 3.10b,d.
  • 116. CHAPTER 3 Methods of Analysis 109 PROBLEMS Sections 3.2 and 3.3 Nodal Analysis 3.1 Determine v1, v2, and the power dissipated in all the resistors in the circuit of Fig. 3.50. 10 A 6 A 8 Ω 4 Ω v1 v2 2 Ω Figure 3.50 For Prob. 3.1. 3.2 For the circuit in Fig. 3.51, obtain v1 and v2. 3 A 6 A 5 Ω 10 Ω 2 Ω v1 v2 4 Ω Figure 3.51 For Prob. 3.2. 3.3 Find the currents i1 through i4 and the voltage vo in the circuit in Fig. 3.52. 10 A 2 A 60 Ω 30 Ω 20 Ω 10 Ω i1 i2 i3 i4 vo Figure 3.52 For Prob. 3.3. 3.4 Given the circuit in Fig. 3.53, calculate the currents i1 through i4. 4 A 5 A 2 A 5 Ω 10 Ω 10 Ω 5 Ω i1 i2 i3 i4 Figure 3.53 For Prob. 3.4. 3.5 Obtain vo in the circuit of Fig. 3.54. 30 V + − 2 kΩ 20 V + − 5 kΩ 4 kΩ vo + − Figure 3.54 For Prob. 3.5. 3.6 Use nodal analysis to obtain vo in the circuit in Fig. 3.55. 6 Ω 2 Ω 12 V 10 V + − + − 4 Ω i3 i2 vo i1 Figure 3.55 For Prob. 3.6. 3.7 Using nodal analysis, find vo in the circuit of Fig. 3.56. 3 V 4vo + − 2 Ω vo + − 1 Ω 3 Ω 5 Ω + − Figure 3.56 For Prob. 3.7. 3.8 Calculate vo in the circuit in Fig. 3.57. 12 V 2vo + − 8 Ω 6 Ω + − 3 Ω vo + − Figure 3.57 For Prob. 3.8. 3.9 Find io in the circuit in Fig. 3.58.
  • 117. 110 PART 1 DC Circuits 2 Ω 4 Ω 8 Ω 1 Ω 4 A 2io io Figure 3.58 For Prob. 3.9. 3.10 Solve for i1 and i2 in the circuit in Fig. 3.22 (Section 3.5) using nodal analysis. 3.11 Use nodal analysis to find currents i1 and i2 in the circuit of Fig. 3.59. 24 V + − 40 Ω 20 Ω 5 A 10 Ω 20 Ω 30 Ω i2 i1 Figure 3.59 For Prob. 3.11. 3.12 Calculate v1 and v2 in the circuit in Fig. 3.60 using nodal analysis. 8 Ω 4 Ω 3 A 2 Ω 2 V v2 v1 + − Figure 3.60 For Prob. 3.12. 3.13 Using nodal analysis, find vo in the circuit of Fig. 3.61. 2 Ω 5 A 8 Ω + − + − 4 Ω 20 V vo + − 1 Ω 40 V Figure 3.61 For Prob. 3.13. 3.14 Apply nodal analysis to find io and the power dissipated in each resistor in the circuit of Fig. 3.62. 5 S 6 S 2 A io 4 A 3 S 10 V + − Figure 3.62 For Prob. 3.14. 3.15 Determine voltages v1 through v3 in the circuit of Fig. 3.63 using nodal analysis. 1 S 13 V 2 S v1 v2 2vo v3 8 S 2 A 4 S vo + − + − + − Figure 3.63 For Prob. 3.15. 3.16 Using nodal analysis, find current io in the circuit of Fig. 3.64. 60 V io 3io 10 Ω 8 Ω 2 Ω + − 4 Ω Figure 3.64 For Prob. 3.16. 3.17 Determine the node voltages in the circuit in Fig. 3.65 using nodal analysis. 5 A 2 3 1 2 Ω 2 Ω 10 V + − 8 Ω 4 Ω Figure 3.65 For Prob. 3.17.
  • 118. CHAPTER 3 Methods of Analysis 111 3.18 For the circuit in Fig. 3.66, find v1 and v2 using nodal analysis. 3 mA v2 v1 2 kΩ 4 kΩ 1 kΩ vo 3vo + − + − Figure 3.66 For Prob. 3.18. 3.19 Determine v1 and v2 in the circuit in Fig. 3.67. 3 A v2 5vo v1 8 Ω 1 Ω 4 Ω 12 V 2 Ω vo + − – + + − Figure 3.67 For Prob. 3.19. 3.20 Obtain v1 and v2 in the circuit of Fig. 3.68. 2 A 5 A 10 Ω v1 v2 5 Ω 8 V + − Figure 3.68 For Prob. 3.20. 3.21 Find vo and io in the circuit in Fig. 3.69. 20 V 2 Ω 2 Ω 1 Ω + − 40 V + − 10 V + − 4 Ω vo + − io Figure 3.69 For Prob. 3.21. 3.22 ∗ Use nodal analysis to determine voltages v1, v2, and v3 in the circuit in Fig. 3.70. 2 S 2 A 4 S 2 S 4 A io 1 S 4 S 1 S v1 3io v2 v3 Figure 3.70 For Prob. 3.22. 3.23 Using nodal analysis, find vo and io in the circuit of Fig. 3.71. + − 100 V 80 Ω vo + − 10 Ω 20 Ω 40 Ω 120 V + − 2io 4vo + − io Figure 3.71 For Prob. 3.23. 3.24 Find the node voltages for the circuit in Fig. 3.72. 4 Ω 1 A 1 Ω 4 Ω 10 V io 1 Ω 2 Ω v1 2vo 4io v2 v3 + − vo + − + − Figure 3.72 For Prob. 3.24. ∗An asterisk indicates a challenging problem.
  • 119. 112 PART 1 DC Circuits 3.25 ∗ Obtain the node voltages v1, v2, and v3 in the circuit of Fig. 3.73. 10 kΩ 4 mA 5 kΩ v1 20 V 10 V v2 v3 12 V + − + − + − Figure 3.73 For Prob. 3.25. Sections 3.4 and 3.5 Mesh Analysis 3.26 Which of the circuits in Fig. 3.74 is planar? For the planar circuit, redraw the circuits with no crossing branches. 2 Ω 6 Ω 5 Ω 2 A (a) 4 Ω 3 Ω 1 Ω (b) 12 V + − 2 Ω 3 Ω 5 Ω 4 Ω 1 Ω Figure 3.74 For Prob. 3.26. 3.27 Determine which of the circuits in Fig. 3.75 is planar and redraw it with no crossing branches. 10 V + − 3 Ω 5 Ω 2 Ω 7 Ω 4 Ω (a) 1 Ω 6 Ω 7 Ω 6 Ω 1 Ω 3 Ω 4 A (b) 8 Ω 2 Ω 5 Ω 4 Ω Figure 3.75 For Prob. 3.27. 3.28 Rework Prob. 3.5 using mesh analysis. 3.29 Rework Prob. 3.6 using mesh analysis. 3.30 Solve Prob. 3.7 using mesh analysis. 3.31 Solve Prob. 3.8 using mesh analysis. 3.32 For the bridge network in Fig. 3.76, find io using mesh analysis. 30 V + − 2 kΩ 2 kΩ 6 kΩ 6 kΩ 4 kΩ 4 kΩ io Figure 3.76 For Prob. 3.32. 3.33 Apply mesh analysis to find i in Fig. 3.77.
  • 120. CHAPTER 3 Methods of Analysis 113 + − + − 10 Ω 2 Ω 5 Ω 1 Ω 8 V 6 V i1 i2 i3 i 4 Ω Figure 3.77 For Prob. 3.33. 3.34 Use mesh analysis to find vab and io in the circuit in Fig. 3.78. + − 20 Ω 20 Ω 30 Ω 30 Ω 20 Ω 80 V + − 80 V 30 Ω vab + − io Figure 3.78 For Prob. 3.34. 3.35 Use mesh analysis to obtain io in the circuit of Fig. 3.79. 3 A 12 V + − 4 Ω io 1 Ω 6 V 2 Ω 5 Ω + − Figure 3.79 For Prob. 3.35. 3.36 Find current i in the circuit in Fig. 3.80. 4 A 30 V i + − 3 Ω 1 Ω 2 Ω 6 Ω 4 Ω 8 Ω Figure 3.80 For Prob. 3.36. 3.37 Find vo and io in the circuit of Fig. 3.81. 16 V 2io 3 Ω 1 Ω 2 Ω 2 Ω + − io vo Figure 3.81 For Prob. 3.37. 3.38 Use mesh analysis to find the current io in the circuit in Fig. 3.82. 3io 10 Ω 4 Ω 60 V + − io 8 Ω 2 Ω Figure 3.82 For Prob. 3.38. 3.39 Apply mesh analysis to find vo in the circuit in Fig. 3.83. 20 V 5 A 2 Ω 8 Ω 1 Ω 40 V vo + − + − 4 Ω Figure 3.83 For Prob. 3.39. 3.40 Use mesh analysis to find i1, i2, and i3 in the circuit of Fig. 3.84.
  • 121. 114 PART 1 DC Circuits 12 V + − 8 Ω 4 Ω + − 2 Ω vo 2vo i2 i3 i1 3 A + − Figure 3.84 For Prob. 3.40. 3.41 Rework Prob. 3.11 using mesh analysis. 3.42 ∗ In the circuit of Fig. 3.85, solve for i1, i2, and i3. 4 A 2 Ω 1 A i3 i1 i2 6 Ω 12 Ω 4 Ω 8 V + − 10 V + − Figure 3.85 For Prob. 3.42. 3.43 Determine v1 and v2 in the circuit of Fig. 3.86. 12 V 2 Ω 2 Ω 2 Ω + − 2 Ω v1 2 Ω v2 + − + − Figure 3.86 For Prob. 3.43. 3.44 Find i1, i2, and i3 in the circuit in Fig. 3.87. 10 Ω 10 Ω 120 V 30 Ω 30 Ω 30 Ω i3 i2 i1 + − Figure 3.87 For Prob. 3.44. 3.45 Rework Prob. 3.23 using mesh analysis. 3.46 Calculate the power dissipated in each resistor in the circuit in Fig. 3.88. 10 V 0.5io 4 Ω 8 Ω 1 Ω 2 Ω + − io Figure 3.88 For Prob. 3.46. 3.47 Calculate the current gain io/is in the circuit of Fig. 3.89. 5vo 20 Ω 10 Ω 40 Ω io is 30 Ω vo + − – + Figure 3.89 For Prob. 3.47. 3.48 Find the mesh currents i1, i2, and i3 in the network of Fig. 3.90. 4 kΩ 8 kΩ 2 kΩ 100 V 4 mA 2i1 40 V + − + − i1 i2 i3 Figure 3.90 For Prob. 3.48. 3.49 Find vx and ix in the circuit shown in Fig. 3.91. ix 2 Ω vx 4ix + − 5 Ω 50 V 3 A + − + − vx 4 10 Ω Figure 3.91 For Prob. 3.49.
  • 122. CHAPTER 3 Methods of Analysis 115 3.50 Find vo and io in the circuit of Fig. 3.92. + − + − io + − 2 A 100 V 40 Ω 10 Ω 50 Ω 10 Ω vo 0.2vo 4io Figure 3.92 For Prob. 3.50. Section 3.6 Nodal and Mesh Analyses by Inspection 3.51 Obtain the node-voltage equations for the circuit in Fig. 3.93 by inspection. Determine the node voltages v1 and v2. 3 A 5 A 4 Ω 2 Ω v1 v2 1 Ω 6 A Figure 3.93 For Prob. 3.51. 3.52 By inspection, write the node-voltage equations for the circuit in Fig. 3.94 and obtain the node voltages. 4 A 3 S 2 A 1 A v1 1 S 2 S 5 S v2 v3 Figure 3.94 For Prob. 3.52. 3.53 For the circuit shown in Fig. 3.95, write the node-voltage equations by inspection. 2 kΩ 2 kΩ 10 mA 20 mA v1 4 kΩ 4 kΩ 1 kΩ 5 mA v2 v3 Figure 3.95 For Prob. 3.53. 3.54 Write the node-voltage equations of the circuit in Fig. 3.96 by inspection. I1 v1 v3 G4 G5 G2 G3 G1 v2 I2 Figure 3.96 For Prob. 3.54. 3.55 Obtain the mesh-current equations for the circuit in Fig. 3.97 by inspection. Calculate the power absorbed by the 8- resistor. + − + − + − 12 A 20 V 2 Ω 2 Ω i1 i2 i3 8 V 4 Ω 8 Ω 5 Ω Figure 3.97 For Prob. 3.55. 3.56 By inspection, write the mesh-current equations for the circuit in Fig. 3.98.
  • 123. 116 PART 1 DC Circuits + − + − + − 10 V 4 Ω 5 Ω 2 Ω 4 Ω i1 i2 i3 8 V 4 V i4 1 Ω Figure 3.98 For Prob. 3.56. 3.57 Write the mesh-current equations for the circuit in Fig. 3.99. + − + − + − + − 6 V 4 V 1 Ω 1 Ω 3 Ω 1 Ω i1 i2 i4 i3 2 V 3 V 2 Ω 4 Ω 5 Ω Figure 3.99 For Prob. 3.57. 3.58 By inspection, obtain the mesh-current equations for the circuit in Fig. 3.100. + − + − + − i1 i3 V1 V3 V2 V4 i2 i4 R1 R2 R3 R4 R5 R6 R7 R8 + − Figure 3.100 For Prob. 3.58. Section 3.8 Circuit Analysis with PSpice 3.59 Use PSpice to solve Prob. 3.44. 3.60 Use PSpice to solve Prob. 3.22. 3.61 Rework Prob. 3.51 using PSpice. 3.62 Find the nodal voltages v1 through v4 in the circuit in Fig. 3.101 using PSpice. + − + − 8 A 20 V 1 Ω v1 2 Ω 4 Ω 10 Ω 12 Ω v2 v3 io 6io v4 Figure 3.101 For Prob. 3.62. 3.63 Use PSpice to solve the problem in Example 3.4. 3.64 If the Schematics Netlist for a network is as follows, draw the network. R_R1 1 2 2K R_R2 2 0 4K R_R3 3 0 8K R_R4 3 4 6K R_R5 1 3 3K V_VS 4 0 DC 100 I_IS 0 1 DC 4 F_F1 1 3 VF_F1 2 VF_F1 5 0 0V E_E1 3 2 1 3 3 3.65 The following program is the Schematics Netlist of a particular circuit. Draw the circuit and determine the voltage at node 2. R_R1 1 2 20 R_R2 2 0 50 R_R3 2 3 70 R_R4 3 0 30 V_VS 1 0 20V I_IS 2 0 DC 2A Section 3.9 Applications 3.66 Calculate vo and io in the circuit of Fig. 3.102. + − + − 3 mV vo + − 4 kΩ 50io io vo 100 20 kΩ Figure 3.102 For Prob. 3.66. 3.67 For the simplified transistor circuit of Fig. 3.103, calculate the voltage vo.
  • 124. CHAPTER 3 Methods of Analysis 117 + − + − i 2 kΩ 5 kΩ 1 kΩ 30 mV vo 400i Figure 3.103 For Prob. 3.67. 3.68 For the circuit in Fig. 3.104, find the gain vo/vs. + − – + + − + − 500 Ω 400 Ω 2 kΩ 200 Ω vs vo v1 60v1 Figure 3.104 For Prob. 3.68. 3.69 ∗ Determine the gain vo/vs of the transistor amplifier circuit in Fig. 3.105. 2 kΩ 100 Ω 200 Ω vs 40io io vo 10 kΩ vo 1000 + − + − + − Figure 3.105 For Prob. 3.69. 3.70 For the simple transistor circuit of Fig. 3.106, let β = 75, VBE = 0.7 V. What value of vi is required to give a collector-emitter voltage of 2 V? + − 5 V 2 kΩ vi 40 kΩ Figure 3.106 For Prob. 3.70. 3.71 Calculate vs for the transistor in Fig. 3.107 given that vo = 4 V, β = 150, VBE = 0.7 V. + − 18 V 1 kΩ vs 10 kΩ + − 500 Ω vo Figure 3.107 For Prob. 3.71. 3.72 For the transistor circuit of Fig. 3.108, find IB , VCE, and vo. Take β = 200, VBE = 0.7 V. + − 9 V 5 kΩ 3 V 6 kΩ + − 400 Ω vo VCE + − IB 2 kΩ Figure 3.108 For Prob. 3.72. 3.73 Find IB and VC for the circuit in Fig. 3.109. Let β = 100, VBE = 0.7 V. + − IB 12 V 4 kΩ VC 10 kΩ 5 kΩ Figure 3.109 For Prob. 3.73. COMPREHENSIVE PROBLEMS 3.74 ∗ Rework Example 3.11 with hand calculation.
  • 125. 119 C H A P T E R CIRCUIT THEOREMS 4 Our schools had better get on with what is their overwhelmingly most important task: teaching their charges to express themselves clearly and with precision in both speech and writing; in other words, leading them toward mastery of their own language. Failing that, all their instruction in mathematics and science is a waste of time. —Joseph Weizenbaum, M.I.T. Enhancing Your Career Enhancing Your Communication Skill Taking a course in circuit analysis is one step in preparing yourself for a career in electrical engineering. Enhancing your commu- nication skill while in school should also be part of that preparation, as a large part of your time will be spent com- municating. People in industry have complained again and again that graduating engineers are ill-prepared in written and oral communication. An engineer who communicates ef- fectively becomes a valuable asset. You can probably speak or write easily and quickly. But how effectively do you communicate? The art of ef- fective communication is of the utmost importance to your success as an engineer. For engineers in industry, communication is key to promotability. Consider the result of a survey of U.S. cor- porations that asked what factors influence managerial pro- motion. Thesurveyincludesalistingof22personalqualities and their importance in advancement. You may be surprised to note that “technical skill based on experience” placed fourth from the bottom. Attributes such as self-confidence, ambition, flexibility, maturity, ability to make sound deci- sions, getting things done with and through people, and ca- pacity for hard work all ranked higher. At the top of the list was “ability to communicate.” The higher your professional career progresses, the more you will need to communicate. Therefore, you should regard effective communication as an important tool in your engineering tool chest. Learning to communicate effectively is a lifelong task you should always work toward. The best time to begin is while still in school. Continually look for opportunities to develop and strengthen your reading, writing, listening, Ability to work hard Working and getting along with people Maturity Appearance Problem-solving skills Self- determination College education Effective communication Ability to communicate effectively is regarded by many as the most important step to an executive promotion. (Adapted from J. Sherlock, A Guide to Technical Communication. Boston, MA: Allyn and Bacon, 1985, p. 7.) and speaking skills. You can do this through classroom presentations, team projects, active participation in student organizations, and enrollment in communication courses. The risks are less now than later in the workplace.
  • 126. 120 PART 1 DC Circuits 4.1 INTRODUCTION A major advantage of analyzing circuits using Kirchhoff’s laws as we did in Chapter 3 is that we can analyze a circuit without tampering with its original configuration. A major disadvantage of this approach is that, for a large, complex circuit, tedious computation is involved. The growth in areas of application of electric circuits has led to an evolution from simple to complex circuits. To handle the complexity, engineers over the years have developed some theorems to simplify cir- cuit analysis. Such theorems include Thevenin’s and Norton’s theorems. Since these theorems are applicable to linear circuits, we first discuss the concept of circuit linearity. In addition to circuit theorems, we discuss the concepts of superposition, source transformation, and maximum power transfer in this chapter. The concepts we develop are applied in the last section to source modeling and resistance measurement. 4.2 LINEARITY PROPERTY Linearity is the property of an element describing a linear relationship between cause and effect. Although the property applies to many circuit elements, we shall limit its applicability to resistors in this chapter. The property is a combination of both the homogeneity (scaling) property and the additivity property. The homogeneity property requires that if the input (also called the excitation) is multiplied by a constant, then the output (also called the response) is multiplied by the same constant. For a resistor, for example, Ohm’s law relates the input i to the output v, v = iR (4.1) If the current is increased by a constant k, then the voltage increases correspondingly by k, that is, kiR = kv (4.2) The additivity property requires that the response to a sum of inputs is the sum of the responses to each input applied separately. Using the voltage-current relationship of a resistor, if v1 = i1R (4.3a) and v2 = i2R (4.3b) then applying (i1 + i2) gives v = (i1 + i2)R = i1R + i2R = v1 + v2 (4.4) We say that a resistor is a linear element because the voltage-current relationship satisfies both the homogeneity and the additivity properties. In general, a circuit is linear if it is both additive and homogeneous. A linear circuit consists of only linear elements, linear dependent sources, and independent sources.
  • 127. CHAPTER 4 Circuit Theorems 121 A linear circuit is one whose output is linearly related (or directly proportional) to its input. Throughout this book we consider only linear circuits. Note that since p = i2 R = v2 /R (making it a quadratic function rather than a linear one), the relationship between power and voltage (or current) is nonlinear. Therefore, the theorems covered in this chapter are not applicable to power. To understand the linearity principle, consider the linear circuit shown in Fig. 4.1. The linear circuit has no independent sources inside it. It is excited by a voltage source vs, which serves as the input. The circuit is terminated by a load R. We may take the current i through R as the output. Suppose vs = 10 V gives i = 2 A. According to the linearity principle, vs = 1 V will give i = 0.2 A. By the same token, i = 1 mA must be due to vs = 5 mV. vs R i + − Linear circuit Figure 4.1 A linear circuit with input vs and output i. E X A M P L E 4 . 1 For the circuit in Fig. 4.2, find io when vs = 12 V and vs = 24 V. + − vs vx 3vx i1 i2 2 Ω 8 Ω 4 Ω 6 Ω 4 Ω – + + − io Figure4.2 For Example 4.1. Solution: Applying KVL to the two loops, we obtain 12i1 − 4i2 + vs = 0 (4.1.1) − 4i1 + 16i2 − 3vx − vs = 0 (4.1.2) But vx = 2i1. Equation (4.1.2) becomes −10i1 + 16i2 − vs = 0 (4.1.3) Adding Eqs. (4.1.1) and (4.1.3) yields 2i1 + 12i2 = 0 ⇒ i1 = −6i2 Substituting this in Eq. (4.1.1), we get −76i2 + vs = 0 ⇒ i2 = vs 76 When vs = 12 V, io = i2 = 12 76 A When vs = 24 V, io = i2 = 24 76 A showing that when the source value is doubled, io doubles. P R A C T I C E P R O B L E M 4 . 1 For the circuit in Fig. 4.3, find vo when is = 15 and is = 30 A. is 6 Ω 4 Ω 2 Ω + − vo Figure4.3 For Practice Prob. 4.1. Answer: 10 V, 20 V.
  • 128. 122 PART 1 DC Circuits E X A M P L E 4 . 2 Assume Io = 1 A and use linearity to find the actual value of Io in the circuit in Fig. 4.4. Io I4 I2 I3 V2 6 Ω 2 Ω 2 5 Ω 7 Ω I1 V1 3 Ω 1 4 Ω Is = 15 A Figure4.4 For Example 4.2. Solution: If Io = 1 A, then V1 = (3 + 5)Io = 8 V and I1 = V1/4 = 2 A. Applying KCL at node 1 gives I2 = I1 + Io = 3 A V2 = V1 + 2I2 = 8 + 6 = 14 V, I3 = V2 7 = 2 A Applying KCL at node 2 gives I4 = I3 + I2 = 5 A Therefore, Is = 5 A. This shows that assuming Io = 1 gives Is = 5 A; the actual source current of 15 A will give Io = 3 A as the actual value. P R A C T I C E P R O B L E M 4 . 2 Assume that Vo = 1 V and use linearity to calculate the actual value of Vo in the circuit of Fig. 4.5. 10 V 12 Ω 8 Ω 5 Ω + − + − Vo Figure4.5 For Practice Prob. 4.2 Answer: 4 V. 4.3 SUPERPOSITION If a circuit has two or more independent sources, one way to determine the value of a specific variable (voltage or current) is to use nodal or mesh analysis as in Chapter 3. Another way is to determine the contribution of each independent source to the variable and then add them up. The latter approach is known as the superposition.
  • 129. CHAPTER 4 Circuit Theorems 123 The idea of superposition rests on the linearity property. Superpositionisnotlimitedtocircuitanalysisbut isapplicableinmanyfieldswherecauseandeffect bear a linear relationship to one another. The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately. However, to apply the superposition prin- ciple, we must keep two things in mind: 1. We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current source by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit. Other terms such as killed, made inactive, dead- ened, or set equal to zero are often used to con- vey the same idea. 2. Dependent sources are left intact because they are controlled by circuit variables. With these in mind, we apply the superposition principle in three steps: Steps to Apply Superposition Principle: 1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis. 2. Repeat step 1 for each of the other independent sources. 3. Find the total contribution by adding algebraically all the contributions due to the independent sources. Analyzing a circuit using superposition has one major disadvan- tage: it may very likely involve more work. If the circuit has three independent sources, we may have to analyze three simpler circuits each providing the contribution due to the respective individual source. How- ever, superposition does help reduce a complex circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits. Keep in mind that superposition is based on linearity. For this reason, it is not applicable to the effect on power due to each source, because the power absorbed by a resistor depends on the square of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposition. For example, when current i1 flows through re- sistorR,thepowerisp1 =Ri2 1,andwhencurrent i2 flows through R, the power is p2 = Ri2 2. If cur- rent i1 + i2 flows through R, the power absorb- ed is p3 = R(i1 + i2)2 = Ri2 1 + Ri2 2 + 2Ri1i2 = p1 + p2. Thus, the power relation is nonlinear. E X A M P L E 4 . 3 Use the superposition theorem to find v in the circuit in Fig. 4.6. 6 V v 3 A 8 Ω 4 Ω + − + − Figure4.6 For Example 4.3. Solution: Since there are two sources, let v = v1 + v2 where v1 and v2 are the contributions due to the 6-V voltage source and
  • 130. 124 PART 1 DC Circuits the 3-A current source, respectively. To obtain v1, we set the current source to zero, as shown in Fig. 4.7(a). Applying KVL to the loop in Fig. 4.7(a) gives 12i1 − 6 = 0 ⇒ i1 = 0.5 A Thus, v1 = 4i1 = 2 V We may also use voltage division to get v1 by writing v1 = 4 4 + 8 (6) = 2 V To get v2, we set the voltage source to zero, as in Fig. 4.7(b). Using current division, i3 = 8 4 + 8 (3) = 2 A Hence, v2 = 4i3 = 8 V And we find v = v1 + v2 = 2 + 8 = 10 V + − 6 V i1 8 Ω v1 4 Ω (a) + − 3 A 8 Ω v2 i2 i3 4 Ω (b) + − Figure4.7 For Example 4.3: (a) calculating v1, (b) calculating v2. P R A C T I C E P R O B L E M 4 . 3 Using the superposition theorem, find vo in the circuit in Fig. 4.8. 3 Ω 5 Ω 2 Ω 8 A 20 V + − + − vo Figure4.8 For Practice Prob. 4.3. Answer: 12 V. E X A M P L E 4 . 4 Find io in the circuit in Fig. 4.9 using superposition. 4 A 20 V 3 Ω 5 Ω 1 Ω 2 Ω 4 Ω + − 5io io + − Figure4.9 For Example 4.4. Solution: The circuit in Fig. 4.9 involves a dependent source, which must be left intact. We let io = i o + i o (4.4.1) where i o and i o are due to the 4-A current source and 20-V voltage source respectively. To obtain i o, we turn off the 20-V source so that we have the circuit in Fig. 4.10(a). We apply mesh analysis in order to obtain i o. For loop 1, i1 = 4 A (4.4.2)
  • 131. CHAPTER 4 Circuit Theorems 125 4 A 3 Ω 5 Ω 1 Ω 2 Ω 4 Ω + − i1 i3 i′ o 5i′ o 0 (a) 3 Ω 5 Ω 1 Ω 2 Ω 4 Ω + − i′′ o 5i′′ o (b) 20 V + − i1 i2 i3 i5 i4 Figure4.10 For Example 4.4: Applying superposition to (a) obtain i 0, (b) obtain i 0 . For loop 2, −3i1 + 6i2 − 1i3 − 5i o = 0 (4.4.3) For loop 3, −5i1 − 1i2 + 10i3 + 5i o = 0 (4.4.4) But at node 0, i3 = i1 − i o = 4 − i o (4.4.5) Substituting Eqs. (4.4.2) and (4.4.5) into Eqs. (4.4.3) and (4.4.4) gives two simultaneous equations 3i2 − 2i o = 8 (4.4.6) i2 + 5i o = 20 (4.4.7) which can be solved to get i o = 52 17 A (4.4.8) To obtain i o , we turn off the 4-A current source so that the circuit becomes that shown in Fig. 4.10(b). For loop 4, KVL gives 6i4 − i5 − 5i o = 0 (4.4.9) and for loop 5, −i4 + 10i5 − 20 + 5i o = 0 (4.4.10) But i5 = −i o . Substituting this in Eqs. (4.4.9) and (4.4.10) gives 6i4 − 4i o = 0 (4.4.11) i4 + 5i o = −20 (4.4.12) which we solve to get
  • 132. 126 PART 1 DC Circuits i o = − 60 17 A (4.4.13) Now substituting Eqs. (4.4.8) and (4.4.13) into Eq. (4.4.1) gives io = − 8 17 = −0.4706 A P R A C T I C E P R O B L E M 4 . 4 Use superposition to find vx in the circuit in Fig. 4.11. vx 20 Ω 0.1vx 4 Ω 10 V 2 A + − Figure4.11 For Practice Prob. 4.4. Answer: vx = 12.5 V. E X A M P L E 4 . 5 For the circuit in Fig. 4.12, use the superposition theorem to find i. + − + − 24 V 8 Ω 4 Ω 3 Ω 3 A 12 V 4 Ω i Figure4.12 For Example 4.5. Solution: In this case, we have three sources. Let i = i1 + i2 + i3 where i1, i2, and i3 are due to the 12-V, 24-V, and 3-A sources respectively. To get i1, consider the circuit in Fig. 4.13(a). Combining 4 (on the right- hand side) in series with 8 gives 12 . The 12 in parallel with 4 gives 12 × 4/16 = 3 . Thus, i1 = 12 6 = 2 A To get i2, consider the circuit in Fig. 4.13(b). Applying mesh analysis, 16ia − 4ib + 24 = 0 ⇒ 4ia − ib = −6 (4.5.1) 7ib − 4ia = 0 ⇒ ia = 7 4 ib (4.5.2) Substituting Eq. (4.5.2) into Eq. (4.5.1) gives i2 = ib = −1 To get i3, consider the circuit in Fig. 4.13(c). Using nodal analysis, 3 = v2 8 + v2 − v1 4 ⇒ 24 = 3v2 − 2v1 (4.5.3) v2 − v1 4 = v1 4 + v1 3 ⇒ v2 = 10 3 v1 (4.5.4) Substituting Eq. (4.5.4) into Eq. (4.5.3) leads to v1 = 3 and
  • 133. CHAPTER 4 Circuit Theorems 127 8 Ω 4 Ω 4 Ω 3 Ω 12 V + − 3 Ω 3 Ω 12 V + − (a) 8 Ω 24 V 4 Ω 4 Ω 3 Ω (b) + − ib ia 8 Ω 4 Ω 4 Ω 3 Ω 3 A v1 v2 (c) i2 i2 i2 i1 Figure4.13 For Example 4.5. i3 = v1 3 = 1 A Thus, i = i1 + i2 + i3 = 2 − 1 + 1 = 2 A P R A C T I C E P R O B L E M 4 . 5 Find i in the circuit in Fig. 4.14 using the superposition principle. 16 V 8 Ω 2 Ω 4 A 6 Ω + − 12 V + − i Figure4.14 For Practice Prob. 4.5. Answer: 0.75 A. 4.4 SOURCE TRANSFORMATION We have noticed that series-parallel combination and wye-delta transfor- mation help simplify circuits. Source transformation is another tool for simplifying circuits. Basic to these tools is the concept of equivalence.
  • 134. 128 PART 1 DC Circuits We recall that an equivalent circuit is one whose v-i characteristics are identical with the original circuit. In Section 3.6, we saw that node-voltage (or mesh-current) equa- tions can be obtained by mere inspection of a circuit when the sources are all independent current (or all independent voltage) sources. It is there- fore expedient in circuit analysis to be able to substitute a voltage source in series with a resistor for a current source in parallel with a resistor, or vice versa, as shown in Fig. 4.15. Either substitution is known as a source transformation. + − vs R a b is R a b Figure4.15 Transformation of independent sources. A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa. The two circuits in Fig. 4.15 are equivalent—provided they have the same voltage-current relation at terminals a-b. It is easy to show that they are indeed equivalent. If the sources are turned off, the equivalent resistance at terminals a-b in both circuits is R. Also, when terminals a-b are short- circuited, the short-circuit current flowing from a to b is isc = vs/R in the circuit on the left-hand side and isc = is for the circuit on the right- hand side. Thus, vs/R = is in order for the two circuits to be equivalent. Hence, source transformation requires that vs = isR or is = vs R (4.5) Source transformation also applies to dependent sources, provided we carefully handle the dependent variable. As shown in Fig. 4.16, a dependent voltage source in series with a resistor can be transformed to a dependent current source in parallel with the resistor or vice versa. vs R a b is R a b + − Figure4.16 Transformation of dependent sources. Like the wye-delta transformation we studied in Chapter 2, a source transformation does not affect the remaining part of the circuit. When
  • 135. CHAPTER 4 Circuit Theorems 129 applicable, source transformation is a powerful tool that allows circuit manipulations to ease circuit analysis. However, we should keep the following points in mind when dealing with source transformation. 1. Note from Fig. 4.15 (or Fig. 4.16) that the arrow of the current source is directed toward the positive terminal of the voltage source. 2. Note from Eq. (4.5) that source transformation is not possible when R = 0, which is the case with an ideal voltage source. However, for a practical, nonideal voltage source, R = 0. Similarly, an ideal current source with R = ∞ cannot be replaced by a finite voltage source. More will be said on ideal and nonideal sources in Section 4.10.1. E X A M P L E 4 . 6 Use source transformation to find vo in the circuit in Fig. 4.17. 2 Ω 3 Ω 12 V 8 Ω 4 Ω 3 A + − + − vo Figure4.17 For Example 4.6. Solution: We first transform the current and voltage sources to obtain the circuit in Fig. 4.18(a). Combining the 4- and 2- resistors in series and trans- forming the 12-V voltage source gives us Fig. 4.18(b). We now combine the 3- and 6- resistors in parallel to get 2-. We also combine the 2-A and 4-A current sources to get a 2-A source. Thus, by repeatedly applying source transformations, we obtain the circuit in Fig. 4.18(c). 4 Ω 2 Ω 4 A 8 Ω 3 Ω 12 V + − (a) + − vo 4 A 8 Ω 6 Ω 3 Ω 2 A (b) 2 A 8 Ω 2 Ω (c) i + − vo + − vo Figure4.18 For Example 4.6. We use current division in Fig. 4.18(c) to get i = 2 2 + 8 (2) = 0.4 and vo = 8i = 8(0.4) = 3.2 V
  • 136. 130 PART 1 DC Circuits Alternatively, since the 8- and 2- resistors in Fig. 4.18(c) are in parallel, they have the same voltage vo across them. Hence, vo = (8 2)(2 A) = 8 × 2 10 (2) = 3.2 V P R A C T I C E P R O B L E M 4 . 6 Find io in the circuit of Fig. 4.19 using source transformation. 4 Ω 5 A 5 V 7 Ω 3 A 3 Ω 1 Ω 6 Ω − + io Figure4.19 For Practice Prob. 4.6. Answer: 1.78 A. E X A M P L E 4 . 7 Find vx in Fig. 4.20 using source transformation. 4 Ω 2 Ω 0.25vx 2 Ω 6 V 18 V + − + − vx + − Figure4.20 For Example 4.7. Solution: The circuit in Fig. 4.20 involves a voltage-controlled dependent current source. We transform this dependent current source as well as the 6-V independent voltage source as shown in Fig. 4.21(a). The 18-V volt- age source is not transformed because it is not connected in series with any resistor. The two 2- resistors in parallel combine to give a 1- resistor, which is in parallel with the 3-A current source. The current is transformed to a voltage source as shown in Fig. 4.21(b). Notice that the terminals for vx are intact. Applying KVL around the loop in Fig. 4.21(b) gives −3 + 5i + vx + 18 = 0 (4.7.1) 18 V 3 A 4 Ω 2 Ω 2 Ω + − + − (a) 18 V 3 V 4 Ω 1 Ω vx vx + − + − + − + − (b) i + − vx Figure4.21 For Example 4.7: Applying source transformation to the circuit in Fig. 4.20.
  • 137. CHAPTER 4 Circuit Theorems 131 Applying KVL to the loop containing only the 3-V voltage source, the 1- resistor, and vx yields −3 + 1i + vx = 0 ⇒ vx = 3 − i (4.7.2) Substituting this into Eq. (4.7.1), we obtain 15 + 5i + 3 − i = 0 ⇒ i = −4.5 A Alternatively, we may apply KVL to the loop containing vx, the 4- resistor, the voltage-controlled dependent voltage source, and the 18-V voltage source in Fig. 4.21(b). We obtain −vx + 4i + vx + 18 = 0 ⇒ i = −4.5 A Thus, vx = 3 − i = 7.5 V. P R A C T I C E P R O B L E M 4 . 7 Use source transformation to find ix in the circuit shown in Fig. 4.22. 2ix 5 Ω 4 A 10 Ω – + ix Figure4.22 For Practice Prob. 4.7. Answer: 1.176 A. 4.5 THEVENIN’S THEOREM It often occurs in practice that a particular element in a circuit is variable (usually called the load) while other elements are fixed. As a typical example, a household outlet terminal may be connected to different ap- pliances constituting a variable load. Each time the variable element is changed, the entire circuit has to be analyzed all over again. To avoid this problem, Thevenin’s theorem provides a technique by which the fixed part of the circuit is replaced by an equivalent circuit. According to Thevenin’s theorem, the linear circuit in Fig. 4.23(a) can be replaced by that in Fig. 4.23(b). (The load in Fig. 4.23 may be a single resistor or another circuit.) The circuit to the left of the terminals a-b in Fig. 4.23(b) is known as the Thevenin equivalent circuit; it was developed in 1883 by M. Leon Thevenin (1857–1926), a French telegraph engineer. Linear two-terminal circuit Load I a b V + − (a) Load I a b V + − (b) + − VTh RTh Figure4.23 Replacing a linear two-terminal circuit by its Thevenin equivalent: (a) original circuit, (b) the Thevenin equivalent circuit. Thevenin’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor RTh, where VTh is the open-circuit voltage at the terminals and RTh is the input or equivalent resistance at the terminals when the independent sources are turned off. The proof of the theorem will be given later, in Section 4.7. Our major concern right now is how to find the Thevenin equivalent voltage
  • 138. 132 PART 1 DC Circuits VTh and resistance RTh. To do so, suppose the two circuits in Fig. 4.23 are equivalent. Two circuits are said to be equivalent if they have the same voltage-current relation at their terminals. Let us find out what will make the two circuits in Fig. 4.23 equivalent. If the terminals a-b are made open-circuited (by removing the load), no current flows, so that the open-circuit voltage across the terminals a-b in Fig. 4.23(a) must be equal to the voltage source VTh in Fig. 4.23(b), since the two circuits are equivalent. Thus VTh is the open-circuit voltage across the terminals as shown in Fig. 4.24(a); that is, VTh = voc (4.6) Linear two-terminal circuit a b voc + − (a) VTh = voc Linear circuit with all independent sources set equal to zero a b Rin (b) RTh = Rin Figure4.24 Finding VTh and RTh. Again, withtheloaddisconnectedandterminalsa-b open-circuited, we turn off all independent sources. The input resistance (or equivalent resistance) of the dead circuit at the terminals a-b in Fig. 4.23(a) must be equal to RTh in Fig. 4.23(b) because the two circuits are equivalent. Thus, RTh is the input resistance at the terminals when the independent sources are turned off, as shown in Fig. 4.24(b); that is, RTh = Rin (4.7) To apply this idea in finding the Thevenin resistance RTh, we need to consider two cases. CASE 1 If the network has no dependent sources, we turn off all in- dependent sources. RTh is the input resistance of the network looking between terminals a and b, as shown in Fig. 4.24(b). CASE 2 If the network has dependent sources, we turn off all inde- pendent sources. As with superposition, dependent sources are not to be turned off because they are controlled by circuit variables. We apply a voltage source vo at terminals a and b and determine the resulting cur- rent io. Then RTh = vo/io, as shown in Fig. 4.25(a). Alternatively, we may insert a current source io at terminals a-b as shown in Fig. 4.25(b) and find the terminal voltage vo. Again RTh = vo/io. Either of the two approaches will give the same result. In either approach we may assume any value of vo and io. For example, we may use vo = 1 V or io = 1 A, or even use unspecified values of vo or io. vo Circuit with all independent sources set equal to zero a b (a) RTh = + − vo io io io vo Circuit with all independent sources set equal to zero a b (b) RTh = vo io + − Figure 4.25 Finding RTh when circuit has dependent sources. Laterwewillseethatanalternativewayoffinding RTh is RTh = voc/isc. It often occurs that RTh takes a negative value. In this case, the negative resistance (v = −iR) implies that the circuit is supplying power.
  • 139. CHAPTER 4 Circuit Theorems 133 This is possible in a circuit with dependent sources; Example 4.10 will illustrate this. Thevenin’s theorem is very important in circuit analysis. It helps simplify a circuit. A large circuit may be replaced by a single independent voltage source and a single resistor. This replacement technique is a powerful tool in circuit design. As mentioned earlier, a linear circuit with a variable load can be re- placed by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same way externally as the original circuit. Con- sider a linear circuit terminated by a load RL, as shown in Fig. 4.26(a). The current IL through the load and the voltage VL across the load are easily determined once the Thevenin equivalent of the circuit at the load’s terminals is obtained, as shown in Fig. 4.26(b). From Fig. 4.26(b), we obtain IL = VTh RTh + RL (4.8a) VL = RLIL = RL RTh + RL VTh (4.8b) Note from Fig. 4.26(b) that the Thevenin equivalent is a simple voltage divider, yielding VL by mere inspection. Linear circuit a b (a) RL IL a b (b) RL IL + − VTh RTh Figure4.26 A circuit with a load: (a) original circuit, (b) Thevenin equivalent. E X A M P L E 4 . 8 Find the Thevenin equivalent circuit of the circuit shown in Fig. 4.27, to the left of the terminals a-b. Then find the current through RL = 6, 16, and 36 . RL 32 V 2 A 4 Ω 1 Ω 12 Ω + − a b Figure4.27 For Example 4.8. Solution: We find RTh by turning off the 32-V voltage source (replacing it with a short circuit) and the 2-A current source (replacing it with an open circuit). The circuit becomes what is shown in Fig. 4.28(a). Thus, RTh = 4 12 + 1 = 4 × 12 16 + 1 = 4 32 V 2 A 4 Ω 1 Ω 12 Ω + − VTh VTh + − (b) 4 Ω 1 Ω 12 Ω (a) RTh i1 i2 Figure4.28 For Example 4.8: (a) finding RTh, (b) finding VTh. To find VTh, consider the circuit in Fig. 4.28(b). Applying mesh analysis to the two loops, we obtain −32 + 4i1 + 12(i1 − i2) = 0, i2 = −2 A
  • 140. 134 PART 1 DC Circuits Solving for i1, we get i1 = 0.5 A. Thus, VTh = 12(i1 − i2) = 12(0.5 + 2.0) = 30 V Alternatively, it is even easier to use nodal analysis. We ignore the 1- resistor since no current flows through it. At the top node, KCL gives 32 − VTh 4 + 2 = VTh 12 or 96 − 3VTh + 24 = VTh ⇒ VTh = 30 V as obtained before. We could also use source transformation to find VTh. The Thevenin equivalent circuit is shown in Fig. 4.29. The current through RL is IL = VTh RTh + RL = 30 4 + RL When RL = 6, IL = 30 10 = 3 A When RL = 16, IL = 30 20 = 1.5 A When RL = 36, IL = 30 40 = 0.75 A RL 30 V 4 Ω + − a b iL Figure4.29 The Thevenin equivalent circuit for Example 4.8. P R A C T I C E P R O B L E M 4 . 8 Using Thevenin’s theorem, find the equivalent circuit to the left of the terminals in the circuit in Fig. 4.30. Then find i. 12 V 2 A 6 Ω 6 Ω 4 Ω 1 Ω + − a b i Figure4.30 For Practice Prob. 4.8. Answer: VTh = 6 V, RTh = 3 , i = 1.5 A. E X A M P L E 4 . 9 Find the Thevenin equivalent of the circuit in Fig. 4.31.
  • 141. CHAPTER 4 Circuit Theorems 135 Solution: This circuit contains a dependent source, unlike the circuit in the previ- ous example. To find RTh, we set the independent source equal to zero but leave the dependent source alone. Because of the presence of the dependent source, however, we excite the network with a voltage source vo connected to the terminals as indicated in Fig. 4.32(a). We may set vo = 1 V to ease calculation, since the circuit is linear. Our goal is to find the current io through the terminals, and then obtain RTh = 1/io. (Al- ternatively, we may insert a 1-A current source, find the corresponding voltage vo, and obtain RTh = vo/1.) 5 A 2 Ω 2vx 2 Ω 6 Ω 4 Ω a b − + + − vx Figure4.31 For Example 4.9. 2 Ω 2vx 2 Ω 6 Ω 4 Ω a b − + + − vo = 1 V io (a) i1 i2 (b) 5 A 2 Ω 2vx 2 Ω 6 Ω 4 Ω a b − + voc + − i3 i1 i2 i3 + − vx + − vx Figure4.32 Finding RTh and VTh for Example 4.9. Applying mesh analysis to loop 1 in the circuit in Fig. 4.32(a) results in −2vx + 2(i1 − i2) = 0 or vx = i1 − i2 But −4i2 = vx = i1 − i2; hence, i1 = −3i2 (4.9.1) For loops 2 and 3, applying KVL produces 4i2 + 2(i2 − i1) + 6(i2 − i3) = 0 (4.9.2) 6(i3 − i2) + 2i3 + 1 = 0 (4.9.3) Solving these equations gives i3 = − 1 6 A But io = −i3 = 1/6 A. Hence, RTh = 1 V io = 6 To get VTh, we find voc in the circuit of Fig. 4.32(b). Applying mesh analysis, we get
  • 142. 136 PART 1 DC Circuits i1 = 5 (4.9.4) − 2vx + 2(i3 − i2) = 0 ⇒ vx = i3 − i2 (4.9.5) 4(i2 − i1) + 2(i2 − i3) + 6i2 = 0 or 12i2 − 4i1 − 2i3 = 0 (4.9.6) But 4(i1 − i2) = vx. Solving these equations leads to i2 = 10/3. Hence, VTh = voc = 6i2 = 20 V The Thevenin equivalent is as shown in Fig. 4.33. 20 V 6 Ω a b + − Figure4.33 The Thevenin equivalent of the circuit in Fig. 4.31. P R A C T I C E P R O B L E M 4 . 9 Find the Thevenin equivalent circuit of the circuit in Fig. 4.34 to the left of the terminals. 6 V 3 Ω 5 Ω 4 Ω a b 1.5Ix + − Ix Figure4.34 For Practice Prob. 4.9. Answer: VTh = 5.33 V, RTh = 0.44 . E X A M P L E 4 . 1 0 Determine the Thevenin equivalent of the circuit in Fig. 4.35(a). 2ix 4 Ω 2 Ω a b ix vo (a) 2ix io 4 Ω 2 Ω a b ix (b) Figure4.35 For Example 4.10. Solution: Since the circuit in Fig. 4.35(a) has no independent sources, VTh = 0 V. To find RTh, it is best to apply a current source io at the terminals as shown in Fig. 4.35(b). Applying nodal analysis gives io + ix = 2ix + vo 4 (4.10.1) But ix = 0 − vo 2 = − vo 2 (4.10.2) Substituting Eq. (4.10.2) into Eq. (4.10.1) yields io = ix + vo 4 = − vo 2 + vo 4 = − vo 4 or vo = −4io Thus, RTh = vo io = −4 The negative value of the resistance tells us that, according to the passive sign convention, the circuit in Fig. 4.35(a) is supplying power. Of course, the resistors in Fig. 4.35(a) cannot supply power (they absorb power); it
  • 143. CHAPTER 4 Circuit Theorems 137 is the dependent source that supplies the power. This is an example of how a dependent source and resistors could be used to simulate negative resistance. P R A C T I C E P R O B L E M 4 . 1 0 Obtain the Thevenin equivalent of the circuit in Fig. 4.36. 5 Ω 15 Ω a b 10 Ω 4vx + − + − vx Figure 4.36 For Practice Prob. 4.10. Answer: VTh = 0 V, RTh = −7.5 . 4.6 NORTON’S THEOREM In 1926, about 43 years after Thevenin published his theorem, E. L. Norton, an American engineer at Bell Telephone Laboratories, proposed a similar theorem. Norton’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistor RN, where IN is the short-circuit current through the terminals and RN is the input or equivalent resistance at the terminals when the independent sources are turned off. Thus, the circuit in Fig. 4.37(a) can be replaced by the one in Fig. 4.37(b). Linear two-terminal circuit a b (a) (b) RN a b IN Figure4.37 (a) Original circuit, (b) Norton equivalent circuit. The proof of Norton’s theorem will be given in the next section. For now, we are mainly concerned with how to get RN and IN . We find RN in the same way we find RTh. In fact, from what we know about source transformation, the Thevenin and Norton resistances are equal; that is, RN = RTh (4.9) TofindtheNortoncurrentIN , wedeterminetheshort-circuitcurrent flowing from terminal a to b in both circuits in Fig. 4.37. It is evident that the short-circuit current in Fig. 4.37(b) is IN . This must be the same short-circuit current from terminal a to b in Fig. 4.37(a), since the two circuits are equivalent. Thus, IN = isc (4.10) shown in Fig. 4.38. Dependent and independent sources are treated the same way as in Thevenin’s theorem. Linear two-terminal circuit a b isc = IN Figure 4.38 Finding Norton current IN . Observe the close relationship between Norton’s and Thevenin’s theorems: RN = RTh as in Eq. (4.9), and IN = VTh RTh (4.11)
  • 144. 138 PART 1 DC Circuits This is essentially source transformation. For this reason, source trans- formation is often called Thevenin-Norton transformation. TheTheveninandNortonequivalentcircuitsare related by a source transformation. Since VTh, IN , and RTh are related according to Eq. (4.11), to de- termine the Thevenin or Norton equivalent circuit requires that we find: • The open-circuit voltage voc across terminals a and b. • The short-circuit current isc at terminals a and b. • The equivalent or input resistance Rin at terminals a and b when all independent sources are turned off. We can calculate any two of the three using the method that takes the least effort and use them to get the third using Ohm’s law. Example 4.11 will illustrate this. Also, since VTh = voc (4.12a) IN = isc (4.12b) RTh = voc isc = RN (4.12c) the open-circuit and short-circuit tests are sufficient to find any Thevenin or Norton equivalent. E X A M P L E 4 . 1 1 Find the Norton equivalent circuit of the circuit in Fig. 4.39. 2 A 8 Ω 8 Ω 5 Ω 4 Ω 12 V a b + − Figure4.39 For Example 4.11. Solution: We find RN in the same way we find RTh in the Thevenin equivalent cir- cuit. Set the independent sources equal to zero. This leads to the circuit in Fig. 4.40(a), from which we find RN . Thus, RN = 5 (8 + 4 + 8) = 5 20 = 20 × 5 25 = 4 To find IN , we short-circuit terminals a and b, as shown in Fig. 4.40(b). We ignore the 5- resistor because it has been short-circuited. Applying mesh analysis, we obtain i1 = 2 A, 20i2 − 4i1 − 12 = 0 From these equations, we obtain i2 = 1 A = isc = IN Alternatively, we may determine IN from VTh/RTh. We obtain VTh as the open-circuit voltage across terminals a and b in Fig. 4.40(c). Using mesh analysis, we obtain i3 = 2 A 25i4 − 4i3 − 12 = 0 ⇒ i4 = 0.8 A and voc = VTh = 5i4 = 4 V
  • 145. CHAPTER 4 Circuit Theorems 139 2 A 5 Ω 4 Ω 12 V a b + − isc = IN (b) 2 A 5 Ω 4 Ω 12 V a b + − (c) 8 Ω 5 Ω a b 4 Ω (a) RN VTh = voc + − i1 i3 i4 i2 8 Ω 8 Ω 8 Ω 8 Ω 8 Ω Figure4.40 For Example 4.11; finding: (a) RN , (b) IN = isc, (c) VTh = voc. Hence, IN = VTh RTh = 4 4 = 1 A as obtained previously. This also serves to confirm Eq. (4.7) that RTh = voc/isc = 4/1 = 4 . Thus, the Norton equivalent circuit is as shown in Fig. 4.41. 1 A 4 Ω a b Figure4.41 Norton equiva- lent of the circuit in Fig. 4.39. P R A C T I C E P R O B L E M 4 . 1 1 Find the Norton equivalent circuit for the circuit in Fig. 4.42. 4 A 15 V 6 Ω a b 3 Ω + − 3 Ω Figure4.42 For Practice Prob. 4.11. Answer: RN = 3 , IN = 4.5 A. E X A M P L E 4 . 1 2 Using Norton’s theorem, find RN and IN of the circuit in Fig. 4.43 at ter- minals a-b. Solution: To find RN , we set the independent voltage source equal to zero and con- nect a voltage source of vo = 1 V (or any unspecified voltage vo) to the
  • 146. 140 PART 1 DC Circuits terminals. WeobtainthecircuitinFig.4.44(a). Weignorethe4-resistor because it is short-circuited. Also due to the short circuit, the 5- resistor, the voltage source, and the dependent current source are all in parallel. Hence, ix = vo/5 = 1/5 = 0.2. At node a, −io = ix + 2ix = 3ix = 0.6, and RN = vo io = 1 −0.6 = −1.67 5 Ω 2Ix 10 V 4 Ω a b + − ix Figure4.43 For Example 4.12. TofindIN , weshort-circuitterminalsa andb andfindthecurrentisc, as indicated in Fig. 4.44(b). Note from this figure that the 4- resistor, the 10-V voltage source, the 5- resistor, and the dependent current source are all in parallel. Hence, ix = 10 − 0 5 = 2 A At node a, KCL gives isc = ix + 2ix = 2 + 4 = 6 A Thus, IN = 6 A 5 Ω 2ix vo = 1 V io 4 Ω a b + − (a) 5 Ω 2ix isc = IN 4 Ω a b (b) 10 V + − ix ix Figure4.44 For Example 4.12: (a) finding RN , (b) finding IN . P R A C T I C E P R O B L E M 4 . 1 2 Find the Norton equivalent circuit of the circuit in Fig. 4.45. 10 A 2vx 6 Ω 2 Ω a b − + + − vx Figure4.45 For Practice Prob. 4.12. Answer: RN = 1 , IN = 10 A. †4.7 DERIVATIONS OF THEVENIN’S AND NORTON’S THEOREMS In this section, we will prove Thevenin’s and Norton’s theorems using the superposition principle.
  • 147. CHAPTER 4 Circuit Theorems 141 Consider the linear circuit in Fig. 4.46(a). It is assumed that the cir- cuit contains resistors, and dependent and independent sources. We have access to the circuit via terminals a and b, through which current from an external source is applied. Our objective is to ensure that the voltage- current relation at terminals a and b is identical to that of the Thevenin equivalent in Fig. 4.46(b). For the sake of simplicity, suppose the linear circuit in Fig. 4.46(a) contains two independent voltage sources vs1 and vs2 and two independent current sources is1 and is2. We may obtain any circuit variable, such as the terminal voltage v, by applying superposition. That is, we consider the contribution due to each independent source in- cluding the external source i. By superposition, the terminal voltage v is i Linear circuit a b (a) i a b (b) v + − v + − VTh + − RTh Figure4.46 Derivation of Thevenin equivalent: (a) a current-driven circuit, (b) its Thevenin equivalent. v = A0i + A1vs1 + A2vs2 + A3is1 + A4is2 (4.13) where A0, A1, A2, A3, and A4 are constants. Each term on the right-hand side of Eq. (4.13) is the contribution of the related independent source; that is, A0i is the contribution to v due to the external current source i, A1vs1 is the contribution due to the voltage source vs1, and so on. We may collect terms for the internal independent sources together as B0, so that Eq. (4.13) becomes v = A0i + B0 (4.14) where B0 = A1vs1 + A2vs2 + A3is1 + A4is2. We now want to evaluate the values of constants A0 and B0. When the terminals a and b are open- circuited, i = 0 and v = B0. Thus B0 is the open-circuit voltage voc, which is the same as VTh, so B0 = VTh (4.15) When all the internal sources are turned off, B0 = 0. The circuit can then be replaced by an equivalent resistance Req, which is the same as RTh, and Eq. (4.14) becomes v = A0i = RThi ⇒ A0 = RTh (4.16) Substituting the values of A0 and B0 in Eq. (4.14) gives v = RThi + VTh (4.17) which expresses the voltage-current relation at terminals a and b of the circuit in Fig. 4.46(b). Thus, the two circuits in Fig. 4.46(a) and 4.46(b) are equivalent. When the same linear circuit is driven by a voltage source v as shown in Fig. 4.47(a), the current flowing into the circuit can be obtained by superposition as v Linear circuit a b (a) v a b (b) IN RN + − + − i i Figure4.47 Derivation of Norton equivalent: (a) a voltage-driven circuit, (b) its Norton equivalent. i = C0v + D0 (4.18) where C0v is the contribution to i due to the external voltage source v and D0 contains the contributions to i due to all internal independent sources. When the terminals a-b are short-circuited, v = 0 so that i = D0 = −isc, where isc is the short-circuit current flowing out of terminal a, which is the same as the Norton current IN , i.e., D0 = −IN (4.19)
  • 148. 142 PART 1 DC Circuits When all the internal independent sources are turned off, D0 = 0 and the circuit can be replaced by an equivalent resistance Req (or an equivalent conductance Geq = 1/Req), which is the same as RTh or RN . Thus Eq. (4.19) becomes i = v RTh − IN (4.20) This expresses the voltage-current relation at terminals a-b of the circuit in Fig. 4.47(b), confirming that the two circuits in Fig. 4.47(a) and 4.47(b) are equivalent. 4.8 MAXIMUM POWER TRANSFER In many practical situations, a circuit is designed to provide power to a load. While for electric utilities, minimizing power losses in the process of transmission and distribution is critical for efficiency and economic reasons, there are other applications in areas such as communications where it is desirable to maximize the power delivered to a load. We now address the problem of delivering the maximum power to a load when given a system with known internal losses. It should be noted that this will result in significant internal losses greater than or equal to the power delivered to the load. The Thevenin equivalent is useful in finding the maximum power a linear circuit can deliver to a load. We assume that we can adjust the load resistance RL. If the entire circuit is replaced by its Thevenin equivalent except for the load, as shown in Fig. 4.48, the power delivered to the load is p = i2 RL = VTh RTh + RL 2 RL (4.21) For a given circuit, VTh and RTh are fixed. By varying the load resistance RL, the power delivered to the load varies as sketched in Fig. 4.49. We notice from Fig. 4.49 that the power is small for small or large values of RL but maximum for some value of RL between 0 and ∞. We now want to show that this maximum power occurs when RL is equal to RTh. This is known as the maximum power theorem. RL VTh RTh + − a b i Figure4.48 The circuit used for maximum power transfer. p RL RTh 0 pmax Figure 4.49 Power delivered to the load as a function of RL. Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen from the load (RL = RTh). To prove the maximum power transfer theorem, we differentiate p in Eq. (4.21) with respect to RL and set the result equal to zero. We obtain dp dRL = V 2 Th (RTh + RL)2 − 2RL(RTh + RL) (RTh + RL)4 = V 2 Th (RTh + RL − 2RL) (RTh + RL)3 = 0
  • 149. CHAPTER 4 Circuit Theorems 143 This implies that 0 = (RTh + RL − 2RL) = (RTh − RL) (4.22) which yields RL = RTh (4.23) showing that the maximum power transfer takes place when the load resistance RL equals the Thevenin resistance RTh. We can readily confirm that Eq. (4.23) gives the maximum power by showing that d2 p/dR2 L 0. Thesourceandloadaresaidtobematched when RL = RTh. The maximum power transferred is obtained by substituting Eq. (4.23) into Eq. (4.21), for pmax = V 2 Th 4RTh (4.24) Equation (4.24) applies only when RL = RTh. When RL = RTh, we compute the power delivered to the load using Eq. (4.21). E X A M P L E 4 . 1 3 Find the value of RL for maximum power transfer in the circuit of Fig. 4.50. Find the maximum power. 12 V 2 A 6 Ω 3 Ω 2 Ω 12 Ω RL + − a b Figure4.50 For Example 4.13. Solution: We need to find the Thevenin resistance RTh and the Thevenin voltage VTh across the terminals a-b. To get RTh, we use the circuit in Fig. 4.51(a) and obtain RTh = 2 + 3 + 6 12 = 5 + 6 × 12 18 = 9 6 Ω 3 Ω 2 Ω 12 Ω RTh 12 V 2 A 6 Ω 3 Ω 2 Ω 12 Ω + − VTh + − (a) (b) i1 i2 Figure4.51 For Example 4.13: (a) finding RTh, (b) finding VTh.
  • 150. 144 PART 1 DC Circuits To get VTh, we consider the circuit in Fig. 4.51(b). Applying mesh anal- ysis, −12 + 18i1 − 12i2 = 0, i2 = −2 A Solving for i1, we get i1 = −2/3. Applying KVL around the outer loop to get VTh across terminals a-b, we obtain −12 + 6i1 + 3i2 + 2(0) + VTh = 0 ⇒ VTh = 22 V For maximum power transfer, RL = RTh = 9 and the maximum power is pmax = V 2 Th 4RL = 222 4 × 9 = 13.44 W P R A C T I C E P R O B L E M 4 . 1 3 Determine the value of RL that will draw the maximum power from the rest of the circuit in Fig. 4.52. Calculate the maximum power. 9 V 4 Ω 2 Ω RL 1 Ω 3vx + − + − + − vx Figure4.52 For Practice Prob. 4.13. Answer: 4.22 , 2.901 W. 4.9 VERIFYING CIRCUIT THEOREMS WITH PSPICE In this section, we learn how to use PSpice to verify the theorems covered in this chapter. Specifically, we will consider using dc sweep analysis to find the Thevenin or Norton equivalent at any pair of nodes in a circuit and the maximum power transfer to a load. The reader is advised to read Section D.3 of Appendix D in preparation for this section. To find the Thevenin equivalent of a circuit at a pair of open ter- minals using PSpice, we use the schematic editor to draw the circuit and insert an independent probing current source, say, Ip, at the terminals. The probing current source must have a part name ISRC. We then per- form a DC Sweep on Ip, as discussed in Section D.3. Typically, we may let the current through Ip vary from 0 to 1 A in 0.1-A increments. After simulating the circuit, we use Probe to display a plot of the voltage across Ip versus the current through Ip. The zero intercept of the plot gives us the Thevenin equivalent voltage, while the slope of the plot is equal to the Thevenin resistance. To find the Norton equivalent involves similar steps except that we insert a probing independent voltage source (with a part name VSRC), say, Vp, at the terminals. We perform a DC Sweep on Vp and let Vp vary from 0 to 1 V in 0.1-V increments. A plot of the current through
  • 151. CHAPTER 4 Circuit Theorems 145 Vp versus the voltage across Vp is obtained using the Probe menu after simulation. The zero intercept is equal to the Norton current, while the slope of the plot is equal to the Norton conductance. To find the maximum power transfer to a load using PSpice involves performing a dc parametric sweep on the component value of RL in Fig. 4.48 and plotting the power delivered to the load as a function of RL. According to Fig. 4.49, the maximum power occurs when RL = RTh. This is best illustrated with an example, and Example 4.15 provides one. We use VSRC and ISRC as part names for the independent voltage and current sources. E X A M P L E 4 . 1 4 Consider the circuit is in Fig. 4.31 (see Example 4.9). Use PSpice to find the Thevenin and Norton equivalent circuits. Solution: (a) To find the Thevenin resistance RTh and Thevenin voltage VTh at the terminals a-b in the circuit in Fig. 4.31, we first use Schematics to draw the circuit as shown in Fig. 4.53(a). Notice that a probing current source I2 is inserted at the terminals. Under Analysis/Setput, we select DC Sweep. In the DC Sweep dialog box, we select Linear for the Sweep Type and Current Source for the Sweep Var. Type. We enter I2 under the Name box, 0 as Start Value, 1 as End Value, and 0.1 as Increment. After simulation, we add trace V(I2:−) from the Probe menu and obtain the plot shown in Fig. 4.53(b). From the plot, we obtain VTh = Zero intercept = 20 V, RTh = Slope = 26 − 20 1 = 6 These agree with what we got analytically in Example 4.9. R2 R4 2 2 GAIN=2 E1 R4 4 R3 6 I2 I1 0 + − (a) (b) 26 V 24 V 22 V 20 V 0 A 0.2 A 0.4 A 0.6 A 0.8 A 1.0 A = V(I2:-) + − Figure4.53 For Example 4.14: (a) schematic and (b) plot for finding RTh and VTh. (b) To find the Norton equivalent, we modify the schematic in Fig. 4.53(a) by replaying the probing current source with a probing voltage source V1. The result is the schematic in Fig. 4.54(a). Again, in the DC Sweep dialog box, we select Linear for the Sweep Type and Voltage Source for the Sweep Var. Type. We enter V1 under Name box, 0 as Start Value, 1 as End Value,
  • 152. 146 PART 1 DC Circuits and 0.1 as Increment. When the Probe is running, we add trace I(V1) and obtain the plot in Fig. 4.54(b). From the plot, we obtain IN = Zero intercept = 3.335 A GN = Slope = 3.335 − 3.165 1 = 0.17 S R2 R1 2 2 GAIN=2 E1 R4 4 R3 6 V1 I1 0 + − (a) (b) 3.4 A 3.3 A 3.2 A 3.1 A 0 V 0.2 V 0.4 V 0.6 V 0.8 V 1.0 V I(V1) V_V1 + − + − Figure4.54 For Example 4.14: (a) schematic and (b) plot for finding GN and IN . P R A C T I C E P R O B L E M 4 . 1 4 Rework Practice Prob. 4.9 using PSpice. Answer: VTh = 5.33 V, RTh = 0.44 . E X A M P L E 4 . 1 5 Refer to the circuit in Fig. 4.55. Use PSpice to find the maximum power transfer to RL. RL 1 V 1 kΩ + − Figure 4.55 For Example 4.15. Solution: We need to perform a dc sweep on RL to determine when the power across it is maximum. We first draw the circuit using Schematics as shown in Fig. 4.56. Once the circuit is drawn, we take the following three steps to further prepare the circuit for a dc sweep. {RL} DC=1 V + − 0 R1 R2 1k V1 PARAMETERS: RL 2k Figure 4.56 Schematic for the circuit in Fig. 4.55. The first step involves defining the value of RL as a parameter, since we want to vary it. To do this: 1. DCLICKL the value 1k of R2 (representing RL) to open up the Set Attribute Value dialog box. 2. Replace 1k with {RL} and click OK to accept the change. Note that the curly brackets are necessary. The second step is to define parameter. To achieve this: 1. Select Draw/Get New Part/Libraries · · ·/special.slb. 2. Type PARAM in the PartName box and click OK. 3. DRAG the box to any position near the circuit. 4. CLICKL to end placement mode.
  • 153. CHAPTER 4 Circuit Theorems 147 5. DCLICKL to open up the PartName: PARAM dialog box. 6. CLICKL on NAME1 = and enter RL (with no curly brackets) in the Value box, and CLICKL Save Attr to accept change. 7. CLICKL on VALUE1 = and enter 2k in the Value box, and CLICKL Save Attr to accept change. 8. Click OK. The value 2k in item 7 is necessary for a bias point calculation; it cannot be left blank. The third step is to set up the DC Sweep to sweep the parameter. To do this: 1. Select Analysis/Setput to bring up the DC Sweep dialog box. 2. For the Sweep Type, select Linear (or Octave for a wide range of RL). 3. For the Sweep Var. Type, select Global Parameter. 4. Under the Name box, enter RL. 5. In the Start Value box, enter 100. 6. In the End Value box, enter 5k. 7. In the Increment box, enter 100. 8. Click OK and Close to accept the parameters. 250 uW 150 uW 200 uW 100 uW 50 uW 0 2.0 K 4.0 K 6.0 K -V(R2:2)*I(R2) RL Figure 4.57 For Example 4.15: the plot of power across PL. After taking these steps and saving the circuit, we are ready to sim- ulate. Select Analysis/Simulate. If there are no errors, we select Add TraceintheProbemenuandtype−V(R2:2)∗ I(R2)intheTraceCommand box. [The negative sign is needed since I(R2) is negative.] This gives the plot of the power delivered to RL as RL varies from 100 to 5 k. We can also obtain the power absorbed by RL by typing V(R2:2)∗ V(R2:2)/RL in the Trace Command box. Either way, we obtain the plot in Fig. 4.57. It is evident from the plot that the maximum power is 250 µW. Notice that the maximum occurs when RL = 1 k, as expected analytically. P R A C T I C E P R O B L E M 4 . 1 5 Find the maximum power transferred to RL if the 1-k resistor in Fig. 4.55 is replaced by a 2-k resistor. Answer: 125 µW. †4.10 APPLICATIONS In this section we will discuss two important practical applications of the concepts covered in this chapter: source modeling and resistance measurement. 4.10.1 Source Modeling Source modeling provides an example of the usefulness of the Thevenin or the Norton equivalent. An active source such as a battery is often characterized by its Thevenin or Norton equivalent circuit. An ideal voltage source provides a constant voltage irrespective of the current
  • 154. 148 PART 1 DC Circuits drawn by the load, while an ideal current source supplies a constant current regardless of the load voltage. As Fig. 4.58 shows, practical voltage and current sources are not ideal, due to their internal resistances or source resistances Rs and Rp. They become ideal as Rs → 0 and Rp → ∞. To show that this is the case, consider the effect of the load on voltage sources, as shown in Fig. 4.59(a). By the voltage division principle, the load voltage is vL = RL Rs + RL vs (4.25) As RL increases, the load voltage approaches a source voltage vs, as illustrated in Fig. 4.59(b). From Eq. (4.25), we should note that: vs Rs + − (a) is Rp (b) Figure4.58 (a) Practical voltage source, (b) practical current source. 1. The load voltage will be constant if the internal resistance Rs of the source is zero or, at least, Rs RL. In other words, the smaller Rs is compared to RL, the closer the voltage source is to being ideal. 2. When the load is disconnected (i.e., the source is open- circuited so that RL → ∞), voc = vs. Thus, vs may be regarded as the unloaded source voltage. The connection of the load causes the terminal voltage to drop in magnitude; this is known as the loading effect. RL vs Rs + − vL + − (a) (b) vL RL 0 vs Practical source Ideal source Figure 4.59 (a) Practical voltage source connected to a load RL, (b) load voltage decreases as RL decreases. The same argument can be made for a practical current source when connected to a load as shown in Fig. 4.60(a). By the current division principle, iL = Rp Rp + RL is (4.26) Figure 4.60(b) shows the variation in the load current as the load re- sistance increases. Again, we notice a drop in current due to the load (loading effect), and load current is constant (ideal current source) when the internal resistance is very large (i.e., Rp → ∞ or, at least, Rp RL). RL (a) is Rp IL (b) IL RL 0 is Practical source Ideal source Figure4.60 (a) Practical current source connected to a load RL, (b) load current decreases as RL increases. Sometimes, we need to know the unloaded source voltage vs and the internal resistance Rs of a voltage source. To find vs and Rs, we follow the procedure illustrated in Fig. 4.61. First, we measure the open-circuit voltage voc as in Fig. 4.61(a) and set vs = voc (4.27)
  • 155. CHAPTER 4 Circuit Theorems 149 Then, we connect a variable load RL across the terminals as in Fig. 4.61(b). We adjust the resistance RL until we measure a load voltage of exactly one-half of the open-circuit voltage, vL = voc/2, because now RL = RTh = Rs. At that point, we disconnect RL and measure it. We set Rs = RL (4.28) For example, a car battery may have vs = 12 V and Rs = 0.05 . Signal source (a) voc + − Signal source (b) vL + − RL Figure4.61 (a) Measuring voc, (b) measuring vL. E X A M P L E 4 . 1 6 The terminal voltage of a voltage source is 12 V when connected to a 2-W load. When the load is disconnected, the terminal voltage rises to 12.4 V. (a) Calculate the source voltage vs and internal resistance Rs. (b) Deter- mine the voltage when an 8- load is connected to the source. Rs (a) (b) RL vs Rs iL + − vL + − 8 Ω 12 V − + v + − 2.4 Ω Figure4.62 For Example 4.16. Solution: (a)WereplacethesourcebyitsTheveninequivalent. Theterminalvoltage when the load is disconnected is the open-circuit voltage, vs = voc = 12.4 V When the load is connected, as shown in Fig. 4.62(a), vL = 12 V and pL = 2 W. Hence, pL = vL2 RL ⇒ RL = v2 L pL = 122 2 = 72 The load current is iL = vL RL = 12 72 = 1 6 A The voltage across Rs is the difference between the source voltage vs and the load voltage vL, or 12.4 − 12 = 0.4 = RsiL, Rs = 0.4 IL = 2.4 (b) Now that we have the Thevenin equivalent of the source, we connect the 8- load across the Thevenin equivalent as shown in Fig. 4.62(b). Using voltage division, we obtain v = 8 8 + 2.4 (12) = 9.231 V
  • 156. 150 PART 1 DC Circuits P R A C T I C E P R O B L E M 4 . 1 6 The measured open-circuit voltage across a certain amplifier is 9 V. The voltage drops to 8 V when a 20- loudspeaker is connected to the am- plifier. Calculate the voltage when a 10- loudspeaker is used instead. Answer: 7.2 V. 4.10.2 Resistance Measurement Although the ohmmeter method provides the simplest way to measure re- sistance, more accurate measurement may be obtained using the Wheat- stone bridge. While ohmmeters are designed to measure resistance in low, mid, or high range, a Wheatstone bridge is used to measure resis- tance in the mid range, say, between 1 and 1 M. Very low values of resistances are measured with a milliohmmeter, while very high values are measured with a Megger tester. Historical note: The bridge was invented by Charles Wheatstone (1802–1875), a British professor who also invented the telegraph, as Samuel Morse did independently in the United States. v R1 R3 R2 Rx + − Galvanometer v1 + − + − v2 Figure 4.63 The Wheatstone bridge; Rx is the resistance to be measured. The Wheatstone bridge (or resistance bridge) circuit is used in a number of applications. Here we will use it to measure an unknown re- sistance. The unknown resistance Rx is connected to the bridge as shown in Fig. 4.63. The variable resistance is adjusted until no current flows through the galvanometer, which is essentially a d’Arsonval movement operating as a sensitive current-indicating device like an ammeter in the microamp range. Under this condition v1 = v2, and the bridge is said to be balanced. Since no current flows through the galvanometer, R1 and R2 behave as though they were in series; so do R3 and Rx. The fact that no current flows through the galvanometer also implies that v1 = v2. Applying the voltage division principle, v1 = R2 R1 + R2 v = v2 = Rx R3 + Rx v (4.29) Hence, no current flows through the galvanometer when R2 R1 + R2 = Rx R3 + Rx ⇒ R2R3 = R1Rx or Rx = R3 R1 R2 (4.30) If R1 = R3, and R2 is adjusted until no current flows through the gal- vanometer, then Rx = R2. How do we find the current through the galvanometer when the Wheatstone bridge is unbalanced? We find the Thevenin equivalent (VTh and RTh) with respect to the galvanometer terminals. If Rm is the resis- tance of the galvanometer, the current through it under the unbalanced condition is I = VTh RTh + Rm (4.31) Example 4.18 will illustrate this.
  • 157. CHAPTER 4 Circuit Theorems 151 E X A M P L E 4 . 1 7 In Fig. 4.63, R1 = 500 and R3 = 200 . The bridge is balanced when R2 is adjusted to be 125 . Determine the unknown resistance Rx. Solution: Using Eq. (4.30), Rx = R3 R1 R2 = 200 500 125 = 50 P R A C T I C E P R O B L E M 4 . 1 7 A Wheatstone bridge has R1 = R3 = 1 k. R2 is adjusted until no cur- rent flows through the galvanometer. At that point, R2 = 3.2 k. What is the value of the unknown resistance? Answer: 3.2 k. E X A M P L E 4 . 1 8 The circuit in Fig. 4.64 represents an unbalanced bridge. If the galvano- meter has a resistance of 40 , find the current through the galvanometer. 220 V 400 Ω 600 Ω + − G 3 kΩ 1 kΩ 40 Ω a b Figure4.64 Unbalanced bridge of Example 4.18. Solution: We first need to replace the circuit by its Thevenin equivalent at termi- nals a and b. The Thevenin resistance is found using the circuit in Fig. 4.65(a). Notice that the 3-k and 1-k resistors are in parallel; so are the 400- and 600- resistors. The two parallel combinations form a series combination with respect to terminals a and b. Hence, RTh = 3000 1000 + 400 600 = 3000 × 1000 3000 + 1000 + 400 × 600 400 + 600 = 750 + 240 = 990 To find the Thevenin voltage, we consider the circuit in Fig. 4.65(b). Using the voltage division principle, v1 = 1000 1000 + 3000 (220) = 55 V, v2 = 600 600 + 400 (220) = 132 V Applying KVL around loop ab gives −v1 + VTh + v2 = 0 or VTh = v1 − v2 = 55 − 132 = −77 V
  • 158. 152 PART 1 DC Circuits 220 V 400 Ω 600 Ω + − 3 kΩ 1 kΩ a b + − VTh (b) VTh 40 Ω + − (c) 400 Ω 600 Ω 3 kΩ 1 kΩ a b RTh (a) RTh a b G IG + − v1 + − v2 Figure 4.65 For Example 4.18: (a) Finding RTh, (b) finding VTh, (c) determining the current through the galvanometer. Having determined the Thevenin equivalent, we find the current through the galvanometer using Fig. 4.65(c). IG = VTh RTh + Rm = −77 990 + 40 = −74.76 mA The negative sign indicates that the current flows in the direction opposite to the one assumed, that is, from terminal b to terminal a. P R A C T I C E P R O B L E M 4 . 1 8 Obtain the current through the galvanometer, having a resistance of 14 , in the Wheatstone bridge shown in Fig. 4.66. 14 Ω 60 Ω 16 V 40 Ω 20 Ω 30 Ω G Figure4.66 For Practice Prob. 4.18. Answer: 64 mA.
  • 159. CHAPTER 4 Circuit Theorems 153 4.11 SUMMARY 1. A linear network consists of linear elements, linear dependent sources, and linear independent sources. 2. Network theorems are used to reduce a complex circuit to a simpler one, thereby making circuit analysis much simpler. 3. The superposition principle states that for a circuit having multiple independent sources, the voltage across (or current through) an element is equal to the algebraic sum of all the individual voltages (or currents) due to each independent source acting one at a time. 4. Source transformation is a procedure for transforming a voltage source in series with a resistor to a current source in parallel with a resistor, or vice versa. 5. Thevenin’s and Norton’s theorems allow us to isolate a portion of a network while the remaining portion of the network is replaced by an equivalent network. The Thevenin equivalent consists of a voltage source VTh in series with a resistor RTh, while the Norton equivalent consists of a current source IN in parallel with a resistor RN . The two theorems are related by source transformation. RN = RTh, IN = VTh RTh 6. For a given Thevenin equivalent circuit, maximum power transfer occurs when RL = RTh, that is, when the load resistance is equal to the Thevenin resistance. 7. PSpice can be used to verify the circuit theorems covered in this chapter. 8. Source modeling and resistance measurement using the Wheatstone bridge provide applications for Thevenin’s theorem. REVIEW QUESTIONS 4.1 The current through a branch in a linear network is 2 A when the input source voltage is 10 V. If the voltage is reduced to 1 V and the polarity is reversed, the current through the branch is: (a) −2 (b) −0.2 (c) 0.2 (d) 2 (e) 20 4.2 For superposition, it is not required that only one independent source be considered at a time; any number of independent sources may be considered simultaneously. (a) True (b) False 4.3 The superposition principle applies to power calculation. (a) True (b) False 4.4 Refer to Fig. 4.67. The Thevenin resistance at terminals a and b is: (a) 25 (b) 20 (c) 5 (d) 4 50 V 20 Ω + − 5 Ω a b Figure 4.67 For Review Questions 4.4 to 4.6. 4.5 The Thevenin voltage across terminals a and b of the circuit in Fig. 4.67 is: (a) 50 V (b) 40 V (c) 20 V (d) 10 V
  • 160. 154 PART 1 DC Circuits 4.6 The Norton current at terminals a and b of the circuit in Fig. 4.67 is: (a) 10 A (b) 2.5 A (c) 2 A (d) 0 A 4.7 The Norton resistance RN is exactly equal to the Thevenin resistance RTh. (a) True (b) False 4.8 Which pair of circuits in Fig. 4.68 are equivalent? (a) a and b (b) b and d (c) a and c (d) c and d + − 20 V 5 Ω (a) 4 A 5 Ω (b) 5 Ω (c) + − 20 V 5 Ω (d) 4 A Figure 4.68 For Review Question 4.8. 4.9 A load is connected to a network. At the terminals to which the load is connected, RTh = 10 and VTh = 40 V. The maximum power supplied to the load is: (a) 160 W (b) 80 W (c) 40 W (d) 1 W 4.10 The source is supplying the maximum power to the load when the load resistance equals the source resistance. (a) True (b) False Answers: 4.1b, 4.2a, 4.3b, 4.4d, 4.5b, 4.6a, 4.7a, 4.8c, 4.9c, 4.10b. PROBLEMS Section 4.2 Linearity Property 4.1 Calculate the current io in the current of Fig. 4.69. What does this current become when the input voltage is raised to 10 V? + − io 1 Ω 5 Ω 3 Ω 8 Ω 1 V Figure 4.69 For Prob. 4.1. 4.2 Find vo in the circuit of Fig. 4.70. If the source current is reduced to 1 µA, what is vo? 5 Ω 4 Ω 6 Ω 8 Ω 1 A 2 Ω + − vo Figure 4.70 For Prob. 4.2. 4.3 (a) In the circuit in Fig. 4.71, calculate vo and io when vs = 1 V. (b) Find vo and io when vs = 10 V. (c) What are vo and io when each of the 1- resistors is replaced by a 10- resistor and vs = 10 V? + − 1 Ω 1 Ω 1 Ω 1 Ω vs 1 Ω io + − vo Figure 4.71 For Prob. 4.3. 4.4 Use linearity to determine io in the circuit of Fig. 4.72. 2 Ω 3 Ω 4 Ω 6 Ω 9 A io Figure 4.72 For Prob. 4.4.
  • 161. CHAPTER 4 Circuit Theorems 155 4.5 For the circuit in Fig. 4.73, assume vo = 1 V, and use linearity to find the actual value of vo. 2 Ω 3 Ω 4 Ω 6 Ω vo 2 Ω 6 Ω 15 V + − Figure 4.73 For Prob. 4.5. Section 4.3 Superposition 4.6 Apply superposition to find i in the circuit of Fig. 4.74. 20 V 5 A 6 Ω 4 Ω + − i Figure 4.74 For Prob. 4.6. 4.7 Given the circuit in Fig. 4.75, calculate ix and the power dissipated by the 10- resistor using superposition. 12 Ω 4 A 10 Ω 40 Ω 15 V − + ix Figure 4.75 For Prob. 4.7. 4.8 For the circuit in Fig. 4.76, find the terminal voltage Vab using superposition. 4 V 2 A a b 10 Ω 3Vab + − + − Vab + − Figure 4.76 For Prob. 4.8. 4.9 Use superposition principle to find i in Fig. 4.77. 6 Ω 4 A 2 Ω 3 Ω 12 V − + i Figure 4.77 For Prob. 4.9. 4.10 Determine vo in the circuit of Fig. 4.78 using the superposition principle. 12 V 5 Ω 6 Ω 2 A 4 Ω 12 Ω 3 Ω + − 19 V + − + − vo Figure 4.78 For Prob. 4.10. 4.11 Apply the superposition principle to find vo in the circuit of Fig. 4.79. + − 6 Ω 2 Ω 3 Ω 1 A 2 A 20 V 4 Ω + − vo Figure 4.79 For Prob. 4.11. 4.12 For the circuit in Fig. 4.80, use superposition to find i. Calculate the power delivered to the 3- resistor. 20 V 2 A 3 Ω 2 Ω 1 Ω 4 Ω 16 V − + i + − Figure 4.80 For Probs. 4.12 and 4.45.
  • 162. 156 PART 1 DC Circuits 4.13 Given the circuit in Fig. 4.81, use superposition to get io. 12 V 3 Ω 4 Ω 4 A 2 Ω 5 Ω 10 Ω + − 2 A io Figure 4.81 For Probs. 4.13 and 4.23. 4.14 Use superposition to obtain vx in the circuit of Fig. 4.82. Check your result using PSpice. 90 V 6 A 30 Ω 10 Ω 20 Ω 60 Ω 30 Ω + − 40 V + − + − vx Figure 4.82 For Prob. 4.14. 4.15 Find vx in Fig. 4.83 by superposition. 2 Ω 1 Ω 5ix 2 A 4 Ω 10 V + − ix + − vx Figure 4.83 For Prob. 4.15. 4.16 Use superposition to solve for ix in the circuit of Fig. 4.84. 8 Ω 2 Ω 6 A 4 A − + ix 4ix + − vx Figure 4.84 For Prob. 4.16. Section 4.4 Source Transformation 4.17 Find i in Prob. 4.9 using source transformation. 4.18 Apply source transformation to determine vo and io in the circuit in Fig. 4.85. 12 V 2 A 6 Ω 3 Ω + − io + − vo Figure 4.85 For Prob. 4.18. 4.19 For the circuit in Fig. 4.86, use source transformation to find i. 5 Ω 10 Ω 4 Ω 5 Ω 2 A 20 V + − i Figure 4.86 For Prob. 4.19. 4.20 Obtain vo in the circuit of Fig. 4.87 using source transformation. Check your result using PSpice. 3 A 9 Ω 2 Ω 2 A 30 V 5 Ω 4 Ω 6 A + − + − vo Figure 4.87 For Prob. 4.20. 4.21 Use source transformation to solve Prob. 4.14. 4.22 Apply source transformation to find vx in the circuit of Fig. 4.88.
  • 163. CHAPTER 4 Circuit Theorems 157 50 V 8 A 10 Ω 12 Ω 20 Ω 40 Ω + − 40 V + − a b + − vx Figure 4.88 For Probs. 4.22 and 4.32. 4.23 Given the circuit in Fig. 4.81, use source transformation to find io. 4.24 Use source transformation to find vo in the circuit of Fig. 4.89. 4 kΩ 1 kΩ 3 mA 2 kΩ 3vo − + + − vo Figure 4.89 For Prob. 4.24. 4.25 Determine vx in the circuit of Fig. 4.90 using source transformation. + − 3 Ω 6 Ω 2vx 8 Ω 12 V + − + − vx Figure 4.90 For Prob. 4.25. 4.26 Use source transformation to find ix in the circuit of Fig. 4.91. 10 Ω 15 Ω 0.5ix 40 Ω 60 V + − 50 Ω ix Figure 4.91 For Prob. 4.26. Sections 4.5 and 4.6 Thevenin’s and Norton’s Theorems 4.27 Determine RTh and VTh at terminals 1-2 of each of the circuits in Fig. 4.92. 10 Ω + − 20 V 40 Ω (a) (b) 1 2 2 A 30 Ω 30 V + − 60 Ω 1 2 Figure 4.92 For Probs. 4.27 and 4.37. 4.28 Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.93. 20 Ω 10 Ω 3 A 40 V 40 Ω a b + − Figure 4.93 For Probs. 4.28 and 4.39. 4.29 Use Thevenin’s theorem to find vo in Prob. 4.10. 4.30 Solve for the current i in the circuit of Fig. 4.94 using Thevenin’s theorem. (Hint: Find the Thevenin equivalent across the 12- resistor.) 12 Ω 30 V 40 Ω + − 10 Ω 50 V + − i Figure 4.94 For Prob. 4.30. 4.31 For Prob. 4.8, obtain the Thevenin equivalent at terminals a-b.
  • 164. 158 PART 1 DC Circuits 4.32 Given the circuit in Fig. 4.88, obtain the Thevenin equivalent at terminals a-b and use the result to get vx. 4.33 ∗ For the circuit in Fig. 4.95, find the Thevenin equivalent between terminals a and b. 20 Ω 20 Ω 10 Ω 10 Ω 5 A 10 Ω 20 V 30 V + − − + 10 Ω a b Figure 4.95 For Prob. 4.33. 4.34 Find the Thevenin equivalent looking into terminals a-b of the circuit in Fig. 4.96 and solve for ix. 20 V 2 A 10 Ω 6 Ω 10 Ω + − 5 Ω a b ix Figure 4.96 For Prob. 4.34. 4.35 For the circuit in Fig. 4.97, obtain the Thevenin equivalent as seen from terminals: (a) a-b (b) b-c 4 Ω 24 V 5 Ω 2 Ω 1 Ω 3 Ω 2 A a b c + − Figure 4.97 For Prob. 4.35. 4.36 Find the Norton equivalent of the circuit in Fig. 4.98. *An asterisk indicates a challenging problem. 4 A 4 Ω a b 6 Ω 6 Ω Figure 4.98 For Prob. 4.36. 4.37 Obtain RN and IN at terminals 1 and 2 of each of the circuits in Fig. 4.92. 4.38 Determine the Norton equivalent at terminals a-b for the circuit in Fig. 4.99. 2 A a b 4 Ω 2 Ω 10io + − io Figure 4.99 For Prob. 4.38. 4.39 Find the Norton equivalent looking into terminals a-b of the circuit in Fig. 4.93. 4.40 Obtain the Norton equivalent of the circuit in Fig. 4.100 to the left of terminals a-b. Use the result to find current i. 2 A a b 4 A 4 Ω 5 Ω 12 V 6 Ω + − i Figure 4.100 For Prob. 4.40. 4.41 Given the circuit in Fig. 4.101, obtain the Norton equivalent as viewed from terminals: (a) a-b (b) c-d 120 V c a b d 6 A 2 Ω 3 Ω 4 Ω 6 Ω + − Figure 4.101 For Prob. 4.41.
  • 165. CHAPTER 4 Circuit Theorems 159 4.42 For the transistor model in Fig. 4.102, obtain the Thevenin equivalent at terminals a-b. 6 V 20io 3 kΩ 2 kΩ + − a b io Figure 4.102 For Prob. 4.42. 4.43 Find the Norton equivalent at terminals a-b of the circuit in Fig. 4.103. 2 Ω 6 Ω 0.25vo 3 Ω 18 V + − vo + − a b Figure 4.103 For Prob. 4.43. 4.44 ∗ Obtain the Norton equivalent at terminals a-b of the circuit in Fig. 4.104. 2 V 80I 8 kΩ 50 kΩ + − a b 0.01Vab + − I + − Vab Figure 4.104 For Prob. 4.44. 4.45 Use Norton’s theorem to find current i in the circuit of Fig. 4.80. 4.46 Obtain the Thevenin and Norton equivalent circuits at the terminals a-b for the circuit in Fig. 4.105. 50 V 3 Ω 2 Ω 10 Ω + − a b 0.5vx 6 Ω + − vx Figure 4.105 For Probs. 4.46 and 4.65. 4.47 The network in Fig. 4.106 models a bipolar transistor common-emitter amplifier connected to a load. Find the Thevenin resistance seen by the load. vs R1 bib RL + − R2 ib Figure 4.106 For Prob. 4.47. 4.48 Determine the Thevenin and Norton equivalents at terminals a-b of the circuit in Fig. 4.107. 8 A 10 Ω 20 Ω 50 Ω 40 Ω a b Figure 4.107 For Probs. 4.48 and 4.66. 4.49 ∗ For the circuit in Fig. 4.108, find the Thevenin and Norton equivalent circuits at terminals a-b. + − 18 V 3 A 4 Ω 6 Ω 5 Ω a b 2 A 10 V + − Figure 4.108 For Probs. 4.49 and 4.67. 4.50 ∗ Obtain the Thevenin and Norton equivalent circuits at terminals a-b of the circuit in Fig. 4.109.
  • 166. 160 PART 1 DC Circuits 12 V 6 Ω 2 Ω 6 Ω 2 Ω 6 Ω + − 12 V 2 Ω + − 12 V + − a b Figure 4.109 For Prob. 4.50. 4.51 ∗ Find the Thevenin equivalent of the circuit in Fig. 4.110. 10 Ω 20 Ω 40 Ω + − io 0.1io 2vo + − vo b a Figure 4.110 For Prob. 4.51. 4.52 Find the Norton equivalent for the circuit in Fig. 4.111. 0.5vo 10 Ω + − vo 20 V Figure 4.111 For Prob. 4.52. 4.53 Obtain the Thevenin equivalent seen at terminals a-b of the circuit in Fig. 4.112. 10ix 4 Ω 2 Ω 1 Ω + − ix a b Figure 4.112 For Prob. 4.53. Section 4.8 Maximum Power Transfer 4.54 Find the maximum power that can be delivered to the resistor R in the circuit in Fig. 4.113. R 3 Ω 2 Ω 5 Ω 20 V 6 A + − − + 10 V Figure 4.113 For Prob. 4.54. 4.55 Refer to Fig. 4.114. For what value of R is the power dissipated in R maximum? Calculate that power. 6 Ω 30 V 4 Ω 8 Ω R 12 Ω + − Figure 4.114 For Prob. 4.55. 4.56 ∗ Compute the value of R that results in maximum power transfer to the 10- resistor in Fig. 4.115. Find the maximum power. + − + − R 20 Ω 10 Ω 8 V 12 V Figure 4.115 For Prob. 4.56. 4.57 Find the maximum power transferred to resistor R in the circuit of Fig. 4.116. R 40 kΩ 30 kΩ 100 V + − 3vo 22 kΩ 10 kΩ + − vo Figure 4.116 For Prob. 4.57.
  • 167. CHAPTER 4 Circuit Theorems 161 4.58 For the circuit in Fig. 4.117, what resistor connected across terminals a-b will absorb maximum power from the circuit? What is that power? 8 V 120vo 3 kΩ 10 kΩ 40 kΩ 1 kΩ + − a b – + + − vo Figure 4.117 For Prob. 4.58. 4.59 (a) For the circuit in Fig. 4.118, obtain the Thevenin equivalent at terminals a-b. (b) Calculate the current in RL = 8 . (c) Find RL for maximum power deliverable to RL. (d) Determine that maximum power. 6 Ω 4 Ω 2 A 20 V 4 A 2 Ω a b RL + − Figure 4.118 For Prob. 4.59. 4.60 For the bridge circuit shown in Fig. 4.119, find the load RL for maximum power transfer and the maximum power absorbed by the load. R3 RL R4 R1 R2 vs + − Figure 4.119 For Prob. 4.60. 4.61 For the circuit in Fig. 4.120, determine the value of R such that the maximum power delivered to the load is 3 mW. R R R 2 V 3 V RL + − 1 V + − + − Figure 4.120 For Prob. 4.61. Section 4.9 Verifying Circuit Theorems with PSpice 4.62 Solve Prob. 4.28 using PSpice. 4.63 Use PSpice to solve Prob. 4.35. 4.64 Use PSpice to solve Prob. 4.42. 4.65 Obtain the Thevenin equivalent of the circuit in Fig. 4.105 using PSpice. 4.66 Use PSpice to find the Thevenin equivalent circuit at terminals a-b of the circuit in Fig. 4.107. 4.67 For the circuit in Fig. 4.108, use PSpice to find the Thevenin equivalent at terminals a-b. Section 4.10 Applications 4.68 A battery has a short-circuit current of 20 A and an open-circuit voltage of 12 V. If the battery is connected to an electric bulb of resistance 2 , calculate the power dissipated by the bulb. 4.69 The following results were obtained from measurements taken between the two terminals of a resistive network. Terminal Voltage 12 V 0 V Terminal Current 0 V 1.5 A Find the Thevenin equivalent of the network. 4.70 A black box with a circuit in it is connected to a variable resistor. An ideal ammeter (with zero resistance) and an ideal voltmeter (with infinite resistance) are used to measure current and voltage as shown in Fig. 4.121. The results are shown in the table below. Black box V A i R Figure 4.121 For Prob. 4.70.
  • 168. 162 PART 1 DC Circuits (a) Find i when R = 4 . (b) Determine the maximum power from the box. R() V (V) i(A) 2 3 1.5 8 8 1.0 14 10.5 0.75 4.71 A transducer is modeled with a current source Is and a parallel resistance Rs. The current at the terminals of the source is measured to be 9.975 mA when an ammeter with an internal resistance of 20 is used. (a) If adding a 2-k resistor across the source terminals causes the ammeter reading to fall to 9.876 mA, calculate Is and Rs. (b) What will the ammeter reading be if the resistance between the source terminals is changed to 4 k? 4.72 The Wheatstone bridge circuit shown in Fig. 4.122 is used to measure the resistance of a strain gauge. The adjustable resistor has a linear taper with a maximum value of 100 . If the resistance of the strain gauge is found to be 42.6 , what fraction of the full slider travel is the slider when the bridge is balanced? 4 kΩ 100 Ω 2 kΩ + − vs Rs Rx G Figure 4.122 For Prob. 4.72. 4.73 (a) In the Wheatstone bridge circuit of Fig. 4.123, select the values of R1 and R3 such that the bridge can measure Rx in the range of 0–10 . R3 Rx R1 V + − G 50 Ω Figure 4.123 For Prob. 4.73. (b) Repeat for the range of 0–100 . 4.74 ∗ Consider the bridge circuit of Fig. 4.124. Is the bridge balanced? If the 10-k resistor is replaced by an 18-k resistor, what resistor connected between terminals a-b absorbs the maximum power? What is this power? 220 V 2 kΩ 3 kΩ 6 kΩ 5 kΩ 10 kΩ + − a b Figure 4.124 For Prob. 4.74. COMPREHENSIVE PROBLEMS 4.75 The circuit in Fig. 4.125 models a common-emitter transistor amplifier. Find ix using source transformation. vs Rs + − bix Ro ix Figure 4.125 For Prob. 4.75. 4.76 An attenuator is an interface circuit that reduces the voltage level without changing the output resistance. (a) By specifying Rs and Rp of the interface circuit in Fig. 4.126, design an attenuator that will meet the following requirements: Vo Vg = 0.125, Req = RTh = Rg = 100 (b) Using the interface designed in part (a), calculate the current through a load of RL = 50 when Vg = 12 V.
  • 169. CHAPTER 4 Circuit Theorems 163 V g Rg Rs RL Req + − Rp Vo + − Attenuator Load Figure 4.126 For Prob. 4.76. 4.77 ∗ A dc voltmeter with a sensitivity of 20 k/V is used to find the Thevenin equivalent of a linear network. Readings on two scales are as follows: (a) 0–10 V scale: 4 V (b) 0–50 V scale: 5 V Obtain the Thevenin voltage and the Thevenin resistance of the network. 4.78 ∗ A resistance array is connected to a load resistor R and a 9-V battery as shown in Fig. 4.127. (a) Find the value of R such that Vo = 1.8 V. (b) Calculate the value of R that will draw the maximum current. What is the maximum current? 60 Ω 10 Ω 10 Ω 8 Ω 8 Ω R 10 Ω 40 Ω 9 V + − 3 4 1 2 + − Vo Figure 4.127 For Prob. 4.78. 4.79 A common-emitter amplifier circuit is shown in Fig. 4.128. Obtain the Thevenin equivalent to the left of points B and E. RL Rc E 4 kΩ 6 kΩ 12 V B + − Figure 4.128 For Prob. 4.79. 4.80 ∗ For Practice Prob. 4.17, determine the current through the 40- resistor and the power dissipated by the resistor.
  • 170. 165 C H A P T E R OPERATIONAL AMPLIFIERS 5 If A is success in life, then A equals X plus Y plus Z. Work is X, Y is play and Z is keeping your mouth shut. —Albert Einstein Enhancing Your Career Career in Electronic Instrumentation Engineering in- volves applying physical principles to design devices for the benefit of humanity. But physical principles cannot be understood without measurement. In fact, physicists often say that physics is the science that measures reality. Just as measurements are a tool for understanding the physical world, instruments are tools for measurement. The opera- tional amplifier introduced in this chapter is a building block of modern electronic instrumentation. Therefore, mastery of operational amplifier fundamentals is paramount to any practical application of electronic circuits. Electronic instruments are used in all fields of sci- ence and engineering. They have proliferated in science and technology to the extent that it would be ridiculous to have a scientific or technical education without exposure to elec- tronic instruments. For example, physicists, physiologists, chemists, and biologists must learn to use electronic instru- ments. For electrical engineering students in particular, the skill in operating digital and analog electronic instruments is crucial. Such instruments include ammeters, voltmeters, ohmmeters, oscilloscopes, spectrum analyzers, and signal generators. Beyond developing the skill for operating the instru- ments, some electrical engineers specialize in designing and constructing electronic instruments. These engineers derive pleasure in building their own instruments. Most of them Electronic Instrumentation used in medical research. Source: Geoff Tompkinson/Science Photo Library. invent and patent their inventions. Specialists in electronic instruments find employment in medical schools, hospitals, research laboratories, aircraft industries, and thousands of other industries where electronic instruments are routinely used.
  • 171. 166 PART 1 DC Circuits 5.1 INTRODUCTION Having learned the basic laws and theorems for circuit analysis, we are now ready to study an active circuit element of paramount importance: the operational amplifier, or op amp for short. The op amp is a versatile circuit building block. The term operational amplifier was introduced in 1947 by John Ragazzini and his colleagues, in theirworkonanalogcomputersfortheNational Defense Research Council during World War II. Thefirstopampsusedvacuumtubesratherthan transistors. The op amp is an electronic unit that behaves like a voltage-controlled voltage source. An op amp may also be regarded as a voltage amplifier with very high gain. It can also be used in making a voltage- or current-controlled current source. An op amp can sum signals, amplify a signal, integrate it, or differentiate it. The ability of the op amp to perform these mathematical operations is the reason it is called an operational amplifier. It is also the reason for the widespread use of op amps in analog design. Op amps are popular in practical circuit designs because they are versatile, inexpensive, easy to use, and fun to work with. We begin by discussing the ideal op amp and later consider the nonideal op amp. Using nodal analysis as a tool, we consider ideal op amp circuits such as the inverter, voltage follower, summer, and difference amplifier. We will analyze op amp circuits with PSpice. Finally, we learn how an op amp is used in digital-to-analog converters and instrumentation amplifiers. 5.2 OPERATIONAL AMPLIFIERS An operational amplifier is designed so that it performs some mathemat- ical operations when external components, such as resistors and capaci- tors, are connected to its terminals. Thus, An op amp is an active circuit element designed to perform mathematical operations of addition, subtraction, multiplication, division, differentiation, and integration. The op amp is an electronic device consisting of a complex arrange- ment of resistors, transistors, capacitors, and diodes. A full discussion of what is inside the op amp is beyond the scope of this book. It will suffice to treat the op amp as a circuit building block and simply study what takes place at its terminals. Figure 5.1 A typical operational amplifier. (Courtesy of Tech America.) Op amps are commercially available in integrated circuit packages in several forms. Figure 5.1 shows a typical op amp package. A typical one is the eight-pin dual in-line package (or DIP), shown in Fig. 5.2(a). Pin or terminal 8 is unused, and terminals 1 and 5 are of little concern to us. The five important terminals are: ThepindiagraminFig.5.2(a)correspondstothe 741 general-purpose op amp made by Fairchild Semiconductor. 1. The inverting input, pin 2. 2. The noninverting input, pin 3. 3. The output, pin 6.
  • 172. CHAPTER 5 Operational Amplifiers 167 4. The positive power supply V + , pin 7. 5. The negative power supply V − , pin 4. The circuit symbol for the op amp is the triangle in Fig. 5.2(b); as shown, the op amp has two inputs and one output. The inputs are marked with minus (−) and plus (+) to specify inverting and noninverting inputs, respectively. An input applied to the noninverting terminal will appear with the same polarity at the output, while an input applied to the inverting terminal will appear inverted at the output. + − 1 Balance 2 Inverting input 3 Noninverting input 4 V− 5 Balance 6 Output 7 V+ 8 No connection (a) (b) 2 Inverting input 3 Noninverting input 4 V− V+ 1 5 Offset Null 6 Output 7 Figure5.2 A typical op amp: (a) pin configuration, (b) circuit symbol. As an active element, the op amp must be powered by a voltage supply as typically shown in Fig. 5.3. Although the power supplies are often ignored in op amp circuit diagrams for the sake of simplicity, the power supply currents must not be overlooked. By KCL, io = i1 + i2 + i+ + i− (5.1) 7 4 6 Vcc Vcc + − + − io i1 i2 i+ i− 2 3 Figure5.3 Powering the op amp. v1 v2 vo + − + − vd Ri Ro Avd Figure 5.4 The equivalent circuit of the non- ideal op amp. The equivalent circuit model of an op amp is shown in Fig. 5.4. The output section consists of a voltage-controlled source in series with the output resistance Ro. It is evident from Fig. 5.4 that the input resistance Ri is the Thevenin equivalent resistance seen at the input terminals, while the output resistance Ro is the Thevenin equivalent resistance seen at the output. The differential input voltage vd is given by vd = v2 − v1 (5.2) where v1 is the voltage between the inverting terminal and ground and v2 is the voltage between the noninverting terminal and ground. The op amp senses the difference between the two inputs, multiplies it by the gain A, and causes the resulting voltage to appear at the output. Thus, the output vo is given by vo = Avd = A(v2 − v1) (5.3) A is called the open-loop voltage gain because it is the gain of the op amp without any external feedback from output to input. Table 5.1 shows typical values of voltage gain A, input resistance Ri, output resistance Ro, and supply voltage VCC. Sometimes, voltage gain is expressed in decibels (dB), as discussed in Chapter 14. A dB = 20 log10 A The concept of feedback is crucial to our understanding of op amp circuits. A negative feedback is achieved when the output is fed back to the inverting terminal of the op amp. As Example 5.1 shows, when there
  • 173. 168 PART 1 DC Circuits TABLE 5.1 Typical ranges for op amp parameters. Parameter Typical range Ideal values Open-loop gain, A 105 to 108 ∞ Input resistance, Ri 106 to 1013 ∞ Output resistance, Ro 10 to 100 0 Supply voltage, Vcc 5 to 24 V is a feedback path from output to input, the ratio of the output voltage to the input voltage is called the closed-loop gain. As a result of the negative feedback, it can be shown that the closed-loop gain is almost insensitive to the open-loop gain A of the op amp. For this reason, op amps are used in circuits with feedback paths. A practical limitation of the op amp is that the magnitude of its output voltage cannot exceed |VCC|. In other words, the output voltage is dependent on and is limited by the power supply voltage. Figure 5.5 illustrates that the op amp can operate in three modes, depending on the differential input voltage vd: 1. Positive saturation, vo = VCC. 2. Linear region, −VCC ≤ vo = Avd ≤ VCC. 3. Negative saturation, vo = −VCC. Positive saturation Negative saturation vd vo VCC −VCC 0 Figure5.5 Op amp output voltage vo as a function of the differential input voltage vd . If we attempt to increase vd beyond the linear range, the op amp becomes saturated and yields vo = VCC or vo = −VCC. Throughout this book, we will assume that our op amps operate in the linear mode. This means that the output voltage is restricted by −VCC ≤ vo ≤ VCC (5.4) Although we shall always operate the op amp in the linear region, the possibility of saturation must be borne in mind when one designs with op amps, to avoid designing op amp circuits that will not work in the laboratory. E X A M P L E 5 . 1 A 741 op amp has an open-loop voltage gain of 2 × 105 , input resistance of 2 M, and output resistance of 50 . The op amp is used in the circuit of Fig. 5.6(a). Find the closed-loop gain vo/vs. Determine current i when vs = 2 V. Solution: Using the op amp model in Fig. 5.4, we obtain the equivalent circuit of Fig. 5.6(a) as shown in Fig. 5.6(b). We now solve the circuit in Fig. 5.6(b) by using nodal analysis. At node 1, KCL gives vs − v1 10 × 103 = v1 2000 × 103 + v1 − vo 20 × 103
  • 174. CHAPTER 5 Operational Amplifiers 169 10 kΩ 20 kΩ vs i vo + − + − 1 O (a) (b) + − 741 10 kΩ 20 kΩ vs i Ro = 50 Ω Ri = 2 MΩ + − 1 O + − Avd v1 vo − + vd Figure5.6 For Example 5.1: (a) original circuit, (b) the equivalent circuit. Multiplying through by 2000 × 103 , we obtain 200vs = 301v1 − 100vo or 2vs 3v1 − vo ⇒ v1 = 2vs + vo 3 (5.1.1) At node O, v1 − vo 20 × 103 = vo − Avd 50 But vd = −v1 and A = 200,000. Then v1 − vo = 400(vo + 200,000v1) (5.1.2) Substituting v1 from Eq. (5.1.1) into Eq. (5.1.2) gives 0 26,667,067vo + 53,333,333vs vo vs = −1.9999699 This is closed-loop gain, because the 20-k feedback resistor closes the loop between the output and input terminals. When vs = 2 V, vo = −3.9999398 V. From Eq. (5.1.1), we obtain v1 = 20.066667 µV. Thus, i = v1 − vo 20 × 103 = 0.1999 mA It is evident that working with a nonideal op amp is tedious, as we are dealing with very large numbers. P R A C T I C E P R O B L E M 5 . 1 If the same 741 op amp in Example 5.1 is used in the circuit of Fig. 5.7, calculate the closed-loop gain vo/vs. Find io when vs = 1 V. 40 kΩ 20 kΩ 5 kΩ vs vo io + − + − + − 741 Figure5.7 For Practice Prob. 5.1. Answer: 9.0041, −362 mA.
  • 175. 170 PART 1 DC Circuits 5.3 IDEAL OP AMP To facilitate the understanding of op amp circuits, we will assume ideal op amps. An op amp is ideal if it has the following characteristics: 1. Infinite open-loop gain, A ∞. 2. Infinite input resistance, Ri ∞. 3. Zero output resistance, Ro 0. An ideal op amp is an amplifier with infinite open-loop gain, infinite input resistance, and zero output resistance. Although assuming an ideal op amp provides only an approxi- mate analysis, most modern amplifiers have such large gains and input impedances that the approximate analysis is a good one. Unless stated otherwise, we will assume from now on that every op amp is ideal. For circuit analysis, the ideal op amp is illustrated in Fig. 5.8, which is derived from the nonideal model in Fig. 5.4. Two important character- istics of the ideal op amp are: i2 = 0 i1 = 0 v1 v2 = v1 + − vo + − vd + − + − + − Figure5.8 Ideal op amp model. 1. The currents into both input terminals are zero: i1 = 0, i2 = 0 (5.5) This is due to infinite input resistance. An infinite resistance between the input terminals implies that an open circuit exists there and current cannot enter the op amp. But the output current is not necessarily zero according to Eq. (5.1). 2. The voltage across the input terminals is negligibly small; i.e., vd = v2 − v1 0 (5.6) or v1 = v2 (5.7) Thus, an ideal op amp has zero current into its two input terminals and negligibly small voltage between the two input terminals. Equations (5.5) and (5.7) are extremely important and should be regarded as the key handles to analyzing op amp circuits. The two characteristics can be exploited by noting that for voltage calculations the input port behaves as a short circuit, while for current calculations the input port behaves as an open circuit. E X A M P L E 5 . 2 Rework Practice Prob. 5.1 using the ideal op amp model. Solution: We may replace the op amp in Fig. 5.7 by its equivalent model in Fig. 5.9 as we did in Example 5.1. But we do not really need to do this. We
  • 176. CHAPTER 5 Operational Amplifiers 171 just need to keep Eqs. (5.5) and (5.7) in mind as we analyze the circuit in Fig. 5.7. Thus, the Fig. 5.7 circuit is presented as in Fig. 5.9. Notice that v2 = vs (5.2.1) Since i1 = 0, the 40-k and 5-k resistors are in series because the same current flows through them. v1 is the voltage across the 5-k resistor. Hence, using the voltage division principle, v1 = 5 5 + 40 vo = vo 9 (5.2.2) According to Eq. (5.7), v2 = v1 (5.2.3) Substituting Eqs. (5.2.1) and (5.2.2) into Eq. (5.2.3) yields the closed-loop gain, vs = vo 9 ⇒ vo vs = 9 (5.2.4) which is very close to the value of 8.99955796 obtained with the nonideal model in Practice Prob. 5.1. This shows that negligibly small error results from assuming ideal op amp characteristics. 40 kΩ 20 kΩ 5 kΩ vs i1 = 0 i2 = 0 i0 + − v1 v2 O + − + − vo Figure5.9 For Example 5.2. At node O, io = vo 40 + 5 + vo 20 mA (5.2.5) From Eq. (5.2.4), when vs = 1 V, vo = 9 V. Substituting for vo = 9 V in Eq. (5.2.5) produces io = 0.2 + 0.45 = 0.65 mA This, again, is close to the value of 0.649 mA obtained in Practice Prob. 5.1 with the nonideal model. P R A C T I C E P R O B L E M 5 . 2 Repeat Example 5.1 using the ideal op amp model. Answer: −2 , 0.2 mA. Throughoutthisbook,weassumethatanopamp operates in the linear range. Keep in mind the voltage constraint on the op amp in this mode. A key feature of the inverting amplifier is that boththeinputsignalandthefeedbackareapplied at the inverting terminal of the op amp. 5.4 INVERTING AMPLIFIER Inthisandthefollowingsections, weconsidersomeusefulopampcircuits that often serve as modules for designing more complex circuits. The first ofsuchopampcircuitsistheinvertingamplifiershowninFig.5.10. Inthis circuit, the noninverting input is grounded, vi is connected to the inverting input through R1, and the feedback resistor Rf is connected between the inverting input and output. Our goal is to obtain the relationship between the input voltage vi and the output voltage vo. Applying KCL at node 1, i1 = i2 ⇒ vi − v1 R1 = v1 − vo Rf (5.8)
  • 177. 172 PART 1 DC Circuits But v1 = v2 = 0 for an ideal op amp, since the noninverting terminal is grounded. Hence, vi R1 = − vo Rf or vo = − Rf R1 vi (5.9) The voltage gain is Av = vo/vi = −Rf /R1. The designation of the circuit in Fig. 5.10 as an inverter arises from the negative sign. Thus, An inverting amplifier reverses the polarity of the input signal while amplifying it. Notice that the gain is the feedback resistance divided by the input resistance which means that the gain depends only on the external ele- ments connected to the op amp. In view of Eq. (5.9), an equivalent circuit for the inverting amplifier is shown in Fig. 5.11. The inverting amplifier is used, for example, in a current-to-voltage converter. Note there are two types of gains: the one here is the closed-loop voltage gain Av, while the op amp itself has an open-loop voltage gain A. R1 Rf vi vo + − + − v2 v1 0 A 0 V + − + − 1 i1 i2 Figure5.10 The inverting amplifier. – + + − vi + − vo R1 Rf R1 vi Figure5.11 An equivalent circuit for the inverter in Fig. 5.10. E X A M P L E 5 . 3 Refer to the op amp in Fig. 5.12. If vi = 0.5 V, calculate: (a) the output voltage vo, and (b) the current in the 10 k resistor. 10 kΩ 25 kΩ vi vo + − + − + − Figure5.12 For Example 5.3. Solution: (a) Using Eq. (5.9), vo vi = − Rf R1 = − 25 10 = −2.5 vo = −2.5vi = −2.5(0.5) = −1.25 V (b) The current through the 10-k resistor is i = vi − 0 R1 = 0.5 − 0 10 × 103 = 50 µA
  • 178. CHAPTER 5 Operational Amplifiers 173 P R A C T I C E P R O B L E M 5 . 3 Find the output of the op amp circuit shown in Fig. 5.13. Calculate the current through the feedback resistor. 5 kΩ 15 kΩ 40 mV vo + − + − + − Figure5.13 For Practice Prob. 5.3. Answer: −120 mV, 8 µA. E X A M P L E 5 . 4 Determine vo in the op amp circuit shown in Fig. 5.14. 20 kΩ 40 kΩ 6 V vo + − 2 V + − + − a b + − Figure5.14 For Example 5.4. Solution: Applying KCL at node a, va − vo 40 = 6 − va 20 va − vo = 12 − 2va ⇒ vo = 3va − 12 But va = vb = 2 V for an ideal op amp, because of the zero voltage drop across the input terminals of the op amp. Hence, vo = 6 − 12 = −6 V Notice that if vb = 0 = va, then vo = −12, as expected from Eq. (5.9). P R A C T I C E P R O B L E M 5 . 4 Two kinds of current-to-voltage converters (also known as transresistance amplifiers) are shown in Fig. 5.15. (a) Show that for the converter in Fig. 5.15(a), vo is = −R (b) Show that for the converter in Fig. 5.15(b), vo is = −R1 1 + R3 R1 + R3 R2 Answer: Proof. R is vo + − (a) + − R1 is R2 vo + − (b) R3 + − Figure5.15 For Practice Prob. 5.4.
  • 179. 174 PART 1 DC Circuits R1 Rf vo + − v1 v2 vi + − i2 i1 + − Figure5.16 The noninverting amplifier. 5.5 NONINVERTING AMPLIFIER Another important application of the op amp is the noninverting amplifier shown in Fig. 5.16. In this case, the input voltage vi is applied directly at the noninverting input terminal, and resistor R1 is connected between the ground and the inverting terminal. We are interested in the output voltage and the voltage gain. Application of KCL at the inverting terminal gives i1 = i2 ⇒ 0 − v1 R1 = v1 − vo Rf (5.10) But v1 = v2 = vi. Equation (5.10) becomes −vi R1 = vi − vo Rf or vo = 1 + Rf R1 vi (5.11) The voltage gain is Av = vo/vi = 1 + Rf /R1, which does not have a negative sign. Thus, the output has the same polarity as the input. A noninverting amplifier is an op amp circuit designed to provide a positive voltage gain. Again we notice that the gain depends only on the external resistors. Notice that if feedback resistor Rf = 0 (short circuit) or R1 = ∞ (open circuit) or both, the gain becomes 1. Under these conditions (Rf = 0 and R1 = ∞), the circuit in Fig. 5.16 becomes that shown in Fig. 5.17, which is called a voltage follower (or unity gain amplifier) because the output follows the input. Thus, for a voltage follower vo = vi (5.12) Such a circuit has a very high input impedance and is therefore useful as an intermediate-stage (or buffer) amplifier to isolate one circuit from another, as portrayed in Fig. 5.18. The voltage follower minimizes interaction between the two stages and eliminates interstage loading. vo = vi + − vi + − + − Figure5.17 The voltage follower. + − vi + − vo + − First stage Second stage Figure5.18 A voltage follower used to isolate two cascaded stages of a circuit.
  • 180. CHAPTER 5 Operational Amplifiers 175 E X A M P L E 5 . 5 For the op amp circuit in Fig. 5.19, calculate the output voltage vo. 4 kΩ 10 kΩ 6 V vo + − + − 4 V + − a b + − Figure5.19 For Example 5.9. Solution: We may solve this in two ways: using superposition and using nodal analysis. METHOD 1 Using superposition, we let vo = vo1 + vo2 where vo1 is due to the 6-V voltage source, and vo2 is due to the 4-V input. To get vo1, we set the 4-V source equal to zero. Under this condition, the circuit becomes an inverter. Hence Eq. (5.9) gives vo1 = − 10 4 (6) = −15 V To get vo2, we set the 6-V source equal to zero. The circuit becomes a noninverting amplifier so that Eq. (5.11) applies. vo2 = 1 + 10 4 4 = 14 V Thus, vo = vo1 + vo2 = −15 + 14 = −1 V METHOD 2 Applying KCL at node a, 6 − va 4 = va − vo 10 But va = vb = 4, and so 6 − 4 4 = 4 − vo 10 ⇒ 5 = 4 − vo or vo = −1 V, as before. P R A C T I C E P R O B L E M 5 . 5 Calculate vo in the circuit in Fig. 5.20. 5 kΩ 4 kΩ 2 kΩ 3 V vo + − + − 8 kΩ + − Figure5.20 For Practice Prob. 5.5. Answer: 7 V.
  • 181. 176 PART 1 DC Circuits 5.6 SUMMING AMPLIFIER Besides amplification, the op amp can perform addition and subtraction. The addition is performed by the summing amplifier covered in this sec- tion; the subtraction is performed by the difference amplifier covered in the next section. A summing amplifier is an op amp circuit that combines several inputs and produces an output that is the weighted sum of the inputs. i1 i2 i3 v1 v2 v3 i i + − vo 0 0 R1 Rf R2 R3 a + − Figure5.21 The summing amplifier. The summing amplifier, shown in Fig. 5.21, is a variation of the inverting amplifier. It takes advantage of the fact that the inverting con- figuration can handle many inputs at the same time. We keep in mind that the current entering each op amp input is zero. Applying KCL at node a gives i = i1 + i2 + i3 (5.13) But i1 = v1 − va R1 , i2 = v2 − va R2 i3 = v3 − va R3 , i = va − vo Rf (5.14) We note that va = 0 and substitute Eq. (5.14) into Eq. (5.13). We get vo = − Rf R1 v1 + Rf R2 v2 + Rf R3 v3 (5.15) indicating that the output voltage is a weighted sum of the inputs. For this reason, the circuit in Fig. 5.21 is called a summer. Needless to say, the summer can have more than three inputs. E X A M P L E 5 . 6 Calculate vo and io in the op amp circuit in Fig. 5.22. 10 kΩ 5 kΩ 2.5 kΩ 2 kΩ io vo a b 1 V + − + − + − 2 V + − Figure5.22 For Example 5.6.
  • 182. CHAPTER 5 Operational Amplifiers 177 Solution: This is a summer with two inputs. Using Eq. (5.15), vo = − 10 5 (2) + 10 2.5 (1) = −(4 + 4) = −8 V The current io is the sum of the currents through the 10-k and 2-k resistors. Both of these resistors have voltage vo = −8 V across them, since va = vb = 0. Hence, io = vo − 0 10 + vo − 0 2 mA = −0.8 − 0.4 = −1.2 mA P R A C T I C E P R O B L E M 5 . 6 Find vo and io in the op amp circuit shown in Fig. 5.23. vo io + − 20 kΩ 10 kΩ 6 kΩ 8 kΩ 4 kΩ + − + − + − 1.2 V 2 V 1.5 V + − Figure5.23 For Practice Prob. 5.6. Answer: −3.8 V, −1.425 mA. 5.7 DIFFERENCE AMPLIFIER Difference (or differential) amplifiers are used in various applications where there is need to amplify the difference between two input signals. They are first cousins of the instrumentation amplifier, the most useful and popular amplifier, which we will discuss in Section 5.10. A difference amplifier is a device that amplifies the difference between two inputs but rejects any signals common to the two inputs. Thedifferenceamplifierisalsoknownasthesub- tractor, for reasons to be shown later. Consider the op amp circuit shown in Fig. 5.24. Keep in mind that zero currents enter the op amp terminals. Applying KCL to node a, v1 − va R1 = va − vo R2 or vo = R2 R1 + 1 va − R2 R1 v1 (5.16)
  • 183. 178 PART 1 DC Circuits v1 v2 + − vo 0 0 + − R1 R2 R3 R4 va vb + − + − Figure5.24 Difference amplifier. Applying KCL to node b, v2 − vb R3 = vb − 0 R4 or vb = R4 R3 + R4 v2 (5.17) But va = vb. Substituting Eq. (5.17) into Eq. (5.16) yields vo = R2 R1 + 1 R4 R3 + R4 v2 − R2 R1 v1 or vo = R2 R1 (1 + R1/R2) (1 + R3/R4) v2 − R2 R1 v1 (5.18) Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the property that vo = 0 when v1 = v2. This property exists when R1 R2 = R3 R4 (5.19) Thus, when the op amp circuit is a difference amplifier, Eq. (5.18) be- comes vo = R2 R1 (v2 − v1) (5.20) If R2 = R1 and R3 = R4, the difference amplifier becomes a subtractor, with the output vo = v2 − v1 (5.21) E X A M P L E 5 . 7 Design an op amp circuit with inputs v1 and v2 such that vo = −5v1 +3v2. Solution: The circuit requires that vo = 3v2 − 5v1 (5.7.1) This circuit can be realized in two ways.
  • 184. CHAPTER 5 Operational Amplifiers 179 DESIGN 1 If we desire to use only one op amp, we can use the op amp circuit of Fig. 5.24. Comparing Eq. (5.7.1) with Eq. (5.18), R2 R1 = 5 ⇒ R2 = 5R1 (5.7.2) Also, 5 (1 + R1/R2) (1 + R3/R4) = 3 ⇒ 6 5 1 + R3/R4 = 3 5 or 2 = 1 + R3 R4 ⇒ R3 = R4 (5.7.3) If we choose R1 = 10 k and R3 = 20 k, then R2 = 50 k and R4 = 20 k. vo 5R1 R6 R1 va + − + − v1 v2 3R3 5R1 Figure5.25 For Example 5.7. DESIGN 2 If we desire to use more than one op amp, we may cascade an inverting amplifier and a two-input inverting summer, as shown in Fig. 5.25. For the summer, vo = −va − 5v1 (5.7.4) and for the inverter, va = −3v2 (5.7.5) Combining Eqs. (5.7.4) and (5.7.5) gives vo = 3v2 − 5v1 which is the desired result. In Fig. 5.25, we may select R1 = 10 k and R2 = 20 k or R1 = R2 = 10 k. P R A C T I C E P R O B L E M 5 . 7 Design a difference amplifier with gain 4. Answer: Typical: R1 = R3 = 10 k, R2 = R4 = 40 k. E X A M P L E 5 . 8 An instrumentation amplifier shown in Fig. 5.26 is an amplifier of low- level signals used in process control or measurement applications and commercially available in single-package units. Show that vo = R2 R1 1 + 2R3 R4 (v2 − v1) Solution: We recognize that the amplifier A3 in Fig. 5.26 is a difference amplifier. Thus, from Eq. (5.20), vo = R2 R1 (vo2 − vo1) (5.8.1)
  • 185. 180 PART 1 DC Circuits vo1 vo2 v1 v2 0 0 vo + − + − + − A1 A2 A3 R3 R4 R1 R1 R2 R2 R3 va vb + − + − i Figure5.26 Instrumentation amplifier; for Example 5.8. Since the op amps A1 and A2 draw no current, current i flows through the three resistors as though they were in series. Hence, vo1 − vo2 = i(R3 + R4 + R3) = i(2R3 + R4) (5.8.2) But i = va − vb R4 and va = v1, vb = v2. Therefore, i = v1 − v2 R4 (5.8.3) Inserting Eqs. (5.8.2) and (5.8.3) into Eq. (5.8.1) gives vo = R2 R1 1 + 2R3 R4 (v2 − v1) as required. We will discuss the instrumentation amplifier in detail in Section 5.10. P R A C T I C E P R O B L E M 5 . 8 Obtain io in the instrumentation amplifier circuit of Fig. 5.27. + − + − + − io 20 kΩ 20 kΩ 40 kΩ 10 kΩ 40 kΩ 8.00 V 8.01 V Figure5.27 Instrumentation amplifier; for Practice Prob. 5.8. Answer: 2 µA.
  • 186. CHAPTER 5 Operational Amplifiers 181 5.8 CASCADED OP AMP CIRCUITS As we know, op amp circuits are modules or building blocks for designing complex circuits. It is often necessary in practical applications to connect op amp circuits in cascade (i.e., head to tail) to achieve a large overall gain. In general, two circuits are cascaded when they are connected in tandem, one behind another in a single file. A cascade connection is a head-to-tail arrangement of two or more op amp circuits such that the output of one is the input of the next. When op amp circuits are cascaded, each circuit in the string is called a stage; the original input signal is increased by the gain of the individual stage. Op amp circuits have the advantage that they can be cascaded without changing their input-output relationships. This is due to the fact that each (ideal) op amp circuit has infinite input resistance and zero output resistance. Figure 5.28 displays a block diagram representa- tion of three op amp circuits in cascade. Since the output of one stage is the input to the next stage, the overall gain of the cascade connection is the product of the gains of the individual op amp circuits, or A = A1A2A3 (5.22) Although the cascade connection does not affect the op amp input-output relationships, care must be exercised in the design of an actual op amp circuit to ensure that the load due to the next stage in the cascade does not saturate the op amp. Stage 1 v2 = A1v1 + − v1 + − + − A1 Stage 2 A2 v3 = A2v2 vo = A3v3 + − Stage 3 A3 Figure5.28 A three-stage cascaded connection. E X A M P L E 5 . 9 Find vo and io in the circuit in Fig. 5.29. 10 kΩ 12 kΩ 4 kΩ 20 mV vo + − + − 3 kΩ a b io + − + − Figure5.29 For Example 5.9. Solution: This circuit consists of two noninverting amplifiers cascaded. At the output of the first op amp, va = 1 + 12 3 (20) = 100 mV At the output of the second op amp, vo = 1 + 10 4 va = (1 + 2.5)100 = 350 mV
  • 187. 182 PART 1 DC Circuits The required current io is the current through the 10-k resistor. io = vo − vb 10 mA But vb = va = 100 mV. Hence, io = (350 − 100) × 10−3 10 × 103 = 25 µA P R A C T I C E P R O B L E M 5 . 9 Determine vo and io in the op amp circuit in Fig. 5.30. 6 kΩ 4 kΩ 4 V vo + − + − io + − + − Figure5.30 For Practice Prob. 5.9. Answer: 10 V, 1 mA. E X A M P L E 5 . 1 0 If v1 = 1 V and v2 = 2 V, find vo in the op amp circuit of Fig. 5.31. + − + − + − A B C 5 kΩ 15 kΩ v1 10 kΩ 2 kΩ 4 kΩ 8 kΩ 6 kΩ v2 vo a b Figure5.31 For Example 5.10. Solution: The circuit consists of two inverters A and B and a summer C as shown in Fig. 5.31. We first find the outputs of the inverters. va = − 6 2 (v1) = −3(1) = −3 V, vb = − 8 4 (v2) = −2(2) = −4 V
  • 188. CHAPTER 5 Operational Amplifiers 183 These become the inputs to the summer so that the output is obtained as vo = − 10 5 va + 10 15 vb = − 2(−3) + 2 3 (−4) = 8.333 V P R A C T I C E P R O B L E M 5 . 1 0 If v1 = 2 V and v2 = 1.5 V, find vo in the op amp circuit of Fig. 5.32. + − + − + − + − + − 10 kΩ v1 v2 vo 50 kΩ 20 kΩ 30 kΩ 60 kΩ Figure5.32 Practice Prob. 5.10. Answer: 9 V. 5.9 OP AMP CIRCUIT ANALYSIS WITH PSPICE PSpice for Windows does not have a model for an ideal op amp, although one may create one as a subcircuit using the Create Subcircuit line in the Tools menu. Rather than creating an ideal op amp, we will use one of the four nonideal, commercially available op amps supplied in the PSpice library eval.slb. The op amp models have the part names LF411, LM111, LM324, and uA471, as shown in Fig. 5.33. Each of them can be obtained from Draw/Get New Part/libraries · · ·/eval.lib or by simply selecting Draw/Get New Part and typing the part name in the PartName dialog box, as usual. Note that each of them requires dc supplies, without which the op amp will not work. The dc supplies should be connected as shown in Fig. 5.3. + − LM324 2 3 1 4 U1A 11 V+ V− + − LM111 3 2 7 V+ V− U2 8 5 6 1 4 G BB ⁄S (c) Five–connection op amp subcircuit (b) Op amp subcircuit + − uA741 2 3 U3 4 V+ V− + − LF411 2 3 6 7 5 1 U4 4 V+ V− 7 5 1 6 052 051 B2 B1 (d) Five–connection op amp subcircuit (a) JFET–input op amp subcircuit Figure5.33 Nonideal op amp model available in PSpice.
  • 189. 184 PART 1 DC Circuits E X A M P L E 5 . 1 1 Use PSpice to solve the op amp circuit for Example 5.1. Solution: Using Schematics, we draw the circuit in Fig. 5.6(a) as shown in Fig. 5.34. Notice that the positive terminal of the voltage source vs is con- nected to the inverting terminal (pin 2) via the 10-k resistor, while the noninverting terminal (pin 3) is grounded as required in Fig. 5.6(a). Also, notice how the op amp is powered; the positive power supply terminal V+ (pin 7) is connected to a 15-V dc voltage source, while the negative power supply terminal V− (pin 4) is connected to −15 V. Pins 1 and 5 are left floating because they are used for offset null adjustment, which does not concern us in this chapter. Besides adding the dc power supplies to the original circuit in Fig. 5.6(a), we have also added pseudocomponents VIEWPOINT and IPROBE to respectively measure the output voltage vo at pin 6 and the required current i through the 20-k resistor. + − uA741 2 3 U1 4 V+ V− 7 5 1 6 052 051 + − + − + − 20 K R2 1.999E–04 V3 –15 V 0 15 V V2 10 K R1 VS 2 V 0 –3.9983 Figure5.34 Schematic for Example 5.11. After saving the schematic, we simulate the circuit by selecting Analysis/Simulate and have the results displayed on VIEWPOINT and IPROBE. From the results, the closed-loop gain is vo vs = −3.9983 2 = −1.99915 and i = 0.1999 mA, in agreement with the results obtained analytically in Example 5.1. P R A C T I C E P R O B L E M 5 . 1 1 Rework Practice Prob. 5.1 using PSpice. Answer: 9.0027, 0.6502 mA.
  • 190. CHAPTER 5 Operational Amplifiers 185 †5.10 APPLICATIONS The op amp is a fundamental building block in modern electronic instru- mentation. It is used extensively in many devices, along with resistors and other passive elements. Its numerous practical applications include instrumentation amplifiers, digital-to-analog converters, analog comput- ers, level shifters, filters, calibration circuits, inverters, summers, inte- grators, differentiators, subtractors, logarithmic amplifiers, comparators, gyrators, oscillators, rectifiers, regulators, voltage-to-current converters, current-to-voltage converters, and clippers. Some of these we have al- ready considered. We will consider two more applications here: the digital-to-analog converter and the instrumentation amplifier. 5.10.1 Digital-to-Analog Converter The digital-to-analog converter (DAC) transforms digital signals into ana- logform. Atypicalexampleofafour-bitDACisillustratedinFig.5.35(a). The four-bit DAC can be realized in many ways. A simple realization is the binary weighted ladder, shown in Fig. 5.35(b). The bits are weights according to the magnitude of their place value, by descending value of Rf /Rn so that each lesser bit has half the weight of the next higher. This is obviously an inverting summing amplifier. The output is related to the inputs as shown in Eq. (5.15). Thus, −Vo = Rf R1 V1 + Rf R2 V2 + Rf R3 V3 + Rf R4 V4 (5.23) Input V1 is called the most significant bit (MSB), while input V4 is the least significant bit (LSB). Each of the four binary inputs V1, . . . , V4 can assume only two voltage levels: 0 or 1 V. By using the proper input and feedback resistor values, the DAC provides a single output that is proportional to the inputs. Analog output Digital input (0000–1111) Four-bit DAC (a) + − V1 V2 V3 V4 R1 R2 R3 R4 Rf V o MSB LSB (b) Figure 5.35 Four-bit DAC: (a) block diagram, (b) binary weighted ladder type. In practice, the voltage levels may be typically 0 and ± 5 V. E X A M P L E 5 . 1 2 In the op amp circuit of Fig. 5.35(b), let Rf = 10 k, R1 = 10 k, R2 = 20 k, R3 = 40 k, and R4 = 80 k. Obtain the analog output for binary inputs [0000], [0001], [0010], . . . , [1111]. Solution: Substituting the given values of the input and feedback resistors in Eq. (5.23) gives −Vo = Rf R1 V1 + Rf R2 V2 + Rf R3 V3 + Rf R4 V4 = V1 + 0.5V2 + 0.25V3 + 0.125V4 Using this equation, a digital input [V1V2V3V4] = [0000] produces an analog output of −Vo = 0 V; [V1V2V3V4] = [0001] gives −Vo = 0.125 V. Similarly,
  • 191. 186 PART 1 DC Circuits [V1V2V3V4] = [0010] ⇒ −Vo = 0.25 V [V1V2V3V4] = [0011] ⇒ −Vo = 0.25 + 0.125 = 0.375 V [V1V2V3V4] = [0100] ⇒ −Vo = 0.5 V . . . [V1V2V3V4] = [1111] ⇒ −Vo = 1 + 0.5 + 0.25 + 0.125 = 1.875 V Table 5.2 summarizes the result of the digital-to-analog conversion. Note that we have assumed that each bit has a value of 0.125 V. Thus, in this system, we cannot represent a voltage between 1.000 and 1.125, for example. This lack of resolution is a major limitation of digital-to- analog conversions. For greater accuracy, a word representation with a greater number of bits is required. Even then a digital representation of an analog voltage is never exact. In spite of this inexact representation, digital representation has been used to accomplish remarkable things such as audio CDs and digital photography. TABLE 5.2 Input and output values of the four-bit DAC. Binary input Output [V1V2V3V4] Decimal value −Vo 0000 0 0 0001 1 0.125 0010 2 0.25 0011 3 0.375 0100 4 0.5 0101 5 0.625 0110 6 0.75 0111 7 0.875 1000 8 1.0 1001 9 0.125 1010 10 0.25 1011 11 1.375 1011 12 1.5 1100 13 1.625 1101 14 1.75 1111 15 1.875 P R A C T I C E P R O B L E M 5 . 1 2 A three-bit DAC is shown in Fig. 5.36. (a) Determine |Vo| for [V1V2V3] = [010]. (b) Find |Vo| if [V1V2V3] = [110].
  • 192. CHAPTER 5 Operational Amplifiers 187 (c) If |Vo| = 1.25 V is desired, what should be [V1V2V3] ? (d) To get |Vo| = 1.75 V, what should be [V1V2V3] ? + − 10 kΩ 20 kΩ 40 kΩ 10 kΩ v1 v2 v3 vo Figure 5.36 Three-bit DAC; for Practice Prob. 5.12. Answer: 0.5 V, 1.5 V, [101], [111]. 5.10.2 Instrumentation Amplifiers One of the most useful and versatile op amp circuits for precision mea- surement and process control is the instrumentation amplifier (IA), so called because of its widespread use in measurement systems. Typical applications of IAs include isolation amplifiers, thermocouple amplifiers, and data acquisition systems. The instrumentation amplifier is an extension of the difference am- plifier in that it amplifies the difference between its input signals. As shown in Fig. 5.26 (see Example 5.8), an instrumentation amplifier typ- ically consists of three op amps and seven resistors. For convenience, the amplifier is shown again in Fig. 5.37(a), where the resistors are made equal except for the external gain-setting resistor RG, connected between the gain set terminals. Figure 5.37(b) shows its schematic symbol. Ex- ample 5.8 showed that vo = Av(v2 − v1) (5.24) where the voltage gain is Av = 1 + 2R RG (5.25) As shown in Fig. 5.38, the instrumentation amplifier amplifies small dif- ferential signal voltages superimposed on larger common-mode voltages. Since the common-mode voltages are equal, they cancel each other. The IA has three major characteristics: + − + − + − 1 2 3 R R R R R R RG v1 v2 vo Inverted input Gain set Gain set Noninverting input Output (a) (b) + − Figure5.37 (a) The instrumentation amplifier with an external resistance to adjust the gain, (b) schematic diagram.
  • 193. 188 PART 1 DC Circuits + − RG Small differential signals riding on larger common–mode signals Instrumentation amplifier Amplified differential signal, No common-mode signal Figure 5.38 The IA rejects common voltages but amplifies small signal voltages. (Source: T. L. Floyd, Electronic Devices, 2nd ed., Englewood Cliffs, NJ: Prentice Hall, 1996, p. 795.) 1. The voltage gain is adjusted by one external resistor RG. 2. The input impedance of both inputs is very high and does not vary as the gain is adjusted. 3. The output vo depends on the difference between the inputs v1 and v2, not on the voltage common to them (common-mode voltage). Due to the widespread use of IAs, manufacturers have developed these amplifiers on single-package units. A typical example is the LH0036, developed by National Semiconductor. The gain can be var- ied from 1 to 1,000 by an external resistor whose value may vary from 100 to 10 k. E X A M P L E 5 . 1 3 In Fig. 5.37, let R = 10 k, v1 = 2.011 V, and v2 = 2.017 V. If RG is ad- justed to 500 , determine: (a) the voltage gain, (b) the output voltage vo. Solution: (a) The voltage gain is Av = 1 + 2R RG = 1 + 2 × 10,000 500 = 41 (b) The output voltage is vo = Av(v2 − v1) = 41(2.017 − 2.011) = 41(6) mV = 246 mV P R A C T I C E P R O B L E M 5 . 1 3 Determine the value of the external gain-setting resistor RG required for the IA in Fig. 5.37 to produce a gain of 142 when R = 25 k. Answer: 354.6 . 5.11 SUMMARY 1. The op amp is a high-gain amplifier that has high input resistance and low output resistance.
  • 194. CHAPTER 5 Operational Amplifiers 189 2. Table 5.3 summarizes the op amp circuits considered in this chapter. The expression for the gain of each amplifier circuit holds whether the inputs are dc, ac, or time-varying in general. TABLE 5.3 Summary of basic op amp circuits. Op amp circuit Name/output-input relationship + − R2 R1 vi vo Inverting amplifier vo = − R2 R1 vi vo R1 + − vi R2 Noninverting amplifier vo = 1 + R2 R1 vi + − vo vi Voltage follower vo = vi v1 v2 v3 vo R1 R2 R3 Rf + − Summer vo = − Rf R1 v1 + Rf R2 v2 + Rf R3 v3 + − R2 R1 v1 R1 R2 v2 vo Difference amplifier vo = R2 R1 (v2 − v1) 3. An ideal op amp has an infinite input resistance, a zero output resistance, and an infinite gain. 4. For an ideal op amp, the current into each of its two input terminals is zero, and the voltage across its input terminals is negligibly small. 5. In an inverting amplifier, the output voltage is a negative multiple of the input. 6. In a noninverting amplifier, the output is a positive multiple of the input. 7. In a voltage follower, the output follows the input. 8. In a summing amplifier, the output is the weighted sum of the inputs.
  • 195. 190 PART 1 DC Circuits 9. In a difference amplifier, the output is proportional to the difference of the two inputs. 10. Op amp circuits may be cascaded without changing their input-output relationships. 11. PSpice can be used to analyze an op amp circuit. 12. Typical applications of the op amp considered in this chapter include the digital-to-analog converter and the instrumentation amplifier. REVIEW QUESTIONS 5.1 The two input terminals of an op amp are labeled as: (a) high and low. (b) positive and negative. (c) inverting and noninverting. (d) differential and nondifferential. 5.2 For an ideal op amp, which of the following statements are not true? (a) The differential voltage across the input terminals is zero. (b) The current into the input terminals is zero. (c) The current from the output terminal is zero. (d) The input resistance is zero. (e) The output resistance is zero. 5.3 For the circuit in Fig. 5.39, voltage vo is: (a) −6 V (b) −5 V (c) −1.2 V (d) −0.2 V + − + − 2 kΩ ix vo 1 V 10 kΩ 3 kΩ + − Figure 5.39 For Reivew Questions 5.3 and 5.4. 5.4 For the circuit in Fig. 5.39, current ix is: (a) 0.6 A (b) 0.5 A (c) 0.2 A (d) 1/12 A 5.5 If vs = 0 in the circuit of Fig. 5.40, current io is: (a) −10 A (b) −2.5 A (c) 10/12 A (d) 10/14 A + − + − + − 4 kΩ io vo 10 V 8 kΩ 2 kΩ + − vs a Figure 5.40 For Review Questions 5.5 to 5.7. 5.6 If vs = 8 V in the circuit of Fig. 5.40, the output voltage is: (a) −44 V (b) −8 V (c) 4 V (d) 7 V 5.7 Refer to Fig. 5.40. If vs = 8 V, voltage va is: (a) −8 V (b) 0 V (c) 10/3 V (d) 8 V 5.8 The power absorbed by the 4-k resistor in Fig. 5.41 is: (a) 9 mW (b) 4 mW (c) 2 mW (d) 1 mW + − 6 V 2 kΩ vo + − 4 kΩ + − Figure 5.41 For Review Question 5.8. 5.9 Which of these amplifiers is used in a digital-to-analog converter? (a) noninverter (b) voltage follower (c) summer (d) difference amplifier
  • 196. CHAPTER 5 Operational Amplifiers 191 5.10 Difference amplifiers are used in: (a) instrumentation amplifiers (b) voltage followers (c) voltage regulators (d) buffers (e) summing amplifiers (f) subtracting amplifiers Answers: 5.1c, 5.2c,d, 5.3b, 5.4b, 5.5a, 5.6c, 5.7d, 5.8b, 5.9c, 5.10a,f. PROBLEMS Section 5.2 Operational Amplifiers 5.1 The equivalent model of a certain op amp is shown in Fig. 5.42. Determine: (a) the input resistance (b) the output resistance (c) the voltage gain in dB. 60 Ω + − vd + − 1.5 MΩ 8 × 10vd Figure 5.42 For Prob. 5.1. 5.2 The open-loop gain of an op amp is 100,000. Calculate the output voltage when there are inputs of +10 µV on the inverting terminal and + 20 µV on the noninverting terminal. 5.3 Determine the output voltage when −20 µV is applied to the inverting terminal of an op amp and +30 µV to its noninverting terminal. Assume that the op amp has an open-loop gain of 200,000. 5.4 The output voltage of an op amp is −4 V when the noninverting input is 1 mV. If the open-loop gain of the op amp is 2 × 106 , what is the inverting input? 5.5 For the op amp circuit of Fig. 5.43, the op amp has an open-loop gain of 100,000, an input resistance of 10 k, and an output resistance of 100 . Find the voltage gain vo/vi using the nonideal model of the op amp. + − + − vo vi + − Figure 5.43 For Prob. 5.5. 5.6 Using the same parameters for the 741 op amp in Example 5.1, find vo in the op amp circuit of Fig. 5.44. + − + − vo 741 1 mV Figure 5.44 For Prob. 5.6. 5.7 The op amp in Fig. 5.45 has Ri = 100 k, Ro = 100 , A = 100,000. Find the differential voltage vd and the output voltage vo. + − + − 10 kΩ 100 kΩ vo vd + − 1 mV + − Figure 5.45 For Prob. 5.7. Section 5.3 Ideal Op Amp 5.8 Obtain vo for each of the op amp circuits in Fig. 5.46. 2 kΩ (a) vo + − 1 mA 2 V 10 kΩ (b) vo + − 1 V 2 kΩ + − + − + − + − Figure 5.46 For Prob. 5.8. 5.9 Determine vo for each of the op amp circuits in Fig. 5.47.
  • 197. 192 PART 1 DC Circuits + − + − 4 V 2 kΩ vo 1 mA + − 1 V 2 kΩ vo + − + − 3 V + − + − Figure 5.47 For Prob. 5.9. 5.10 Find the gain vo/vs of the circuit in Fig. 5.48. 10 kΩ 10 kΩ vo + − + − 20 kΩ + − vs Figure 5.48 For Prob. 5.10. 5.11 Find vo and io in the circuit in Fig. 5.49. + − 5 kΩ 2 kΩ + − io + − 8 kΩ 10 kΩ 4 kΩ 3 V vo Figure 5.49 For Prob. 5.11. 5.12 Refer to the op amp circuit in Fig. 5.50. Determine the power supplied by the voltage source. 1.2 V 2 kΩ 4 kΩ + − vo 1 kΩ 4 kΩ + − Figure 5.50 For Prob. 5.12. 5.13 Find vo and io in the circuit of Fig. 5.51. 50 kΩ vo + − + − 1 V 100 kΩ 90 kΩ 10 kΩ io 10 kΩ + − Figure 5.51 For Prob. 5.13. 5.14 Determine the output voltage vo in the circuit of Fig. 5.52. 5 kΩ vo + − 2 mA 20 kΩ 10 kΩ 10 kΩ + − Figure 5.52 For Prob. 5.14. Section 5.4 Inverting Amplifier 5.15 (a) For the circuit shown in Fig. 5.53, show that the gain is vo vi = − 1 R R1 + R2 + R1R2 R3 (b) Evaluate the gain when R = 10 k, R1 = 100 k, R2 = 50 k, R3 = 25 k.
  • 198. CHAPTER 5 Operational Amplifiers 193 vo vi R R1 R2 R3 + − Figure 5.53 For Prob. 5.15. 5.16 Calculate the gain vo/vi when the switch in Fig. 5.54 is in: (a) position 1 (b) position 2 (c) position 3 10 kΩ 1 vo + − 5 kΩ + − vi 2 MΩ 80 kΩ 12 kΩ 2 3 + − Figure 5.54 For Prob. 5.16. 5.17 Calculate the gain vo/vi of the op amp circuit in Fig. 5.55. 20 kΩ vo + − + − vi 10 kΩ 50 kΩ 1 MΩ + − Figure 5.55 For Prob. 5.17. 5.18 Determine io in the circuit of Fig. 5.56. 1 V 5 kΩ 4 kΩ io + − 4 kΩ 10 kΩ 2 kΩ + − Figure 5.56 For Prob. 5.18. 5.19 In the circuit in Fig. 5.57, calculate vo if vs = 0. + − + − 9 V 4 kΩ 4 kΩ 2 kΩ 8 kΩ vo + − vs + − Figure 5.57 For Prob. 5.19. 5.20 Repeat the previous problem if vs = 3 V. 5.21 Design an inverting amplifier with a gain of −15. Section 5.5 Noninverting Amplifier 5.22 Find va and vo in the op amp circuit of Fig. 5.58. 2 V vo + − 3 V va + − Figure 5.58 For Prob. 5.22. 5.23 Refer to Fig. 5.59. (a) Determine the overall gain vo/vi of the circuit. (b) What value of vi will result in vo = 15 cos 120πt? + − 2 kΩ + − 1 MΩ vo 8 kΩ 20 kΩ vi + − Figure 5.59 For Prob. 5.23.
  • 199. 194 PART 1 DC Circuits 5.24 Find io in the op amp circuit of Fig. 5.60. + − 0.4 V 20 kΩ 10 kΩ io 50 kΩ + − Figure 5.60 For Prob. 5.24. 5.25 In the circuit shown in Fig. 5.61, find ix and the power absorbed by the 20- resistor. + − 1.2 V 30 kΩ 20 kΩ ix 60 kΩ + − Figure 5.61 For Prob. 5.25. 5.26 For the circuit in Fig. 5.62, find ix. + − + − 6 kΩ 6 kΩ 3 kΩ 4 mA vo 12 kΩ ix Figure 5.62 For Prob. 5.26. 5.27 Calculate ix and vo in the circuit of Fig. 5.63. Find the power dissipated by the 60-k resistor. + − vo + − 30 kΩ 60 kΩ ix 4 mV 20 kΩ 50 kΩ 10 kΩ + − Figure 5.63 For Prob. 5.27. 5.28 Refer to the op amp circuit in Fig. 5.64. Calculate ix and the power dissipated by the 3-k resistor. + − 4 kΩ 2 kΩ ix 1 mA 3 kΩ 1 kΩ Figure 5.64 For Prob. 5.28. 5.29 Design a noninverting amplifier with a gain of 10. Section 5.6 Summing Amplifier 5.30 Determine the output of the summing amplifier in Fig. 5.65. 30 kΩ 10 kΩ 1 V 20 kΩ 2 V 30 kΩ 3 V + − + − + − vo + − + − Figure 5.65 For Prob. 5.30. 5.31 Calculate the output voltage due to the summing amplifier shown in Fig. 5.66. 50 kΩ 25 kΩ 10 mV 20 kΩ 20 mV 50 kΩ 100 mV + − + − 10 kΩ 50 mV + − + − vo + − + − Figure 5.66 For Prob. 5.31. 5.32 An averaging amplifier is a summer that provides an output equal to the average of the inputs. By using
  • 200. CHAPTER 5 Operational Amplifiers 195 proper input and feedback resistor values, one can get −vout = 1 4 (v1 + v2 + v3 + v4) Using a feedback resistor of 10 k, design an averaging amplifier with four inputs. 5.33 A four-input summing amplifier has R1 = R2 = R3 = R4 = 12 k. What value of feedback resistor is needed to make it an averaging amplifier? 5.34 Show that the output voltage vo of the circuit in Fig. 5.67 is vo = (R3 + R4) R3(R1 + R2) (R2v1 + R1v2) R4 R3 R1 R2 vo v1 v2 + − Figure 5.67 For Prob. 5.34. 5.35 Design an op amp circuit to perform the following operation: vo = 3v1 − 2v2 All resistances must be ≤ 100 k. 5.36 Using only two op amps, design a circuit to solve −vout = v1 − v2 3 + v3 2 Section 5.7 Difference Amplifier 5.37 Find vo and io in the differential amplifier of Fig. 5.68. 10 V io 2 kΩ 1 kΩ 4 kΩ 3 kΩ 5 kΩ vo + − + − 8 V + − + − Figure 5.68 For Prob. 5.37. 5.38 The circuit in Fig. 5.69 is a differential amplifier driven by a bridge. Find vo. 20 kΩ 80 kΩ 20 kΩ 80 kΩ vo + 5 mV 40 kΩ 10 kΩ 60 kΩ 30 kΩ + − Figure 5.69 For Prob. 5.38. 5.39 Design a difference amplifier to have a gain of 2 and a common mode input resistance of 10 k at each input. 5.40 Design a circuit to amplify the difference between two inputs by 2. (a) Use only one op amp. (b) Use two op amps. 5.41 Using two op amps, design a subtractor. 5.42 ∗ The ordinary difference amplifier for fixed-gain operation is shown in Fig. 5.70(a). It is simple and reliable unless gain is made variable. One way of providing gain adjustment without losing simplicity and accuracy is to use the circuit in Fig. 5.70(b). Another way is to use the circuit in Fig. 5.70(c). Show that: (a) for the circuit in Fig. 5.70(a), vo vi = R2 R1 (b) for the circuit in Fig. 5.70(b), vo vi = R2 R1 1 1 + R1 2RG (c) for the circuit in Fig. 5.70(c), vo vi = R2 R1 1 + R2 2RG ∗An asterisk indicates a challenging problem.
  • 201. 196 PART 1 DC Circuits R1 R2 R2 R2 R1 (a) vo R2 RG (b) R1 2 R1 2 R2 2 R2 2 R2 2 R2 2 R1 2 R1 2 vi (c) R1 R1 RG + − + − + − + − vi + − vi + − + − vo + − vo + − Figure 5.70 For Prob. 5.42. Section 5.8 Cascaded Op Amp Circuits 5.43 The individual gains of the stages in a multistage amplifier are shown in Fig. 5.71. (a) Calculate the overall voltage gain vo/vi. (b) Find the voltage gain that would be needed in a fourth stage which would make the overall gain to be 60 dB when added. vo vi –20 –12.5 +0.8 Figure 5.71 For Prob. 5.43. 5.44 In a certain electronic device, a three-stage amplifier is desired, whose overall voltage gain is 42 dB. The individual voltage gains of the first two stages are to be equal, while the gain of the third is to be one-fourth of each of the first two. Calculate the voltage gain of each. 5.45 Refer to the circuit in Fig. 5.72. Calculate io if: (a) vs = 12 mV (b) vs = 10 cos 377t mV. vs + − 2 kΩ io 12 kΩ 6 kΩ 12 kΩ 4 kΩ + − + − Figure 5.72 For Prob. 5.45. 5.46 Calculate io in the op amp circuit of Fig. 5.73. 0.6 V + − 4 kΩ io 1 kΩ 10 kΩ 5 kΩ 2 kΩ + − + − 3 kΩ Figure 5.73 For Prob. 5.46. 5.47 Find the voltage gain vo/vs of the circuit in Fig. 5.74. vs + − 10 kΩ 5 kΩ 20 kΩ vo + − + − + − Figure 5.74 For Prob. 5.47. 5.48 Calculate the current gain io/is of the op amp circuit in Fig. 5.75.
  • 202. CHAPTER 5 Operational Amplifiers 197 is 10 kΩ 3 kΩ 4 kΩ + − + − 5 kΩ 3 kΩ 2 kΩ io Figure 5.75 For Prob. 5.48. 5.49 Find vo in terms of v1 and v2 in the circuit in Fig. 5.76. R3 R2 R1 R4 + − + − v1 v2 vo R5 Figure 5.76 For Prob. 5.49. 5.50 Obtain the closed-loop voltage gain vo/vi of the circuit in Fig. 5.77. Rf R2 R1 + − + − vi R3 vo R4 + − + − Figure 5.77 For Prob. 5.50. 5.51 Determine the gain vo/vi of the circuit in Fig. 5.78. + − + + − − R3 R2 R1 R4 R5 R6 vo vi + − Figure 5.78 For Prob. 5.51. 5.52 For the circuit in Fig. 5.79, find vo. + − + − + − + − + − 25 kΩ 10 kΩ 40 kΩ 100 kΩ 20 kΩ 6 V 4 V 2 V 20 kΩ vo + − Figure 5.79 For Prob. 5.52. 5.53 Obtain the output vo in the circuit of Fig. 5.80. + − + − + − 80 kΩ 80 kΩ 20 kΩ 0.4 V 40 kΩ 20 kΩ vo + − + − 0.2 V + − Figure 5.80 For Prob. 5.53. 5.54 Find vo in the circuit in Fig. 5.81, assuming that Rf = ∞ (open circuit).
  • 203. 198 PART 1 DC Circuits 10 mV + − 15 kΩ 6 kΩ 5 kΩ Rf + − + − 1 kΩ 2 kΩ + − vo Figure 5.81 For Probs. 5.54 and 5.55. 5.55 Repeat the previous problem if Rf = 10 k. 5.56 Determine vo in the op amp circuit of Fig. 5.82. + − 30 kΩ A C 40 kΩ 10 kΩ 1 V 20 kΩ 60 kΩ + − 10 kΩ 2 V + − 20 kΩ B 3 V + − 10 kΩ 4 V 10 kΩ vo + − + − + − 10 kΩ Figure 5.82 For Prob. 5.56. 5.57 Find the load voltage vL in the circuit of Fig. 5.83. + − + − + − 100 kΩ 250 kΩ 0.4 V 2 kΩ + − + − + − vL 20 kΩ Figure 5.83 For Prob. 5.57. 5.58 Determine the load voltage vL in the circuit of Fig. 5.84. 50 kΩ 10 kΩ 5 kΩ 1.8 V 4 kΩ vL + − + − + − + − Figure 5.84 For Prob. 5.58. 5.59 Find io in the op amp circuit of Fig. 5.85. + − + − + − 100 kΩ 32 kΩ 10 kΩ 20 kΩ 1.6 kΩ 0.6 V + − 0.4 V + − + − io Figure 5.85 For Prob. 5.59. Section 5.9 Op Amp Circuit Analysis with PSpice 5.60 Rework Example 5.11 using the nonideal op amp LM324 instead of uA741. 5.61 Solve Prob. 5.18 using PSpice and op amp uA741. 5.62 Solve Prob. 5.38 using PSpice and op amp LM324. 5.63 Use PSpice to obtain vo in the circuit of Fig. 5.86. + − 20 kΩ 30 kΩ 10 kΩ 1 V + − + − + − 40 kΩ 2 V + − + − vo + − Figure 5.86 For Prob. 5.63. 5.64 Determine vo in the op amp circuit of Fig. 5.87 using PSpice.
  • 204. CHAPTER 5 Operational Amplifiers 199 vo + − 20 kΩ 5 V 1 V 10 kΩ + − + − 20 kΩ 10 kΩ 40 kΩ + − 100 kΩ + − Figure 5.87 For Prob. 5.64. 5.65 Use PSpice to solve Prob. 5.56, assuming that the op amps are uA741. 5.66 Use PSpice to verify the results in Example 5.9. Assume nonideal op amps LM324. Section 5.10 Applications 5.67 A five-bit DAC covers a voltage range of 0 to 7.75 V. Calculate how much voltage each bit is worth. 5.68 Design a six-bit digital-to-analog converter. (a) If |Vo| = 1.1875 V is desired, what should [V1V2V3V4V5V6] be? (b) Calculate |Vo| if [V1V2V3V4V5V6] = [011011]. (c) What is the maximum value |Vo| can assume? 5.69 ∗ A four-bit R-2R ladder DAC is presented in Fig. 5.88. (a) Show that the output voltage is given by −Vo = Rf V1 2R + V2 4R + V3 8R + V4 16R (b) If Rf = 12 k and R = 10 k, find |Vo| for [V1V2V3V4] = [1011] and [V1V2V3V4] = [0101]. R R R R V o + − V1 V2 V3 V4 2R 2R 2R 2R Rf Figure 5.88 For Prob. 5.69. 5.70 If RG = 100 and R = 20 k, calculate the voltage gain of the IA in Fig. 5.37. 5.71 Assuming a gain of 200 for an IA, find its output voltage for: (a) v1 = 0.402 V and v2 = 0.386 V (b) v1 = 1.002 V and v2 = 1.011 V. 5.72 Figure 5.89 displays a two-op-amp instrumentation amplifier. Derive an expression for vo in terms of v1 and v2. How can this amplifier be used as a subtractor? v2 v1 vo R4 R3 R2 R1 + − + − Figure 5.89 For Prob. 5.72. 5.73 ∗ Figure 5.90 shows an instrumentation amplifier driven by a bridge. Obtain the gain vo/vi of the amplifier. 25 kΩ 10 kΩ 10 kΩ 500 kΩ vo 25 kΩ 2 kΩ 30 kΩ 20 kΩ vi 80 kΩ 40 kΩ 500 kΩ + − + − + − Figure 5.90 For Prob. 5.73.
  • 205. 200 PART 1 DC Circuits COMPREHENSIVE PROBLEMS 5.74 A gain of 6 (+ or −, it does not matter) is required in an audio system. Design an op amp circuit to provide the gain with an input resistance of 2 k. 5.75 The op amp circuit in Fig. 5.91 is a current amplifier. Find the current gain io/is of the amplifier. + − 20 kΩ 4 kΩ 5 kΩ 2 kΩ is io Figure 5.91 For Prob. 5.75. 5.76 A noninverting current amplifier is portrayed in Fig. 5.92. Calculate the gain io/is. Take R1 = 8 k and R2 = 1 k. + − R1 R2 R2 is io Figure 5.92 For Prob. 5.76. 5.77 Refer to the bridge amplifier shown in Fig. 5.93. Determine the voltage gain vo/vi. + − 60 kΩ vi vo RL + − + − + − 50 kΩ 20 kΩ 30 kΩ Figure 5.93 For Prob. 5.77. 5.78 ∗ A voltage-to-current converter is shown in Fig. 5.94, which means that iL = Avi if R1R2 = R3R4. Find the constant term A. + − R3 R1 iL R2 vi RL R4 + − Figure 5.94 For Prob. 5.78.
  • 206. 201 C H A P T E R CAPACITORS AND INDUCTORS 6 The important thing about a problem is not its solution, but the strength we gain in finding the solution. —Anonymous Historical Profiles Michael Faraday (1791–1867), an English chemist and physicist, was probably the greatest experimentalist who ever lived. Born near London, Faraday realized his boyhood dream by working with the great chemist Sir Humphry Davy at the Royal Institution, where he worked for 54 years. He made several contributions in all areas of physical science and coined such words as electrolysis, anode, and cathode. His discovery of electromagnetic induction in 1831 was a major breakthrough in engineering because it provided a way of generating electricity. The electric motor and generator operate on this principle. The unit of capacitance, the farad, was named in his honor. Joseph Henry (1797–1878), an American physicist, discovered inductance and con- structed an electric motor. Born in Albany, New York, Henry graduated from Albany Academy and taught philosophy at Princeton University from 1832 to 1846. He was the first secretary of the Smithsonian Institution. He conducted several experiments on electromagnetism and developed powerful electromagnets that could lift objects weighing thousands of pounds. Interestingly, Joseph Henry discovered electromagnetic induction before Faraday but failed to publish his findings. The unit of inductance, the henry, was named after him.
  • 207. 202 PART 1 DC Circuits 6.1 INTRODUCTION So far we have limited our study to resistive circuits. In this chapter, we shall introduce two new and important passive linear circuit elements: the capacitor and the inductor. Unlike resistors, which dissipate energy, capacitors and inductors do not dissipate but store energy, which can be retrieved at a later time. For this reason, capacitors and inductors are called storage elements. In contrast to a resistor, which spends or dis- sipates energy irreversibly, an inductor or ca- pacitor stores or releases energy (i.e., has a memory). The application of resistive circuits is quite limited. With the in- troduction of capacitors and inductors in this chapter, we will be able to analyze more important and practical circuits. Be assured that the circuit analysis techniques covered in Chapters 3 and 4 are equally applicable to circuits with capacitors and inductors. We begin by introducing capacitors and describing how to combine them in series or in parallel. Later, we do the same for inductors. As typical applications, we explore how capacitors are combined with op amps to form integrators, differentiators, and analog computers. 6.2 CAPACITORS A capacitor is a passive element designed to store energy in its electric field. Besides resistors, capacitors are the most common electrical com- ponents. Capacitors are used extensively in electronics, communications, computers, and power systems. For example, they are used in the tuning circuits of radio receivers and as dynamic memory elements in computer systems. A capacitor is typically constructed as depicted in Fig. 6.1. Metal plates, each with area A d Dielectric with permittivity e Figure6.1 A typical capacitor. A capacitor consists of two conducting plates separated by an insulator (or dielectric). In many practical applications, the plates may be aluminum foil while the dielectric may be air, ceramic, paper, or mica. When a voltage source v is connected to the capacitor, as in Fig. 6.2, the source deposits a positive charge q on one plate and a negative charge −q on the other. The capacitor is said to store the electric charge. The amount of charge stored, represented by q, is directly proportional to the applied voltage v so that q = Cv (6.1) where C, the constant of proportionality, is known as the capacitance of the capacitor. The unit of capacitance is the farad (F), in honor of the English physicist Michael Faraday (1791–1867). From Eq. (6.1), we may derive the following definition. Alternatively,capacitanceistheamountofcharge stored per plate for a unit voltage difference in a capacitor. − − − −q +q + + + + + + − + v Figure6.2 A capacitor with applied voltage v. Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in farads (F). Note from Eq. (6.1) that 1 farad = 1 coulomb/volt.
  • 208. CHAPTER 6 Capacitors and Inductors 203 Although the capacitance C of a capacitor is the ratio of the charge q per plate to the applied voltage v, it does not depend on q or v. It depends on the physical dimensions of the capacitor. For example, for the parallel-plate capacitor shown in Fig. 6.1, the capacitance is given by C = A d (6.2) where A is the surface area of each plate, d is the distance between the plates, and is the permittivity of the dielectric material between the plates. Although Eq. (6.2) applies to only parallel-plate capacitors, we may infer from it that, in general, three factors determine the value of the capacitance: Capacitorvoltageratingandcapacitancearetyp- ically inversely rated due to the relationships in Eqs. (6.1) and (6.2). Arcing occurs if d is small and V is high. 1. The surface area of the plates—the larger the area, the greater the capacitance. 2. The spacing between the plates—the smaller the spacing, the greater the capacitance. 3. The permittivity of the material—the higher the permittivity, the greater the capacitance. Capacitors are commercially available in different values and types. Typically, capacitors have values in the picofarad (pF) to microfarad (µF) range. They are described by the dielectric material they are made of and by whether they are of fixed or variable type. Figure 6.3 shows the circuit symbols for fixed and variable capacitors. Note that according to the passive sign convention, current is considered to flow into the positive terminal of the capacitor when the capacitor is being charged, and out of the positive terminal when the capacitor is discharging. i i C v + − C v + − Figure6.3 Circuit symbols for capacitors: (a) fixed capacitor, (b) variable capacitor. Figure6.4showscommontypesoffixed-valuecapacitors. Polyester capacitors are light in weight, stable, and their change with temperature is predictable. Instead of polyester, other dielectric materials such as mica and polystyrene may be used. Film capacitors are rolled and housed in metal or plastic films. Electrolytic capacitors produce very high capaci- tance. Figure 6.5 shows the most common types of variable capacitors. The capacitance of a trimmer (or padder) capacitor or a glass piston capac- itor is varied by turning the screw. The trimmer capacitor is often placed in parallel with another capacitor so that the equivalent capacitance can be varied slightly. The capacitance of the variable air capacitor (meshed plates) is varied by turning the shaft. Variable capacitors are used in radio (a) (b) (c) Figure 6.4 Fixed capacitors: (a) polyester capacitor, (b) ceramic capacitor, (c) electrolytic capacitor. (Courtesy of Tech America.)
  • 209. 204 PART 1 DC Circuits receivers allowing one to tune to various stations. In addition, capacitors are used to block dc, pass ac, shift phase, store energy, start motors, and suppress noise. (a) (b) Figure6.5 Variable capacitors: (a) trimmer capacitor, (b) filmtrim capacitor. (Courtesy of Johanson.) To obtain the current-voltage relationship of the capacitor, we take the derivative of both sides of Eq. (6.1). Since i = dq dt (6.3) differentiating both sides of Eq. (6.1) gives i = C dv dt (6.4) According to Eq. (6.4), for a capacitor to carry current, its voltage must vary with time. Hence, for constant voltage, i = 0 . This is the current-voltage relationship for a capacitor, assuming the pos- itive sign convention. The relationship is illustrated in Fig. 6.6 for a capacitor whose capacitance is independent of voltage. Capacitors that satisfy Eq. (6.4) are said to be linear. For a nonlinear capacitor, the plot of the current-voltage relationship is not a straight line. Although some capacitors are nonlinear, most are linear. We will assume linear capacitors in this book. Slope = C dv⁄dt 0 i Figure6.6 Current-voltage relationship of a capacitor. The voltage-current relation of the capacitor can be obtained by integrating both sides of Eq. (6.4). We get v = 1 C t −∞ i dt (6.5) or v = 1 C t t0 i dt + v(t0) (6.6) where v(t0) = q(t0)/C is the voltage across the capacitor at time t0. Equation (6.6) shows that capacitor voltage depends on the past history of the capacitor current. Hence, the capacitor has memory—a property that is often exploited. The instantaneous power delivered to the capacitor is p = vi = Cv dv dt (6.7) The energy stored in the capacitor is therefore w = t −∞ p dt = C t −∞ v dv dt dt = C t −∞ v dv = 1 2 Cv2 t t=−∞ (6.8) We note that v(−∞) = 0, because the capacitor was uncharged at t = −∞. Thus, w = 1 2 Cv2 (6.9) Using Eq. (6.1), we may rewrite Eq. (6.9) as w = q2 2C (6.10)
  • 210. CHAPTER 6 Capacitors and Inductors 205 Equation (6.9) or (6.10) represents the energy stored in the electric field that exists between the plates of the capacitor. This energy can be re- trieved, since an ideal capacitor cannot dissipate energy. In fact, the word capacitor is derived from this element’s capacity to store energy in an electric field. We should note the following important properties of a capacitor: 1. Note from Eq. (6.4) that when the voltage across a capacitor is not changing with time (i.e., dc voltage), the current through the capacitor is zero. Thus, A capacitor is an open circuit to dc. However, if a battery (dc voltage) is connected across a capacitor, the capacitor charges. 2. The voltage on the capacitor must be continuous. The voltage on a capacitor cannot change abruptly. The capacitor resists an abrupt change in the voltage across it. According to Eq. (6.4), a discontinuous change in voltage requires an infinite current, which is physically impossible. For example, the voltage across a capacitor may take the form shown in Fig. 6.7(a), whereas it is not physically possible for the capacitor voltage to take the form shown in Fig. 6.7(b) because of the abrupt change. Conversely, the current through a capacitor can change instantaneously. 3. The ideal capacitor does not dissipate energy. It takes power from the circuit when storing energy in its field and returns previously stored energy when delivering power to the circuit. 4. A real, nonideal capacitor has a parallel-model leakage resistance, as shown in Fig. 6.8. The leakage resistance may be as high as 100 M and can be neglected for most practical applications. For this reason, we will assume ideal capacitors in this book. An alternative way of looking at this is using Eq. (6.9), whichindicatesthatenergyisproportional to voltage squared. Since injecting or extracting energy can only be done over some finite time, voltage cannot change instantaneously across a capacitor. v t (a) v t (b) Figure6.7 Voltage across a capacitor: (a) allowed, (b) not allowable; an abrupt change is not possible. Leakage resistance Capacitance Figure 6.8 Circuit model of a nonideal capacitor. E X A M P L E 6 . 1 (a) Calculate the charge stored on a 3-pF capacitor with 20 V across it. (b) Find the energy stored in the capacitor. Solution: (a) Since q = Cv, q = 3 × 10−12 × 20 = 60 pC (b) The energy stored is w = 1 2 Cv2 = 1 2 × 3 × 10−12 × 400 = 600 pJ
  • 211. 206 PART 1 DC Circuits P R A C T I C E P R O B L E M 6 . 1 What is the voltage across a 3-µF capacitor if the charge on one plate is 0.12 mC? How much energy is stored? Answer: 40 V, 2.4 mJ. E X A M P L E 6 . 2 The voltage across a 5-µF capacitor is v(t) = 10 cos 6000t V Calculate the current through it. Solution: By definition, the current is i(t) = C dv dt = 5 × 10−6 d dt (10 cos 6000t) = −5 × 10−6 × 6000 × 10 sin 6000t = −0.3 sin 6000t A P R A C T I C E P R O B L E M 6 . 2 If a 10-µF capacitor is connected to a voltage source with v(t) = 50 sin 2000t V determine the current through the capacitor. Answer: cos 2000t A. E X A M P L E 6 . 3 Determine the voltage across a 2-µF capacitor if the current through it is i(t) = 6e−3000t mA Assume that the initial capacitor voltage is zero. Solution: Since v = 1 C t 0 i dt + v(0) and v(0) = 0, v = 1 2 × 10−6 t 0 6e−3000t dt·10−3 = 3 × 103 −3000 e−3000t t 0 = (1 − e−3000t ) V P R A C T I C E P R O B L E M 6 . 3 The current through a 100-µF capacitor is i(t) = 50 sin 120πt mA. Cal- culate the voltage across it at t = 1 ms and t = 5 ms. Take v(0) = 0. Answer: −93.137 V, −1.736 V.
  • 212. CHAPTER 6 Capacitors and Inductors 207 E X A M P L E 6 . 4 Determinethecurrentthrougha200-µFcapacitorwhosevoltageisshown in Fig. 6.9. v(t) 0 4 3 2 1 50 −50 t Figure6.9 For Example 6.4. Solution: The voltage waveform can be described mathematically as v(t) =        50t V 0 t 1 100 − 50t V 1 t 3 −200 + 50t V 3 t 4 0 otherwise Since i = C dv/dt and C = 200 µF, we take the derivative of v to obtain i(t) = 200 × 10−6 ×        50 0 t 1 −50 1 t 3 50 3 t 4 0 otherwise =        10 mA 0 t 1 −10 mA 1 t 3 10 mA 3 t 4 0 otherwise Thus the current waveform is as shown in Fig. 6.10. i (mA) 0 4 3 2 1 10 −10 t Figure6.10 For Example 6.4. P R A C T I C E P R O B L E M 6 . 4 An initially uncharged 1-mF capacitor has the current shown in Fig. 6.11 across it. Calculate the voltage across it at t = 2 ms and t = 5 ms. i (mA) 0 6 4 2 100 t (ms) Figure6.11 For Practice Prob. 6.4. Answer: 100 mV, 400 mV. E X A M P L E 6 . 5 Obtain the energy stored in each capacitor in Fig. 6.12(a) under dc con- ditions. Solution: Under dc conditions, we replace each capacitor with an open circuit, as shown in Fig. 6.12(b). The current through the series combination of the 2-k and 4-k resistors is obtained by current division as i = 3 3 + 2 + 4 (6 mA) = 2 mA Hence, the voltages v1 and v2 across the capacitors are v1 = 2000i = 4 V v2 = 4000i = 8 V
  • 213. 208 PART 1 DC Circuits v1 + − v2 + − 6 mA 3 kΩ 5 kΩ 4 kΩ 2 kΩ 2 mF 4 mF (a) 6 mA 3 kΩ 5 kΩ 4 kΩ 2 kΩ (b) i Figure6.12 For Example 6.5. and the energies stored in them are w1 = 1 2 C1v2 1 = 1 2 (2 × 10−3 )(4)2 = 16 mJ w2 = 1 2 C2v2 2 = 1 2 (4 × 10−3 )(8)2 = 128 mJ P R A C T I C E P R O B L E M 6 . 5 Under dc conditions, find the energy stored in the capacitors in Fig. 6.13. 10 V + − 6 kΩ 1 kΩ 20 mF 10 mF 3 kΩ Figure6.13 For Practice Prob. 6.5. Answer: 405 µJ, 90 µJ. 6.3 SERIES AND PARALLEL CAPACITORS We know from resistive circuits that series-parallel combination is a pow- erful tool for reducing circuits. This technique can be extended to series- parallel connections of capacitors, which are sometimes encountered. We desire to replace these capacitors by a single equivalent capacitor Ceq. i C1 (a) i1 C2 C3 CN iN v i (b) Ceq v + − + − i2 i3 Figure6.14 (a) Parallel-connected N capacitors, (b) equivalent circuit for the parallel capacitors. In order to obtain the equivalent capacitor Ceq of N capacitors in parallel, consider the circuit in Fig. 6.14(a). The equivalent circuit is in Fig. 6.14(b). Note that the capacitors have the same voltage v across them. Applying KCL to Fig. 6.14(a), i = i1 + i2 + i3 + · · · + iN (6.11) But ik = Ck dv/dt. Hence, i = C1 dv dt + C2 dv dt + C3 dv dt + · · · + CN dv dt = N k=1 Ck dv dt = Ceq dv dt (6.12)
  • 214. CHAPTER 6 Capacitors and Inductors 209 where Ceq = C1 + C2 + C3 + · · · + CN (6.13) The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances. We observe that capacitors in parallel combine in the same manner as resistors in series. v C1 (a) C2 C3 CN v (b) Ceq v v1 v2 v3 vN + − + − i i + − + − + − + − + − Figure6.15 (a) Series-connected N capacitors, (b) equivalent circuit for the series capacitor. We now obtain Ceq of N capacitors connected in series by compar- ing the circuit in Fig. 6.15(a) with the equivalent circuit in Fig. 6.15(b). Note that the same current i flows (and consequently the same charge) through the capacitors. Applying KVL to the loop in Fig. 6.15(a), v = v1 + v2 + v3 + · · · + vN (6.14) But vk = 1 Ck t t0 i(t) dt + vk(t0). Therefore, v = 1 C1 t t0 i(t) dt + v1(t0) + 1 C2 t t0 i(t) dt + v2(t0) + · · · + 1 CN t t0 i(t) dt + vN (t0) = 1 C1 + 1 C2 + · · · + 1 CN t t0 i(t) dt + v1(t0) + v2(t0) + · · · + vN (t0) = 1 Ceq t t0 i(t) dt + v(t0) (6.15) where 1 Ceq = 1 C1 + 1 C2 + 1 C3 + · · · + 1 CN (6.16) The initial voltage v(t0) across Ceq is required by KVL to be the sum of the capacitor voltages at t0. Or according to Eq. (6.15), v(t0) = v1(t0) + v2(t0) + · · · + vN (t0) Thus, according to Eq. (6.16), The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances. Note that capacitors in series combine in the same manner as resistors in parallel. For N = 2 (i.e., two capacitors in series), Eq. (6.16) becomes 1 Ceq = 1 C1 + 1 C2
  • 215. 210 PART 1 DC Circuits or Ceq = C1C2 C1 + C2 (6.17) E X A M P L E 6 . 6 Find the equivalent capacitance seen between terminals a and b of the circuit in Fig. 6.16. a b Ceq 5 mF 20 mF 20 mF 6 mF 60 mF Figure6.16 For Example 6.6. Solution: The 20-µF and 5-µF capacitors are in series; their equivalent capacitance is 20 × 5 20 + 5 = 4 µF This 4-µF capacitor is in parallel with the 6-µF and 20-µF capacitors; their combined capacitance is 4 + 6 + 20 = 30 µF This 30-µF capacitor is in series with the 60-µF capacitor. Hence, the equivalent capacitance for the entire circuit is Ceq = 30 × 60 30 + 60 = 20 µF P R A C T I C E P R O B L E M 6 . 6 Find the equivalent capacitance seen at the terminals of the circuit in Fig. 6.17. Answer: 40 µF. Ceq 120 mF 20 mF 70 F 60 mF 50 mF Figure6.17 For Practice Prob. 6.6. E X A M P L E 6 . 7 For the circuit in Fig. 6.18, find the voltage across each capacitor.
  • 216. CHAPTER 6 Capacitors and Inductors 211 Solution: We first find the equivalent capacitance Ceq, shown in Fig. 6.19. The two parallel capacitors in Fig. 6.18 can be combined to get 40+20 = 60 mF. This 60-mF capacitor is in series with the 20-mF and 30-mF capacitors. Thus, Ceq = 1 1 60 + 1 30 + 1 20 mF = 10 mF The total charge is q = Ceqv = 10 × 10−3 × 30 = 0.3 C This is the charge on the 20-mF and 30-mF capacitors, because they are in series with the 30-V source. (A crude way to see this is to imagine that charge acts like current, since i = dq/dt.) Therefore, v1 = q C1 = 0.3 20 × 10−3 = 15 V v2 = q C2 = 0.3 30 × 10−3 = 10 V Having determined v1 and v2, we now use KVL to determine v3 by v3 = 30 − v1 − v2 = 5 V 20 mF 40 mF 30 mF 20 mF 30 V + − v1 v2 v3 + − + − + − Figure6.18 For Example 6.7. Ceq 30 V + − q + − Figure6.19 Equivalent circuit for Fig. 6.18. Alternatively, since the 40-mF and 20-mF capacitors are in parallel, they have the same voltage v3 and their combined capacitance is 40 + 20 = 60 mF. This combined capacitance is in series with the 20-mF and 30-mF capacitors and consequently has the same charge on it. Hence, v3 = q 60 mF = 0.3 60 × 10−3 = 5 V P R A C T I C E P R O B L E M 6 . 7 Find the voltage across each of the capacitors in Fig. 6.20. 30 mF 20 mF 60 mF 40 mF 60 V + − v1 v3 v2 v4 + − + − + − + − Figure6.20 For Practice Prob. 6.7. Answer: v1 = 30 V, v2 = 30 V, v3 = 10 V, v4 = 20 V. 6.4 INDUCTORS An inductor is a passive element designed to store energy in its magnetic field. Inductors find numerous applications in electronic and power sys- tems. They are used in power supplies, transformers, radios, TVs, radars, and electric motors. Any conductor of electric current has inductive properties and may be regarded as an inductor. But in order to enhance the inductive effect, a practical inductor is usually formed into a cylindrical coil with many turns of conducting wire, as shown in Fig. 6.21.
  • 217. 212 PART 1 DC Circuits An inductor consists of a coil of conducting wire. If current is allowed to pass through an inductor, it is found that the voltage across the inductor is directly proportional to the time rate of change of the current. Using the passive sign convention, v = L di dt (6.18) where L is the constant of proportionality called the inductance of the inductor. The unit of inductance is the henry (H), named in honor of the American inventor Joseph Henry (1797–1878). It is clear from Eq. (6.18) that 1 henry equals 1 volt-second per ampere. Length, l Cross-sectional area, A Core material Number of turns, N Figure6.21 Typical form of an inductor. In view of Eq. (6.18), for an inductor to have voltageacrossitsterminals,itscurrentmustvary with time. Hence, v = 0 for constant current through the inductor. Inductance is the property whereby an inductor exhibits opposition to the change of current flowing through it, measured in henrys (H). The inductance of an inductor depends on its physical dimension and construction. Formulas for calculating the inductance of inductors of different shapes are derived from electromagnetic theory and can be found in standard electrical engineering handbooks. For example, for the inductor (solenoid) shown in Fig. 6.21, L = N2 µA (6.19) where N is the number of turns, is the length, A is the cross-sectional area, and µ is the permeability of the core. We can see from Eq. (6.19) that inductance can be increased by increasing the number of turns of coil, using material with higher permeability as the core, increasing the cross-sectional area, or reducing the length of the coil. Like capacitors, commercially available inductors come in different values and types. Typical practical inductors have inductance values ranging from a few microhenrys (µH), as in communication systems, to tens of henrys (H) as in power systems. Inductors may be fixed or variable. The core may be made of iron, steel, plastic, or air. The terms coil and choke are also used for inductors. Common inductors are shown in Fig. 6.22. The circuit symbols for inductors are shown in Fig. 6.23, following the passive sign convention. (a) (b) (c) Figure6.22 Various types of inductors: (a) solenoidal wound inductor, (b) toroidal inductor, (c) chip inductor. (Courtesy of Tech America.) Equation (6.18) is the voltage-current relationship for an inductor. Figure 6.24 shows this relationship graphically for an inductor whose inductance is independent of current. Such an inductor is known as a linear inductor. For a nonlinear inductor, the plot of Eq. (6.18) will not be a straight line because its inductance varies with current. We will assume linear inductors in this textbook unless stated otherwise. The current-voltage relationship is obtained from Eq. (6.18) as di = 1 L v dt
  • 218. CHAPTER 6 Capacitors and Inductors 213 Integrating gives i = 1 L t −∞ v(t) dt (6.20) i i i (a) v L + − (b) v L + − (c) v L + − Figure6.23 Circuit symbols for inductors: (a) air-core, (b) iron-core, (c) variable iron-core. or i = 1 L t t0 v(t) dt + i(t0) (6.21) where i(t0) is the total current for −∞ t t0 and i(−∞) = 0. The idea of making i(−∞) = 0 is practical and reasonable, because there must be a time in the past when there was no current in the inductor. The inductor is designed to store energy in its magnetic field. The energy stored can be obtained from Eqs. (6.18) and (6.20). The power delivered to the inductor is p = vi = L di dt i (6.22) The energy stored is w = t −∞ p dt = t −∞ L di dt i dt = L t −∞ i di = 1 2 Li2 (t) − 1 2 Li2 (−∞) (6.23) Since i(−∞) = 0, w = 1 2 Li2 (6.24) Slope = L di ⁄dt 0 v Figure6.24 Voltage-current relationship of an inductor. We should note the following important properties of an inductor. i t (a) i t (b) Figure6.25 Current through an inductor: (a) allowed, (b) not allowable; an abrupt change is not possible. 1. Note from Eq. (6.18) that the voltage across an inductor is zero when the current is constant. Thus, An inductor acts like a short circuit to dc. 2. An important property of the inductor is its opposition to the change in current flowing through it. The current through an inductor cannot change instantaneously. According to Eq. (6.18), a discontinuous change in the current through an inductor requires an infinite voltage, which is not physically possible. Thus, an inductor opposes an abrupt change in the current through it. For example, the current through an inductor may take the form shown in Fig. 6.25(a), whereas the inductor current cannot take the form shown in Fig. 6.25(b) in real-life situations due to the discontinuities. However, the voltage across an inductor can change abruptly.
  • 219. 214 PART 1 DC Circuits 3. Like the ideal capacitor, the ideal inductor does not dissipate energy. The energy stored in it can be retrieved at a later time. The inductor takes power from the circuit when storing energy and delivers power to the circuit when returning previously stored energy. 4. A practical, nonideal inductor has a significant resistive component, as shown in Fig. 6.26. This is due to the fact that the inductor is made of a conducting material such as copper, which has some resistance. This resistance is called the winding resistance Rw, and it appears in series with the inductance of the inductor. The presence of Rw makes it both an energy storage device and an energy dissipation device. Since Rw is usually very small, it is ignored in most cases. The nonideal inductor also has a winding capacitance Cw due to the capacitive coupling between the conducting coils. Cw is very small and can be ignored in most cases, except at high frequencies. We will assume ideal inductors in this book. Since an inductor is often made of a highly con- ducting wire, it has a very small resistance. L Rw Cw Figure6.26 Circuit model for a practical inductor. E X A M P L E 6 . 8 The current through a 0.1-H inductor is i(t) = 10te−5t A. Find the voltage across the inductor and the energy stored in it. Solution: Since v = L di/dt and L = 0.1 H, v = 0.1 d dt (10te−5t ) = e−5t + t(−5)e−5t = e−5t (1 − 5t) V The energy stored is w = 1 2 Li2 = 1 2 (0.1)100t2 e−10t = 5t2 e−10t J P R A C T I C E P R O B L E M 6 . 8 If the current through a 1-mH inductor is i(t) = 20 cos 100t mA, find the terminal voltage and the energy stored. Answer: −2 sin 100t mV, 0.2 cos2 100t µJ. E X A M P L E 6 . 9 Find the current through a 5-H inductor if the voltage across it is v(t) = 30t2 , t 0 0, t 0 Also find the energy stored within 0 t 5 s. Solution: Since i = 1 L t t0 v(t) dt + i(t0) and L = 5 H, i = 1 5 t 0 30t2 dt + 0 = 6 × t3 3 = 2t3 A
  • 220. CHAPTER 6 Capacitors and Inductors 215 The power p = vi = 60t5 , and the energy stored is then w = p dt = 5 0 60t5 dt = 60 t6 6 5 0 = 156.25 kJ Alternatively, we can obtain the energy stored using Eq. (6.13), by writing w 5 0 = 1 2 Li2 (5) − 1 2 Li(0) = 1 2 (5)(2 × 53 )2 − 0 = 156.25 kJ as obtained before. P R A C T I C E P R O B L E M 6 . 9 The terminal voltage of a 2-H inductor is v = 10(1−t) V. Find the current flowing through it at t = 4 s and the energy stored in it within 0 t 4 s. Assume i(0) = 2 A. Answer: −18 A, 320 J. E X A M P L E 6 . 1 0 Consider the circuit in Fig. 6.27(a). Under dc conditions, find: (a) i, vC, and iL, (b) the energy stored in the capacitor and inductor. 12 V 1 F + − 4 Ω 5 Ω 1 Ω 2 H i iL vC + − vC + − (a) (a) 12 V + − 4 Ω 5 Ω 1 Ω i iL (b) Figure6.27 For Example 6.10. Solution: (a) Under dc conditions, we replace the capacitor with an open circuit and the inductor with a short circuit, as in Fig. 6.27(b). It is evident from Fig. 6.27(b) that i = iL = 12 1 + 5 = 2 A The voltage vC is the same as the voltage across the 5- resistor. Hence, vC = 5i = 10 V (b) The energy in the capacitor is wC = 1 2 Cv2 C = 1 2 (1)(102 ) = 50 J and that in the inductor is wL = 1 2 Li2 L = 1 2 (2)(22 ) = 4 J P R A C T I C E P R O B L E M 6 . 1 0 Determine vC, iL, and the energy stored in the capacitor and inductor in the circuit of Fig. 6.28 under dc conditions. 4 A 2 F 3 Ω 1 Ω 0.25 H iL vC + − Figure6.28 For Practice Prob. 6.10. Answer: 3 V, 3 A, 9 J, 1.125 J.
  • 221. 216 PART 1 DC Circuits 6.5 SERIES AND PARALLEL INDUCTORS Now that the inductor has been added to our list of passive elements, it is necessary to extend the powerful tool of series-parallel combination. We need to know how to find the equivalent inductance of a series-connected or parallel-connected set of inductors found in practical circuits. Consider a series connection of N inductors, as shown in Fig. 6.29(a), with the equivalent circuit shown in Fig. 6.29(b). The inductors have the same current through them. Applying KVL to the loop, v = v1 + v2 + v3 + · · · + vN (6.25) L1 (a) L2 L3 LN (b) Leq i i v + − v + − + − v1 + − v2 + − v3 + − vN . . . Figure6.29 (a) A series connection of N inductors, (b) equivalent circuit for the series inductors. Substituting vk = Lk di/dt results in v = L1 di dt + L2 di dt + L3 di dt + · · · + LN di dt = (L1 + L2 + L3 + · · · + LN ) di dt = N k=1 Lk di dt = Leq di dt (6.26) where Leq = L1 + L2 + L3 + · · · + LN (6.27) Thus, The equivalent inductance of series-connected inductors is the sum of the individual inductances. Inductors in series are combined in exactly the same way as resistors in series. (a) (b) Leq i v + − v + − L1 L2 L3 LN i i1 i2 i3 iN Figure6.30 (a) A parallel connection of N inductors, (b) equivalent circuit for the parallel inductors. We now consider a parallel connection of N inductors, as shown in Fig. 6.30(a), with the equivalent circuit in Fig. 6.30(b). The inductors have the same voltage across them. Using KCL, i = i1 + i2 + i3 + · · · + iN (6.28) But ik = 1 Lk t t0 v dt + ik(t0); hence, i = 1 L1 t t0 v dt + i1(t0) + 1 L2 t t0 v dt + i2(t0) + · · · + 1 LN t t0 v dt + iN (t0) = 1 L1 + 1 L2 + · · · + 1 LN t t0 v dt + i1(t0) + i2(t0) + · · · + iN (t0) = N k=1 1 Lk t t0 v dt + N k=1 ik(t0) = 1 Leq t t0 v dt + i(t0) (6.29)
  • 222. CHAPTER 6 Capacitors and Inductors 217 where 1 Leq = 1 L1 + 1 L2 + 1 L3 + · · · + 1 LN (6.30) The initial current i(t0) through Leq at t = t0 is expected by KCL to be the sum of the inductor currents at t0. Thus, according to Eq. (6.29), i(t0) = i1(t0) + i2(t0) + · · · + iN (t0) According to Eq. (6.30), The equivalent inductance of parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances. Note that the inductors in parallel are combined in the same way as resis- tors in parallel. For two inductors in parallel (N = 2), Eq. (6.30) becomes 1 Leq = 1 L1 + 1 L2 or Leq = L1L2 L1 + L2 (6.31) It is appropriate at this point to summarize the most important character- istics of the three basic circuit elements we have studied. The summary is given in Table 6.1. TABLE 6.1 Important characteristics of the basic elements.† Relation Resistor (R) Capacitor (C) Inductor (L) v-i: v = iR v = 1 C t t0 i dt + v(t0) v = L di dt i-v: i = v/R i = C dv dt i = 1 L t t0 i dt + i(t0) p or w: p = i2 R = v2 R w = 1 2 Cv2 w = 1 2 Li2 Series: Req = R1 + R2 Ceq = C1C2 C1 + C2 Leq = L1 + L2 Parallel: Req = R1R2 R1 + R2 Ceq = C1 + C2 Leq = L1L2 L1 + L2 At dc: Same Open circuit Short circuit Circuit variable that cannot change abruptly: Not applicable v i †Passive sign convention is assumed. E X A M P L E 6 . 1 1 Find the equivalent inductance of the circuit shown in Fig. 6.31.
  • 223. 218 PART 1 DC Circuits 4 H 20 H 8 H 10 H 12 H 7 H Leq Figure6.31 For Example 6.11. Solution: The 10-H, 12-H, and 20-H inductors are in series; thus, combining them gives a 42-H inductance. This 42-H inductor is in parallel with the 7-H inductor so that they are combined, to give 7 × 42 7 + 42 = 6 H This 6-H inductor is in series with the 4-H and 8-H inductors. Hence, Leq = 4 + 6 + 8 = 18 H P R A C T I C E P R O B L E M 6 . 1 1 Calculate the equivalent inductance for the inductive ladder network in Fig. 6.32. 20 mH 100 mH 40 mH 30 mH 20 mH 40 mH 50 mH Leq Figure6.32 For Practice Prob. 6.11. Answer: 25 mH. E X A M P L E 6 . 1 2 For the circuit in Fig. 6.33, i(t) = 4(2 − e−10t ) mA. If i2(0) = −1 mA, find: (a) i1(0); (b) v(t), v1(t), and v2(t); (c) i1(t) and i2(t). 4 H 12 H 4 H v + − v2 v1 + + − − i i1 i2 Figure6.33 For Example 6.12. Solution: (a) From i(t) = 4(2 − e−10t ) mA, i(0) = 4(2 − 1) = 4 mA. Since i = i1 + i2, i1(0) = i(0) − i2(0) = 4 − (−1) = 5 mA (b) The equivalent inductance is Leq = 2 + 4 12 = 2 + 3 = 5 H Thus, v(t) = Leq di dt = 5(4)(−1)(−10)e−10t mV = 200e−10t mV and v1(t) = 2 di dt = 2(−4)(−10)e−10t mV = 80e−10t mV Since v = v1 + v2, v2(t) = v(t) − v1(t) = 120e−10t mV
  • 224. CHAPTER 6 Capacitors and Inductors 219 (c) The current i1 is obtained as i1(t) = 1 4 t 0 v2 dt + i1(0) = 120 4 t 0 e−10t dt + 5 mA = −3e−10t t 0 + 5 mA = −3e−10t + 3 + 5 = 8 − 3e−10t mA Similarly, i2(t) = 1 12 t 0 v2 dt + i2(0) = 120 12 t 0 e−10t dt − 1 mA = −e−10t t 0 − 1 mA = −e−10t + 1 − 1 = −e−10t mA Note that i1(t) + i2(t) = i(t). P R A C T I C E P R O B L E M 6 . 1 2 In the circuit of Fig. 6.34, i1(t) = 0.6e−2t A. If i(0) = 1.4 A, find: (a) i2(0); (b) i2(t) and i(t); (c) v(t), v1(t), and v2(t). 3 H 6 H 8 H v + − v2 + − i i1 i2 + − v1 Figure 6.34 For Practice Prob. 6.12. Answer: (a) 0.8 A, (b) (−0.4 + 1.2e−2t ) A, (−0.4 + 1.8e−2t ) A, (c) −7.2e−2t V, −28.8e−2t V, −36e−2t V. †6.6 APPLICATIONS Circuit elements such as resistors and capacitors are commercially avail- able in either discrete form or integrated-circuit (IC) form. Unlike ca- pacitors and resistors, inductors with appreciable inductance are difficult to produce on IC substrates. Therefore, inductors (coils) usually come in discrete form and tend to be more bulky and expensive. For this rea- son, inductors are not as versatile as capacitors and resistors, and they are more limited in applications. However, there are several applications in which inductors have no practical substitute. They are routinely used in relays, delays, sensing devices, pick-up heads, telephone circuits, ra- dio and TV receivers, power supplies, electric motors, microphones, and loudspeakers, to mention a few. Capacitors and inductors possess the following three special prop- erties that make them very useful in electric circuits: 1. The capacity to store energy makes them useful as temporary voltage or current sources. Thus, they can be used for generating a large amount of current or voltage for a short period of time. 2. Capacitors oppose any abrupt change in voltage, while inductors oppose any abrupt change in current. This property
  • 225. 220 PART 1 DC Circuits makes inductors useful for spark or arc suppression and for converting pulsating dc voltage into relatively smooth dc voltage. 3. Capacitors and inductors are frequency sensitive. This property makes them useful for frequency discrimination. The first two properties are put to use in dc circuits, while the third one is taken advantage of in ac circuits. We will see how useful these properties are in later chapters. For now, consider three applications involving capacitors and op amps: integrator, differentiator, and analog computer. 6.6.1 Integrator Important op amp circuits that use energy-storage elements include inte- grators and differentiators. These op amp circuits often involve resistors and capacitors; inductors (coils) tend to be more bulky and expensive. The op amp integrator is used in numerous applications, especially in analog computers, to be discussed in Section 6.6.3. An integrator is an op amp circuit whose output is proportional to the integral of the input signal. R1 Rf i1 v1 i2 vi + − vo + − v2 0 A 0 V + − + − (a) R a C iR iC vi + − vo + − + − (b) 1 Figure6.35 Replacing the feedback resistor in the inverting amplifier in (a) produces an integrator in (b). If the feedback resistor Rf in the familiar inverting amplifier of Fig. 6.35(a) is replaced by a capacitor, we obtain an ideal integrator, as shown in Fig. 6.35(b). It is interesting that we can obtain a mathematical representation of integration this way. At node a in Fig. 6.35(b), iR = iC (6.32) But iR = vi R , iC = −C dvo dt Substituting these in Eq. (6.32), we obtain vi R = −C dvo dt (6.33a) dvo = − 1 RC vi dt (6.33b) Integrating both sides gives vo(t) − vo(0) = − 1 RC t 0 vi(t) dt (6.34) To ensure that vo(0) = 0, it is always necessary to discharge the integra- tor’s capacitor prior to the application of a signal. Assuming vo(0) = 0, vo = − 1 RC t 0 vi(t) dt (6.35) which shows that the circuit in Fig. 6.35(b) provides an output voltage proportional to the integral of the input. In practice, the op amp integrator
  • 226. CHAPTER 6 Capacitors and Inductors 221 requires a feedback resistor to reduce dc gain and prevent saturation. Care must be taken that the op amp operates within the linear range so that it does not saturate. E X A M P L E 6 . 1 3 If v1 = 10 cos 2t mV and v2 = 0.5t mV, find vo in the op amp circuit in Fig. 6.36. Assume that the voltage across the capacitor is initially zero. vo v1 v2 2 mF 3 MΩ 100 kΩ + − Figure6.36 For Example 6.13. Solution: This is a summing integrator, and vo = − 1 R1C v1 dt − 1 R2C v2 dt = − 1 3 × 106 × 2 × 10−6 t 0 10 cos 2t dt − 1 100 × 103 × 2 × 10−6 t 0 0.5t dt = − 1 6 10 2 sin 2t − 1 0.2 0.5t2 2 = −0.833 sin 2t − 1.25t2 mV P R A C T I C E P R O B L E M 6 . 1 3 The integrator in Fig. 6.35 has R = 25 k, C = 10 µF. Determine the output voltage when a dc voltage of 10 mV is applied at t = 0. Assume that the op amp is initially nulled. Answer: −40t mV. 6.6.2 Differentiator A differentiator is an op amp circuit whose output is proportional to the rate of change of the input signal. R a C iC iR vi + − vo + − + − Figure6.37 An op amp differentiator. In Fig. 6.35(a), if the input resistor is replaced by a capacitor, the resulting circuit is a differentiator, shown in Fig. 6.37. Applying KCL at node a, iR = iC (6.36) But iR = − vo R , iC = C dvi dt Substituting these in Eq. (6.36) yields vo = −RC dvi dt (6.37) showing that the output is the derivative of the input. Differentiator cir- cuits are electronically unstable because any electrical noise within the
  • 227. 222 PART 1 DC Circuits circuit is exaggerated by the differentiator. For this reason, the differen- tiator circuit in Fig. 6.37 is not as useful and popular as the integrator. It is seldom used in practice. E X A M P L E 6 . 1 4 Sketch the output voltage for the circuit in Fig. 6.38(a), given the input voltage in Fig. 6.38(b). Take vo = 0 at t = 0. vo vi + − (a) (b) + − 0.2 mF 5 kΩ vi 8 6 4 2 0 4 t (ms) + − Figure6.38 For Example 6.14. Solution: This is a differentiator with RC = 5 × 103 × 0.2 × 10−6 = 10−3 s For 0 t 4 ms, we can express the input voltage in Fig. 6.38(b) as vi = 2t 0 t 2 ms 8 − 2t 2 t 4 ms This is repeated for 4 t 8. Using Eq. (6.37), the output is obtained as vo = −RC dvi dt = −2 mV 0 t 2 ms 2 mV 2 t 4 ms Thus, the output is as sketched in Fig. 6.39. vi (mV) 8 6 4 2 2 0 −2 t (ms) Figure6.39 Output of the circuit in Fig. 6.38(a). P R A C T I C E P R O B L E M 6 . 1 4 The differentiator in Fig. 6.37 has R = 10 k and C = 2 µF. Given that vi = 3t V, determine the output vo. Answer: −60 mV. 6.6.3 Analog Computer Op amps were initially developed for electronic analog computers. Ana- log computers can be programmed to solve mathematical models of me- chanical or electrical systems. These models are usually expressed in terms of differential equations. To solve simple differential equations using the analog computer requires cascading three types of op amp circuits: integrator circuits, summing amplifiers, and inverting/noninverting amplifiers for negative/
  • 228. CHAPTER 6 Capacitors and Inductors 223 positive scaling. The best way to illustrate how an analog computer solves a differential equation is with an example. Suppose we desire the solution x(t) of the equation a d2 x dt2 + b dx dt + cx = f (t), t 0 (6.38) where a, b, and c are constants, and f (t) is an arbitrary forcing function. The solution is obtained by first solving the highest-order derivative term. Solving for d2 x/dt2 yields d2 x dt2 = f (t) a − b a dx dt − c a x (6.39) To obtain dx/dt, the d2 x/dt2 term is integrated and inverted. Finally, to obtain x, the dx/dt term is integrated and inverted. The forcing function is injected at the proper point. Thus, the analog computer for solving Eq. (6.38) is implemented by connecting the necessary summers, inverters, and integrators. A plotter or oscilloscope may be used to view the output x, or dx/dt, or d2 x/dt2 , depending on where it is connected in the system. Although the above example is on a second-order differential equa- tion, any differential equation can be simulated by an analog computer comprising integrators, inverters, and inverting summers. But care must be exercised in selecting the values of the resistors and capacitors, to ensure that the op amps do not saturate during the solution time interval. The analog computers with vacuum tubes were built in the 1950s and 1960s. Recently their use has declined. They have been superseded by modern digital computers. However, we still study analog computers for two reasons. First, the availability of integrated op amps has made it possible to build analog computers easily and cheaply. Second, un- derstanding analog computers helps with the appreciation of the digital computers. E X A M P L E 6 . 1 5 Design an analog computer circuit to solve the differential equation: d2 vo dt2 + 2 dvo dt + vo = 10 sin 4t t 0 subject to vo(0) = −4, v o(0) = 1, where the prime refers to the time derivative. Solution: We first solve for the second derivative as d2 vo dt2 = 10 sin 4t − 2 dvo dt − vo (6.15.1) Solving this requires some mathematical operations, including summing, scaling, and integration. Integrating both sides of Eq. (6.15.1) gives dvo dt = − t 0 −10 sin 4t + 2 dvo dt + vo + v o(0) (6.15.2)
  • 229. 224 PART 1 DC Circuits where v o(0) = 1. We implement Eq. (6.15.2) using the summing inte- grator shown in Fig. 6.40(a). The values of the resistors and capacitors have been chosen so that RC = 1 for the term − 1 RC t 0 vo dt Other terms in the summing integrator of Eq. (6.15.2) are implemented accordingly. The initial condition dvo(0)/dt = 1 is implemented by connecting a 1-V battery with a switch across the capacitor as shown in Fig. 6.40(a). (a) 1 mF 1 MΩ 1 V 0.6 MΩ 1 MΩ dvo dt dvo dt dvo dt t = 0 –10 sin (4t) vo 1 mF 1 MΩ 1 V 0.5 MΩ 1 MΩ t = 0 10 sin (4t) vo (b) 1 mF 4 V 1 MΩ 1 MΩ dvo dt t = 0 −vo vo 1 MΩ (c) 1 mF 4 V 1 MΩ 1 MΩ t = 0 vo 1 MΩ + − + − + − + − + − + − + − + − + − + − + − Figure6.40 For Example 6.15. The next step is to obtain vo by integrating dvo/dt and inverting the result, vo = − t 0 − dvo dt dt + v(0) (6.15.3) This is implemented with the circuit in Fig. 6.40(b) with the battery giving the initial condition of −4 V. We now combine the two circuits in Fig. 6.40 (a) and (b) to obtain the complete circuit shown in Fig. 6.40(c). When theinputsignal10 sin 4t isapplied, weopentheswitchesatt = 0toobtain the output waveform vo, which may be viewed on an oscilloscope.
  • 230. CHAPTER 6 Capacitors and Inductors 225 P R A C T I C E P R O B L E M 6 . 1 5 Design an analog computer circuit to solve the differential equation: d2 vo dt2 + 3 dvo dt + 2vo = 4 cos 10t t 0 subject to vo(0) = 2, v o(0) = 0. Answer: See Fig. 6.41, where RC = 1 s. d2 v dt2 d2 v dt2 cos (10t) 2 V t = 0 v + − C R R 2 R C R R R R 3 R 4 + − + − + − + − + − R R Figure6.41 For Practice Prob. 6.15. 6.7 SUMMARY 1. The current through a capacitor is directly proportional to the time rate of change of the voltage across it. i = C dv dt The current through a capacitor is zero unless the voltage is changing. Thus, a capacitor acts like an open circuit to a dc source. 2. The voltage across a capacitor is directly proportional to the time integral of the current through it. v = 1 C t −∞ i dt = 1 C t t0 i dt + i(t0) The voltage across a capacitor cannot change instantly. 3. Capacitors in series and in parallel are combined in the same way as conductances.
  • 231. 226 PART 1 DC Circuits 4. The voltage across an inductor is directly proportional to the time rate of change of the current through it. v = L di dt The voltage across the inductor is zero unless the current is chang- ing. Thus an inductor acts like a short circuit to a dc source. 5. The current through an inductor is directly proportional to the time integral of the voltage across it. i = 1 L t −∞ v dt = 1 L t t0 v dt + v(t0) The current through an inductor cannot change instantly. 6. Inductors in series and in parallel are combined in the same way resistors in series and in parallel are combined. 7. At any given time t, the energy stored in a capacitor is 1 2 Cv2 , while the energy stored in an inductor is 1 2 Li2 . 8. Three application circuits, the integrator, the differentiator, and the analog computer, can be realized using resistors, capacitors, and op amps. REVIEW QUESTIONS 6.1 What charge is on a 5-F capacitor when it is connected across a 120-V source? (a) 600 C (b) 300 C (c) 24 C (d) 12 C 6.2 Capacitance is measured in: (a) coulombs (b) joules (c) henrys (d) farads 6.3 When the total charge in a capacitor is doubled, the energy stored: (a) remains the same (b) is halved (c) is doubled (d) is quadrupled 6.4 Can the voltage waveform in Fig. 6.42 be associated with a capacitor? (a) Yes (b) No 0 2 1 10 −10 t v(t) Figure 6.42 For Review Question 6.4. 6.5 The total capacitance of two 40-mF series-connected capacitors in parallel with a 4-mF capacitor is: (a) 3.8 mF (b) 5 mF (c) 24 mF (d) 44 mF (e) 84 mF 6.6 In Fig. 6.43, if i = cos 4t and v = sin 4t, the element is: (a) a resistor (b) a capacitor (c) an inductor v + − i Element Figure 6.43 For Review Question 6.6. 6.7 A 5-H inductor changes its current by 3 A in 0.2 s. The voltage produced at the terminals of the inductor is: (a) 75 V (b) 8.888 V (c) 3 V (d) 1.2 V 6.8 If the current through a 10-mH inductor increases from zero to 2 A, how much energy is stored in the inductor? (a) 40 mJ (b) 20 mJ (c) 10 mJ (d) 5 mJ
  • 232. CHAPTER 6 Capacitors and Inductors 227 6.9 Inductors in parallel can be combined just like resistors in parallel. (a) True (b) False 6.10 For the circuit in Fig. 6.44, the voltage divider formula is: (a) v1 = L1 + L2 L1 vs (b) v1 = L1 + L2 L2 vs (c) v1 = L2 L1 + L2 vs (d) v1 = L1 L1 + L2 vs vs + − v2 v1 L1 L2 + − + − Figure 6.44 For Review Question 6.10. Answers: 6.1a, 6.2d, 6.3d, 6.4b, 6.5c, 6.6b, 6.7a, 6.8b, 6.9a, 6.10d. PROBLEMS Section 6.2 Capacitors 6.1 If the voltage across a 5-F capacitor is 2te−3t V, find the current and the power. 6.2 A 40-µF capacitor is charged to 120 V and is then allowed to discharge to 80 V. How much energy is lost? 6.3 In 5 s, the voltage across a 40-mF capacitor changes from 160 V to 220 V. Calculate the average current through the capacitor. 6.4 A current of 6 sin 4t A flows through a 2-F capacitor. Find the voltage v(t) across the capacitor given that v(0) = 1 V. 6.5 If the current waveform in Fig. 6.45 is applied to a 20-µF capacitor, find the voltage v(t) across the capacitor. Assume that v(0) = 0. 0 2 1 4 t i(t) Figure 6.45 For Prob. 6.5. 6.6 The voltage waveform in Fig. 6.46 is applied across a 30-µF capacitor. Draw the current waveform through it. v(t) V 0 6 8 10 12 4 2 10 −10 t (ms) Figure 6.46 For Prob. 6.6. 6.7 At t = 0, the voltage across a 50-mF capacitor is 10 V. Calculate the voltage across the capacitor for t 0 when current 4t mA flows through it. 6.8 The current through a 0.5-F capacitor is 6(1 − e−t ) A. Determine the voltage and power at t = 2 s. Assume v(0) = 0. 6.9 If the voltage across a 2-F capacitor is as shown in Fig. 6.47, find the current through the capacitor. v(t) (V) 0 5 3 4 5 6 7 2 1 10 t (s) Figure 6.47 For Prob. 6.9. 6.10 The current through an initially uncharged 4-µF capacitor is shown in Fig. 6.48. Find the voltage across the capacitor for 0 t 3.
  • 233. 228 PART 1 DC Circuits 0 2 3 1 40 −40 t (s) i(t) (mA) Figure 6.48 For Prob. 6.10. 6.11 A voltage of 60 cos 4πt V appears across the terminals of a 3-mF capacitor. Calculate the current through the capacitor and the energy stored in it from t = 0 to t = 0.125 s. 6.12 Find the voltage across the capacitors in the circuit of Fig. 6.49 under dc conditions. 3 Ω 60 V 20 Ω 10 Ω 50 Ω v2 v1 C1 C2 + − + − + − Figure 6.49 For Prob. 6.12. Section 6.3 Series and Parallel Capacitors 6.13 What is the total capacitance of four 30-mF capacitors connected in: (a) parallel (b) series 6.14 Two capacitors (20 µF and 30 µF) are connected to a 100-V source. Find the energy stored in each capacitor if they are connected in: (a) parallel (b) series 6.15 Determine the equivalent capacitance for each of the circuits in Fig. 6.50. 4 F 4 F 6 F 3 F 12 F (a) 6 F 4 F 2 F 5 F (b) 2 F 3 F (c) 6 F 3 F 4 F Figure 6.50 For Prob. 6.15. 6.16 Find Ceq for the circuit in Fig. 6.51. 30 mF 5 mF 40 mF 15 mF 20 mF Ceq Figure 6.51 For Prob. 6.16. 6.17 Calculate the equivalent capacitance for the circuit in Fig. 6.52. All capacitances are in mF. 4 8 1 15 5 6 6 2 3 Ceq Figure 6.52 For Prob. 6.17. 6.18 Determine the equivalent capacitance at terminals a-b of the circuit in Fig. 6.53. 6 mF 4 mF 5 mF 3 mF 12 mF 2 mF a b Figure 6.53 For Prob. 6.18.
  • 234. CHAPTER 6 Capacitors and Inductors 229 6.19 Obtain the equivalent capacitance of the circuit in Fig. 6.54. 40 mF 20 mF a b 35 mF 5 mF 10 mF 15 mF 15 mF 10 mF Figure 6.54 For Prob. 6.19. 6.20 For the circuit in Fig. 6.55, determine: (a) the voltage across each capacitor, (b) the energy stored in each capacitor. 2 mF 6 mF 4 mF 3 mF + − 120 V Figure 6.55 For Prob. 6.20. 6.21 Repeat Prob. 6.20 for the circuit in Fig. 6.56. 60 mF 20 mF 14 mF 80 mF 30 mF + − 90 V Figure 6.56 For Prob. 6.21. 6.22 (a) Show that the voltage-division rule for two capacitors in series as in Fig. 6.57(a) is v1 = C2 C1 + C2 vs, v2 = C1 C1 + C2 vs assuming that the initial conditions are zero. C1 is C2 (b) C1 vs v1 v2 C2 (a) + − + − + − i1 i2 Figure 6.57 For Prob. 6.22. (b) For two capacitors in parallel as in Fig. 6.57(b), show that the current-division rule is i1 = C1 C1 + C2 is, i2 = C2 C1 + C2 is assuming that the initial conditions are zero. 6.23 Three capacitors, C1 = 5 µF, C2 = 10 µF, and C3 = 20 µF, are connected in parallel across a 150-V source. Determine: (a) the total capacitance, (b) the charge on each capacitor, (c) the total energy stored in the parallel combination. 6.24 The three capacitors in the previous problem are placed in series with a 200-V source. Compute: (a) the total capacitance, (b) the charge on each capacitor, (c) the total energy stored in the series combination. 6.25 ∗ Obtain the equivalent capacitance of the network shown in Fig. 6.58. 30 mF 20 mF 10 mF 50 mF 40 mF Figure 6.58 For Prob. 6.25. 6.26 Determine Ceq for each circuit in Fig. 6.59. ∗An asterisk indicates a challenging problem.
  • 235. 230 PART 1 DC Circuits C C C C C Ceq (a) C C C C Ceq (b) Figure 6.59 For Prob. 6.26. 6.27 Assuming that the capacitors are initially uncharged, find vo(t) in the circuit in Fig. 6.60. is + − vo(t) 6 mF 3 mF is (mA) 0 2 1 60 t (s) Figure 6.60 For Prob. 6.27. 6.28 If v(0) = 0, find v(t), i1(t), and i2(t) in the circuit in Fig. 6.61. i1 is i2 v 6 mF 4 mF is (mA) 5 3 4 1 2 20 0 −20 t + − Figure 6.61 For Prob. 6.28. 6.29 For the circuit in Fig. 6.62, let v = 10e−3t V and v1(0) = 2 V. Find: (a) v2(0) (b) v1(t) and v2(t) (c) i(t), i1(t), and i2(t) 50 mF 30 mF 20 mF v + − v2 v1 i − + + − i1 i2 Figure 6.62 For Prob. 6.29. Section 6.4 Inductors 6.30 The current through a 10-mH inductor is 6e−t/2 A. Find the voltage and the power at t = 3 s. 6.31 The current in a coil increases uniformly from 0.4 to 1 A in 2 s so that the voltage across the coil is 60 mV. Calculate the inductance of the coil. 6.32 The current through a 0.25-mH inductor is 12 cos 2t A. Determine the terminal voltage and the power. 6.33 The current through a 12-mH inductor is 4 sin 100t A. Find the voltage, and also the energy stored in the inductor for 0 t π/200 s. 6.34 The current through a 40-mH inductor is i(t) = 0, t 0 te−2t A, t 0 Find the voltage v(t). 6.35 The voltage across a 2-H inductor is 20(1 − e−2t ) V. If the initial current through the inductor is 0.3 A, find the current and the energy stored in the inductor at t = 1 s. 6.36 If the voltage waveform in Fig. 6.63 is applied across the terminals of a 5-H inductor, calculate the current through the inductor. Assume i(0) = −1 A. v(t) (V) 5 4 2 1 3 10 0 t Figure 6.63 For Prob. 6.36. 6.37 The current in an 80-mH inductor increases from 0 to 60 mA. How much energy is stored in the inductor? 6.38 A voltage of (4 + 10 cos 2t) V is applied to a 5-H inductor. Find the current i(t) through the inductor if i(0) = −1 A.
  • 236. CHAPTER 6 Capacitors and Inductors 231 6.39 If the voltage waveform in Fig. 6.64 is applied to a 10-mH inductor, find the inductor current i(t). Assume i(0) = 0. v(t) 0 2 1 5 –5 t Figure 6.64 For Prob. 6.39. 6.40 Find vC, iL, and the energy stored in the capacitor and inductor in the circuit of Fig. 6.65 under dc conditions. 5 Ω 2 Ω 4 Ω 2 F 3 A 0.5 H vC + − iL Figure 6.65 For Prob. 6.40. 6.41 For the circuit in Fig. 6.66, calculate the value of R that will make the energy stored in the capacitor the same as that stored in the inductor under dc conditions. R 2 Ω 5 A 4 mH 160 mF Figure 6.66 For Prob. 6.41. 6.42 Under dc conditions, find the voltage across the capacitors and the current through the inductors in the circuit of Fig. 6.67. 6 Ω 4 Ω 30 V + − C1 L1 C2 L2 Figure 6.67 For Prob. 6.42. Section 6.5 Series and Parallel Inductors 6.43 Find the equivalent inductance for each circuit in Fig. 6.68. 5 H 1 H 4 H 4 H 6 H (a) 1 H 2 H 6 H 4 H 12 H (b) 6 H 2 H 4 H 3 H (c) Figure 6.68 For Prob. 6.43. 6.44 Obtain Leq for the inductive circuit of Fig. 6.69. All inductances are in mH. 6 5 4 12 10 3 Figure 6.69 For Prob. 6.44.
  • 237. 232 PART 1 DC Circuits 6.45 Determine Leq at terminals a-b of the circuit in Fig. 6.70. 60 mH 20 mH 30 mH 25 mH 10 mH a b Figure 6.70 For Prob. 6.45. 6.46 Find Leq at the terminals of the circuit in Fig. 6.71. 8 mH 6 mH 8 mH 12 mH 4 mH 6 mH 5 mH 8 mH 10 mH a b Figure 6.71 For Prob. 6.46. 6.47 Find the equivalent inductance looking into the terminals of the circuit in Fig. 6.72. 9 H 6 H 4 H 3 H 12 H 10 H a b Figure 6.72 For Prob. 6.47. 6.48 Determine Leq in the circuit in Fig. 6.73. L L L L L L Leq Figure 6.73 For Prob. 6.48. 6.49 Find Leq in the circuit in Fig. 6.74. L L L L L Leq L L L Figure 6.74 For Prob. 6.49. 6.50 ∗ Determine Leq that may be used to represent the inductive network of Fig. 6.75 at the terminals. 3 H 4 H 5 H Leq + − i a b dt di 2 Figure 6.75 For Prob. 6.50. 6.51 The current waveform in Fig. 6.76 flows through a 3-H inductor. Sketch the voltage across the inductor over the interval 0 t 6 s. i(t) 0 2 3 4 5 6 2 1 t Figure 6.76 For Prob. 6.51.
  • 238. CHAPTER 6 Capacitors and Inductors 233 6.52 (a) For two inductors in series as in Fig. 6.77(b), show that the current-division principle is v1 = L1 L1 + L2 vs, v2 = L2 L1 + L2 vs assuming that the initial conditions are zero. (b) For two inductors in parallel as in Fig. 6.77(b), show that the current-division principle is i1 = L2 L1 + L2 is, i2 = L1 L1 + L2 is assuming that the initial conditions are zero. vs + − + − v2 + − v1 L1 L2 (a) (a) is L1 L2 (b) i1 i2 Figure 6.77 For Prob. 6.52. 6.53 In the circuit of Fig. 6.78, let is(t) = 6e−2t mA, t ≥ 0 and i1(0) = 4 mA. Find: (a) i2(0), (b) i1(t) and i2(t), t 0, (c) v1(t) and v2(t), t 0, (d) the energy in each inductor at t = 0.5 s. is(t) i1 i2 20 mH 30 mH 10 mH + − v1 v2 + − Figure 6.78 For Prob. 6.53. 6.54 The inductors in Fig. 6.79 are initially charged and are connected to the black box at t = 0. If i1(0) = 4 A, i2(0) = −2 A, and v(t) = 50e−200t mV, t ≥ 0, find: (a) the energy initially stored in each inductor, (b) the total energy delivered to the black box from t = 0 to t = ∞, (c) i1(t) and i2(t), t ≥ 0, (d) i(t), t ≥ 0. i1 i2 20 H 5 H v + − Black box i(t) t = 0 Figure 6.79 For Prob. 6.54. 6.55 Find i and v in the circuit of Fig. 6.80 assuming that i(0) = 0 = v(0). 40 mH 60 mH 20 mH 16 mH v + − 12 sin 4t mV + − i Figure 6.80 For Prob. 6.55. Section 6.6 Applications 6.56 An op amp integrator has R = 50 k and C = 0.04 µF. If the input voltage is vi = 10 sin 50t mV, obtain the output voltage. 6.57 A 10-V dc voltage is applied to an integrator with R = 50 k, C = 100 µF at t = 0. How long will it take for the op amp to saturate if the saturation voltages are +12 V and −12 V? Assume that the initial capacitor voltage was zero. 6.58 An op amp integrator with R = 4 M and C = 1 µF has the input waveform shown in Fig. 6.81. Plot the output waveform. vi (mV) 0 20 10 –10 –20 3 4 5 6 2 1 t (ms) Figure 6.81 For Prob. 6.58. 6.59 Using a single op amp, a capacitor, and resistors of 100 k or less, design a circuit to implement vo = −50 t 0 vi(t) dt Assume vo = 0 at t = 0.
  • 239. 234 PART 1 DC Circuits 6.60 Show how you would use a single op amp to generate vo = − t 0 (v1 + 4v2 + 10v3) dt If the integrating capacitor is C = 2 µF, obtain other component values. 6.61 At t = 1.5 ms, calculate vo due to the cascaded integrators in Fig. 6.82. Assume that the integrators are reset to 0 V at t = 0. 1 V 2 mF 10 kΩ 20 kΩ vo + − 0.5 mF + − + − + − Figure 6.82 For Prob. 6.61. 6.62 Show that the circuit in Fig. 6.83 is a noninverting integrator. vo vi + − + − R R R C R + − Figure 6.83 For Prob. 6.62. 6.63 The triangular waveform in Fig. 6.84(a) is applied to the input of the op amp differentiator in Fig. 6.84(b). Plot the output. (a) vi(t) 0 10 3 4 2 1 t (ms) –10 vo vi + − + − 20 kΩ 0.01 mF (b) + − Figure 6.84 For Prob. 6.63. 6.64 An op amp differentiator has R = 250 k and C = 10 µF. The input voltage is a ramp r(t) = 12t mV. Find the output voltage. 6.65 A voltage waveform has the following characteristics: a positive slope of 20 V/s for 5 ms followed by a negative slope of 10 V/s for 10 ms. If the waveform is applied to a differentiator with R = 50 k, C = 10 µF, sketch the output voltage waveform. 6.66 ∗ The output vo of the op amp circuit of Fig. 6.85(a) is shown in Fig. 6.85(b). Let Ri = Rf = 1 M and C = 1 µF. Determine the input voltage waveform and sketch it. (b) (a) 0 4 3 4 2 1 t (ms) −4 vo vi vo Ri C Rf + − + − + − Figure 6.85 For Prob. 6.66. 6.67 Design an analog computer to simulate d2 vo dt2 + 2 dvo dt + vo = 10 sin 2t where v0(0) = 2 and v 0(0) = 0.
  • 240. CHAPTER 6 Capacitors and Inductors 235 6.68 Design an analog computer to solve the differential equation di(t) dt + 3i(t) = 2 t 0 and assume that i(0) = 1 mA. 6.69 Figure 6.86 presents an analog computer designed to solve a differential equation. Assuming f (t) is known, set up the equation for f (t). vo(t) −f(t) 1 mF 1 mF 1 MΩ 1 MΩ 1 MΩ 100 kΩ 200 kΩ 500 kΩ 100 kΩ + − + − + − + − Figure 6.86 For Prob. 6.69. COMPREHENSIVE PROBLEMS 6.70 Your laboratory has available a large number of 10-µF capacitors rated at 300 V. To design a capacitor bank of 40-µF rated at 600 V, how many 10-µF capacitors are needed and how would you connect them? 6.71 When a capacitor is connected to a dc source, its voltage rises from 20 V to 36 V in 4 µs with an average charging current of 0.6 A. Determine the value of the capacitance. 6.72 A square-wave generator produces the voltage waveform shown in Fig. 6.87(a). What kind of a circuit component is needed to convert the voltage waveform to the triangular current waveform shown in Fig. 6.87(b)? Calculate the value of the component, assuming that it is initially uncharged. v (V) 0 5 −5 3 4 2 1 t (ms) (a) (b) i (A) 4 3 4 2 1 0 t (ms) Figure 6.87 For Prob. 6.72. 6.73 In an electric power plant substation, a capacitor bank is made of 10 capacitor strings connected in parallel. Each string consists of eight 1000-µF capacitors connected in series, with each capacitor charged to 100 V. (a) Calculate the total capacitance of the bank. (b) Determine the total energy stored in the bank.
  • 241. 237 C H A P T E R FIRST-ORDER CIRCUITS 7 I often say that when you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of a science, whatever the matter may be. —Lord Kelvin Enhancing Your Career Careers in Computer Engineering Electrical engineer- ing education has gone through drastic changes in recent decades. Most departments have come to be known as De- partment of Electrical and Computer Engineering, empha- sizing the rapid changes due to computers. Computers oc- cupy a prominent place in modern society and education. They have become commonplace and are helping to change the face of research, development, production, business, and entertainment. The scientist, engineer, doctor, attor- ney, teacher, airline pilot, businessperson—almost anyone benefits from a computer’s abilities to store large amounts of information and to process that information in very short periods of time. The internet, a computer communication network, is becoming essential in business, education, and library science. Computer usage is growing by leaps and bounds. Three major disciplines study computer systems: computer science, computer engineering, and information management science. Computer engineering has grown so fast and wide that it is divorcing itself from electrical en- gineering. But, in many schools of engineering, computer engineering is still an integral part of electrical engineering. An education in computer engineering should provide breadth in software, hardware design, and basic modeling techniques. It should include courses in data structures, dig- ital systems, computer architecture, microprocessors, inter- facing, software engineering, and operating systems. Elec- trical engineers who specialize in computer engineering find Computer design of very large scale integrated (VLSI) circuits. Source: M. E. Hazen, Fundamentals of DC and AC Circuits, Philadelphia: Saunders, 1990, p. 20A4. jobs in computer industries and in numerous fields where computers are being used. Companies that produce soft- ware are growing rapidly in number and size and providing employment for those who are skilled in programming. An excellent way to advance one’s knowledge of computers is to join the IEEE Computer Society, which sponsors diverse magazines, journals, and conferences.
  • 242. 238 PART 1 DC Circuits 7.1 INTRODUCTION Now that we have considered the three passive elements (resistors, ca- pacitors, and inductors) and one active element (the op amp) individually, we are prepared to consider circuits that contain various combinations of two or three of the passive elements. In this chapter, we shall examine two types of simple circuits: a circuit comprising a resistor and capaci- tor and a circuit comprising a resistor and an inductor. These are called RC and RL circuits, respectively. As simple as these circuits are, they find continual applications in electronics, communications, and control systems, as we shall see. We carry out the analysis of RC and RL circuits by applying Kirch- hoff’s laws, as we did for resistive circuits. The only difference is that applying Kirchhoff’s laws to purely resistive circuits results in algebraic equations, while applying the laws to RC and RL circuits produces dif- ferential equations, which are more difficult to solve than algebraic equa- tions. The differential equations resulting from analyzing RC and RL circuits are of the first order. Hence, the circuits are collectively known as first-order circuits. A first-order circuit is characterized by a first-order differential equation. In addition to there being two types of first-order circuits (RC and RL), there are two ways to excite the circuits. The first way is by initial conditions of the storage elements in the circuits. In these so- called source-free circuits, we assume that energy is initially stored in the capacitive or inductive element. The energy causes current to flow in the circuit and is gradually dissipated in the resistors. Although source- free circuits are by definition free of independent sources, they may have dependent sources. The second way of exciting first-order circuits is by independent sources. In this chapter, the independent sources we will consider are dc sources. (In later chapters, we shall consider sinusoidal and exponential sources.) The two types of first-order circuits and the two ways of exciting them add up to the four possible situations we will study in this chapter. Finally, we consider four typical applications of RC and RL cir- cuits: delay and relay circuits, a photoflash unit, and an automobile igni- tion circuit. 7.2 THE SOURCE-FREE RC CIRCUIT A source-free RC circuit occurs when its dc source is suddenly discon- nected. The energy already stored in the capacitor is released to the resistors. v + − iR iC R C Figure 7.1 A source-free RC circuit. A circuit response is the manner in which the circuit reacts to an excitation. Consider a series combination of a resistor and an initially charged capacitor, as shown in Fig. 7.1. (The resistor and capacitor may be the equivalent resistance and equivalent capacitance of combinations of re- sistors and capacitors.) Our objective is to determine the circuit response, which, for pedagogic reasons, we assume to be the voltage v(t) across
  • 243. CHAPTER 7 First-Order Circuits 239 the capacitor. Since the capacitor is initially charged, we can assume that at time t = 0, the initial voltage is v(0) = V0 (7.1) with the corresponding value of the energy stored as w(0) = 1 2 CV 2 0 (7.2) Applying KCL at the top node of the circuit in Fig. 7.1, iC + iR = 0 (7.3) By definition, iC = C dv/dt and iR = v/R. Thus, C dv dt + v R = 0 (7.4a) or dv dt + v RC = 0 (7.4b) This is a first-order differential equation, since only the first derivative of v is involved. To solve it, we rearrange the terms as dv v = − 1 RC dt (7.5) Integrating both sides, we get ln v = − t RC + ln A where ln A is the integration constant. Thus, ln v A = − t RC (7.6) Taking powers of e produces v(t) = Ae−t/RC But from the initial conditions, v(0) = A = V0. Hence, v(t) = V0e−t/RC (7.7) This shows that the voltage response of the RC circuit is an exponential decay of the initial voltage. Since the response is due to the initial energy stored and the physical characteristics of the circuit and not due to some external voltage or current source, it is called the natural response of the circuit. The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation. The natural response depends on the nature of the circuit alone, with no external sources. In fact, the circuit has a response only because of the energy initially stored in the capacitor. The natural response is illustrated graphically in Fig. 7.2. Note that at t = 0, we have the correct initial condition as in Eq. (7.1). As t increases, the voltage decreases toward zero. The rapidity with which the voltage decreases is expressed in terms of the time constant, denoted by the lower case Greek letter tau, τ.
  • 244. 240 PART 1 DC Circuits The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.8 percent of its initial value.1 Voe−t ⁄ t t t 0.368Vo Vo v 0 Figure 7.2 The voltage response of the RC circuit. t 2t 3t 4t 5t t (s) 0 v Vo 0.37 0.25 0.75 1.0 0.50 Tangent at t = 0 Figure7.3 Graphical determination of the time constant τ from the response curve. This implies that at t = τ, Eq. (7.7) becomes V0e−τ/RC = V0e−1 = 0.368V0 or τ = RC (7.8) In terms of the time constant, Eq. (7.7) can be written as v(t) = V0e−t/τ (7.9) With a calculator it is easy to show that the value of v(t)/V0 is as shown in Table 7.1. It is evident from Table 7.1 that the voltage v(t) is less than 1 percent of V0 after 5τ (five time constants). Thus, it is customary to assume that the capacitor is fully discharged (or charged) after five time constants. In other words, it takes 5τ for the circuit to reach its final state or steady state when no changes take place with time. Notice that for every time interval of τ, the voltage is reduced by 36.8 percent of its previous value, v(t + τ) = v(t)/e = 0.368v(t), regardless of the value of t. TABLE 7.1 Values of v(t)/V0 = e−t/τ . t v(t)/V0 τ 0.36788 2τ 0.13534 3τ 0.04979 4τ 0.01832 5τ 0.00674 Observe from Eq. (7.8) that the smaller the time constant, the more rapidly the voltage decreases, that is, the faster the response. This is illustrated in Fig. 7.4. A circuit with a small time constant gives a fast response in that it reaches the steady state (or final state) quickly due to quick dissipation of energy stored, whereas a circuit with a large time 1The time constant may be viewed from another perspective. Evaluating the derivative of v(t) in Eq. (7.7) at t = 0, we obtain d dt v V0 t = 0 = − 1 τ e−t/τ t=0 = − 1 τ Thus the time constant is the initial rate of decay, or the time taken for v/V0 to decay from unity to zero, assuming a constant rate of decay. This initial slope interpretation of the time constant is often used in the laboratory to find τ graphically from the response curve displayed on an oscilloscope. To find τ from the response curve, draw the tangent to the curve, as shown in Fig. 7.3. The tangent intercepts with the time axis at t = τ.
  • 245. CHAPTER 7 First-Order Circuits 241 0 t 1 3 4 5 1 2 v Vo e−t ⁄t = t = 0.5 t = 1 t = 2 Figure7.4 Plot of v/V0 = e−t/τ for various values of the time constant. constant gives a slow response because it takes longer to reach steady state. At any rate, whether the time constant is small or large, the circuit reaches steady state in five time constants. With the voltage v(t) in Eq. (7.9), we can find the current iR(t), iR(t) = v(t) R = V0 R e−t/τ (7.10) The power dissipated in the resistor is p(t) = viR = V 2 0 R e−2t/τ (7.11) The energy absorbed by the resistor up to time t is wR(t) = t 0 p dt = t 0 V 2 0 R e−2t/τ dt = − τV 2 0 2R e−2t/τ t 0 = 1 2 CV 2 0 (1 − e−2t/τ ), τ = RC (7.12) Notice that as t → ∞, wR(∞) → 1 2 CV 2 0 , which is the same as wC(0), the energy initially stored in the capacitor. The energy that was initially stored in the capacitor is eventually dissipated in the resistor. In summary: The Key to Working with a Source-free RC Circuit is Finding: 1. The initial voltage v(0) = V0 across the capacitor. 2. The time constant τ. Thetimeconstantisthesameregardlessofwhat the output is defined to be. With these two items, we obtain the response as the capacitor voltage vC(t) = v(t) = v(0)e−t/τ . Once the capacitor voltage is first obtained, other variables (capacitor current iC, resistor voltage vR, and resistor current iR) can be determined. In finding the time constant τ = RC, R is often the Thevenin equivalent resistance at the terminals of the capacitor; that is, we take out the capacitor C and find R = RTh at its terminals. When a circuit contains a single capacitor and several resistors and dependent sources, the Thevenin equivalent can be found at the termi- nals of the capacitor to form a simple RC circuit. Also, onecanuseThevenin’stheoremwhensev- eral capacitors can be combined to form a single equivalent capacitor.
  • 246. 242 PART 1 DC Circuits E X A M P L E 7 . 1 In Fig. 7.5, let vC(0) = 15 V. Find vC, vx, and ix for t 0. 5 Ω 8 Ω 12 Ω vC vx ix + − + − 0.1 F Figure7.5 For Example 7.1. Solution: We first need to make the circuit in Fig. 7.5 conform with the standard RC circuit in Fig. 7.1. We find the equivalent resistance or the Thevenin resistance at the capacitor terminals. Our objective is always to first obtain capacitor voltage vC. From this, we can determine vx and ix. The 8- and 12- resistors in series can be combined to give a 20- resistor. This 20- resistor in parallel with the 5- resistor can be combined so that the equivalent resistance is Req = 20 × 5 20 + 5 = 4 Hence, the equivalent circuit is as shown in Fig. 7.6, which is analogous to Fig. 7.1. The time constant is τ = ReqC = 4(0.1) = 0.4 s Thus, v = v(0)e−t/τ = 15e−t/0.4 V, vC = v = 15e−2.5t V From Fig. 7.5, we can use voltage division to get vx; so vx = 12 12 + 8 v = 0.6(15e−2.5t ) = 9e−2.5t V Finally, ix = vx 12 = 0.75e−2.5t A v + − Req 0.1 F Figure7.6 Equivalent circuit for the circuit in Fig. 7.5. P R A C T I C E P R O B L E M 7 . 1 Refer to the circuit in Fig. 7.7. Let vC(0) = 30 V. Determine vC, vx, and io for t ≥ 0. 12 Ω 8 Ω vC F 6 Ω io + − vx + − 1 3 Figure7.7 For Practice Prob. 7.1. Answer: 30e−0.25t V, 10e−0.25t V, −2.5e−0.25t A. E X A M P L E 7 . 2 The switch in the circuit in Fig. 7.8 has been closed for a long time, and it is opened at t = 0. Find v(t) for t ≥ 0. Calculate the initial energy stored in the capacitor. 3 Ω 20 V + − v 9 Ω t = 0 1 Ω 20 mF + − Figure7.8 For Example 7.2. Solution: For t 0, the switch is closed; the capacitor is an open circuit to dc, as represented in Fig. 7.9(a). Using voltage division
  • 247. CHAPTER 7 First-Order Circuits 243 vC(t) = 9 9 + 3 (20) = 15 V, t 0 Since the voltage across a capacitor cannot change instantaneously, the voltage across the capacitor at t = 0− is the same at t = 0, or vC(0) = V0 = 15 V 9 Ω 1 Ω vC(0) 3 Ω 9 Ω 1 Ω + − + − 20 V (a) (b) + − Vo = 15 V 20 mF Figure 7.9 For Example 7.2: (a) t 0, (b) t 0. For t 0, the switch is opened, and we have the RC circuit shown in Fig. 7.9(b). [Notice that the RC circuit in Fig. 7.9(b) is source free; the independent source in Fig. 7.8 is needed to provide V0 or the initial energy in the capacitor.] The 1- and 9- resistors in series give Req = 1 + 9 = 10 The time constant is τ = ReqC = 10 × 20 × 10−3 = 0.2 s Thus, the voltage across the capacitor for t ≥ 0 is v(t) = vC(0)e−t/τ = 15e−t/0.2 V or v(t) = 15e−5t V The initial energy stored in the capacitor is wC(0) = 1 2 Cv2 C(0) = 1 2 × 20 × 10−3 × 152 = 2.25 J P R A C T I C E P R O B L E M 7 . 2 If the switch in Fig. 7.10 opens at t = 0, find v(t) for t ≥ 0 and wC(0). Answer: 8e−2t V, 5.33 J. 6 Ω + − 24 V + − v 12 Ω 4 Ω t = 0 F 1 6 Figure7.10 For Practice Prob. 7.2. 7.3 THE SOURCE-FREE RL CIRCUIT vL + − R L i vR + − Figure 7.11 A source- free RL circuit. Consider the series connection of a resistor and an inductor, as shown in Fig. 7.11. Our goal is to determine the circuit response, which we will assume to be the current i(t) through the inductor. We select the inductor current as the response in order to take advantage of the idea that the inductor current cannot change instantaneously. At t = 0, we assume that the inductor has an initial current I0, or i(0) = I0 (7.13) with the corresponding energy stored in the inductor as w(0) = 1 2 LI2 0 (7.14)
  • 248. 244 PART 1 DC Circuits Applying KVL around the loop in Fig. 7.11, vL + vR = 0 (7.15) But vL = L di/dt and vR = iR. Thus, L di dt + Ri = 0 or di dt + R L i = 0 (7.16) Rearranging terms and integrating gives i(t) I0 di i = − t 0 R L dt ln i i(t) I0 = − Rt L t 0 ⇒ ln i(t) − ln I0 = − Rt L + 0 or ln i(t) I0 = − Rt L (7.17) Taking the powers of e, we have i(t) = I0e−Rt/L (7.18) This shows that the natural response of the RL circuit is an exponential decay of the initial current. The current response is shown in Fig. 7.12. It is evident from Eq. (7.18) that the time constant for the RL circuit is τ = L R (7.19) with τ again having the unit of seconds. Thus, Eq. (7.18) may be written as i(t) = I0e−t/τ (7.20) Tangent at t = 0 Ioe−t ⁄t t t 0.368Io Io i(t) 0 Figure 7.12 The current response of the RL circuit. The smaller the time constant τ of a circuit, the faster the rate of decay of the response. The larger the time constant, the slower the rate of decayoftheresponse. Atanyrate, theresponse decays to less than 1 percent of its initial value (i.e., reaches steady state) after 5τ. Figure 7.12 shows an initial slope interpretation may be given to τ. With the current in Eq. (7.20), we can find the voltage across the resistor as vR(t) = iR = I0Re−t/τ (7.21) The power dissipated in the resistor is p = vRi = I2 0 Re−2t/τ (7.22) The energy absorbed by the resistor is wR(t) = t 0 p dt = t 0 I2 0 Re−2t/τ dt = − 1 2 τI2 0 Re−2t/τ t 0 , τ = L R or wR(t) = 1 2 LI2 0 (1 − e−2t/τ ) (7.23) Note that as t → ∞, wR(∞) → 1 2 LI2 0 , which is the same as wL(0), the initial energy stored in the inductor as in Eq. (7.14). Again, the energy initially stored in the inductor is eventually dissipated in the resistor.
  • 249. CHAPTER 7 First-Order Circuits 245 In summary: The Key to Working with a Source-free RL Circuit is to Find: 1. The initial current i(0) = I0 through the inductor. 2. The time constant τ of the circuit. With the two items, we obtain the response as the inductor current iL(t) = i(t) = i(0)e−t/τ . Once we determine the inductor current iL, other vari- ables (inductor voltage vL, resistor voltage vR, and resistor current iR) can be obtained. Note that in general, R in Eq. (7.19) is the Thevenin resistance at the terminals of the inductor. When a circuit has a single inductor and several resistors and dependent sources, the Thevenin equivalent can be found at the terminals of the inductor to form a simple RL circuit. Also, one can use Thevenin’s theorem when several induc- torscanbecombinedtoformasingleequivalent inductor. E X A M P L E 7 . 3 Assuming that i(0) = 10 A, calculate i(t) and ix(t) in the circuit in Fig. 7.13. 2 Ω 4 Ω 0.5 H + − i 3i ix Figure7.13 For Example 7.3. Solution: There are two ways we can solve this problem. One way is to obtain the equivalent resistance at the inductor terminals and then use Eq. (7.20). The other way is to start from scratch by using Kirchhoff’s voltage law. Whichever approach is taken, it is always better to first obtain the inductor current. METHOD 1 The equivalent resistance is the same as the Thevenin resistance at the inductor terminals. Because of the dependent source, we insert a voltage source with vo = 1 V at the inductor terminals a-b, as in Fig. 7.14(a). (We could also insert a 1-A current source at the ter- minals.) Applying KVL to the two loops results in 2(i1 − i2) + 1 = 0 ⇒ i1 − i2 = − 1 2 (7.3.1) 6i2 − 2i1 − 3i1 = 0 ⇒ i2 = 5 6 i1 (7.3.2) Substituting Eq. (7.3.2) into Eq. (7.3.1) gives 4 Ω 2 Ω vo = 1 V + − + − io i1 i2 3i (a) a b 4 Ω 2 Ω + − i1 i2 3i (b) 0.5 H Figure7.14 Solving the circuit in Fig. 7.13.
  • 250. 246 PART 1 DC Circuits i1 = −3 A, io = −i1 = 3 A Hence, Req = RTh = vo io = 1 3 The time constant is τ = L Req = 1 2 1 3 = 3 2 s Thus, the current through the inductor is i(t) = i(0)e−t/τ = 10e−(2/3)t A, t 0 METHOD 2 We may directly apply KVL to the circuit as in Fig. 7.14(b). For loop 1, 1 2 di1 dt + 2(i1 − i2) = 0 or di1 dt + 4i1 − 4i2 = 0 (7.3.3) For loop 2, 6i2 − 2i1 − 3i1 = 0 ⇒ i2 = 5 6 i1 (7.3.4) Substituting Eq. (7.3.4) into Eq. (7.3.3) gives di1 dt + 2 3 i1 = 0 Rearranging terms, di1 i1 = − 2 3 dt Since i1 = i, we may replace i1 with i and integrate: ln i i(t) i(0) = − 2 3 t t 0 or ln i(t) i(0) = − 2 3 t Taking the powers of e, we finally obtain i(t) = i(0)e−(2/3)t = 10e−(2/3)t A, t 0 which is the same as by Method 1. The voltage across the inductor is v = L di dt = 0.5(10) − 2 3 e−(2/3)t = − 10 3 e−(2/3)t V Since the inductor and the 2- resistor are in parallel, ix(t) = v 2 = −1.667e−(2/3)t A, t 0
  • 251. CHAPTER 7 First-Order Circuits 247 P R A C T I C E P R O B L E M 7 . 3 Find i and vx in the circuit in Fig. 7.15. Let i(0) = 5 A. 1 Ω 5 Ω 3 Ω + − 2vx H i + − vx 1 6 Figure7.15 For Practice Prob. 7.3. Answer: 5e−53t A, −15e−53t V. E X A M P L E 7 . 4 The switch in the circuit of Fig. 7.16 has been closed for a long time. At t = 0, the switch is opened. Calculate i(t) for t 0. 2 Ω 4 Ω + − 40 V 16 Ω 12 Ω 2 H t = 0 i(t) Figure7.16 For Example 7.4. Solution: When t 0, the switch is closed, and the inductor acts as a short circuit to dc. The 16- resistor is short-circuited; the resulting circuit is shown in Fig. 7.17(a). To get i1 in Fig. 7.17(a), we combine the 4- and 12- resistors in parallel to get 4 × 12 4 + 12 = 3 Hence, i1 = 40 2 + 3 = 8 A We obtain i(t) from i1 in Fig. 7.17(a) using current division, by writing i(t) = 12 12 + 4 i1 = 6 A, t 0 Since the current through an inductor cannot change instantaneously, i(0) = i(0− ) = 6 A When t 0, the switch is open and the voltage source is discon- nected. We now have the RL circuit in Fig. 7.17(b). Combining the re- sistors, we have Req = (12 + 4) 16 = 8 The time constant is τ = L Req = 2 8 = 1 4 s Thus, i(t) = i(0)e−t/τ = 6e−4t A 4 Ω 12 Ω 2 Ω + − i1 2 H i(t) 40 V i(t) (a) 16 Ω 12 Ω 4 Ω (b) Figure 7.17 Solving the circuit of Fig. 7.16: (a) for t 0, (b) for t 0.
  • 252. 248 PART 1 DC Circuits P R A C T I C E P R O B L E M 7 . 4 For the circuit in Fig. 7.18, find i(t) for t 0. 5 Ω 5 A 12 Ω 8 Ω 2 H t = 0 i(t) Figure7.18 For Practice Prob. 7.4. Answer: 2e−2t A, t 0. E X A M P L E 7 . 5 In the circuit shown in Fig. 7.19, find io, vo, and i for all time, assuming that the switch was open for a long time. 10 V 6 Ω 2 H t = 0 i io + − vo 3 Ω 2 Ω + − Figure7.19 For Example 7.5. Solution: Itisbettertofirstfindtheinductorcurrenti andthenobtainotherquantities from it. 2 Ω 3 Ω + − 10 V 6 Ω i io + − vo + − vo (a) (b) 6 Ω 3 Ω 2 H i io vL + − Figure 7.20 The circuit in Fig. 7.19 for: (a) t 0, (b) t 0. For t 0, the switch is open. Since the inductor acts like a short circuit to dc, the 6- resistor is short-circuited, so that we have the circuit shown in Fig. 7.20(a). Hence, io = 0, and i(t) = 10 2 + 3 = 2 A, t 0 vo(t) = 3i(t) = 6 V, t 0 Thus, i(0) = 2. For t 0, the switch is closed, so that the voltage source is short- circuited. We now have a source-free RL circuit as shown in Fig. 7.20(b). At the inductor terminals, RTh = 3 6 = 2 so that the time constant is τ = L RTh = 1 s Hence, i(t) = i(0)e−t/τ = 2e−t A, t 0 Since the inductor is in parallel with the 6- and 3- resistors, vo(t) = −vL = −L di dt = −2(−2e−t ) = 4e−t V, t 0 and io(t) = vL 6 = − 2 3 e−t A, t 0
  • 253. CHAPTER 7 First-Order Circuits 249 Thus, for all time, io(t) =    0 A, t 0 − 2 3 e−t A, t 0 , vo(t) = 6 V, t 0 4e−t V, t 0 i(t) = 2 A, t 0 2e−t A, t ≥ 0 We notice that the inductor current is continuous at t = 0, while the current through the 6- resistor drops from 0 to −2/3 at t = 0, and the voltage across the 3- resistor drops from 6 to 4 at t = 0. We also notice that the time constant is the same regardless of what the output is defined to be. Figure 7.21 plots i and io. t 2 i(t) 2 3 − io(t) Figure7.21 A plot of i and i0. P R A C T I C E P R O B L E M 7 . 5 Determine i, io, and vo for all t in the circuit shown in Fig. 7.22. Assume that the switch was closed for a long time. 1 H 4 Ω 2 Ω 3 Ω 6 A i t = 0 io vo + − Figure7.22 For Practice Prob. 7.5. Answer: i = 4 A, t 0 4e−2t A, t ≥ 0 , io = 2 A, t 0 −(4/3)e−2t A, t 0 , vo = 4 V, t 0 −(8/3)e−2t V, t 0 7.4 SINGULARITY FUNCTIONS Before going on with the second half of this chapter, we need to digress and consider some mathematical concepts that will aid our understanding of transient analysis. A basic understanding of singularity functions will help us make sense of the response of first-order circuits to a sudden application of an independent dc voltage or current source. Singularity functions (also called switching functions) are very use- fulincircuitanalysis. Theyserveasgoodapproximationstotheswitching signals that arise in circuits with switching operations. They are helpful in the neat, compact description of some circuit phenomena, especially the step response of RC or RL circuits to be discussed in the next sections. By definition, Singularity functions are functions that either are discontinuous or have discontinuous derivatives. The three most widely used singularity functions in circuit analysis are the unit step, the unit impulse, and the unit ramp functions.
  • 254. 250 PART 1 DC Circuits The unit step function u(t) is 0 for negative values of t and 1 for positive values of t. In mathematical terms, u(t) = 0, t 0 1, t 0 (7.24) 0 t 1 u(t) Figure 7.23 The unit step function. The unit step function is undefined at t = 0, where it changes abruptly from 0 to 1. It is dimensionless, like other mathematical functions such as sine and cosine. Figure 7.23 depicts the unit step function. If the abrupt change occurs at t = t0 (where t0 0) instead of t = 0, the unit step function becomes u(t − t0) = 0, t t0 1, t t0 (7.25) which is the same as saying that u(t) is delayed by t0 seconds, as shown in Fig. 7.24(a). To get Eq. (7.25) from Eq. (7.24), we simply replace every t by t − t0. If the change is at t = −t0, the unit step function becomes u(t + t0) = 0, t −t0 1, t −t0 (7.26) meaning that u(t) is advanced by t0 seconds, as shown in Fig. 7.24(b). 0 t 1 u(t − to) to (a) 0 t u(t + to) −to (b) 1 Figure7.24 (a) The unit step function delayed by t0, (b) the unit step advanced by t0. Alternatively,wemayderiveEqs.(7.25)and(7.26) from Eq. (7.24) by writing u[f(t)] = 1, f(t) 0, where f(t) may be t − t0 or t + t0. We use the step function to represent an abrupt change in voltage or current, like the changes that occur in the circuits of control systems and digital computers. For example, the voltage v(t) = 0, t t0 V0, t t0 (7.27) may be expressed in terms of the unit step function as v(t) = V0u(t − t0) (7.28) If we let t0 = 0, then v(t) is simply the step voltage V0u(t). A voltage source of V0u(t) is shown in Fig. 7.25(a); its equivalent circuit is shown in Fig. 7.25(b). It is evident in Fig. 7.25(b) that terminals a-b are short- circuited (v = 0) for t 0 and that v = V0 appears at the terminals for t 0. Similarly, a current source of I0u(t) is shown in Fig. 7.26(a), while its equivalent circuit is in Fig. 7.26(b). Notice that for t 0, there is an open circuit (i = 0), and that i = I0 flows for t 0. + − (a) Vou(t) + − (b) Vo b a b a t = 0 = Figure7.25 (a) Voltage source of V0u(t), (b) its equivalent circuit.
  • 255. CHAPTER 7 First-Order Circuits 251 (a) Iou(t) (b) Io b a b a t = 0 i = Figure7.26 (a) Current source of I0u(t), (b) its equivalent circuit. The derivative of the unit step function u(t) is the unit impulse function δ(t), which we write as δ(t) = d dt u(t) =    0, t 0 Undefined, t = 0 0, t 0 (7.29) The unit impulse function—also known as the delta function—is shown in Fig. 7.27. The unit impulse function δ(t) is zero everywhere except at t = 0, where it is undefined. Impulsive currents and voltages occur in electric circuits as a result of switching operations or impulsive sources. Although the unit impulse functionisnotphysicallyrealizable(justlikeidealsources, idealresistors, etc.), it is a very useful mathematical tool. The unit impulse may be regarded as an applied or resulting shock. It may be visualized as a very short duration pulse of unit area. This may be expressed mathematically as 0+ 0− δ(t) dt = 1 (7.30) where t = 0− denotes the time just before t = 0 and t = 0+ is the time just after t = 0. For this reason, it is customary to write 1 (denoting unit area) beside the arrow that is used to symbolize the unit impulse function, as in Fig. 7.27. The unit area is known as the strength of the impulse function. When an impulse function has a strength other than unity, the area of the impulse is equal to its strength. For example, an impulse function 10δ(t) has an area of 10. Figure 7.28 shows the impulse functions 5δ(t + 2), 10δ(t), and −4δ(t − 3). 0 t (1) d(t) Figure7.27 The unit impulse function. 5d(t + 2) 10d(t) −4d(t − 3) 1 0 2 3 t −1 −2 Figure7.28 Three impulse functions. To illustrate how the impulse function affects other functions, let us evaluate the integral b a f (t)δ(t − t0) dt (7.31) where a t0 b. Since δ(t − t0) = 0 except at t = t0, the integrand is
  • 256. 252 PART 1 DC Circuits zero except at t0. Thus, b a f (t)δ(t − t0) dt = b a f (t0)δ(t − t0) dt = f (t0) b a δ(t − t0) dt = f (t0) or b a f (t)δ(t − t0) dt = f (t0) (7.32) This shows that when a function is integrated with the impulse function, we obtain the value of the function at the point where the impulse occurs. This is a highly useful property of the impulse function known as the sampling or sifting property. The special case of Eq. (7.31) is for t0 = 0. Then Eq. (7.32) becomes 0+ 0− f (t)δ(t) dt = f (0) (7.33) Integrating the unit step function u(t) results in the unit ramp func- tion r(t); we write r(t) = t −∞ u(t) dt = tu(t) (7.34) or r(t) = 0, t ≤ 0 t, t ≥ 0 (7.35) The unit ramp function is zero for negative values of t and has a unit slope for positive values of t. Figure 7.29 shows the unit ramp function. In general, a ramp is a function that changes at a constant rate. 0 t 1 r(t) 1 Figure 7.29 The unit ramp function. The unit ramp function may be delayed or advanced as shown in Fig. 7.30. For the delayed unit ramp function, r(t − t0) = 0, t ≤ t0 t − t0, t ≥ t0 (7.36) and for the advanced unit ramp function, r(t + t0) = 0, t ≤ −t0 t − t0, t ≥ −t0 (7.37) (a) 0 t −to + 1 −to 1 r(t + to) r(t − to) (b) 0 t to + 1 to 1 Figure7.30 The unit ramp function: (a) delayed by t0, (b) advanced by t0. We should keep in mind that the three singularity functions (im- pulse, step, and ramp) are related by differentiation as δ(t) = du(t) dt , u(t) = dr(t) dt (7.38) or by integration as u(t) = t −∞ δ(t) dt, r(t) = t −∞ u(t) dt (7.39)
  • 257. CHAPTER 7 First-Order Circuits 253 Although there are many more singularity functions, we are only inter- ested in these three (the impulse function, the unit step function, and the ramp function) at this point. E X A M P L E 7 . 6 Express the voltage pulse in Fig. 7.31 in terms of the unit step. Calculate its derivative and sketch it. 0 t 10 v(t) 3 4 5 1 2 Figure7.31 For Example 7.6. Solution: The type of pulse in Fig. 7.31 is called the gate function. It may be re- garded as a step function that switches on at one value of t and switches off at another value of t. The gate function shown in Fig. 7.31 switches on at t = 2 s and switches off at t = 5 s. It consists of the sum of two unit step functions as shown in Fig. 7.32(a). From the figure, it is evident that v(t) = 10u(t − 2) − 10u(t − 5) = 10[u(t − 2) − u(t − 5)] Taking the derivative of this gives dv dt = 10[δ(t − 2) − δ(t − 5)] which is shown in Fig. 7.32(b). We can obtain Fig. 7.32(b) directly from Fig. 7.31 by simply observing that there is a sudden increase by 10 V at t = 2 s leading to 10δ(t − 2). At t = 5 s, there is a sudden decrease by 10 V leading to −10 V δ(t − 5). Gate functions are used along with switches to pass or block another signal. 0 t 2 1 10 10u(t − 2) −10u(t − 5) (a) 1 2 0 3 4 5 t 10 −10 + (b) 10 3 4 5 t 1 2 0 −10 dv dt Figure7.32 (a) Decomposition of the pulse in Fig. 7.31, (b) derivative of the pulse in Fig. 7.31.
  • 258. 254 PART 1 DC Circuits P R A C T I C E P R O B L E M 7 . 6 Express the current pulse in Fig. 7.33 in terms of the unit step. Find its integral and sketch it. Answer: 10[u(t)−2u(t −2)+u(t −4)], 10[r(t)−2r(t −2)+r(t −4)]. See Fig. 7.34. 0 t 10 −10 i(t) 2 4 Figure7.33 For Practice Prob. 7.6. 2 0 4 t 20 i dt ∫ Figure7.34 Integral of i(t) in Fig. 7.33. E X A M P L E 7 . 7 Express the sawtooth function shown in Fig. 7.35 in terms of singularity functions. 0 t 10 v(t) 2 Figure 7.35 For Example 7.7. Solution: There are three ways of solving this problem. The first method is by mere observation of the given function, while the other methods involve some graphical manipulations of the function. METHOD 1 By looking at the sketch of v(t) in Fig. 7.35, it is not hard to notice that the given function v(t) is a combination of singularity functions. So we let v(t) = v1(t) + v2(t) + · · · (7.7.1) The function v1(t) is the ramp function of slope 5, shown in Fig. 7.36(a); that is, v1(t) = 5r(t) (7.7.2) 0 t 10 v1(t) 2 0 t 10 v1 + v2 2 0 t −10 v2(t) 2 + (a) (b) (c) = Figure7.36 Partial decomposition of v(t) in Fig. 7.35.
  • 259. CHAPTER 7 First-Order Circuits 255 Since v1(t) goes to infinity, we need another function at t = 2 s in order to get v(t). We let this function be v2, which is a ramp function of slope −5, as shown in Fig. 7.36(b); that is, v2(t) = −5r(t − 2) (7.7.3) Adding v1 and v2 gives us the signal in Fig. 7.36(c). Obviously, this is not the same as v(t) in Fig. 7.35. But the difference is simply a constant 10 units for t 2 s. By adding a third signal v3, where v3 = −10u(t − 2) (7.7.4) we get v(t), as shown in Fig. 7.37. Substituting Eqs. (7.7.2) through (7.7.4) into Eq. (7.7.1) gives v(t) = 5r(t) − 5r(t − 2) − 10u(t − 2) 0 t 10 v1 + v2 2 + (c) (a) = 0 t 10 v(t) 2 0 t −10 v3(t) 2 (b) Figure7.37 Complete decomposition of v(t) in Fig. 7.35. METHOD 2 A close observation of Fig. 7.35 reveals that v(t) is a mul- tiplication of two functions: a ramp function and a gate function. Thus, v(t) = 5t[u(t) − u(t − 2)] = 5tu(t) − 5tu(t − 2) = 5r(t) − 5(t − 2 + 2)u(t − 2) = 5r(t) − 5(t − 2)u(t − 2) − 10u(t − 2) = 5r(t) − 5r(t − 2) − 10u(t − 2) the same as before. METHOD 3 This method is similar to Method 2. We observe from Fig. 7.35 that v(t) is a multiplication of a ramp function and a unit step function, as shown in Fig. 7.38. Thus, v(t) = 5r(t)u(−t + 2) If we replace u(−t) by 1 − u(t), then we can replace u(−t + 2) by 1 − u(t − 2). Hence, v(t) = 5r(t)[1 − u(t − 2)] which can be simplified as in Method 2 to get the same result.
  • 260. 256 PART 1 DC Circuits 0 t 10 5r(t) 2 × 0 t u(−t + 2) 2 1 Figure7.38 Decomposition of v(t) in Fig. 7.35. P R A C T I C E P R O B L E M 7 . 7 Refer to Fig. 7.39. Express i(t) in terms of singularity functions. i(t) (A) 1 0 2 3 t (s) 2 −2 Figure7.39 For Practice Prob. 7.7. Answer: 2u(t) − 2r(t) + 4r(t − 2) − 2r(t − 3). E X A M P L E 7 . 8 Given the signal g(t) =    3, t 0 −2, 0 t 1 2t − 4, t 1 express g(t) in terms of step and ramp functions. Solution: The signal g(t) may be regarded as the sum of three functions specified within the three intervals t 0, 0 t 1, and t 1. For t 0, g(t) may be regarded as 3 multiplied by u(−t), where u(−t) = 1 for t 0 and 0 for t 0. Within the time interval 0 t 1, the function may be considered as −2 multiplied by a gated function [u(t) − u(t − 1)]. For t 1, the function may be regarded as 2t − 4 multiplied by the unit step function u(t − 1). Thus, g(t) = 3u(−t) − 2[u(t) − u(t − 1)] + (2t − 4)u(t − 1) = 3u(−t) − 2u(t) + (2t − 4 + 2)u(t − 1) = 3u(−t) − 2u(t) + 2(t − 1)u(t − 1) = 3u(−t) − 2u(t) + 2r(t − 1) One may avoid the trouble of using u(−t) by replacing it with 1 − u(t). Then
  • 261. CHAPTER 7 First-Order Circuits 257 g(t) = 3[1 − u(t)] − 2u(t) + 2r(t − 1) = 3 − 5u(t) + 2r(t − 1) Alternatively, we may plot g(t) and apply Method 1 from Example 7.7. P R A C T I C E P R O B L E M 7 . 8 If h(t) =        0, t 0 4, 0 t 2 6 − t, 2 t 6 0, t 6 express h(t) in terms of the singularity functions. Answer: 4u(t) − r(t − 2) + r(t − 6). E X A M P L E 7 . 9 Evaluate the following integrals involving the impulse function: 10 0 (t2 + 4t − 2)δ(t − 2) dt ∞ −∞ (δ(t − 1)e−t cos t + δ(t + 1)e−t sin t)dt Solution: For the first integral, we apply the sifting property in Eq. (7.32). 10 0 (t2 + 4t − 2)δ(t − 2)dt = (t2 + 4t − 2)|t=2 = 4 + 8 − 2 = 10 Similarly, for the second integral, ∞ −∞ (δ(t − 1)e−t cos t + δ(t + 1)e−t sin t)dt = e−t cos t|t=1 + e−t sin t|t=−1 = e−1 cos 1 + e1 sin(−1) = 0.1988 − 2.2873 = −2.0885 P R A C T I C E P R O B L E M 7 . 9 Evaluate the following integrals: ∞ −∞ (t3 + 5t2 + 10)δ(t + 3) dt, 10 0 δ(t − π) cos 3t dt Answer: 28, −1. 7.5 STEP RESPONSE OF AN RC CIRCUIT When the dc source of an RC circuit is suddenly applied, the voltage or current source can be modeled as a step function, and the response is known as a step response.
  • 262. 258 PART 1 DC Circuits The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source. The step response is the response of the circuit due to a sudden application of a dc voltage or current source. R C t = 0 + − Vsu(t) Vs + − v (a) R C + − + − v (b) Figure 7.40 An RC circuit with voltage step input. Consider the RC circuit in Fig. 7.40(a) which can be replaced by the circuit in Fig. 7.40(b), where Vs is a constant, dc voltage source. Again, we select the capacitor voltage as the circuit response to be determined. We assume an initial voltage V0 on the capacitor, although this is not necessary for the step response. Since the voltage of a capacitor cannot change instantaneously, v(0− ) = v(0+ ) = V0 (7.40) where v(0− ) is the voltage across the capacitor just before switching and v(0+ ) is its voltage immediately after switching. Applying KCL, we have C dv dt + v − Vsu(t) R = 0 or dv dt + v RC = Vs RC u(t) (7.41) where v is the voltage across the capacitor. For t 0, Eq. (7.41) becomes dv dt + v RC = Vs RC (7.42) Rearranging terms gives dv dt = − v − Vs RC or dv v − Vs = − dt RC (7.43) Integrating both sides and introducing the initial conditions, ln(v − Vs) v(t) V0 = − t RC t 0 ln(v(t) − Vs) − ln(V0 − Vs) = − t RC + 0 or ln v − Vs V0 − Vs = − t RC (7.44) Taking the exponential of both sides v − Vs V0 − Vs = e−t/τ , τ = RC v − Vs = (V0 − Vs)e−t/τ or v(t) = Vs + (V0 − Vs)e−t/τ , t 0 (7.45)
  • 263. CHAPTER 7 First-Order Circuits 259 Thus, v(t) = V0, t 0 Vs + (V0 − Vs)e−t/τ , t 0 (7.46) This is known as the complete response of the RC circuit to a sudden application of a dc voltage source, assuming the capacitor is initially charged. The reason for the term “complete” will become evident a little later. Assuming that Vs V0, a plot of v(t) is shown in Fig. 7.41. 0 t Vs v(t) Vo Figure7.41 Response of an RC circuit with initially charged capacitor. If we assume that the capacitor is uncharged initially, we set V0 = 0 in Eq. (7.46) so that v(t) = 0, t 0 Vs(1 − e−t/τ ), t 0 (7.47) which can be written alternatively as v(t) = Vs(1 − e−t/τ )u(t) (7.48) This is the complete step response of the RC circuit when the capacitor is initially uncharged. The current through the capacitor is obtained from Eq. (7.47) using i(t) = C dv/dt. We get i(t) = C dv dt = C τ Vse−t/τ , τ = RC, t 0 or i(t) = Vs R e−t/τ u(t) (7.49) Figure 7.42 shows the plots of capacitor voltage v(t) and capacitor current i(t). 0 t Vs v(t) 0 t i(t) Vs R (a) (b) Figure7.42 Step response of an RC circuit with initially uncharged capacitor: (a) voltage response, (b) current response. Rather than going through the derivations above, there is a sys- tematic approach—or rather, a short-cut method—for finding the step response of an RC or RL circuit. Let us reexamine Eq. (7.45), which is more general than Eq. (7.48). It is evident that v(t) has two components. Thus, we may write v = vf + vn (7.50) where vf = Vs (7.51) and vn = (V0 − Vs)e−t/τ (7.52) We know that vn is the natural response of the circuit, as discussed in Section 7.2. Since this part of the response will decay to almost zero after five time constants, it is also called the transient response because it is a temporary response that will die out with time. Now, vf is known as the forced response because it is produced by the circuit when an external “force” is applied (a voltage source in this case). It represents what the circuit is forced to do by the input excitation. It is also known as the steady-state response, because it remains a long time after the circuit is excited.
  • 264. 260 PART 1 DC Circuits The natural response or transient response is the circuit’s temporary response that will die out with time. The forced response or steady-state response is the behavior of the circuit a long time after an external excitation is applied. The complete response of the circuit is the sum of the natural response and the forced response. Therefore, we may write Eq. (7.45) as v(t) = v(∞) + [v(0) − v(∞)]e−t/τ (7.53) where v(0) is the initial voltage at t = 0+ and v(∞) is the final or steady- state value. Thus, to find the step response of an RC circuit requires three things: This is the same as saying that the complete re- sponse is the sum of the transient response and the steady-state response. Once we know x(0), x(∞), and τ, almost all the circuit problems in this chapter can be solved using the formula x(t) = x(∞)+ [x(0) − x(∞)] e-t/τ 1. The initial capacitor voltage v(0). 2. The final capacitor voltage v(∞). 3. The time constant τ. We obtain item 1 from the given circuit for t 0 and items 2 and 3 from the circuit for t 0. Once these items are determined, we obtain the response using Eq. (7.53). This technique equally applies to RL circuits, as we shall see in the next section. Note that if the switch changes position at time t = t0 instead of at t = 0, there is a time delay in the response so that Eq. (7.53) becomes v(t) = v(∞) + [v(t0) − v(∞)]e−(t−t0)/τ (7.54) where v(t0) is the initial value at t = t+ 0 . Keep in mind that Eq. (7.53) or (7.54) applies only to step responses, that is, when the input excitation is constant. E X A M P L E 7 . 1 0 The switch in Fig. 7.43 has been in position A for a long time. At t = 0, the switch moves to B. Determine v(t) for t 0 and calculate its value at t = 1 s and 4 s. 3 kΩ 24 V 30 V v 5 kΩ 0.5 mF 4 kΩ + − + − t = 0 A B + − Figure7.43 For Example 7.10.
  • 265. CHAPTER 7 First-Order Circuits 261 Solution: For t 0, the switch is at position A. Since v is the same as the voltage across the 5-k resistor, the voltage across the capacitor just before t = 0 is obtained by voltage division as v(0− ) = 5 5 + 3 (24) = 15 V Using the fact that the capacitor voltage cannot change instantaneously, v(0) = v(0− ) = v(0+ ) = 15 V For t 0, the switch is in position B. The Thevenin resistance connected to the capacitor is RTh = 4 k, and the time constant is τ = RThC = 4 × 103 × 0.5 × 10−3 = 2 s Since the capacitor acts like an open circuit to dc at steady state, v(∞) = 30 V. Thus, v(t) = v(∞) + [v(0) − v(∞)]e−t/τ = 30 + (15 − 30)e−t/2 = (30 − 15e−0.5t ) V At t = 1, v(1) = 30 − 15e−0.5 = 20.902 V At t = 4, v(4) = 30 − 15e−2 = 27.97 V P R A C T I C E P R O B L E M 7 . 1 0 Find v(t) for t 0 in the circuit in Fig. 7.44. Assume the switch has been open for a long time and is closed at t = 0. Calculate v(t) at t = 0.5. 2 Ω 10 V 50 V v 6 Ω + − + − t = 0 F 1 3 + − Figure7.44 For Practice Prob. 7.10. Answer: −5 + 15e−2t V, 0.5182 V. E X A M P L E 7 . 1 1 In Fig. 7.45, the switch has been closed for a long time and is opened at t = 0. Find i and v for all time. 10 Ω 30u(t) V 10 V v 20 Ω + − + − i t = 0 F 1 4 + − Figure7.45 For Example 7.11.
  • 266. 262 PART 1 DC Circuits Solution: The resistor current i can be discontinuous at t = 0, while the capacitor voltage v cannot. Hence, it is always better to find v and then obtain i from v. By definition of the unit step function, 30u(t) = 0, t 0 30, t 0 For t 0, the switch is closed and 30u(t) = 0, so that the 30u(t) voltage source is replaced by a short circuit and should be regarded as contributing nothing to v. Since the switch has been closed for a long time, the capacitor voltage has reached steady state and the capacitor acts like an open circuit. Hence, the circuit becomes that shown in Fig. 7.46(a) for t 0. From this circuit we obtain v = 10 V, i = − v 10 = −1 A Since the capacitor voltage cannot change instantaneously, v(0) = v(0− ) = 10 V 10 Ω 10 V + − v 20 Ω + − i (a) 10 Ω 30 V + − v 20 Ω + − i (b) F 1 4 Figure 7.46 Solution of Example 7.11: (a) for t 0, (b) for t 0. For t 0, the switch is opened and the 10-V voltage source is disconnected from the circuit. The 30u(t) voltage source is now opera- tive, so the circuit becomes that shown in Fig. 7.46(b). After a long time, the circuit reaches steady state and the capacitor acts like an open circuit again. We obtain v(∞) by using voltage division, writing v(∞) = 20 20 + 10 (30) = 20 V The Thevenin resistance at the capacitor terminals is RTh = 10 20 = 10 × 20 30 = 20 3 and the time constant is τ = RThC = 20 3 · 1 4 = 5 3 s Thus, v(t) = v(∞) + [v(0) − v(∞)]e−t/τ = 20 + (10 − 20)e−(3/5)t = (20 − 10e−0.6t ) V To obtain i, we notice from Fig. 7.46(b) that i is the sum of the currents through the 20- resistor and the capacitor; that is, i = v 20 + C dv dt = 1 − 0.5e−0.6t + 0.25(−0.6)(−10)e−0.6t = (1 + e−0.6t ) A Notice from Fig. 7.46(b) that v + 10i = 30 is satisfied, as expected. Hence, v = 10 V, t 0 (20 − 10e−0.6t ) V, t ≥ 0
  • 267. CHAPTER 7 First-Order Circuits 263 i = −1 A, t 0 (1 + e−0.6t ) A, t 0 Notice that the capacitor voltage is continuous while the resistor current is not. P R A C T I C E P R O B L E M 7 . 1 1 The switch in Fig. 7.47 is closed at t = 0. Find i(t) and v(t) for all time. Note that u(−t) = 1 for t 0 and 0 for t 0. Also, u(−t) = 1 − u(t). 5 Ω + − 20u(−t) V 10 Ω 0.2 F 3 A v i t = 0 + − Figure7.47 For Practice Prob. 7.11. Answer: i(t) = 0, t 0 −2(1 + e−1.5t ) A, t 0 , v = 20 V, t 0 10(1 + e−1.5t ) V, t 0 7.6 STEP RESPONSE OF AN RL CIRCUIT Consider the RL circuit in Fig. 7.48(a), which may be replaced by the circuit in Fig. 7.48(b). Again, our goal is to find the inductor current i as the circuit response. Rather than apply Kirchhoff’s laws, we will use the simple technique in Eqs. (7.50) through (7.53). Let the response be the sum of the natural current and the forced current, i = in + if (7.55) We know that the natural response is always a decaying exponential, that is, in = Ae−t/τ , τ = L R (7.56) where A is a constant to be determined. R Vs t = 0 i i + − + − v(t) L (a) R Vsu(t) + − + − v(t) L (b) Figure 7.48 An RL circuit with a step input voltage. The forced response is the value of the current a long time after the switch in Fig. 7.48(a) is closed. We know that the natural response essentially dies out after five time constants. At that time, the inductor becomes a short circuit, and the voltage across it is zero. The entire source voltage Vs appears across R. Thus, the forced response is if = Vs R (7.57)
  • 268. 264 PART 1 DC Circuits Substituting Eqs. (7.56) and (7.57) into Eq. (7.55) gives i = Ae−t/τ + Vs R (7.58) We now determine the constant A from the initial value of i. Let I0 be the initial current through the inductor, which may come from a source other than Vs. Since the current through the inductor cannot change instantaneously, i(0+ ) = i(0− ) = I0 (7.59) Thus at t = 0, Eq. (7.58) becomes I0 = A + Vs R From this, we obtain A as A = I0 − Vs R Substituting for A in Eq. (7.58), we get i(t) = Vs R + I0 − Vs R e−t/τ (7.60) This is the complete response of the RL circuit. It is illustrated in Fig. 7.49. The response in Eq. (7.60) may be written as i(t) = i(∞) + [i(0) − i(∞)]e−t/τ (7.61) where i(0) and i(∞) are the initial and final values of i. Thus, to find the step response of an RL circuit requires three things: 0 t i(t) Vs R Io Figure7.49 Total response of the RL circuit with initial inductor current I0. 1. The initial inductor current i(0) at t = 0+ . 2. The final inductor current i(∞). 3. The time constant τ. We obtain item 1 from the given circuit for t 0 and items 2 and 3 from the circuit for t 0. Once these items are determined, we obtain the response using Eq. (7.61). Keep in mind that this technique applies only for step responses. Again, if the switching takes place at time t = t0 instead of t = 0, Eq. (7.61) becomes i(t) = i(∞) + [i(t0) − i(∞)]e−(t−t0)/τ (7.62) If I0 = 0, then i(t) =    0, t 0 Vs R (1 − e−t/τ ), t 0 (7.63a) or i(t) = Vs R (1 − e−t/τ )u(t) (7.63b) This is the step response of the RL circuit. The voltage across the inductor is obtained from Eq. (7.63) using v = L di/dt. We get
  • 269. CHAPTER 7 First-Order Circuits 265 v(t) = L di dt = Vs L τR e−t/τ , τ = L R , t 0 or v(t) = Vse−t/τ u(t) (7.64) Figure 7.50 shows the step responses in Eqs. (7.63) and (7.64). 0 t v(t) 0 t i(t) Vs R (a) (b) Vs Figure7.50 Step responses of an RL circuit with no initial inductor current: (a) current response, (b) voltage response. E X A M P L E 7 . 1 2 Find i(t) in the circuit in Fig. 7.51 for t 0. Assume that the switch has been closed for a long time. 2 Ω 3 Ω + − 10 V i t = 0 H 1 3 Figure7.51 For Example 7.12. Solution: When t 0, the 3- resistor is short-circuited, and the inductor acts like a short circuit. The current through the inductor at t = 0− (i.e., just before t = 0) is i(0− ) = 10 2 = 5 A Since the inductor current cannot change instantaneously, i(0) = i(0+ ) = i(0− ) = 5 A When t 0, the switch is open. The 2- and 3- resistors are in series, so that i(∞) = 10 2 + 3 = 2 A The Thevenin resistance across the inductor terminals is RTh = 2 + 3 = 5 For the time constant, τ = L RTh = 1 3 5 = 1 15 s Thus, i(t) = i(∞) + [i(0) − i(∞)]e−t/τ = 2 + (5 − 2)e−15t = 2 + 3e−15t A, t 0
  • 270. 266 PART 1 DC Circuits Check: In Fig. 7.51, for t 0, KVL must be satisfied; that is, 10 = 5i + L di dt 5i + L di dt = [10 + 15e−15t ] + 1 3 (3)(−15)e−15t = 10 This confirms the result. P R A C T I C E P R O B L E M 7 . 1 2 The switch in Fig. 7.52 has been closed for a long time. It opens at t = 0. Find i(t) for t 0. 1.5 H 10 Ω 5 Ω 3 A t = 0 i Figure7.52 For Practice Prob. 7.12. Answer: (2 + e−10t ) A, t 0. E X A M P L E 7 . 1 3 At t = 0, switch 1 in Fig. 7.53 is closed, and switch 2 is closed 4 s later. Find i(t) for t 0. Calculate i for t = 2 s and t = 5 s. 4 Ω 6 Ω + − + − 40 V 10 V 2 Ω 5 H i t = 0 t = 4 S1 S2 P Figure7.53 For Example 7.13. Solution: We need to consider the three time intervals t ≤ 0, 0 ≤ t ≤ 4, and t ≥ 4 separately. For t 0, switches S1 and S2 are open so that i = 0. Since the inductor current cannot change instantly, i(0− ) = i(0) = i(0+ ) = 0 For 0 ≤ t ≤ 4, S1 is closed so that the 4- and 6- resistors are in series. Hence, assuming for now that S1 is closed forever, i(∞) = 40 4 + 6 = 4 A, RTh = 4 + 6 = 10 τ = L RTh = 5 10 = 1 2 s
  • 271. CHAPTER 7 First-Order Circuits 267 Thus, i(t) = i(∞) + [i(0) − i(∞)]e−t/τ = 4 + (0 − 4)e−2t = 4(1 − e−2t ) A, 0 ≤ t ≤ 4 For t ≥ 4, S2 is closed; the 10-V voltage source is connected, and the circuit changes. This sudden change does not affect the inductor current because the current cannot change abruptly. Thus, the initial current is i(4) = i(4− ) = 4(1 − e−8 ) 4 A To find i(∞), let v be the voltage at node P in Fig. 7.53. Using KCL, 40 − v 4 + 10 − v 2 = v 6 ⇒ v = 180 11 V i(∞) = v 6 = 30 11 = 2.727 A The Thevenin resistance at the inductor terminals is RTh = 4 2 + 6 = 4 × 2 6 + 6 = 22 3 and τ = L RTh = 5 22 3 = 15 22 s Hence, i(t) = i(∞) + [i(4) − i(∞)]e−(t−4)/τ , t ≥ 4 We need (t − 4) in the exponential because of the time delay. Thus, i(t) = 2.727 + (4 − 2.727)e−(t−4)/τ , τ = 15 22 = 2.727 + 1.273e−1.4667(t−4) , t ≥ 4 Putting all this together, i(t) =    0, t ≤ 0 4(1 − e−2t ), 0 ≤ t ≤ 4 2.727 + 1.273e−1.4667(t−4) , t ≥ 4 At t = 2, i(2) = 4(1 − e−4 ) = 3.93 A At t = 5, i(5) = 2.727 + 1.273e−1.4667 = 3.02 A P R A C T I C E P R O B L E M 7 . 1 3 Switch S1 in Fig. 7.54 is closed at t = 0, and switch S2 is closed at t = 2 s. Calculate i(t) for all t. Find i(1) and i(3).
  • 272. 268 PART 1 DC Circuits Answer: i(t) =    0, t 0 2(1 − e−9t ), 0 t 2 3.6 − 1.6e−5(t−2) , t 2 i(1) = 1.9997 A, i(3) = 3.589 A. 10 Ω 15 Ω 20 Ω 6 A 5 H t = 0 S1 t = 2 S2 i(t) Figure7.54 For Practice Prob. 7.13. †7.7 FIRST-ORDER OP AMP CIRCUITS An op amp circuit containing a storage element will exhibit first-order behavior. Differentiators and integrators treated in Section 6.6 are exam- ples of first-order op amp circuits. Again, for practical reasons, inductors are hardly ever used in op amp circuits; therefore, the op amp circuits we consider here are of the RC type. As usual, we analyze op amp circuits using nodal analysis. Some- times, the Thevenin equivalent circuit is used to reduce the op amp circuit to one that we can easily handle. The following three examples illustrate the concepts. The first one deals with a source-free op amp circuit, while the other two involve step responses. The three examples have been care- fully selected to cover all possible RC types of op amp circuits, depending on the location of the capacitor with respect to the op amp; that is, the capacitor can be located in the input, the output, or the feedback loop. E X A M P L E 7 . 1 4 For the op amp circuit in Fig. 7.55(a) , find vo for t 0, given that v(0) = 3 V. Let Rf = 80 k, R1 = 20 k, and C = 5 µF. vo v + − R1 Rf (a) + − 3 2 1 1 vo (0+ ) 3 V + − (b) + − 3 2 vo v + − (c) 80 kΩ 80 kΩ 20 kΩ 20 kΩ 1 A C − + C + − + − + − Figure7.55 For Example 7.14.
  • 273. CHAPTER 7 First-Order Circuits 269 Solution: This problem can be solved in two ways: METHOD 1 Consider the circuit in Fig. 7.55(a). Let us derive the appropriate differential equation using nodal analysis. If v1 is the voltage at node 1, at that node, KCL gives 0 − v1 R1 = C dv dt (7.14.1) Since nodes 2 and 3 must be at the same potential, the potential at node 2 is zero. Thus, v1 − 0 = v or v1 = v and Eq. (7.14.1) becomes dv dt + v CR1 = 0 (7.14.2) This is similar to Eq. (7.4b) so that the solution is obtained the same way as in Section 7.2, i.e., v(t) = V0e−t/τ , τ = R1C (7.14.3) where V0 is the initial voltage across the capacitor. But v(0) = 3 = V0 and τ = 20 × 103 × 5 × 10−6 = 0.1. Hence, v(t) = 3e−10t (7.14.4) Applying KCL at node 2 gives C dv dt = 0 − vo Rf or vo = −Rf C dv dt (7.14.5) Now we can find v0 as vo = −80 × 103 × 5 × 10−6 (−30e−10t ) = 12e−10t V, t 0 METHOD 2 Let us now apply the short-cut method from Eq. (7.53). We need to find vo(0+ ), vo(∞), and τ. Since v(0+ ) = v(0− ) = 3 V, we apply KCL at node 2 in the circuit of Fig. 7.55(b) to obtain 3 20,000 + 0 − vo(0+ ) 80,000 = 0 or vo(0+ ) = 12 V. Since the circuit is source free, v(∞) = 0 V. To find τ, we need the equivalent resistance Req across the capacitor terminals. If we remove the capacitor and replace it by a 1-A current source, we have the circuit shown in Fig. 7.55(c). Applying KVL to the input loop yields 20,000(1) − v = 0 ⇒ v = 20 kV Then Req = v 1 = 20 k and τ = ReqC = 0.1. Thus, vo(t) = vo(∞) + [vo(0) − vo(∞)]e−t/τ = 0 + (12 − 0)e−10t = 12e−10t V, t 0 as before.
  • 274. 270 PART 1 DC Circuits P R A C T I C E P R O B L E M 7 . 1 4 For the op amp circuit in Fig. 7.56, find vo for t 0 if v(0) = 4 V. Assume that Rf = 50 k, R1 = 10 k, and C = 10 µF. vo + − R1 Rf v + − C + − Figure7.56 For Practice Prob. 7.14. Answer: −4e−2t V, t 0. E X A M P L E 7 . 1 5 Determine v(t) and vo(t) in the circuit of Fig. 7.57. vo v1 + − 3 V v + − 1 mF 50 kΩ 20 kΩ 20 kΩ + − t = 0 10 kΩ + − Figure7.57 For Example 7.15. Solution: This problem can be solved in two ways, just like the previous example. However, we will apply only the second method. Since what we are looking for is the step response, we can apply Eq. (7.53) and write v(t) = v(∞) + [v(0) − v(∞)]e−t/τ , t 0 (7.15.1) where we need only find the time constant τ, the initial value v(0), and the final value v(∞). Notice that this applies strictly to the capacitor voltage due a step input. Since no current enters the input terminals of the op amp, the elements on the feedback loop of the op amp constitute an RC circuit, with τ = RC = 50 × 103 × 10−6 = 0.05 (7.15.2) For t 0, the switch is open and there is no voltage across the capacitor. Hence, v(0) = 0. For t 0, we obtain the voltage at node 1 by voltage division as v1 = 20 20 + 10 3 = 2 V (7.15.3) Since there is no storage element in the input loop, v1 remains constant for all t. At steady state, the capacitor acts like an open circuit so that the op amp circuit is a noninverting amplifier. Thus, vo(∞) = 1 + 50 20 v1 = 3.5 × 2 = 7 V (7.15.4) But v1 − vo = v (7.15.5)
  • 275. CHAPTER 7 First-Order Circuits 271 so that v(∞) = 2 − 7 = −5 V Substituting τ, v(0), and v(∞) into Eq. (7.15.1) gives v(t) = −5 + [0 − (−5)]e−20t = 5(e−20t − 1) V, t 0 (7.15.6) From Eqs. (7.15.3), (7.15.5), and (7.15.6), we obtain vo(t) = v1(t) − v(t) = 7 − 5e−20t V, t 0 (7.15.7) P R A C T I C E P R O B L E M 7 . 1 5 Find v(t) and vo(t) in the op amp circuit of Fig. 7.58. vo + − 4 mV v + − 1 mF 100 kΩ + − t = 0 10 kΩ + − Figure7.58 For Practice Prob. 7.15. Answer: 40(1 − e−10t ) mV, 40(e−10t − 1) mV. E X A M P L E 7 . 1 6 Find the step response vo(t) for t 0 in the op amp circuit of Fig. 7.59. Let vi = 2u(t) V, R1 = 20 k, Rf = 50 k, R2 = R3 = 10 k, C = 2 µF. vi vo + − C + − R1 Rf R2 R3 + − Figure7.59 For Example 7.16. Solution: Notice that the capacitor in Example 7.14 is located in the input loop, while the capacitor in Example 7.15 is located in the feedback loop. In this example, the capacitor is located in the output of the op amp. Again, we can solve this problem directly using nodal analysis. However, using the Thevenin equivalent circuit may simplify the problem. We temporarily remove the capacitor and find the Thevenin equiv- alent at its terminals. To obtain VTh, consider the circuit in Fig. 7.60(a). Since the circuit is an inverting amplifier, Vab = − Rf R1 vi By voltage division, VTh = R3 R2 + R3 Vab = − R3 R2 + R3 Rf R1 vi
  • 276. 272 PART 1 DC Circuits vi + − R1 Rf R2 R3 + − Vab VTh + − a b (a) (b) RTh Ro R2 R3 + − Figure7.60 Obtaining VTh and RTh across the capacitor in Fig. 7.59. To obtain RTh, consider the circuit in Fig. 7.60(b), where Ro is the output resistance of the op amp. Since we are assuming an ideal op amp, Ro = 0, and RTh = R2 R3 = R2R3 R2 + R3 Substituting the given numerical values, VTh = − R3 R2 + R3 Rf R1 vi = − 10 20 50 20 2u(t) = −2.5u(t) RTh = R2R3 R2 + R3 = 5 k The Thevenin equivalent circuit is shown in Fig. 7.61, which is similar to Fig. 7.40. Hence, the solution is similar to that in Eq. (7.48); that is, vo(t) = −2.5(1 − e−t/τ ) u(t) where τ = RThC = 5 × 103 × 2 × 10−6 = 0.01. Thus, the step response for t 0 is vo(t) = 2.5(e−100t − 1) u(t) V 5 kΩ + − −2.5u(t) 2 mF Figure 7.61 Theveninequivalentcircuitof the circuit in Fig. 7.59. P R A C T I C E P R O B L E M 7 . 1 6 Obtain the step response vo(t) for the circuit of Fig. 7.62. Let vi = 2u(t) V, R1 = 20 k, Rf = 40 k, R2 = R3 = 10 k, C = 2 µF. Rf + − R1 R2 R3 vo vi + − C + − Figure7.62 For Practice Prob. 7.16. Answer: 6(1 − e−50t )u(t) V.
  • 277. CHAPTER 7 First-Order Circuits 273 7.8 TRANSIENT ANALYSIS WITH PSPICE As we discussed in Section 7.5, the transient response is the tem- porary response of the circuit that soon disappears. PSpice can be used to obtain the transient response of a circuit with storage elements. Sec- tion D.4 in Appendix D provides a review of transient analysis using PSpice for Windows. It is recommended that you read Section D.4 before continuing with this section. PSpice uses “transient” to mean “function of time.” Therefore, the transient response in PSpice may not actually die out as expected. If necessary, dc PSpice analysis is first carried out to determine the initial conditions. Then the initial conditions are used in the transient PSpice analysis to obtain the transient responses. It is recommended but not necessary that during this dc analysis, all capacitors should be open-circuited while all inductors should be short-circuited. E X A M P L E 7 . 1 7 Use PSpice to find the response i(t) for t 0 in the circuit of Fig. 7.63. 4 Ω 2 Ω 6 A 3 H t = 0 i(t) Figure7.63 For Example 7.17. Solution: Solving this problem by hand gives i(0) = 0, i(∞) = 2 A, RTh = 6, τ = 3/6 = 0.5 s, so that i(t) = i(∞) + [i(0) − i(∞)]e−t/τ = 2(1 − e−2t ), t 0 To use PSpice, we first draw the schematic as shown in Fig. 7.64. We recall from Appendix D that the part name for a close switch is Sw−tclose. We do not need to specify the initial condition of the in- ductor because PSpice will determine that from the circuit. By select- ing Analysis/Setup/Transient, we set Print Step to 25 ms and Final Step to 5τ = 2.5 s. After saving the circuit, we simulate by selecting Analysis/Simulate. In the Probe menu, we select Trace/Add and display −I(L1) as the current through the inductor. Figure 7.65 shows the plot of i(t), which agrees with that obtained by hand calculation. R2 2 6 A 3 H IDC R1 L1 tClose = 0 1 2 U1 4 0 Figure 7.64 The schematic of the circuit in Fig. 7.63. 1.5 A 0.5 A 2.0 A 1.0 A 0 A 0 s 1.0 s 2.0 s 3.0 s -I(L1) Time Figure 7.65 For Example 7.17; the response of the circuit in Fig. 7.63.
  • 278. 274 PART 1 DC Circuits Note that the negative sign on I(L1) is needed because the current enters through the upper terminal of the inductor, which happens to be the negative terminal after one counterclockwise rotation. A way to avoid the negative sign is to ensure that current enters pin 1 of the inductor. To obtain this desired direction of positive current flow, the initially horizon- tal inductor symbol should be rotated counterclockwise 270◦ and placed in the desired location. P R A C T I C E P R O B L E M 7 . 1 7 For the circuit in Fig. 7.66, use PSpice to find v(t) for t 0. 3 Ω + − 12 V 6 Ω 0.5 F + − v(t) t = 0 Figure7.66 For Practice Prob. 7.17. Answer: v(t) = 8(1 − e−t ) V, t 0. The response is similar in shape to that in Fig. 7.65. E X A M P L E 7 . 1 8 In the circuit in Fig. 7.67, determine the response v(t). 12 Ω + − 30 V 3 Ω 6 Ω 6 Ω 0.1 F 4 A + − v(t) t = 0 t = 0 Figure7.67 For Example 7.18. Solution: There are two ways of solving this problem using PSpice. METHOD 1 One way is to first do the dc PSpice analysis to determine the initial capacitor voltage. The schematic of the revelant circuit is in Fig. 7.68(a). Two pseudocomponent VIEWPOINTs are inserted to mea- sure the voltages at nodes 1 and 2. When the circuit is simulated, we obtain the displayed values in Fig. 7.68(a) as V1 = 0 V and V2 = 8 V. Thus the initial capacitor voltage is v(0) = V1 − V2 = −8 V. The PSpice transient analysis uses this value along with the schematic in Fig. 7.68(b). Once the circuit in Fig. 7.68(b) is drawn, we insert the capacitor initial voltage as IC = −8. We select Analysis/Setup/Transient and set Print Step to 0.1 s and Final Step to 4τ = 4 s. After saving the circuit, we select Analysis/Simulate to simulate the circuit. In the Probe menu, we select
  • 279. CHAPTER 7 First-Order Circuits 275 Trace/Add and display V(R2:2) - V(R3:2) or V(C1:1) - V(C1:2) as the capacitor voltage v(t). The plot of v(t) is shown in Fig. 7.69. This agrees with the result obtained by hand calculation, v(t) = 10 − 18e−t . 0.0000 8.0000 6 4A 0.1 1 R3 3 R4 I1 2 6 R2 0 C1 (a) 6 12 R2 6 R3 30 V 0 R1 (b) + − 0.1 C1 V1 Figure7.68 (a) Schematic for dc analysis to get v(0), (b) schematic for transient analysis used in getting the response v(t). 5 V -5 V 10 V 0 V -10 V 0 s 1.0 s 2.0 s 3.0 s 4.0 s V(R2:2) - V(R3:2) Time Figure7.69 Response v(t) for the circuit in Fig. 7.67. METHOD 2 We can simulate the circuit in Fig. 7.67 directly, since PSpice can handle the open and close switches and determine the initial conditions automatically. Using this approach, the schematic is drawn as shown in Fig. 7.70. After drawing the circuit, we select Analysis/ Setup/Transient and set Print Step to 0.1 s and Final Step to 4τ = 4 s. We save the circuit, then select Analysis/Simulate to simulate the circuit. In the Probe menu, we select Trace/Add and display V(R2:2) - V(R3:2) as the capacitor voltage v(t). The plot of v(t) is the same as that shown in Fig. 7.69. R1 6 30 V 4 A R2 6 R3 3 R4 I1 tClose = 0 1 2 12 U1 1 2 U2 0 + − tOpen = 0 0.1 C1 V1 Figure7.70 For Example 7.18.
  • 280. 276 PART 1 DC Circuits P R A C T I C E P R O B L E M 7 . 1 8 The switch in Fig. 7.71 was open for a long time but closed at t = 0. If i(0) = 10 A, find i(t) for t 0 by hand and also by PSpice. 5 Ω 30 Ω 12 A 2 H t = 0 6 Ω i(t) Figure7.71 For Practice Prob. 7.18. Answer: i(t) = 6 + 4e−5t A. The plot of i(t) obtained by PSpice analysis is shown in Fig. 7.72. 9 A 10 A 7 A 8 A 6 A 0 s 0.5 s 1.0 s I(L1) Time Figure7.72 For Practice Prob. 7.18. †7.9 APPLICATIONS The various devices in which RC and RL circuits find applications in- clude filtering in dc power supplies, smoothing circuits in digital com- munications, differentiators, integrators, delay circuits, and relay circuits. Some of these applications take advantage of the short or long time con- stants of the RC or RL circuits. We will consider four simple applications here. The first two are RC circuits, the last two are RL circuits. 7.9.1 Delay Circuits An RC circuit can be used to provide various time delays. Figure 7.73 shows such a circuit. It basically consists of an RC circuit with the capacitor connected in parallel with a neon lamp. The voltage source can provide enough voltage to fire the lamp. When the switch is closed, the capacitor voltage increases gradually toward 110 V at a rate determined R1 R2 110 V C 0.1 mF S + − 70 V Neon lamp Figure7.73 An RC delay circuit.
  • 281. CHAPTER 7 First-Order Circuits 277 by the circuit’s time constant, (R1 + R2)C. The lamp will act as an open circuit and not emit light until the voltage across it exceeds a particular level, say 70 V. When the voltage level is reached, the lamp fires (goes on), and the capacitor discharges through it. Due to the low resistance of the lamp when on, the capacitor voltage drops fast and the lamp turns off. The lamp acts again as an open circuit and the capacitor recharges. By adjusting R2, we can introduce either short or long time delays into the circuit and make the lamp fire, recharge, and fire repeatedly every time constant τ = (R1 + R2)C, because it takes a time period τ to get the capacitor voltage high enough to fire or low enough to turn off. The warning blinkers commonly found on road construction sites are one example of the usefulness of such an RC delay circuit. E X A M P L E 7 . 1 9 Consider the circuit in Fig. 7.73, and assume that R1 = 1.5 M, 0 R 2.5 M. (a) Calculate the extreme limits of the time constant of the cir- cuit. (b) How long does it take for the lamp to glow for the first time after the switch is closed? Let R2 assume its largest value. Solution: (a) The smallest value for R2 is 0 , and the corresponding time constant for the circuit is τ = (R1 + R2)C = (1.5 × 106 + 0) × 0.1 × 10−6 = 0.15 s The largest value for R2 is 2.5 M, and the corresponding time constant for the circuit is τ = (R1 + R2)C = (1.5 + 2.5) × 106 × 0.1 × 10−6 = 0.4 s Thus, by proper circuit design, the time constant can be adjusted to in- troduce a proper time delay in the circuit. (b) Assuming that the capacitor is initially uncharged, vC(0) = 0, while vC(∞) = 110. But vC(t) = vC(∞) + [vC(0) − vC(∞)]e−t/τ = 110[1 − e−t/τ ] where τ = 0.4 s, as calculated in part (a). The lamp glows when vC = 70 V. If vC(t) = 70 V at t = t0, then 70 = 110[1 − e−t0/τ ] ⇒ 7 11 = 1 − e−t0/τ or e−t0/τ = 4 11 ⇒ et0/τ = 11 4 Taking the natural logarithm of both sides gives t0 = τ ln 11 4 = 0.4 ln 2.75 = 0.4046 s A more general formula for finding t0 is t0 = τ ln v(0) − v(∞) v(t0) − v(∞)
  • 282. 278 PART 1 DC Circuits The lamp will fire repeatedly every τ seconds if and only if t0 τ. In this example, that condition is not satisfied. P R A C T I C E P R O B L E M 7 . 1 9 The RC circuit in Fig. 7.74 is designed to operate an alarm which acti- vates when the current through it exceeds 120 µA. If 0 ≤ R ≤ 6 k, find the range of the time delay that the circuit can cause. 10 kΩ R 9 V 80 mF 4 kΩ S + − Alarm Figure7.74 For Practice Prob. 7.19. Answer: Between 47.23 ms and 124 ms. 7.9.2 Photoflash Unit R1 + − High voltage dc supply R2 C v vs 1 2 i + − Figure7.75 Circuit for a flash unit providing slow charge in position 1 and fast discharge in position 2. An electronic flash unit provides a common example of an RC circuit. This application exploits the ability of the capacitor to oppose any abrupt change in voltage. Figure 7.75 shows a simplified circuit. It consists essentially of a high-voltage dc supply, a current-limiting large resistor R1, and a capacitor C in parallel with the flashlamp of low resistance R2. When the switch is in position 1, the capacitor charges slowly due to the large time constant (τ1 = R1C). As shown in Fig. 7.76, the capacitor voltage rises gradually from zero to Vs, while its current decreases grad- ually from I1 = Vs/R1 to zero. The charging time is approximately five times the time constant, tcharge = 5R1C (7.65) With the switch in position 2, the capacitor voltage is discharged. The low resistance R2 of the photolamp permits a high discharge current with peak I2 = Vs/R2 in a short duration, as depicted in Fig. 7.76(b). Discharging takes place in approximately five times the time constant, 0 t Vs v 0 (a) (b) −I2 I1 i Figure7.76 (a) Capacitor voltage showing slow charge and fast discharge, (b) capacitor current showing low charging current I1 = Vs/R1 and high discharge current I2 = Vs/R2.
  • 283. CHAPTER 7 First-Order Circuits 279 tdischarge = 5R2C (7.66) Thus, the simple RC circuit of Fig. 7.75 provides a short-duration, high- current pulse. Such a circuit also finds applications in electric spot weld- ing and the radar transmitter tube. E X A M P L E 7 . 2 0 An electronic flashgun has a current-limiting 6-k resistor and 2000-µF electrolytic capacitor charged to 240 V. If the lamp resistance is 12 , find: (a) the peak charging current, (b) the time required for the capaci- tor to fully charge, (c) the peak discharging current, (d) the total energy stored in the capacitor, and (e) the average power dissipated by the lamp. Solution: (a) The peak charging current is I1 = Vs R1 = 240 6 × 103 = 40 mA (b) From Eq. (7.65), tcharge = 5R1C = 5 × 6 × 103 × 2000 × 10−6 = 60 s = 1 minute (c) The peak discharging current is I2 = Vs R2 = 240 12 = 20 A (d) The energy stored is W = 1 2 CV 2 s = 1 2 × 2000 × 10−6 × 2402 = 57.6 J (e) The energy stored in the capacitor is dissipated across the lamp during the discharging period. From Eq. (7.66), tdischarge = 5R2C = 5 × 12 × 2000 × 10−6 = 0.12 s Thus, the average power dissipated is p = W tdischarge = 57.6 0.12 = 480 W P R A C T I C E P R O B L E M 7 . 2 0 The flash unit of a camera has a 2-mF capacitor charged to 80 V. (a) How much charge is on the capacitor? (b) What is the energy stored in the capacitor? (c) If the flash fires in 0.8 ms, what is the average current through the flashtube? (d) How much power is delivered to the flashtube? (e) After a picture has been taken, the capacitor needs to be recharged by a power unit which supplies a maximum of 5 mA. How much time does it take to charge the capacitor? Answer: (a) 0.16 C, (b) 6.4 J, (c) 200 A, (d) 8 kW, (e) 32 s.
  • 284. 280 PART 1 DC Circuits 7.9.3 Relay Circuits Amagneticallycontrolledswitchiscalledarelay. Arelayisessentiallyan electromagneticdeviceusedtoopenorcloseaswitchthatcontrolsanother circuit. Figure 7.77(a) shows a typical relay circuit. The coil circuit is an RL circuit like that in Fig. 7.77(b), where R and L are the resistance and inductance of the coil. When switch S1 in Fig. 7.77(a) is closed, the coil circuit is energized. The coil current gradually increases and produces a magnetic field. Eventually the magnetic field is sufficiently strong to pull the movable contact in the other circuit and close switch S2. At this point, the relay is said to be pulled in. The time interval td between the closure of switches S1 and S2 is called the relay delay time. Relays were used in the earliest digital circuits and are still used for switching high-power circuits. S2 Coil Magnetic field S1 S1 Vs (a) (b) Vs R L Figure7.77 A relay circuit. E X A M P L E 7 . 2 1 The coil of a certain relay is operated by a 12-V battery. If the coil has a resistance of 150 and an inductance of 30 mH and the current needed to pull in is 50 mA, calculate the relay delay time. Solution: The current through the coil is given by i(t) = i(∞) + [i(0) − i(∞)]e−t/τ where i(0) = 0, i(∞) = 12 150 = 80 mA τ = L R = 30 × 10−3 150 = 0.2 ms Thus, i(t) = 80[1 − e−t/τ ] mA If i(td) = 50 mA, then 50 = 80[1 − e−td /τ ] ⇒ 5 8 = 1 − e−td /τ
  • 285. CHAPTER 7 First-Order Circuits 281 or e−td /τ = 3 8 ⇒ etd /τ = 8 3 By taking the natural logarithm of both sides, we get td = τ ln 8 3 = 0.2 ln 8 3 ms = 0.1962 ms P R A C T I C E P R O B L E M 7 . 2 1 A relay has a resistance of 200 and an inductance of 500 mH. The relay contacts close when the current through the coil reaches 350 mA. What time elapses between the application of 110 V to the coil and contact closure? Answer: 2.529 ms. 7.9.4 Automobile Ignition Circuit The ability of inductors to oppose rapid change in current makes them useful for arc or spark generation. An automobile ignition system takes advantage of this feature. R Vs v + − i Spark plug Air gap L Figure 7.78 Circuit for an automobile ignition system. The gasoline engine of an automobile requires that the fuel-air mixture in each cylinder be ignited at proper times. This is achieved by means of a spark plug (Fig. 7.78), which essentially consists of a pair of electrodes separated by an air gap. By creating a large voltage (thousands of volts) between the electrodes, a spark is formed across the air gap, thereby igniting the fuel. But how can such a large voltage be obtained from the car battery, which supplies only 12 V? This is achieved by means of an inductor (the spark coil) L. Since the voltage across the inductor is v = L di/dt, we can make di/dt large by creating a large change in current in a very short time. When the ignition switch in Fig. 7.78 is closed, the current through the inductor increases gradually and reaches the final value of i = Vs/R, where Vs = 12 V. Again, the time taken for the inductor to charge is five times the time constant of the circuit (τ = L/R), tcharge = 5 L R (7.67) Since at steady state, i is constant, di/dt = 0 and the inductor voltage v = 0. When the switch suddenly opens, a large voltage is developed across the inductor (due to the rapidly collapsing field) causing a spark or arc in the air gap. The spark continues until the energy stored in the inductor is dissipated in the spark discharge. In laboratories, when one is working with inductive circuits, this same effect causes a very nasty shock, and one must exercise caution. E X A M P L E 7 . 2 2 A solenoid with resistance 4 and inductance 6 mH is used in an auto- mobile ignition circuit similar to that in Fig. 7.78. If the battery supplies 12 V, determine: the final current through the solenoid when the switch
  • 286. 282 PART 1 DC Circuits is closed, the energy stored in the coil, and the voltage across the air gap, assuming that the switch takes 1 µs to open. Solution: The final current through the coil is I = Vs R = 12 4 = 3 A The energy stored in the coil is W = 1 2 LI2 = 1 2 × 6 × 10−3 × 32 = 27 mJ The voltage across the gap is V = L )I )t = 6 × 10−3 × 3 1 × 10−6 = 18 kV P R A C T I C E P R O B L E M 7 . 2 2 The spark coil of an automobile ignition system has a 20-mH inductance and a 5- resistance. With a supply voltage of 12 V, calculate: the time needed for the coil to fully charge, the energy stored in the coil, and the voltage developed at the spark gap if the switch opens in 2 µs. Answer: 20 ms, 57.6 mJ, and 24 kV. 7.10 SUMMARY 1. The analysis in this chapter is applicable to any circuit that can be reduced to an equivalent circuit comprising a resistor and a single energy-storage element (inductor or capacitor). Such a circuit is first-order because its behavior is described by a first-order differen- tial equation. When analyzing RC and RL circuits, one must always keep in mind that the capacitor is an open circuit to steady-state dc conditions while the inductor is a short circuit to steady-state dc conditions. 2. The natural response is obtained when no independent source is present. It has the general form x(t) = x(0)e−t/τ where x represents current through (or voltage across) a resistor, a capacitor, or an inductor, and x(0) is the initial value of x. The natural response is also called the transient response because it is the temporary response that vanishes with time. 3. The time constant τ is the time required for a response to decay to 1/e of its initial value. For RC circuits, τ = RC and for RL circuits, τ = L/R. 4. The singularity functions include the unit step, the unit ramp func- tion, and the unit impulse functions. The unit step function u(t) is u(t) = 0, t 0 1, t 0
  • 287. CHAPTER 7 First-Order Circuits 283 The unit impulse function is δ(t) =    0, t 0 Undefined, t = 0 0, t 0 The unit ramp function is r(t) = 0, t ≤ 0 t, t ≥ 0 5. The forced (or steady-state) response is the behavior of the circuit after an independent source has been applied for a long time. 6. The total or complete response consists of the natural response and the forced response. 7. The step response is the response of the circuit to a sudden applica- tion of a dc current or voltage. Finding the step response of a first- order circuit requires the initial value x(0+ ), the final value x(∞), and the time constant τ. With these three items, we obtain the step response as x(t) = x(∞) + [x(0+ ) − x(∞)]e−t/τ A more general form of this equation is x(t) = x(∞) + [x(t+ 0 ) − x(∞)]e−(t−t0)/τ Or we may write it as Instantaneous value = Final + [Initial − Final]e−(t−t0)/τ 8. PSpice is very useful for obtaining the transient response of a circuit. 9. Four practical applications of RC and RL circuits are: a delay circuit, a photoflash unit, a relay circuit, and an automobile ignition circuit. REVIEW QUESTIONS 7.1 An RC circuit has R = 2 and C = 4 F. The time constant is: (a) 0.5 s (b) 2 s (c) 4 s (d) 8 s (e) 15 s 7.2 The time constant for an RL circuit with R = 2 and L = 4 H is: (a) 0.5 s (b) 2 s (c) 4 s (d) 8 s (e) 15 s 7.3 A capacitor in an RC circuit with R = 2 and C = 4 F is being charged. The time required for the capacitor voltage to reach 63.2 percent of its steady-state value is: (a) 2 s (b) 4 s (c) 8 s (d) 16 s (e) none of the above 7.4 An RL circuit has R = 2 and L = 4 H. The time needed for the inductor current to reach 40 percent of its steady-state value is: (a) 0.5 s (b) 1 s (c) 2 s (d) 4 s (e) none of the above 7.5 In the circuit of Fig. 7.79, the capacitor voltage just before t = 0 is: (a) 10 V (b) 7 V (c) 6 V (d) 4 V (e) 0 V v(t) 10 V 2 Ω 3 Ω + − + − t = 0 7 F Figure 7.79 For Review Questions 7.5 and 7.6.
  • 288. 284 PART 1 DC Circuits 7.6 In the circuit of Fig. 7.79, v(∞) is: (a) 10 V (b) 7 V (c) 6 V (d) 4 V (e) 0 V 7.7 For the circuit of Fig. 7.80, the inductor current just before t = 0 is: (a) 8 A (b) 6 A (c) 4 A (d) 2 A (e) 0 A 10 A 3 Ω 2 Ω 5 H i(t) t = 0 Figure 7.80 For Review Questions 7.7 and 7.8. 7.8 In the circuit of Fig. 7.80, i(∞) is: (a) 8 A (b) 6 A (c) 4 A (d) 2 A (e) 0 A 7.9 If vs changes from 2 V to 4 V at t = 0, we may express vs as: (a) δ(t) V (b) 2u(t) V (c) 2u(−t) + 4u(t) V (d) 2 + 2u(t) V (e) 4u(t) − 2 V 7.10 The pulse in Fig. 7.110(a) can be expressed in terms of singularity functions as: (a) 2u(t) + 2u(t − 1) V (b) 2u(t) − 2u(t − 1) V (c) 2u(t) − 4u(t − 1) V (d) 2u(t) + 4u(t − 1) V Answers: 7.1d, 7.2b, 7.3c, 7.4b, 7.5d, 7.6a, 7.7c, 7.8e, 7.9c,d, 7.10b. PROBLEMS Section 7.2 The Source-Free RC Circuit 7.1 Show that Eq. (7.9) can be obtained by working with the current i in the RC circuit rather than working with the voltage v. 7.2 Find the time constant for the RC circuit in Fig. 7.81. + − 80 Ω 120 Ω 12 Ω 50 V 0.5 mF Figure 7.81 For Prob. 7.2. 7.3 Determine the time constant of the circuit in Fig. 7.82. 4 kΩ 12 kΩ 3 mF 1 mF 5 kΩ Figure 7.82 For Prob. 7.3. 7.4 Obtain the time constant of the circuit in Fig. 7.83. + − R2 R1 vs C2 C1 Figure 7.83 For Prob. 7.4. 7.5 The switch in Fig. 7.84 has been in position a for a long time, until t = 4 s when it is moved to position b and left there. Determine v(t) at t = 10 s. v(t) 24 V 20 Ω 80 Ω + − + − 0.1 F t = 4 a b Figure 7.84 For Prob. 7.5. 7.6 If v(0) = 20 V in the circuit in Fig. 7.85, obtain v(t) for t 0. 10 Ω 8 Ω 0.5 V 0.1 F + − v + − Figure 7.85 For Prob. 7.6.
  • 289. CHAPTER 7 First-Order Circuits 285 7.7 For the circuit in Fig. 7.86, if v = 10e−4t V and i = 0.2e−4t A, t 0 (a) Find R and C. (b) Determine the time constant. (c) Calculate the initial energy in the capacitor. (d) Obtain the time it takes to dissipate 50 percent of the initial energy. R v i C + − Figure 7.86 For Prob. 7.7. 7.8 In the circuit of Fig. 7.87, v(0) = 20 V. Find v(t) for t 0. 2 Ω 0.25 F 8 Ω 6 Ω 3 Ω 8 Ω + − v Figure 7.87 For Prob. 7.8. 7.9 Given that i(0) = 3 A, find i(t) for t 0 in the circuit in Fig. 7.88. i 10 mF 10 Ω 4 Ω 15 Ω Figure 7.88 For Prob. 7.9. Section 7.3 The Source-Free RL Circuit 7.10 Derive Eq. (7.20) by working with voltage v across the inductor of the RL circuit instead of working with the current i. 7.11 The switch in the circuit in Fig. 7.89 has been closed for a long time. At t = 0, the switch is opened. Calculate i(t) for t 0. 3 Ω + − 12 V 4 Ω i t = 0 2 H Figure 7.89 For Prob. 7.11. 7.12 For the circuit shown in Fig. 7.90, calculate the time constant. 70 Ω 2 mH + − 20 V 80 Ω 20 Ω 30 Ω Figure 7.90 For Prob. 7.12. 7.13 What is the time constant of the circuit in Fig. 7.91? 10 kΩ 10 mH 30 kΩ 6 kΩ 20 mH Figure 7.91 For Prob. 7.13. 7.14 Determine the time constant for each of the circuits in Fig. 7.92. L R1 R2 R3 (a) R1 R2 L2 L1 R3 (b) Figure 7.92 For Prob. 7.14. 7.15 Consider the circuit of Fig. 7.93. Find vo(t) if i(0) = 2 A and v(t) = 0.
  • 290. 286 PART 1 DC Circuits vo(t) v(t) 1 Ω 3 Ω + − + − i(t) H 1 4 Figure 7.93 For Prob. 7.15. 7.16 For the circuit in Fig. 7.94, determine vo(t) when i(0) = 1 A and v(t) = 0. vo(t) v(t) 3 Ω + − + − i(t) 2 Ω 0.4 H Figure 7.94 For Prob. 7.16. 7.17 In the circuit of Fig. 7.95, find i(t) for t 0 if i(0) = 2 A. 40 Ω 10 Ω 0.5i 6 H i Figure 7.95 For Prob. 7.17. 7.18 For the circuit in Fig. 7.96, v = 120e−50t V and i = 30e−50t A, t 0 (a) Find L and R. (b) Determine the time constant. (c) Calculate the initial energy in the inductor. (d) What fraction of the initial energy is dissipated in 10 ms? R i + − v L Figure 7.96 For Prob. 7.18. 7.19 In the circuit in Fig. 7.97, find the value of R for which energy stored in the inductor will be 1 J. 40 Ω R + − 60 V 2 H 80 Ω Figure 7.97 For Prob. 7.19. 7.20 Find i(t) and v(t) for t 0 in the circuit of Fig. 7.98 if i(0) = 10 A. 5 Ω 20 Ω 1 Ω 2 H + − v(t) i(t) Figure 7.98 For Prob. 7.20. 7.21 Consider the circuit in Fig. 7.99. Given that vo(0) = 2 V, find vo and vx for t 0. 3 Ω 1 Ω 2 Ω vo + − vx H 1 3 + − Figure 7.99 For Prob. 7.21. Section 7.4 Singularity Functions 7.22 Express the following signals in terms of singularity functions. (a) v(t) = 0, t 0 −5, t 0 (b) i(t) =      0, t 1 −10, 1 t 3 10, 3 t 5 0, t 5
  • 291. CHAPTER 7 First-Order Circuits 287 (c) x(t) =      t − 1, 1 t 2 1, 2 t 3 4 − t, 3 t 4 0, Otherwise (d) y(t) =    2, t 0 −5, 0 t 1 0, t 1 7.23 Express the signals in Fig. 7.100 in terms of singularity functions. 0 t 1 −1 v1(t) 1 −1 (a) 0 1 2 t −1 −2 v4(t) (d) 0 2 4 6 t 2 4 v3(t) (c) 0 2 4 t 2 v2(t) (b) Figure 7.100 For Prob. 7.23. 7.24 Sketch the waveform that is represented by v(t) = u(t) + u(t − 1) − 3u(t − 2) + 2u(t − 3) 7.25 Sketch the waveform represented by i(t) = r(t) + r(t − 1) − u(t − 2) − r(t − 2) + r(t − 3) + u(t − 4) 7.26 Evaluate the following integrals involving the impulse functions: (a) ∞ −∞ 4t2 δ(t − 1) dt (b) ∞ −∞ 4t2 cos 2πtδ(t − 0.5) dt 7.27 Evaluate the following integrals: (a) ∞ −∞ e−4t2 δ(t − 2) dt (b) ∞ −∞ [5δ(t) + e−t δ(t) + cos 2πtδ(t)]dt 7.28 The voltage across a 10-mH inductor is 20δ(t − 2) mV. Find the inductor current, assuming that the inductor is initially uncharged. 7.29 Find the solution of the following first-order differential equations subject to the specified initial conditions. (a) 5 dv/dt + 3v = 0, v(0) = −2 (b) 4 dv/dt − 6v = 0, v(0) = 5 7.30 Solve for v in the following differential equations, subject to the stated initial condition. (a) dv/dt + v = u(t), v(0) = 0 (b) 2 dv/dt − v = 3u(t), v(0) = −6 Section 7.5 Step Response of an RC Circuit 7.31 Calculate the capacitor voltage for t 0 and t 0 for each of the circuits in Fig. 7.101. + − 1 Ω 4 Ω 20 V 12 V + − t = 0 v 2 F (a) (b) 3 Ω 2 A 4 Ω + − + − t = 0 2 F v Figure 7.101 For Prob. 7.31. 7.32 Find the capacitor voltage for t 0 and t 0 for each of the circuits in Fig. 7.102.
  • 292. 288 PART 1 DC Circuits 3 Ω 2 Ω + − 3 F + − v 12 V 4 V + − t = 0 (a) (b) 4 Ω 2 Ω 5 F 6 A + − v t = 0 Figure 7.102 For Prob. 7.32. 7.33 For the circuit in Fig. 7.103, find v(t) for t 0. 1 F + − v 6 Ω 30 Ω 12 V t = 0 + − Figure 7.103 For Prob. 7.33. 7.34 (a) If the switch in Fig. 7.104 has been open for a long time and is closed at t = 0, find vo(t). (b) Suppose that the switch has been closed for a long time and is opened at t = 0. Find vo(t). 3 F + − vo 2 Ω 4 Ω 12 V + − t = 0 Figure 7.104 For Prob. 7.34. 7.35 Consider the circuit in Fig. 7.105. Find i(t) for t 0 and t 0. 3 F 40 Ω 30 Ω 50 Ω 0.5i 80 V + − t = 0 i Figure 7.105 For Prob. 7.35. 7.36 The switch in Fig. 7.106 has been in position a for a long time. At t = 0, it moves to position b. Calculate i(t) for all t 0. 2 F 6 Ω 3 Ω 30 V + − 12 V + − i t = 0 a b Figure 7.106 For Prob. 7.36. 7.37 Find the step responses v(t) and i(t) to vs = 5u(t) V in the circuit of Fig. 7.107. v(t) vs 4 Ω 12 Ω + − + − 0.5 F 7 Ω i(t) Figure 7.107 For Prob. 7.37. 7.38 Determine v(t) for t 0 in the circuit in Fig. 7.108 if v(0) = 0. 3u(t − 1) A 3u(t) A 8 Ω 2 Ω + − 0.1 F v Figure 7.108 For Prob. 7.38. 7.39 Find v(t) and i(t) in the circuit of Fig. 7.109.
  • 293. CHAPTER 7 First-Order Circuits 289 v u(−t) A 10 Ω + − 0.1 F 20 Ω i Figure 7.109 For Prob. 7.39. 7.40 If the waveform in Fig. 7.110(a) is applied to the circuit of Fig. 7.110(b), find v(t). Assume v(0) = 0. v is 4 Ω + − 0.5 F 6 Ω (b) 0 1 t (s) 2 is (A) (a) Figure 7.110 For Prob. 7.40 and Review Question 7.10. 7.41 ∗ In the circuit in Fig. 7.111, find ix for t 0. Let R1 = R2 = 1 k, R3 = 2 k, and C = 0.25 mF. R2 R1 30 mA t = 0 R3 ix C Figure 7.111 For Prob. 7.41. Section 7.6 Step Response of an RL Circuit 7.42 Rather than applying the short-cut technique used in Section 7.6, use KVL to obtain Eq. (7.60). 7.43 For the circuit in Fig. 7.112, find i(t) for t 0. 40 Ω 20 V 5 H i + − t = 0 10 Ω Figure 7.112 For Prob. 7.43. 7.44 Determine the inductor current i(t) for both t 0 and t 0 for each of the circuits in Fig. 7.113. 4 Ω 6 A 2 Ω 3 H i t = 0 (b) 25 V 4 H i (a) + − t = 0 2 Ω 3 Ω Figure 7.113 For Prob. 7.44. 7.45 Obtain the inductor current for both t 0 and t 0 in each of the circuits in Fig. 7.114. ∗An asterisk indicates a challenging problem.
  • 294. 290 PART 1 DC Circuits 4 Ω 2 A 2 Ω 3 Ω 6 Ω 12 Ω 3.5 H i 4 Ω (a) 10 V 2 H i (b) + − 24 V + − t = 0 t = 0 Figure 7.114 For Prob. 7.45. 7.46 Find v(t) for t 0 and t 0 in the circuit in Fig. 7.115. 8 Ω 4io 3 Ω 0.5 H 2 Ω 20 V + − 24 V + − t = 0 + − io + − v Figure 7.115 For Prob. 7.46. 7.47 For the network shown in Fig. 7.116, find v(t) for t 0. 6 Ω 12 Ω 2 A 0.5 H 20 Ω 5 Ω + − v + − 20 V t = 0 Figure 7.116 For Prob. 7.47. 7.48 ∗ Find i1(t) and i2(t) for t 0 in the circuit of Fig. 7.117. 6 Ω 5 A 2.5 H 5 Ω 20 Ω 4 H i1 i2 t = 0 Figure 7.117 For Prob. 7.48. 7.49 Rework Prob. 7.15 if i(0) = 10 A and v(t) = 20u(t) V. 7.50 Determine the step response vo(t) to vs = 18u(t) in the circuit of Fig. 7.118. 3 Ω 6 Ω vs 1.5 H 4 Ω + − + − vo Figure 7.118 For Prob. 7.50. 7.51 Find v(t) for t 0 in the circuit of Fig. 7.119 if the initial current in the inductor is zero. 5 Ω 20 Ω 4u(t) 8 H + − v Figure 7.119 For Prob. 7.51. 7.52 In the circuit in Fig. 7.120, is changes from 5 A to 10 A at t = 0; that is, is = 5u(−t) + 10u(t). Find v and i. 4 Ω is 0.5 H + − v i Figure 7.120 For Prob. 7.52. 7.53 For the circuit in Fig. 7.121, calculate i(t) if i(0) = 0.
  • 295. CHAPTER 7 First-Order Circuits 291 3 Ω 6 Ω + − u(t − 1) V u(t) V 2 H i + − Figure 7.121 For Prob. 7.53. 7.54 Obtain v(t) and i(t) in the circuit of Fig. 7.122. 5 Ω + − 10u(−t) V 20 Ω 0.5 H i + − v Figure 7.122 For Prob. 7.54. 7.55 Find vo(t) for t 0 in the circuit of Fig. 7.123. 6 Ω 2 Ω 3 Ω + − + − vo t = 0 4 H 10 V Figure 7.123 For Prob. 7.55. 7.56 If the input pulse in Fig. 7.124(a) is applied to the circuit in Fig. 7.124(b), determine the response i(t). 5 Ω + − vs 20 Ω 2 H i (b) (a) 0 t (s) vs (V) 10 1 Figure 7.124 For Prob. 7.56. Section 7.7 First-order Op Amp Circuits 7.57 Find the output current io for t 0 in the op amp circuit of Fig. 7.125. Let v(0) = −4 V. 10 kΩ 10 kΩ v 20 kΩ io + − 2 mF + − Figure 7.125 For Prob. 7.57. 7.58 If v(0) = 5 V, find vo(t) for t 0 in the op amp circuit in Fig. 7.126. Let R = 10 k and C = 1 µF. R R R v vo + − C + − Figure 7.126 For Prob. 7.58. 7.59 Obtain vo for t 0 in the circuit of Fig. 7.127. 10 kΩ 10 kΩ + − vo + − 25 mF t = 0 4 V + − Figure 7.127 For Prob. 7.59. 7.60 For the op amp circuit in Fig. 7.128, find vo(t) for t 0.
  • 296. 292 PART 1 DC Circuits 20 kΩ 100 kΩ 10 kΩ + − vo + − 25 mF t = 0 4 V + − Figure 7.128 For Prob. 7.60. 7.61 Determine vo for t 0 when vs = 20 mV in the op amp circuit of Fig. 7.129. 20 kΩ + − vo vs 5 mF t = 0 + − Figure 7.129 For Prob. 7.61. 7.62 For the op amp circuit in Fig. 7.130, find io for t 2. 10 kΩ 10 kΩ 20 kΩ + − 100 mF t = 2 4 V io + − Figure 7.130 For Prob. 7.62. 7.63 Find io in the op amp circuit in Fig. 7.131. Assume that v(0) = −2 V, R = 10 k, and C = 10 µF. R + − v 3u(t) io + − C + − Figure 7.131 For Prob. 7.63. 7.64 For the op amp circuit of Fig. 7.132, let R1 = 10 k, Rf = 20 k, C = 20 µF, and v(0) = 1 V. Find vo. Rf R1 + − vo + − 4u(t) v + − C + − Figure 7.132 For Prob. 7.64. 7.65 Determine vo(t) for t 0 in the circuit of Fig. 7.133. Let is = 10u(t) µA and assume that the capacitor is initially uncharged. 10 kΩ 50 kΩ vo + − is 2 mF + − Figure 7.133 For Prob. 7.65. 7.66 In the circuit of Fig. 7.134, find vo and io, given that vs = 4u(t) V and v(0) = 1 V. vo vs 2 mF 10 kΩ 20 kΩ + − v + − io + − Figure 7.134 For Prob. 7.66.
  • 297. CHAPTER 7 First-Order Circuits 293 Section 7.8 Transient Analysis with PSpice 7.67 Repeat Prob. 7.40 using PSpice. 7.68 The switch in Fig. 7.135 opens at t = 0. Use PSpice to determine v(t) for t 0. 5 Ω 4 Ω 5 A 6 Ω 20 Ω + − 30 V t = 0 + − v 100 mF Figure 7.135 For Prob. 7.68. 7.69 The switch in Fig. 7.136 moves from position a to b at t = 0. Use PSpice to find i(t) for t 0. 4 Ω 6 Ω 3 Ω + − 108 V 6 Ω 2 H i(t) t = 0 a b Figure 7.136 For Prob. 7.69. 7.70 Repeat Prob. 7.56 using PSpice. Section 7.9 Applications 7.71 In designing a signal-switching circuit, it was found that a 100-µF capacitor was needed for a time constant of 3 ms. What value resistor is necessary for the circuit? 7.72 A simple relaxation oscillator circuit is shown in Fig. 7.137. The neon lamp fires when its voltage reaches 75 V and turns off when its voltage drops to 30 V. Its resistance is 120 when on and infinitely high when off. (a) For how long is the lamp on each time the capacitor discharges? (b) What is the time interval between light flashes? 120 V 4 MΩ Neon lamp 6 mF + − Figure 7.137 For Prob. 7.72. 7.73 Figure 7.138 shows a circuit for setting the length of time voltage is applied to the electrodes of a welding machine. The time is taken as how long it takes the capacitor to charge from 0 to 8 V. What is the time range covered by the variable resistor? 100 kΩ to 1 MΩ 12 V 2 mF Welding control unit Electrode Figure 7.138 For Prob. 7.73. 7.74 A 120-V dc generator energizes a motor whose coil has an inductance of 50 H and a resistance of 100 . A field discharge resistor of 400 is connected in parallel with the motor to avoid damage to the motor, as shown in Fig. 7.139. The system is at steady state. Find the current through the discharge resistor 100 ms after the breaker is tripped. + − 120 V 400 Ω Circuit breaker Motor Figure 7.139 For Prob. 7.74. COMPREHENSIVE PROBLEMS 7.75 The circuit in Fig. 7.140(a) can be designed as an approximate differentiator or an integrator, depending on whether the output is taken across the resistor or the capacitor, and also on the time constant τ = RC of the circuit and the width T of the input pulse in Fig. 7.140(b). The circuit is a differentiator if τ T , say τ 0.1T, or an integrator if τ T , say τ 10T . (a) What is the minimum pulse width that will allow a differentiator output to appear across the capacitor?
  • 298. 294 PART 1 DC Circuits (b) If the output is to be an integrated form of the input, what is the maximum value the pulse width can assume? 300 kΩ + − 200 pF vi (a) 0 T t Vm vi (b) Figure 7.140 For Prob. 7.75. 7.76 An RL circuit may be used as a differentiator if the output is taken across the inductor and τ T (say τ 0.1T ), where T is the width of the input pulse. If R is fixed at 200 k, determine the maximum value of L required to differentiate a pulse with T = 10 µs. 7.77 An attenuator probe employed with oscilloscopes was designed to reduce the magnitude of the input voltage vi by a factor of 10. As shown in Fig. 7.141, the oscilloscope has internal resistance Rs and capacitance Cs, while the probe has an internal resistance Rp. If Rp is fixed at 6 M, find Rs and Cs for the circuit to have a time constant of 15 µs. vo vi Probe Scope Rp Cs + − + − Rs Figure 7.141 For Prob. 7.77. 7.78 The circuit in Fig. 7.142 is used by a biology student to study “frog kick.” She noticed that the frog kicked a little when the switch was closed but kicked violently for 5 s when the switch was opened. Model the frog as a resistor and calculate its resistance. Assume that it takes 10 mA for the frog to kick violently. 50 Ω 2 H + − 12 V Switch Frog Figure 7.142 For Prob. 7.78. 7.79 To move a spot of a cathode-ray tube across the screen requires a linear increase in the voltage across the deflection plates, as shown in Fig. 7.143. Given that the capacitance of the plates is 4 nF, sketch the current flowing through the plates. Rise time = 2 ms Drop time = 5 ms t 10 v (V) (not to scale) Figure 7.143 For Prob. 7.79.
  • 299. 295 C H A P T E R SECOND-ORDER CIRCUITS 8 “Engineering is not only a learned profession, it is also a learning pro- fession, one whose practitioners first become and then remain students throughout their active careers.” —William L. Everitt Enhancing Your Career To increase your engineering career opportunities after grad- uation, develop a strong fundamental understanding in a broad set of engineering areas. When possible, this might best be accomplished by working toward a graduate degree immediately upon receiving your undergraduate degree. Each degree in engineering represents certain skills the students acquire. At the Bachelor degree level, you learn the language of engineering and the fundamentals of engi- neering and design. At the Master’s level, you acquire the ability to do advanced engineering projects and to commu- nicate your work effectively both orally and in writing. The Ph.D. represents a thorough understanding of the fundamen- tals of electrical engineering and a mastery of the skills nec- essary both for working at the frontiers of an engineering area and for communicating one’s effort to others. If you have no idea what career you should pursue af- ter graduation, a graduate degree program will enhance your ability to explore career options. Since your undergraduate degree will only provide you with the fundamentals of en- gineering, a Master’s degree in engineering supplemented by business courses benefits more engineering students than does getting a Master’s of Business Administration (MBA). The best time to get your MBA is after you have been a prac- ticing engineer for some years and decide your career path would be enhanced by strengthening your business skills. Engineers should constantly educate themselves, formally and informally, taking advantage of all means of education. Perhaps there is no better way to enhance your career than to join a professional society such as IEEE and be an active member. Key career plot points Networking worldwide Professional organization Technical information Career resources Networking the World TM Enhancing your career involves understanding your goals, adapting to changes, anticipating opportunities, and planning your own niche. (Courtesy of IEEE.)
  • 300. 296 PART 1 DC Circuits 8.1 INTRODUCTION In the previous chapter we considered circuits with a single storage ele- ment (a capacitor or an inductor). Such circuits are first-order because the differential equations describing them are first-order. In this chap- ter we will consider circuits containing two storage elements. These are known as second-order circuits because their responses are described by differential equations that contain second derivatives. Typical examples of second-order circuits are RLC circuits, in which the three kinds of passive elements are present. Examples of such circuits are shown in Fig. 8.1(a) and (b). Other examples are RC and RL circuits, as shown in Fig. 8.1(c) and (d). It is apparent from Fig. 8.1 that a second-order circuit may have two storage elements of different type or the same type (provided elements of the same type cannot be represented by an equivalent single element). An op amp circuit with two storage elements may also be a second-order circuit. As with first-order circuits, a second-order circuit may contain several resistors and dependent and independent sources. A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements. Our analysis of second-order circuits will be similar to that used for first-order. We will first consider circuits that are excited by the initial conditions of the storage elements. Although these circuits may contain dependent sources, they are free of independent sources. These source- free circuits will give natural responses as expected. Later we will con- sider circuits that are excited by independent sources. These circuits will give both the natural response and the forced response. We consider only dc independent sources in this chapter. The case of sinusoidal and exponential sources is deferred to later chapters. vs R R L C + − (a) is C L R (b) vs R1 R2 + − (c) is C2 C1 (d) L1 L2 Figure8.1 Typical examples of second-order circuits: (a) series RLC circuit, (b) parallel RLC circuit, (c) RL circuit, (d) RC circuit. We begin by learning how to obtain the initial conditions for the cir- cuit variables and their derivatives, as this is crucial to analyzing second- order circuits. Then we consider series and parallel RLC circuits such as shown in Fig. 8.1 for the two cases of excitation: by initial conditions of the energy storage elements and by step inputs. Later we examine other types of second-order circuits, including op amp circuits. We will con- sider PSpice analysis of second-order circuits. Finally, we will consider the automobile ignition system and smoothing circuits as typical appli- cations of the circuits treated in this chapter. Other applications such as resonant circuits and filters will be covered in Chapter 14. 8.2 FINDING INITIAL AND FINAL VALUES Perhaps the major problem students face in handling second-order circuits is finding the initial and final conditions on circuit variables. Students are
  • 301. CHAPTER 8 Second-Order Circuits 297 usually comfortable getting the initial and final values of v and i but often have difficulty finding the initial values of their derivatives: dv/dt and di/dt. For this reason, this section is explicitly devoted to the subtleties of getting v(0), i(0), dv(0)/dt, di(0)/dt, i(∞), and v(∞). Unless otherwise stated in this chapter, v denotes capacitor voltage, while i is the inductor current. There are two key points to keep in mind in determining the initial conditions. First—as always in circuit analysis—we must carefully handle the polarity of voltage v(t) across the capacitor and the direction of the cur- rent i(t) through the inductor. Keep in mind that v and i are defined strictly according to the passive sign convention (see Figs. 6.3 and 6.23). One should carefully observe how these are defined and apply them ac- cordingly. Second, keep in mind that the capacitor voltage is always continu- ous so that v(0+ ) = v(0− ) (8.1a) and the inductor current is always continuous so that i(0+ ) = i(0− ) (8.1b) where t = 0− denotes the time just before a switching event and t = 0+ is the time just after the switching event, assuming that the switching event takes place at t = 0. Thus, in finding initial conditions, we first focus on those variables that cannot change abruptly, capacitor voltage and inductor current, by applying Eq. (8.1). The following examples illustrate these ideas. E X A M P L E 8 . 1 The switch in Fig. 8.2 has been closed for a long time. It is open at t = 0. Find: (a) i(0+ ), v(0+ ), (b) di(0+ )dt, dv(0+ )/dt, (c) i(∞), v(∞). 12 V 4 Ω 0.25 H + − 0.1 F i v + − 2 Ω t = 0 Figure8.2 For Example 8.1. Solution: (a) If the switch is closed a long time before t = 0, it means that the circuit has reached dc steady state at t = 0. At dc steady state, the inductor acts like a short circuit, while the capacitor acts like an open circuit, so we have the circuit in Fig. 8.3(a) at t = 0− . Thus, 12 V 4 Ω 0.25 H + − 0.1 F i (b) 12 V 4 Ω + − i v + − 2 Ω (a) 12 V 4 Ω + − i v + − (c) + − vL v + − Figure8.3 Equivalent circuit of that in Fig. 8.2 for: (a) t = 0−, (b) t = 0+, (c) t → ∞.
  • 302. 298 PART 1 DC Circuits i(0− ) = 12 4 + 2 = 2 A, v(0− ) = 2i(0− ) = 4 V As the inductor current and the capacitor voltage cannot change abruptly, i(0+ ) = i(0− ) = 2 A, v(0+ ) = v(0− ) = 4 V (b) At t = 0+ , the switch is open; the equivalent circuit is as shown in Fig. 8.3(b). The same current flows through both the inductor and capacitor. Hence, iC(0+ ) = i(0+ ) = 2 A Since C dv/dt = iC, dv/dt = iC/C, and dv(0+ ) dt = iC(0+ ) C = 2 0.1 = 20 V/s Similarly, since L di/dt = vL, di/dt = vL/L. We now obtain vL by applying KVL to the loop in Fig. 8.3(b). The result is −12 + 4i(0+ ) + vL(0+ ) + v(0+ ) = 0 or vL(0+ ) = 12 − 8 − 4 = 0 Thus, di(0+ ) dt = vL(0+ ) L = 0 0.25 = 0 A/s (c) For t 0, the circuit undergoes transience. But as t → ∞, the circuit reaches steady state again. The inductor acts like a short circuit and the capacitor like an open circuit, so that the circuit becomes that shown in Fig. 8.3(c), from which we have i(∞) = 0 A, v(∞) = 12 V P R A C T I C E P R O B L E M 8 . 1 The switch in Fig. 8.4 was open for a long time but closed at t = 0. De- termine: (a) i(0+ ), v(0+ ), (b) di(0+ )dt, dv(0+ )/dt, (c) i(∞), v(∞). 10 Ω 24 V v + − 2 Ω + − i t = 0 0.4 H F 1 20 Figure8.4 For Practice Prob. 8.1. Answer: (a) 2 A, 4 V, (b) 50 A/s, 0 V/s, (c) 12 A, 24 V.
  • 303. CHAPTER 8 Second-Order Circuits 299 E X A M P L E 8 . 2 In the circuit of Fig. 8.5, calculate: (a) iL(0+ ), vC(0+ ), vR(0+ ), (b) diL(0+ )dt, dvC(0+ )/dt, dvR(0+ )/dt, (c) iL(∞), vC(∞), vR(∞). 3u(t) A 4 Ω 20 V 0.6 H vC + − vR + − 2 Ω + − iL F 1 2 Figure8.5 For Example 8.2. Solution: (a) For t 0, 3u(t) = 0. At t = 0− , since the circuit has reached steady state, the inductor can be replaced by a short circuit, while the capacitor is replaced by an open circuit as shown in Fig. 8.6(a). From this figure we obtain iL(0− ) = 0, vR(0− ) = 0, vC(0− ) = −20 V (8.2.1) Although the derivatives of these quantities at t = 0− are not required, it is evident that they are all zero, since the circuit has reached steady state and nothing changes. 3 A 4 Ω 20 V 0.6 H vR + − 2 Ω + − iL iC vL (b) a b 4 Ω 20 V vC + − vR + − 2 Ω + − iL (a) vo vC + − + − + − F 1 2 Figure8.6 The circuit in Fig. 8.5 for: (a) t = 0−, (b) t = 0+. For t 0, 3u(t) = 3, so that the circuit is now equivalent to that in Fig. 8.6(b). Since the inductor current and capacitor voltage cannot change abruptly, iL(0+ ) = iL(0− ) = 0, vC(0+ ) = vC(0− ) = −20 V (8.2.2) Although the voltage across the 4- resistor is not required, we will use it to apply KVL and KCL; let it be called vo. Applying KCL at node a in Fig. 8.6(b) gives 3 = vR(0+ ) 2 + vo(0+ ) 4 (8.2.3)
  • 304. 300 PART 1 DC Circuits Applying KVL to the middle mesh in Fig. 8.6(b) yields −vR(0+ ) + vo(0+ ) + vC(0+ ) + 20 = 0 (8.2.4) Since vC(0+) = −20 V from Eq. (8.2.2), Eq. (8.2.4) implies that vR(0+ ) = vo(0+ ) (8.2.5) From Eqs. (8.2.3) and (8.2.5), we obtain vR(0+ ) = vo(0+ ) = 4 V (8.2.6) (b) Since L diL/dt = vL, diL(0+ ) dt = vL(0+ ) L But applying KVL to the right mesh in Fig. 8.6(b) gives vL(0+ ) = vC(0+ ) + 20 = 0 Hence, diL(0+ ) dt = 0 (8.2.7) Similarly, since C dvC/dt = iC, then dvC/dt = iC/C. We apply KCL at node b in Fig. 8.6(b) to get iC: vo(0+ ) 4 = iC(0+ ) + iL(0+ ) (8.2.8) Since vo(0+ ) = 4 and iL(0+ ) = 0, iC(0+ ) = 4/4 = 1 A. Then dvC(0+ ) dt = iC(0+ ) C = 1 0.5 = 2 V/s (8.2.9) To get dvR(0+ )/dt, we apply KCL to node a and obtain 3 = vR 2 + vo 4 Taking the derivative of each term and setting t = 0+ gives 0 = 2 dvR(0+ ) dt + dvo(0+ ) dt (8.2.10) We also apply KVL to the middle mesh in Fig. 8.6(b) and obtain −vR + vC + 20 + vo = 0 Again, taking the derivative of each term and setting t = 0+ yields − dvR(0+ ) dt + dvC(0+ ) dt + dvo(0+ ) dt = 0 Substituting for dvC(0+ )/dt = 2 gives dvR(0+ ) dt = 2 + dvo(0+ ) dt (8.2.11) From Eqs. (8.2.10) and (8.2.11), we get
  • 305. CHAPTER 8 Second-Order Circuits 301 dvR(0+ ) dt = 2 3 V/s We can find diR(0+ )/dt although it is not required. Since vR = 5iR, diR(0+ ) dt = 1 5 dvR(0+ ) dt = 1 5 2 3 = 2 15 A/s (c) As t → ∞, the circuit reaches steady state. We have the equivalent circuit in Fig. 8.6(a) except that the 3-A current source is now operative. By current division principle, iL(∞) = 2 2 + 4 3 A = 1 A vR(∞) = 4 2 + 4 3 A × 2 = 4 V, vC(∞) = −20 V (8.2.12) P R A C T I C E P R O B L E M 8 . 2 For the circuit in Fig. 8.7, find: (a) iL(0+ ), vC(0+ ), vR(0+ ), (b) diL(0+ )/dt, dvC(0+ )/dt, dvR(0+ )/dt, (c) iL(∞), vC(∞), vR(∞). 2u(t) A 3 A 5 Ω 2 H iC iL vC + − iR vL vR + − + − F 1 5 Figure8.7 For Practice Prob. 8.2. Answer: (a) −3 A, 0, 0, (b) 0, 10 V/s, 0, (c) −1 A, 10 V, 10 V. 8.3 THE SOURCE-FREE SERIES RLC CIRCUIT An understanding of the natural response of the series RLC circuit is a necessary background for future studies in filter design and communica- tions networks. Consider the series RLC circuit shown in Fig. 8.8. The circuit is being excited by the energy initially stored in the capacitor and inductor. The energy is represented by the initial capacitor voltage V0 and initial inductor current I0. Thus, at t = 0, v(0) = 1 C 0 −∞ i dt = V0 (8.2a) i(0) = I0 (8.2b) i R L Io Vo C + − Figure 8.8 A source-free series RLC circuit. Applying KVL around the loop in Fig. 8.8, Ri + L di dt + 1 C t −∞ i dt = 0 (8.3)
  • 306. 302 PART 1 DC Circuits To eliminate the integral, we differentiate with respect to t and rearrange terms. We get d2 i dt2 + R L di dt + i LC = 0 (8.4) This is a second-order differential equation and is the reason for calling the RLC circuits in this chapter second-order circuits. Our goal is to solve Eq. (8.4). To solve such a second-order differential equation requires that we have two initial conditions, such as the initial value of i and its first derivative or initial values of some i and v. The initial value of i is given in Eq. (8.2b). We get the initial value of the derivative of i from Eqs. (8.2a) and (8.3); that is, Ri(0) + L di(0) dt + V0 = 0 or di(0) dt = − 1 L (RI0 + V0) (8.5) With the two initial conditions in Eqs. (8.2b) and (8.5), we can now solve Eq. (8.4). Our experience in the preceding chapter on first-order circuits suggests that the solution is of exponential form. So we let i = Aest (8.6) where A and s are constants to be determined. Substituting Eq. (8.6) into Eq. (8.4) and carrying out the necessary differentiations, we obtain As2 est + AR L sest + A LC est = 0 or Aest s2 + R L s + 1 LC = 0 (8.7) Since i = Aest is the assumed solution we are trying to find, only the expression in parentheses can be zero: s2 + R L s + 1 LC = 0 (8.8) See Appendix C.1 for the formula to find the roots of a quadratic equation. This quadratic equation is known as the characteristic equation of the differential Eq. (8.4), since the roots of the equation dictate the character of i. The two roots of Eq. (8.8) are s1 = − R 2L + R 2L 2 − 1 LC (8.9a) s2 = − R 2L − R 2L 2 − 1 LC (8.9b) A more compact way of expressing the roots is s1 = −α + α2 − ω2 0, s2 = −α − α2 − ω2 0 (8.10)
  • 307. CHAPTER 8 Second-Order Circuits 303 where α = R 2L , ω0 = 1 √ LC (8.11) The roots s1 and s2 are called natural frequencies, measured in nepers per second (Np/s), because they are associated with the natural response of the circuit; ω0 is known as the resonant frequency or strictly as the undamped natural frequency, expressed in radians per second (rad/s); and α is the neper frequency or the damping factor, expressed in nepers per second. In terms of α and ω0, Eq. (8.8) can be written as s2 + 2αs + ω2 0 = 0 (8.8a) The variables s and ω are important quantities we will be discussing throughout the rest of the text. The neper (Np) is a dimensionless unit named after John Napier (1550–1617), a Scottish math- ematician. The ratio α/ω0 is known as the damping ratio ζ. The two values of s in Eq. (8.10) indicate that there are two possible solutions for i, each of which is of the form of the assumed solution in Eq. (8.6); that is, i1 = A1es1t , i2 = A2es2t (8.12) Since Eq. (8.4) is a linear equation, any linear combination of the two distinct solutions i1 and i2 is also a solution of Eq. (8.4). A complete or total solution of Eq. (8.4) would therefore require a linear combination of i1 and i2. Thus, the natural response of the series RLC circuit is i(t) = A1es1t + A2es2t (8.13) where the constants A1 and A2 are determined from the initial values i(0) and di(0)/dt in Eqs. (8.2b) and (8.5). From Eq. (8.10), we can infer that there are three types of solutions: 1. If α ω0, we have the overdamped case. 2. If α = ω0, we have the critically damped case. 3. If α ω0, we have the underdamped case. We will consider each of these cases separately. The response is overdamped when the roots of the circuit’s characteristic equation are unequal and real, critically damped when the roots are equal and real, and underdamped when the roots are complex. Overdamped Case (α ω0) From Eqs. (8.9) and (8.10), α ω0 when C 4L/R2 . When this hap- pens, both roots s1 and s2 are negative and real. The response is i(t) = A1es1t + A2es2t (8.14) which decays and approaches zero as t increases. Figure 8.9(a) illustrates a typical overdamped response. Critically Damped Case (α = ω0) When α = ω0, C = 4L/R2 and s1 = s2 = −α = − R 2L (8.15) For this case, Eq. (8.13) yields i(t) = A1e−αt + A2e−αt = A3e−αt
  • 308. 304 PART 1 DC Circuits where A3 = A1 + A2. This cannot be the solution, because the two initial conditions cannot be satisfied with the single constant A3. What then could be wrong? Our assumption of an exponential solution is incorrect for the special case of critical damping. Let us go back to Eq. (8.4). When α = ω0 = R/2L, Eq. (8.4) becomes d2 i dt2 + 2α di dt + α2 i = 0 or d dt di dt + αi + α di dt + αi = 0 (8.16) If we let f = di dt + αi (8.17) then Eq. (8.16) becomes df dt + αf = 0 which is a first-order differential equation with solution f = A1e−αt , where A1 is a constant. Equation (8.17) then becomes di dt + αi = A1e−αt or eαt di dt + eαt αi = A1 (8.18) This can be written as d dt (eαt i) = A1 (8.19) Integrating both sides yields eαt i = A1t + A2 or i = (A1t + A2)eαt (8.20) where A2 is another constant. Hence, the natural response of the critically damped circuit is a sum of two terms: a negative exponential and a negative exponential multiplied by a linear term, or i(t) = (A2 + A1t)e−αt (8.21) A typical critically damped response is shown in Fig. 8.9(b). In fact, Fig. 8.9(b) is a sketch of i(t) = te−αt , which reaches a maximum value of e−1 /α at t = 1/α, one time constant, and then decays all the way to zero. t i(t) 0 e–t (c) t 1 a i(t) 0 (b) t i(t) 0 (a) 2p vd Figure 8.9 (a) Overdamped response, (b) critically damped response, (c) underdamped response. Underdamped Case (α ω0) For α ω0, C 4L/R2 . The roots may be written as s1 = −α + −(ω2 0 − α2) = −α + jωd (8.22a) s2 = −α − −(ω2 0 − α2) = −α − jωd (8.22b)
  • 309. CHAPTER 8 Second-Order Circuits 305 where j = √ −1 and ωd = √ ω2 0 − α2 , which is called the damping frequency. Both ω0 and ωd are natural frequencies because they help determine the natural response; while ω0 is often called the undamped natural frequency, ωd is called the damped natural frequency. The natural response is i(t) = A1e−(α−jωd )t + A2e−(α+jωd )t = e−αt (A1ejωd t + A2e−jωd t ) (8.23) Using Euler’s identities, ejθ = cos θ + j sin θ, e−jθ = cos θ − j sin θ (8.24) we get i(t) = e−αt [A1(cos ωdt + j sin ωdt) + A2(cos ωdt − j sin ωdt)] = e−αt [(A1 + A2) cos ωdt + j(A1 − A2) sin ωdt] (8.25) Replacing constants (A1 + A2) and j(A1 − A2) with constants B1 and B2, we write i(t) = e−αt (B1 cos ωdt + B2 sin ωdt) (8.26) With the presence of sine and cosine functions, it is clear that the natural response for this case is exponentially damped and oscillatory in nature. The response has a time constant of 1/α and a period of T = 2π/ωd. Fig- ure 8.9(c) depicts a typical underdamped response. [Figure 8.9 assumes for each case that i(0) = 0.] Once the inductor current i(t) is found for the RLC series circuit as shown above, other circuit quantities such as individual element voltages can easily be found. For example, the resistor voltage is vR = Ri, and the inductor voltage is vL = L di/dt. The inductor current i(t) is selected as the key variable to be determined first in order to take advantage of Eq. (8.1b). We conclude this section by noting the following interesting, pe- culiar properties of an RLC network: R = 0 produces a perfectly sinusoidal response. Thisresponsecannotbepracticallyaccomplished with L and C because of the inherent losses in them. See Figs. 6.8 and 6.26. An electronic de- vice called an oscillator can produce a perfectly sinusoidal response. Examples 8.5 and 8.7 demonstrate the effect of varying R. Theresponseofasecond-ordercircuitwithtwo storage elements of the same type, as in Fig. 8.1(c) and (d), cannot be oscillatory. 1. The behavior of such a network is captured by the idea of damping, which is the gradual loss of the initial stored energy, as evidenced by the continuous decrease in the amplitude of the response. The damping effect is due to the presence of resistance R. The damping factor α determines the rate at which the response is damped. If R = 0, then α = 0, and we have an LC circuit with 1/ √ LC as the undamped natural frequency. Since α ω0 in this case, the response is not only undamped but also oscillatory. The circuit is said to be loss- less, because the dissipating or damping element (R) is absent. By adjusting the value of R, the response may be made undamped, overdamped, critically damped, or underdamped. 2. Oscillatory response is possible due to the presence of the two types of storage elements. Having both L and C allows the flow of energy back and forth between the two. The damped oscillation exhibited by the underdamped response is known as ringing. It stems from the ability of the storage elements L and C to transfer energy back and forth between them.
  • 310. 306 PART 1 DC Circuits 3. Observe from Fig. 8.9 that the waveforms of the responses differ. In general, it is difficult to tell from the waveforms the difference between the overdamped and critically damped responses. The critically damped case is the borderline between the underdamped and overdamped cases and it decays the fastest. With the same initial conditions, the overdamped case has the longest settling time, because it takes the longest time to dissipate the initial stored energy. If we desire the fastest response without oscillation or ringing, the critically damped circuit is the right choice. Whatthismeansinmostpracticalcircuitsisthat weseekanoverdampedcircuitthatisascloseas possible to a critically damped circuit. E X A M P L E 8 . 3 In Fig. 8.8, R = 40 , L = 4 H, and C = 1/4 F. Calculate the char- acteristic roots of the circuit. Is the natural response overdamped, under- damped, or critically damped? Solution: We first calculate α = R 2L = 40 2(4) = 5, ω0 = 1 √ LC = 1 4 × 1 4 = 1 The roots are s1,2 = −α ± α2 − ω2 0 = −5 ± √ 25 − 1 or s1 = −0.101, s2 = −9.899 Since α ω0, we conclude that the response is overdamped. This is also evident from the fact that the roots are real and negative. P R A C T I C E P R O B L E M 8 . 3 If R = 10 , L = 5 H, and C = 2 mF in Fig. 8.8, find α, ω0, s1, and s2. What type of natural response will the circuit have? Answer: 1, 10, −1 ± j9.95, underdamped. E X A M P L E 8 . 4 Find i(t) in the circuit in Fig. 8.10. Assume that the circuit has reached steady state at t = 0− . t = 0 10 V 4 Ω 0.5 H 0.02 F v + − 3 Ω + − 6 Ω i Figure8.10 For Example 8.4. Solution: For t 0, the switch is closed. The capacitor acts like an open circuit while the inductor acts like a shunted circuit. The equivalent circuit is shown in Fig. 8.11(a). Thus, at t = 0, i(0) = 10 4 + 6 = 1 A, v(0) = 6i(0) = 6 V where i(0) is the initial current through the inductor and v(0) is the initial voltage across the capacitor.
  • 311. CHAPTER 8 Second-Order Circuits 307 0.5 H 0.02 F 9 Ω i (b) 10 V 4 Ω v + − 6 Ω + − i (a) v + − Figure8.11 The circuit in Fig. 8.10: (a) for t 0, (b) for t 0. For t 0, the switch is opened and the voltage source is dis- connected. The equivalent circuit is shown in Fig. 8.11(b), which is a source-free series RLC circuit. Notice that the 3- and 6- resistors, which are in series in Fig. 8.10 when the switch is opened, have been combined to give R = 9 in Fig. 8.11(b). The roots are calculated as follows: α = R 2L = 9 2 1 2 = 9, ω0 = 1 √ LC = 1 1 2 × 1 50 = 10 s1,2 = −α ± α2 − ω2 0 = −9 ± √ 81 − 100 or s1,2 = −9 ± j4.359 Hence, the response is underdamped (α ω); that is, i(t) = e−9t (A1 cos 4.359t + A2 sin 4.359t) (8.4.1) We now obtain A1 and A2 using the initial conditions. At t = 0, i(0) = 1 = A1 (8.4.2) From Eq. (8.5), di dt t=0 = − 1 L [Ri(0) + v(0)] = −2[9(1) − 6] = −6 A/s (8.4.3) Note that v(0) = V0 = −6 V is used, because the polarity of v in Fig. 8.11(b) is opposite that in Fig. 8.8. Taking the derivative of i(t) in Eq. (8.4.1), di dt = −9e−9t (A1 cos 4.359t + A2 sin 4.359t) + e−9t (4.359)(−A1 sin 4.359t + A2 cos 4.359t) Imposing the condition in Eq. (8.4.3) at t = 0 gives −6 = −9(A1 + 0) + 4.359(−0 + A2) But A1 = 1 from Eq. (8.4.2). Then −6 = −9 + 4.359A2 ⇒ A2 = 0.6882
  • 312. 308 PART 1 DC Circuits Substituting the values of A1 and A2 in Eq. (8.4.1) yields the com- plete solution as i(t) = e−9t (cos 4.359t + 0.6882 sin 4.359t) A P R A C T I C E P R O B L E M 8 . 4 The circuit in Fig. 8.12 has reached steady state at t = 0− . If the make- before-break switch moves to position b at t = 0, calculate i(t) for t 0. t = 0 a b 50 V 10 Ω 1 H + − 5 Ω i(t) F 1 9 Figure8.12 For Practice Prob. 8.4. Answer: e−2.5t (5 cos 1.6583t − 7.5378 sin 1.6583t) A. 8.4 THE SOURCE-FREE PARALLEL RLC CIRCUIT Parallel RLC circuits find many practical applications, notably in com- munications networks and filter designs. v R L C I0 v + − v + − V0 + − Figure 8.13 A source-free parallel RLC circuit. Consider the parallel RLC circuit shown in Fig. 8.13. Assume initial inductor current I0 and initial capacitor voltage V0, i(0) = I0 = 1 L 0 ∞ v(t) dt (8.27a) v(0) = V0 (8.27b) Since the three elements are in parallel, they have the same voltage v across them. According to passive sign convention, the current is entering each element; that is, the current through each element is leaving the top node. Thus, applying KCL at the top node gives v R + 1 L t −∞ v dt + C dv dt = 0 (8.28) Taking the derivative with respect to t and dividing by C results in d2 v dt2 + 1 RC dv dt + 1 LC v = 0 (8.29) We obtain the characteristic equation by replacing the first derivative by s and the second derivative by s2 . By following the same reasoning used in establishing Eqs. (8.4) through (8.8), the characteristic equation is obtained as s2 + 1 RC s + 1 LC = 0 (8.30) The roots of the characteristic equation are s1,2 = − 1 2RC ± 1 2RC 2 − 1 LC
  • 313. CHAPTER 8 Second-Order Circuits 309 or s1,2 = −α ± α2 − ω2 0 (8.31) where α = 1 2RC , ω0 = 1 √ LC (8.32) The names of these terms remain the same as in the preceding section, as they play the same role in the solution. Again, there are three possible solutions, depending on whether α ω0, α = ω0, or α ω0. Let us consider these cases separately. Overdamped Case (α ω0) From Eq. (8.32), α ω0 when L 4R2 C. The roots of the characteristic equation are real and negative. The response is v(t) = A1es1t + A2es2t (8.33) Critically Damped Case (α = ω0) For α = ω, L = 4R2 C. The roots are real and equal so that the response is v(t) = (A1 + A2t)e−αt (8.34) Underdamped Case (α ω0) When α ω0, L 4R2 C. In this case the roots are complex and may be expressed as s1,2 = −α ± jωd (8.35) where ωd = ω2 0 − α2 (8.36) The response is v(t) = e−αt (A1 cos ωdt + A2 sin ωdt) (8.37) The constants A1 and A2 in each case can be determined from the initial conditions. We need v(0) and dv(0)/dt. The first term is known from Eq. (8.27b). We find the second term by combining Eqs. (8.27) and (8.28), as V0 R + I0 + C dv(0) dt = 0 or dv(0) dt = − (V0 + RI0) RC (8.38) The voltage waveforms are similar to those shown in Fig. 8.9 and will depend on whether the circuit is overdamped, underdamped, or critically damped.
  • 314. 310 PART 1 DC Circuits Having found the capacitor voltage v(t) for the parallel RLC circuit as shown above, we can readily obtain other circuit quantities such as individual element currents. For example, the resistor current is iR = v/R and the capacitor voltage is vC = C dv/dt. We have selected the capacitor voltage v(t) as the key variable to be determined first in order to take advantage of Eq. (8.1a). Notice that we first found the inductor current i(t) for the RLC series circuit, whereas we first found the capacitor voltage v(t) for the parallel RLC circuit. E X A M P L E 8 . 5 In the parallel circuit of Fig. 8.13, find v(t) for t 0, assuming v(0) = 5 V, i(0) = 0, L = 1 H, and C = 10 mF. Consider these cases: R = 1.923 , R = 5 , and R = 6.25 . Solution: CASE 1 If R = 1.923 , α = 1 2RC = 1 2 × 1.923 × 10 × 10−3 = 26 ω0 = 1 √ LC = 1 √ 1 × 10 × 10−3 = 10 Since α ω0 in this case, the response is overdamped. The roots of the characteristic equation are s1,2 = −α ± α2 − ω2 0 = −2, −50 and the corresponding response is v(t) = A1e−2t + A2e−50t (8.5.1) We now apply the initial conditions to get A1 and A2. v(0) = 5 = A1 + A2 (8.5.2) dv(0) dt = − v(0) + Ri(0) RC = − 5 + 0 1.923 × 10 × 10−3 = 260 But differentiating Eq. (8.5.1), dv dt = −2A1e−2t − 50A2e−50t At t = 0, 260 = −2A1 − 50A2 (8.5.3) From Eqs. (8.5.2) and (8.5.3), we obtain A1 = 10.625 and A2 = −5.625. Substituting A1 and A2 in Eq. (8.5.1) yields v(t) = 10.625e−2t − 5.625e−50t (8.5.4) CASE 2 When R = 5 , α = 1 2RC = 1 2 × 5 × 10 × 10−3 = 10
  • 315. CHAPTER 8 Second-Order Circuits 311 while ω0 = 10 remains the same. Since α = ω0 = 10, the response is critically damped. Hence, s1 = s2 = −10, and v(t) = (A1 + A2t)e−10t (8.5.5) To get A1 and A2, we apply the initial conditions v(0) = 5 = A1 (8.5.6) dv(0) dt = − v(0) + Ri(0) RC = − 5 + 0 5 × 10 × 10−3 = 100 But differentiating Eq. (8.5.5), dv dt = (−10A1 − 10A2t + A2)e−10t At t = 0, 100 = −10A1 + A2 (8.5.7) From Eqs. (8.5.6) and (8.5.7), A1 = 5 and A2 = 150. Thus, v(t) = (5 + 150t)e−10t V (8.5.8) CASE 3 When R = 6.25 , α = 1 2RC = 1 2 × 6.25 × 10 × 10−3 = 8 while ω0 = 10 remains the same. As α ω0 in this case, the response is underdamped. The roots of the characteristic equation are s1,2 = −α ± α2 − ω2 0 = −8 ± j6 Hence, v(t) = (A1 cos 6t + A2 sin 6t)e−8t (8.5.9) We now obtain A1 and A2, as v(0) = 5 = A1 (8.5.10) dv(0) dt = − v(0) + Ri(0) RC = − 5 + 0 6.25 × 10 × 10−3 = 80 But differentiating Eq. (8.5.9), dv dt = (−8A1 cos 6t − 8A2 sin 6t − 6A1 sin 6t + 6A2 cos 6t)e−8t At t = 0, 80 = −8A1 + 6A2 (8.5.11) From Eqs. (8.5.10) and (8.5.11), A1 = 5 and A2 = 20. Thus, v(t) = (5 cos 6t + 20 sin 6t)e−8t (8.5.12) Notice that by increasing the value of R, the degree of damping decreases and the responses differ. Figure 8.14 plots the three cases.
  • 316. 312 PART 1 DC Circuits 0.2 0.4 0.6 1 2 3 4 5 6 7 8 9 10 0.8 1 –1 t (s) v(t) V Overdamped Critically damped Underdamped Figure8.14 For Example 8.5: responses for three degrees of damping. P R A C T I C E P R O B L E M 8 . 5 In Fig. 8.13, let R = 2 , L = 0.4 H, C = 25 mF, v(0) = 0, i(0) = 3 A. Find v(t) for t 0. Answer: −120te−10t V. E X A M P L E 8 . 6 Find v(t) for t 0 in the RLC circuit of Fig. 8.15. 40 V 0.4 H 50 Ω 20 mF 30 Ω + − i t = 0 v + − Figure8.15 For Example 8.6. Solution: When t 0, the switch is open; the inductor acts like a short circuit while the capacitor behaves like an open circuit. The initial voltage across the capacitor is the same as the voltage across the 50- resistor; that is, v(0) = 50 30 + 50 (40) = 5 8 × 40 = 25 V (8.6.1) The initial current through the inductor is
  • 317. CHAPTER 8 Second-Order Circuits 313 i(0) = − 40 30 + 50 = −0.5 A The direction of i is as indicated in Fig. 8.15 to conform with the direction of I0 in Fig. 8.13, which is in agreement with the convention that current flows into the positive terminal of an inductor (see Fig. 6.23). We need to express this in terms of dv/dt, since we are looking for v. dv(0) dt = − v(0) + Ri(0) RC = − 25 − 50 × 0.5 50 × 20 × 10−6 = 0 (8.6.2) When t 0, the switch is closed. The voltage source along with the 30- resistor is separated from the rest of the circuit. The parallel RLC circuit acts independently of the voltage source, as illustrated in Fig. 8.16. Next, we determine that the roots of the characteristic equation are α = 1 2RC = 1 2 × 50 × 20 × 10−6 = 500 ω0 = 1 √ LC = 1 √ 0.4 × 20 × 10−6 = 354 s1,2 = −α ± α2 − ω2 0 = −500 ± √ 250,000 − 124,997.6 = −500 ± 354 or s1 = −854, s2 = −146 Since α ω0, we have the overdamped response v(t) = A1e−854t + A2e−164t (8.6.3) At t = 0, we impose the condition in Eq. (8.6.1), v(0) = 25 = A1 + A2 ⇒ A2 = 25 − A1 (8.6.4) Taking the derivative of v(t) in Eq. (8.6.3), dv dt = −854A1e−854t − 164A2e−164t Imposing the condition in Eq. (8.6.2), 40 V 0.4 H 50 Ω 20 mF 30 Ω + − Figure8.16 The circuit in Fig. 8.15 when t 0. The parallel RLC circuit on the left-hand side acts inde- pendently of the circuit on the right-hand side of the junction.
  • 318. 314 PART 1 DC Circuits dv(0) dt = 0 = −854A1 − 164A2 or 0 = 854A1 + 164A2 (8.6.5) Solving Eqs. (8.6.4) and (8.6.5) gives A1 = −5.16, A2 = 30.16 Thus, the complete solution in Eq. (8.6.3) becomes v(t) = −5.16e−854t + 30.16e−164t V P R A C T I C E P R O B L E M 8 . 6 Refer to the circuit in Fig. 8.17. Find v(t) for t 0. 2 A 4 mF 20 Ω 10 H t = 0 v + − Figure8.17 For Practice Prob. 8.6. Answer: 66.67(e−10t − e−2.5t ) V. 8.5 STEP RESPONSE OF A SERIES RLC CIRCUIT As we learned in the preceding chapter, the step response is obtained by the sudden application of a dc source. Consider the series RLC circuit shown in Fig. 8.18. Applying KVL around the loop for t 0, L di dt + Ri + v = Vs (8.39) But i = C dv dt Substituting for i in Eq. (8.39) and rearranging terms, d2 v dt2 + R L dv dt + v LC = Vs LC (8.40) which has the same form as Eq. (8.4). More specifically, the coefficients are the same (and that is important in determining the frequency param- eters) but the variable is different. (Likewise, see Eq. (8.47).) Hence, the characteristic equation for the series RLC circuit is not affected by the presence of the dc source. Vs R L C + − i t = 0 v + − Figure 8.18 Step voltage applied to a series RLC circuit. ThesolutiontoEq.(8.40)hastwocomponents: thenaturalresponse vn(t) and the forced response vf (t); that is, v(t) = vn(t) + vf (t) (8.41) The natural response is the solution when we set Vs = 0 in Eq. (8.40) and is the same as the one obtained in Section 8.3. The natural response vn for the overdamped, underdamped, and critically damped cases are:
  • 319. CHAPTER 8 Second-Order Circuits 315 vn(t) = A1es1t + A2es2t (Overdamped) (8.42a) vn(t) = (A1 + A2t)e−αt (Critically damped) (8.42b) vn(t) = (A1 cos ωdt + A2 sin ωdt)e−αt (Underdamped) (8.42c) The forced response is the steady state or final value of v(t). In the circuit in Fig. 8.18, the final value of the capacitor voltage is the same as the source voltage Vs. Hence, vf (t) = v(∞) = Vs (8.43) Thus, the complete solutions for the overdamped, underdamped, and critically damped cases are: v(t) = Vs + A1es1t + A2es2t (Overdamped) (8.44a) v(t) = Vs + (A1 + A2t)e−αt (Critically damped) (8.44b) v(t) = Vs + (A1 cos ωdt + A2 sin ωdt)e−αt (Underdamped) (8.44c) The values of the constants A1 and A2 are obtained from the initial con- ditions: v(0) and dv(0)/dt. Keep in mind that v and i are, respectively, the voltage across the capacitor and the current through the inductor. Therefore, Eq. (8.44) only applies for finding v. But once the capaci- tor voltage vC = v is known, we can determine i = C dv/dt, which is the same current through the capacitor, inductor, and resistor. Hence, the voltage across the resistor is vR = iR, while the inductor voltage is vL = L di/dt. Alternatively, the complete response for any variable x(t) can be found directly, because it has the general form x(t) = xf (t) + xn(t) (8.45) where the xf = x(∞) is the final value and xn(t) is the natural response. The final value is found as in Section 8.2. The natural response has the same form as in Eq. (8.42), and the associated constants are determined from Eq. (8.44) based on the values of x(0) and dx(0)/dt. E X A M P L E 8 . 7 For the circuit in Fig. 8.19, find v(t) and i(t) for t 0. Consider these cases: R = 5 , R = 4 , andR = 1 . 24 V R 1 H + − 0.5 F 1 Ω i t = 0 v + − Figure8.19 For Example 8.7. Solution: CASE 1 When R = 5 . For t 0, the switch is closed. The capa- citor behaves like an open circuit while the inductor acts like a short cir- cuit. The initial current through the inductor is i(0) = 24 5 + 1 = 4 A and the initial voltage across the capacitor is the same as the voltage across the 1- resistor; that is,
  • 320. 316 PART 1 DC Circuits v(0) = 1i(0) = 4 V For t 0, the switch is opened, so that we have the 1- resistor disconnected. What remains is the series RLC circuit with the voltage source. The characteristic roots are determined as follows. α = R 2L = 5 2 × 1 = 2.5, ω0 = 1 √ LC = 1 √ 1 × 0.25 = 2 s1,2 = −α ± α2 − ω2 0 = −1, −4 Since α ω0, we have the overdamped natural response. The total response is therefore v(t) = vf + (A1e−t + A2e−4t ) where vf is the forced or steady-state response. It is the final value of the capacitor voltage. In Fig. 8.19, vf = 24 V. Thus, v(t) = 24 + (A1e−t + A2e−4t ) (8.7.1) We now need to find A1 and A2 using the initial conditions. v(0) = 4 = 24 + A1 + A2 or −20 = A1 + A2 (8.7.2) The current through the inductor cannot change abruptly and is the same current through the capacitor at t = 0+ because the inductor and capacitor are now in series. Hence, i(0) = C dv(0) dt = 4 ⇒ dv(0) dt = 4 C = 4 0.25 = 16 Before we use this condition, we need to take the derivative of v in Eq. (8.7.1). dv dt = −A1e−t − 4A2e−4t (8.7.3) At t = 0, dv(0) dt = 16 = −A1 − 4A2 (8.7.4) From Eqs. (8.7.2) and (8.7.4), A1 = −64/3 and A2 = 4/3. Substituting A1 and A2 in Eq. (8.7.1), we get v(t) = 24 + 4 3 (−16e−t + e−4t ) V (8.7.5) Since the inductor and capacitor are in series for t 0, the inductor current is the same as the capacitor current. Hence, i(t) = C dv dt Multiplying Eq. (8.7.3) by C = 0.25 and substituting the values of A1 and A2 gives
  • 321. CHAPTER 8 Second-Order Circuits 317 i(t) = 4 3 (4e−t − e−4t ) A (8.7.6) Note that i(0) = 4 A, as expected. CASE 2 WhenR = 4 . Again, theinitialcurrentthroughtheinductor is i(0) = 24 4 + 1 = 4.5 A and the initial capacitor voltage is v(0) = 1i(0) = 4.5 V For the characteristic roots, α = R 2L = 4 2 × 1 = 2 while ω0 = 2 remains the same. In this case, s1 = s2 = −α = −2, and we have the critically damped natural response. The total response is therefore v(t) = vf + (A1 + A2t)e−2t and, as vf = 24 V, v(t) = 24 + (A1 + A2t)e−2t (8.7.7) To find A1 and A2, we use the initial conditions. We write v(0) = 4.5 = 24 + A1 ⇒ A1 = −19.5 (8.7.8) Since i(0) = C dv(0)/dt = 4.5 or dv(0) dt = 4.5 C = 18 From Eq. (8.7.7), dv dt = (−2A1 − 2tA2 + A2)e−2t (8.7.9) At t = 0, dv(0) dt = 18 = −2A1 + A2 (8.7.10) From Eqs. (8.7.8) and (8.7.10), A1 = −19.5 and A2 = 57. Thus, Eq. (8.7.7) becomes v(t) = 24 + (−19.5 + 57t)e−2t V (8.7.11) The inductor current is the same as the capacitor current, that is, i(t) = C dv dt Multiplying Eq. (8.7.9) by C = 0.25 and substituting the values of A1 and A2 gives i(t) = (4.5 − 28.5t)e−2t A (8.7.12) Note that i(0) = 4.5 A, as expected.
  • 322. 318 PART 1 DC Circuits CASE 3 When R = 1 . The initial inductor current is i(0) = 24 1 + 1 = 12 A and the initial voltage across the capacitor is the same as the voltage across the 1- resistor, v(0) = 1i(0) = 12 V α = R 2L = 1 2 × 1 = 0.5 Since α = 0.5 ω0 = 2, we have the underdamped response s1,2 = −α ± α2 − ω2 0 = −0.5 ± j1.936 The total response is therefore v(t) = 24 + (A1 cos 1.936t + A2 sin 1.936t)e−0.5t (8.7.13) We now determine A1 and A2. We write v(0) = 12 = 24 + A1 ⇒ A1 = −12 (8.7.14) Since i(0) = C dv(0)/dt = 12, dv(0) dt = 12 C = 48 (8.7.15) But dv dt = e−0.5t (−1.936A1 sin 1.936t + 1.936A2 cos 1.936t) − 0.5e−0.5t (A1 cos 1.936t + A2 sin 1.936t) (8.7.16) At t = 0, dv(0) dt = 48 = (−0 + 1.936A2) − 0.5(A1 + 0) Substituting A1 = −12 gives A2 = 21.694, and Eq. (8.7.13) becomes v(t) = 24 + (21.694 sin 1.936t − 12 cos 1.936t)e−0.5t V (8.7.17) The inductor current is i(t) = C dv dt Multiplying Eq. (8.7.16) by C = 0.25 and substituting the values of A1 and A2 gives i(t) = (3.1 sin 1.936t + 12 cos 1.936t)e−0.5t A (8.7.18) Note that i(0) = 12 A, as expected. Figure 8.20 plots the responses for the three cases. From this figure, we observe that the critically damped response approaches the step input of 24 V the fastest.
  • 323. CHAPTER 8 Second-Order Circuits 319 t (s) v(t) V 0 1 8 16 24 32 40 2 3 4 5 6 7 Overdamped Critically damped Underdamped Figure 8.20 For Example 8.7: response for three degrees of damping. P R A C T I C E P R O B L E M 8 . 7 Having been in position a for a long time, the switch in Fig. 8.21 is moved to position b at t = 0. Find v(t) and vR(t) for t 0. t = 0 a b 12 V 1 Ω + − 10 V + − 10 Ω 2 Ω 2.5 H − + vR v + − F 1 40 Figure8.21 For Practice Prob. 8.7. Answer: 10 − (1.1547 sin 3.464t + 2 cos 3.464t)e−2t V, 2.31e−2t sin 3.464t V. 8.6 STEP RESPONSE OF A PARALLEL RLC CIRCUIT Is C R L t = 0 i v + − Figure 8.22 Parallel RLC circuit with an applied current. Consider the parallel RLC circuit shown in Fig. 8.22. We want to find i due to a sudden application of a dc current. Applying KCL at the top node for t 0, v R + i + C dv dt = Is (8.46) But v = L di dt Substituting for v in Eq. (8.46) and dividing by LC, we get
  • 324. 320 PART 1 DC Circuits d2 i dt2 + 1 RC di dt + i LC = Is LC (8.47) which has the same characteristic equation as Eq. (8.29). The complete solution to Eq. (8.47) consists of the natural response in(t) and the forced response if ; that is, i(t) = in(t) + if (t) (8.48) The natural response is the same as what we had in Section 8.3. The forced response is the steady state or final value of i. In the circuit in Fig. 8.22, the final value of the current through the inductor is the same as the source current Is. Thus, i(t) = Is + A1es1t + A2es2t (Overdamped) i(t) = Is + (A1 + A2t)e−αt (Critically damped) (8.49) i(t) = Is + (A1 cos ωdt + A2 sin ωdt)e−αt (Underdamped) The constants A1 and A2 in each case can be determined from the initial conditions for i and di/dt. Again, we should keep in mind that Eq. (8.49) only applies for finding the inductor current i. But once the inductor cur- rent iL = i is known, we can find v = L di/dt, which is the same voltage across inductor, capacitor, and resistor. Hence, the current through the resistor is iR = v/R, while the capacitor current is iC = C dv/dt. Al- ternatively, the complete response for any variable x(t) may be found directly, using x(t) = xf (t) + xn(t) (8.50) where xf and xn are its final value and natural response, respectively. E X A M P L E 8 . 8 In the circuit in Fig. 8.23, find i(t) and iR(t) for t 0. 4 A 20 Ω 20 H iR i + − 30u(–t) V t = 0 8 mF 20 Ω v + − Figure8.23 For Example 8.8. Solution: For t 0, the switch is open, and the circuit is partitioned into two independent subcircuits. The 4-A current flows through the inductor, so that i(0) = 4 A
  • 325. CHAPTER 8 Second-Order Circuits 321 Since 30u(−t) = 30 when t 0 and 0 when t 0, the voltage source is operative for t 0 under consideration. The capacitor acts like an open circuit and the voltage across it is the same as the voltage across the 20- resistor connected in parallel with it. By voltage division, the initial capacitor voltage is v(0) = 20 20 + 20 (30) = 15 V For t 0, the switch is closed, and we have a parallel RLC circuit with a current source. The voltage source is off or short-circuited. The two 20- resistors are now in parallel. They are combined to give R = 20 20 = 10 . The characteristic roots are determined as follows: α = 1 2RC = 1 2 × 10 × 8 × 10−3 = 6.25 ω0 = 1 √ LC = 1 √ 20 × 8 × 10−3 = 2.5 s1,2 = −α ± α2 − ω2 0 = −6.25 ± √ 39.0625 − 6.25 = −6.25 ± 5.7282 or s1 = −11.978, s2 = −0.5218 Since α ω0, we have the overdamped case. Hence, i(t) = Is + A1e−11.978t + A2e−0.5218t (8.8.1) where Is = 4 is the final value of i(t). We now use the initial conditions to determine A1 and A2. At t = 0, i(0) = 4 = 4 + A1 + A2 ⇒ A2 = −A1 (8.8.2) Taking the derivative of i(t) in Eq. (8.8.1), di dt = −11.978A1e−11.978t − 0.5218A2e−0.5218t so that at t = 0, di(0) dt = −11.978A1 − 0.5218A2 (8.8.3) But L di(0) dt = v(0) = 15 ⇒ di(0) dt = 15 L = 15 20 = 0.75 Substituting this into Eq. (8.8.3) and incorporating Eq. (8.8.2), we get 0.75 = (11.978 − 0.5218)A2 ⇒ A2 = 0.0655 Thus, A1 = −0.0655 and A2 = 0.0655. Inserting A1 and A2 in Eq. (8.8.1) gives the complete solution as
  • 326. 322 PART 1 DC Circuits i(t) = 4 + 0.0655(e−0.5218t − e−11.978t ) A From i(t), we obtain v(t) = L di/dt and iR(t) = v(t) 20 = L 20 di dt = 0.785e−11.978t − 0.0342e−0.5218t A P R A C T I C E P R O B L E M 8 . 8 Find i(t) and v(t) for t 0 in the circuit in Fig. 8.24. 20u(t) A 5 H i 0.2 F v + − Figure8.24 For Practice Prob. 8.8. Answer: 20(1 − cos t) A, 100 sin t V. 8.7 GENERAL SECOND-ORDER CIRCUITS Now that we have mastered series and parallel RLC circuits, we are prepared to apply the ideas to any second-order circuit. Although the series and parallel RLC circuits are the second-order circuits of greatest interest, other second-order circuits including op amps are also useful. Given a second-order circuit, we determine its step response x(t) (which may be voltage or current) by taking the following four steps: Acircuitmaylookcomplicatedatfirst. Butonce the sources are turned off in an attempt to find the natural response, it may be reducible to a first-order circuit, when the storage elements can be combined, or to a parallel/series RLC cir- cuit. If it is reducible to a first-order circuit, the solution becomes simply what we had in Chap- ter 7. If it is reducible to a parallel or series RLC circuit, we apply the techniques of previous sections in this chapter. 1. We first determine the initial conditions x(0) and dx(0)/dt and the final value x(∞), as discussed in Section 8.2. 2. We find the natural response xn(t) by turning off independent sources and applying KCL and KVL. Once a second-order differential equation is obtained, we determine its characteristic roots. Depending on whether the response is overdamped, critically damped, or underdamped, we obtain xn(t) with two unknown constants as we did in the previous sections. 3. We obtain the forced response as xf (t) = x(∞) (8.51) where x(∞) is the final value of x, obtained in step 1. 4. The total response is now found as the sum of the natural response and forced response x(t) = xn(t) + xf (t) (8.52) We finally determine the constants associated with the natural response by imposing the initial conditions x(0) and dx(0)/dt, determined in step 1. We can apply this general procedure to find the step response of any second-order circuit, including those with op amps. The following examples illustrate the four steps.
  • 327. CHAPTER 8 Second-Order Circuits 323 E X A M P L E 8 . 9 Find the complete response v and then i for t 0 in the circuit of Fig. 8.25. 12 V + − 4 Ω 2 Ω t = 0 1 H i v + − F 1 2 Figure8.25 For Example 8.9. Solution: We first find the initial and final values. At t = 0− , the circuit is at steady state. The switch is open, the equivalent circuit is shown in Fig. 8.26(a). It is evident from the figure that v(0− ) = 12 V, i(0− ) = 0 At t = 0+ , the switch is closed; the equivalent circuit is in Fig. 8.26(b). By the continuity of capacitor voltage and inductor current, we know that v(0+ ) = v(0− ) = 12 V, i(0+ ) = i(0− ) = 0 (8.9.1) To get dv(0+ )/dt, we use C dv/dt = iC or dv/dt = iC/C. Applying KCL at node a in Fig. 8.26(b), i(0+ ) = iC(0+ ) + v(0+ ) 2 0 = iC(0+ ) + 12 2 ⇒ iC(0+ ) = −6 A 12 V + − 4 Ω 2 Ω 1 H i 0.5 F v + − iC (b) 12 V + − 4 Ω i v + − (a) a Figure8.26 Equivalent circuit of the circuit in Fig. 8.25 for: (a) t = 0, (b) t 0. Hence, dv(0+ ) dt = −6 0.5 = −12 V/s (8.9.2) The final values are obtained when the inductor is replaced by a short circuit and the capacitor by an open circuit in Fig. 8.26(b), giving i(∞) = 12 4 + 2 = 2 A, v(∞) = 2i(∞) = 4 V (8.9.3) 4 Ω 2 Ω 1 H i a v v + − F 1 2 Figure 8.27 Obtaining the natural response for Example 8.9. Next, we obtain the natural response for t 0. By turning off the 12-V voltage source, we have the circuit in Fig. 8.27. Applying KCL at node a in Fig. 8.27 gives i = v 2 + 1 2 dv dt (8.9.4) Applying KVL to the left mesh results in 4i + 1 di dt + v = 0 (8.9.5) Since we are interested in v for the moment, we substitute i from Eq. (8.9.4) into Eq. (8.9.5). We obtain 2v + 2 dv dt + 1 2 dv dt + 1 2 d2 v dt2 + v = 0 or d2 v dt2 + 5 dv dt + 6v = 0 From this, we obtain the characteristic equation as
  • 328. 324 PART 1 DC Circuits s2 + 5s + 6 = 0 with roots s = −2 and s = −3. Thus, the natural response is vn(t) = Ae−2t + Be−3t (8.9.6) where A and B are unknown constants to be determined later. The forced response is vf (t) = v(∞) = 4 (8.9.7) The complete response is v(t) = vn + vf = 4 + Ae−2t + Be−3t (8.9.8) We now determine A and B using the initial values. From Eq. (8.9.1), v(0) = 12. Substituting this into Eq. (8.9.8) at t = 0 gives 12 = 4 + A + B ⇒ A + B = 8 (8.9.9) Taking the derivative of v in Eq. (8.9.8), dv dt = −2Ae−2t − 3Be−3t (8.9.10) Substituting Eq. (8.9.2) into Eq. (8.9.10) at t = 0 gives −12 = −2A − 3B ⇒ 2A + 3B = 12 (8.9.11) From Eqs. (8.9.9) and (8.9.11), we obtain A = 12, B = −4 so that Eq. (8.9.8) becomes v(t) = 4 + 12e−2t − 4e−3t V, t 0 (8.9.12) From v, we can obtain other quantities of interest by referring to Fig. 8.26(b). To obtain i, for example, i = v 2 + 1 2 dv dt = 2 + 6e−2t − 2e−3t − 12e−2t + 6e−3t = 2 − 6e−2t + 4e−3t A, t 0 (8.9.13) Notice that i(0) = 0, in agreement with Eq. (8.9.1). P R A C T I C E P R O B L E M 8 . 9 Determine v and i for t 0 in the circuit of Fig. 8.28. t = 0 2 A 10 Ω 4 Ω 2 H i v + − F 1 20 Figure8.28 For Practice Prob. 8.9. Answer: 8(1 − e−5t ) V, 2(1 − e−5t ) A.
  • 329. CHAPTER 8 Second-Order Circuits 325 E X A M P L E 8 . 1 0 Find vo(t) for t 0 in the circuit of Fig. 8.29. 7u(t) V + − 3 Ω 1 Ω vo + − i1 i2 H 1 2 H 1 5 Figure8.29 For Example 8.10. Solution: This is an example of a second-order circuit with two inductors. We first obtain the mesh currents i1 and i2, which happen to be the currents through the inductors. We need to obtain the initial and final values of these currents. For t 0, 7u(t) = 0, so that i1(0− ) = 0 = i2(0− ). For t 0, 7u(t) = 7, so that the equivalent circuit is as shown in Fig. 8.30(a). Due to the continuity of inductor current, i1(0+ ) = i1(0− ) = 0, i2(0+ ) = i2(0− ) = 0 (8.10.1) vL2(0+ ) = vo(0+ ) = 1[(i1(0+ ) − i2(0+ )] = 0 (8.10.2) Applying KVL to the left loop in Fig. 8.30(a) at t = 0+ , 7 = 3i1(0+ ) + vL1(0+) + vo(0+ ) or vL1(0+ ) = 7 V Since L1 di1/dt = vL1, di1(0+ ) dt = vL1 L1 = 7 1 2 = 14 V/s (8.10.3) Similarly, since L2 di2/dt = vL2, di2(0+ ) dt = vL2 L2 = 0 (8.10.4) As t → ∞, the circuit reaches steady state, and the inductors can be replaced by short circuits, as shown in Fig. 8.30(b). From this figure, i1(∞) = i2(∞) = 7 3 A (8.10.5) 7 V + − 3 Ω 1 Ω vo vL2 + − + − vL1 i1 + − i2 (a) 7 V + − 3 Ω 1 Ω i1 i2 (b) L1 = 1 2 H L2 = 1 5 H Figure8.30 Equivalent circuit of that in Fig. 8.29 for: (a) t 0, (b) t → ∞. Next, we obtain the natural responses by removing the voltage source, as shown in Fig. 8.31. Applying KVL to the two meshes yields 4i1 − i2 + 1 2 di1 dt = 0 (8.10.6)
  • 330. 326 PART 1 DC Circuits and i2 + 1 5 di2 dt − i1 = 0 (8.10.7) From Eq. (8.10.6), i2 = 4i1 + 1 2 di1 dt (8.10.8) Substituting Eq. (12.8.8) into Eq. (8.10.7) gives 4i1 + 1 2 di1 dt + 4 5 di1 dt + 1 10 d2 i1 dt2 − i1 = 0 d2 i1 dt2 + 13 di1 dt + 30i1 = 0 From this we obtain the characteristic equation as s2 + 13s + 30 = 0 which has roots s = −3 and s = −10. Hence, the natural response is i1n = Ae−3t + Be−10t (8.10.9) where A and B are constants. The forced response is i1f = i1(∞) = 7 3 A (8.10.10) From Eqs. (8.10.9) and (8.10.10), we obtain the complete response as i1(t) = 7 3 + Ae−3t + Be−10t (8.10.11) We finally obtain A and B from the initial values. From Eqs. (8.10.1) and (8.10.11), 0 = 7 3 + A + B (8.10.12) Taking the derivative of Eq. (8.10.11), setting t = 0 in the derivative, and enforcing Eq. (8.10.3), we obtain 14 = −3A − 10B (8.10.13) From Eqs. (8.10.12) and (8.10.13), A = −4/3 and B = −1. Thus, i1(t) = 7 3 − 4 3 e−3t − e−10t (8.10.14) 3 Ω 1 Ω i1 i2 H 1 2 H 1 5 Figure 8.31 Obtaining the natural response for Example 8.10. We now obtain i2 from i1. Applying KVL to the left loop in Fig. 8.30(a) gives 7 = 4i1 − i2 + 1 2 di1 dt ⇒ i2 = −7 + 4i1 + 1 2 di1 dt Substituting for i1 in Eq. (8.10.14) gives i2(t) = −7 + 28 3 − 16 3 e−3t − 4e−10t + 2e−3t + 5e−10t = 7 3 − 10 3 e−3t + e−10t (8.10.15)
  • 331. CHAPTER 8 Second-Order Circuits 327 From Fig. 8.29, vo(t) = 1[i1(t) − i2(t)] (8.10.16) Substituting Eqs. (8.10.14) and (8.10.15) into Eq. (8.10.16) yields vo(t) = 2(e−3t − e−10t ) (8.10.17) Note that vo(0) = 0, as expected from Eq. (8.10.2). P R A C T I C E P R O B L E M 8 . 1 0 For t 0, obtain vo(t) in the circuit of Fig. 8.32. (Hint: First find v1 and v2.) 5u(t) V + − 1 Ω 1 Ω + − vo v1 v2 F 1 2 F 1 3 Figure8.32 For Practice Prob. 8.10. Answer: 2(e−t − e−6t ) V, t 0. 8.8 SECOND-ORDER OP AMP CIRCUITS An op amp circuit with two storage elements that cannot be combined into a single equivalent element is second-order. Because inductors are bulky and heavy, they are rarely used in practical op amp circuits. For this reason, we will only consider RC second-order op amp circuits here. Such circuits find a wide range of applications in devices such as filters and oscillators. The use of op amps in second-order circuits avoidstheuseofinductors, whicharesomewhat undesirable in some applications. The analysis of a second-order op amp circuit follows the same four steps given and demonstrated in the previous section. E X A M P L E 8 . 1 1 In the op amp circuit of Fig. 8.33, find vo(t) for t 0 when vs = 10u(t) mV. Let R1 = R2 = 10 k, C1 = 20 µF, and C2 = 100 µF. vs R1 v1 + − C1 vo R2 – + C2 v2 + − 1 2 vo + − Figure8.33 For Example 8.11.
  • 332. 328 PART 1 DC Circuits Solution: Although we could follow the same four steps given in the previous section to solve this problem, we will solve it a little differently. Due to the voltage follower configuration, the voltage across C1 is vo. Applying KCL at node 1, vs − v1 R1 = C2 dv2 dt + v1 − vo R2 (8.11.1) At node 2, KCL gives v1 − vo R2 = C1 dvo dt (8.11.2) But v2 = v1 − vo (8.11.3) We now try to eliminate v1 and v2 in Eqs. (8.11.1) to (8.11.3). Substituting Eqs. (8.11.2) and (8.11.3) into Eq. (8.11.1) yields vs − v1 R1 = C2 dv1 dt − C2 dvo dt + C1 dvo dt (8.11.4) From Eq. (8.11.2), v1 = vo + R2C1 dvo dt (8.11.5) Substituting Eq. (8.11.5) into Eq. (8.11.4), we obtain vs R1 = vo R1 + R2C1 R1 dvo dt + C2 dvo dt + R2C1C2 d2 vo dt2 − C2 dvo dt + C1 dvo dt or d2 vo dt2 + 1 R1C2 + 1 R2C2 dvo dt + vo R1R2C1C2 = vs R1R2C1C2 (8.11.6) With the given values of R1, R2, C1, and C2, Eq. (8.11.6) becomes d2 vo dt2 + 2 dvo dt + 5vo = 5vs (8.11.7) To obtain the natural response, set vs = 0 in Eq. (8.11.7), which is the same as turning off the source. The characteristic equation is s2 + 2s + 5 = 0 which has complex roots s1,2 = −1 ± j2. Hence, the natural response is von = e−t (A cos 2t + B sin 2t) (8.11.8) where A and B are unknown constants to be determined. As t → ∞, the circuit reaches the steady-state condition, and the capacitors can be replaced by open circuits. Since no current flows through C1 and C2 under steady-state conditions and no current can enter the input terminals of the ideal op amp, current does not flow through R1 and R2. Thus, vo(∞) = v1(∞) = vs
  • 333. CHAPTER 8 Second-Order Circuits 329 The forced response is then vof = vo(∞) = vs = 10 mV, t 0 (8.11.9) The complete response is vo(t) = von + vof = 10 + e−t (A cos 2t + B sin 2t) mV (8.11.10) To determine A and B, we need the initial conditions. For t 0, vs = 0, so that vo(0− ) = v2(0− ) = 0 For t 0, the source is operative. However, due to capacitor voltage continuity, vo(0+ ) = v2(0+ ) = 0 (8.11.11) From Eq. (8.11.3), v1(0+ ) = v2(0+ ) + vo(0+ ) = 0 and hence, from Eq. (8.11.2), dvo(0+ ) dt = v1 − vo R2C1 = 0 (8.11.12) We now impose Eq. (8.11.11) on the complete response in Eq. (8.11.10) at t = 0, for 0 = 10 + A ⇒ A = −10 (8.11.13) Taking the derivative of Eq. (8.11.10), dvo dt = e−t (−A cos 2t − B sin 2t − 2A sin 2t + 2B cos 2t) Setting t = 0 and incorporating Eq. (8.11.12), we obtain 0 = −A + 2B (8.11.14) From Eqs. (8.11.13) and (8.11.14), A = −10 and B = −5. Thus the step response becomes vo(t) = 10 − e−t (10 cos 2t + 5 sin 2t) mV, t 0 P R A C T I C E P R O B L E M 8 . 1 1 In the op amp circuit shown in Fig. 8.34, vs = 4u(t) V, find vo(t) for t 0. Assume that R1 = R2 = 10 k, C1 = 20 µF, and C2 = 100 µF. vs R1 + − C2 vo + − R2 C1 – + Figure8.34 For Practice Prob. 8.11. Answer: 4 − 5e−t + e−5t V, t 0.
  • 334. 330 PART 1 DC Circuits 8.9 PSPICE ANALYSIS OF RLC CIRCUITS RLC circuits can be analyzed with great ease using PSpice, just like the RC or RL circuits of Chapter 7. The following two examples will illustrate this. The reader may review Section D.4 in Appendix D on PSpice for transient analysis. E X A M P L E 8 . 1 2 The input voltage in Fig. 8.35(a) is applied to the circuit in Fig. 8.35(b). Use PSpice to plot v(t) for 0 t 4s. 2 0 t (s) 12 vs (a) (b) vs 3 H 60 Ω 60 Ω + − v + − F 1 27 Figure8.35 For Example 8.12. Solution: The given circuit is drawn using Schematics as in Fig. 8.36. The pulse is specified using VPWL voltage source, but VPULSE could be used instead. Using the piecewise linear function, we set the attributes of VPWL as T1 = 0, V1 = 0, T2 = 0.001, V2 = 12, and so forth, as shown in Fig. 8.36. Two voltage markers are inserted to plot the input and output voltages. Once the circuit is drawn and the attributes are set, we select Analysis/Setup/Transient to open up the Transient Analysis dialog box. As a parallel RLC circuit, the roots of the characteristic equation are −1 and −9. Thus, we may set Final Time as 4 s (four times the magnitude of the lower root). When the schematic is saved, we select Analysis/Simulate and obtain the plots for the input and output voltages under the Probe window as shown in Fig. 8.37. T1=0 T2=0.001 T3=2 T4=2.001 V1=0 V2=12 V3=12 V4=0 V1 R1 60 R2 60 0.037 C1 0 3H L1 + − V V Figure8.36 Schematic for the circuit in Fig. 8.35(b). 12 V 4 V 8 V 0 V 0 s 1.0 s 2.0 s 3.0 s 4.0 s V(L1:2) Time V(V1:+) Figure8.37 For Example 8.12: the input and output voltages. P R A C T I C E P R O B L E M 8 . 1 2 Find i(t) using PSpice for 0 t 4 s if the pulse voltage in Fig. 8.35(a) is applied to the circuit in Fig. 8.38. Answer: See Fig. 8.39.
  • 335. CHAPTER 8 Second-Order Circuits 331 vs 5 Ω 1 mF 2 H + − i Figure8.38 For Practice Prob. 8.12. 3.0 A 1.0 A 2.0 A 0 A 0 s 1.0 s 2.0 s 3.0 s 4.0 s I(L1) Time Figure8.39 Plot of i(t) for Practice Prob. 8.12. E X A M P L E 8 . 1 3 For the circuit in Fig. 8.40, use PSpice to obtain i(t) for 0 t 3 s. 4 A 7 H 5 Ω 6 Ω i(t) t = 0 a b F 1 42 Figure8.40 For Example 8.13. Solution: When the switch is in position a, the 6- resistor is redundant. The schematic for this case is shown in Fig. 8.41(a). To ensure that current i(t) enters pin 1, the inductor is rotated three times before it is placed in the circuit. The same applies for the capacitor. We insert pseudocomponents IDC 4 A R1 5 7 H L1 0 23.81m C1 (a) R2 6 7 H L1 0 23.81m C1 IC=0 IC=4A (b) I 0.0000 4.000E+00 Figure8.41 For Example 8.13: (a) for dc analysis, (b) for transient analysis.
  • 336. 332 PART 1 DC Circuits VIEWPOINT and IPROBE to determine the initial capacitor voltage and initial inductor current. We carry out a dc PSpice analysis by selecting Analysis/Simulate. As shown in Fig. 8.41(a), we obtain the initial ca- pacitor voltage as 0 V and the initial inductor current i(0) as 4 A from the dc analysis. These initial values will be used in the transient analysis. When the switch is moved to position b, the circuit becomes a source-free parallel RLC circuit with the schematic in Fig. 8.41(b). We set the initial condition IC = 0 for the capacitor and IC = 4 A for the inductor. A current marker is inserted at pin 1 of the inductor. We select Analysis/Setup/Transient to open up the Transient Analysis dialog box and set Final Time to 3 s. After saving the schematic, we select Analysis/Transient. Figure 8.42 shows the plot of i(t). The plot agrees withi(t) = 4.8e−t − 0.8e−6t A,whichisthesolutionbyhandcalculation. 4.00 A 3.92 A 3.96 A 3.88 A 0 s 1.0 s 2.0 s 3.0 s I(L1) Time Figure8.42 Plot of i(t) for Example 8.13. P R A C T I C E P R O B L E M 8 . 1 3 Refer to the circuit in Fig. 8.21 (see Practice Prob. 8.7). Use PSpice to obtain v(t) for 0 t 2. Answer: See Fig. 8.43. 11 V 9 V 10 V 8 V 0 s 0.5 s 1.0 s 1.5 s 2.0 s V(C1:1) Time Figure8.43 Plot of v(t) for Practice Prob. 8.13. †8.10 DUALITY The concept of duality is a time-saving, effort-effective measure of solv- ing circuit problems. Consider the similarity between Eq. (8.4) and Eq. (8.29). The two equations are the same, except that we must interchange the following quantities: (1) voltage and current, (2) resistance and con- ductance, (3) capacitance and inductance. Thus, it sometimes occurs in circuit analysis that two different circuits have the same equations and solutions, except that the roles of certain complementary elements are in- terchanged. This interchangeability is known as the principle of duality.
  • 337. CHAPTER 8 Second-Order Circuits 333 The duality principle asserts a parallelism between pairs of characterizing equations and theorems of electric circuits. Dual pairs are shown in Table 8.1. Note that power does not appear in Table 8.1, because power has no dual. The reason for this is the principle of linearity; since power is not linear, duality does not apply. Also notice from Table 8.1 that the principle of duality extends to circuit elements, configurations, and theorems. TABLE 8.1 Dual pairs. Resistance R Conductance G Inductance L Capacitance C Voltage v Current i Voltage source Current source Node Mesh Series path Parallel path Open circuit Short circuit KVL KCL Thevenin Norton Even when the principle of linearity applies, a circuit element or variable may not have a dual. For example, mutual inductance (to be covered in Chapter 13) has no dual. Two circuits that are described by equations of the same form, but in which the variables are interchanged, are said to be dual to each other. Two circuits are said to be duals of one another if they are described by the same characterizing equations with dual quantities interchanged. The usefulness of the duality principle is self-evident. Once we know the solution to one circuit, we automatically have the solution for the dual circuit. It is obvious that the circuits in Figs. 8.8 and 8.13 are dual. Consequently, the result in Eq. (8.32) is the dual of that in Eq. (8.11). We must keep in mind that the principle of duality is limited to planar circuits. Nonplanar circuits have no duals, as they cannot be described by a system of mesh equations. To find the dual of a given circuit, we do not need to write down the mesh or node equations. We can use a graphical technique. Given a planar circuit, we construct the dual circuit by taking the following three steps: 1. Place a node at the center of each mesh of the given circuit. Place the reference node (the ground) of the dual circuit outside the given circuit. 2. Draw lines between the nodes such that each line crosses an element. Replace that element by its dual (see Table 8.1). 3. To determine the polarity of voltage sources and direction of current sources, follow this rule: A voltage source that pro- duces a positive (clockwise) mesh current has as its dual a cur- rent source whose reference direction is from the ground to the nonreference node. In case of doubt, one may verify the dual circuit by writing the nodal or mesh equations. The mesh (or nodal) equations of the original circuit are similar to the nodal (or mesh) equations of the dual circuit. The duality principle is illustrated with the following two examples. E X A M P L E 8 . 1 4 Construct the dual of the circuit in Fig. 8.44.
  • 338. 334 PART 1 DC Circuits Solution: As shown in Fig. 8.45(a), we first locate nodes 1 and 2 in the two meshes and also the ground node 0 for the dual circuit. We draw a line between one node and another crossing an element. We replace the line joining the nodes by the duals of the elements which it crosses. For example, a line between nodes 1 and 2 crosses a 2-H inductor, and we place a 2-F capacitor (an inductor’s dual) on the line. A line between nodes 1 and 0 crossing the 6-V voltage source will contain a 6-A current source. By drawing lines crossing all the elements, we construct the dual circuit on the given circuit as in Fig. 8.45(a). The dual circuit is redrawn in Fig. 8.45(b) for clarity. 6 V 2 Ω 10 mF 2 H + − t = 0 Figure8.44 For Example 8.14. 6 V 6 A 10 mF 10 mH 2 H 2 F + − 2 F t = 0 2 0 1 1 2 2 Ω 0.5 Ω t = 0 6 A 10 mH 0.5 Ω t = 0 0 (a) (b) Figure8.45 (a) Construction of the dual circuit of Fig. 8.44, (b) dual circuit redrawn. P R A C T I C E P R O B L E M 8 . 1 4 Draw the dual circuit of the one in Fig. 8.46. Answer: See Fig. 8.47. 50 mA 4 H 3 F 10 Ω Figure8.46 For Practice Prob. 8.14. 50 mV 4 F + − 0.1 Ω 3 H Figure8.47 Dual of the circuit in Fig. 8.46. E X A M P L E 8 . 1 5 Obtain the dual of the circuit in Fig. 8.48. Solution: The dual circuit is constructed on the original circuit as in Fig. 8.49(a). We first locate nodes 1 to 3 and the reference node 0. Joining nodes 1 and 2, we cross the 2-F capacitor, which is replaced by a 2-H inductor.
  • 339. CHAPTER 8 Second-Order Circuits 335 10 V + − 20 Ω 5 H 3 A i2 i3 i1 2 F Figure8.48 For Example 8.15. Joining nodes 2 and 3, we cross the 20- resistor, which is replaced by a 1/20- resistor. We keep doing this until all the elements are crossed. The result is in Fig. 8.49(a). The dual circuit is redrawn in Fig. 8.49(b). 1 2 3 0 10 A 3 V 5 F 0 2 H + − 1 2 3 (b) (a) 10 V 10 A + − 20 Ω 5 H 3 A 3 V 2 F 2 H 5 F + − Ω 1 20 Ω 1 20 Figure8.49 For Example 8.15: (a) construction of the dual circuit of Fig. 8.48, (b) dual circuit redrawn. To verify the polarity of the voltage source and the direction of the current source, we may apply mesh currents i1, i2, and i3 (all in the clockwise direction) in the original circuit in Fig. 8.48. The 10-V voltage source produces positive mesh current i1, so that its dual is a 10-A current source directed from 0 to 1. Also, i3 = −3 A in Fig. 8.48 has as its dual v3 = −3 V in Fig. 8.49(b). P R A C T I C E P R O B L E M 8 . 1 5 For the circuit in Fig. 8.50, obtain the dual circuit. Answer: See Fig. 8.51. 2 A 20 V 3 Ω 0.2 F 4 H + − 5 Ω Figure8.50 For Practice Prob. 8.15. 2 V 20 A 4 F 0.2 H + − Ω 1 3 Ω 1 5 Figure8.51 Dual of the circuit in Fig. 8.50.
  • 340. 336 PART 1 DC Circuits †8.11 APPLICATIONS Practical applications of RLC circuits are found in control and com- munications circuits such as ringing circuits, peaking circuits, resonant circuits, smoothing circuits, and filters. Most of the circuits cannot be covered until we treat ac sources. For now, we will limit ourselves to two simple applications: automobile ignition and smoothing circuits. 8.11.1 Automobile Ignition System In Section 7.9.4, we considered the automobile ignition system as a charg- ing system. That was only a part of the system. Here, we consider another part—thevoltagegeneratingsystem. Thesystemismodeledbythecircuit shown in Fig. 8.52. The 12-V source is due to the battery and alternator. The 4- resistor represents the resistance of the wiring. The ignition coil is modeled by the 8-mH inductor. The 1-µF capacitor (known as the condenser to automechanics) is in parallel with the switch (known as the breaking points or electronic ignition). In the following example, we determine how the RLC circuit in Fig. 8.52 is used in generating high voltage. 12 V 4 Ω 8 mH i vL + − t = 0 1 mF vC + − Ignition coil Spark plug Figure8.52 Automobile ignition circuit. E X A M P L E 8 . 1 6 Assuming that the switch in Fig. 8.52 is closed prior to t = 0− , find the inductor voltage vL for t 0. Solution: If the switch is closed prior to t = 0− and the circuit is in steady state, then i(0− ) = 12 4 = 3 A, vC(0− ) = 0 At t = 0+ , the switch is opened. The continuity conditions require that i(0+ ) = 3 A, vC(0+ ) = 0 (8.16.1) We obtain di(0+ )/dt from vL(0+ ). Applying KVL to the mesh at t = 0+ yields −12 + 4i(0+ ) + vL(0+ ) + vC(0+ ) = 0 −12 + 4 × 3 + vL(0+ ) + 0 = 0 ⇒ vL(0+ ) = 0
  • 341. CHAPTER 8 Second-Order Circuits 337 Hence, di(0+ ) dt = vL(0+ ) L = 0 (8.16.2) As t → ∞, the system reaches steady state, so that the capacitor acts like an open circuit. Then i(∞) = 0 (8.16.3) If we apply KVL to the mesh for t 0, we obtain 12 = Ri + L di dt + 1 C t 0 i dt + vC(0) Taking the derivative of each term yields d2 i dt2 + R L di dt + i LC = 0 (8.16.4) We obtain the natural response by following the procedure in Section 8.3. Substituting R = 4 , L = 8 mH, and C = 1 µF, we get α = R 2L = 250, ω0 = 1 √ LC = 1.118 × 104 Since α ω0, the response is underdamped. The damped natural fre- quency is ωd = ω2 0 − α2 ω0 = 1.118 × 104 The natural response is in(t) = e−α (A cos ωdt + B sin ωdt) (8.16.5) where A and B are constants. The forced response is if (t) = i(∞) = 0 (8.16.6) so that the complete response is i(t) = in(t) + if (t) = e−250t (A cos 11,180t + B sin 11,180t) (8.16.7) We now determine A and B. i(0) = 3 = A + 0 ⇒ A = 3 Taking the derivative of Eq. (8.16.7), di dt = −250e−250t (A cos 11,180t + B sin 11,180t) + e−250t (−11,180A sin 11,180t + 11,180B cos 11,180t) Setting t = 0 and incorporating Eq. (8.16.2), 0 = −250A + 11,180B ⇒ B = 0.0671 Thus i(t) = e−250t (3 cos 11,180t + 0.0671 sin 11,180t) (8.16.8)
  • 342. 338 PART 1 DC Circuits The voltage across the inductor is then vL(t) = L di dt = −268e−250t sin 11,180t (8.16.9) This has a maximum value when sine is unity, that is, at 11,180t0 = π/2 or t0 = 140.5 µs. At time = t0, the inductor voltage reaches its peak, which is vL(t0) = −268e−250t0 = −259 V (8.16.10) Although this is far less than the voltage range of 6000 to 10,000 V required to fire the spark plug in a typical automobile, a device known as a transformer (to be discussed in Chapter 13) is used to step up the inductor voltage to the required level. P R A C T I C E P R O B L E M 8 . 1 6 In Fig. 8.52, find the capacitor voltage vC for t 0. Answer: 12 − 12e−250t cos 11,180t + 267.7e−250t sin 11,180t V. 8.11.2 Smoothing Circuits In a typical digital communication system, the signal to be transmitted is first sampled. Sampling refers to the procedure of selecting samples of a signal for processing, as opposed to processing the entire signal. Each sample is converted into a binary number represented by a series of pulses. The pulses are transmitted by a transmission line such as a coaxial cable, twisted pair, or optical fiber. At the receiving end, the signal is applied to a digital-to-analog (D/A) converter whose output is a “staircase” function, that is, constant at each time interval. In order to recover the transmitted analog signal, the output is smoothed by letting it pass through a “smoothing” circuit, as illustrated in Fig. 8.53. An RLC circuit may be used as the smoothing circuit. vs(t) Smoothing circuit p(t) D/A v0(t) Figure8.53 A series of pulses is applied to the digital-to-analog (D/A) converter, whose output is applied to the smoothing circuit. E X A M P L E 8 . 1 7 The output of a D/A converter is shown in Fig. 8.54(a). If the RLC cir- cuit in Fig. 8.54(b) is used as the smoothing circuit, determine the output voltage vo(t). vs 1 Ω 1 H 1 F + − 1 3 0 0 2 (b) (a) t (s) –2 0 4 10 v0 + − vs Figure 8.54 For Example 8.17: (a) output of a D/A converter, (b) an RLC smoothing circuit.
  • 343. CHAPTER 8 Second-Order Circuits 339 Solution: This problem is best solved using PSpice. The schematic is shown in Fig. 8.55(a). The pulse in Fig. 8.54(a) is specified using the piecewise linear function. The attributes of V1 are set as T1 = 0, V1 = 0, T2 = 0.001, V2 = 4, T3 = 1, V3 = 4, and so on. To be able to plot both input and output voltages, we insert two voltage markers as shown. We select Analysis/Setup/Transient to open up the Transient Analysis dialog box and set Final Time as 6 s. Once the schematic is saved, we select Anal- ysis/Simulate to run Probe and obtain the plots shown in Fig. 8.55(b). T1=0 T2=0.001 T3=1 T4=1.001 T5=2 T6=2.001 T7=3 T8=3.001 V1=0 V2=4 V3=4 V4=10 V5=10 V6=-2 V7=-2 V8=0 V1 R1 1 1 C1 0 1H L1 + − V V 10 V 0 V 5 V -5 V 0 s 2.0 s 4.0 s 6.0 s V(V1:+) Time V(C1:1) (a) (b) Figure8.55 For Example 8.17: (a) schematic, (b) input and output voltages. P R A C T I C E P R O B L E M 8 . 1 7 Rework Example 8.17 if the output of the D/A converter is as shown in Fig. 8.56. Answer: See Fig. 8.57. t (s) –3 –1 0 8 7 1 2 3 4 vs Figure8.56 For Practice Prob. 8.17. 8.0 V 0 V 4.0 V -4.0 V 0 s 2.0 s 4.0 s 6.0 s V(V1:+) Time V(C1:1) Figure8.57 Result of Practice Prob. 8.17.
  • 344. 340 PART 1 DC Circuits 8.12 SUMMARY 1. The determination of the initial values x(0) and dx(0)/dt and final value x(∞) is crucial to analyzing second-order circuits. 2. The RLC circuit is second-order because it is described by a second-order differential equation. Its characteristic equation is s2 + 2αs + ω2 0 = 0, where α is the damping factor and ω0 is the undamped natural frequency. For a series circuit, α = R/2L, for a parallel circuit α = 1/2RC, and for both cases ω0 = 1/ √ LC. 3. If there are no independent sources in the circuit after switching (or sudden change), we regard the circuit as source-free. The complete solution is the natural response. 4. The natural response of an RLC circuit is overdamped, under- damped, or critically damped, depending on the roots of the char- acteristic equation. The response is critically damped when the roots are equal (s1 = s2 or α = ω0), overdamped when the roots are real and unequal (s1 = s2 or α ω0), or underdamped when the roots are complex conjugate (s1 = s∗ 2 or α ω0). 5. If independent sources are present in the circuit after switching, the complete response is the sum of the natural response and the forced or steady-state response. 6. PSpice is used to analyze RLC circuits in the same way as for RC or RL circuits. 7. Two circuits are dual if the mesh equations that describe one circuit have the same form as the nodal equations that describe the other. The analysis of one circuit gives the analysis of its dual circuit. 8. The automobile ignition circuit and the smoothing circuit are typical applications of the material covered in this chapter. REVIEW QUESTIONS 8.1 For the circuit in Fig. 8.58, the capacitor voltage at t = 0− (just before the switch is closed) is: (a) 0 V (b) 4 V (c) 8 V (d) 12 V 4 Ω 2 F 1 H 12 V + − t = 0 2 Ω Figure 8.58 For Review Questions 8.1 and 8.2. 8.2 For the circuit in Fig. 8.58, the initial inductor current (at t = 0) is: (a) 0 A (b) 2 A (c) 6 A (d) 12 A 8.3 When a step input is applied to a second-order circuit, the final values of the circuit variables are found by: (a) Replacing capacitors with closed circuits and inductors with open circuits. (b) Replacing capacitors with open circuits and inductors with closed circuits. (c) Doing neither of the above. 8.4 If the roots of the characteristic equation of an RLC circuit are −2 and −3, the response is: (a) (A cos 2t + B sin 2t)e−3t (b) (A + 2Bt)e−3t (c) Ae−2t + Bte−3t (d) Ae−2t + Be−3t where A and B are constants. 8.5 In a series RLC circuit, setting R = 0 will produce: (a) an overdamped response
  • 345. CHAPTER 8 Second-Order Circuits 341 (b) a critically damped response (c) an underdamped response (d) an undamped response (e) none of the above 8.6 A parallel RLC circuit has L = 2 H and C = 0.25 F. The value of R that will produce unity damping factor is: (a) 0.5 (b) 1 (c) 2 (d) 4 8.7 Refer to the series RLC circuit in Fig. 8.59. What kind of response will it produce? (a) overdamped (b) underdamped (c) critically damped (d) none of the above 1 H 1 F 1 Ω Figure 8.59 For Review Question 8.7. 8.8 Consider the parallel RLC circuit in Fig. 8.60. What type of response will it produce? (a) overdamped (b) underdamped (c) critically damped (d) none of the above 1 F 1 H 1 Ω Figure 8.60 For Review Question 8.8. 8.9 Match the circuits in Fig. 8.61 with the following items: (i) first-order circuit (ii) second-order series circuit (iii) second-order parallel circuit (iv) none of the above vs R C1 (c) is C2 C1 L R1 L (d) (e) is C (f) R1 C2 R2 vs R L + − (a) is C (b) R C + − vs R1 R2 + − L L C R2 Figure 8.61 For Review Question 8.9. 8.10 In an electric circuit, the dual of resistance is: (a) conductance (b) inductance (c) capacitance (d) open circuit (e) short circuit Answers: 8.1a, 8.2c, 8.3b, 8.4d, 8.5d, 8.6c, 8.7b, 8.8b, 8.9 (i)-c, (ii)-b,e, (iii)-a, (iv)-d,f, 8.10a. PROBLEMS Section 8.2 Finding Initial and Final Values 8.1 For the circuit in Fig. 8.62, find: (a) i(0+ ) and v(0+ ), (b) di(0+ )/dt and dv(0+ )/dt, (c) i(∞) and v(∞). 12 V 0.4 F 6 Ω + − 2 H 4 Ω i t = 0 v + − Figure 8.62 For Prob. 8.1.
  • 346. 342 PART 1 DC Circuits 8.2 In the circuit of Fig. 8.63, determine: (a) iR(0+ ), iL(0+ ), and iC(0+ ), (b) diR(0+ )/dt, diL(0+ )/dt, and diC(0+ )/dt, (c) iR(∞), iL(∞), and iC(∞). 80 V 20 kΩ 2 mH 1 mF 60 kΩ + − iL iC 25 kΩ iR t = 0 Figure 8.63 For Prob. 8.2. 8.3 Refer to the circuit shown in Fig. 8.64. Calculate: (a) iL(0+ ), vC(0+ ), and vR(0+ ), (b) diL(0+ )/dt, dvC(0+ )/dt, and dvR(0+ )/dt, (c) iL(∞), vC(∞), and vR(∞). 2u(t) A 40 Ω 10 V vR + − 10 Ω + − IL vC + − F 1 4 H 1 8 Figure 8.64 For Prob. 8.3. 8.4 In the circuit of Fig. 8.65, find: (a) v(0+ ) and i(0+ ), (b) dv(0+ )/dt and di(0+ )/dt, (c) v(∞) and i(∞). 4u(–t) V 4u(t) A 3 Ω 0.25 H 0.1 F 5 Ω + − i v + − Figure 8.65 For Prob. 8.4. 8.5 Refer to the circuit in Fig. 8.66. Determine: (a) i(0+ ) and v(0+ ), (b) di(0+ )/dt and dv(0+ )/dt, (c) i(∞) and v(∞). 4u(t) A 1 H 4 Ω v + − 6 Ω i F 1 4 Figure 8.66 For Prob. 8.5. 8.6 In the circuit of Fig. 8.67, find: (a) vR(0+ ) and vL(0+ ), (b) dvR(0+ )/dt and dvL(0+ )/dt, (c) vR(∞) and vL(∞). Vsu(t) Rs R + − L C + − vR + − vL Figure 8.67 For Prob. 8.6. Section 8.3 Source-Free Series RLC Circuit 8.7 The voltage in an RLC network is described by the differential equation d2 v dt2 + 4 dv dt + 4v = 0 subject to the initial conditions v(0) = 1 and dv(0)/dt = −1. Determine the characteristic equation. Find v(t) for t 0. 8.8 The branch current in an RLC circuit is described by the differential equation d2 i dt2 + 6 di dt + 9i = 0 and the initial conditions are i(0) = 0, di(0)/dt = 4. Obtain the characteristic equation and determine i(t) for t 0. 8.9 The current in an RLC circuit is described by d2 i dt2 + 10 di dt + 25i = 0 If i(0) = 10 and di(0)/dt = 0, find i(t) for t 0. 8.10 The differential equation that describes the voltage in an RLC network is d2 v dt2 + 5 dv dt + 4v = 0 Given that v(0) = 0, dv(0)/dt = 10, obtain v(t). 8.11 The natural response of an RLC circuit is described by the differential equation d2 v dt2 + 2 dv dt + v = 0
  • 347. CHAPTER 8 Second-Order Circuits 343 for which the initial conditions are v(0) = 10 and dv(0)/dt = 0. Solve for v(t). 8.12 If R = 20 , L = 0.6 H, what value of C will make an RLC series circuit: (a) overdamped, (b) critically damped, (c) underdamped? 8.13 For the circuit in Fig. 8.68, calculate the value of R needed to have a critically damped response. R 4 H 0.01 F 60 Ω Figure 8.68 For Prob. 8.13. 8.14 Find v(t) for t 0 if v(0) = 6 V and i(0) = 2 A in the circuit shown in Fig. 8.69. 2 H 30 Ω 60 Ω v(t) + − 0.02 F i(t) Figure 8.69 For Prob. 8.14. 8.15 The responses of a series RLC circuit are vC(t) = 30 − 10e−20t + 30e−10t V iL(t) = 40e−20t − 60e−10t mA where vC and iL are the capacitor voltage and inductor current, respectively. Determine the values of R, L, and C. 8.16 Find i(t) for t 0 in the circuit of Fig. 8.70. t = 0 30 V 10 Ω 2.5 H 1 mF 40 Ω + − 60 Ω i(t) Figure 8.70 For Prob. 8.16. 8.17 Obtain v(t) for t 0 in the circuit of Fig. 8.71. t = 0 120 V 10 Ω 4 H 1 F v + − + − Figure 8.71 For Prob. 8.17. 8.18 The switch in the circuit of Fig. 8.72 has been closed for a long time but is opened at t = 0. Determine i(t) for t 0. 2 Ω 12 V i(t) + − t = 0 H 1 2 F 1 4 Figure 8.72 For Prob. 8.18. 8.19 ∗ Calculate v(t) for t 0 in the circuit of Fig. 8.73. t = 0 24 V 12 Ω 60 Ω + − 3 H 6 Ω 15 Ω 25 Ω v + − F 1 27 Figure 8.73 For Prob. 8.19. Section 8.4 Source-Free Parallel RLC Circuit 8.20 For a parallel RLC circuit, the responses are vL(t) = 4e−20t cos 50t − 10e−20t sin 50t V iC(t) = −6.5e−20t cos 50t mA where iC and vL are the capacitor current and inductor voltage, respectively. Determine the values of R, L, and C. 8.21 For the network in Fig. 8.74, what value of C is needed to make the response underdamped with unity damping factor (α = 1)? ∗An asterisk indicates a challenging problem.
  • 348. 344 PART 1 DC Circuits 0.5 H 10 mF C 10 Ω Figure 8.74 For Prob. 8.21. 8.22 Find v(t) for t 0 in the circuit in Fig. 8.75. 25u(–t) 5 Ω + − 1 mF i 0.1 H v + − Figure 8.75 For Prob. 8.22. 8.23 In the circuit in Fig. 8.76, calculate io(t) and vo(t) for t 0. t = 0 30 V 2 Ω vo(t) 8 Ω + − io(t) 1 H + − F 1 4 Figure 8.76 For Prob. 8.23. Section 8.5 Step Response of a Series RLC Circuit 8.24 The step response of an RLC circuit is given by d2 i dt2 + 2 di dt + 5i = 10 Given that i(0) = 2 and di(0)/dt = 4, solve for i(t). 8.25 A branch voltage in an RLC circuit is described by d2 v dt2 + 4 dv dt + 8v = 24 If the initial conditions are v(0) = 0 = dv(0)/dt, find v(t). 8.26 The current in an RLC network is governed by the differential equation d2 i dt2 + 3 di dt + 2i = 4 subject to i(0) = 1, di(0)/dt = −1. Solve for i(t). 8.27 Solve the following differential equations subject to the specified initial conditions (a) d2 v/dt2 + 4v = 12, v(0) = 0, dv(0)/dt = 2 (b) d2 i/dt2 + 5 di/dt + 4i = 8, i(0) = −1, di(0)/dt = 0 (c) d2 v/dt2 + 2 dv/dt + v = 3, v(0) = 5, dv(0)/dt = 1 (d) d2 i/dt2 + 2 di/dt + 5i = 10, i(0) = 4, di(0)/dt = −2 8.28 Consider the circuit in Fig. 8.77. Find vL(0) and vC(0). 2u(t) 40 Ω 50 V 1 F vL + − 0.5 H + − 10 Ω vC + − Figure 8.77 For Prob. 8.28. 8.29 For the circuit in Fig. 8.78, find v(t) for t 0. 1 H 4 Ω 50u(t) V 2u(–t) A 0.04 F + − 2 Ω v + − Figure 8.78 For Prob. 8.29. 8.30 Find v(t) for t 0 in the circuit in Fig. 8.79. 3 A 1 H 10 Ω 5 Ω 4 F t = 0 4u(t) A v + − Figure 8.79 For Prob. 8.30. 8.31 Calculate i(t) for t 0 in the circuit in Fig. 8.80.
  • 349. CHAPTER 8 Second-Order Circuits 345 20 V 5 Ω + − t = 0 i v + − F 1 16 H 1 4 Figure 8.80 For Prob. 8.31. 8.32 Determine v(t) for t 0 in the circuit in Fig. 8.81. t = 0 8 V + − 12 V + − 1 H 2 Ω v + − F 1 5 Figure 8.81 For Prob. 8.32. 8.33 Obtain v(t) and i(t) for t 0 in the circuit in Fig. 8.82. 3u(t) A 5 H 0.2 F 2 Ω 1 Ω 20 V 5 Ω + − i(t) v(t) + − Figure 8.82 For Prob. 8.33. 8.34 ∗ For the network in Fig. 8.83, solve for i(t) for t 0. 30 V 6 Ω + − 10 V + − 6 Ω t = 0 6 Ω i(t) H 1 2 F 1 8 Figure 8.83 For Prob. 8.34. 8.35 Refer to the circuit in Fig. 8.84. Calculate i(t) for t 0. 10 Ω 2 A t = 0 10 Ω 5 Ω i(t) F 1 3 H 3 4 Figure 8.84 For Prob. 8.35. 8.36 Determine v(t) for t 0 in the circuit in Fig. 8.85. 60u(t) V + − 30u(t) V + − 20 Ω 0.25 H 30 Ω 0.5 F v + − Figure 8.85 For Prob. 8.36. 8.37 The switch in the circuit of Fig. 8.86 is moved from position a to b at t = 0. Determine i(t) for t 0. 12 V 2 H + − 2 Ω 14 Ω 6 Ω 4 A i(t) a b 0.02 F t = 0 Figure 8.86 For Prob. 8.37.
  • 350. 346 PART 1 DC Circuits 8.38 ∗ For the network in Fig. 8.87, find i(t) for t 0. 5 Ω 1 H 100 V 5 Ω + − t = 0 20 Ω i F 1 25 Figure 8.87 For Prob. 8.38. 8.39 ∗ Given the network in Fig. 8.88, find v(t) for t 0. 4 A 1 Ω t = 0 2 A 6 Ω 1 H v + − F 1 25 Figure 8.88 For Prob. 8.39. Section 8.6 Step Response of a Parallel RLC Circuit 8.40 In the circuit of Fig. 8.89, find v(t) and i(t) for t 0. Assume v(0) = 0 V and i(0) = 1 A. 4u(t) A 0.5 F 1 H 2 Ω i v + − Figure 8.89 For Prob. 8.40. 8.41 Find i(t) for t 0 in the circuit in Fig. 8.90. 12u(t) V + − 8 mH 2 kΩ i(t) 5 mF Figure 8.90 For Prob. 8.41. 8.42 Find the output voltage vo(t) in the circuit of Fig. 8.91. 3 A 10 mF 5 Ω 1 H 10 Ω t = 0 vo + − Figure 8.91 For Prob. 8.42. 8.43 Given the circuit in Fig. 8.92, find i(t) and v(t) for t 0. 1 Ω 6 V + − 2 Ω t = 0 1 H i(t) v(t) + − F 1 4 Figure 8.92 For Prob. 8.43. 8.44 Determine i(t) for t 0 in the circuit of Fig. 8.93. 3 A 5 Ω 5 H i(t) 12 V t = 0 4 Ω F 1 20 + − Figure 8.93 For Prob. 8.44. 8.45 For the circuit in Fig. 8.94, find i(t) for t 0. 6u(t) A 40 Ω 10 mF 4 H i(t) 30 V + − 10 Ω Figure 8.94 For Prob. 8.45.
  • 351. CHAPTER 8 Second-Order Circuits 347 8.46 Find v(t) for t 0 in the circuit in Fig. 8.95. io C L R t = 0 v + − Figure 8.95 For Prob. 8.46. Section 8.7 General Second-Order Circuits 8.47 Derive the second-order differential equation for vo in the circuit of Fig. 8.96. R2 vs C2 + − R1 C1 vo + − Figure 8.96 For Prob. 8.47. 8.48 Obtain the differential equation for vo in the circuit in Fig. 8.97. R1 vs C + − R2 L vo + − Figure 8.97 For Prob. 8.48. 8.49 For the circuit in Fig. 8.98, find v(t) for t 0. Assume that v(0+ ) = 4 V and i(0+ ) = 2 A. 2 Ω 0.5 F 0.1 F i 4 v + − i Figure 8.98 For Prob. 8.49. 8.50 In the circuit of Fig. 8.99, find i(t) for t 0. 20 V 6 Ω 4 Ω t = 0 + − i F 1 25 H 1 4 Figure 8.99 For Prob. 8.50. 8.51 If the switch in Fig. 8.100 has been closed for a long time before t = 0 but is opened at t = 0, determine: (a) the characteristic equation of the circuit, (b) ix and vR for t 0. t = 0 16 V 1 H + − 8 Ω 12 Ω vR + − ix F 1 36 Figure 8.100 For Prob. 8.51. 8.52 Obtain i1 and i2 for t 0 in the circuit of Fig. 8.101. 4u(t) A 1 H 2 Ω i2 i1 1 H 3 Ω Figure 8.101 For Prob. 8.52. 8.53 For the circuit in Prob. 8.5, find i and v for t 0. 8.54 Find the response vR(t) for t 0 in the circuit in Fig. 8.102. Let R = 3 , L = 2 H, and C = 1/18 F. 10u(t) V R + − L C + − vR Figure 8.102 For Prob. 8.54.
  • 352. 348 PART 1 DC Circuits Section 8.8 Second-Order Op Amp Circuits 8.55 Derive the differential equation relating vo to vs in the op amp circuit of Fig. 8.103. R2 C2 R1 C1 vs vo + − Figure 8.103 For Prob. 8.55. 8.56 Obtain the differential equation for vo(t) in the network of Fig. 8.104. R2 C2 R1 C1 vs vo + − Figure 8.104 For Prob. 8.56. 8.57 Determine the differential equation for the op amp circuit in Fig. 8.105. If v1(0+ ) = 2 V and v2(0+ ) = 0 V, find vo for t 0. Let R = 100 k and C = 1 µF. R vo + − − C v2 + − C v1 + − R + + − Figure 8.105 For Prob. 8.57. 8.58 Given that vs = 2u(t) V in the op amp circuit of Fig. 8.106, find vo(t) for t 0. Let R1 = R2 = 10 k, R3 = 20 k, R4 = 40 k, C1 = C2 = 100 µF. vs R1 C1 R4 vo R2 – + C2 R3 Figure 8.106 For Prob. 8.58. 8.59 ∗ In the op amp circuit of Fig. 8.107, determine vo(t) for t 0. Let vin = u(t) V, R1 = R2 = 10 k, C1 = C2 = 100 µF. R2 C1 R1 C2 vin vo + − Figure 8.107 For Prob. 8.59. Section 8.9 PSpice Analysis of RLC Circuit 8.60 For the step function vs = u(t), use PSpice to find the response v(t) for 0 t 6 s in the circuit of Fig. 8.108. 2 Ω vs + − 1 H 1 F v(t) + − Figure 8.108 For Prob. 8.60. 8.61 Given the source-free circuit in Fig. 8.109, use PSpice to get i(t) for 0 t 20 s. Take v(0) = 30 V and i(0) = 2 A.
  • 353. CHAPTER 8 Second-Order Circuits 349 1 Ω 10 H 2.5 F i v + − Figure 8.109 For Prob. 8.61. 8.62 Obtain v(t) for 0 t 4 s in the circuit of Fig. 8.110 using PSpice. 13u(t) A 39u(t) V 6 Ω 6 Ω + − 1 H v(t) + − 20 Ω 0.4 F Figure 8.110 For Prob. 8.62. 8.63 Rework Prob. 8.23 using PSpice. Plot vo(t) for 0 t 4 s. Section 8.10 Duality 8.64 Draw the dual of the network in Fig. 8.111. 4 A 5 mH 10 mH 20 Ω 2 mF Figure 8.111 For Prob. 8.64. 8.65 Obtain the dual of the circuit in Fig. 8.112. 12 V + − 24 V + − 4 Ω 10 Ω 2 H 0.5 F Figure 8.112 For Prob. 8.65. 8.66 Find the dual of the circuit in Fig. 8.113. 20 Ω 10 Ω 30 Ω 4 H 60 V 1 F 2 A + − 120 V − + Figure 8.113 For Prob. 8.66. 8.67 Draw the dual of the circuit in Fig. 8.114. + − 2 Ω 3 Ω 12 V 5 A 1 Ω 0.25 H 1 F Figure 8.114 For Prob. 8.67. Section 8.11 Applications 8.68 An automobile airbag igniter is modeled by the circuit in Fig. 8.115. Determine the time it takes the voltage across the igniter to reach its first peak after switching from A to B. Let R = 3 , C = 1/30 F, and L = 60 mH. t = 0 A B 12 V + − L R C Airbag igniter Figure 8.115 For Prob. 8.68. 8.69 A passive interface is to be designed to connect an electric motor to an ideal voltage source. If the motor is modeled as a 40-mH inductor in parallel with a 16- resistor, design the interface circuit so that the overall circuit is critically damped at the natural frequency of 60 Hz.
  • 354. 350 PART 1 DC Circuits COMPREHENSIVE PROBLEMS 8.70 A mechanical system is modeled by a series RLC circuit. It is desired to produce an overdamped response with time constants 0.1 ms and 0.5 ms. If a series 50-k resistor is used, find the values of L and C. 8.71 An oscillogram can be adequately modeled by a second-order system in the form of a parallel RLC circuit. It is desired to give an underdamped voltage across a 200- resistor. If the damping frequency is 4 kHz and the time constant of the envelope is 0.25 s, find the necessary values of L and C. 8.72 The circuit in Fig. 8.116 is the electrical analog of body functions used in medical schools to study convulsions. The analog is as follows: C1 = Volume of fluid in a drug C2 = Volume of blood stream in a specified region R1 = Resistance in the passage of the drug from the input to the blood stream R2 = Resistance of the excretion mechanism, such as kidney, etc. v0 = Initial concentration of the drug dosage v(t) = Percentage of the drug in the blood stream Find v(t) for t 0 given that C1 = 0.5 µF, C2 = 5 µF, R1 = 5 M, R2 = 2.5 M, and v0 = 60u(t) V. R1 t = 0 C2 C1 vo + − R2 v(t) + − Figure 8.116 For Prob. 8.72. 8.73 Figure 8.117 shows a typical tunnel-diode oscillator circuit. The diode is modeled as a nonlinear resistor with iD = f (vD), i.e., the diode current is a nonlinear function of the voltage across the diode. Derive the differential equation for the circuit in terms of v and iD. R L i C v + − + − vs ID vD + − Figure 8.117 For Prob. 8.73.
  • 355. 351 AC CIRCUITS P A R T 2 C h a p t e r 9 Sinusoids and Phasors C h a p t e r 1 0 Sinusoidal Steady-State Analysis C h a p t e r 1 1 AC Power Analysis C h a p t e r 1 2 Three-Phase Circuits C h a p t e r 1 3 Magnetically Coupled Circuits C h a p t e r 1 4 Frequency Response
  • 356. 352
  • 357. 353 C H A P T E R SINUSOIDS AND PHASORS 9 The desire to understand the world and the desire to reform it are the two great engines of progress. — Bertrand Russell Historical Profiles Heinrich Rudorf Hertz (1857–1894), a German experimental physicist, demonstrated thatelectromagneticwavesobeythesamefundamentallawsaslight. Hisworkconfirmed James Clerk Maxwell’s celebrated 1864 theory and prediction that such waves existed. Hertz was born into a prosperous family in Hamburg, Germany. He attended the University of Berlin and did his doctorate under the prominent physicist Hermann von Helmholtz. He became a professor at Karlsruhe, where he began his quest for electromagnetic waves. Hertz successfully generated and detected electromagnetic waves; he was the first to show that light is electromagnetic energy. In 1887, Hertz noted for the first time the photoelectric effect of electrons in a molecular structure. Although Hertz only lived to the age of 37, his discovery of electromagnetic waves paved the way for the practical use of such waves in radio, television, and other communication systems. The unit of frequency, the hertz, bears his name. Charles Proteus Steinmetz (1865–1923), a German-Austrian mathematician and en- gineer, introduced the phasor method (covered in this chapter) in ac circuit analysis. He is also noted for his work on the theory of hysteresis. Steinmetz was born in Breslau, Germany, and lost his mother at the age of one. As a youth, he was forced to leave Germany because of his political activities just as he was about to complete his doctoral dissertation in mathematics at the University of Breslau. He migrated to Switzerland and later to the United States, where he was employed by General Electric in 1893. That same year, he published a paper in which complex numbers were used to analyze ac circuits for the first time. This led to one of his many textbooks, Theory and Calculation of ac Phenomena, published by McGraw-Hill in 1897. In 1901, he became the president of the American Institute of Electrical Engineers, which later became the IEEE.
  • 358. 354 PART 2 AC Circuits 9.1 INTRODUCTION Thus far our analysis has been limited for the most part to dc circuits: those circuits excited by constant or time-invariant sources. We have restricted the forcing function to dc sources for the sake of simplicity, for pedagogic reasons, and also for historic reasons. Historically, dc sources were the main means of providing electric power up until the late 1800s. At the end of that century, the battle of direct current versus alternating current began. Both had their advocates among the electrical engineers of the time. Because ac is more efficient and economical to transmit over long distances, ac systems ended up the winner. Thus, it is in keeping with the historical sequence of events that we considered dc sources first. We now begin the analysis of circuits in which the source voltage or current is time-varying. In this chapter, we are particularly interested in sinusoidally time-varying excitation, or simply, excitation by a sinusoid. A sinusoid is a signal that has the form of the sine or cosine function. A sinusoidal current is usually referred to as alternating current (ac). Such a current reverses at regular time intervals and has alternately posi- tive and negative values. Circuits driven by sinusoidal current or voltage sources are called ac circuits. We are interested in sinusoids for a number of reasons. First, nature itself is characteristically sinusoidal. We experience sinusoidal variation in the motion of a pendulum, the vibration of a string, the ripples on the ocean surface, the political events of a nation, the economic fluctuations of the stock market, and the natural response of underdamped second- order systems, to mention but a few. Second, a sinusoidal signal is easy to generate and transmit. It is the form of voltage generated throughout the world and supplied to homes, factories, laboratories, and so on. It is the dominant form of signal in the communications and electric power industries. Third, through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids. Sinusoids, therefore, play an important role in the analysis of periodic signals. Lastly, a sinusoid is easy to handle mathematically. The derivative and integral of a sinusoid are themselves sinusoids. For these and other reasons, the sinusoid is an extremely important function in circuit analysis. A sinusoidal forcing function produces both a natural (or transient) response and a forced (or steady-state) response, much like the step func- tion, which we studied in Chapters 7 and 8. The natural response of a circuit is dictated by the nature of the circuit, while the steady-state re- sponse always has a form similar to the forcing function. However, the natural response dies out with time so that only the steady-state response remains after a long time. When the natural response has become negligi- bly small compared with the steady-state response, we say that the circuit is operating at sinusoidal steady state. It is this sinusoidal steady-state response that is of main interest to us in this chapter.
  • 359. CHAPTER 9 Sinusoids and Phasors 355 We begin with a basic discussion of sinusoids and phasors. We then introduce the concepts of impedance and admittance. The basic circuit laws, Kirchhoff’s and Ohm’s, introduced for dc circuits, will be applied to ac circuits. Finally, we consider applications of ac circuits in phase-shifters and bridges. 9.2 SINUSOIDS Consider the sinusoidal voltage v(t) = Vm sin ωt (9.1) where Vm = the amplitude of the sinusoid ω = the angular frequency in radians/s ωt = the argument of the sinusoid The sinusoid is shown in Fig. 9.1(a) as a function of its argument and in Fig. 9.1(b) as a function of time. It is evident that the sinusoid repeats itself every T seconds; thus, T is called the period of the sinusoid. From the two plots in Fig. 9.1, we observe that ωT = 2π, T = 2π ω (9.2) The fact that v(t) repeats itself every T seconds is shown by replacing t by t + T in Eq. (9.1). We get v(t + T ) = Vm sin ω(t + T ) = Vm sin ω t + 2π ω = Vm sin(ωt + 2π) = Vm sin ωt = v(t) (9.3) Hence, v(t + T ) = v(t) (9.4) that is, v has the same value at t + T as it does at t and v(t) is said to be periodic. In general, A periodic function is one that satisfies f(t) = f(t + nT), for all t and for all integers n. 0 Vm –Vm π 2π 4π vt (a) v(t) 0 Vm –Vm (b) v(t) T 2 T 2T t 3π 3T 2 Figure9.1 A sketch of Vm sin ωt: (a) as a function of ωt, (b) as a function of t.
  • 360. 356 PART 2 AC Circuits As mentioned, the period T of the periodic function is the time of one complete cycle or the number of seconds per cycle. The reciprocal of this quantity is the number of cycles per second, known as the cyclic frequency f of the sinusoid. Thus, f = 1 T (9.5) From Eqs. (9.2) and (9.5), it is clear that ω = 2πf (9.6) While ω is in radians per second (rad/s), f is in hertz (Hz). Theunitoff isnamedaftertheGermanphysicist Heinrich R. Hertz (1857–1894). Let us now consider a more general expression for the sinusoid, v(t) = Vm sin(ωt + φ) (9.7) where (ωt + φ) is the argument and φ is the phase. Both argument and phase can be in radians or degrees. Let us examine the two sinusoids v1(t) = Vm sin ωt and v2(t) = Vm sin(ωt + φ) (9.8) shown in Fig. 9.2. The starting point of v2 in Fig. 9.2 occurs first in time. Therefore, we say that v2 leads v1 by φ or that v1 lags v2 by φ. If φ = 0, we also say that v1 and v2 are out of phase. If φ = 0, then v1 and v2 are said to be in phase; they reach their minima and maxima at exactly the same time. We can compare v1 and v2 in this manner because they operate at the same frequency; they do not need to have the same amplitude. Vm –Vm vt f v2 = Vm sin(vt + f) v1 = Vm sin vt π 2π Figure9.2 Two sinusoids with different phases. A sinusoid can be expressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes. This is achieved by using the following trigonometric identities: sin(A ± B) = sin A cos B ± cos A sin B cos(A ± B) = cos A cos B ∓ sin A sin B (9.9)
  • 361. CHAPTER 9 Sinusoids and Phasors 357 With these identities, it is easy to show that sin(ωt ± 180◦ ) = − sin ωt cos(ωt ± 180◦ ) = − cos ωt sin(ωt ± 90◦ ) = ± cos ωt cos(ωt ± 90◦ ) = ∓ sin ωt (9.10) Using these relationships, we can transform a sinusoid from sine form to cosine form or vice versa. A graphical approach may be used to relate or compare sinusoids as an alternative to using the trigonometric identities in Eqs. (9.9) and (9.10). Consider the set of axes shown in Fig. 9.3(a). The horizontal axis represents the magnitude of cosine, while the vertical axis (pointing down) denotes the magnitude of sine. Angles are measured positively counterclockwise from the horizontal, as usual in polar coordinates. This graphical technique can be used to relate two sinusoids. For example, we see in Fig. 9.3(a) that subtracting 90◦ from the argument of cos ωt gives sin ωt, or cos(ωt −90◦ ) = sin ωt. Similarly, adding 180◦ to the argument of sin ωt gives − sin ωt, or sin(ωt − 180◦ ) = − sin ωt, as shown in Fig. 9.3(b). –90° 180° + sin vt + sin vt + cos vt + cos vt (a) (b) Figure9.3 A graphical means of relating cosine and sine: (a) cos(ωt − 90◦) = sin ωt, (b) sin(ωt + 180◦) = − sin ωt. The graphical technique can also be used to add two sinusoids of the same frequency when one is in sine form and the other is in cosine form. To add A cos ωt and B sin ωt, we note that A is the magnitude of cos ωt while B is the magnitude of sin ωt, as shown in Fig. 9.4(a). The magnitude and argument of the resultant sinusoid in cosine form is readily obtained from the triangle. Thus, A cos ωt + B sin ωt = C cos(ωt − θ) (9.11) where C = A2 + B2, θ = tan−1 B A (9.12) For example, we may add 3 cos ωt and −4 sin ωt as shown in Fig. 9.4(b) and obtain 3 cos ωt − 4 sin ωt = 5 cos(ωt + 53.1◦ ) (9.13) A C B –u sin vt cos vt (a) sin vt cos vt 0 53.1° +3 –4 5 (b) Figure 9.4 (a) Adding A cos ωt and B sin ωt, (b) adding 3 cos ωt and −4 sin ωt.
  • 362. 358 PART 2 AC Circuits Compared with the trigonometric identities in Eqs. (9.9) and (9.10), the graphical approach eliminates memorization. However, we must not confuse the sine and cosine axes with the axes for complex numbers to be discussed in the next section. Something else to note in Figs. 9.3 and 9.4 is that although the natural tendency is to have the vertical axis point up, the positive direction of the sine function is down in the present case. E X A M P L E 9 . 1 Find the amplitude, phase, period, and frequency of the sinusoid v(t) = 12 cos(50t + 10◦ ) Solution: The amplitude is Vm = 12 V. The phase is φ = 10◦ . The angular frequency is ω = 50 rad/s. The period T = 2π ω = 2π 50 = 0.1257 s. The frequency is f = 1 T = 7.958 Hz. P R A C T I C E P R O B L E M 9 . 1 Given the sinusoid 5 sin(4πt − 60◦ ), calculate its amplitude, phase, an- gular frequency, period, and frequency. Answer: 5, −60◦ , 12.57 rad/s, 0.5 s, 2 Hz. E X A M P L E 9 . 2 Calculate the phase angle between v1 = −10 cos(ωt + 50◦ ) and v2 = 12 sin(ωt − 10◦ ). State which sinusoid is leading. Solution: Let us calculate the phase in three ways. The first two methods use trigo- nometric identities, while the third method uses the graphical approach. METHOD 1 In order to compare v1 and v2, we must express them in the same form. If we express them in cosine form with positive amplitudes, v1 = −10 cos(ωt + 50◦ ) = 10 cos(ωt + 50◦ − 180◦ ) v1 = 10 cos(ωt − 130◦ ) or v1 = 10 cos(ωt + 230◦ ) (9.2.1) and v2 = 12 sin(ωt − 10◦ ) = 12 cos(ωt − 10◦ − 90◦ ) v2 = 12 cos(ωt − 100◦ ) (9.2.2) It can be deduced from Eqs. (9.2.1) and (9.2.2) that the phase difference between v1 and v2 is 30◦ . We can write v2 as
  • 363. CHAPTER 9 Sinusoids and Phasors 359 v2 = 12 cos(ωt − 130◦ + 30◦ ) or v2 = 12 cos(ωt + 260◦ ) (9.2.3) Comparing Eqs. (9.2.1) and (9.2.3) shows clearly that v2 leads v1 by 30◦ . METHOD 2 Alternatively, we may express v1 in sine form: v1 = −10 cos(ωt + 50◦ ) = 10 sin(ωt + 50◦ − 90◦ ) = 10 sin(ωt − 40◦ ) = 10 sin(ωt − 10◦ − 30◦ ) But v2 = 12 sin(ωt − 10◦ ). Comparing the two shows that v1 lags v2 by 30◦ . This is the same as saying that v2 leads v1 by 30◦ . 50° 10° v1 v2 sin vt cos vt Figure9.5 For Example 9.2. METHOD 3 We may regard v1 as simply −10 cos ωt with a phase shift of +50◦ . Hence, v1 is as shown in Fig. 9.5. Similarly, v2 is 12 sin ωt with a phase shift of −10◦ , as shown in Fig. 9.5. It is easy to see from Fig. 9.5 that v2 leads v1 by 30◦ , that is, 90◦ − 50◦ − 10◦ . P R A C T I C E P R O B L E M 9 . 2 Find the phase angle between i1 = −4 sin(377t + 25◦ ) and i2 = 5 cos(377t − 40◦ ) Does i1 lead or lag i2? Answer: 155◦ , i1 leads i2. 9.3 PHASORS Sinusoids are easily expressed in terms of phasors, which are more con- venient to work with than sine and cosine functions. A phasor is a complex number that represents the amplitude and phase of a sinusoid. Phasors provide a simple means of analyzing linear circuits excited by sinusoidal sources; solutions of such circuits would be intractable other- wise. The notion of solving ac circuits using phasors was first introduced by Charles Steinmetz in 1893. Before we completely define phasors and apply them to circuit analysis, we need to be thoroughly familiar with complex numbers. Charles Proteus Steinmetz (1865–1923) was a German-Austrian mathematician and electrical engineer. AppendixBpresentsashorttutorialoncomplex numbers. A complex number z can be written in rectangular form as z = x + jy (9.14a) where j = √ −1; x is the real part of z; y is the imaginary part of z. In this context, the variables x and y do not represent a location as in two-dimensional vector analysis but rather the real and imaginary parts of z in the complex plane. Nevertheless, we note that there are some
  • 364. 360 PART 2 AC Circuits resemblances between manipulating complex numbers and manipulating two-dimensional vectors. The complex number z can also be written in polar or exponential form as z = r φ = rejφ (9.14b) where r is the magnitude of z, and φ is the phase of z. We notice that z can be represented in three ways: z = x + jy Rectangular form z = r φ Polar form z = rejφ Exponential form (9.15) The relationship between the rectangular form and the polar form is shown in Fig. 9.6, where the x axis represents the real part and the y axis represents the imaginary part of a complex number. Given x and y, we can get r and φ as r = x2 + y2, φ = tan−1 y x (9.16a) On the other hand, if we know r and φ, we can obtain x and y as x = r cos φ, y = r sin φ (9.16b) Thus, z may be written as z = x + jy = r φ = r(cos φ + j sin φ) (9.17) 0 2j j –2j –j z y r x Real axis Imaginary axis f Figure9.6 Representation of a complex number z = x + jy = r φ. Addition and subtraction of complex numbers are better performed in rectangular form; multiplication and division are better done in polar form. Given the complex numbers z = x + jy = r φ, z1 = x1 + jy1 = r1 φ1 z2 = x2 + jy2 = r2 φ2 the following operations are important. Addition: z1 + z2 = (x1 + x2) + j(y1 + y2) (9.18a) Subtraction: z1 − z2 = (x1 − x2) + j(y1 − y2) (9.18b) Multiplication: z1z2 = r1r2 φ1 + φ2 (9.18c) Division: z1 z2 = r1 r2 φ1 − φ2 (9.18d) Reciprocal: 1 z = 1 r − φ (9.18e) Square Root: √ z = √ r φ/2 (9.18f)
  • 365. CHAPTER 9 Sinusoids and Phasors 361 Complex Conjugate: z∗ = x − jy = r − φ = re−jφ (9.18g) Note that from Eq. (9.18e), 1 j = −j (9.18h) These are the basic properties of complex numbers we need. Other prop- erties of complex numbers can be found in Appendix B. The idea of phasor representation is based on Euler’s identity. In general, e±jφ = cos φ ± j sin φ (9.19) which shows that we may regard cos φ and sin φ as the real and imaginary parts of ejφ ; we may write cos φ = Re(ejφ ) (9.20a) sin φ = Im(ejφ ) (9.20b) where Re and Im stand for the real part of and the imaginary part of. Given a sinusoid v(t) = Vm cos(ωt + φ), we use Eq. (9.20a) to express v(t) as v(t) = Vm cos(ωt + φ) = Re(Vmej(ωt+φ) ) (9.21) or v(t) = Re(Vmejφ ejωt ) (9.22) Thus, v(t) = Re(Vejωt ) (9.23) where V = Vmejφ = Vm φ (9.24) V is thus the phasor representation of the sinusoid v(t), as we said earlier. In other words, a phasor is a complex representation of the magnitude and phase of a sinusoid. Either Eq. (9.20a) or Eq. (9.20b) can be used to develop the phasor, but the standard convention is to use Eq. (9.20a). A phasor may be regarded as a mathematical equivalent of a sinusoid with the time depen- dence dropped. If we use sine for the phasor instead of cosine, then v(t) = Vm sin (ωt + φ) = Im (Vmej(ωt + φ) ) andthecorrespondingphasoristhesameasthat in Eq. (9.24). One way of looking at Eqs. (9.23) and (9.24) is to consider the plot of the sinor Vejωt = Vmej(ωt+φ) on the complex plane. As time increases, the sinor rotates on a circle of radius Vm at an angular velocity ω in the counterclockwise direction, as shown in Fig. 9.7(a). In other words, the entire complex plane is rotating at an angular velocity of ω. We may regard v(t) as the projection of the sinor Vejωt on the real axis, as shown in Fig. 9.7(b). The value of the sinor at time t = 0 is the phasor V of the sinusoid v(t). The sinor may be regarded as a rotating phasor. Thus, whenever a sinusoid is expressed as a phasor, the term ejωt is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the frequency ω of the phasor; otherwise we can make serious mistakes.
  • 366. 362 PART 2 AC Circuits Rotation at v rad ⁄s at t = to at t = 0 f Vm Im Re 0 to t Vm v(t) = Re(Ve jvt ) (a) (b) Figure 9.7 Representation of Vejωt : (a) sinor rotating counterclockwise, (b) its projection on the real axis, as a function of time. Equation (9.23) states that to obtain the sinusoid corresponding to a given phasor V, multiply the phasor by the time factor ejωt and take the real part. As a complex quantity, a phasor may be expressed in rectangular form, polar form, or exponential form. Since a phasor has magnitude and phase (“direction”), it behaves as a vector and is printed in boldface. For example, phasors V = Vm φ and I = Im − θ are graphically represented in Fig. 9.8. Such a graphical representation of phasors is known as a phasor diagram. We use lightface italic letters such as z to repre- sent complex numbers but boldface letters such as V to represent phasors, because phasors are vectorlike quantities. Lagging direction Leading direction Real axis Imaginary axis Vm Im v v V I –u f Figure 9.8 A phasor diagram showing V = Vm φ and I = Im − θ . Equations (9.21) through (9.23) reveal that to get the phasor corre- sponding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number. Then we take out the time factor ejωt , andwhatever is left is the pha-
  • 367. CHAPTER 9 Sinusoids and Phasors 363 sor corresponding to the sinusoid. By suppressing the time factor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows: v(t) = Vm cos(ωt + φ) (Time-domain representation) ⇐⇒ V = Vm φ (Phasor-domain representation) (9.25) Givenasinusoidv(t) = Vm cos(ωt + φ),weobtainthecorrespond- ing phasor as V = Vm φ. Equation (9.25) is also demonstrated in Table 9.1, where the sine function is considered in addition to the cosine func- tion. From Eq. (9.25), we see that to get the phasor representation of a sinusoid, we express it in cosine form and take the magnitude and phase. Given a phasor, we obtain the time-domain representation as the cosine function with the same magnitude as the phasor and the argument as ωt plus the phase of the phasor. The idea of expressing information in alternate domains is fundamental to all areas of engineering. TABLE 9.1 Sinusoid-phasor transformation. Time-domain representation Phasor-domain representation Vm cos(ωt + φ) Vm φ Vm sin(ωt + φ) Vm φ − 90◦ Im cos(ωt + θ) Im θ Im sin(ωt + θ) Im θ − 90◦ Note that in Eq. (9.25) the frequency (or time) factor ejωt is sup- pressed, and the frequency is not explicitly shown in the phasor-domain representation because ω is constant. However, the response depends on ω. For this reason, the phasor domain is also known as the frequency domain. From Eqs. (9.23) and (9.24), v(t) = Re(Vejωt ) = Vm cos (ωt +φ), so that dv dt = −ωVm sin(ωt + φ) = ωVm cos(ωt + φ + 90◦ ) = Re(ωVmejωt ejφ ej90◦ ) = Re(jωVejωt ) (9.26) This shows that the derivative v(t) is transformed to the phasor domain as jωV dv dt (Time domain) ⇐⇒ jωV (Phasor domain) (9.27) Similarly, the integral of v(t) is transformed to the phasor domain as V/jω v dt (Time domain) ⇐⇒ V jω (Phasor domain) (9.28) Differentiating a sinusoid is equivalent to multi- plying its corresponding phasor by jω. Integrating a sinusoid is equivalent to dividing its corresponding phasor by jω.
  • 368. 364 PART 2 AC Circuits Equation (9.27) allows the replacement of a derivative with respect to time with multiplication of jω in the phasor domain, whereas Eq. (9.28) allows the replacement of an integral with respect to time with division by jω in the phasor domain. Equations (9.27) and (9.28) are usefulinfindingthesteady-statesolution, whichdoesnotrequireknowing the initial values of the variable involved. This is one of the important applications of phasors. Besides time differentiation and integration, another important use of phasors is found in summing sinusoids of the same frequency. This is best illustrated with an example, and Example 9.6 provides one. Adding sinusoids of the same frequency is equiv- alent to adding their corresponding phasors. The differences between v(t) and V should be emphasized: 1. v(t) is the instantaneous or time-domain representation, while V is the frequency or phasor-domain representation. 2. v(t) is time dependent, while V is not. (This fact is often forgotten by students.) 3. v(t) is always real with no complex term, while V is generally complex. Finally, we should bear in mind that phasor analysis applies only when frequency is constant; it applies in manipulating two or more sinusoidal signals only if they are of the same frequency. E X A M P L E 9 . 3 Evaluate these complex numbers: (a) (40 50◦ + 20 − 30◦ )1/2 (b) 10 − 30◦ + (3 − j4) (2 + j4)(3 − j5)∗ Solution: (a) Using polar to rectangular transformation, 40 50◦ = 40(cos 50◦ + j sin 50◦ ) = 25.71 + j30.64 20 − 30◦ = 20[cos(−30◦ ) + j sin(−30◦ )] = 17.32 − j10 Adding them up gives 40 50◦ + 20 − 30◦ = 43.03 + j20.64 = 47.72 25.63◦ Taking the square root of this, (40 50◦ + 20 − 30◦ )1/2 = 6.91 12.81◦ (b) Using polar-rectangular transformation, addition, multiplication, and division, 10 − 30◦ + (3 − j4) (2 + j4)(3 − j5)∗ = 8.66 − j5 + (3 − j4) (2 + j4)(3 + j5) = 11.66 − j9 −14 + j22 = 14.73 − 37.66◦ 26.08 122.47◦ = 0.565 − 160.31◦
  • 369. CHAPTER 9 Sinusoids and Phasors 365 P R A C T I C E P R O B L E M 9 . 3 Evaluate the following complex numbers: (a) [(5 + j2)(−1 + j4) − 5 60◦ ]∗ (b) 10 + j5 + 3 40◦ −3 + j4 + 10 30◦ Answer: (a) −15.5 − j13.67, (b) 8.293 + j2.2. E X A M P L E 9 . 4 Transform these sinusoids to phasors: (a) v = −4 sin(30t + 50◦ ) (b) i = 6 cos(50t − 40◦ ) Solution: (a) Since − sin A = cos(A + 90◦ ), v = −4 sin(30t + 50◦ ) = 4 cos(30t + 50◦ + 90◦ ) = 4 cos(30t + 140◦ ) The phasor form of v is V = 4 140◦ (b) i = 6 cos(50t − 40◦ ) has the phasor I = 6 − 40◦ P R A C T I C E P R O B L E M 9 . 4 Express these sinusoids as phasors: (a) v = −7 cos(2t + 40◦ ) (b) i = 4 sin(10t + 10◦ ) Answer: (a) V = 7 220◦ , (b) I = 4 − 80◦ . E X A M P L E 9 . 5 Find the sinusoids represented by these phasors: (a) V = j8e−j20◦ (b) I = −3 + j4 Solution: (a) Since j = 1 90◦ , V = j8 − 20◦ = (1 90◦ )(8 − 20◦ ) = 8 90◦ − 20◦ = 8 70◦ V Converting this to the time domain gives v(t) = 8 cos(ωt + 70◦ ) V (b) I = −3 + j4 = 5 126.87◦ . Transforming this to the time domain gives i(t) = 5 cos(ωt + 126.87◦ ) A
  • 370. 366 PART 2 AC Circuits P R A C T I C E P R O B L E M 9 . 5 Find the sinusoids corresponding to these phasors: (a) V = −10 30◦ (b) I = j(5 − j12) Answer: (a) v(t) = 10 cos(ωt +210◦ ), (b) i(t) = 13 cos(ωt +22.62◦ ). E X A M P L E 9 . 6 Given i1(t) = 4 cos(ωt + 30◦ ) and i2(t) = 5 sin(ωt − 20◦ ), find their sum. Solution: Here is an important use of phasors—for summing sinusoids of the same frequency. Current i1(t) is in the standard form. Its phasor is I1 = 4 30◦ We need to express i2(t) in cosine form. The rule for converting sine to cosine is to subtract 90◦ . Hence, i2 = 5 cos(ωt − 20◦ − 90◦ ) = 5 cos(ωt − 110◦ ) and its phasor is I2 = 5 − 110◦ If we let i = i1 + i2, then I = I1 + I2 = 4 30◦ + 5 − 110◦ = 3.464 + j2 − 1.71 − j4.698 = 1.754 − j2.698 = 3.218 − 56.97◦ A Transforming this to the time domain, we get i(t) = 3.218 cos(ωt − 56.97◦ ) A Of course, we can find i1 + i2 using Eqs. (9.9), but that is the hard way. P R A C T I C E P R O B L E M 9 . 6 If v1 = −10 sin(ωt +30◦ ) and v2 = 20 cos(ωt −45◦ ), find V = v1 +v2. Answer: v(t) = 10.66 cos(ωt − 30.95◦ ). E X A M P L E 9 . 7 Using the phasor approach, determine the current i(t) in a circuit de- scribed by the integrodifferential equation 4i + 8 i dt − 3 di dt = 50 cos(2t + 75◦ )
  • 371. CHAPTER 9 Sinusoids and Phasors 367 Solution: We transform each term in the equation from time domain to phasor domain. Keeping Eqs. (9.27) and (9.28) in mind, we obtain the phasor form of the given equation as 4I + 8I jω − 3jωI = 50 75◦ But ω = 2, so I(4 − j4 − j6) = 50 75◦ I = 50 75◦ 4 − j10 = 50 75◦ 10.77 − 68.2◦ = 4.642 143.2◦ A Converting this to the time domain, i(t) = 4.642 cos(2t + 143.2◦ ) A Keep in mind that this is only the steady-state solution, and it does not require knowing the initial values. P R A C T I C E P R O B L E M 9 . 7 Find the voltage v(t) in a circuit described by the integrodifferential equa- tion 2 dv dt + 5v + 10 v dt = 20 cos(5t − 30◦ ) using the phasor approach. Answer: v(t) = 2.12 cos(5t − 88◦ ). 9.4 PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS Now that we know how to represent a voltage or current in the phasor or frequency domain, one may legitimately ask how we apply this to circuits involving the passive elements R, L, and C. What we need to do is to transform the voltage-current relationship from the time domain to the frequency domain for each element. Again, we will assume the passive sign convention. We begin with the resistor. If the current through a resistor R is i = Im cos(ωt + φ), the voltage across it is given by Ohm’s law as v = iR = RIm cos(ωt + φ) (9.29) The phasor form of this voltage is V = RIm φ (9.30) But the phasor representation of the current is I = Im φ. Hence, V = RI (9.31)
  • 372. 368 PART 2 AC Circuits showing that the voltage-current relation for the resistor in the phasor domain continues to be Ohm’s law, as in the time domain. Figure 9.9 illustrates the voltage-current relations of a resistor. We should note from Eq. (9.31) that voltage and current are in phase, as illustrated in the phasor diagram in Fig. 9.10. (a) i v + − R v = iR (b) I V + − R V = IR Figure9.9 Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain. I f V 0 Re Im Figure 9.10 Phasor diagram for the resistor. For the inductor L, assume the current through it is i = Im cos(ωt + φ). The voltage across the inductor is v = L di dt = −ωLIm sin(ωt + φ) (9.32) Recall from Eq. (9.10) that − sin A = cos(A + 90◦ ). We can write the voltage as v = ωLIm cos(ωt + φ + 90◦ ) (9.33) which transforms to the phasor V = ωLImej(φ+90◦ ) = ωLImejφ ej90◦ = ωLIm φej90◦ (9.34) But Im φ = I, and from Eq. (9.19), ej90◦ = j. Thus, V = jωLI (9.35) showing that the voltage has a magnitude of ωLIm and a phase of φ+90◦ . The voltage and current are 90◦ out of phase. Specifically, the current lags the voltage by 90◦ . Figure 9.11 shows the voltage-current relations for the inductor. Figure 9.12 shows the phasor diagram. Although it is equally correct to say that the in- ductor voltage leads the current by 90◦ , con- vention gives the current phase relative to the voltage. i v + − L v = L di dt (a) I V + − L V = jvLI (b) Figure9.11 Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain. For the capacitor C, assume the voltage across it is v = Vm cos(ωt + φ). The current through the capacitor is i = C dv dt (9.36) By following the same steps as we took for the inductor or by applying Eq. (9.27) on Eq. (9.36), we obtain I = jωCV ⇒ V = I jωC (9.37) showing that the current and voltage are 90◦ out of phase. To be specific, thecurrentleadsthevoltageby90◦ . Figure9.13showsthevoltage-current v Re Im V I 0 f Figure9.12 Phasor diagram for the inductor; I lags V. i v + − C (a) i = C dv dt I V + − C (b) I = jvCV Figure9.13 Voltage-current relations for a capacitor in the: (a) time domain, (b) frequency domain.
  • 373. CHAPTER 9 Sinusoids and Phasors 369 relations for the capacitor; Fig. 9.14 gives the phasor diagram. Table 9.2 summarizes the time-domain and phasor-domain representations of the circuit elements. v Re Im I V 0 f Figure 9.14 Phasor diagram for the capa- citor; I leads V. TABLE 9.2 Summary of voltage-current relationships. Element Time domain Frequency domain R v = Ri V = RI L v = L di dt V = jωLI C i = C dv dt V = I jωC E X A M P L E 9 . 8 The voltage v = 12 cos(60t + 45◦ ) is applied to a 0.1-H inductor. Find the steady-state current through the inductor. Solution: For the inductor, V = jωLI, where ω = 60 rad/s and V = 12 45◦ V. Hence I = V jωL = 12 45◦ j60 × 0.1 = 12 45◦ 6 90◦ = 2 − 45◦ A Converting this to the time domain, i(t) = 2 cos(60t − 45◦ ) A P R A C T I C E P R O B L E M 9 . 8 If voltage v = 6 cos(100t −30◦ ) is applied to a 50 µF capacitor, calculate the current through the capacitor. Answer: 30 cos(100t + 60◦ ) mA. 9.5 IMPEDANCE AND ADMITTANCE In the preceding section, we obtained the voltage-current relations for the three passive elements as V = RI, V = jωLI, V = I jωC (9.38) These equations may be written in terms of the ratio of the phasor voltage to the phasor current as V I = R, V I = jωL, V I = 1 jωC (9.39) From these three expressions, we obtain Ohm’s law in phasor form for any type of element as
  • 374. 370 PART 2 AC Circuits Z = V I or V = ZI (9.40) where Z is a frequency-dependent quantity known as impedance, mea- sured in ohms. The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms (). The impedance represents the opposition which the circuit exhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity. The impedances of resistors, inductors, and capacitors can be read- ily obtained from Eq. (9.39). Table 9.3 summarizes their impedances and admittance. From the table we notice that ZL = jωL and ZC = −j/ωC. Consider two extreme cases of angular frequency. When ω = 0 (i.e., for dc sources), ZL = 0 and ZC → ∞, confirming what we already know— that the inductor acts like a short circuit, while the capacitor acts like an open circuit. When ω → ∞ (i.e., for high frequencies), ZL → ∞ and ZC = 0, indicating that the inductor is an open circuit to high frequencies, while the capacitor is a short circuit. Figure 9.15 illustrates this. TABLE 9.3 Impedances and admittances of passive elements. Element Impedance Admittance R Z = R Y = 1 R L Z = jωL Y = 1 jωL C Z = 1 jωC Y = jωC Short circuit at dc Open circuit at high frequencies (a) Open circuit at dc Short circuit at high frequencies (b) L C Figure9.15 Equivalent circuits at dc and high frequencies: (a) inductor, (b) capacitor. As a complex quantity, the impedance may be expressed in rectan- gular form as Z = R + jX (9.41) where R = Re Z is the resistance and X = Im Z is the reactance. The reactance X may be positive or negative. We say that the impedance is inductive when X is positive or capacitive when X is negative. Thus, impedance Z = R + jX is said to be inductive or lagging since current lags voltage, while impedance Z = R − jX is capacitive or leading because current leads voltage. The impedance, resistance, and reactance are all measured in ohms. The impedance may also be expressed in polar form as Z = |Z| θ (9.42) Comparing Eqs. (9.41) and (9.42), we infer that Z = R + jX = |Z| θ (9.43) where |Z| = R2 + X2, θ = tan−1 X R (9.44) and R = |Z| cos θ, X = |Z| sin θ (9.45) Itissometimesconvenienttoworkwiththereciprocalofimpedance, known as admittance.
  • 375. CHAPTER 9 Sinusoids and Phasors 371 The admittance Y is the reciprocal of impedance, measured in siemens (S). The admittance Y of an element (or a circuit) is the ratio of the phasor current through it to the phasor voltage across it, or Y = 1 Z = I V (9.46) The admittances of resistors, inductors, and capacitors can be obtained from Eq. (9.39). They are also summarized in Table 9.3. As a complex quantity, we may write Y as Y = G + jB (9.47) where G = Re Y is called the conductance and B = Im Y is called the sus- ceptance. Admittance, conductance, and susceptance are all expressed in the unit of siemens (or mhos). From Eqs. (9.41) and (9.47), G + jB = 1 R + jX (9.48) By rationalization, G + jB = 1 R + jX · R − jX R − jX = R − jX R2 + X2 (9.49) Equating the real and imaginary parts gives G = R R2 + X2 , B = − X R2 + X2 (9.50) showing that G = 1/R as it is in resistive circuits. Of course, if X = 0, then G = 1/R. E X A M P L E 9 . 9 Find v(t) and i(t) in the circuit shown in Fig. 9.16. + − i + − 5 Ω v 0.1 F vs = 10 cos 4t Figure9.16 For Example 9.9. Solution: From the voltage source 10 cos 4t, ω = 4, Vs = 10 0◦ V The impedance is Z = 5 + 1 jωC = 5 + 1 j4 × 0.1 = 5 − j2.5 Hence the current I = Vs Z = 10 0◦ 5 − j2.5 = 10(5 + j2.5) 52 + 2.52 = 1.6 + j0.8 = 1.789 26.57◦ A (9.9.1) The voltage across the capacitor is
  • 376. 372 PART 2 AC Circuits V = IZC = I jωC = 1.789 26.57◦ j4 × 0.1 = 1.789 26.57◦ 0.4 90◦ = 4.47 − 63.43◦ V (9.9.2) Converting I and V in Eqs. (9.9.1) and (9.9.2) to the time domain, we get i(t) = 1.789 cos(4t + 26.57◦ ) A v(t) = 4.47 cos(4t − 63.43◦ ) V Notice that i(t) leads v(t) by 90◦ as expected. P R A C T I C E P R O B L E M 9 . 9 Refer to Fig. 9.17. Determine v(t) and i(t). + − i 4 Ω v 0.2 H vs = 5 sin 10t + − Figure9.17 For Practice Prob. 9.9. Answer: 2.236 sin(10t + 63.43◦ ) V, 1.118 sin(10t − 26.57◦ ) A. †9.6 KIRCHHOFF’S LAWS IN THE FREQUENCY DOMAIN We cannot do circuit analysis in the frequency domain without Kirch- hoff’s current and voltage laws. Therefore, we need to express them in the frequency domain. For KVL, let v1, v2, . . . , vn be the voltages around a closed loop. Then v1 + v2 + · · · + vn = 0 (9.51) In the sinusoidal steady state, each voltage may be written in cosine form, so that Eq. (9.51) becomes Vm1 cos(ωt + θ1) + Vm2 cos(ωt + θ2) + · · · + Vmn cos(ωt + θn) = 0 (9.52) This can be written as Re(Vm1ejθ1 ejωt ) + Re(Vm2ejθ2 ejωt ) + · · · + Re(Vmnejθn ejωt ) = 0 or Re[(Vm1ejθ1 + Vm2ejθ2 + · · · + Vmnejθn )ejωt ] = 0 (9.53) If we let Vk = Vmkejθk , then Re[(V1 + V2 + · · · + Vn)ejωt ] = 0 (9.54) Since ejωt = 0, V1 + V2 + · · · + Vn = 0 (9.55) indicating that Kirchhoff’s voltage law holds for phasors.
  • 377. CHAPTER 9 Sinusoids and Phasors 373 By following a similar procedure, we can show that Kirchhoff’s current law holds for phasors. If we let i1, i2, . . . , in be the current leaving or entering a closed surface in a network at time t, then i1 + i2 + · · · + in = 0 (9.56) If I1, I2, . . . , In are the phasor forms of the sinusoids i1, i2, . . . , in, then I1 + I2 + · · · + In = 0 (9.57) which is Kirchhoff’s current law in the frequency domain. Once we have shown that both KVL and KCL hold in the frequency domain, it is easy to do many things, such as impedance combination, nodal and mesh analyses, superposition, and source transformation. 9.7 IMPEDANCE COMBINATIONS Consider the N series-connected impedances shown in Fig. 9.18. The same current I flows through the impedances. Applying KVL around the loop gives V = V1 + V2 + · · · + VN = I(Z1 + Z2 + · · · + ZN ) (9.58) The equivalent impedance at the input terminals is Zeq = V I = Z1 + Z2 + · · · + ZN or Zeq = Z1 + Z2 + · · · + ZN (9.59) showingthatthetotalorequivalentimpedanceofseries-connectedimped- ances is the sum of the individual impedances. This is similar to the series connection of resistances. + − + − + − + − I Z1 Zeq Z2 ZN V1 V V2 V N Figure9.18 N impedances in series. + − + − I + − Z1 V 1 Z2 V2 V Figure9.19 Voltage division. If N = 2, as shown in Fig. 9.19, the current through the impedances is I = V Z1 + Z2 (9.60) Since V1 = Z1I and V2 = Z2I, then V1 = Z1 Z1 + Z2 V, V2 = Z2 Z1 + Z2 V (9.61) which is the voltage-division relationship.
  • 378. 374 PART 2 AC Circuits In the same manner, we can obtain the equivalent impedance or admittance of the N parallel-connected impedances shown in Fig. 9.20. The voltage across each impedance is the same. Applying KCL at the top node, I = I1 + I2 + · · · + IN = V 1 Z1 + 1 Z2 + · · · + 1 ZN (9.62) The equivalent impedance is 1 Zeq = I V = 1 Z1 + 1 Z2 + · · · + 1 ZN (9.63) and the equivalent admittance is Yeq = Y1 + Y2 + · · · + YN (9.64) This indicates that the equivalent admittance of a parallel connection of admittances is the sum of the individual admittances. I + − I1 I2 IN V I Z1 Z2 ZN Zeq Figure9.20 N impedances in parallel. When N = 2, as shown in Fig. 9.21, the equivalent impedance becomes Zeq = 1 Yeq = 1 Y1 + Y2 = 1 1/Z1 + 1/Z2 = Z1Z2 Z1 + Z2 (9.65) Also, since V = IZeq = I1Z1 = I2Z2 the currents in the impedances are I1 = Z2 Z1 + Z2 I, I2 = Z1 Z1 + Z2 I (9.66) which is the current-division principle. I1 I2 + − I Z1 Z2 V Figure9.21 Current division. The delta-to-wye and wye-to-delta transformations that we applied to resistive circuits are also valid for impedances. With reference to Fig. 9.22, the conversion formulas are as follows.
  • 379. CHAPTER 9 Sinusoids and Phasors 375 a b c n Z1 Zb Zc Za Z2 Z3 Figure9.22 Superimposed Y and ' networks. Y-' Conversion: Za = Z1Z2 + Z2Z3 + Z3Z1 Z1 Zb = Z1Z2 + Z2Z3 + Z3Z1 Z2 Zc = Z1Z2 + Z2Z3 + Z3Z1 Z3 (9.67) '-Y Conversion: Z1 = ZbZc Za + Zb + Zc Z2 = ZcZa Za + Zb + Zc Z3 = ZaZb Za + Zb + Zc (9.68) A delta or wye circuit is said to be balanced if it has equal impedances in all three branches. When a '-Y circuit is balanced, Eqs. (9.67) and (9.68) become Z' = 3ZY or ZY = 1 3 Z' (9.69) where ZY = Z1 = Z2 = Z3 and Z' = Za = Zb = Zc. As you see in this section, the principles of voltage division, cur- rent division, circuit reduction, impedance equivalence, and Y-' trans- formation all apply to ac circuits. Chapter 10 will show that other circuit techniques—such as superposition, nodal analysis, mesh analysis, source transformation, the Thevenin theorem, and the Norton theorem—are all applied to ac circuits in a manner similar to their application in dc circuits.
  • 380. 376 PART 2 AC Circuits E X A M P L E 9 . 1 0 Find the input impedance of the circuit in Fig. 9.23. Assume that the cir- cuit operates at ω = 50 rad/s. 3 Ω 10 mF Zin 8 Ω 2 mF 0.2 H Figure9.23 For Example 9.10. Solution: Let Z1 = Impedance of the 2-mF capacitor Z2 = Impedance of the 3- resistor in series with the 10-mF capacitor Z3 = Impedance of the 0.2-H inductor in series with the 8- resistor Then Z1 = 1 jωC = 1 j50 × 2 × 10−3 = −j10 Z2 = 3 + 1 jωC = 3 + 1 j50 × 10 × 10−3 = (3 − j2) Z3 = 8 + jωL = 8 + j50 × 0.2 = (8 + j10) The input impedance is Zin = Z1 + Z2 Z3 = −j10 + (3 − j2)(8 + j10) 11 + j8 = −j10 + (44 + j14)(11 − j8) 112 + 82 = −j10 + 3.22 − j1.07 Thus, Zin = 3.22 − j11.07 P R A C T I C E P R O B L E M 9 . 1 0 Determine the input impedance of the circuit in Fig. 9.24 at ω = 10 rad/s. 20 Ω 4 mF 2 mF Zin 50 Ω 2 H Figure9.24 For Practice Prob. 9.10. Answer: 32.38 − j73.76 . E X A M P L E 9 . 1 1 Determine vo(t) in the circuit in Fig. 9.25. + − + − 60 Ω 10 mF vo 20 cos(4t − 15°) 5 H Figure9.25 For Example 9.11. Solution: To do the analysis in the frequency domain, we must first transform the time-domain circuit in Fig. 9.25 to the phasor-domain equivalent in Fig. 9.26. The transformation produces
  • 381. CHAPTER 9 Sinusoids and Phasors 377 vs = 20 cos(4t − 15◦ ) ⇒ Vs = 20 − 15◦ V, ω = 4 10 mF ⇒ 1 jωC = 1 j4 × 10 × 10−3 = −j25 5 H ⇒ jωL = j4 × 5 = j20 Let Z1 = Impedance of the 60- resistor Z2 = Impedance of the parallel combination of the 10-mF capacitor and the 5-H inductor Then Z1 = 60 and Z2 = −j25 j20 = −j25 × j20 −j25 + j20 = j100 By the voltage-division principle, Vo = Z2 Z1 + Z2 Vs = j100 60 + j100 (20 − 15◦ ) = (0.8575 30.96◦ )(20 − 15◦ ) = 17.15 15.96◦ V. We convert this to the time domain and obtain vo(t) = 17.15 cos(4t + 15.96◦ )V + − + − −j25 Ω j20 Ω 60 Ω 20 −15° V o Figure9.26 The frequency-domain equivalent of the circuit in Fig. 9.25. P R A C T I C E P R O B L E M 9 . 1 1 Calculate vo in the circuit in Fig. 9.27. + − + − 10 Ω vo 0.5 H F 10 cos (10t + 75°) 1 20 Figure9.27 For Practice Prob. 9.11. Answer: vo(t) = 7.071 cos(10t − 60◦ ) V. E X A M P L E 9 . 1 2 Find current I in the circuit in Fig. 9.28. + − 12 Ω 8 Ω 8 Ω j4 Ω j6 Ω −j4 Ω −j3 Ω 2 Ω 50 0° I a b c Figure9.28 For Example 9.12.
  • 382. 378 PART 2 AC Circuits Solution: The delta network connected to nodes a, b, and c can be converted to the Y network of Fig. 9.29. We obtain the Y impedances as follows using Eq. (9.68): Zan = j4(2 − j4) j4 + 2 − j4 + 8 = 4(4 + j2) 10 = (1.6 + j0.8) Zbn = j4(8) 10 = j3.2 , Zcn = 8(2 − j4) 10 = (1.6 − j3.2) The total impedance at the source terminals is Z = 12 + Zan + (Zbn − j3) (Zcn + j6 + 8) = 12 + 1.6 + j0.8 + (j0.2) (9.6 + j2.8) = 13.6 + j0.8 + j0.2(9.6 + j2.8) 9.6 + j3 = 13.6 + j1 = 13.64 4.204◦ The desired current is I = V Z = 50 0◦ 13.64 4.204◦ = 3.666 − 4.204◦ A Zcn + − I Zan Zcn 50 0° 12 Ω 8 Ω j6 Ω −j3 Ω c b a n Zbn Figure 9.29 The circuit in Fig. 9.28 after delta-to-wye transformation. P R A C T I C E P R O B L E M 9 . 1 2 Find I in the circuit in Fig. 9.30. + − I −j2 Ω −j3 Ω j5 Ω j4 Ω 5 Ω 10 Ω 8 Ω 30 0° V Figure9.30 For Practice Prob. 9.12. Answer: 6.364 3.802◦ A.
  • 383. CHAPTER 9 Sinusoids and Phasors 379 †9.8 APPLICATIONS In Chapters 7 and 8, we saw certain uses of RC, RL, and RLC circuits in dc applications. These circuits also have ac applications; among them are coupling circuits, phase-shifting circuits, filters, resonant circuits, ac bridge circuits, and transformers. This list of applications is inexhaustive. We will consider some of them later. It will suffice here to observe two simple ones: RC phase-shifting circuits, and ac bridge circuits. 9.8.1 Phase-Shifters A phase-shifting circuit is often employed to correct an undesirable phase shift already present in a circuit or to produce special desired effects. An RC circuit is suitable for this purpose because its capacitor causes the circuit current to lead the applied voltage. Two commonly used RC circuits are shown in Fig. 9.31. (RL circuits or any reactive circuits could also serve the same purpose.) (a) I + − + − V o Vi R C (b) I + − V o Vi + − R C Figure9.31 Series RC shift circuits: (a) leading output, (b) lagging output. In Fig. 9.31(a), the circuit current I leads the applied voltage Vi by some phase angle θ, where 0 θ 90◦ , depending on the values of R and C. If XC = −1/ωC, then the total impedance is Z = R + jXC, and the phase shift is given by θ = tan−1 XC R (9.70) This shows that the amount of phase shift depends on the values of R, C, and the operating frequency. Since the output voltage Vo across the resistor is in phase with the current, Vo leads (positive phase shift) Vi as shown in Fig. 9.32(a). vo t vi (a) t vi vo (b) u Phase shift u Phase shift Figure9.32 Phase shift in RC circuits: (a) leading output, (b) lagging output. In Fig. 9.31(b), the output is taken across the capacitor. The current I leads the input voltage Vi by θ, but the output voltage vo(t) across the capacitor lags (negative phase shift) the input voltage vi(t) as illustrated in Fig. 9.32(b). We should keep in mind that the simple RC circuits in Fig. 9.31 also act as voltage dividers. Therefore, as the phase shift θ approaches 90◦ , the output voltage Vo approaches zero. For this reason, these simple RC circuits are used only when small amounts of phase shift are required.
  • 384. 380 PART 2 AC Circuits If it is desired to have phase shifts greater than 60◦ , simple RC networks are cascaded, thereby providing a total phase shift equal to the sum of the individual phase shifts. In practice, the phase shifts due to the stages are not equal, because the succeeding stages load down the earlier stages unless op amps are used to separate the stages. E X A M P L E 9 . 1 3 Design an RC circuit to provide a phase of 90◦ leading. + − + − 20 Ω 20 Ω V i −j20 Ω −j20 Ω V o Z V1 Figure9.33 An RC phase shift circuit with 90◦ leading phase shift; for Example 9.13. Solution: If we select circuit components of equal ohmic value, say R = |XC| = 20 , at a particular frequency, according to Eq. (9.70), the phase shift is exactly 45◦ . By cascading two similar RC circuits in Fig. 9.31(a), we obtain the circuit in Fig. 9.33, providing a positive or leading phase shift of 90◦ , as we shall soon show. Using the series-parallel combination technique, Z in Fig. 9.33 is obtained as Z = 20 (20 − j20) = 20(20 − j20) 40 − j20 = 12 − j4 (9.13.1) Using voltage division, V1 = Z Z − j20 Vi = 12 − j4 12 − j24 Vi = √ 2 3 45◦ Vi (9.13.2) and Vo = 20 20 − j20 V1 = √ 2 2 45◦ V1 (9.13.3) Substituting Eq. (9.13.2) into Eq. (9.13.3) yields Vo = √ 2 2 45◦ √ 2 3 45◦ Vi = 1 3 90◦ Vi Thus, the output leads the input by 90◦ but its magnitude is only about 33 percent of the input. P R A C T I C E P R O B L E M 9 . 1 3 Design an RC circuit to provide a 90◦ lagging phase shift. If a voltage of 10 V is applied, what is the output voltage? Answer: Figure 9.34 shows a typical design; 3.33 V. + − + − 10 Ω 10 Ω −j10 Ω −j10 Ω V o V i Figure9.34 For Practice Prob. 9.13. E X A M P L E 9 . 1 4 For the RL circuit shown in Fig. 9.35(a), calculate the amount of phase shift produced at 2 kHz.
  • 385. CHAPTER 9 Sinusoids and Phasors 381 Solution: At 2 kHz, we transform the 10-mH and 5-mH inductances to the corre- sponding impedances. 10 mH ⇒ XL = ωL = 2π × 2 × 103 × 10 × 10−3 = 40π = 125.7 5 mH ⇒ XL = ωL = 2π × 2 × 103 × 5 × 10−3 = 20π = 62.83 Consider the circuit in Fig. 9.35(b). The impedance Z is the parallel combination of j125.7 and 100 + j62.83 . Hence, Z = j125.7 (100 + j62.83) = j125.7(100 + j62.83) 100 + j188.5 = 69.56 60.1◦ (9.14.1) Using voltage division, V1 = Z Z + 150 Vi = 69.56 60.1◦ 184.7 + j60.3 Vi = 0.3582 42.02◦ Vi (9.14.2) and Vo = j62.832 100 + j62.832 V1 = 0.532 57.86◦ V1 (9.14.3) Combining Eqs. (9.14.2) and (9.14.3), Vo = (0.532 57.86◦ )(0.3582 42.02◦ ) Vi = 0.1906 100◦ Vi showing that the output is about 19 percent of the input in magnitude but leading the input by 100◦ . If the circuit is terminated by a load, the load will affect the phase shift. 150 Ω 100 Ω 10 mH 5 mH (a) 150 Ω 100 Ω (b) + − + − Z Vi V1 V o j125.7 Ω j62.83 Ω Figure9.35 For Example 9.14. P R A C T I C E P R O B L E M 9 . 1 4 Refer to the RL circuit in Fig. 9.36. If 1 V is applied, find the magnitude and the phase shift produced at 5 kHz. Specify whether the phase shift is leading or lagging. 10 Ω 50 Ω + − + − Vi V o 1 mH 2 mH Figure9.36 For Practice Prob. 9.14. Answer: 0.172, 120.4◦ , lagging. 9.8.2 AC Bridges An ac bridge circuit is used in measuring the inductance L of an inductor or the capacitance C of a capacitor. It is similar in form to the Wheatstone bridge for measuring an unknown resistance (discussed in Section 4.10) and follows the same principle. To measure L and C, however, an ac source is needed as well as an ac meter instead of the galvanometer. The ac meter may be a sensitive ac ammeter or voltmeter.
  • 386. 382 PART 2 AC Circuits Consider the general ac bridge circuit displayed in Fig. 9.37. The bridge is balanced when no current flows through the meter. This means that V1 = V2. Applying the voltage division principle, V1 = Z2 Z1 + Z2 Vs = V2 = Zx Z3 + Zx Vs (9.71) Thus, Z2 Z1 + Z2 = Zx Z3 + Zx ⇒ Z2Z3 = Z1Zx (9.72) or Zx = Z3 Z1 Z2 (9.73) This is the balanced equation for the ac bridge and is similar to Eq. (4.30) for the resistance bridge except that the R’s are replaced by Z’s. AC meter + − + − ≈ Vs Z1 Z3 Z2 V1 V2 Zx Figure9.37 A general ac bridge. Specific ac bridges for measuring L and C are shown in Fig. 9.38, where Lx and Cx are the unknown inductance and capacitance to be measured while Ls and Cs are a standard inductance and capacitance (the values of which are known to great precision). In each case, two resistors, R1 and R2, are varied until the ac meter reads zero. Then the bridge is balanced. From Eq. (9.73), we obtain Lx = R2 R1 Ls (9.74) and Cx = R1 R2 Cs (9.75) Notice that the balancing of the ac bridges in Fig. 9.38 does not depend on the frequency f of the ac source, since f does not appear in the relation- ships in Eqs. (9.74) and (9.75). AC meter ≈ R1 R2 Ls Lx (a) AC meter ≈ R1 R2 Cs Cx (b) Figure 9.38 Specific ac bridges: (a) for measuring L, (b) for measuring C. E X A M P L E 9 . 1 5 The ac bridge circuit of Fig. 9.37 balances when Z1 is a 1-k resistor, Z2 is a 4.2-k resistor, Z3 is a parallel combination of a 1.5-M resistor
  • 387. CHAPTER 9 Sinusoids and Phasors 383 and a 12-pF capacitor, and f = 2 kHz. Find: (a) the series components that make up Zx, and (b) the parallel components that make up Zx. Solution: From Eq. (9.73), Zx = Z3 Z1 Z2 (9.15.1) where Zx = Rx + jXx, Z1 = 1000 , Z2 = 4200 (9.15.2) and Z3 = R3 1 jωC3 = R3 jωC3 R3 + 1/jωC3 = R3 1 + jωR3C3 Since R3 = 1.5 M and C3 = 12 pF, Z3 = 1.5 × 106 1 + j2π × 2 × 103 × 1.5 × 106 × 12 × 10−12 = 1.5 × 106 1 + j0.2262 or Z3 = 1.427 − j0.3228 M (9.15.3) (a) Assuming that Zx is made up of series components, we substitute Eqs. (9.15.2) and (9.15.3) in Eq. (9.15.1) and obtain Rx + jXx = 4200 1000 (1.427 − j0.3228) × 106 = (5.993 − j1.356) M Equating the real and imaginary parts yields Rx = 5.993 M and a capacitive reactance Xx = 1 ωC = 1.356 × 106 or C = 1 ωXx = 1 2π × 2 × 103 × 1.356 × 106 = 58.69 pF (b) If Zx is made up of parallel components, we notice that Z3 is also a parallel combination. Hence, Eq. (9.15.1) becomes Zx = 4200 1000 R3 1 jωC3 = 4.2R3 1 jωC3 = 4.2Z3 (9.15.4) This simply means that the unknown impedance Zx is 4.2 times Z3. Since Z3 consists of R3 and X3 = 1/ωC3, there are many ways we can get 4.2Z3. Therefore, there is no unique answer to the problem. If we suppose that 4.2 = 3 × 1.4 and we decide to multiply R3 by 1.4 while multiplying X3 by 3, then the answer is Rx = 1.4R3 = 2.1 M and Xx = 1 ωCx = 3X3 = 3 ωC3 ⇒ Cx = 1 3 C3 = 4 pF
  • 388. 384 PART 2 AC Circuits Alternatively, we may decide to multiply R3 by 3 while multiplying Xx by 1.4 and obtain Rx = 4.5 M and Cx = C3/1.4 = 8.571 pF. Of course, there are several other possibilities. In a situation like this when there is no unique solution, care must be taken to select reasonably sized component values whenever possible. P R A C T I C E P R O B L E M 9 . 1 5 In the ac bridge circuit of Fig. 9.37, suppose that balance is achieved when Z1 is a 4.8-k resistor, Z2 is a 10- resistor in series with a 0.25-µH inductor, Z3 is a 12-k resistor, and f = 6 MHz. Determine the series components that make up Zx. Answer: A 25- resistor in series with a 0.625-µH inductor. 9.9 SUMMARY 1. A sinusoid is a signal in the form of the sine or cosine function. It has the general form v(t) = Vm cos(ωt + φ) where Vm is the amplitude, ω = 2πf is the angular frequency, (ωt + φ) is the argument, and φ is the phase. 2. A phasor is a complex quantity that represents both the magnitude and the phase of a sinusoid. Given the sinusoid v(t) = Vm cos(ωt + φ), its phasor V is V = Vm φ 3. In ac circuits, voltage and current phasors always have a fixed relation to one another at any moment of time. If v(t) = Vm cos(ωt + φv) represents the voltage through an element and i(t) = Im cos(ωt + φi) represents the current through the element, then φi = φv if the element is a resistor, φi leads φv by 90◦ if the element is a capacitor, and φi lags φv by 90◦ if the element is an inductor. 4. The impedance Z of a circuit is the ratio of the phasor voltage across it to the phasor current through it: Z = V I = R(ω) + jX(ω) The admittance Y is the reciprocal of impedance: Z = 1 Y = G(ω) + jB(ω) Impedances are combined in series or in parallel the same way as resistances in series or parallel; that is, impedances in series add while admittances in parallel add. 5. For a resistor Z = R, for an inductor Z = jX = jωL, and for a capacitor Z = −jX = 1/jωC. 6. Basic circuit laws (Ohm’s and Kirchhoff’s) apply to ac circuits in the same manner as they do for dc circuits; that is,
  • 389. CHAPTER 9 Sinusoids and Phasors 385 V = ZI Ik = 0 (KCL) Vk = 0 (KVL) 7. The techniques of voltage/current division, series/parallel combina- tion of impedance/admittance, circuit reduction, and Y-' trans- formation all apply to ac circuit analysis. 8. AC circuits are applied in phase-shifters and bridges. REVIEW QUESTIONS 9.1 Which of the following is not a right way to express the sinusoid A cos ωt? (a) A cos 2πf t (b) A cos(2πt/T ) (c) A cos ω(t − T ) (d) A sin(ωt − 90◦ ) 9.2 A function that repeats itself after fixed intervals is said to be: (a) a phasor (b) harmonic (c) periodic (d) reactive 9.3 Which of these frequencies has the shorter period? (a) 1 krad/s (b) 1 kHz 9.4 If v1 = 30 sin(ωt + 10◦ ) and v2 = 20 sin(ωt + 50◦ ), which of these statements are true? (a) v1 leads v2 (b) v2 leads v1 (c) v2 lags v1 (d) v1 lags v2 (e) v1 and v2 are in phase 9.5 The voltage across an inductor leads the current through it by 90◦ . (a) True (b) False 9.6 The imaginary part of impedance is called: (a) resistance (b) admittance (c) susceptance (d) conductance (e) reactance 9.7 The impedance of a capacitor increases with increasing frequency. (a) True (b) False 9.8 At what frequency will the output voltage vo(t) in Fig. 9.39 be equal to the input voltage v(t)? (a) 0 rad/s (b) 1 rad/s (c) 4 rad/s (d) ∞ rad/s (e) none of the above + − + − 1 Ω H v(t) vo(t) 1 4 Figure 9.39 For Review Question 9.8. 9.9 A series RC circuit has VR = 12 V and VC = 5 V. The supply voltage is: (a) −7 V (b) 7 V (c) 13 V (d) 17 V 9.10 A series RCL circuit has R = 30 , XC = −50 , and XL = 90 . The impedance of the circuit is: (a) 30 + j140 (b) 30 + j40 (c) 30 − j40 (d) −30 − j40 (e) −30 + j40 Answers: 9.1d, 9.2c, 9.3b, 9.4b,d, 9.5a, 9.6e, 9.7b, 9.8d, 9.9c, 9.10b. PROBLEMS Section 9.2 Sinusoids 9.1 In a linear circuit, the voltage source is vs = 12 sin(103 t + 24◦ ) V (a) What is the angular frequency of the voltage? (b) What is the frequency of the source? (c) Find the period of the voltage. (d) Express vs in cosine form. (e) Determine vs at t = 2.5 ms. 9.2 A current source in a linear circuit has is = 8 cos(500πt − 25◦ ) A (a) What is the amplitude of the current? (b) What is the angular frequency?
  • 390. 386 PART 2 AC Circuits (c) Find the frequency of the current. (d) Calculate is at t = 2 ms. 9.3 Express the following functions in cosine form: (a) 4 sin(ωt − 30◦ ) (b) −2 sin 6t (c) −10 sin(ωt + 20◦ ) 9.4 (a) Express v = 8 cos(7t + 15◦ ) in sine form. (b) Convert i = −10 sin(3t − 85◦ ) to cosine form. 9.5 Given v1 = 20 sin(ωt + 60◦ ) and v2 = 60 cos(ωt − 10◦ ), determine the phase angle between the two sinusoids and which one lags the other. 9.6 For the following pairs of sinusoids, determine which one leads and by how much. (a) v(t) = 10 cos(4t − 60◦ ) and i(t) = 4 sin(4t + 50◦ ) (b) v1(t) = 4 cos(377t + 10◦ ) and v2(t) = −20 cos 377t (c) x(t) = 13 cos 2t + 5 sin 2t and y(t) = 15 cos(2t − 11.8◦ ) Section 9.3 Phasors 9.7 If f (φ) = cos φ + j sin φ, show that f (φ) = ejφ . 9.8 Calculate these complex numbers and express your results in rectangular form: (a) 15 45◦ 3 − j4 + j2 (b) 8 − 20◦ (2 + j)(3 − j4) + 10 −5 + j12 (c) 10 + (8 50◦ )(5 − j12) 9.9 Evaluate the following complex numbers and express your results in rectangular form: (a) 2 + 3 + j4 5 − j8 (b) 4 − 10◦ + 1 − j2 3 6◦ (c) 8 10◦ + 6 − 20◦ 9 80◦ − 4 50◦ 9.10 Given the complex numbers z1 = −3 + j4 and z2 = 12 + j5, find: (a) z1z2 (b) z1 z∗ 2 (c) z1 + z2 z1 − z2 9.11 Let X = 8 40◦ and Y = 10 − 30◦ . Evaluate the following quantities and express your results in polar form. (a) (X + Y)X∗ (b) (X − Y)∗ (c) (X + Y)/X 9.12 Evaluate these determinants: (a) 10 + j6 −5 2 − j3 −1 + j (b) 20 − 30◦ 16 0◦ −4 − 10◦ 3 45◦ (c) 1 − j j 1 −j 1 j 0 −j 1 + j 9.13 Transform the following sinusoids to phasors: (a) −10 cos(4t + 75◦ ) (b) 5 sin(20t − 10◦ ) (c) 4 cos 2t + 3 sin 2t 9.14 Express the sum of the following sinusoidal signals in the form of A cos(ωt + θ) with A 0 and 0 θ 360◦ . (a) 8 cos(5t − 30◦ ) + 6 cos 5t (b) 20 cos(120πt + 45◦ ) − 30 sin(120πt + 20◦ ) (c) 4 sin 8t + 3 sin(8t − 10◦ ) 9.15 Obtain the sinusoids corresponding to each of the following phasors: (a) V1 = 60 15◦ , ω = 1 (b) V2 = 6 + j8, ω = 40 (c) I1 = 2.8e−jπ/3 , ω = 377 (d) I2 = −0.5 − j1.2, ω = 103 9.16 Using phasors, find: (a) 3 cos(20t + 10◦ ) − 5 cos(20t − 30◦ ) (b) 40 sin 50t + 30 cos(50t − 45◦ ) (c) 20 sin 400t + 10 cos(400t + 60◦ ) − 5 sin(400t − 20◦ ) 9.17 Find a single sinusoid corresponding to each of these phasors: (a) V = 40 − 60◦ (b) V = −30 10◦ + 50 60◦ (c) I = j6e−j10◦ (d) I = 2 j + 10 − 45◦ 9.18 Find v(t) in the following integrodifferential equations using the phasor approach: (a) v(t) + v dt = 10 cos t (b) dv dt + 5v(t) + 4 v dt = 20 sin(4t + 10◦ ) 9.19 Using phasors, determine i(t) in the following equations: (a) 2 di dt + 3i(t) = 4 cos(2t − 45◦ ) (b) 10 i dt + di dt + 6i(t) = 5 cos(5t + 22◦ ) 9.20 The loop equation for a series RLC circuit gives di dt + 2i + t −∞ i dt = cos 2t Assuming that the value of the integral at t = −∞ is zero, find i(t) using the phasor method.
  • 391. CHAPTER 9 Sinusoids and Phasors 387 9.21 A parallel RLC circuit has the node equation dv dt + 50v + 100 v dt = 110 cos(377t − 10◦ ) Determine v(t) using the phasor method. You may assume that the value of the integral at t = −∞ is zero. Section 9.4 Phasor Relationships for Circuit Elements 9.22 Determine the current that flows through an 8- resistor connected to a voltage source vs = 110 cos 377t V. 9.23 What is the instantaneous voltage across a 2-µF capacitor when the current through it is i = 4 sin(106 t + 25◦ ) A? 9.24 The voltage across a 4-mH inductor is v = 60 cos(500t − 65◦ ) V. Find the instantaneous current through it. 9.25 A current source of i(t) = 10 sin(377t + 30◦ ) A is applied to a single-element load. The resulting voltage across the element is v(t) = −65 cos(377t + 120◦ ) V. What type of element is this? Calculate its value. 9.26 Two elements are connected in series as shown in Fig. 9.40. If i = 12 cos(2t − 30◦ ) A, find the element values. + − i 180 cos(2t + 10°) V Figure 9.40 For Prob. 9.26. 9.27 A series RL circuit is connected to a 110-V ac source. If the voltage across the resistor is 85 V, find the voltage across the inductor. 9.28 What value of ω will cause the forced response vo in Fig. 9.41 to be zero? + − 2 Ω + − 5 mF vo 50 cos vt V 20 mH Figure 9.41 For Prob. 9.28. Section 9.5 Impedance and Admittance 9.29 If vs = 5 cos 2t V in the circuit of Fig. 9.42, find vo. + − + − 2 Ω vo 0.25 F vs 1 H Figure 9.42 For Prob. 9.29. 9.30 Find ix when is = 2 sin 5t A is supplied to the circuit in Fig. 9.43. 2 Ω 1 H is ix 0.2 F Figure 9.43 For Prob. 9.30. 9.31 Find i(t) and v(t) in each of the circuits of Fig. 9.44. + − v i 4 Ω F 10 cos(3t + 45°) A (a) i 4 Ω 8 Ω F 50 cos 4t V + − + − 3 H (b) v 1 12 1 6 Figure 9.44 For Prob. 9.31. 9.32 Calculate i1(t) and i2(t) in the circuit of Fig. 9.45 if the source frequency is 60 Hz. + − 8 Ω 40 0° V j5 Ω −j10 Ω i1 i2 Figure 9.45 For Prob. 9.32.
  • 392. 388 PART 2 AC Circuits 9.33 In the circuit of Fig. 9.46, find io when: (a) ω = 1 rad/s (b) ω = 5 rad/s (c) ω = 10 rad/s + − 2 Ω 4 cos vt V 0.05 F io 1 H Figure 9.46 For Prob. 9.33. 9.34 Find v(t) in the RLC circuit of Fig. 9.47. + − + − 1 Ω 1 Ω 1 H 1 F v 10 cos t V Figure 9.47 For Prob. 9.34. 9.35 Calculate vo(t) in the circuit in Fig. 9.48. + − + − 30 Ω vo(t) 50 Ω 0.1 H 60 sin 200t V 50 mF Figure 9.48 For Prob. 9.35. 9.36 Determine io(t) in the RLC circuit of Fig. 9.49. io 1 Ω 4 cos 2t A 1 F 1 H Figure 9.49 For Prob. 9.36. 9.37 Calculate i(t) in the circuit of Fig. 9.50. + − 3 Ω 10 mH 5 mF 6 cos 200t V 4 Ω 5 Ω i Figure 9.50 For Prob. 9.37. 9.38 Find current Io in the network of Fig. 9.51. 2 Ω 2 Ω Io −j2 Ω j4 Ω −j2 Ω 5 0° A Figure 9.51 For Prob. 9.38. 9.39 If is = 5 cos(10t + 40◦ ) A in the circuit in Fig. 9.52, find io. 0.2 H 0.1 F 4 Ω 3 Ω io is Figure 9.52 For Prob. 9.39. 9.40 Find vs(t) in the circuit of Fig. 9.53 if the current ix through the 1- resistor is 0.5 sin 200t A. + − 1 Ω 2 Ω vs j2 Ω −j1 Ω ix Figure 9.53 For Prob. 9.40. 9.41 If the voltage vo across the 2- resistor in the circuit of Fig. 9.54 is 10 cos 2t V, obtain is. + − vo 0.1 F 1 Ω 2 Ω is 0.5 H Figure 9.54 For Prob. 9.41.
  • 393. CHAPTER 9 Sinusoids and Phasors 389 9.42 If Vo = 8 30◦ V in the circuit of Fig. 9.55, find Is. + − 5 Ω 10 Ω V o −j5 Ω j5 Ω Is Figure 9.55 For Prob. 9.42. 9.43 In the circuit of Fig. 9.56, find Vs if Io = 2 0◦ A. + − Io 1 Ω 2 Ω V s j2 Ω j4 Ω −j2 Ω −j1 Ω Figure 9.56 For Prob. 9.43. 9.44 Find Z in the network of Fig. 9.57, given that Vo = 4 0◦ V. + − + − Z 12 Ω V o 20 −90° V j8 Ω −j4 Ω Figure 9.57 For Prob. 9.44. Section 9.7 Impedance Combinations 9.45 At ω = 50 rad/s, determine Zin for each of the circuits in Fig. 9.58. 10 Ω 20 Ω 0.4 H 0.2 H Zin 1 mF (b) 1 Ω 1 Ω 10 mH 10 mF Zin (a) Figure 9.58 For Prob. 9.45. 9.46 Calculate Zeq for the circuit in Fig. 9.59. 1 Ω 2 Ω 6 Ω Zeq j4 Ω −j2 Ω Figure 9.59 For Prob. 9.46. 9.47 Find Zeq in the circuit of Fig. 9.60. Zeq 1 − j Ω 1 + j2 Ω j5 Ω 1 + j3 Ω Figure 9.60 For Prob. 9.47. 9.48 For the circuit in Fig. 9.61, find the input impedance Zin at 10 krad/s. + − + − v 2v 50 Ω 2 mH Zin 1 mF Figure 9.61 For Prob. 9.48. 9.49 Determine I and ZT for the circuit in Fig. 9.62. + − 2 Ω 3 Ω 4 Ω ZT 120 10° V j4 Ω −j6 Ω I Figure 9.62 For Prob. 9.49.
  • 394. 390 PART 2 AC Circuits 9.50 For the circuit in Fig. 9.63, calculate ZT and Vab. + − 20 Ω + − ZT V ab 60 90° V j10 Ω −j5 Ω 40 Ω a b Figure 9.63 For Prob. 9.50. 9.51 At ω = 103 rad/s, find the input admittance of each of the circuits in Fig. 9.64. Yin (a) 20 mH 12.5 mF 60 Ω 60 Ω Yin (b) 30 Ω 10 mH 20 mF 60 Ω 40 Ω Figure 9.64 For Prob. 9.51. 9.52 Determine Yeq for the circuit in Fig. 9.65. Yeq 3 Ω 5 Ω j1 Ω −j2 Ω −j4 Ω Figure 9.65 For Prob. 9.52. 9.53 Find the equivalent admittance Yeq of the circuit in Fig. 9.66. 2 S 4 S 1 S j5 S j1 S −j3 S −j2 S Figure 9.66 For Prob. 9.53. 9.54 Find the equivalent impedance of the circuit in Fig. 9.67. 10 Ω Zeq j15 Ω −j5 Ω −j10 Ω 2 Ω 5 Ω 8 Ω Figure 9.67 For Prob. 9.54. 9.55 Obtain the equivalent impedance of the circuit in Fig. 9.68. Zeq 1 Ω j2 Ω j4 Ω −j2 Ω −j Ω 2 Ω Figure 9.68 For Prob. 9.55. 9.56 Calculate the value of Zab in the network of Fig. 9.69. 20 Ω 20 Ω j6 Ω −j9 Ω 10 Ω −j9 Ω −j9 Ω j6 Ω j6 Ω a b Figure 9.69 For Prob. 9.56. 9.57 Determine the equivalent impedance of the circuit in Fig. 9.70. 2 Ω 4 Ω j6 Ω j8 Ω j8 Ω j12 Ω −j4 Ω −j6 Ω a b Figure 9.70 For Prob. 9.57.
  • 395. CHAPTER 9 Sinusoids and Phasors 391 Section 9.8 Applications 9.58 Design an RL circuit to provide a 90◦ leading phase shift. 9.59 Design a circuit that will transform a sinusoidal input to a cosinusoidal output. 9.60 Refer to the RC circuit in Fig. 9.71. (a) Calculate the phase shift at 2 MHz. (b) Find the frequency where the phase shift is 45◦ . + − + − 5 Ω 20 nF V o V i Figure 9.71 For Prob. 9.60. 9.61 (a) Calculate the phase shift of the circuit in Fig. 9.72. (b) State whether the phase shift is leading or lagging (output with respect to input). (c) Determine the magnitude of the output when the input is 120 V. + − + − 20 Ω 40 Ω 30 Ω V o j10 Ω j30 Ω j60 Ω V i Figure 9.72 For Prob. 9.61. 9.62 Consider the phase-shifting circuit in Fig. 9.73. Let Vi = 120 V operating at 60 Hz. Find: (a) Vo when R is maximum (b) Vo when R is minimum (c) the value of R that will produce a phase shift of 45◦ + − + − 50 Ω 200 mH vo vi 0 R 100 Ω Figure 9.73 For Prob. 9.62. 9.63 The ac bridge in Fig. 9.37 is balanced when R1 = 400 , R2 = 600 , R3 = 1.2 k, and C2 = 0.3 µF. Find Rx and Cx. 9.64 A capacitance bridge balances when R1 = 100 , R2 = 2 k, and Cs = 40 µF. What is Cx, the capacitance of the capacitor under test? 9.65 An inductive bridge balances when R1 = 1.2 k, R2 = 500 , and Ls = 250 mH. What is the value of Lx, the inductance of the inductor under test? 9.66 The ac bridge shown in Fig. 9.74 is known as a Maxwell bridge and is used for accurate measurement of inductance and resistance of a coil in terms of a standard capacitance Cs. Show that when the bridge is balanced, Lx = R2R3Cs and Rx = R2 R1 R3 Find Lx and Rx for R1 = 40 k, R2 = 1.6 k, R3 = 4 k, and Cs = 0.45 µF. AC meter R3 Lx Rx R2 R1 Cs Figure 9.74 Maxwell bridge; for Prob. 9.66. 9.67 The ac bridge circuit of Fig. 9.75 is called a Wien bridge. It is used for measuring the frequency of a source. Show that when the bridge is balanced, f = 1 2π √ R2R4C2C4 AC meter R3 R2 R1 C4 C2 R4 Figure 9.75 Wein bridge; for Prob. 9.67.
  • 396. 392 PART 2 AC Circuits COMPREHENSIVE PROBLEMS 9.68 The circuit shown in Fig. 9.76 is used in a television receiver. What is the total impedance of this circuit? 240 Ω j95 Ω −j84 Ω Figure 9.76 For Prob. 9.68. 9.69 The network in Fig. 9.77 is part of the schematic describing an industrial electronic sensing device. What is the total impedance of the circuit at 2 kHz? 50 Ω 10 mH 2 mF 80 Ω 100 Ω Figure 9.77 For Prob. 9.69. 9.70 A series audio circuit is shown in Fig. 9.78. (a) What is the impedance of the circuit? (b) If the frequency were halved, what would be the impedance of the circuit? 250 Hz ≈ j30 Ω 120 Ω −j20 Ω −j20 Ω Figure 9.78 For Prob. 9.70. 9.71 An industrial load is modeled as a series combination of a capacitance and a resistance as shown in Fig. 9.79. Calculate the value of an inductance L across the series combination so that the net impedance is resistive at a frequency of 5 MHz. 200 Ω 50 nF L Figure 9.79 For Prob. 9.71. 9.72 An industrial coil is modeled as a series combination of an inductance L and resistance R, as shown in Fig. 9.80. Since an ac voltmeter measures only the magnitude of a sinusoid, the following measurements are taken at 60 Hz when the circuit operates in the steady state: |Vs| = 145 V, |V1| = 50 V, |Vo| = 110 V Use these measurements to determine the values of L and R. 80 Ω + − + − V 1 V s + − V o R L Coil Figure 9.80 For Prob. 9.72. 9.73 Figure 9.81 shows a parallel combination of an inductance and a resistance. If it is desired to connect a capacitor in series with the parallel combination such that the net impedance is resistive at 10 MHz, what is the required value of C? 300 Ω 20 mH C Figure 9.81 For Prob. 9.73. 9.74 A power transmission system is modeled as shown in Fig. 9.82. Given the source voltage Vs = 115 0◦ V, source impedance Zs = 1 + j0.5 , line impedance Z. = 0.4 + j0.3 , and load impedance ZL = 23.2 + j18.9 , find the load current IL. + − vs Zᐉ Zs Zᐉ ZL IL Source Transmission line Load Figure 9.82 For Prob. 9.74.
  • 397. 393 C H A P T E R SINUSOIDAL STEADY-STATE ANALYSIS 1 0 An expert problem solver must be endowed with two incompatible quan- tities, a restless imagination and a patient pertinacity. —Howard W. Eves Enhancing Your Career Career in Software Engineering Software engineering is that aspect of engineering that deals with the practical ap- plication of scientific knowledge in the design, construction, and validation of computer programs and the associated doc- umentation required to develop, operate, and maintain them. It is a branch of electrical engineering that is becoming in- creasingly important as more and more disciplines require one form of software package or another to perform rou- tine tasks and as programmable microelectronic systems are used in more and more applications. The role of a software engineer should not be con- fused with that of a computer scientist; the software engi- neer is a practitioner, not a theoretician. A software engineer should have good computer-programming skill and be famil- iar with programming languages, in particular C++ , which is becoming increasingly popular. Because hardware and software are closely interlinked, it is essential that a soft- ware engineer have a thorough understanding of hardware design. Most important, the software engineer should have some specialized knowledge of the area in which the soft- ware development skill is to be applied. All in all, the field of software engineering offers a great career to those who enjoy programming and devel- oping software packages. The higher rewards will go to those having the best preparation, with the most interesting and challenging opportunities going to those with graduate education. Output of a modeling software. (Courtesy of National Instruments.)
  • 398. 394 PART 2 AC Circuits 10.1 INTRODUCTION In Chapter 9, we learned that the forced or steady-state response of cir- cuits to sinusoidal inputs can be obtained by using phasors. We also know that Ohm’s and Kirchhoff’s laws are applicable to ac circuits. In this chapter, we want to see how nodal analysis, mesh analysis, Thevenin’s theorem, Norton’s theorem, superposition, and source transformations are applied in analyzing ac circuits. Since these techniques were already introduced for dc circuits, our major effort here will be to illustrate with examples. Analyzing ac circuits usually requires three steps. Steps to Analyze ac Circuits: 1. Transform the circuit to the phasor or frequency domain. 2. Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.). 3. Transform the resulting phasor to the time domain. Step 1 is not necessary if the problem is specified in the frequency domain. In step 2, the analysis is performed in the same manner as dc circuit analysis except that complex numbers are involved. Having read Chapter 9, we are adept at handling step 3. Frequency-domain analysis of an ac circuit via phasors is much easier than analysis of the cir- cuit in the time domain. Toward the end of the chapter, we learn how to apply PSpice in solving ac circuit problems. We finally apply ac circuit analysis to two practical ac circuits: oscillators and ac transistor circuits. 10.2 NODAL ANALYSIS The basis of nodal analysis is Kirchhoff’s current law. Since KCL is valid for phasors, as demonstrated in Section 9.6, we can analyze ac circuits by nodal analysis. The following examples illustrate this. E X A M P L E 1 0 . 1 Find ix in the circuit of Fig. 10.1 using nodal analysis. 0.5 H 0.1 F 1 H 10 Ω 2ix ix + − 20 cos 4t V Figure10.1 For Example 10.1. Solution: We first convert the circuit to the frequency domain:
  • 399. CHAPTER 10 Sinusoidal Steady-State Analysis 395 20 cos 4t ⇒ 20 0◦ , ω = 4 rad/s 1 H ⇒ jωL = j4 0.5 H ⇒ jωL = j2 0.1 F ⇒ 1 jωC = −j2.5 Thus, the frequency-domain equivalent circuit is as shown in Fig. 10.2. –j2.5 Ω j2 Ω j4 Ω 10 Ω 2Ix Ix + − V 1 V2 20 0° V Figure10.2 Frequency-domain equivalent of the circuit in Fig. 10.1. Applying KCL at node 1, 20 − V1 10 = V1 −j2.5 + V1 − V2 j4 or (1 + j1.5)V1 + j2.5V2 = 20 (10.1.1) At node 2, 2Ix + V1 − V2 j4 = V2 j2 But Ix = V1/−j2.5. Substituting this gives 2V1 −j2.5 + V1 − V2 j4 = V2 j2 By simplifying, we get 11V1 + 15V2 = 0 (10.1.2) Equations (10.1.1) and (10.1.2) can be put in matrix form as 1 + j1.5 j2.5 11 15 V1 V2 = 20 0 We obtain the determinants as = 1 + j1.5 j2.5 11 15 = 15 − j5 1 = 20 j2.5 0 15 = 300, 2 = 1 + j1.5 20 11 0 = −220 V1 = 1 = 300 15 − j5 = 18.97 18.43◦ V V2 = 2 = −220 15 − j5 = 13.91 198.3◦ V
  • 400. 396 PART 2 AC Circuits The current Ix is given by Ix = V1 −j2.5 = 18.97 18.43◦ 2.5 − 90◦ = 7.59 108.4◦ A Transforming this to the time domain, ix = 7.59 cos(4t + 108.4◦ ) A P R A C T I C E P R O B L E M 1 0 . 1 Using nodal analysis, find v1 and v2 in the circuit of Fig. 10.3. 4 Ω 2 Ω 3vx vx 2 H 0.2 F v1 v2 + − + − 10 sin 2t A Figure10.3 For Practice Prob. 10.1. Answer: v1(t) = 20.96 sin(2t + 58◦ ) V, v2(t) = 44.11 sin(2t + 41◦ ) V. E X A M P L E 1 0 . 2 Compute V1 and V2 in the circuit of Fig. 10.4. 4 Ω 12 Ω 1 2 V1 V2 –j3 Ω j6 Ω + − 10 45° V 3 0° A Figure10.4 For Example 10.2. Solution: Nodes 1 and 2 form a supernode as shown in Fig. 10.5. Applying KCL at the supernode gives 3 = V1 −j3 + V2 j6 + V2 12 or 36 = j4V1 + (1 − j2)V2 (10.2.1) But a voltage source is connected between nodes 1 and 2, so that
  • 401. CHAPTER 10 Sinusoidal Steady-State Analysis 397 –j3 Ω j6 Ω 12 Ω 3 A Supernode V1 V2 Figure10.5 A supernode in the circuit of Fig. 10.4. V1 = V2 + 10 45◦ (10.2.2) Substituting Eq. (10.2.2) in Eq. (10.2.1) results in 36 − 40 135◦ = (1 + j2)V2 ⇒ V2 = 31.41 − 87.18◦ V From Eq. (10.2.2), V1 = V2 + 10 45◦ = 25.78 − 70.48◦ V P R A C T I C E P R O B L E M 1 0 . 2 Calculate V1 and V2 in the circuit shown in Fig. 10.6. 4 Ω 2 Ω j4 Ω –j1 Ω + − + − V1 V2 15 0° V 20 60° V Figure10.6 For Practice Prob. 10.2. Answer: V1 = 19.36 69.67◦ V, V2 = 3.376 165.7◦ V. 10.3 MESH ANALYSIS Kirchhoff’s voltage law (KVL) forms the basis of mesh analysis. The validity of KVL for ac circuits was shown in Section 9.6 and is illustrated in the following examples. E X A M P L E 1 0 . 3 Determine current Io in the circuit of Fig. 10.7 using mesh analysis. Solution: Applying KVL to mesh 1, we obtain (8 + j10 − j2)I1 − (−j2)I2 − j10I3 = 0 (10.3.1)
  • 402. 398 PART 2 AC Circuits 4 Ω 8 Ω –j2 Ω –j2 Ω j10 Ω + − Io I2 I3 I1 5 0° A 20 90° V Figure10.7 For Example 10.3. For mesh 2, (4 − j2 − j2)I2 − (−j2)I1 − (−j2)I3 + 20 90◦ = 0 (10.3.2) For mesh 3, I3 = 5. Substituting this in Eqs. (10.3.1) and (10.3.2), we get (8 + j8)I1 + j2I2 = j50 (10.3.3) j2I1 + (4 − j4)I2 = −j20 − j10 (10.3.4) Equations (10.3.3) and (10.3.4) can be put in matrix form as 8 + j8 j2 j2 4 − j4 I1 I2 = j50 −j30 from which we obtain the determinants = 8 + j8 j2 j2 4 − j4 = 32(1 + j)(1 − j) + 4 = 68 2 = 8 + j8 j50 j2 −j30 = 340 − j240 = 416.17 − 35.22◦ I2 = 2 = 416.17 − 35.22◦ 68 = 6.12 − 35.22◦ A The desired current is Io = −I2 = 6.12 144.78◦ A P R A C T I C E P R O B L E M 1 0 . 3 Find Io in Fig. 10.8 using mesh analysis. 8 Ω j4 Ω –j2 Ω 6 Ω + − Io 2 0° A 10 30° V Figure10.8 For Practice Prob. 10.3. Answer: 1.194 65.45◦ A.
  • 403. CHAPTER 10 Sinusoidal Steady-State Analysis 399 E X A M P L E 1 0 . 4 Solve for Vo in the circuit in Fig. 10.9 using mesh analysis. 8 Ω 6 Ω –j2 Ω –j4 Ω j5 Ω 4 0˚ A + − Vo + − 3 0° A 10 0° V Figure10.9 For Example 10.4. Solution: As shown in Fig. 10.10, meshes 3 and 4 form a supermesh due to the current source between the meshes. For mesh 1, KVL gives −10 + (8 − j2)I1 − (−j2)I2 − 8I3 = 0 or (8 − j2)I1 + j2I2 − 8I3 = 10 (10.4.1) For mesh 2, I2 = −3 (10.4.2) For the supermesh, (8 − j4)I3 − 8I1 + (6 + j5)I4 − j5I2 = 0 (10.4.3) Due to the current source between meshes 3 and 4, at node A, I4 = I3 + 4 (10.4.4) Combining Eqs. (10.4.1) and (10.4.2), (8 − j2)I1 − 8I3 = 10 + j6 (10.4.5) Combining Eqs. (10.4.2) to (10.4.4), −8I1 + (14 + j)I3 = −24 − j35 (10.4.6) 8 Ω 6 Ω –j2 Ω –j4 Ω j5 Ω 10 V 3 A 4 A A + − + − I2 I3 I3 I4 I4 I1 Supermesh Vo Figure10.10 Analysis of the circuit in Fig. 10.9.
  • 404. 400 PART 2 AC Circuits From Eqs. (10.4.5) and (10.4.6), we obtain the matrix equation 8 − j2 −8 −8 14 + j I1 I3 = 10 + j6 −24 − j35 We obtain the following determinants = 8 − j2 −8 −8 14 + j = 112 + j8 − j28 + 2 − 64 = 50 − j20 1 = 10 + j6 −8 −24 − j35 14 + j = 140 + j10 + j84 − 6 − 192 − j280 = −58 − j186 Current I1 is obtained as I1 = 1 = −58 − j186 50 − j20 = 3.618 274.5◦ A The required voltage Vo is Vo = −j2(I1 − I2) = −j2(3.618 274.5◦ + 3) = −7.2134 − j6.568 = 9.756 222.32◦ V P R A C T I C E P R O B L E M 1 0 . 4 Calculate current Io in the circuit of Fig. 10.11. j8 Ω –j6 Ω –j4 Ω 5 Ω 10 Ω Io + − 50 0° V 2 0° A Figure10.11 For Practice Prob. 10.4. Answer: 5.075 5.943◦ A. 10.4 SUPERPOSITION THEOREM Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to dc circuits. The theorem becomes important if the circuit has sources operating at different frequencies. In this case, since the impedances depend on frequency, we must have a different frequency-domain circuit for each frequency. The total response must be obtained by adding the individual responses in the time domain. It is incorrect to try to add the responses in the phasor or frequency domain. Why? Because the exponential factor ejωt is implicit in sinusoidal analy- sis, and that factor would change for every angular frequency ω. It would therefore not make sense to add responses at different frequencies in the phasor domain. Thus, when a circuit has sources operating at different
  • 405. CHAPTER 10 Sinusoidal Steady-State Analysis 401 frequencies, one must add the responses due to the individual frequencies in the time domain. E X A M P L E 1 0 . 5 Use the superposition theorem to find Io in the circuit in Fig. 10.7. Solution: Let Io = I o + I o (10.5.1) where I o and I o are due to the voltage and current sources, respectively. To find I o, consider the circuit in Fig. 10.12(a). If we let Z be the parallel combination of −j2 and 8 + j10, then Z = −j2(8 + j10) −2j + 8 + j10 = 0.25 − j2.25 and current I o is I o = j20 4 − j2 + Z = j20 4.25 − j4.25 or I o = −2.353 + j2.353 (10.5.2) 4 Ω 8 Ω –j2 Ω –j2 Ω j10 Ω j20 V + − I'o (a) (b) 4 Ω 8 Ω –j2 Ω –j2 Ω j10 Ω 5 A I'' o I2 I3 I1 Figure10.12 Solution of Example 10.5. To get I o, consider the circuit in Fig. 10.12(b). For mesh 1, (8 + j8)I1 − j10I3 + j2I2 = 0 (10.5.3) For mesh 2, (4 − j4)I2 + j2I1 + j2I3 = 0 (10.5.4) For mesh 3, I3 = 5 (10.5.5) From Eqs. (10.5.4) and (10.5.5), (4 − j4)I2 + j2I1 + j10 = 0 Expressing I1 in terms of I2 gives I1 = (2 + j2)I2 − 5 (10.5.6) Substituting Eqs. (10.5.5) and (10.5.6) into Eq. (10.5.3), we get (8 + j8)[(2 + j2)I2 − 5] − j50 + j2I2 = 0 or I2 = 90 − j40 34 = 2.647 − j1.176 Current I o is obtained as I o = −I2 = −2.647 + j1.176 (10.5.7) From Eqs. (10.5.2) and (10.5.7), we write Io = I o + I o = −5 + j3.529 = 6.12 144.78◦ A
  • 406. 402 PART 2 AC Circuits which agrees with what we got in Example 10.3. It should be noted that applying the superposition theorem is not the best way to solve this problem. It seems that we have made the problem twice as hard as the original one by using superposition. However, in Example 10.6, superposition is clearly the easiest approach. P R A C T I C E P R O B L E M 1 0 . 5 Find current Io in the circuit of Fig. 10.8 using the superposition theorem. Answer: 1.194 65.45◦ A. E X A M P L E 1 0 . 6 Find vo in the circuit in Fig. 10.13 using the superposition theorem. 2 H 1 Ω 4 Ω 0.1 F 5 V + − + − 10 cos 2t V 2 sin 5t A − + vo Figure10.13 For Example 10.6. Solution: Since the circuit operates at three different frequencies (ω = 0 for the dc voltage source), one way to obtain a solution is to use superposition, which breaks the problem into single-frequency problems. So we let vo = v1 + v2 + v3 (10.6.1) where v1 is due to the 5-V dc voltage source, v2 is due to the 10 cos 2t V voltage source, and v3 is due to the 2 sin 5t A current source. To find v1, we set to zero all sources except the 5-V dc source. We recall that at steady state, a capacitor is an open circuit to dc while an inductor is a short circuit to dc. There is an alternative way of looking at this. Since ω = 0, jωL = 0, 1/jωC = ∞. Either way, the equivalent circuit is as shown in Fig. 10.14(a). By voltage division, −v1 = 1 1 + 4 (5) = 1 V (10.6.2) To find v2, we set to zero both the 5-V source and the 2 sin 5t current source and transform the circuit to the frequency domain. 10 cos 2t ⇒ 10 0◦ , ω = 2 rad/s 2 H ⇒ jωL = j4 0.1 F ⇒ 1 jωC = −j5
  • 407. CHAPTER 10 Sinusoidal Steady-State Analysis 403 The equivalent circuit is now as shown in Fig. 10.14(b). Let Z = −j5 4 = −j5 × 4 4 − j5 = 2.439 − j1.951 By voltage division, V2 = 1 1 + j4 + Z (10 0◦ ) = 10 3.439 + j2.049 = 2.498 − 30.79◦ In the time domain, v2 = 2.498 cos(2t − 30.79◦ ) (10.6.3) 1 Ω 4 Ω 5 V + − − + v1 (a) (b) (c) 1 Ω j4 Ω –j5 Ω 4 Ω + − 1 Ω 4 Ω –j2 Ω j10 Ω I1 10 0° V 2 –90° A + − V2 + − V3 Figure 10.14 Solution of Example 10.6: (a) setting all sources to zero except the 5-V dc source, (b) setting all sources to zero except the ac voltage source, (c) setting all sources to zero except the ac current source. To obtain v3, we set the voltage sources to zero and transform what is left to the frequency domain. 2 sin 5t ⇒ 2 − 90◦ , ω = 5 rad/s 2 H ⇒ jωL = j10 0.1 F ⇒ 1 jωC = −j2 The equivalent circuit is in Fig. 10.14(c). Let Z1 = −j2 4 = −j2 × 4 4 − j2 = 0.8 − j1.6 By current division, I1 = j10 j10 + 1 + Z1 (2 − 90◦ ) A V3 = I1 × 1 = j10 1.8 + j8.4 (−j2) = 2.328 − 77.91◦ V In the time domain, v3 = 2.33 cos(5t − 80◦ ) = 2.33 sin(5t + 10◦ ) V (10.6.4) Substituting Eqs. (10.6.2) to (10.6.4) into Eq. (10.6.1), we have vo(t) = −1 + 2.498 cos(2t − 30.79◦ ) + 2.33 sin(5t + 10◦ ) V P R A C T I C E P R O B L E M 1 0 . 6 Calculate vo in the circuit of Fig. 10.15 using the superposition theorem.
  • 408. 404 PART 2 AC Circuits 8 Ω 0.2 F 1 H + − 30 sin 5t V 2 cos 10t A + − vo Figure10.15 For Practice Prob. 10.6. Answer: 4.631 sin(5t − 81.12◦ ) + 1.051 cos(10t − 86.24◦ ) V. 10.5 SOURCE TRANSFORMATION As Fig. 10.16 shows, source transformation in the frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance, or vice versa. As we go from one source type to another, we must keep the following relationship in mind: Vs = ZsIs ⇐⇒ Is = Vs Zs (10.1) a b Vs Vs = ZsIs Zs Zs + − a b Is Is = Zs Vs Figure10.16 Source transformation. E X A M P L E 1 0 . 7 Calculate Vx in the circuit of Fig. 10.17 using the method of source trans- formation. 5 Ω j4 Ω –j13 Ω 3 Ω 10 Ω 4 Ω + − + − Vx 20 –90° V Figure10.17 For Example 10.7.
  • 409. CHAPTER 10 Sinusoidal Steady-State Analysis 405 Solution: We transform the voltage source to a current source and obtain the circuit in Fig. 10.18(a), where Is = 20 − 90◦ 5 = 4 − 90◦ = −j4 A The parallel combination of 5- resistance and (3+j4) impedance gives Z1 = 5(3 + j4) 8 + j4 = 2.5 + j1.25 Converting the current source to a voltage source yields the circuit in Fig. 10.18(b), where Vs = IsZ1 = −j4(2.5 + j1.25) = 5 − j10 V By voltage division, Vx = 10 10 + 2.5 + j1.25 + 4 − j13 (5 − j10) = 5.519 − 28◦ V 5 Ω j4 Ω –j13 Ω 3 Ω 10 Ω 4 Ω + − + − Vx Is = –j4 Α –j13 Ω 10 Ω 4 Ω 2.5 Ω j1.25 Ω Vx Vs = 5 – j10 V + − (a) (b) Figure10.18 Solution of the circuit in Fig. 10.17. P R A C T I C E P R O B L E M 1 0 . 7 Find Io in the circuit of Fig. 10.19 using the concept of source transfor- mation. –j3 Ω j5 Ω j1 Ω 2 Ω Io –j2 Ω 4 90° Α 4 Ω 1 Ω Figure10.19 For Practice Prob. 10.7. Answer: 3.288 99.46◦ A.
  • 410. 406 PART 2 AC Circuits 10.6 THEVENIN AND NORTON EQUIVALENT CIRCUITS Thevenin’s and Norton’s theorems are applied to ac circuits in the same way as they are to dc circuits. The only additional effort arises from the need to manipulate complex numbers. The frequency-domain version of a Thevenin equivalent circuit is depicted in Fig. 10.20, where a linear circuit is replaced by a voltage source in series with an impedance. The Norton equivalent circuit is illustrated in Fig. 10.21, where a linear circuit is replaced by a current source in parallel with an impedance. Keep in mind that the two equivalent circuits are related as VTh = ZN IN , ZTh = ZN (10.2) just as in source transformation. VTh is the open-circuit voltage while IN is the short-circuit current. a b ZTh a b VTh Linear circuit + − Figure10.20 Thevenin equivalent. a b ZN a b IN Linear circuit Figure10.21 Norton equivalent. If the circuit has sources operating at different frequencies (see Example 10.6, for example), the Thevenin or Norton equivalent circuit must be determined at each frequency. This leads to entirely different equivalent circuits, one for each frequency, not one equivalent circuit with equivalent sources and equivalent impedances. E X A M P L E 1 0 . 8 ObtaintheTheveninequivalentatterminalsa-b ofthecircuitinFig.10.22. 4 Ω d a b f c e –j6 Ω j12 Ω 8 Ω + − a b 120 75° V Figure10.22 For Example 10.8. Solution: We find ZTh by setting the voltage source to zero. As shown in Fig. 10.23(a), the 8- resistance is now in parallel with the −j6 reactance, so that their combination gives Z1 = −j6 8 = −j6 × 8 8 − j6 = 2.88 − j3.84 Similarly, the 4- resistance is in parallel with the j12 reactance, and their combination gives Z2 = 4 j12 = j12 × 4 4 + j12 = 3.6 + j1.2
  • 411. CHAPTER 10 Sinusoidal Steady-State Analysis 407 4 Ω 8 Ω –j6 Ω j12 Ω ZTh VTh a e c f,d f,d b (a) (b) 8 Ω 4 Ω j12 Ω –j6 Ω + − I2 I1 d e a b c f − + 120 75° V Figure10.23 Solution of the circuit in Fig. 10.22: (a) finding ZTh, (b) finding VTh. The Thevenin impedance is the series combination of Z1 and Z2; that is, ZTh = Z1 + Z2 = 6.48 − j2.64 To find VTh, consider the circuit in Fig. 10.23(b). Currents I1 and I2 are obtained as I1 = 120 75◦ 8 − j6 A, I2 = 120 75◦ 4 + j12 A Applying KVL around loop bcdeab in Fig. 10.23(b) gives VTh − 4I2 + (−j6)I1 = 0 or VTh = 4I2 + j6I1 = 480 75◦ 4 + j12 + 720 75◦ + 90◦ 8 − j6 = 37.95 3.43◦ + 72 201.87◦ = −28.936 − j24.55 = 37.95 220.31◦ V P R A C T I C E P R O B L E M 1 0 . 8 Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 10.24. –j4 Ω j2 Ω 6 Ω 10 Ω + − a b 30 20° V Figure10.24 For Practice Prob. 10.8. Answer: ZTh = 12.4 − j3.2 , VTh = 18.97 − 51.57◦ V.
  • 412. 408 PART 2 AC Circuits E X A M P L E 1 0 . 9 Find the Thevenin equivalent of the circuit in Fig. 10.25 as seen from ter- minals a-b. –j4 Ω j3 Ω 4 Ω 2 Ω a b Io 0.5 Io 15 0° A Figure10.25 For Example 10.9. Solution: To find VTh, we apply KCL at node 1 in Fig. 10.26(a). 15 = Io + 0.5Io ⇒ Io = 10 A Applying KVL to the loop on the right-hand side in Fig. 10.26(a), we obtain −Io(2 − j4) + 0.5Io(4 + j3) + VTh = 0 or VTh = 10(2 − j4) − 5(4 + j3) = −j55 Thus, the Thevenin voltage is VTh = 55 − 90◦ V 4 + j3 Ω 2 – j4 Ω a b Io 0.5Io 0.5Io VTh 15 A + − 2 1 4 + j3 Ω 2 – j4 Ω a b Vs Is 0.5Io Io Is = 3 0° A (a) (b) + − V s Figure10.26 Solution of the problem in Fig. 10.25: (a) finding VTh, (b) finding ZTh. To obtain ZTh, we remove the independent source. Due to the presence of the dependent current source, we connect a 3-A current source (3 is an arbitrary value chosen for convenience here, a number divisible by the sum of currents leaving the node) to terminals a-b as shown in Fig. 10.26(b). At the node, KCL gives 3 = Io + 0.5Io ⇒ Io = 2 A Applying KVL to the outer loop in Fig. 10.26(b) gives Vs = Io(4 + j3 + 2 − j4) = 2(6 − j)
  • 413. CHAPTER 10 Sinusoidal Steady-State Analysis 409 The Thevenin impedance is ZTh = Vs Is = 2(6 − j) 3 = 4 − j0.6667 P R A C T I C E P R O B L E M 1 0 . 9 DeterminetheTheveninequivalentofthecircuitinFig.10.27asseenfrom the terminals a-b. –j2 Ω j4 Ω 8 Ω 4 Ω a b 0.2Vo 5 0° A − + V o Figure10.27 For Practice Prob. 10.9. Answer: ZTh = 12.166 136.3◦ , VTh = 7.35 72.9◦ V. E X A M P L E 1 0 . 1 0 Obtain current Io in Fig. 10.28 using Norton’s theorem. 3 0° A 40 90° V 8 Ω 5 Ω 20 Ω 10 Ω –j2 Ω j4 Ω j15 Ω + − Io a b Figure10.28 For Example 10.10. Solution: Our first objective is to find the Norton equivalent at terminals a-b. ZN is found in the same way as ZTh. We set the sources to zero as shown in Fig. 10.29(a). As evident from the figure, the (8 − j2) and (10 + j4) impedances are short-circuited, so that ZN = 5 To get IN , we short-circuit terminals a-b as in Fig. 10.29(b) and apply mesh analysis. Notice that meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1, −j40 + (18 + j2)I1 − (8 − j2)I2 − (10 + j4)I3 = 0 (10.10.1) For the supermesh, (13 − j2)I2 + (10 + j4)I3 − (18 + j2)I1 = 0 (10.10.2)
  • 414. 410 PART 2 AC Circuits 3 8 5 10 –j2 j4 j40 + − IN I3 I2 I3 I2 I1 a b (b) 5 20 j15 3 + j8 Io (c) 8 5 10 –j2 j4 ZN (a) Figure10.29 Solution of the circuit in Fig. 10.28: (a) finding ZN , (b) finding VN , (c) calculating Io. At node a, due to the current source between meshes 2 and 3, I3 = I2 + 3 (10.10.3) Adding Eqs. (10.10.1) and (10.10.2) gives −j40 + 5I2 = 0 ⇒ I2 = j8 From Eq. (10.10.3), I3 = I2 + 3 = 3 + j8 The Norton current is IN = I3 = (3 + j8) A Figure 10.29(c) shows the Norton equivalent circuit along with the imped- ance at terminals a-b. By current division, Io = 5 5 + 20 + j15 IN = 3 + j8 5 + j3 = 1.465 38.48◦ A P R A C T I C E P R O B L E M 1 0 . 1 0 Determine the Norton equivalent of the circuit in Fig. 10.30 as seen from terminals a-b. Use the equivalent to find Io. j2 Ω a b Io –j3 Ω –j5 Ω + − 8 Ω 4 Ω 1 Ω 10 Ω 20 0° V 4 –90° A Figure10.30 For Practice Prob. 10.10. Answer: ZN = 3.176 + j0.706 , IN = 8.396 − 32.68◦ A, Io = 1.971 − 2.101◦ A.
  • 415. CHAPTER 10 Sinusoidal Steady-State Analysis 411 10.7 OP AMP AC CIRCUITS The three steps stated in Section 10.1 also apply to op amp circuits, as long as the op amp is operating in the linear region. As usual, we will assume ideal op amps. (See Section 5.2.) As discussed in Chapter 5, the key to analyzing op amp circuits is to keep two important properties of an ideal op amp in mind: 1. No current enters either of its input terminals. 2. The voltage across its input terminals is zero. The following examples will illustrate these ideas. E X A M P L E 1 0 . 1 1 Determine vo(t) for the op amp circuit in Fig. 10.31(a) if vs = 3 cos 1000t V. + − + − Vo Vo V1 –j5 kΩ –j10 kΩ 10 kΩ 10 kΩ 20 kΩ 3 0° V + − + − vs vo 10 kΩ 10 kΩ 0.1 mF 0.2 mF 20 kΩ 1 2 0 V (a) (b) Figure10.31 For Example 10.11: (a) the original circuit in the time domain, (b) its frequency-domain equivalent. Solution: We first transform the circuit to the frequency domain, as shown in Fig. 10.31(b), where Vs = 3 0◦ , ω = 1000 rad/s. Applying KCL at node 1, we obtain 3 0◦ − V1 10 = V1 −j5 + V1 − 0 10 + V1 − Vo 20 or 6 = (5 + j4)V1 − Vo (10.11.1) At node 2, KCL gives V1 − 0 10 = 0 − Vo −j10 which leads to V1 = −jVo (10.11.2) Substituting Eq. (10.11.2) into Eq. (10.11.1) yields 6 = −j(5 + j4)Vo − Vo = (3 − j5)Vo Vo = 6 3 − j5 = 1.029 59.04◦ Hence, vo(t) = 1.029 cos(1000t + 59.04◦ ) V
  • 416. 412 PART 2 AC Circuits P R A C T I C E P R O B L E M 1 0 . 1 1 Find vo and io in the op amp circuit of Fig. 10.32. Let vs = 2 cos 5000t V. + − + − vs vo 10 kΩ 20 kΩ 20 nF 10 nF io Figure10.32 For Practice Prob. 10.11. Answer: 0.667 sin 5000t V, 66.67 sin 5000t µA. E X A M P L E 1 0 . 1 2 Compute the closed-loop gain and phase shift for the circuit in Fig. 10.33. Assume that R1 = R2 = 10 k, C1 = 2 µF, C2 = 1 µF, and ω = 200 rad/s. + − + − vs vo R1 R2 C2 C1 + − Figure10.33 For Example 10.12. Solution: The feedback and input impedances are calculated as Zf = R2 1 jωC2 = R2 1 + jωR2C2 Zi = R1 + 1 jωC1 = 1 + jωR1C1 jωC1 Since the circuit in Fig. 10.33 is an inverting amplifier, the closed-loop gain is given by G = Vo Vs = − Zf Zi = jωC1R2 (1 + jωR1C1)(1 + jωR2C2) Substituting the given values of R1, R2, C1, C2, and ω, we obtain G = j4 (1 + j4)(1 + j2) = 0.434 − 49.4◦ Thus the closed-loop gain is 0.434 and the phase shift is −49.4◦ . P R A C T I C E P R O B L E M 1 0 . 1 2 Obtain the closed-loop gain and phase shift for the circuit in Fig. 10.34. Let R = 10 k, C = 1 µF, and ω = 1000 rad/s. + − + − vs vo R R C Figure10.34 For Practice Prob. 10.12. Answer: 1.015, −5.599◦ .
  • 417. CHAPTER 10 Sinusoidal Steady-State Analysis 413 10.8 AC ANALYSIS USING PSPICE PSpice affords a big relief from the tedious task of manipulating complex numbers in ac circuit analysis. The procedure for using PSpice for ac analysis is quite similar to that required for dc analysis. The reader should read Section D.5 in Appendix D for a review of PSpice concepts for ac analysis. AC circuit analysis is done in the phasor or frequency domain, and all sources must have the same frequency. Although AC analysis with PSpice involves using AC Sweep, our analysis in this chapter requires a single frequency f = ω/2π. The output file of PSpice contains voltage and current phasors. If necessary, the impedances can be calculated using the voltages and currents in the output file. E X A M P L E 1 0 . 1 3 Obtain vo and io in the circuit of Fig. 10.35 using PSpice. 2 mF 50 mH 4 kΩ 2 kΩ io 0.5io + − 8 sin(1000t + 50°) V vo + − Figure10.35 For Example 10.13. Solution: We first convert the sine function to cosine. 8 sin(1000t + 50◦ ) = 8 cos(1000t + 50◦ − 90◦ ) = 8 cos(1000t − 40◦ ) The frequency f is obtained from ω as f = ω 2π = 1000 2π = 159.155 Hz The schematic for the circuit is shown in Fig. 10.36. Notice the current- controlled current source F1 is connected such that its current flows from ACMAG=8 ACPHASE=-40 AC=ok MAG=ok PHASE=ok AC=yes MAG=yes PHASE=ok V R1 C1 2u L1 F1 4k IPRINT 50mH GAIN=0.5 2k R2 0 2 3 + − Figure10.36 The schematic of the circuit in Fig. 10.35.
  • 418. 414 PART 2 AC Circuits node0tonode3inconformitywiththeoriginalcircuitinFig.10.35. Since we only want the magnitude and phase of vo and io, we set the attributes of IPRINT AND VPRINT1 each to AC = yes, MAG = yes, PHASE = yes. As a single-frequency analysis, we select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq = 159.155, and Final Freq = 159.155. Af- ter saving the schematic, we simulate it by selecting Analysis/Simulate. The output file includes the source frequency in addition to the attributes checked for the pseudocomponents IPRINT and VPRINT1, FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592E+02 3.264E-03 -3.743E+01 FREQ VM(3) VP(3) 1.592E+02 1.550E+00 -9.518E+01 From this output file, we obtain Vo = 1.55 − 95.18◦ V, Io = 3.264 − 37.43◦ mA which are the phasors for vo = 1.55 cos(1000t − 95.18◦ ) = 1.55 sin(1000t − 5.18◦ ) V and io = 3.264 cos(1000t − 37.43◦ ) mA P R A C T I C E P R O B L E M 1 0 . 1 3 Use PSpice to obtain vo and io in the circuit of Fig. 10.37. 1 mF 2 H 2 kΩ 3 kΩ 1 kΩ io 2vo + − 10 cos 3000t A vo + − + − Figure10.37 For Practice Prob. 10.13. Answer: 0.2682 cos(3000t−154.6◦ ) V, 0.544 cos(3000t−55.12◦ ) mA. E X A M P L E 1 0 . 1 4 Find V1 and V2 in the circuit of Fig. 10.38. Solution: The circuit in Fig. 10.35 is in the time domain, whereas the one in Fig. 10.38 is in the frequency domain. Since we are not given a particular
  • 419. CHAPTER 10 Sinusoidal Steady-State Analysis 415 2 Ω 2 Ω V1 V2 –j1 Ω –j2 j2 Ω –j1 Ω 1 Ω 3 0° A 18 30° V + − j2 Ω 0.2Vx − + Vx Figure10.38 For Example 10.14. frequency and PSpice requires one, we select any frequency consistent with the given impedances. For example, if we select ω = 1 rad/s, the corresponding frequency is f = ω/2π = 0.159155 Hz. We obtain the values of the capacitance (C = 1/ωXC) and inductances (L = XL/ω). Making these changes results in the schematic in Fig. 10.39. To ease wiring, we have exchanged the positions of the voltage-controlled cur- rent source G1 and the 2 + j2 impedance. Notice that the current of G1 flows from node 1 to node 3, while the controlling voltage is across the capacitor c2, as required in Fig. 10.38. The attributes of pseudocom- ponents VPRINT1 are set as shown. As a single-frequency analysis, we select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq = 0.159155, and Final Freq = 0.159155. After saving the schematic, we select Analysis/Simulate to simulate the circuit. When this is done, the output file includes FREQ VM(1) VP(1) 1.592E-01 2.708E+00 -5.673E+01 FREQ VM(3) VP(3) 1.592E-01 4.468E+00 -1.026E+02 AC=3 AC R1 I1 R2 L1 L2 R3 C1 C3 G1 V1 C2 1 2 3 4 5 0.5 2 2H 2H 2 1 1 1 AC=ok MAG=ok PHASE=ok AC=ok MAG=ok PHASE=yes ACMAG=18 ACPHASE=30 0 GAIN=0.2 + − + − − − Figure10.39 Schematic for the circuit in Fig. 10.38.
  • 420. 416 PART 2 AC Circuits from which we obtain V1 = 2.708 − 56.73◦ V, V2 = 4.468 − 102.6◦ V P R A C T I C E P R O B L E M 1 0 . 1 4 Obtain Vx and Ix in the circuit depicted in Fig. 10.40. 1 Ω 1 Ω Vx Ix 4Ix –j1 Ω j2 Ω 2 Ω 4 60° A 12 0° V j2 Ω –j0.25 + − + − Figure10.40 For Practice Prob. 10.14. Answer: 13.02 − 76.08◦ V, 8.234 − 4.516◦ A. †10.9 APPLICATIONS The concepts learned in this chapter will be applied in later chapters to calculate electric power and determine frequency response. The concepts are also used in analyzing coupled circuits, three-phase circuits, ac tran- sistor circuits, filters, oscillators, and other ac circuits. In this section, we apply the concepts to develop two practical ac circuits: the capacitance multiplier and the sine wave oscillators. 10.9.1 Capacitance Multiplier The op amp circuit in Fig. 10.41 is known as a capacitance multiplier, for reasons that will become obvious. Such a circuit is used in integrated- circuit technology to produce a multiple of a small physical capacitance C when a large capacitance is needed. The circuit in Fig. 10.41 can be used to multiply capacitance values by a factor up to 1000. For example, a 10-pF capacitor can be made to behave like a 100-nF capacitor. In Fig. 10.41, the first op amp operates as a voltage follower, while the second one is an inverting amplifier. The voltage follower isolates the capacitance formed by the circuit from the loading imposed by the inverting amplifier. Since no current enters the input terminals of the op amp, the input current Ii flows through the feedback capacitor. Hence, at node 1, Ii = Vi − Vo 1/jωC = jωC(Vi − Vo) (10.3)
  • 421. CHAPTER 10 Sinusoidal Steady-State Analysis 417 + − R1 A1 R2 + − Vo Ii Zi Vi A2 C 0 V 2 1 + − Vi Figure10.41 Capacitance multiplier. Applying KCL at node 2 gives Vi − 0 R1 = 0 − Vo R2 or Vo = − R2 R1 Vi (10.4) Substituting Eq. (10.4) into (10.3) gives Ii = jωC 1 + R2 R1 Vi or Ii Vi = jω 1 + R2 R1 C (10.5) The input impedance is Zi = Vi Ii = 1 jωCeq (10.6) where Ceq = 1 + R2 R1 C (10.7) Thus, by a proper selection of the values of R1 and R2, the op amp circuit in Fig. 10.41 can be made to produce an effective capacitance between the input terminal and ground, which is a multiple of the physical capacitance C. The size of the effective capacitance is practically limited by the inverted output voltage limitation. Thus, the larger the capacitance multiplication, the smaller is the allowable input voltage to prevent the op amps from reaching saturation. A similar op amp circuit can be designed to simulate inductance. (See Prob. 10.69.) There is also an op amp circuit configuration to create a resistance multiplier. E X A M P L E 1 0 . 1 5 Calculate Ceq in Fig. 10.41 when R1 = 10 k, R2 = 1 M, and C = 1 nF.
  • 422. 418 PART 2 AC Circuits Solution: From Eq. (10.7) Ceq = 1 + R2 R1 C = 1 + 1 × 106 10 × 103 1 nF = 101 nF P R A C T I C E P R O B L E M 1 0 . 1 5 Determine the equivalent capacitance of the op amp circuit in Fig. 10.41 if R1 = 10 k, R2 = 10 M, and C = 10 nF. Answer: 10 µF. 10.9.2 Oscillators We know that dc is produced by batteries. But how do we produce ac? One way is using oscillators, which are circuits that convert dc to ac. An oscillator is a circuit that produces an ac waveform as output when powered by a dc input. The only external source an oscillator needs is the dc power supply. Ironically, the dc power supply is usually obtained by converting the ac supplied by the electric utility company to dc. Having gone through the trouble of conversion, one may wonder why we need to use the oscillator to convert the dc to ac again. The problem is that the ac supplied by the utility company operates at a preset frequency of 60 Hz in the United States (50 Hz in some other nations), whereas many applications such as electronic circuits, communication systems, and microwave devices require internally generated frequencies that range from 0 to 10 GHz or higher. Oscillators are used for generating these frequencies. This corresponds to ω = 2πf = 377 rad/s. In order for sine wave oscillators to sustain oscillations, they must meet the Barkhausen criteria: 1. The overall gain of the oscillator must be unity or greater. Therefore, losses must be compensated for by an amplifying device. 2. The overall phase shift (from input to output and back to the input) must be zero. Three common types of sine wave oscillators are phase-shift, twin T , and Wien-bridge oscillators. Here we consider only the Wien-bridge oscillator. + − Rf Rg R1 R2 C1 C2 + − v2 + − vo Positive feedback path to create oscillations Negative feedback path to control gain Figure10.42 Wien-bridge oscillator. The Wien-bridge oscillator is widely used for generating sinusoids in the frequency range below 1 MHz. It is an RC op amp circuit with only a few components, easily tunable and easy to design. As shown in Fig. 10.42, the oscillator essentially consists of a noninverting amplifier with two feedback paths: the positive feedback path to the noninverting input creates oscillations, while the negative feedback path to the inverting
  • 423. CHAPTER 10 Sinusoidal Steady-State Analysis 419 input controls the gain. If we define the impedances of the RC series and parallel combinations as Zs and Zp, then Zs = R1 + 1 jωC1 = R1 − j ωC1 (10.8) Zp = R2 1 jωC2 = R2 1 + jωR2C2 (10.9) The feedback ratio is V2 Vo = Zp Zs + Zp (10.10) Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives V2 Vo = R2 R2 + R1 − j ωC1 (1 + jωR2C2) = ωR2C1 ω(R2C1 + R1C1 + R2C2) + j(ω2R1C1R2C2 − 1) (10.11) To satisfy the second Barkhausen criterion, V2 must be in phase with Vo, which implies that the ratio in Eq. (10.11) must be purely real. Hence, the imaginary part must be zero. Setting the imaginary part equal to zero gives the oscillation frequency ωo as ω2 oR1C1R2C2 − 1 = 0 or ωo = 1 √ R1R2C1C2 (10.12) In most practical applications, R1 = R2 = R and C1 = C2 = C, so that ωo = 1 RC = 2πfo (10.13) or fo = 1 2πRC (10.14) Substituting Eq. (10.13) and R1 = R2 = R, C1 = C2 = C into Eq. (10.11) yields V2 Vo = 1 3 (10.15) Thus in order to satisfy the first Barkhausen criterion, the op amp must compensate by providing a gain of 3 or greater so that the overall gain is at least 1 or unity. We recall that for a noninverting amplifier, Vo V2 = 1 + Rf Rg = 3 (10.16) or Rf = 2Rg (10.17)
  • 424. 420 PART 2 AC Circuits Due to the inherent delay caused by the op amp, Wien-bridge oscil- lators are limited to operating in the frequency range of 1 MHz or less. E X A M P L E 1 0 . 1 6 Design a Wien-bridge circuit to oscillate at 100 kHz. Solution: Using Eq. (10.14), we obtain the time constant of the circuit as RC = 1 2πfo = 1 2π × 100 × 103 = 1.59 × 10−6 (10.16.1) If we select R = 10 k, then we can select C = 159 pF to satisfy Eq. (10.16.1). Since the gain must be 3, Rf /Rg = 2. We could select Rf = 20 k while Rg = 10 k. P R A C T I C E P R O B L E M 1 0 . 1 6 In the Wien-bridge oscillator circuit in Fig. 10.42, let R1 = R2 = 2.5 k, C1 = C2 = 1 nF. Determine the frequency fo of the oscillator. Answer: 63.66 kHz. 10.10 SUMMARY 1. We apply nodal and mesh analysis to ac circuits by applying KCL and KVL to the phasor form of the circuits. 2. In solving for the steady-state response of a circuit that has indepen- dent sources with different frequencies, each independent source must be considered separately. The most natural approach to analyz- ing such circuits is to apply the superposition theorem. A separate phasor circuit for each frequency must be solved independently, and the corresponding response should be obtained in the time domain. The overall response is the sum of the time-domain responses of all the individual phasor circuits. 3. The concept of source transformation is also applicable in the fre- quency domain. 4. The Thevenin equivalent of an ac circuit consists of a voltage source VTh in series with the Thevenin impedance ZTh. 5. The Norton equivalent of an ac circuit consists of a current source IN in parallel with the Norton impedance ZN (= ZTh). 6. PSpice is a simple and powerful tool for solving ac circuit problems. It relieves us of the tedious task of working with the complex num- bers involved in steady-state analysis. 7. The capacitance multiplier and the ac oscillator provide two typical applications for the concepts presented in this chapter. A capacitance multiplier is an op amp circuit used in producing a multiple of a physical capacitance. An oscillator is a device that uses a dc input to generate an ac output.
  • 425. CHAPTER 10 Sinusoidal Steady-State Analysis 421 REVIEW QUESTIONS 10.1 The voltage Vo across the capacitor in Fig. 10.43 is: (a) 5 0◦ V (b) 7.071 45◦ V (c) 7.071 − 45◦ V (d) 5 − 45◦ V 1 Ω + − Vo + − –j1 Ω 10 0° V Figure 10.43 For Review Question 10.1. 10.2 The value of the current Io in the circuit in Fig. 10.44 is: (a) 4 0◦ A (b) 2.4 − 90◦ A (c) 0.6 0◦ A (d) −1 A j8 Ω –j2 Ω 3 0° A Io Figure 10.44 For Review Question 10.2. 10.3 Using nodal analysis, the value of Vo in the circuit of Fig. 10.45 is: (a) −24 V (b) −8 V (c) 8 V (d) 24 V –j3 Ω j6 Ω 4 90° A V o Figure 10.45 For Review Question 10.3. 10.4 In the circuit of Fig. 10.46, current i(t) is: (a) 10 cos t A (b) 10 sin t A (c) 5 cos t A (d) 5 sin t A (e) 4.472 cos(t − 63.43◦ ) A 1 H 1 F + − 1 Ω 10 cos t V i(t) Figure 10.46 For Review Question 10.4. 10.5 Refer to the circuit in Fig. 10.47 and observe that the two sources do not have the same frequency. The current ix(t) can be obtained by: (a) source transformation (b) the superposition theorem (c) PSpice 1 F + − + − sin 2t V sin 10t V 1 H 1 Ω ix Figure 10.47 For Review Question 10.5. 10.6 For the circuit in Fig. 10.48, the Thevenin impedance at terminals a-b is: (a) 1 (b) 0.5 − j0.5 (c) 0.5 + j0.5 (d) 1 + j2 (e) 1 − j2 1 Ω 1 H + − 1 F a b 5 cos t V Figure 10.48 For Review Questions 10.6 and 10.7. 10.7 In the circuit of Fig. 10.48, the Thevenin voltage at terminals a-b is: (a) 3.535 − 45◦ V (b) 3.535 45◦ V (c) 7.071 − 45◦ V (d) 7.071 45◦ V 10.8 Refer to the circuit in Fig. 10.49. The Norton equivalent impedance at terminals a-b is: (a) −j4 (b) −j2 (c) j2 (d) j4 –j2 Ω j4 Ω + − a b 6 0° V Figure 10.49 For Review Questions 10.8 and 10.9.
  • 426. 422 PART 2 AC Circuits 10.9 The Norton current at terminals a-b in the circuit of Fig. 10.49 is: (a) 1 0◦ A (b) 1.5 − 90◦ A (c) 1.5 90◦ A (d) 3 90◦ A 10.10 PSpice can handle a circuit with two independent sources of different frequencies. (a) True (b) False Answers: 10.1c, 10.2a, 10.3d, 10.4a, 10.5b, 10.6c, 10.7a, 10.8a, 10.9d, 10.10b. PROBLEMS Section 10.2 Nodal Analysis 10.1 Find vo in the circuit in Fig. 10.50. 1 F + − + − 1 H 3 Ω vo 10 cos(t – 45°) V 5 sin(t + 30°) V + − Figure 10.50 For Prob. 10.1. 10.2 For the circuit depicted in Fig. 10.51 below, determine io. 10.3 Determine vo in the circuit of Fig. 10.52. + − 2 H 4 Ω vo 16 sin 4t V 2 cos 4t A + − 1 Ω 6 Ω F 1 12 Figure 10.52 For Prob. 10.3. 10.4 Compute vo(t) in the circuit of Fig. 10.53. + − 1 H 0.25 F 1 Ω 0.5ix vo + − 16 sin (4t – 10°) V ix Figure 10.53 For Prob. 10.4. 10.5 Use nodal analysis to find vo in the circuit of Fig. 10.54. + − 10 mH 50 mF 20 Ω 20 Ω 30 Ω 10 cos 103 t V io 4io vo + − Figure 10.54 For Prob. 10.5. 10.6 Using nodal analysis, find io(t) in the circuit in Fig. 10.55. 0.02 F + − 1 H 10 Ω 20 sin (10t – 4)V 4 cos (10t – 3)A io Figure10.51 For Prob. 10.2.
  • 427. CHAPTER 10 Sinusoidal Steady-State Analysis 423 0.5 F + − 1 H 2 H 2 Ω 0.25 F 8 sin (2t + 30°) V cos 2t A io Figure 10.55 For Prob. 10.6. 10.7 By nodal analysis, find io in the circuit in Fig. 10.56. 10 Ω 20 Ω 50 mF 10 mH 20 sin1000t A 2io io Figure 10.56 For Prob. 10.7. 10.8 Calculate the voltage at nodes 1 and 2 in the circuit of Fig. 10.57 using nodal analysis. 10 Ω 1 2 –j2 Ω –j5 Ω j2 Ω j4 Ω 20 30° A Figure 10.57 For Prob. 10.8. 10.9 Solve for the current I in the circuit of Fig. 10.58 using nodal analysis. 2 Ω 4 Ω –j2 Ω j1 Ω 2I 5 0° A 20 –90° V + − I Figure 10.58 For Prob. 10.9. 10.10 Using nodal analysis, find V1 and V2 in the circuit of Fig. 10.59. 20 Ω 10 Ω j2 A 1 + j A –j5 Ω j10 Ω V1 V2 Figure 10.59 For Prob. 10.10. 10.11 By nodal analysis, obtain current Io in the circuit in Fig. 10.60. 3 Ω 2 Ω 1 Ω j4 Ω –j2 Ω + − 100 20° V Io Figure 10.60 For Prob. 10.11. 10.12 Use nodal analysis to obtain Vo in the circuit of Fig. 10.61 below. 8 Ω 2 Ω –j1 Ω –j2 Ω j6 Ω 4 Ω j5 Ω 2Vx Vo 4 45° A + − Vx + − Figure10.61 For Prob. 10.12.
  • 428. 424 PART 2 AC Circuits 10.13 Obtain Vo in Fig. 10.62 using nodal analysis. 4 Ω 2 Ω –j4 Ω j2 Ω Vo 0.2Vo + − + − 12 0° V Figure 10.62 For Prob. 10.13. 10.14 Refer to Fig. 10.63. If vs(t) = Vm sin ωt and vo(t) = A sin(ωt + φ), derive the expressions for A and φ. + − vo(t) vs(t) + − L R C Figure 10.63 For Prob. 10.14. 10.15 For each of the circuits in Fig. 10.64, find Vo/Vi for ω = 0, ω → ∞, and ω2 = 1/LC. Vo + − Vo + − Vi + − Vi + − C R R C L L (b) (a) Figure 10.64 For Prob. 10.15. 10.16 For the circuit in Fig. 10.65, determine Vo/Vs. + − Vs Vo + − L R1 R2 C Figure 10.65 For Prob. 10.16. Section 10.3 Mesh Analysis 10.17 Obtain the mesh currents I1 and I2 in the circuit of Fig. 10.66. + − Vs L R C2 C1 I2 I1 Figure 10.66 For Prob. 10.17. 10.18 Solve for io in Fig. 10.67 using mesh analysis. + − + − 2 H 0.25 F 4 Ω 10 cos 2t V 6 sin 2t V io Figure 10.67 For Prob. 10.18. 10.19 Rework Prob. 10.5 using mesh analysis. 10.20 Using mesh analysis, find I1 and I2 in the circuit of Fig. 10.68. + − + − I2 I1 j10 Ω –j20 Ω 40 Ω 50 0° V 40 30° V Figure 10.68 For Prob. 10.20. 10.21 By using mesh analysis, find I1 and I2 in the circuit depicted in Fig. 10.69. I2 I1 j4 Ω j2 Ω j1 Ω –j6 Ω 3 Ω 2 Ω 30 20° V 3 Ω + − Figure 10.69 For Prob. 10.21.
  • 429. CHAPTER 10 Sinusoidal Steady-State Analysis 425 10.22 Repeat Prob. 10.11 using mesh analysis. 10.23 Use mesh analysis to determine current Io in the circuit of Fig. 10.70 below. 10.24 Determine Vo and Io in the circuit of Fig. 10.71 using mesh analysis. j4 Ω Io 3Vo –j2 Ω 4 –30° A 2 Ω + − Vo + − Figure 10.71 For Prob. 10.24. 10.25 Compute I in Prob. 10.9 using mesh analysis. 10.26 Use mesh analysis to find Io in Fig. 10.28 (for Example 10.10). 10.27 Calculate Io in Fig. 10.30 (for Practice Prob. 10.10) using mesh analysis. 10.28 Compute Vo in the circuit of Fig. 10.72 using mesh analysis. –j3 Ω 2 Ω j4 Ω + − 2 Ω 2 Ω 12 0° V 2 0° A 4 90° A V o + − Figure 10.72 For Prob. 10.28. 10.29 Using mesh analysis, obtain Io in the circuit shown in Fig. 10.73. –j4 Ω j2 Ω 2 Ω 1 Ω 1 Ω Io + − 10 90° V 4 0° A 2 0° A Figure 10.73 For Prob. 10.29. Section 10.4 Superposition Theorem 10.30 Find io in the circuit shown in Fig. 10.74 using superposition. 4 Ω + − + − 2 Ω 8 V 1 H 10 cos 4t V io Figure 10.74 For Prob. 10.30. 10.31 Using the superposition principle, find ix in the circuit of Fig. 10.75. + − 3 Ω 4 H 10 cos(2t – 60°) V 5 cos(2t + 10°) A ix F 1 8 Figure 10.75 For Prob. 10.31. 10.32 Rework Prob. 10.2 using the superposition theorem. 10.33 Solve for vo(t) in the circuit of Fig. 10.76 using the superposition principle. –j40 Ω –j40 Ω j60 Ω 80 Ω 20 Ω Io + − + − 100 120° V 60 –30° V Figure10.70 For Prob. 10.23.
  • 430. 426 PART 2 AC Circuits + − + − 6 Ω 2 H 10 V 12 cos 3t V 4 sin 2t A + − vo F 1 12 Figure 10.76 For Prob. 10.33. 10.34 Determine io in the circuit of Fig. 10.77. + − 1 Ω 2 H 24 V 2 cos 3t 2 Ω 4 Ω + − io 10 sin(3t – 30°) V F 1 6 Figure 10.77 For Prob. 10.34. 10.35 Find io in the circuit in Fig. 10.78 using superposition. 80 Ω 60 Ω 40 mH 20 mF 24 V 100 Ω + − + − 50 cos 2000t V 2 sin 4000t A io Figure 10.78 For Prob. 10.35. Section 10.5 Source Transformation 10.36 Using source transformation, find i in the circuit of Fig. 10.79. 3 Ω 5 Ω 5 mH 1 mF 8 sin(200t + 30°) A i Figure 10.79 For Prob. 10.36. 10.37 Use source transformation to find vo in the circuit in Fig. 10.80. 20 Ω 80 Ω 0.4 mH 0.2 mF + − 5 cos 105 t V vo + − Figure 10.80 For Prob. 10.37. 10.38 Solve Prob. 10.20 using source transformation. 10.39 Use the method of source transformation to find Ix in the circuit of Fig. 10.81. + − 2 Ω j4 Ω –j2 Ω –j3 Ω 6 Ω 4 Ω Ix 60 0° V 5 90° A Figure 10.81 For Prob. 10.39. 10.40 Use the concept of source transformation to find Vo in the circuit of Fig. 10.82. + − 4 Ω j4 Ω –j3 Ω –j2 Ω j2 Ω 2 Ω 20 0° V Vo + − Figure 10.82 For Prob. 10.40. Section 10.6 Thevenin and Norton Equivalent Circuits 10.41 Find the Thevenin and Norton equivalent circuits at terminals a-b for each of the circuits in Fig. 10.83.
  • 431. CHAPTER 10 Sinusoidal Steady-State Analysis 427 –j10 Ω j20 Ω 10 Ω a b 50 30° V + − a b 4 0° A –j5 Ω j10 Ω 8 Ω (b) (a) Figure 10.83 For Prob. 10.41. 10.42 For each of the circuits in Fig. 10.84, obtain Thevenin and Norton equivalent circuits at terminals a-b. –j5 Ω j4 Ω 6 Ω 30 Ω a b 2 0° A a b 120 45° V –j2 Ω j10 Ω 60 Ω (b) (a) + − Figure 10.84 For Prob. 10.42. 10.43 Find the Thevenin and Norton equivalent circuits for the circuit shown in Fig. 10.85. j20 Ω 5 Ω 2 Ω 60 120° V + − –j10 Ω Figure 10.85 For Prob. 10.43. 10.44 For the circuit depicted in Fig. 10.86, find the Thevenin equivalent circuit at terminals a-b. a b 5 45° A j10 Ω 8 Ω –j6 Ω Figure 10.86 For Prob. 10.44. 10.45 Repeat Prob. 10.1 using Thevenin’s theorem. 10.46 Find the Thevenin equivalent of the circuit in Fig. 10.87 as seen from: (a) terminals a-b (b) terminals c-d 10 Ω a b 4 0° A 20 0° V –j4 Ω j5 Ω 4 Ω + − c d Figure 10.87 For Prob. 10.46. 10.47 Solve Prob. 10.3 using Thevenin’s theorem. 10.48 Using Thevenin’s theorem, find vo in the circuit in Fig. 10.88. 2 H 4 Ω 2 Ω vo io 3io + − 12 cos t V + − F 1 4 F 1 8 Figure 10.88 For Prob. 10.48. 10.49 Obtain the Norton equivalent of the circuit depicted in Fig. 10.89 at terminals a-b. a b 5 mF 10 H 2 kΩ 4 cos(200t + 30°) V Figure 10.89 For Prob. 10.49.
  • 432. 428 PART 2 AC Circuits 10.50 For the circuit shown in Fig. 10.90, find the Norton equivalent circuit at terminals a-b. 60 Ω 40 Ω –j30 Ω j80 Ω a b 3 60° A Figure 10.90 For Prob. 10.50. 10.51 Compute io in Fig. 10.91 using Norton’s theorem. 2 Ω 4 H 5 cos 2t V + − io F 1 4 F 1 2 Figure 10.91 For Prob. 10.51. 10.52 At terminals a-b, obtain Thevenin and Norton equivalent circuits for the network depicted in Fig. 10.92. Take ω = 10 rad/s. a b 10 mF 10 Ω 2 sin vt V 12 cos vt + − vo 2vo + − H 1 2 Figure 10.92 For Prob. 10.52. Section 10.7 Op Amp AC Circuits 10.53 For the differentiator shown in Fig. 10.93, obtain Vo/Vs. Find vo(t) when vs(t) = Vm sin ωt and ω = 1/RC. + − vs vo R C + − + − Figure 10.93 For Prob. 10.53. 10.54 The circuit in Fig. 10.94 is an integrator with a feedback resistor. Calculate vo(t) if vs = 2 cos 4 × 104 t V. + − vs vo + − 10 nF 100 kΩ 50 kΩ + − Figure 10.94 For Prob. 10.54. 10.55 Compute io(t) in the op amp circuit in Fig. 10.95 if vs = 4 cos 104 t V. + − vs 50 kΩ 1 nF 100 kΩ io + − Figure 10.95 For Prob. 10.55. 10.56 If the input impedance is defined as Zin = Vs/Is, find the input impedance of the op amp circuit in Fig. 10.96 when R1 = 10 k, R2 = 20 k, C1 = 10 nF, C2 = 20 nF, and ω = 5000 rad/s. Vs C2 C1 R1 R2 Is Zin Vo + − + − Figure 10.96 For Prob. 10.56. 10.57 Evaluate the voltage gain Av = Vo/Vs in the op amp circuit of Fig. 10.97. Find Av at ω = 0, ω → ∞, ω = 1/R1C1, and ω = 1/R2C2. + − Vs Vo + − C1 R1 C2 R2 + − Figure 10.97 For Prob. 10.57.
  • 433. CHAPTER 10 Sinusoidal Steady-State Analysis 429 10.58 In the op amp circuit of Fig. 10.98, find the closed-loop gain and phase shift if C1 = C2 = 1 nF, R1 = R2 = 100 k, R3 = 20 k, R4 = 40 k, and ω = 2000 rad/s. vs vo C1 R1 R2 + − C2 R4 R3 + − + − Figure 10.98 For Prob. 10.58. 10.59 Compute the closed-loop gain Vo/Vs for the op amp circuit of Fig. 10.99. + − + − vs vo + − C1 R1 R3 C2 R2 Figure 10.99 For Prob. 10.59. 10.60 Determine vo(t) in the op amp circuit in Fig. 10.100 below. 10.61 For the op amp circuit in Fig. 10.101, obtain vo(t). vo + − 10 kΩ 20 kΩ 40 kΩ 0.1 mF 0.2 mF + − + − + − 5 cos 103 t V Figure 10.101 For Prob. 10.61. 10.62 Obtain vo(t) for the op amp circuit in Fig. 10.102 if vs = 4 cos(1000t − 60◦ ) V. vo vs + − 10 kΩ 50 kΩ 20 kΩ 0.2 mF 0.1 mF + − + − + − Figure 10.102 For Prob. 10.62. Section 10.8 AC Analysis Using PSpice 10.63 Use PSpice to solve Example 10.10. 10.64 Solve Prob. 10.13 using PSpice. vo + − 10 kΩ 20 kΩ 20 kΩ 40 kΩ 10 kΩ 0.25 mF 0.5 mF + − 2 sin 400t V Figure10.100 For Prob. 10.60.
  • 434. 430 PART 2 AC Circuits 10.65 Obtain Vo in the circuit of Fig. 10.103 using PSpice. 1 Ω j4 Ω –j2 Ω 2 Ω + − Vx 2Vx + − Vo 3 0° A Figure 10.103 For Prob. 10.65. 10.66 Use PSpice to find V1, V2, and V3 in the network of Fig. 10.104. + − 8 Ω j10 Ω j10 Ω –j4 Ω –j4 Ω V1 V3 V2 60 30° V 4 0° A Figure 10.104 For Prob. 10.66. 10.67 Determine V1, V2, and V3 in the circuit of Fig. 10.105 using PSpice. 8 Ω j10 Ω 1 Ω 2 Ω j6 Ω –j2 Ω –j4 Ω V1 V3 V2 4 0° A 2 0° A Figure 10.105 For Prob. 10.67. 10.68 Use PSpice to find vo and io in the circuit of Fig. 10.106 below. Section 10.9 Applications 10.69 The op amp circuit in Fig. 10.107 is called an inductance simulator. Show that the input impedance is given by Zin = Vin Iin = jωLeq where Leq = R1R3R4 R2 C Vin Iin + − + − R1 R2 R3 C R4 + − Figure 10.107 For Prob. 10.69. 10.70 Figure 10.108 shows a Wien-bridge network. Show that the frequency at which the phase shift between the input and output signals is zero is f = 1 2 πRC, and that the necessary gain is Av = Vo/Vi = 3 at that frequency. Vi + − R R1 R2 R C C + − Vo Figure 10.108 For Prob. 10.70. 10.71 Consider the oscillator in Fig. 10.109. (a) Determine the oscillation frequency. 20 mF 25 mF 2 H 4 Ω 10 Ω vo 0.5vo io 4io + − 6 cos 4t V + − + − Figure10.106 For Prob. 10.68.
  • 435. CHAPTER 10 Sinusoidal Steady-State Analysis 431 (b) Obtain the minimum value of R for which oscillation takes place. + − R 10 kΩ 20 kΩ 80 kΩ 0.4 mH 2 nF Figure 10.109 For Prob. 10.71. 10.72 The oscillator circuit in Fig. 10.110 uses an ideal op amp. (a) Calculate the minimum value of Ro that will cause oscillation to occur. (b) Find the frequency of oscillation. + − 10 kΩ 100 kΩ 1 MΩ 10 mH 2 nF Ro Figure 10.110 For Prob. 10.72. 10.73 Figure 10.111 shows a Colpitts oscillator. Show that the oscillation frequency is fo = 1 2π √ LCT where CT = C1C2/(C1 + C2). Assume Ri XC2 . + − Rf Ri C2 C1 L V o Figure 10.111 A Colpitts oscillator; for Prob. 10.73. (Hint: Set the imaginary part of the impedance in the feedback circuit equal to zero.) 10.74 Design a Colpitts oscillator that will operate at 50 kHz. 10.75 Figure 10.112 shows a Hartley oscillator. Show that the frequency of oscillation is fo = 1 2π √ C(L1 + L2) + − Rf Ri L2 L1 C Vo Figure 10.112 A Hartley oscillator; for Prob. 10.75. 10.76 Refer to the oscillator in Fig. 10.113. (a) Show that V2 Vo = 1 3 + j(ωL/R − R/ωL) (b) Determine the oscillation frequency fo. (c) Obtain the relationship between R1 and R2 in order for oscillation to occur. + − R L R L R1 R2 V o V2 Figure 10.113 For Prob. 10.76.
  • 436. 433 C H A P T E R AC POWER ANALYSIS 1 1 An engineer is an unordinary person who can do for one dollar what any ordinary person can do for two dollars. —Anonymous Enhancing Your Career Career in Power Systems The discovery of the principle of an ac generator by Michael Faraday in 1831 was a major breakthrough in engineering; it provided a convenient way of generating the electric power that is needed in every elec- tronic, electrical, or electromechanical device we use now. Electric power is obtained by converting energy from sources such as fossil fuels (gas, oil, and coal), nuclear fuel (uranium), hydro energy (water falling through a head), geothermal energy (hot water, steam), wind energy, tidal en- ergy, and biomass energy (wastes). These various ways of generating electric power are studied in detail in the field of power engineering, which has become an indispensable sub- discipline of electrical engineering. An electrical engineer should be familiar with the analysis, generation, transmis- sion, distribution, and cost of electric power. The electric power industry is a very large employer of electrical engineers. The industry includes thousands of electric utility systems ranging from large, interconnected systems serving large regional areas to small power companies serving individual communities or factories. Due to the complexity of the power industry, there are numerous electrical engineering jobs in different areas of the industry: power plant (generation), transmission and distribution, maintenance, research, data acquisition and flow control, and management. Since electric power is used everywhere, electric utility companies are everywhere, of- fering exciting training and steady employment for men and women in thousands of communities throughout the world. A pole-type transformer with a low-voltage, three-wire distribution system. Source: W. N. Alerich, Electricity, 3rd ed. Albany, NY: Delmar Publishers, 1981, p. 152. (Courtesy of General Electric.)
  • 437. 434 PART 2 AC Circuits 11.1 INTRODUCTION Our effort in ac circuit analysis so far has been focused mainly on cal- culating voltage and current. Our major concern in this chapter is power analysis. Power analysis is of paramount importance. Power is the most important quantity in electric utilities, electronic, and communication systems, because such systems involve transmission of power from one point to another. Also, every industrial and household electrical device— every fan, motor, lamp, pressing iron, TV, personal computer—has a power rating that indicates how much power the equipment requires; exceeding the power rating can do permanent damage to an appliance. The most common form of electric power is 50- or 60-Hz ac power. The choice of ac over dc allowed high-voltage power transmission from the power generating plant to the consumer. We will begin by defining and deriving instantaneous power and average power. We will then introduce other power concepts. As practi- cal applications of these concepts, we will discuss how power is measured and reconsider how electric utility companies charge their customers. 11.2 INSTANTANEOUS AND AVERAGE POWER As mentioned in Chapter 2, the instantaneous power p(t) absorbed by an element is the product of the instantaneous voltage v(t) across the element and the instantaneous current i(t) through it. Assuming the passive sign convention, p(t) = v(t)i(t) (11.1) The instantaneous power is the power at any instant of time. It is the rate at which an element absorbs energy. We can also think of the instantaneous power as the power absorbed by the element at a spe- cificinstantoftime. Instantaneousquantitiesare denoted by lowercase letters. Sinusoidal source Passive linear network i(t) + − v(t) Figure 11.1 Sinusoidal source and passive linear circuit. Consider the general case of instantaneous power absorbed by an arbitrary combination of circuit elements under sinusoidal excitation, as shown in Fig. 11.1. Let the voltage and current at the terminals of the circuit be v(t) = Vm cos(ωt + θv) (11.2a) i(t) = Im cos(ωt + θi) (11.2b) where Vm and Im are the amplitudes (or peak values), and θv and θi are the phase angles of the voltage and current, respectively. The instantaneous power absorbed by the circuit is p(t) = v(t)i(t) = VmIm cos(ωt + θv) cos(ωt + θi) (11.3) We apply the trigonometric identity cos A cos B = 1 2 [cos(A − B) + cos(A + B)] (11.4) and express Eq. (11.3) as p(t) = 1 2 VmIm cos(θv − θi) + 1 2 VmIm cos(2ωt + θv + θi) (11.5)
  • 438. CHAPTER 11 AC Power Analysis 435 This shows us that the instantaneous power has two parts. The first part is constant or time independent. Its value depends on the phase difference between the voltage and the current. The second part is a sinusoidal function whose frequency is 2ω, which is twice the angular frequency of the voltage or current. A sketch of p(t) in Eq. (11.5) is shown in Fig. 11.2, where T = 2π/ω is the period of voltage or current. We observe that p(t) is periodic, p(t) = p(t + T0), and has a period of T0 = T/2, since its frequency is twice that of voltage or current. We also observe that p(t) is positive for some part of each cycle and negative for the rest of the cycle. When p(t) is positive, power is absorbed by the circuit. When p(t) is negative, power is absorbed by the source; that is, power is transferred from the circuit to the source. This is possible because of the storage elements (capacitors and inductors) in the circuit. 0 VmIm cos(uv − ui) VmIm p(t) T 2 T t 1 2 1 2 Figure11.2 The instantaneous power p(t) entering a circuit. Theinstantaneouspowerchangeswithtimeandisthereforedifficult to measure. The average power is more convenient to measure. In fact, the wattmeter, the instrument for measuring power, responds to average power. The average power is the average of the instantaneous power over one period. Thus, the average power is given by P = 1 T T 0 p(t) dt (11.6) Although Eq. (11.6) shows the averaging done over T , we would get the same result if we performed the integration over the actual period of p(t) which is T0 = T/2. Substituting p(t) in Eq. (11.5) into Eq. (11.6) gives P = 1 T T 0 1 2 VmIm cos(θv − θi) dt + 1 T T 0 1 2 VmIm cos(2ωt + θv + θi) dt
  • 439. 436 PART 2 AC Circuits = 1 2 VmIm cos(θv − θi) 1 T T 0 dt + 1 2 VmIm 1 T T 0 cos(2ωt + θv + θi) dt (11.7) The first integrand is constant, and the average of a constant is the same constant. The second integrand is a sinusoid. We know that the average of asinusoidoveritsperiodiszerobecausetheareaunderthesinusoidduring a positive half-cycle is canceled by the area under it during the following negative half-cycle. Thus, the second term in Eq. (11.7) vanishes and the average power becomes P = 1 2 VmIm cos(θv − θi) (11.8) Since cos(θv − θi) = cos(θi − θv), what is important is the difference in the phases of the voltage and current. Note that p(t) is time-varying while P does not depend on time. To find the instantaneous power, we must necessarily have v(t) and i(t) in the time domain. But we can find the average power when voltage and current are expressed in the time domain, as in Eq. (11.2), or when they are expressed in the frequency domain. The phasor forms of v(t) and i(t) in Eq. (11.2) are V = Vm θv and I = Im θi, respectively. P is calculated using Eq. (11.8) or using phasors V and I. To use phasors, we notice that 1 2 VI∗ = 1 2 VmIm θv − θi = 1 2 VmIm [cos(θv − θi) + j sin(θv − θi)] (11.9) We recognize the real part of this expression as the average power P according to Eq. (11.8). Thus, P = 1 2 Re VI∗ = 1 2 VmIm cos(θv − θi) (11.10) Consider two special cases of Eq. (11.10). When θv = θi, the voltage and current are in phase. This implies a purely resistive circuit or resistive load R, and P = 1 2 VmIm = 1 2 I2 mR = 1 2 |I|2 R (11.11) where |I|2 = I × I∗ . Equation (11.11) shows that a purely resistive circuit absorbs power at all times. When θv − θi = ±90◦ , we have a purely reactive circuit, and P = 1 2 VmIm cos 90◦ = 0 (11.12)
  • 440. CHAPTER 11 AC Power Analysis 437 showing that a purely reactive circuit absorbs no average power. In sum- mary, A resistive load (R) absorbs power at all times, while a reactive load (L or C) absorbs zero average power. E X A M P L E 1 1 . 1 Given that v(t) = 120 cos(377t +45◦ ) V and i(t) = 10 cos(377t −10◦ ) A find the instantaneous power and the average power absorbed by the passive linear network of Fig. 11.1. Solution: The instantaneous power is given by p = vi = 1200 cos(377t + 45◦ ) cos(377t − 10◦ ) Applying the trigonometric identity cos A cos B = 1 2 [cos(A + B) + cos(A − B)] gives p = 600[cos(754t + 35◦ ) + cos 55◦ ] or p(t) = 344.2 + 600 cos(754t + 35◦ ) W The average power is P = 1 2 VmIm cos(θv − θi) = 1 2 120(10) cos[45◦ − (−10◦ )] = 600 cos 55◦ = 344.2 W which is the constant part of p(t) above. P R A C T I C E P R O B L E M 1 1 . 1 Calculate the instantaneous power and average power absorbed by the passive linear network of Fig. 11.1 if v(t) = 80 cos(10t + 20◦ ) V and i(t) = 15 sin(10t + 60◦ ) A Answer: 385.7 + 600 cos(20t − 10◦ ) W, 385.7 W. E X A M P L E 1 1 . 2 Calculate the average power absorbed by an impedance Z = 30 − j70 when a voltage V = 120 0◦ is applied across it.
  • 441. 438 PART 2 AC Circuits Solution: The current through the impedance is I = V Z = 120 0◦ 30 − j70 = 120 0◦ 76.16 − 66.8◦ = 1.576 66.8◦ A The average power is P = 1 2 VmIm cos(θv − θi) = 1 2 (120)(1.576) cos(0 − 66.8◦ ) = 37.24 W P R A C T I C E P R O B L E M 1 1 . 2 A current I = 10 30◦ flows through an impedance Z = 20 − 22◦ . Find the average power delivered to the impedance. Answer: 927.2 W. E X A M P L E 1 1 . 3 For the circuit shown in Fig. 11.3, find the average power supplied by the source and the average power absorbed by the resistor. 4 Ω + − I −j2 Ω 5 30° V Figure11.3 For Example 11.3. Solution: The current I is given by I = 5 30◦ 4 − j2 = 5 30◦ 4.472 − 26.57◦ = 1.118 56.57◦ A The average power supplied by the voltage source is P = 1 2 (5)(1.118) cos(30◦ − 56.57◦ ) = 2.5 W The current through the resistor is I = IR = 1.118 56.57◦ A and the voltage across it is VR = 4IR = 4.472 56.57◦ V The average power absorbed by the resistor is P = 1 2 (4.472)(1.118) = 2.5 W which is the same as the average power supplied. Zero average power is absorbed by the capacitor. P R A C T I C E P R O B L E M 1 1 . 3 In the circuit of Fig. 11.4, calculate the average power absorbed by the resistor and inductor. Find the average power supplied by the voltage source. 3 Ω + − j1 Ω 8 45° V Figure11.4 For Practice Prob. 11.3. Answer: 9.6 W, 0 W, 9.6 W.
  • 442. CHAPTER 11 AC Power Analysis 439 E X A M P L E 1 1 . 4 Determine the power generated by each source and the average power ab- sorbed by each passive element in the circuit of Fig. 11.5(a). 20 Ω + − j10 Ω −j5 Ω 4 0° Α 60 30° V 1 3 5 4 2 (a) 20 Ω + − j10 Ω −j5 Ω 4 0° Α 60 30° V (b) + − + − V2 V1 I1 I2 Figure11.5 For Example 11.4. Solution: We apply mesh analysis as shown in Fig. 11.5(b). For mesh 1, I1 = 4 A For mesh 2, (j10 − j5)I2 − j10I1 + 60 30◦ = 0, I1 = 4 A or j5I2 = −60 30◦ + j40 ⇒ I2 = −12 − 60◦ + 8 = 10.58 79.1◦ A For the voltage source, the current flowing from it is I2 = 10.58 79.1◦ A and the voltage across it is 60 30◦ V, so that the average power is P5 = 1 2 (60)(10.58) cos(30◦ − 79.1◦ ) = 207.8 W Following the passive sign convention (see Fig. 1.8), this average power is absorbed by the source, in view of the direction of I2 and the polarity of the voltage source. That is, the circuit is delivering average power to the voltage source. For the current source, the current through it is I1 = 4 0◦ and the voltage across it is V1 = 20I1 + j10(I1 − I2) = 80 + j10(4 − 2 − j10.39) = 183.9 + j20 = 184.984 6.21◦ V The average power supplied by the current source is P1 = − 1 2 (184.984)(4) cos(6.21◦ − 0) = −367.8 W It is negative according to the passive sign convention, meaning that the current source is supplying power to the circuit. For the resistor, the current through it is I1 = 4 0◦ and the voltage across it is 20I1 = 80 0◦ , so that the power absorbed by the resistor is P2 = 1 2 (80)(4) = 160 W
  • 443. 440 PART 2 AC Circuits For the capacitor, the current through it is I2 = 10.58 79.1◦ and the voltage across it is −j5I2 = (5 − 90◦ )(10.58 79.1◦ ) = 52.9 79.1◦ − 90◦ . The average power absorbed by the capacitor is P4 = 1 2 (52.9)(10.58) cos(−90◦ ) = 0 For the inductor, the current through it is I1 − I2 = 2 − j10.39 = 10.58 − 79.1◦ . The voltage across it is j10(I1 − I2) = 105.8 − 79.1◦ + 90◦ . Hence, the average power absorbed by the in- ductor is P3 = 1 2 (105.8)(10.58) cos 90◦ = 0 Notice that the inductor and the capacitor absorb zero average power and that the total power supplied by the current source equals the power absorbed by the resistor and the voltage source, or P1 + P2 + P3 + P4 + P5 = −367.8 + 160 + 0 + 0 + 207.8 = 0 indicating that power is conserved. P R A C T I C E P R O B L E M 1 1 . 4 Calculate the average power absorbed by each of the five elements in the circuit of Fig. 11.6. 8 Ω + − + − −j2 Ω j4 Ω 40 0° V 20 90° V Figure11.6 For Practice Prob. 11.4. Answer: 40-V Voltage source: −100 W; resistor: 100 W; others: 0 W. 11.3 MAXIMUM AVERAGE POWER TRANSFER In Section 4.8 we solved the problem of maximizing the power deliv- ered by a power-supplying resistive network to a load RL. Represent- ing the circuit by its Thevenin equivalent, we proved that the maximum power would be delivered to the load if the load resistance is equal to the Thevenin resistance RL = RTh. We now extend that result to ac circuits. Consider the circuit in Fig. 11.7, where an ac circuit is connected to a load ZL and is represented by its Thevenin equivalent. The load is usually represented by an impedance, which may model an electric motor, an antenna, a TV, and so forth. In rectangular form, the Thevenin impedance ZTh and the load impedance ZL are ZTh = RTh + jXTh (11.13a) ZL = RL + jXL (11.13b)
  • 444. CHAPTER 11 AC Power Analysis 441 The current through the load is I = VTh ZTh + ZL = VTh (RTh + jXTh) + (RL + jXL) (11.14) From Eq. (11.11), the average power delivered to the load is P = 1 2 |I|2 RL = |VTh|2 RL/2 (RTh + RL)2 + (XTh + XL)2 (11.15) Our objective is to adjust the load parameters RL and XL so that P is maximum. To do this we set ∂P/∂RL and ∂P/∂XL equal to zero. From Eq. (11.15), we obtain ∂P ∂XL = − |VTh|2 RL(XTh + XL) [(RTh + RL)2 + (XTh + XL)2]2 (11.16a) ∂P ∂RL = |VTh|2 [(RTh + RL)2 + (XTh + XL)2 − 2RL(RTh + RL)] 2[(RTh + RL)2 + (XTh + XL)2]2 (11.16b) Setting ∂P/∂XL to zero gives XL = −XTh (11.17) and setting ∂P/∂RL to zero results in RL = R2 Th + (XTh + XL)2 (11.18) Combining Eqs. (11.17) and (11.18) leads to the conclusion that for max- imum average power transfer, ZL must be selected so that XL = −XTh and RL = RTh, i.e., ZL = RL + jXL = RTh − jXTh = Z∗ Th (11.19) I ZL (a) VTh ZTh (b) ZL + − Linear circuit Figure11.7 Finding the maximum average power transfer: (a) circuit with a load, (b) the Thevenin equivalent. When ZL =Z* Th,wesaythattheloadismatched to the source. For maximum average power transfer, the load impedance ZL must be equal to the complex conjugate of the Thevenin impedance ZTh. This result is known as the maximum average power transfer theorem for the sinusoidal steady state. Setting RL = RTh and XL = −XTh in Eq. (11.15) gives us the maximum average power as Pmax = |VTh|2 8RTh (11.20) In a situation in which the load is purely real, the condition for maximum power transfer is obtained from Eq. (11.18) by setting XL = 0; that is, RL = R2 Th + X2 Th = |ZTh| (11.21) This means that for maximum average power transfer to a purely resistive load, the load impedance (or resistance) is equal to the magnitude of the Thevenin impedance.
  • 445. 442 PART 2 AC Circuits E X A M P L E 1 1 . 5 Determine the load impedance ZL that maximizes the average power drawn from the circuit of Fig. 11.8. What is the maximum average power? 4 Ω 8 Ω + − −j6 Ω j5 Ω 10 0° V ZL Figure11.8 For Example 11.5. Solution: First we obtain the Thevenin equivalent at the load terminals. To get ZTh, consider the circuit shown in Fig. 11.9(a). We find ZTh = j5 + 4 (8 − j6) = j5 + 4(8 − j6) 4 + 8 − j6 = 2.933 + j4.467 To find VTh, consider the circuit in Fig. 11.8(b). By voltage division, VTh = 8 − j6 4 + 8 − j6 (10) = 7.454 − 10.3◦ V The load impedance draws the maximum power from the circuit when ZL = Z∗ Th = 2.933 − j4.467 According to Eq. (11.20), the maximum average power is Pmax = |VTh|2 8RTh = (7.454)2 8(2.933) = 2.368 W 4 Ω 8 Ω −j6 Ω j5 Ω 10 V ZTh (a) 4 Ω 8 Ω −j6 Ω j5 Ω VTh (b) + − + − Figure11.9 Finding the Thevenin equivalent of the circuit in Fig. 11.8. P R A C T I C E P R O B L E M 1 1 . 5 For the circuit shown in Fig. 11.10, find the load impedance ZL that ab- sorbs the maximum average power. Calculate that maximum average power. 5 Ω 8 Ω −j4 Ω j10 Ω ZL 2 A Figure11.10 For Practice Prob. 11.5. Answer: 3.415 − j0.7317 , 1.429 W. E X A M P L E 1 1 . 6 In the circuit in Fig. 11.11, find the value of RL that will absorb the max- imum average power. Calculate that power.
  • 446. CHAPTER 11 AC Power Analysis 443 Solution: We first find the Thevenin equivalent at the terminals of RL. ZTh = (40 − j30) j20 = j20(40 − j30) j20 + 40 − j30 = 9.412 + j22.35 By voltage division, VTh = j20 j20 + 40 − j30 (150 30◦ ) = 72.76 134◦ V The value of RL that will absorb the maximum average power is RL = |ZTh| = 9.4122 + 22.352 = 24.25 The current through the load is I = VTh ZTh + RL = 72.76 134◦ 33.39 + j22.35 = 1.8 100.2◦ A The maximum average power absorbed by RL is Pmax = 1 2 |I|2 RL = 1 2 (1.8)2 (24.25) = 39.29 W 40 Ω + − j20 Ω −j30 Ω 150 30° V RL Figure11.11 For Example 11.6. P R A C T I C E P R O B L E M 1 1 . 6 In Fig. 11.12, the resistor RL is adjusted until it absorbs the maximum average power. Calculate RL and the maximum average power absorbed by it. 80 Ω + − 90 Ω j60 Ω 120 60° V RL −j30 Ω Figure11.12 For Practice Prob. 11.6. Answer: 30 , 9.883 W. 11.4 EFFECTIVE OR RMS VALUE The idea of effective value arises from the need to measure the effec- tiveness of a voltage or current source in delivering power to a resistive load. The effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.
  • 447. 444 PART 2 AC Circuits In Fig. 11.13, the circuit in (a) is ac while that of (b) is dc. Our objective is to find Ieff that will transfer the same power to resistor R as the sinusoid i. The average power absorbed by the resistor in the ac circuit is P = 1 T T 0 i2 R dt = R T T 0 i2 dt (11.22) while the power absorbed by the resistor in the dc circuit is P = I2 effR (11.23) Equating the expressions in Eqs. (11.22) and (11.23) and solving for Ieff, we obtain Ieff = 1 T T 0 i2 dt (11.24) The effective value of the voltage is found in the same way as current; that is, Veff = 1 T T 0 v2 dt (11.25) This indicates that the effective value is the (square) root of the mean (or average) of the square of the periodic signal. Thus, the effective value is often known as the root-mean-square value, or rms value for short; and we write Ieff = Irms, Veff = Vrms (11.26) For any periodic function x(t) in general, the rms value is given by Xrms = 1 T T 0 x2 dt (11.27) R + − i(t) v(t) (a) R Ieff Veff (b) + − Figure11.13 Finding the effective current: (a) ac circuit, (b) dc circuit. The effective value of a periodic signal is its root mean square (rms) value. Equation 11.27 states that to find the rms value of x(t), we first find its square x2 and then find the mean of that, or 1 T T 0 x2 dt and then the square root ( √ ) of that mean. The rms value of a constant is the constant itself. For the sinusoid i(t) = Im cos ωt, the effective or rms value is Irms = 1 T T 0 I2 m cos2 ωt dt = I2 m T T 0 1 2 (1 + cos 2ωt) dt = Im √ 2 (11.28)
  • 448. CHAPTER 11 AC Power Analysis 445 Similarly, for v(t) = Vm cos ωt, Vrms = Vm √ 2 (11.29) Keep in mind that Eqs. (11.28) and (11.29) are only valid for sinusoidal signals. The average power in Eq. (11.8) can be written in terms of the rms values. P = 1 2 VmIm cos(θv − θi) = Vm √ 2 Im √ 2 cos(θv − θi) = VrmsIrms cos(θv − θi) (11.30) Similarly, the average power absorbed by a resistor R in Eq. (11.11) can be written as P = I2 rmsR = V 2 rms R (11.31) When a sinusoidal voltage or current is specified, it is often in terms of its maximum (or peak) value or its rms value, since its average value is zero. The power industries specify phasor magnitudes in terms of their rms values rather than peak values. For instance, the 110 V available at every household is the rms value of the voltage from the power company. It is convenient in power analysis to express voltage and current in their rms values. Also, analog voltmeters and ammeters are designed to read directly the rms value of voltage and current, respectively. E X A M P L E 1 1 . 7 Determine the rms value of the current waveform in Fig. 11.14. If the current is passed through a 2- resistor, find the average power absorbed by the resistor. 0 t 10 −10 i(t) 2 4 6 8 10 Figure11.14 For Example 11.7. Solution: The period of the waveform is T = 4. Over a period, we can write the current waveform as i(t) = 5t, 0 t 2 −10, 2 t 4 The rms value is Irms = 1 T T 0 i2 dt = 1 4 2 0 (5t)2 dt + 4 2 (−10)2 dt = 1 4 25 t3 3 2 0 + 100t 4 2 = 1 4 200 3 + 200 = 8.165 A The power absorbed by a 2- resistor is P = I2 rmsR = (8.165)2 (2) = 133.3 W
  • 449. 446 PART 2 AC Circuits P R A C T I C E P R O B L E M 1 1 . 7 Find the rms value of the current waveform of Fig. 11.15. If the current flows through a 9- resistor, calculate the average power absorbed by the resistor. 2 3 1 0 4 5 6 t 4 i(t) Figure11.15 For Practice Prob. 11.7. Answer: 2.309 A, 48 W. E X A M P L E 1 1 . 8 The waveform shown in Fig. 11.16 is a half-wave rectified sine wave. Find the rms value and the amount of average power dissipated in a 10- resistor. 0 t 10 v(t) p 2p 3p Figure11.16 For Example 11.8. Solution: The period of the voltage waveform is T = 2π, and v(t) = 10 sin t, 0 t π 0, π t 2π The rms value is obtained as V 2 rms = 1 T T 0 v2 (t) dt = 1 2π π 0 (10 sin t)2 dt + 2π π 02 dt But sin2 t = 1 2 (1 − cos 2t). Hence V 2 rms = 1 2π π 0 100 2 (1 − cos 2t) dt = 50 2π t − sin 2t 2 π 0 = 50 2π π − 1 2 sin 2π − 0 = 25, Vrms = 5 V The average power absorbed is P = V 2 rms R = 52 10 = 2.5 W P R A C T I C E P R O B L E M 1 1 . 8 Find the rms value of the full-wave rectified sine wave in Fig. 11.17. Cal- culate the average power dissipated in a 6- resistor. 0 t 8 v(t) p 2p 3p Figure11.17 For Practice Prob. 11.8. Answer: 5.657 V, 5.334 W.
  • 450. CHAPTER 11 AC Power Analysis 447 11.5 APPARENT POWER AND POWER FACTOR In Section 11.2 we see that if the voltage and current at the terminals of a circuit are v(t) = Vm cos(ωt + θv) and i(t) = Im cos(ωt + θi) (11.32) or, in phasor form, V = Vm θv and I = Im θi, the average power is P = 1 2 VmIm cos(θv − θi) (11.33) In Section 11.4, we saw that P = VrmsIrms cos(θv − θi) = S cos(θv − θi) (11.34) We have added a new term to the equation: S = VrmsIrms (11.35) The average power is a product of two terms. The product VrmsIrms is known as the apparent power S. The factor cos(θv − θi) is called the power factor (pf). The apparent power (in VA) is the product of the rms values of voltage and current. The apparent power is so called because it seems apparent that the power should be the voltage-current product, by analogy with dc resistive cir- cuits. It is measured in volt-amperes or VA to distinguish it from the average or real power, which is measured in watts. The power factor is dimensionless, since it is the ratio of the average power to the apparent power, pf = P S = cos(θv − θi) (11.36) The angle θv − θi is called the power factor angle, since it is the angle whose cosine is the power factor. The power factor angle is equal to the angle of the load impedance if V is the voltage across the load and I is the current through it. This is evident from the fact that Z = V I = Vm θv Im θi = Vm Im θv − θi (11.37) Alternatively, since Vrms = V √ 2 = Vrms θv (11.38a) and Irms = I √ 2 = Irms θi (11.38b) the impedance is Z = V I = Vrms Irms = Vrms Irms θv − θi (11.39)
  • 451. 448 PART 2 AC Circuits The power factor is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of the load impedance. From Eq. (11.36), the power factor may also be regardedastheratiooftherealpowerdissipated in the load to the apparent power of the load. From Eq. (11.36), the power factor may be seen as that factor by which the apparent power must be multiplied to obtain the real or average power. The value of pf ranges between zero and unity. For a purely resistive load, the voltage and current are in phase, so that θv − θi = 0 and pf = 1. This implies that the apparent power is equal to the average power. For a purely reactive load, θv − θi = ±90◦ and pf = 0. In this case the average power is zero. In between these two extreme cases, pf is said to be leading or lagging. Leading power factor means that current leads voltage, which implies a capacitive load. Lagging power factor means that current lags voltage, implying an inductive load. Power factor affects the electric bills consumers pay the electric utility companies, as we will see in Section 11.9.2. E X A M P L E 1 1 . 9 A series-connected load draws a current i(t) = 4 cos(100πt + 10◦ ) A when the applied voltage is v(t) = 120 cos(100πt − 20◦ ) V. Find the apparent power and the power factor of the load. Determine the element values that form the series-connected load. Solution: The apparent power is S = VrmsIrms = 120 √ 2 4 √ 2 = 240 VA The power factor is pf = cos(θv − θi) = cos(−20◦ − 10◦ ) = 0.866 (leading) The pf is leading because the current leads the voltage. The pf may also be obtained from the load impedance. Z = V I = 120 − 20◦ 4 10◦ = 30 − 30◦ = 25.98 − j15 pf = cos(−30◦ ) = 0.866 (leading) The load impedance Z can be modeled by a 25.98- resistor in series with a capacitor with XC = −15 = − 1 ωC or C = 1 15ω = 1 15 × 100π = 212.2 µF P R A C T I C E P R O B L E M 1 1 . 9 Obtain the power factor and the apparent power of a load whose imped- ance is Z = 60 + j40 when the applied voltage is v(t) = 150 cos(377t + 10◦ ) V. Answer: 0.832 lagging, 156 VA.
  • 452. CHAPTER 11 AC Power Analysis 449 E X A M P L E 1 1 . 1 0 Determine the power factor of the entire circuit of Fig. 11.18 as seen by the source. Calculate the average power delivered by the source. 6 Ω 4 Ω + − 30 0° V rms −j2 Ω Figure11.18 For Example 11.10. Solution: The total impedance is Z = 6 + 4 (−j2) = 6 + −j2 × 4 4 − j2 = 6.8 − j1.6 = 7 − 13.24 The power factor is pf = cos(−13.24) = 0.9734 (leading) since the impedance is capacitive. The rms value of the current is Irms = Vrms Z = 30 0◦ 7 − 13.24◦ = 4.286 13.24◦ A The average power supplied by the source is P = VrmsIrms pf = (30)(4.286)0.9734 = 125 W or P = I2 rmsR = (4.286)2 (6.8) = 125 W where R is the resistive part of Z. P R A C T I C E P R O B L E M 1 1 . 1 0 Calculate the power factor of the entire circuit of Fig. 11.19 as seen by the source. What is the average power supplied by the source? 10 Ω + − 8 Ω j4 Ω −j6 Ω 40 0° V rms Figure11.19 For Practice Prob. 11.10. Answer: 0.936 lagging, 118 W. 11.6 COMPLEX POWER Considerable effort has been expended over the years to express power relations as simply as possible. Power engineers have coined the term complex power, which they use to find the total effect of parallel loads. Complex power is important in power analysis because it contains all the information pertaining to the power absorbed by a given load. V I + − Load Z Figure11.20 The voltage and current phasors associated with a load. Consider the ac load in Fig. 11.20. Given the phasor form V = Vm θv and I = Im θi of voltage v(t) and current i(t), the complex power S absorbed by the ac load is the product of the voltage and the complex conjugate of the current, or S = 1 2 VI∗ (11.40) assuming the passive sign convention (see Fig. 11.20). In terms of the rms values, S = VrmsI∗ rms (11.41)
  • 453. 450 PART 2 AC Circuits where Vrms = V √ 2 = Vrms θv (11.42) and Irms = I √ 2 = Irms θi (11.43) Thus we may write Eq. (11.41) as S = VrmsIrms θv − θi = VrmsIrms cos(θv − θi) + jVrmsIrms sin(θv − θi) (11.44) This equation can also be obtained from Eq. (11.9). We notice from Eq. (11.44) that the magnitude of the complex power is the apparent power; hence, the complex power is measured in volt-amperes (VA). Also, we notice that the angle of the complex power is the power factor angle. When working with the rms values of currents or voltages, we may drop the subscript rms if no confusion will be caused by doing so. Thecomplexpowermaybeexpressedintermsoftheloadimpedance Z. From Eq. (11.37), the load impedance Z may be written as Z = V I = Vrms Irms = Vrms Irms θv − θi (11.45) Thus, Vrms = ZIrms. Substituting this into Eq. (11.41) gives S = I2 rmsZ = V 2 rms Z∗ (11.46) Since Z = R + jX, Eq. (11.46) becomes S = I2 rms(R + jX) = P + jQ (11.47) where P and Q are the real and imaginary parts of the complex power; that is, P = Re(S) = I2 rmsR (11.48) Q = Im(S) = I2 rmsX (11.49) P is the average or real power and it depends on the load’s resistance R. Q depends on the load’s reactance X and is called the reactive (or quadrature) power. Comparing Eq. (11.44) with Eq. (11.47), we notice that P = VrmsIrms cos(θv − θi), Q = VrmsIrms sin(θv − θi) (11.50) The real power P is the average power in watts delivered to a load; it is the only useful power. It is the actual power dissipated by the load. The reactive power Q is a measure of the energy exchange between the source and the reactive part of the load. The unit of Q is the volt-ampere reactive(VAR)todistinguishitfromtherealpower, whoseunitisthewatt. We know from Chapter 6 that energy storage elements neither dissipate nor supply power, but exchange power back and forth with the rest of the network. In the same way, the reactive power is being transferred back and forth between the load and the source. It represents a lossless interchange between the load and the source. Notice that:
  • 454. CHAPTER 11 AC Power Analysis 451 1. Q = 0 for resistive loads (unity pf). 2. Q 0 for capacitive loads (leading pf). 3. Q 0 for inductive loads (lagging pf). Thus, Complex power (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. As a complex quantity, its real part is real power P and its imaginary part is reactive power Q. Introducing the complex power enables us to obtain the real and reactive powers directly from voltage and current phasors. Complex Power = S = P + jQ = 1 2 VI∗ = VrmsIrms θv − θi Apparent Power = S = |S| = VrmsIrms = P 2 + Q2 Real Power = P = Re(S) = S cos(θv − θi) Reactive Power = Q = Im(S) = S sin(θv − θi) Power Factor = P S = cos(θv − θi) (11.51) This shows how the complex power contains all the relevant power in- formation in a given load. S contains all power information of a load. The real part of S is the real power P; its imaginary part is the reactive power Q; its magnitude is the apparent power S; and the cosine of its phase angle is the power factor pf. It is a standard practice to represent S, P, and Q in the form of a triangle, known as the power triangle, shown in Fig. 11.21(a). This is similar to the impedance triangle showing the relationship between Z, R, and X, illustrated in Fig. 11.21(b). The power triangle has four items—the apparent/complex power, real power, reactive power, and the power factor angle. Given two of these items, the other two can easily be obtained from the triangle. As shown in Fig. 11.22, when S lies in the first quadrant, we have an inductive load and a lagging pf. When S lies in the fourth quadrant, the load is capacitive and the pf is leading. It is also possible for the complex power to lie in the second or third quadrant. This requires that the load impedance have a negative resistance, which is possible with active circuits. P Re Im S S +Q (lagging pf) −Q (leading pf) uv − ui uv − ui Figure11.22 Power triangle. S Q P u (a) |Z| X R u (b) Figure 11.21 (a) Power triangle, (b) impedance triangle.
  • 455. 452 PART 2 AC Circuits E X A M P L E 1 1 . 1 1 The voltage across a load is v(t) = 60 cos(ωt − 10◦ ) V and the cur- rent through the element in the direction of the voltage drop is i(t) = 1.5 cos(ωt + 50◦ ) A. Find: (a) the complex and apparent powers, (b) the real and reactive powers, and (c) the power factor and the load impedance. Solution: (a) For the rms values of the voltage and current, we write Vrms = 60 √ 2 − 10◦ , Irms = 1.5 √ 2 + 50◦ The complex power is S = VrmsI∗ rms = 60 √ 2 − 10◦ 1.5 √ 2 − 50◦ = 45 − 60◦ VA The apparent power is S = |S| = 45 VA (b) We can express the complex power in rectangular form as S = 45 − 60◦ = 45[cos(−60◦ ) + j sin(−60◦ )] = 22.5 − j38.97 Since S = P + jQ, the real power is P = 22.5 W while the reactive power is Q = −38.97 VAR (c) The power factor is pf = cos(−60◦ ) = 0.5 (leading) It is leading, because the reactive power is negative. The load impedance is Z = V I = 60 − 10◦ 1.5 + 50◦ = 40 − 60◦ which is a capacitive impedance. P R A C T I C E P R O B L E M 1 1 . 1 1 For a load, Vrms = 110 85◦ V, Irms = 0.4 15◦ A. Determine: (a) the complex and apparent powers, (b) the real and reactive powers, and (c) the power factor and the load impedance. Answer: (a) 44 70◦ VA, 44 VA, (b) 15.05 W, 41.35 VAR, (c) 0.342 lagging, 94.06 + j258.4 . E X A M P L E 1 1 . 1 2 A load Z draws 12 kVA at a power factor of 0.856 lagging from a 120-V rms sinusoidal source. Calculate: (a) the average and reactive powers delivered to the load, (b) the peak current, and (c) the load impedance.
  • 456. CHAPTER 11 AC Power Analysis 453 Solution: (a) Given that pf = cos θ = 0.856, we obtain the power angle as θ = cos−1 0.856 = 31.13◦ . If the apparent power is S = 12,000 VA, then the average or real power is P = S cos θ = 12,000 × 0.856 = 10.272 kW while the reactive power is Q = S sin θ = 12,000 × 0.517 = 6.204 kVA (b) Since the pf is lagging, the complex power is S = P + jQ = 10.272 + j6.204 kVA From S = VrmsI∗ rms, we obtain I∗ rms = S Vrms = 10,272 + j6204 120 0◦ = 85.6 + j51.7 A = 100 31.13◦ A Thus Irms = 100 − 31.13◦ and the peak current is Im = √ 2Irms = √ 2(100) = 141.4 A (c) The load impedance Z = Vrms Irms = 120 0◦ 100 − 31.13◦ = 1.2 31.13◦ which is an inductive impedance. P R A C T I C E P R O B L E M 1 1 . 1 2 A sinusoidal source supplies 10 kVA reactive power to load Z = 250 − 75◦ . Determine: (a) the power factor, (b) the apparent power delivered to the load, and (c) the peak voltage. Answer: (a) 0.2588 leading, (b) −10.35 kVAR, (c) 2.275 kV. †11.7 CONSERVATION OF AC POWER Infact, wealreadysawinExamples11.3and11.4 that average power is conserved in ac circuits. The principle of conservation of power applies to ac circuits as well as to dc circuits (see Section 1.5). To see this, consider the circuit in Fig. 11.23(a), where two load impedances Z1 and Z2 are connected in parallel across an ac source V. KCL gives I = I1 + I2 (11.52) The complex power supplied by the source is S = 1 2 VI∗ = 1 2 V(I∗ 1 + I∗ 2) = 1 2 VI∗ 1 + 1 2 VI∗ 2 = S1 + S2 (11.53) where S1 and S2 denote the complex powers delivered to loads Z1 and Z2, respectively.
  • 457. 454 PART 2 AC Circuits (a) (b) I V Z2 Z1 Z2 Z1 + − I1 I2 I V + − + − V1 + − V2 Figure 11.23 An ac voltage source supplied loads connected in: (a) parallel, (b) series. If the loads are connected in series with the voltage source, as shown in Fig. 11.23(b), KVL yields V = V1 + V2 (11.54) The complex power supplied by the source is S = 1 2 VI∗ = 1 2 (V1 + V2)I∗ = 1 2 V1I∗ + 1 2 V2I∗ = S1 + S2 (11.55) where S1 and S2 denote the complex powers delivered to loads Z1 and Z2, respectively. We conclude from Eqs. (11.53) and (11.55) that whether the loads are connected in series or in parallel (or in general), the total power supplied by the source equals the total power delivered to the load. Thus, in general, for a source connected to N loads, S = S1 + S1 + · · · + SN (11.56) This means that the total complex power in a network is the sum of the complex powers of the individual components. (This is also true of real power and reactive power, but not true of apparent power.) This expresses the principle of conservation of ac power: In fact, all forms of ac power are conserved: in- stantaneous, real, reactive, and complex. The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads. From this we imply that the real (or reactive) power flow from sources in a network equals the real (or reactive) power flow into the other elements in the network. E X A M P L E 1 1 . 1 3 Figure 11.24 shows a load being fed by a voltage source through a trans- mission line. The impedance of the line is represented by the (4 + j2) impedance and a return path. Find the real power and reactive power absorbed by: (a) the source, (b) the line, and (c) the load.
  • 458. CHAPTER 11 AC Power Analysis 455 4 Ω + − j2 Ω 220 0° V rms −j10 Ω 15 Ω I Source Line Load Figure11.24 For Example 11.13. Solution: The total impedance is Z = (4 + j2) + (15 − j10) = 19 − j8 = 20.62 − 22.83◦ The current through the circuit is I = Vs Z = 220 0◦ 20.62 − 22.83◦ = 10.67 22.83◦ A rms (a) For the source, the complex power is Ss = VsI∗ = (220 0◦ )(10.67 − 22.83◦ ) = 2347.4 − 22.83◦ = (2163.5 − j910.8) VA From this, we obtain the real power as 2163.5 W and the reactive power as 910.8 VAR (leading). (b) For the line, the voltage is Vline = (4 + j2)I = (4.472 26.57◦ )(10.67 22.83◦ ) = 47.72 49.4◦ V rms The complex power absorbed by the line is Sline = VlineI∗ = (47.72 49.4◦ )(10.67 − 22.83◦ ) = 509.2 26.57◦ = 455.4 + j227.7 VA or Sline = |I|2 Zline = (10.67)2 (4 + j2) = 455.4 + j227.7 VA That is, the real power is 455.4 W and the reactive power is 227.76 VAR (lagging). (c) For the load, the voltage is VL = (15 − j10)I = (18.03 − 33.7◦ )(10.67 22.83◦ ) = 192.38 − 10.87◦ V rms The complex power absorbed by the load is SL = VLI∗ = (192.38 − 10.87◦ )(10.67 − 22.83◦ ) = 2053 − 33.7◦ = (1708 − j1139) VA
  • 459. 456 PART 2 AC Circuits The real power is 1708 W and the reactive power is 1139 VAR (leading). Note that Ss = Sline + SL, as expected. We have used the rms values of voltages and currents. P R A C T I C E P R O B L E M 1 1 . 1 3 In the circuit in Fig. 11.25, the 60- resistor absorbs an average power of 240 W. Find V and the complex power of each branch of the circuit. What is the overall complex power of the circuit? 20 Ω 30 Ω + − −j10 Ω j20 Ω V 60 Ω Figure11.25 For Practice Prob. 11.13. Answer: 240.67 21.45◦ V (rms); the 20- resistor: 656 VA; the (30 − j10) impedance: 480 − j160 VA; the (60 + j20) impedance: 240 + j80 VA; overall: 1376 − j80 VA. E X A M P L E 1 1 . 1 4 In the circuit of Fig. 11.26, Z1 = 60 − 30◦ and Z2 = 40 45◦ . Calculate the total: (a) apparent power, (b) real power, (c) reactive power, and (d) pf. It Z2 Z1 + − I1 I2 120 10° V rms Figure11.26 For Example 11.14. Solution: The current through Z1 is I1 = V Z1 = 120 10◦ 60 − 30◦ = 2 40◦ A rms while the current through Z2 is I2 = V Z2 = 120 10◦ 40 45◦ = 3 − 35◦ A rms The complex powers absorbed by the impedances are S1 = V 2 rms Z∗ 1 = (120)2 60 30◦ = 240 − 30◦ = 207.85 − j120 VA S2 = V 2 rms Z∗ 2 = (120)2 40 − 45◦ = 360 45◦ = 254.6 + j254.6 VA The total complex power is St = S1 + S2 = 462.4 + j134.6 VA (a) The total apparent power is |St | = √ 462.42 + 134.62 = 481.6 VA. (b) The total real power is Pt = Re(St ) = 462.4 W or Pt = P1 + P2. (c) The total reactive power is Qt = Im(St ) = 134.6 VAR or Qt = Q1 + Q2. (d) The pf = Pt /|St | = 462.4/481.6 = 0.96 (lagging).
  • 460. CHAPTER 11 AC Power Analysis 457 We may cross check the result by finding the complex power Ss supplied by the source. It = I1 + I2 = (1.532 + j1.286) + (2.457 − j1.721) = 4 − j0.435 = 4.024 − 6.21◦ A rms Ss = VI∗ t = (120 10◦ )(4.024 6.21◦ ) = 482.88 16.21◦ = 463 + j135 VA which is the same as before. P R A C T I C E P R O B L E M 1 1 . 1 4 Two loads connected in parallel are respectively 2 kW at a pf of 0.75 lead- ing and 4 kW at a pf of 0.95 lagging. Calculate the pf of the two loads. Find the complex power supplied by the source. Answer: 0.9972 (leading), 6 − j0.4495 kVA. 11.8 POWER FACTOR CORRECTION Most domestic loads (such as washing machines, air conditioners, and refrigerators) and industrial loads (such as induction motors) are inductive and operate at a low lagging power factor. Although the inductive nature of the load cannot be changed, we can increase its power factor. The process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction. Alternatively, power factor correction may be viewedastheadditionofareactiveelement(usu- ally a capacitor) in parallel with the load in order to make the power factor closer to unity. An inductive load is modeled as a series combi- nation of an inductor and a resistor. Since most loads are inductive, as shown in Fig. 11.27(a), a load’s powerfactorisimprovedorcorrectedbydeliberatelyinstallingacapacitor in parallel with the load, as shown in Fig. 11.27(b). The effect of adding the capacitor can be illustrated using either the power triangle or the phasor diagram of the currents involved. Figure 11.28 shows the latter, where it is assumed that the circuit in Fig. 11.27(a) has a power factor of cos θ1, while the one in Fig. 11.27(b) has a power factor of cos θ2. It is V + − (a) IL Inductive load V + − (b) IL IC Inductive load C I Figure11.27 Power factor correction: (a) original inductive load, (b) inductive load with improved power factor. V IC IC IL I u1 u2 Figure11.28 Phasor diagram showing the effect of adding a capacitor in parallel with the inductive load.
  • 461. 458 PART 2 AC Circuits evident from Fig. 11.28 that adding the capacitor has caused the phase angle between the supplied voltage and current to reduce from θ1 to θ2, thereby increasing the power factor. We also notice from the magnitudes of the vectors in Fig. 11.28 that with the same supplied voltage, the circuit in Fig. 11.27(a) draws larger current IL than the current I drawn by the circuit in Fig. 11.27(b). Power companies charge more for larger currents, because they result in increased power losses (by a squared factor, since P = I2 LR). Therefore, it is beneficial to both the power company and the consumer that every effort is made to minimize current level or keep the power factor as close to unity as possible. By choosing a suitable size for the capacitor, the current can be made to be completely in phase with the voltage, implying unity power factor. S1 S2 QC Q2 Q1 u1 u2 P Figure 11.29 Powertriangleillustratingpower factor correction. Wecanlookatthepowerfactorcorrectionfromanotherperspective. Consider the power triangle in Fig. 11.29. If the original inductive load has apparent power S1, then P = S1 cos θ1, Q1 = S1 sin θ1 = P tan θ1 (11.57) If we desire to increase the power factor from cos θ1 to cos θ2 without altering the real power (i.e., P = S2 cos θ2), then the new reactive power is Q2 = P tan θ2 (11.58) The reduction in the reactive power is caused by the shunt capacitor, that is, QC = Q1 − Q2 = P(tan θ1 − tan θ2) (11.59) But from Eq. (11.49), QC = V 2 rms/XC = ωCV 2 rms. The value of the required shunt capacitance C is determined as C = QC ωV 2 rms = P(tan θ1 − tan θ2) ωV 2 rms (11.60) Note that the real power P dissipated by the load is not affected by the power factor correction because the average power due to the capacitance is zero. Although the most common situation in practice is that of an in- ductive load, it is also possible that the load is capacitive, that is, the load is operating at a leading power factor. In this case, an inductor should be connected across the load for power factor correction. The required shunt inductance L can be calculated from QL = V 2 rms XL = V 2 rms ωL ⇒ L = V 2 rms ωQL (11.61) where QL = Q1 − Q2, the difference between the new and old reactive powers. E X A M P L E 1 1 . 1 5 When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.
  • 462. CHAPTER 11 AC Power Analysis 459 Solution: If the pf = 0.8, then cos θ1 = 0.8 ⇒ θ1 = 36.87◦ where θ1 is the phase difference between voltage and current. We obtain the apparent power from the real power and the pf as S1 = P cos θ1 = 4000 0.8 = 5000 VA The reactive power is Q1 = S1 sin θ = 5000 sin 36.87 = 3000 VAR When the pf is raised to 0.95, cos θ2 = 0.95 ⇒ θ2 = 18.19◦ The real power P has not changed. But the apparent power has changed; its new value is S2 = P cos θ2 = 4000 0.95 = 4210.5 VA The new reactive power is Q2 = S2 sin θ2 = 1314.4 VAR The difference between the new and old reactive powers is due to the parallel addition of the capacitor to the load. The reactive power due to the capacitor is QC = Q1 − Q2 = 3000 − 1314.4 = 1685.6 VAR and C = QC ωV 2 rms = 1685.6 2π × 60 × 1202 = 310.5 µF P R A C T I C E P R O B L E M 1 1 . 1 5 Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume that the load is supplied by a 110-V (rms), 60-Hz line. Answer: 30.69 mF. Reactive power is measured by an instrument called the varmeter. The varmeter is often con- nected to the load in the same way as the wattmeter. †11.9 APPLICATIONS In this section, we consider two important application areas: how power is measured and how electric utility companies determine the cost of electricity consumption. 11.9.1 Power Measurement The average power absorbed by a load is measured by an instrument called the wattmeter.
  • 463. 460 PART 2 AC Circuits The wattmeter is the instrument for measuring the average power. Figure 11.30 shows a wattmeter that consists essentially of two coils: the current coil and the voltage coil. A current coil with very low impedance (ideally zero) is connected in series with the load (Fig. 11.31) and responds to the load current. The voltage coil with very high impedance(ideallyinfinite)isconnectedinparallelwiththeloadasshown in Fig. 11.31 and responds to the load voltage. The current coil acts like a short circuit because of its low impedance; the voltage coil behaves like an open circuit because of its high impedance. As a result, the presence of the wattmeter does not disturb the circuit or have an effect on the power measurement. Some wattmeters do not have coils; the watt- meter considered here is the electromagnetic type. i + − v R ± ± Figure11.30 A wattmeter. i i + − v Current coil Voltage coil ± ± ZL Figure11.31 The wattmeter connected to the load. When the two coils are energized, the mechanical inertia of the moving system produces a deflection angle that is proportional to the average value of the product v(t)i(t). If the current and voltage of the load are v(t) = Vm cos(ωt + θv) and i(t) = Im cos(ωt + θi), their corre- sponding rms phasors are Vrms = Vm √ 2 θv and Irms = Im √ 2 θi (11.62) and the wattmeter measures the average power given by P = |Vrms||Irms| cos(θv − θi) = 1 2 VmIm cos(θv − θi) (11.63) As shown in Fig. 11.31, each wattmeter coil has two terminals with one marked ±. To ensure upscale deflection, the ± terminal of the current coil is toward the source, while the ± terminal of the voltage coil is connected to the same line as the current coil. Reversing both coil connections still results in upscale deflection. However, reversing one coil and not the other results in downscale deflection and no wattmeter reading.
  • 464. CHAPTER 11 AC Power Analysis 461 E X A M P L E 1 1 . 1 6 Find the wattmeter reading of the circuit in Fig. 11.32. ± ± + − 12 Ω 150 0° V rms j10 Ω −j6 Ω 8 Ω Figure11.32 For Example 11.16. Solution: In Fig. 11.32, the wattmeter reads the average power absorbed by the (8 − j6) impedance because the current coil is in series with the impedance while the voltage coil is in parallel with it. The current through the circuit is I = 150 0◦ (12 + j10) + (8 − j6) = 150 20 + j4 A rms The voltage across the (8 − j6) impedance is V = I(8 − j6) = 150(8 − j6) 20 + j4 V rms The complex power is S = VI∗ = 150(8 − j6) 20 + j4 · 150 20 − j4 = 1502 (8 − j6) 202 + 42 = 423.7 − j324.6 VA The wattmeter reads P = Re(S) = 432.7 W P R A C T I C E P R O B L E M 1 1 . 1 6 For the circuit in Fig. 11.33, find the wattmeter reading. ± ± + − 4 Ω 120 30° V rms j9 Ω −j2 Ω 12 Ω Figure11.33 For Practice Prob. 11.16. Answer: 1437 W.
  • 465. 462 PART 2 AC Circuits 11.9.2 Electricity Consumption Cost In Section 1.7, we considered a simplified model of the way the cost of electricity consumption is determined. But the concept of power factor was not included in the calculations. Now we consider the importance of power factor in electricity consumption cost. Loads with low power factors are costly to serve because they re- quire large currents, as explained in Section 11.8. The ideal situation would be to draw minimum current from a supply so that S = P, Q = 0, andpf = 1. AloadwithnonzeroQmeansthatenergyflowsforthandback between the load and the source, giving rise to additional power losses. In view of this, power companies often encourage their customers to have power factors as close to unity as possible and penalize some customers who do not improve their load power factors. Utility companies divide their customers into categories: as resi- dential(domestic), commercial, andindustrial, orassmallpower, medium power, and large power. They have different rate structures for each category. The amount of energy consumed in units of kilowatt-hours (kWh) is measured using a kilowatt-hour meter installed at the customer’s premises. Although utility companies use different methods for charging cus- tomers, the tariff or charge to a consumer is often two-part. The first part is fixed and corresponds to the cost of generation, transmission, and dis- tribution of electricity to meet the load requirements of the consumers. This part of the tariff is generally expressed as a certain price per kW of maximum demand. Or it may instead be based on kVA of maximum de- mand, to account for the power factor (pf) of the consumer. A pf penalty charge may be imposed on the consumer whereby a certain percentage of kW or kVA maximum demand is charged for every 0.01 fall in pf below a prescribed value, say 0.85 or 0.9. On the other hand, a pf credit may be given for every 0.01 that the pf exceeds the prescribed value. The second part is proportional to the energy consumed in kWh; it may be in graded form, for example, the first 100 kWh at 16 cents/kWh, the next 200 kWh at 10 cents/kWh and so forth. Thus, the bill is deter- mined based on the following equation: Total Cost = Fixed Cost + Cost of Energy (11.64) E X A M P L E 1 1 . 1 7 A manufacturing industry consumes 200 MWh in one month. If the maximum demand is 1600 kW, calculate the electricity bill based on the following two-part rate: Demand charge: $5.00 per month per kW of billing demand. Energy charge: 8 cents per kWh for the first 50,000 kWh, 5 cents per kWh for the remaining energy. Solution: The demand charge is $5.00 × 1600 = $8000 (11.17.1) The energy charge for the first 50,000 kWh is
  • 466. CHAPTER 11 AC Power Analysis 463 $0.08 × 50,000 = $4000 (11.17.2) The remaining energy is 200,000 kWh − 50,000 kWh = 150,000 kWh, and the corresponding energy charge is $0.05 × 150,000 = $7500 (11.17.3) Adding Eqs. (11.17.1) to (11.17.3) gives Total bill for the month = $8000 + $4000 + $7500 = $19,500 It may appear that the cost of electricity is too high. But this is often a small fraction of the overall cost of production of the goods manufactured or the selling price of the finished product. P R A C T I C E P R O B L E M 1 1 . 1 7 The monthly reading of a paper mill’s meter is as follows: Maximum demand: 32,000 kW Energy consumed: 500 MWh Using the two-part rate in Example 11.17, calculate the monthly bill for the paper mill. Answer: $186,500. E X A M P L E 1 1 . 1 8 A 300-kW load supplied at 13 kV (rms) operates 520 hours a month at 80 percent power factor. Calculate the average cost per month based on this simplified tariff: Energy charge: 6 cents per kWh Power-factor penalty: 0.1 percent of energy charge for every 0.01 that pf falls below 0.85. Power-factor credit: 0.1 percent of energy charge for every 0.01 that pf exceeds 0.85. Solution: The energy consumed is W = 300 kW × 520 h = 156,000 kWh The operating power factor pf = 80% = 0.8 is 5 × 0.01 below the prescribed power factor of 0.85. Since there is 0.1 percent energy charge for every 0.01, there is a power-factor penalty charge of 0.5 percent. This amounts to an energy charge of %W = 156,000 × 5 × 0.1 100 = 780 kWh The total energy is
  • 467. 464 PART 2 AC Circuits Wt = W + %W = 156,000 + 780 = 156,780 kWh The cost per month is given by Cost = 6 cents × Wt = $0.06 × 156,780 = $9406.80 P R A C T I C E P R O B L E M 1 1 . 1 8 An 800-kW induction furnace at 0.88 power factor operates 20 hours per day for 26 days in a month. Determine the electricity bill per month based on the tariff in Example 11.16. Answer: $24,885.12. 11.10 SUMMARY 1. The instantaneous power absorbed by an element is the product of the element’s terminal voltage and the current through the element: p = vi. 2. Average or real power P (in watts) is the average of instantaneous power p: P = 1 T T 0 p dt If v(t) = Vm cos(ωt + θv) and i(t) = Im cos(ωt + θi), then Vrms = Vm/ √ 2, Irms = Im/ √ 2, and P = 1 2 VmIm cos(θv − θi) = VrmsIrms cos(θv − θi) Inductors and capacitors absorb no average power, while the aver- age power absorbed by a resistor is 1/2 I2 mR = I2 rmsR. 3. Maximum average power is transferred to a load when the load impedance is the complex conjugate of the Thevenin impedance as seen from the load terminals, ZL = Z∗ Th. 4. The effective value of a periodic signal x(t) is its root-mean-square (rms) value. Xeff = Xrms = 1 T T 0 x2 dt For a sinusoid, the effective or rms value is its amplitude divided by √ 2. 5. The power factor is the cosine of the phase difference between volt- age and current: pf = cos(θv − θi) It is also the cosine of the angle of the load impedance or the ratio of real power to apparent power. The pf is lagging if the current lags voltage (inductive load) and is leading when the current leads voltage (capacitive load).
  • 468. CHAPTER 11 AC Power Analysis 465 6. Apparent power S (in VA) is the product of the rms values of volt- age and current: S = VrmsIrms It is also given by S = |S| = P 2 + Q2, where Q is reactive power. 7. Reactive power (in VAR) is: Q = 1 2 VmIm sin(θv − θi) = VrmsIrms sin(θv − θi) 8. Complex power S (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. It is also the complex sum of real power P and reactive power Q. S = VrmsI∗ rms = VrmsIrms θv − θi = P + jQ Also, S = I2 rmsZ = V 2 rms Z∗ 9. The total complex power in a network is the sum of the complex powers of the individual components. Total real power and reactive power are also, respectively, the sums of the individual real powers and the reactive powers, but the total apparent power is not calcu- lated by the process. 10. Power factor correction is necessary for economic reasons; it is the process of improving the power factor of a load by reducing the overall reactive power. 11. The wattmeter is the instrument for measuring the average power. Energy consumed is measured with a kilowatt-hour meter. REVIEW QUESTIONS 11.1 The average power absorbed by an inductor is zero. (a) True (b) False 11.2 The Thevenin impedance of a network seen from the load terminals is 80 + j55 . For maximum power transfer, the load impedance must be: (a) −80 + j55 (b) −80 − j55 (c) 80 − j55 (d) 80 + j55 11.3 The amplitude of the voltage available in the 60-Hz, 120-V power outlet in your home is: (a) 110 V (b) 120 V (c) 170 V (d) 210 V 11.4 If the load impedance is 20 − j20, the power factor is (a) − 45◦ (b) 0 (c) 1 (d) 0.7071 (e) noneofthese 11.5 A quantity that contains all the power information in a given load is the (a) power factor (b) apparent power (c) average power (d) reactive power (e) complex power 11.6 Reactive power is measured in: (a) watts (b) VA (c) VAR (d) none of these 11.7 In the power triangle shown in Fig. 11.34(a), the reactive power is: (a) 1000 VAR leading (b) 1000 VAR lagging (c) 866 VAR leading (d) 866 VAR lagging
  • 469. 466 PART 2 AC Circuits (a) (b) 60° 500 W 30° 1000 VAR Figure 11.34 For Review Questions 11.7 and 11.8. 11.8 For the power triangle in Fig. 11.34(b), the apparent power is: (a) 2000 VA (b) 1000 VAR (c) 866 VAR (d) 500 VAR 11.9 A source is connected to three loads Z1, Z2, and Z3 in parallel. Which of these is not true? (a) P = P1 + P2 + P3 (b) Q = Q1 + Q2 + Q3 (c) S = S1 + S2 + S3 (d) S = S1 + S2 + S3 11.10 The instrument for measuring average power is the: (a) voltmeter (b) ammeter (c) wattmeter (d) varmeter (e) kilowatt-hour meter Answers: 11.1a, 11.2c, 11.3c, 11.4d, 11.5e, 11.6c, 11.7d, 11.8a, 11.9c, 11.10c. PROBLEMS Section 11.2 Instantaneous and Average Power 11.1 If v(t) = 160 cos 50tV and i(t) = −20 sin(50t − 30◦ ) A, calculate the instantaneous power and the average power. 11.2 At t = 2 s, find the instantaneous power on each of the elements in the circuit of Fig. 11.35. + − 30 cos 500t A 0.3 H 20 mF 200 Ω Figure 11.35 For Prob. 11.2. 11.3 Refer to the circuit depicted in Fig. 11.36. Find the average power absorbed by each element. 4 Ω + − 2 Ω 1 H 0.25 F 10 cos(2t + 30°) V Figure 11.36 For Prob. 11.3. 11.4 Given the circuit in Fig. 11.37, find the average power absorbed by each of the elements. 20 Ω + − 10 Ω j5 Ω −j10 Ω 50 0° V Figure 11.37 For Prob. 11.4. 11.5 Compute the average power absorbed by the 4- resistor in the circuit of Fig. 11.38. 4 Ω j2 Ω −j1 Ω 2 Ω + − + − Vo 4 60° Α 4Vo Figure 11.38 For Prob. 11.5. 11.6 Given the circuit of Fig. 11.39, find the average power absorbed by the 10- resistor. + − 10 Ω + − Io 8 20° V 0.1V o 4 Ω j5 Ω −j5 Ω 8Io + − Vo Figure 11.39 For Prob. 11.6.
  • 470. CHAPTER 11 AC Power Analysis 467 11.7 In the circuit of Fig. 11.40, determine the average power absorbed by the 40- resistor. Io 6 0° A j10 Ω 0.5Io 40 Ω −j20 Ω Figure 11.40 For Prob. 11.7. 11.8 Calculate the average power absorbed by each resistor in the op amp circuit of Fig. 11.41 if the rms value of vs is 2 V. + − 6 kΩ 2 kΩ + − vs 10 kΩ Figure 11.41 For Prob. 11.8. 11.9 In the op amp circuit in Fig. 11.42, find the total average power absorbed by the resistors. + − R R R Vo cos vt V + − + − Figure 11.42 For Prob. 11.9. 11.10 For the network in Fig. 11.43, assume that the port impedance is Zab = R √ 1 + ω2R2C2 − tan−1 ωRC Find the average power consumed by the network when R = 10 k, C = 200 nF, and i = 2 sin(377t + 22◦ ) mA. v i + − Linear network a b Figure 11.43 For Prob. 11.10. Section 11.3 Maximum Average Power Transfer 11.11 For each of the circuits in Fig. 11.44, determine the value of load Z for maximum power transfer and the maximum average power transferred. −j2 Ω 8 Ω 4 0° Α Z (a) + − −j3 Ω j2 Ω 4 Ω 5 Ω 10 30° V Z (b) Figure 11.44 For Prob. 11.11. 11.12 For the circuit in Fig. 11.45, find: (a) the value of the load impedance that absorbs the maximum average power (b) the value of the maximum average power absorbed −j40 Ω j100 Ω 80 Ω 3 20° Α Load Figure 11.45 For Prob. 11.12. 11.13 In the circuit of Fig. 11.46, find the value of ZL that will absorb the maximum power and the value of the maximum power. j1 Ω 1 Ω −j1 Ω 12 0° V ZL + − + − Vo 2Vo Figure 11.46 For Prob. 11.13.
  • 471. 468 PART 2 AC Circuits 11.14 Calculate the value of ZL in the circuit of Fig. 11.47 in order for ZL to receive maximum average power. What is the maximum average power received by Z? −j10 Ω j20 Ω 30 Ω 40 Ω 5 90° A ZL Figure 11.47 For Prob. 11.14. 11.15 Find the value of ZL in the circuit of Fig. 11.48 for maximum power transfer. 40 Ω 40 Ω −j10 Ω j20 Ω + − 80 Ω 60 0° V 5 0° A ZL Figure 11.48 For Prob. 11.15. 11.16 The variable resistor R in the circuit of Fig. 11.49 is adjusted until it absorbs the maximum average power. Find R and the maximum average power absorbed. j1 Ω −j2 Ω 3 Ω 6 Ω 4 0° A R Figure 11.49 For Prob. 11.16. 11.17 The load resistance RL in Fig. 11.50 is adjusted until it absorbs the maximum average power. Calculate the value of RL and the maximum average power. −j10 Ω −j10 Ω + − Io 120 0° V 40 Ω j20 Ω 4Io RL + − Figure 11.50 For Prob. 11.17. 11.18 Assuming that the load impedance is to be purely resistive, what load should be connected to terminals a-b of the circuits in Fig. 11.51 so that the maximum power is transferred to the load? 100 Ω 40 Ω −j10 Ω j30 Ω 50 Ω + − 120 60° V 2 90° A a b Figure 11.51 For Prob. 11.18. Section 11.4 Effective or RMS Value 11.19 Find the rms value of the periodic signal in Fig. 11.52. 2 0 4 6 8 t 15 5 v(t) Figure 11.52 For Prob. 11.19. 11.20 Determine the rms value of the waveform in Fig. 11.53. 1 0 2 3 4 t 5 −5 v(t) Figure 11.53 For Prob. 11.20. 11.21 Find the effective value of the voltage waveform in Fig. 11.54. 2 0 4 6 8 10 t 5 10 v(t) Figure 11.54 For Prob. 11.21. 11.22 Calculate the rms value of the current waveform of Fig. 11.55.
  • 472. CHAPTER 11 AC Power Analysis 469 5 0 10 15 20 25 t 5 i(t) Figure 11.55 For Prob. 11.22. 11.23 Find the rms value of the voltage waveform of Fig. 11.56 as well as the average power absorbed by a 2- resistor when the voltage is applied across the resistor. 2 0 5 7 10 12 t 8 v(t) Figure 11.56 For Prob. 11.23. 11.24 Calculate the effective value of the current waveform in Fig. 11.57 and the average power delivered to a 12- resistor when the current runs through the resistor. 5 0 15 25 t 10 −10 i(t) 10 20 30 Figure 11.57 For Prob. 11.24. 11.25 Compute the rms value of the waveform depicted in Fig. 11.58. 2 −1 0 4 6 8 10 t 2 v(t) Figure 11.58 For Prob. 11.25. 11.26 Obtain the rms value of the current waveform shown in Fig. 11.59. 1 0 2 3 4 5 t 10 i(t) 10t2 Figure 11.59 For Prob. 11.26. 11.27 Determine the effective value of the periodic waveform in Fig. 11.60. 1 0 2 3 4 5 t 10 i(t) Figure 11.60 For Prob. 11.27. 11.28 One cycle of a periodic voltage waveform is depicted in Fig. 11.61. Find the effective value of the voltage. 1 0 2 3 4 6 5 t 10 20 30 v(t) Figure 11.61 For Prob. 11.28. Section 11.5 Apparent Power and Power Factor 11.29 A relay coil is connected to a 210-V, 50-Hz supply. If it has a resistance of 30 and an inductance of 0.5 H, calculate the apparent power and the power factor. 11.30 A certain load comprises 12 − j8 in parallel with j4 . Determine the overall power factor. 11.31 Obtain the power factor for each of the circuits in Fig. 11.62. Specify each power factor as leading or lagging.
  • 473. 470 PART 2 AC Circuits −j2 Ω j2 Ω j1 Ω −j1 Ω 1 Ω −j2 Ω (a) 4 Ω (b) j5 Ω 4 Ω Figure 11.62 For Prob. 11.31. Section 11.6 Complex Power 11.32 A load draws 5 kVAR at a power factor of 0.86 (leading) from a 220-V rms source. Calculate the peak current and the apparent power supplied to the load. 11.33 For the following voltage and current phasors, calculate the complex power, apparent power, real power, and reactive power. Specify whether the pf is leading or lagging. (a) V = 220 30◦ V rms, I = 0.5 60◦ A rms (b) V = 250 − 10◦ V rms, I = 6.2 − 25◦ A rms (c) V = 120 0◦ V rms, I = 2.4 − 15◦ A rms (d) V = 160 45◦ V rms, I = 8.5 90◦ A rms 11.34 For each of the following cases, find the complex power, the average power, and the reactive power: (a) v(t) = 112 cos(ωt + 10◦ ) V, i(t) = 4 cos(ωt − 50◦ ) A (b) v(t) = 160 cos 377t V, i(t) = 4 cos(377t + 45◦ ) A (c) V = 80 60◦ V rms, Z = 50 30◦ (d) I = 10 60◦ V rms, Z = 100 45◦ 11.35 Determine the complex power for the following cases: (a) P = 269 W, Q = 150 VAR (capacitive) (b) Q = 2000 VAR, pf = 0.9 (leading) (c) S = 600 VA, Q = 450 VAR (inductive) (d) Vrms = 220 V, P = 1 kW, |Z| = 40 (inductive) 11.36 Find the complex power for the following cases: (a) P = 4 kW, pf = 0.86 (lagging) (b) S = 2 kVA, P = 1.6 kW (capacitive) (c) Vrms = 208 20◦ V, Irms = 6.5 − 50◦ A (d) Vrms = 120 30◦ V, Z = 40 + j60 11.37 Obtain the overall impedance for the following cases: (a) P = 1000 W, pf = 0.8 (leading), Vrms = 220 V (b) P = 1500 W, Q = 2000 VAR (inductive), Irms = 12 A (c) S = 4500 60◦ VA, V = 120 45◦ V 11.38 For the entire circuit in Fig. 11.63, calculate: (a) the power factor (b) the average power delivered by the source (c) the reactive power (d) the apparent power (e) the complex power 2 Ω 10 Ω + − −j5 Ω j6 Ω 8 Ω 16 45° V Figure 11.63 For Prob. 11.38. Section 11.7 Conservation of AC Power 11.39 For the network in Fig. 11.64, find the complex power absorbed by each element. 4 Ω + − −j3 Ω j5 Ω 8 −20° V Figure 11.64 For Prob. 11.39. 11.40 Find the complex power absorbed by each of the five elements in the circuit of Fig. 11.65. + − + − 20 Ω 40 0° V rms j10 Ω −j20 Ω 50 90° V rms Figure 11.65 For Prob. 11.40. 11.41 Obtain the complex power delivered by the source in the circuit of Fig. 11.66.
  • 474. CHAPTER 11 AC Power Analysis 471 −j2 Ω j4 Ω 5 Ω 3 Ω 2 30° Α 6 Ω Figure 11.66 For Prob. 11.41. 11.42 For the circuit in Fig. 11.67, find the average, reactive, and complex power delivered by the dependent voltage source. 4 Ω 1 Ω −j1 Ω j2 Ω 24 0° V 2 Ω + − 2Vo + − Vo Figure 11.67 For Prob. 11.42. 11.43 Obtain the complex power delivered to the 10-k resistor in Fig. 11.68 below. 11.44 Calculate the reactive power in the inductor and capacitor in the circuit of Fig. 11.69. −j20 Ω j30 Ω 50 Ω 240 0° V 4 0° A 40 Ω + − Figure 11.69 For Prob. 11.44. 11.45 For the circuit in Fig. 11.70, find Vo and the input power factor. 6 0° A rms + − V o 20 kW 0.8 pf lagging 16 kW 0.9 pf lagging Figure 11.70 For Prob. 11.45. 11.46 Given the circuit in Fig. 11.71, find Io and the overall complex power supplied. 100 90° V 2 kVA 0.707 pf leading 1.2 kW 0.8 kVAR (cap) 4 kW 0.9 pf lagging Io + − Figure 11.71 For Prob. 11.46. 11.47 For the circuit in Fig. 11.72, find Vs. V s + − 120 V rms 10 W 0.9 pf lagging 15 W 0.8 pf leading 0.2 Ω 0.3 Ω j0.04 Ω j0.15 Ω + − Figure 11.72 For Prob. 11.47. 11.48 Find Io in the circuit of Fig. 11.73 on the bottom of the next page. 11.49 In the op amp circuit of Fig. 11.74, vs = 4 cos 104 t V. Find the average power delivered to the 50-k resistor. + − 100 kΩ + − vs 50 kΩ 1 nF Figure 11.74 For Prob. 11.49. 11.50 Obtain the average power absorbed by the 6-k resistor in the op amp circuit in Fig. 11.75. + − 10 kΩ Io 0.2 0° V rms 500 Ω −j3 kΩ j1 kΩ 20Io 4 kΩ Figure11.68 For Prob. 11.43.
  • 475. 472 PART 2 AC Circuits + − 4 kΩ j3 kΩ −j2 kΩ j4 kΩ + − 6 kΩ 2 kΩ 4 45° V Figure 11.75 For Prob. 11.50. 11.51 Calculate the complex power delivered to each resistor and capacitor in the op amp circuit of Fig. 11.76. Let vs = 2 cos 103 t V. + − 10 kΩ + − 40 kΩ 20 kΩ vs 0.1 mF 0.2 mF Figure 11.76 For Prob. 11.51. 11.52 Compute the complex power supplied by the current source in the series RLC circuit in Fig. 11.77. L R Io cos vt C Figure 11.77 For Prob. 11.52. Section 11.8 Power Factor Correction 11.53 Refer to the circuit shown in Fig. 11.78. (a) What is the power factor? (b) What is the average power dissipated? (c) What is the value of the capacitance that will give a unity power factor when connected to the load? Z = 10 + j12 Ω + − C 120 V 60 Hz Figure 11.78 For Prob. 11.53. 11.54 An 880-VA, 220-V, 50-Hz load has a power factor of 0.8 lagging. What value of parallel capacitance will correct the load power factor to unity? 11.55 An 40-kW induction motor, with a lagging power factor of 0.76, is supplied by a 120-V rms 60-Hz sinusoidal voltage source. Find the capacitance needed in parallel with the motor to raise the power factor to: (a) 0.9 lagging (b) 1.0. 11.56 A 240-V rms 60-Hz supply serves a load that is 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR (inductive). Find: (a) the apparent power (b) the current drawn from the supply (c) the kVAR rating and capacitance required to improve the power factor to 0.96 lagging (d) the current drawn from the supply under the new power-factor conditions 11.57 A 120-V rms 60-Hz source supplies two loads connected in parallel, as shown in Fig. 11.79. (a) Find the power factor of the parallel combination. (b) Calculate the value of the capacitance connected in parallel that will raise the power factor to unity. 220 0° V 12 kW 0.866 pf leading 20 kVAR 0.6 pf lagging 16 kW 0.85 pf lagging Io + − Figure11.73 For Prob. 11.48.
  • 476. CHAPTER 11 AC Power Analysis 473 Load 1 24 kW pf = 0.8 lagging Load 2 40 kW pf = 0.95 lagging Figure 11.79 For Prob. 11.57. 11.58 Consider the power system shown in Fig. 11.80. Calculate: (a) the total complex power (b) the power factor (c) the capacitance necessary to establish a unity power factor + − 240 V rms, 50 Hz 80 − j50 Ω 120 + j70 Ω 60 + j0 Figure 11.80 For Prob. 11.58. Section 11.9 Applications 11.59 Obtain the wattmeter reading of the circuit in Fig. 11.81 below. 11.60 What is the reading of the wattmeter in the network of Fig. 11.82? 6 Ω 4 H 15 Ω 0.1 F + − 120 cos 2t V ± ± Figure 11.82 For Prob. 11.60. 11.61 Find the wattmeter reading of the circuit shown in Fig. 11.83 below. 11.62 The circuit of Fig. 11.84 portrays a wattmeter connected into an ac network. (a) Find the load current. (b) Calculate the wattmeter reading. ZL = 6.4 Ω pf = 0.825 + − 110 V WM Figure 11.84 For Prob. 11.62. 11.63 The kilowatthour-meter of a home is read once a month. For a particular month, the previous and present readings are as follows: Previous reading: 3246 kWh Present reading: 4017 kWh Calculate the electricity bill for that month based on the following residential rate schedule: 4 Ω −j3 Ω j2 Ω 8 Ω + − 12 0° V 3 30° A ± ± Figure11.81 For Prob. 11.59. 10 Ω 4 Ω + − 5 Ω 1 H 20 cos 4t V ± ± F 1 12 Figure11.83 For Prob. 11.61.
  • 477. 474 PART 2 AC Circuits Minimum monthly charge—$12.00 First 100 kWh per month at 16 cents/kWh Next 200 kWh per month at 10 cents/kWh Over 300 kWh per month at 6 cents/kWh 11.64 A consumer has an annual consumption of 1200 MWh with a maximum demand of 2.4 MVA. The maximum demand charge is $30 per kVA per annum, and the energy charge per kWh is 4 cents. (a) Determine the annual cost of energy. (b) Calculate the charge per kWh with a flat-rate tariff if the revenue to the utility company is to remain the same as for the two-part tariff. COMPREHENSIVE PROBLEMS 11.65 A transmitter delivers maximum power to an antenna when the antenna is adjusted to represent a load of 75- resistance in series with an inductance of 4 µH. If the transmitter operates at 4.12 MHz, find its internal impedance. 11.66 In a TV transmitter, a series circuit has an impedance of 3 k and a total current of 50 mA. If the voltage across the resistor is 80 V, what is the power factor of the circuit? 11.67 A certain electronic circuit is connected to a 110-V ac line. The root-mean-square value of the current drawn is 2 A, with a phase angle of 55◦ . (a) Find the true power drawn by the circuit. (b) Calculate the apparent power. 11.68 An industrial heater has a nameplate which reads: 210 V 60 Hz 12 kVA 0.78 pf lagging Determine: (a) the apparent and the complex power (b) the impedance of the heater 11.69 ∗ A 2000-kW turbine-generator of 0.85 power factor operates at the rated load. An additional load of 300 kW at 0.8 power factor is added. What kVAR of capacitors is required to operate the turbine -generator but keep it from being overloaded? 11.70 The nameplate of an electric motor has the following information: Line voltage: 220 V rms Line current: 15 A rms Line frequency: 60 Hz Power: 2700 W Determine the power factor (lagging) of the motor. Find the value of the capacitance C that must be connected across the motor to raise the pf to unity. 11.71 As shown in Fig. 11.85, a 550-V feeder line supplies an industrial plant consisting of a motor drawing 60 kW at 0.75 pf (inductive), a capacitor with a rating of 20 kVAR, and lighting drawing 20 kW. (a) Calculate the total reactive power and apparent power absorbed by the plant. *An asterisk indicates a challenging problem. (b) Determine the overall pf. (c) Find the current in the feeder line. 60 kW pf = 0.75 550 V 20 kVAR 10 kW + − Figure 11.85 For Prob. 11.71. 11.72 A factory has the following four major loads: • A motor rated at 5 hp, 0.8 pf lagging (1 hp = 0.7457 kW). • A heater rated at 1.2 kW, 1.0 pf. • Ten 120-W lightbulbs. • A synchronous motor rated at 1.6 kVA, 0.6 pf leading. (a) Calculate the total real and reactive power. (b) Find the overall power factor. 11.73 A 1-MVA substation operates at full load at 0.7 power factor. It is desired to improve the power factor to 0.95 by installing capacitors. Assume that new substation and distribution facilities cost $120 per kVA installed, and capacitors cost $30 per kVA installed. (a) Calculate the cost of capacitors needed. (b) Find the savings in substation capacity released. (c) Are capacitors economical for releasing the amount of substation capacity? 11.74 A coupling capacitor is used to block dc current from an amplifier as shown in Fig. 11.86(a). The amplifier and the capacitor act as the source, while the speaker is the load as in Fig. 11.86(b). (a) At what frequency is maximum power transferred to the speaker? (b) If Vs = 4.6 V rms, how much power is delivered to the speaker at that frequency?
  • 478. CHAPTER 11 AC Power Analysis 475 10 Ω 40 nF 80 mH 4 Ω vs Amplifier Vin Speaker Coupling capacitor Amplifier Speaker (a) (b) Figure 11.86 For Prob. 11.74. 11.75 A power amplifier has an output impedance of 40 + j8 . It produces a no-load output voltage of 146 V at 300 Hz. (a) Determine the impedance of the load that achieves maximum power transfer. (b) Calculate the load power under this matching condition. 11.76 A power transmission system is modeled as shown in Fig. 11.87. If Vs = 240 0◦ rms, find the average power absorbed by the load. 0.1 Ω 0.1 Ω j1 Ω j1 Ω + − j1 Ω 100 Ω Vs Source Line Load Figure 11.87 For Prob. 11.76.
  • 479. 477 C H A P T E R THREE-PHASE CIRCUITS 1 2 Society is never prepared to receive any invention. Every new thing is resisted, and it takes years for the inventor to get people to listen to him and years more before it can be introduced. —Thomas Alva Edison Historical Profiles Thomas Alva Edison (1847–1931) was perhaps the greatest American inventor. He patented 1093 inventions, including such history-making inventions as the incandescent electric bulb, the phonograph, and the first commercial motion pictures. Born in Milan, Ohio, the youngest of seven children, Edison received only three months of formal education because he hated school. He was home-schooled by his mother and quickly began to read on his own. In 1868, Edison read one of Faraday’s books and found his calling. He moved to Menlo Park, New Jersey, in 1876, where he managed a well-staffed research laboratory. Most of his inventions came out of this lab- oratory. His laboratory served as a model for modern research organizations. Because of his diverse interests and the overwhelming number of his inventions and patents, Edison began to establish manufacturing companies for making the devices he invented. He designed the first electric power station to supply electric light. Formal electri- cal engineering education began in the mid-1880s with Edison as a role model and leader. Nikola Tesla (1856–1943) was a Croatian-American engineer whose inventions— among them the induction motor and the first polyphase ac power system—greatly influenced the settlement of the ac versus dc debate in favor of ac. He was also re- sponsible for the adoption of 60 Hz as the standard for ac power systems in the United States. Born in Austria-Hungary (now Croatia), to a clergyman, Tesla had an incredible memory and a keen affinity for mathematics. He moved to the United States in 1884 and first worked for Thomas Edison. At that time, the country was in the “battle of the currents” with George Westinghouse (1846–1914) promoting ac and Thomas Edison rigidly leading the dc forces. Tesla left Edison and joined Westinghouse because of his interest in ac. Through Westinghouse, Tesla gained the reputation and acceptance of his polyphase ac generation, transmission, and distribution system. He held 700 patents in his lifetime. His other inventions include high-voltage apparatus (the tesla coil) and a wireless transmission system. The unit of magnetic flux density, the tesla, was named in honor of him.
  • 480. 478 PART 2 AC Circuits 12.1 INTRODUCTION So far in this text, we have dealt with single-phase circuits. A single- phase ac power system consists of a generator connected through a pair of wires (a transmission line) to a load. Figure 12.1(a) depicts a single- phase two-wire system, where Vp is the magnitude of the source voltage and φ is the phase. What is more common in practice is a single-phase three-wiresystem, showninFig.12.1(b). Itcontainstwoidenticalsources (equal magnitude and the same phase) which are connected to two loads by two outer wires and the neutral. For example, the normal household system is a single-phase three-wire system because the terminal voltages have the same magnitude and the same phase. Such a system allows the connection of both 120-V and 240-V appliances. Historical note: Thomas Edison invented a three- wire system, using three wires instead of four. ZL Vp + − (a) f ZL1 Vp a A n N b B + − (b) f ZL2 Vp + − f Figure12.1 Single-phase systems: (a) two-wire type, (b) three-wire type. Circuits or systems in which the ac sources operate at the same frequencybutdifferentphasesareknownaspolyphase. Figure12.2shows a two-phase three-wire system, and Fig. 12.3 shows a three-phase four- wire system. As distinct from a single-phase system, a two-phase system is produced by a generator consisting of two coils placed perpendicular to each other so that the voltage generated by one lags the other by 90◦ . By the same token, a three-phase system is produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 120◦ . Since the three-phase system is by far the most prevalent and most economical polyphase system, discussion in this chapter is mainly on three-phase systems. ZL1 Vp a A n N b B + − ZL2 + − −90° Vp 0° Figure12.2 Two-phase three-wire system. ZL1 a A b B c C n N Vp 0° − + ZL2 Vp −120° Vp +120° − + ZL3 − + Figure12.3 Three-phase four-wire system. Three-phase systems are important for at least three reasons. First, nearly all electric power is generated and distributed in three-phase, at the operating frequency of 60 Hz (or ω = 377 rad/s) in the United States or 50 Hz (or ω = 314 rad/s) in some other parts of the world. When one- phase or two-phase inputs are required, they are taken from the three- phase system rather than generated independently. Even when more than three phases are needed—such as in the aluminum industry, where 48 phases are required for melting purposes—they can be provided by manipulating the three phases supplied. Second, the instantaneous power in a three-phase system can be constant (not pulsating), as we will see in Section 12.7. This results in uniform power transmission and less vibration of three-phase machines. Third, for the same amount of power, the three-phase system is more economical than the single-phase. The
  • 481. CHAPTER 12 Three-Phase Circuits 479 amount of wire required for a three-phase system is less than that required for an equivalent single-phase system. We begin with a discussion of balanced three-phase voltages. Then we analyze each of the four possible configurations of balanced three- phase systems. We also discuss the analysis of unbalanced three-phase systems. We learn how to use PSpice for Windows to analyze a balanced or unbalanced three-phase system. Finally, we apply the concepts devel- oped in this chapter to three-phase power measurement and residential electrical wiring. 12.2 BALANCED THREE-PHASE VOLTAGES Three-phase voltages are often produced with a three-phase ac generator (or alternator) whose cross-sectional view is shown in Fig. 12.4. The gen- erator basically consists of a rotating magnet (called the rotor) surrounded by a stationary winding (called the stator). Three separate windings or coils with terminals a-a , b-b , and c-c are physically placed 120◦ apart around the stator. Terminals a and a , for example, stand for one of the ends of coils going into and the other end coming out of the page. As the rotor rotates, its magnetic field “cuts” the flux from the three coils and induces voltages in the coils. Because the coils are placed 120◦ apart, the induced voltages in the coils are equal in magnitude but out of phase by 120◦ (Fig. 12.5). Since each coil can be regarded as a single-phase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads. Stator Three- phase output a b c n c N S b′ c′ b a a′ Rotor Figure12.4 A three-phase generator. 0 120° V an vt V bn V cn 240° Figure12.5 The generated voltages are 120◦ apart from each other. A typical three-phase system consists of three voltage sources con- nected to loads by three or four wires (or transmission lines). (Three- phase current sources are very scarce.) A three-phase system is equiv- alent to three single-phase circuits. The voltage sources can be either wye-connected as shown in Fig. 12.6(a) or delta-connected as in Fig. 12.6(b). Let us consider the wye-connected voltages in Fig. 12.6(a) for now. The voltages Van, Vbn, and Vcn are respectively between lines a, b, and
  • 482. 480 PART 2 AC Circuits + − + − + − (a) a V an V bn V cn V ca V ab V bc n b c + − + − (b) a b c − + Figure 12.6 Three-phase voltage sources: (a) Y-connected source, (b) -connected source. c, and the neutral line n. These voltages are called phase voltages. If the voltage sources have the same amplitude and frequency ω and are out of phase with each other by 120◦ , the voltages are said to be balanced. This implies that Van + Vbn + Vcn = 0 (12.1) |Van| = |Vbn| = |Vcn| (12.2) Thus, Balanced phase voltages are equal in magnitude and are out of phase with each other by 120◦ . 120° Vcn Van Vbn 120° −120° (a) 120° Vbn Van Vcn 120° −120° (b) v v Figure12.7 Phase sequences: (a) abc or positive sequence, (b) acb or negative sequence. As a common tradition in power systems, volt- age and current in this chapter are in rms values unless otherwise stated. Since the three-phase voltages are 120◦ out of phase with each other, there are two possible combinations. One possibility is shown in Fig. 12.7(a) and expressed mathematically as Van = Vp 0◦ Vbn = Vp − 120◦ Vcn = Vp − 240◦ = Vp + 120◦ (12.3) where Vp is the effective or rms value. This is known as the abc sequence or positive sequence. In this phase sequence, Van leads Vbn, which in turn leads Vcn. This sequence is produced when the rotor in Fig. 12.4 rotates counterclockwise. The other possibility is shown in Fig. 12.7(b) and is given by Van = Vp 0◦ Vcn = Vp − 120◦ Vbn = Vp − 240◦ = Vp + 120◦ (12.4) This is called the acb sequence or negative sequence. For this phase sequence, Van leads Vcn, which in turn leads Vbn. The acb sequence is produced when the rotor in Fig. 12.4 rotates in the clockwise direction.
  • 483. CHAPTER 12 Three-Phase Circuits 481 It is easy to show that the voltages in Eqs. (12.3) or (12.4) satisfy Eqs. (12.1) and (12.2). For example, from Eq. (12.3), Van + Vbn + Vcn = Vp 0◦ + Vp − 120◦ + Vp + 120◦ = Vp(1.0 − 0.5 − j0.866 − 0.5 + j0.866) = 0 (12.5) The phase sequence is the time order in which the voltages pass through their respective maximum values. The phase sequence may also be regarded as the order in which the phase voltages reach their peak (or maximum) values with respect to time. The phase sequence is determined by the order in which the phasors pass through a fixed point in the phase diagram. Reminder: As time increases, each phasor (or sinor) rotates at an angular velocity ω. In Fig. 12.7(a), as the phasors rotate in the counterclockwise direc- tion with frequency ω, they pass through the horizontal axis in a sequence abcabca . . . . Thus, the sequence is abc or bca or cab. Similarly, for the phasors in Fig. 12.7(b), as they rotate in the counterclockwise direction, they pass the horizontal axis in a sequence acbacba . . . . This describes the acb sequence. The phase sequence is important in three-phase power distribution. It determines the direction of the rotation of a motor con- nected to the power source, for example. Like the generator connections, a three-phase load can be either wye-connected or delta-connected, depending on the end application. Figure 12.8(a) shows a wye-connected load, and Fig. 12.8(b) shows a delta-connected load. The neutral line in Fig. 12.8(a) may or may not be there, depending on whether the system is four- or three-wire. (And, of course, a neutral connection is topologically impossible for a delta connection.) A wye- or delta-connected load is said to be unbalanced if the phase impedances are not equal in magnitude or phase. a b n c (a) Z2 Z1 Z3 a b c (b) Zb Zc Zb Figure12.8 Two possible three- phase load configurations: (a) a Y-connected load, (b) a -connected load A balanced load is one in which the phase impedances are equal in magnitude and in phase. Reminder: A Y-connected load consists of three impedancesconnectedtoaneutralnode,whilea -connected load consists of three impedances connected around a loop. The load is balanced when the three impedances are equal in either case. For a balanced wye-connected load, Z1 = Z2 = Z3 = ZY (12.6) whereZY istheloadimpedanceperphase. Forabalanced delta-connected load, Za = Zb = Zc = Z (12.7) where Z is the load impedance per phase in this case. We recall from Eq. (9.69) that Z = 3ZY or ZY = 1 3 Z (12.8) so we know that a wye-connected load can be transformed into a delta- connected load, or vice versa, using Eq. (12.8). Since both the three-phase source and the three-phase load can be either wye- or delta-connected, we have four possible connections:
  • 484. 482 PART 2 AC Circuits • Y-Y connection (i.e., Y-connected source with a Y-connected load). • Y- connection. • - connection. • -Y connection. In subsequent sections, we will consider each of these possible configu- rations. It is appropriate to mention here that a balanced delta-connected load is more common than a balanced wye-connected load. This is due to the ease with which loads may be added or removed from each phase of a delta-connected load. This is very difficult with a wye-connected load because the neutral may not be accessible. On the other hand, delta- connected sources are not common in practice because of the circulating current that will result in the delta-mesh if the three-phase voltages are slightly unbalanced. E X A M P L E 1 2 . 1 Determine the phase sequence of the set of voltages van = 200 cos(ωt + 10◦ ) vbn = 200 cos(ωt − 230◦ ), vcn = 200 cos(ωt − 110◦ ) Solution: The voltages can be expressed in phasor form as Van = 200 10◦ , Vbn = 200 − 230◦ , Vcn = 200 − 110◦ We notice that Van leads Vcn by 120◦ and Vcn in turn leads Vbn by 120◦ . Hence, we have an acb sequence. P R A C T I C E P R O B L E M 1 2 . 1 Given that Vbn = 110 30◦ , find Van and Vcn, assuming a positive (abc) sequence. Answer: 110 150◦ , 110 − 90◦ . 12.3 BALANCED WYE-WYE CONNECTION We begin with the Y-Y system, because any balanced three-phase system can be reduced to an equivalent Y-Y system. Therefore, analysis of this system should be regarded as the key to solving all balanced three-phase systems. A balanced Y-Y system is a three-phase system with a balanced Y-connected source and a balanced Y-connected load. Consider the balanced four-wire Y-Y system of Fig. 12.9, where a Y-connected load is connected to a Y-connected source. We assume a
  • 485. CHAPTER 12 Three-Phase Circuits 483 balanced load so that load impedances are equal. Although the impedance ZY is the total load impedance per phase, it may also be regarded as the sum of the source impedance Zs, line impedance Z, and load impedance ZL for each phase, since these impedances are in series. As illustrated in Fig. 12.9, Zs denotes the internal impedance of the phase winding of the generator; Z is the impedance of the line joining a phase of the source with a phase of the load; ZL is the impedance of each phase of the load; and Zn is the impedance of the neutral line. Thus, in general ZY = Zs + Z + ZL (12.9) Zs and Z are often very small compared with ZL, so one can assume that ZY = ZL if no source or line impedance is given. In any event, by lumping the impedances together, the Y-Y system in Fig. 12.9 can be simplified to that shown in Fig. 12.10. + − Zl Zl Zl Zn Zs Zs Zs ZL ZL ZL V an V cn V bn A N B C b n c a + − + − Figure12.9 A balanced Y-Y system, showing the source, line, and load impedances. + − ZY ZY ZY V an V cn V bn A N B C b n c a In Ib Ia Ic + − + − Figure12.10 Balanced Y-Y connection. Assuming the positive sequence, the phase voltages (or line-to- neutral voltages) are Van = Vp 0◦ Vbn = Vp − 120◦ , Vcn = Vp + 120◦ (12.10) The line-to-line voltages or simply line voltages Vab, Vbc, and Vca are related to the phase voltages. For example, Vab = Van + Vnb = Van − Vbn = Vp 0◦ − Vp − 120◦ = Vp 1 + 1 2 + j √ 3 2 = √ 3Vp 30◦ (12.11a) Similarly, we can obtain Vbc = Vbn − Vcn = √ 3Vp − 90◦ (12.11b) Vca = Vcn − Van = √ 3Vp − 210◦ (12.11c)
  • 486. 484 PART 2 AC Circuits Thus, the magnitude of the line voltages VL is √ 3 times the magnitude of the phase voltages Vp, or VL = √ 3Vp (12.12) where Vp = |Van| = |Vbn| = |Vcn| (12.13) and VL = |Vab| = |Vbc| = |Vca| (12.14) Also the line voltages lead their corresponding phase voltages by 30◦ . Figure 12.11(a) illustrates this. Figure 12.11(a) also shows how to deter- mine Vab from the phase voltages, while Fig. 12.11(b) shows the same for the three line voltages. Notice that Vab leads Vbc by 120◦ , and Vbc leads Vca by 120◦ , so that the line voltages sum up to zero as do the phase voltages. (a) 30° Vcn Vnb Vab = Van + Vnb Van Vbn (b) Vca Vcn Vab Van Vbc Vbn Figure12.11 Phasor diagrams illustrating the relationship between line voltages and phase voltages. Applying KVL to each phase in Fig. 12.10, we obtain the line cur- rents as Ia = Van ZY , Ib = Vbn ZY = Van − 120◦ ZY = Ia − 120◦ Ic = Vcn ZY = Van − 240◦ ZY = Ia − 240◦ (12.15) We can readily infer that the line currents add up to zero, Ia + Ib + Ic = 0 (12.16) so that In = −(Ia + Ib + Ic) = 0 (12.17a) or VnN = ZnIn = 0 (12.17b) that is, the voltage across the neutral wire is zero. The neutral line can thus be removed without affecting the system. In fact, in long distance power transmission, conductors in multiples of three are used with the earth itself acting as the neutral conductor. Power systems designed in this way are well grounded at all critical points to ensure safety. While the line current is the current in each line, the phase current is the current in each phase of the source or load. In the Y-Y system, the line current is the same as the phase current. We will use single subscripts for line currents because it is natural and conventional to assume that line currents flow from the source to the load. ZY Van + − a A n N Ia Figure 12.12 A single-phase equivalent circuit. An alternative way of analyzing a balanced Y-Y system is to do so on a “per phase” basis. We look at one phase, say phase a, and analyze the single-phase equivalent circuit in Fig. 12.12. The single-phase analysis yields the line current Ia as Ia = Van ZY (12.18)
  • 487. CHAPTER 12 Three-Phase Circuits 485 From Ia, we use the phase sequence to obtain other line currents. Thus, as long as the system is balanced, we need only analyze one phase. We may do this even if the neutral line is absent, as in the three-wire system. E X A M P L E 1 2 . 2 Calculate the line currents in the three-wire Y-Y system of Fig. 12.13. + − 5 – j2Ω 10 + j8Ω 10 + j8Ω A B c b a 5 – j2Ω 10 + j8Ω C 5 – j2Ω 110 −120° V 110 −240° V 110 0° V + − + − Figure12.13 Three-wire Y-Y system; for Example 12.2. Solution: The three-phase circuit in Fig. 12.13 is balanced; we may replace it with its single-phase equivalent circuit such as in Fig. 12.12. We obtain Ia from the single-phase analysis as Ia = Van ZY where ZY = (5 − j2) + (10 + j8) = 15 + j6 = 16.155 21.8◦ . Hence, Ia = 110 0◦ 16.155 21.8◦ = 6.81 − 21.8◦ A Since the source voltages in Fig. 12.13 are in positive sequence and the line currents are also in positive sequence, Ib = Ia − 120◦ = 6.81 − 141.8◦ A Ic = Ia − 240◦ = 6.81 − 261.8◦ A = 6.81 98.2◦ A P R A C T I C E P R O B L E M 1 2 . 2 A Y-connected balanced three-phase generator with an impedance of 0.4+j0.3 per phase is connected to a Y-connected balanced load with an impedance of 24 + j19 per phase. The line joining the generator and the load has an impedance of 0.6 + j0.7 per phase. Assuming a positive sequence for the source voltages and that Van = 120 30◦ V, find: (a) the line voltages, (b) the line currents.
  • 488. 486 PART 2 AC Circuits Answer: (a) 207.85 60◦ V, 207.85 − 60◦ V, 207.85 − 180◦ V, (b) 3.75 − 8.66◦ A, 3.75 − 128.66◦ A, 3.75 − 248.66◦ A. 12.4 BALANCED WYE-DELTA CONNECTION A balanced Y-∆ system consists of a balanced Y-connected source feeding a balanced -connected load. This is perhaps the most practical three-phase system, asthethree-phasesourcesareusuallyY- connected while the three-phase loads are usu- ally -connected. The balanced Y-delta system is shown in Fig. 12.14, where the source is wye-connected and the load is -connected. There is, of course, no neutral connection from source to load for this case. Assuming the positive sequence, the phase voltages are again Van = Vp 0◦ Vbn = Vp − 120◦ , Vcn = Vp + 120◦ (12.19) As shown in Section 12.3, the line voltages are Vab = √ 3Vp 30◦ = VAB, Vbc = √ 3Vp − 90◦ = VBC Vca = √ 3Vp − 210◦ = VCA (12.20) showing that the line voltages are equal to the voltages across the load impedances for this system configuration. From these voltages, we can obtain the phase currents as IAB = VAB Z , IBC = VBC Z , ICA = VCA Z (12.21) These currents have the same magnitude but are out of phase with each other by 120◦ . + − Z∆ Z∆ Z∆ V an V cn V bn IAB ICA A C b c B n a + − + − Ia Ib IBC Ic Figure12.14 Balanced Y- connection.
  • 489. CHAPTER 12 Three-Phase Circuits 487 Another way to get these phase currents is to apply KVL. For ex- ample, applying KVL around loop aABbna gives −Van + Z IAB + Vbn = 0 or IAB = Van − Vbn Z = Vab Z = VAB Z (12.22) which is the same as Eq. (12.21). This is the more general way of finding the phase currents. The line currents are obtained from the phase currents by applying KCL at nodes A, B, and C. Thus, Ia = IAB − ICA, Ib = IBC − IAB, Ic = ICA − IBC (12.23) Since ICA = IAB − 240◦ , Ia = IAB − ICA = IAB(1 − 1 − 240◦ ) = IAB(1 + 0.5 − j0.866) = IAB √ 3 − 30◦ (12.24) showing that the magnitude IL of the line current is √ 3 times the magni- tude Ip of the phase current, or IL = √ 3Ip (12.25) where IL = |Ia| = |Ib| = |Ic| (12.26) and Ip = |IAB| = |IBC| = |ICA| (12.27) Also, the line currents lag the corresponding phase currents by 30◦ , as- suming the positive sequence. Figure 12.15 is a phasor diagram illustrat- ing the relationship between the phase and line currents. 30° 30° 30° ICA IAB Ib IBC Ia Ic Figure12.15 Phasor diagram illustrating the relationship between phase and line currents. An alternative way of analyzing the Y- circuit is to transform the -connected load to an equivalent Y-connected load. Using the -Y transformation formula in Eq. (9.69), ZY = Z 3 (12.28) After this transformation, we now have a Y-Y system as in Fig. 12.10. The three-phase Y- system in Fig. 12.14 can be replaced by the single- phase equivalent circuit in Fig. 12.16. This allows us to calculate only the line currents. The phase currents are obtained using Eq. (12.25) and utilizing the fact that each of the phase currents leads the corresponding line current by 30◦ . V an + − Ia Z∆ 3 Figure12.16 A single-phase equivalent circuit of a balanced Y- circuit. E X A M P L E 1 2 . 3 A balanced abc-sequence Y-connected source with Van = 100 10◦ V is connected to a -connected balanced load (8 + j4) per phase. Cal- culate the phase and line currents.
  • 490. 488 PART 2 AC Circuits Solution: This can be solved in two ways. METHOD 1 The load impedance is Z = 8 + j4 = 8.944 26.57◦ If the phase voltage Van = 100 10◦ , then the line voltage is Vab = Van √ 3 30◦ = 100 √ 3 10◦ + 30◦ = VAB or VAB = 173.2 40◦ V The phase currents are IAB = VAB Z = 173.2 40◦ 8.944 26.57◦ = 19.36 13.43◦ A IBC = IAB − 120◦ = 19.36 − 106.57◦ A ICA = IAB + 120◦ = 19.36 133.43◦ A The line currents are Ia = IAB √ 3 − 30◦ = √ 3(19.36) 13.43◦ − 30◦ = 33.53 − 16.57◦ A Ib = Ia − 120◦ = 33.53 − 136.57◦ A Ic = Ia + 120◦ = 33.53 103.43◦ A METHOD 2 Alternatively, using single-phase analysis, Ia = Van Z /3 = 100 10◦ 2.981 26.57◦ = 33.54 − 16.57◦ A as above. Other line currents are obtained using the abc phase sequence. P R A C T I C E P R O B L E M 1 2 . 3 One line voltage of a balanced Y-connected source is VAB = 180 − 20◦ V. If the source is connected to a -connected load of 20 40◦ , find the phase and line currents. Assume the abc sequence. Answer: 9 − 60◦ , 9 − 180◦ , 9 60◦ , 15.59 − 90◦ , 15.59 − 210◦ , 15.59 30◦ A. 12.5 BALANCED DELTA-DELTA CONNECTION A balanced ∆-∆ system is one in which both the balanced source and balanced load are -connected.
  • 491. CHAPTER 12 Three-Phase Circuits 489 The source as well as the load may be delta-connected as shown in Fig. 12.17. Our goal is to obtain the phase and line currents as usual. Assuming a positive sequence, the phase voltages for a delta-connected source are Vab = Vp 0◦ Vbc = Vp − 120◦ , Vca = Vp + 120◦ (12.29) The line voltages are the same as the phase voltages. From Fig. 12.17, assuming there is no line impedances, the phase voltages of the delta- connected source are equal to the voltages across the impedances; that is, Vab = VAB, Vbc = VBC, Vca = VCA (12.30) Hence, the phase currents are IAB = VAB Z = Vab Z , IBC = VBC Z = Vbc Z ICA = VCA Z = Vca Z (12.31) Since the load is delta-connected just as in the previous section, some of the formulas derived there apply here. The line currents are obtained from the phase currents by applying KCL at nodes A, B, and C, as we did in the previous section: Ia = IAB − ICA, Ib = IBC − IAB, Ic = ICA − IBC (12.32) Also, as shown in the last section, each line current lags the corresponding phase current by 30◦ ; the magnitude IL of the line current is √ 3 times the magnitude Ip of the phase current, IL = √ 3Ip (12.33) Z∆ Z∆ Z∆ V ca V bc V ab IAB ICA A C b c B a + − Ia Ib IBC Ic + − − + Figure12.17 A balanced - connection. An alternative way of analyzing the - circuit is to convert both the source and the load to their Y equivalents. We already know that ZY = Z /3. To convert a -connected source to a Y-connected source, see the next section.
  • 492. 490 PART 2 AC Circuits E X A M P L E 1 2 . 4 A balanced -connected load having an impedance 20 − j15 is con- nected to a -connected, positive-sequence generator having Vab = 330 0◦ V. Calculate the phase currents of the load and the line currents. Solution: The load impedance per phase is Z = 20 − j15 = 25 − 36.87◦ The phase currents are IAB = VAB Z = 330 0◦ 25 − 36.87 = 13.2 36.87◦ A IBC = IAB − 120◦ = 13.2 − 83.13◦ A ICA = IAB + 120◦ = 13.2 156.87◦ A For a delta load, the line current always lags the corresponding phase current by 30◦ and has a magnitude √ 3 times that of the phase current. Hence, the line currents are Ia = IAB √ 3 − 30◦ = (13.2 36.87◦ )( √ 3 − 30◦ ) = 22.86 6.87◦ A Ib = Ia − 120◦ = 22.86 − 113.13◦ A Ic = Ia + 120◦ = 22.86 126.87◦ A P R A C T I C E P R O B L E M 1 2 . 4 A positive-sequence, balanced -connected source supplies a balanced -connected load. If the impedance per phase of the load is 18 + j12 and Ia = 22.5 35◦ A, find IAB and VAB. Answer: 13 65◦ A, 281.2 98.69◦ V. 12.6 BALANCED DELTA-WYE CONNECTION A balanced ∆-Y system consists of a balanced -connected source feeding a balanced Y-connected load. Consider the -Y circuit in Fig. 12.18. Again, assuming the abc sequence, the phase voltages of a delta-connected source are Vab = Vp 0◦ , Vbc = Vp − 120◦ Vca = Vp + 120◦ (12.34) These are also the line voltages as well as the phase voltages.
  • 493. CHAPTER 12 Three-Phase Circuits 491 ZY ZY ZY V ca V bc V ab A C b c B N a + − Ia Ib Ic + − − + Figure12.18 A balanced -Y connection. We can obtain the line currents in many ways. One way is to apply KVL to loop aANBba in Fig. 12.18, writing −Vab + ZY Ia − ZY Ib = 0 or ZY (Ia − Ib) = Vab = Vp 0◦ Thus, Ia − Ib = Vp 0◦ ZY (12.35) But Ib lags Ia by 120◦ , since we assumed the abc sequence; that is, Ib = Ia − 120◦ . Hence, Ia − Ib = Ia(1 − 1 − 120◦ ) = Ia 1 + 1 2 + j √ 3 2 = Ia √ 3 30◦ (12.36) Substituting Eq. (12.36) into Eq. (12.35) gives Ia = Vp/ √ 3 − 30◦ ZY (12.37) From this, we obtain the other line currents Ib and Ic using the positive phase sequence, i.e., Ib = Ia − 120◦ , Ic = Ia + 120◦ . The phase currents are equal to the line currents. V an V ab V ca V bn V bc V cn a c b n + − + − + − − + + − + − Figure12.19 Transforming a -connected source to an equivalent Y-connected source. Another way to obtain the line currents is to replace the delta- connected source with its equivalent wye-connected source, as shown in Fig. 12.19. In Section 12.3, we found that the line-to-line voltages of a wye-connected source lead their corresponding phase voltages by 30◦ . Therefore, we obtain each phase voltage of the equivalent wye-connected source by dividing the corresponding line voltage of the delta-connected source by √ 3 and shifting its phase by −30◦ . Thus, the equivalent wye- connected source has the phase voltages Van = Vp √ 3 − 30◦ Vbn = Vp √ 3 − 150◦ , Vcn = Vp √ 3 + 90◦ (12.38)
  • 494. 492 PART 2 AC Circuits If the delta-connected source has source impedance Zs per phase, the equivalent wye-connected source will have a source impedance of Zs/3 per phase, according to Eq. (9.69). Once the source is transformed to wye, the circuit becomes a wye- wye system. Therefore, we can use the equivalent single-phase circuit shown in Fig. 12.20, from which the line current for phase a is Ia = Vp/ √ 3 − 30◦ ZY (12.39) which is the same as Eq. (12.37). ZY + − Ia Vp −30° √3 Figure 12.20 The single-phase equivalent circuit. Alternatively, we may transform the wye-connected load to an equivalent delta-connected load. This results in a delta-delta system, which can be analyzed as in Section 12.5. Note that VAN = IaZY = Vp √ 3 − 30◦ (12.40) VBN = VAN − 120◦ , VCN = VAN + 120◦ As stated earlier, the delta-connected load is more desirable than the wye-connected load. It is easier to alter the loads in any one phase of the delta-connected loads, as the individual loads are connected directly across the lines. However, the delta-connected source is hardly used in practice, because any slight imbalance in the phase voltages will result in unwanted circulating currents. Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four connections. Students are advised not to memorize the formulas but to understand how they are derived. The formulas can always be obtained by directly applying KCL and KVL to the appropriate three-phase circuits. TABLE 12.1 Summary of phase and line voltages/currents for balanced three-phase systems1 . Connection Phase voltages/currents Line voltages/currents Y-Y Van = Vp 0◦ Vab = √ 3Vp 30◦ Vbn = Vp − 120◦ Vbc = Vab − 120◦ Vcn = Vp + 120◦ Vca = Vab + 120◦ Same as line currents Ia = Van/ZY Ib = Ia − 120◦ Ic = Ia + 120◦ Y- Van = Vp 0◦ Vab = VAB = √ 3Vp 30◦ Vbn = Vp − 120◦ Vbc = VBC = Vab − 120◦ Vcn = Vp + 120◦ Vca = VCA = Vab + 120◦ IAB = VAB /Z Ia = IAB √ 3 − 30◦ IBC = VBC/Z Ib = Ia − 120◦ ICA = VCA/Z Ic = Ia + 120◦ 1Positive or abc sequence is assumed.
  • 495. CHAPTER 12 Three-Phase Circuits 493 TABLE 12.1 (continued) Connection Phase voltages/currents Line voltages/currents - Vab = Vp 0◦ Same as phase voltages Vbc = Vp − 120◦ Vca = Vp + 120◦ IAB = Vab/Z Ia = IAB √ 3 − 30◦ IBC = Vbc/Z Ib = Ia − 120◦ ICA = Vca/Z Ic = Ia + 120◦ -Y Vab = Vp 0◦ Same as phase voltages Vbc = Vp − 120◦ Vca = Vp + 120◦ Same as line currents Ia = Vp − 30◦ √ 3ZY Ib = Ia − 120◦ Ic = Ia + 120◦ E X A M P L E 1 2 . 5 A balanced Y-connected load with a phase resistance of 40 and a reac- tance of 25 is supplied by a balanced, positive sequence -connected source with a line voltage of 210 V. Calculate the phase currents. Use Vab as reference. Solution: The load impedance is ZY = 40 + j25 = 47.17 32◦ and the source voltage is Vab = 210 0◦ V When the -connected source is transformed to a Y-connected source, Van = Vab √ 3 − 30◦ = 121.2 − 30◦ V The line currents are Ia = Van ZY = 121.2 − 30◦ 47.12 32◦ = 2.57 − 62◦ A Ib = Ia − 120◦ = 2.57 − 182◦ A Ic = Ia 120◦ = 2.57 58◦ A which are the same as the phase currents.
  • 496. 494 PART 2 AC Circuits P R A C T I C E P R O B L E M 1 2 . 5 In a balanced -Y circuit, Vab = 240 15◦ and ZY = (12 + j15) . Calculate the line currents. Answer: 7.21 − 66.34◦ , 7.21 − 186.34◦ , 7.21 53.66◦ A. 12.7 POWER IN A BALANCED SYSTEM Let us now consider the power in a balanced three-phase system. We begin by examining the instantaneous power absorbed by the load. This requires that the analysis be done in the time domain. For a Y-connected load, the phase voltages are vAN = √ 2Vp cos ωt, vBN = √ 2Vp cos(ωt − 120◦ ) vCN = √ 2Vp cos(ωt + 120◦ ) (12.41) where the factor √ 2 is necessary because Vp has been defined as the rms value of the phase voltage. If ZY = Z θ, the phase currents lag behind their corresponding phase voltages by θ. Thus, ia = √ 2Ip cos(ωt − θ), ib = √ 2Ip cos(ωt − θ − 120◦ ) ic = √ 2Ip cos(ωt − θ + 120◦ ) (12.42) where Ip is the rms value of the phase current. The total instantaneous power in the load is the sum of the instantaneous powers in the three phases; that is, p = pa + pb + pc = vAN ia + vBN ib + vCN ic = 2VpIp[cos ωt cos(ωt − θ) + cos(ωt − 120◦ ) cos(ωt − θ − 120◦ ) + cos(ωt + 120◦ ) cos(ωt − θ + 120◦ )] (12.43) Applying the trigonometric identity cos A cos B = 1 2 [cos(A + B) + cos(A − B)] (12.44) gives p = VpIp[3 cos θ + cos(2ωt − θ) + cos(2ωt − θ − 240◦ ) + cos(2ωt − θ + 240◦ )] = VpIp[3 cos θ + cos α + cos α cos 240◦ + sin α sin 240◦ + cos α cos 240◦ − sin α sin 240◦ ] where α = 2ωt − θ = VpIp 3 cos θ + cos α + 2 − 1 2 cos α = 3VpIp cos θ (12.45) Thus the total instantaneous power in a balanced three-phase system is constant—it does not change with time as the instantaneous power of each phase does. This result is true whether the load is Y- or -connected.
  • 497. CHAPTER 12 Three-Phase Circuits 495 This is one important reason for using a three-phase system to generate and distribute power. We will look into another reason a little later. Since the total instantaneous power is independent of time, the average power per phase Pp for either the -connected load or the Y- connected load is p/3, or Pp = VpIp cos θ (12.46) and the reactive power per phase is Qp = VpIp sin θ (12.47) The apparent power per phase is Sp = VpIp (12.48) The complex power per phase is Sp = Pp + jQp = VpI∗ p (12.49) where Vp and Ip are the phase voltage and phase current with magnitudes Vp and Ip, respectively. The total average power is the sum of the average powers in the phases: P = Pa + Pb + Pc = 3Pp = 3VpIp cos θ = √ 3VLIL cos θ (12.50) For a Y-connected load, IL = Ip but VL = √ 3Vp, whereas for a - connected load, IL = √ 3Ip but VL = Vp. Thus, Eq. (12.50) applies for both Y-connected and -connected loads. Similarly, the total reactive power is Q = 3VpIp sin θ = 3Qp = √ 3VLIL sin θ (12.51) and the total complex power is S = 3Sp = 3VpI∗ p = 3I2 pZp = 3V 2 p Z∗ p (12.52) where Zp = Zp θ is the load impedance per phase. (Zp could be ZY or Z .) Alternatively, we may write Eq. (12.52) as S = P + jQ = √ 3VLIL θ (12.53) Remember that Vp, Ip, VL, and IL are all rms values and that θ is the angle of the load impedance or the angle between the phase voltage and the phase current. A second major advantage of three-phase systems for power dis- tribution is that the three-phase system uses a lesser amount of wire than the single-phase system for the same line voltage VL and the same ab- sorbed power PL. We will compare these cases and assume in both that the wires are of the same material (e.g., copper with resistivity ρ), of the same length , and that the loads are resistive (i.e., unity power factor). For the two-wire single-phase system in Fig. 12.21(a), IL = PL/VL, so the power loss in the two wires is Ploss = 2I2 LR = 2R P 2 L V 2 L (12.54)
  • 498. 496 PART 2 AC Circuits Single- phase source R R Transmission lines (a) PL IL Load − + VL Three- phase balanced load Three- phase balanced source R′ R′ R′ Transmission lines (b) Ia Ib Ic − + − + VL VL −120° 0° Figure12.21 Comparing the power loss in (a) a single-phase system, and (b) a three-phase system. For the three-wire three-phase system in Fig. 12.21(b), I L = |Ia| = |Ib| = |Ic| = PL/ √ 3VL from Eq. (12.50). The power loss in the three wires is P loss = 3(I L)2 R = 3R P 2 L 3V 2 L = R P 2 L V 2 L (12.55) Equations(12.54)and(12.55)showthatforthesametotalpowerdelivered PL and same line voltage VL, Ploss P loss = 2R R (12.56) But from Chapter 2, R = ρ/πr2 and R = ρ/πr2 , where r and r are the radii of the wires. Thus, Ploss P loss = 2r2 r2 (12.57) If the same power loss is tolerated in both systems, then r2 = 2r2 . The ratio of material required is determined by the number of wires and their volumes, so Material for single-phase Material for three-phase = 2(πr2 ) 3(πr2) = 2r2 3r2 = 2 3 (2) = 1.333 (12.58) sincer2 = 2r2 . Equation(12.58)showsthatthesingle-phasesystemuses 33 percent more material than the three-phase system or that the three- phase system uses only 75 percent of the material used in the equivalent single-phase system. In other words, considerably less material is needed to deliver the same power with a three-phase system than is required for a single-phase system. E X A M P L E 1 2 . 6 Refer to the circuit in Fig. 12.13 (in Example 12.2). Determine the total average power, reactive power, and complex power at the source and at the load.
  • 499. CHAPTER 12 Three-Phase Circuits 497 Solution: It is sufficient to consider one phase, as the system is balanced. For phase a, Vp = 110 0◦ V and Ip = 6.81 − 21.8◦ A Thus, at the source, the complex power supplied is Ss = −3VpI∗ p = 3(110 0◦ )(6.81 21.8◦ ) = −2247 21.8◦ = −(2087 + j834.6) VA The real or average power supplied is −2087 W and the reactive power is −834.6 VAR. At the load, the complex power absorbed is SL = 3|Ip|2 Zp where Zp = 10 + j8 = 12.81 38.66◦ and Ip = Ia = 6.81 − 21.8◦ . Hence SL = 3(6.81)2 12.81 38.66◦ = 1782 38.66 = (1392 + j1113) VA The real power absorbed is 1391.7 W and the reactive power absorbed is 1113.3 VAR. The difference between the two complex powers is absorbed by the line impedance (5 − j2) . To show that this is the case, we find the complex power absorbed by the line as S = 3|Ip|2 Z = 3(6.81)2 (5 − j2) = 695.6 − j278.3 VA which is the difference between Ss and SL, that is, Ss + S + SL = 0, as expected. P R A C T I C E P R O B L E M 1 2 . 6 For the Y-Y circuit in Practice Prob. 12.2, calculate the complex power at the source and at the load. Answer: (1054 + j843.3) VA, (1012 + j801.6) VA. E X A M P L E 1 2 . 7 A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor. Solution: The apparent power is S = √ 3VLIL = √ 3(220)(18.2) = 6935.13 VA Since the real power is P = S cos θ = 5600 W
  • 500. 498 PART 2 AC Circuits the power factor is pf = cos θ = P S = 5600 6935.13 = 0.8075 P R A C T I C E P R O B L E M 1 2 . 7 Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced source with a line voltage of 440 V. Answer: 50.94 A. E X A M P L E 1 2 . 8 Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown in Fig. 12.22(a). Load 1 draws 30 kW at a power factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging. Assuming the abc sequence, determine: (a) the complex, real, and reactive powers absorbed by the combined load, (b) the line currents, and (c) the kVAR rating of the three capacitors -connected in parallel with the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor. (a) C C C Balanced load 1 Balanced load 2 (b) Combined load Figure12.22 For Example 12.8: (a) The original balanced loads, (b) the combined load with improved power factor. Solution: (a) For load 1, given that P1 = 30 kW and cos θ1 = 0.6, then sin θ1 = 0.8. Hence, S1 = P1 cos θ1 = 30 kW 0.6 = 50 kVA and Q1 = S1 sin θ1 = 50(0.8) = 40 kVAR. Thus, the complex power due to load 1 is S1 = P1 + jQ1 = 30 + j40 kVA (12.8.1) For load 2, if Q2 = 45 kVAR and cos θ2 = 0.8, then sin θ2 = 0.6. We find S2 = Q2 sin θ2 = 45 kVA 0.6 = 75 kVA and P2 = S2 cos θ2 = 75(0.8) = 60 kW. Therefore the complex power due to load 2 is S2 = P2 + jQ2 = 60 + j45 kVA (12.8.2) From Eqs. (12.8.1) and (12.8.2), the total complex power absorbed by the load is S = S1 + S2 = 90 + j85 kVA = 123.8 43.36◦ kVA (12.8.3) which has a power factor of cos 43.36◦ = 0.727 lagging. The real power is then 90 kW, while the reactive power is 85 kVAR. (b) Since S = √ 3VLIL, the line current is IL = S √ 3VL (12.8.4)
  • 501. CHAPTER 12 Three-Phase Circuits 499 We apply this to each load, keeping in mind that for both loads, VL = 240 kV. For load 1, IL1 = 50,000 √ 3 240,000 = 120.28 mA Since the power factor is lagging, the line current lags the line voltage by θ1 = cos−1 0.6 = 53.13◦ . Thus, Ia1 = 120.28 − 53.13◦ For load 2, IL2 = 75,000 √ 3 240,000 = 180.42 mA and the line current lags the line voltage by θ2 = cos−1 0.8 = 36.87◦ . Hence, Ia2 = 180.42 − 36.87◦ The total line current is Ia = Ia1 + Ia2 = 120.28 − 53.13◦ + 180.42 − 36.87◦ = (72.168 − j96.224) + (144.336 − j108.252) = 216.5 − j204.472 = 297.8 − 43.36◦ mA Alternatively, we could obtain the current from the total complex power using Eq. (12.8.4) as IL = 123,800 √ 3 240,000 = 297.82 mA and Ia = 297.82 − 43.36◦ mA which is the same as before. The other line currents, Ib2 and Ica, can be obtainedaccordingtotheabc sequence(i.e., Ib = 297.82 −163.36◦ mA and Ic = 297.82 76.64◦ mA). (c) We can find the reactive power needed to bring the power factor to 0.9 lagging using Eq. (11.59), QC = P(tan θold − tan θnew) where P = 90 kW, θold = 43.36◦ , and θnew = cos−1 0.9 = 25.84◦ . Hence, QC = 90,000(tan 43.36◦ − tan 25.04◦ ) = 41.4 kVAR This reactive power is for the three capacitors. For each capacitor, the rating Q C = 13.8 kVAR. From Eq. (11.60), the required capacitance is C = Q C ωV 2 rms Since the capacitors are -connected as shown in Fig. 12.22(b), Vrms in the above formula is the line-to-line or line voltage, which is 240 kV. Thus, C = 13,800 (2π60)(240,000)2 = 635.5 pF
  • 502. 500 PART 2 AC Circuits P R A C T I C E P R O B L E M 1 2 . 8 Assume that the two balanced loads in Fig. 12.22(a) are supplied by an 840-V rms 60-Hz line. Load 1 is Y-connected with 30+j40 per phase, while load 2 is a balanced three-phase motor drawing 48 kW at a power factor of 0.8 lagging. Assuming the abc sequence, calculate: (a) the complex power absorbed by the combined load, (b) the kVAR rating of each of the three capacitors -connected in parallel with the load to raise the power factor to unity, and (c) the current drawn from the supply at unity power factor condition. Answer: (a) 56.47 + j47.29 kVA, (b) 15.7 kVAR, (c) 38.813 A. †12.8 UNBALANCED THREE-PHASE SYSTEMS This chapter would be incomplete without mentioning unbalanced three- phase systems. An unbalanced system is caused by two possible situa- tions: (1) the source voltages are not equal in magnitude and/or differ in phase by angles that are unequal, or (2) load impedances are unequal. Thus, An unbalanced system is due to unbalanced voltage sources or an unbalanced load. To simplify analysis, we will assume balanced source voltages, but an unbalanced load. A special technique for handling unbalanced three-phase systems is the method of symmet- rical components, which is beyond the scope of this text. Unbalanced three-phase systems are solved by direct application of mesh and nodal analysis. Figure 12.23 shows an example of an unbal- anced three-phase system that consists of balanced source voltages (not shown in the figure) and an unbalanced Y-connected load (shown in the figure). Since the load is unbalanced, ZA, ZB, and ZC are not equal. The line currents are determined by Ohm’s law as Ia = VAN ZA , Ib = VBN ZB , Ic = VCN ZC (12.59) This set of unbalanced line currents produces current in the neutral line, which is not zero as in a balanced system. Applying KCL at node N gives the neutral line current as In = −(Ia + Ib + Ic) (12.60) ZA ZC ZB A N C B Ia In Ib Ic V AN V BN V CN Figure 12.23 Unbalanced three-phase Y-connected load. In a three-wire system where the neutral line is absent, we can still find the line currents Ia, Ib, and Ic using mesh analysis. At node N, KCL must be satisfied so that Ia + Ib + Ic = 0 in this case. The same could be done for a -Y, Y- , or - three-wire system. As mentioned earlier, in long distance power transmission, conductors in multiples of three (multiple three-wire systems) are used, with the earth itself acting as the neutral conductor.
  • 503. CHAPTER 12 Three-Phase Circuits 501 To calculate power in an unbalanced three-phase system requires that we find the power in each phase using Eqs. (12.46) to (12.49). The total power is not simply three times the power in one phase but the sum of the powers in the three phases. E X A M P L E 1 2 . 9 The unbalanced Y-load of Fig. 12.23 has balanced voltages of 100 V and the acb sequence. Calculate the line currents and the neutral current. Take ZA = 15 , ZB = 10 + j5 , ZC = 6 − j8 . Solution: Using Eq. (12.59), the line currents are Ia = 100 0◦ 15 = 6.67 0◦ A Ib = 100 120◦ 10 + j5 = 100 120◦ 11.18 26.56◦ = 8.94 93.44◦ A Ic = 100 − 120◦ 6 − j8 = 100 − 120◦ 10 − 53.13◦ = 10 − 66.87◦ A Using Eq. (12.60), the current in the neutral line is In = −(Ia + Ib + Ic) = −(6.67 − 0.54 + j8.92 + 3.93 − j9.2) = −10.06 + j0.28 = 10.06 178.4◦ A P R A C T I C E P R O B L E M 1 2 . 9 The unbalanced -load of Fig. 12.24 is supplied by balanced voltages of 200 V in the positive sequence. Find the line currents. Take Vab as reference. 16 Ω 8 Ω j6 Ω 10 Ω –j5 Ω A C B Ia Ib Ic Figure 12.24 Unbalanced -load; for Practice Prob. 12.9. Answer: 18.05 − 41.06◦ , 29.15 220.2◦ , 31.87 74.27◦ A. E X A M P L E 1 2 . 1 0 For the unbalanced circuit in Fig. 12.25, find: (a) the line currents, (b)thetotalcomplexpowerabsorbedbytheload, and(c)thetotalcomplex power supplied by the source. Solution: (a) We use mesh analysis to find the required currents. For mesh 1, 120 − 120◦ − 120 0◦ + (10 + j5)I1 − 10I2 = 0
  • 504. 502 PART 2 AC Circuits + − j5 Ω A N 10 Ω –j10 Ω C b B n c a Ib Ic Ia 120 0° rms 120 120° rms 120 −120° rms + − − + I2 I1 Figure12.25 For Example 12.10. or (10 + j5)I1 − 10I2 = 120 √ 3 30◦ (12.10.1) For mesh 2, 120 120◦ − 120 − 120◦ + (10 − j10)I2 − 10I1 = 0 or −10I1 + (10 − j10)I2 = 120 √ 3 − 90◦ (12.10.2) Equations (12.10.1) and (12.10.2) form a matrix equation: 10 + j5 −10 −10 10 − j10 I1 I2 = 120 √ 3 30◦ 120 √ 3 − 90◦ The determinants are = 10 + j5 −10 −10 10 − j10 = 50 − j50 = 70.71 − 45◦ 1 = 120 √ 3 30◦ −10 120 √ 3 − 90◦ 10 − j10 = 207.85(13.66 − j13.66) = 4015 − 45◦ 2 = 10 + j5 120 √ 3 30◦ −10 120 √ 3 − 90◦ = 207.85(13.66 − j5) = 3023 − 20.1◦
  • 505. CHAPTER 12 Three-Phase Circuits 503 The mesh currents are I1 = 1 = 4015.23 − 45◦ 70.71 − 45◦ = 56.78 A I2 = 2 = 3023.4 − 20.1◦ 70.71 − 45◦ = 42.75 24.9◦ A The line currents are Ia = I1 = 56.78 A, Ic = −I2 = 42.75 − 155.1◦ A Ib = I2 − I1 = 38.78 + j18 − 56.78 = 25.46 135◦ A (b) We can now calculate the complex power absorbed by the load. For phase A, SA = |Ia|2 ZA = (56.78)2 (j5) = j16,120 VA For phase B, SB = |Ib|2 ZB = (25.46)2 (10) = 6480 VA For phase C, SC = |Ic|2 ZC = (42.75)2 (−j10) = −j18,276 VA The total complex power absorbed by the load is SL = SA + SB + SC = 6480 − j2156 VA (c) We check the result above by finding the power supplied by the source. For the voltage source in phase a, Sa = −VanI∗ a = −(120 0◦ )(56.78) = −6813.6 VA For the source in phase b, Sb = −VbnI∗ b = −(120 − 120◦ )(25.46 − 135◦ ) = −3055.2 105◦ = 790 − j2951.1 VA For the source in phase c, Sc = −VbnI∗ c = −(120 120◦ )(42.75 155.1◦ ) = −5130 275.1◦ = −456.03 + j5109.7 VA The total complex power supplied by the three-phase source is Ss = Sa + Sb + Sc = −6480 + j2156 VA showing that Ss + SL = 0 and confirming the conservation principle of ac power. P R A C T I C E P R O B L E M 1 2 . 1 0 Find the line currents in the unbalanced three-phase circuit of Fig. 12.26 and the real power absorbed by the load.
  • 506. 504 PART 2 AC Circuits 10 Ω A j10 Ω −j5 Ω C b B c a 220 −0° rms V 220 120° rms V + − + − − + 220 −120° rms V Figure12.26 For Practice Prob. 12.10. Answer: 64 80.1◦ , 38.1 − 60◦ , 42.5 225◦ A, 4.84 kW. 12.9 PSPICE FOR THREE-PHASE CIRCUITS PSpice can be used to analyze three-phase balanced or unbalanced circuits in the same way it is used to analyze single-phase ac circuits. However, a delta-connected source presents two major problems to PSpice. First, a delta-connected source is a loop of voltage sources—which PSpice does not like. To avoid this problem, we insert a resistor of negligible resistance (say, 1 µ per phase) into each phase of the delta-connected source. Second, the delta-connected source does not provide a convenient node for the ground node, which is necessary to run PSpice. This problem can be eliminated by inserting balanced wye-connected large resistors (say, 1 M per phase) in the delta-connected source so that the neutral node of the wye-connected resistors serves as the ground node 0. Example 12.12 will illustrate this. E X A M P L E 1 2 . 1 1 For the balanced Y- circuit in Fig. 12.27, use PSpice to find the line cur- rent IaA, the phase voltage VAB, and the phase current IAC. Assume that the source frequency is 60 Hz. 100 0° V a n A C B 100 Ω 100 Ω 1 Ω 0.2 H 0.2 H 100 Ω 0.2 H − + 100 −120° V b 1 Ω − + 100 120° V c 1 Ω − + Figure12.27 For Example 12.10.
  • 507. CHAPTER 12 Three-Phase Circuits 505 Solution: The schematic is shown in Fig. 12.28. The pseudocomponents IPRINT are inserted in the appropriate lines to obtain IaA and IAC, while VPRINT2 is inserted between nodes A and B to print differential voltage VAB. We set the attributes of IPRINT and VPRINT2 each to AC = yes, MAG = yes, PHASE = yes, to print only the magnitude and phase of the currents and voltages. As a single-frequency analysis, we select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq = 60, and Final Freq = 60. Once the circuit is saved, it is simulated by selecting Analysis/Simulate. The output file includes the following: FREQ V(A,B) VP(A,B) 6.000E+01 1.699E+02 3.081E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 6.000E+01 2.350E+00 -3.620E+01 FREQ IM(V_PRINT3) IP(V_PRINT3) 6.000E+01 1.357E+00 -6.620E+01 From this, we obtain IaA = 2.35 − 36.2◦ A VAB = 169.9 30.81◦ V, IAC = 1.357 − 66.2◦ A A B C R4 R6 100 0.2H L1 AC = yes MAG = yes PHASE = yes AC = yes MAG = yes PHASE = yes AC = yes MAG = yes PHASE = yes ACMAG = 100 V ACPHASE = 0 1 IPRINT IPRINT ACMAG = 100 ACPHASE = −120 1 R2 R1 V2 V1 ACMAG = 100 V ACPHASE = 120 1 R3 V3 R5 100 0.2H 0.2H L3 L2 100 − + − + − + 0 Figure12.28 Schematic for the circuit in Fig. 12.27. P R A C T I C E P R O B L E M 1 2 . 1 1 Refer to the balanced Y-Y circuit of Fig. 12.29. Use PSpice to find the line current IbB and the phase voltage VAN . Take f = 100 Hz.
  • 508. 506 PART 2 AC Circuits 120 60° V a n A C N 10 Ω 2 Ω 10 mH − + 120 −60° V b 2 Ω − + 120 180° V c 2 Ω 1.6 mH 1.6 mH 1.6 mH − + 10 Ω 10 mH 10 Ω 10 mH B Figure12.29 For Practice Prob. 12.11. Answer: 100.9 60.87◦ V, 8.547 − 91.27◦ A. E X A M P L E 1 2 . 1 2 Consider the unbalanced - circuit in Fig. 12.30. Use PSpice to find the generator current Iab, the line current IbB, and the phase current IBC. 208 130° V 208 −110° V 208 10° V A B b C c a 50 Ω 2 Ω j30 Ω j5 Ω −j40 Ω 2 Ω j5 Ω 2 Ω j5 Ω + − + − + − Figure12.30 For Example 12.12. Solution: As mentioned above, we avoid the loop of voltage sources by inserting a 1-µ series resistor in the delta-connected source. To provide a ground node 0, we insert balanced wye-connected resistors (1 M per phase) in the delta-connected source, as shown in the schematic in Fig. 12.31. Three IPRINT pseudocomponents with their attributes are inserted to be able to get the required currents Iab, IbB, and IBC. Since the operating frequency is not given and the inductances and capacitances should be specified instead of impedances, we assume ω = 1 rad/s so that f = 1/2π = 0.159155 Hz. Thus,
  • 509. CHAPTER 12 Three-Phase Circuits 507 L = XL ω and C = 1 ωXC We select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq = 0.159155, and Final Freq = 0.159155. Once the schematic is saved, we select Analysis/Simulate to simulate the circuit. The output file includes: FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592E-01 9.106E+00 1.685E+02 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592E-01 5.959E+00 2.821E+00 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592E-01 5.500E+00 -7.532E+00 from which we get Iab = 5.96 2.82◦ A IbB = 9.106 168.5◦ A, IBC = 5.5 − 7.53◦ A IPRINT ACMAG = 208V ACPHASE = 130 ACMAG = 208V ACPHASE = -110 V1 V2 30H L4 25m C1 R1 2 1Meg R8 L1 5H R2 2 L2 5H R3 2 L3 5H R5 1Meg R4 1u R6 1Meg R7 1u R9 1u − + − + ACMAG = 208V ACPHASE = 110 V3 − + IPRINT R10 50 AC = yes MAG = yes PHASE = yes AC = yes MAG = yes PHASE = yes AC = yes MAG = yes PHASE = yes IPRINT 0 Figure12.31 Schematic for the circuit in Fig. 12.30. P R A C T I C E P R O B L E M 1 2 . 1 2 For the unbalanced circuit in Fig. 12.32, use PSpice to find the generator current Ica, the line current IcC, and the phase current IAB.
  • 510. 508 PART 2 AC Circuits 220 90° V 220 −150° V 220 −30° V A B b C c a 10 Ω j10 Ω 10 Ω + − + − + − 10 Ω −j10 Ω Figure12.32 For Practice Prob. 12.12. Answer: 24.68 − 90◦ A, 15.56 105◦ A, 37.24 83.79◦ A. †12.10 APPLICATIONS Both wye and delta source connections have important practical applica- tions. The wye source connection is used for long distance transmission of electric power, where resistive losses (I2 R) should be minimal. This is due to the fact that the wye connection gives a line voltage that is √ 3 greater than the delta connection; hence, for the same power, the line current is √ 3 smaller. The delta source connection is used when three single-phase circuits are desired from a three-phase source. This conver- sion from three-phase to single-phase is required in residential wiring, be- cause household lighting and appliances use single-phase power. Three- phase power is used in industrial wiring where a large power is required. In some applications, it is immaterial whether the load is wye- or delta- connected. For example, both connections are satisfactory with induction motors. In fact, some manufacturers connect a motor in delta for 220 V and in wye for 440 V so that one line of motors can be readily adapted to two different voltages. Here we consider two practical applications of those concepts cov- ered in this chapter: power measurement in three-phase circuits and res- idential wiring. Three-phase load (wye or delta, balanced or unbalanced) W1 a b c W3 ± ± W2 ± ± ± ± o Figure 12.33 Three-wattmeter method for measuring three-phase power. 12.10.1 Three-Phase Power Measurement Section 11.9 presented the wattmeter as the instrument for measuring the average (or real) power in single-phase circuits. A single wattmeter can also measure the average power in a three-phase system that is balanced, so that P1 = P2 = P3; the total power is three times the reading of that one wattmeter. However, two or three single-phase wattmeters are neces- sary to measure power if the system is unbalanced. The three-wattmeter method of power measurement, shown in Fig. 12.33, will work regardless of whether the load is balanced or unbalanced, wye- or delta-connected.
  • 511. CHAPTER 12 Three-Phase Circuits 509 The three-wattmeter method is well suited for power measurement in a three-phase system where the power factor is constantly changing. The total average power is the algebraic sum of the three wattmeter readings, PT = P1 + P2 + P3 (12.61) where P1, P2, and P3 correspond to the readings of wattmeters W1, W2, and W3, respectively. Notice that the common or reference point o in Fig. 12.33 is selected arbitrarily. If the load is wye-connected, point o can be connected to the neutral point n. For a delta-connected load, point o can be connected to any point. If point o is connected to point b, for example, the voltage coil in wattmeter W2 reads zero and P2 = 0, indicating that wattmeter W2 is not necessary. Thus, two wattmeters are sufficient to measure the total power. The two-wattmeter method is the most commonly used method for three-phase power measurement. The two wattmeters must be properly connected to any two phases, as shown typically in Fig. 12.34. Notice that the current coil of each wattmeter measures the line current, while the respective voltage coil is connected between the line and the third line and measures the line voltage. Also notice that the ± terminal of the voltage coil is connected to the line to which the corresponding current coil is connected. Although the individual wattmeters no longer read the power taken by any particular phase, the algebraic sum of the two wattmeter readingsequalsthetotalaveragepowerabsorbedbytheload, regardlessof whether it is wye- or delta-connected, balanced or unbalanced. The total real power is equal to the algebraic sum of the two wattmeter readings, PT = P1 + P2 (12.62) We will show here that the method works for a balanced three-phase system. Three-phase load (wye or delta, balanced or unbalanced) W1 a b c W2 ± ± ± ± Figure 12.34 Two-wattmeter method for measuring three-phase power. Consider the balanced, wye-connected load in Fig. 12.35. Our objective is to apply the two-wattmeter method to find the average power absorbed by the load. Assume the source is in the abc sequence and the load impedance ZY = ZY θ. Due to the load impedance, each voltage coil leads its current coil by θ, so that the power factor is cos θ. We recall that each line voltage leads the corresponding phase voltage by 30◦ . Thus, the total phase difference between the phase current Ia and line voltage W1 a b c W2 ± ± ± ± Ib Ic Ia ZY ZY ZY + − V ab − + V cb Figure12.35 Two-wattmeter method applied to a balanced wye load.
  • 512. 510 PART 2 AC Circuits Vab is θ + 30◦ , and the average power read by wattmeter W1 is P1 = Re[VabI∗ a] = VabIa cos(θ + 30◦ ) = VLIL cos(θ + 30◦ ) (12.63) Similarly, we can show that the average power read by wattmeter 2 is P2 = Re[VcbI∗ c] = VcbIc cos(θ − 30◦ ) = VLIL cos(θ − 30◦ ) (12.64) We now use the trigonometric identities cos(A + B) = cos A cos B − sin A sin B cos(A − B) = cos A cos B + sin A sin B (12.65) to find the sum and the difference of the two wattmeter readings in Eqs. (12.63) and (12.64): P1 + P2 = VLIL[cos(θ + 30◦ ) + cos(θ − 30◦ )] = VlIL(cos θ cos 30◦ − sin θ sin 30◦ + cos θ cos 30◦ + sin θ sin 30◦ ) = VLIL2 cos 30◦ cos θ = √ 3VLIL cos θ (12.66) since 2 cos 30◦ = √ 3. Comparing Eq. (12.66) with Eq. (12.50) shows that the sum of the wattmeter readings gives the total average power, PT = P1 + P2 (12.67) Similarly, P1 − P2 = VLIL[cos(θ + 30◦ ) − cos(θ − 30◦ )] = VlIL(cos θ cos 30◦ − sin θ sin 30◦ − cos θ cos 30◦ − sin θ sin 30◦ ) = −VLIL2 sin 30◦ sin θ P2 − P1 = VLIL sin θ (12.68) since 2 sin 30◦ = 1. Comparing Eq. (12.68) with Eq. (12.51) shows that thedifferenceofthewattmeterreadingsisproportionaltothetotalreactive power, or QT = √ 3(P2 − P1) (12.69) From Eqs. (12.67) and (12.69), the total apparent power can be obtained as ST = P 2 T + Q2 T (12.70) Dividing Eq. (12.69) by Eq. (12.67) gives the tangent of the power factor angle as tan θ = QT PT = √ 3 P2 − P1 P2 + P1 (12.71) from which we can obtain the power factor as pf = cos θ. Thus, the two- wattmeter method not only provides the total real and reactive powers, it can also be used to compute the power factor. From Eqs. (12.67), (12.69), and (12.71), we conclude that:
  • 513. CHAPTER 12 Three-Phase Circuits 511 1. If P2 = P1, the load is resistive. 2. If P2 P1, the load is inductive. 3. If P2 P1, the load is capacitive. Although these results are derived from a balanced wye-connected load, they are equally valid for a balanced delta-connected load. However, the two-wattmeter method cannot be used for power measurement in a three-phase four-wire system unless the current through the neutral line is zero. We use the three-wattmeter method to measure the real power in a three-phase four-wire system. E X A M P L E 1 2 . 1 3 Three wattmeters W1, W2, and W3 are connected, respectively, to phases a, b, and c to measure the total power absorbed by the unbalanced wye- connectedloadinExample12.9(seeFig.12.23). (a)Predictthewattmeter readings. (b) Find the total power absorbed. Solution: Part of the problem is already solved in Example 12.9. Assume that the wattmeters are properly connected as in Fig. 12.36. − + V CN − + V AN − + V BN W3 Ia Ib Ic In W1 A N C B W2 j5 Ω −j8 Ω 10 Ω 6 Ω 15 Ω Figure12.36 For Example 12.13. (a) From Example 12.9, VAN = 100 0◦ , VBN = 100 120◦ , VCN = 100 − 120◦ V while Ia = 6.67 0◦ , Ib = 8.94 93.44◦ , Ic = 10 − 66.87◦ A We calculate the wattmeter readings as follows: P1 = Re(VAN I∗ a) = VAN Ia cos(θVAN − θIa ) = 100 × 6.67 × cos(0◦ − 0◦ ) = 667 W P2 = Re(VBN I∗ b) = VBN Ib cos(θVBN − θIb ) = 100 × 8.94 × cos(120◦ − 93.44◦ ) = 800 W P3 = Re(VCN I∗ c) = VCN Ic cos(θVCN − θIc ) = 100 × 10 × cos(−120◦ + 66.87◦ ) = 600 W
  • 514. 512 PART 2 AC Circuits (b) The total power absorbed is PT = P1 + P2 + P3 = 667 + 800 + 600 = 2067 W We can find the power absorbed by the resistors in Fig. 12.36 and use that to check or confirm this result. PT = |Ia|2 (15) + |Ib|2 (10) + |Ic|2 (6) = 6.672 (15) + 8.942 (10) + 102 (6) = 667 + 800 + 600 = 2067 W which is exactly the same thing. P R A C T I C E P R O B L E M 1 2 . 1 3 Repeat Example 12.13 for the network in Fig. 12.24 (see Practice Prob. 12.9). Hint: Connect the reference point o in Fig. 12.33 to point B. Answer: (a) 2961 W, 0 W, 4339 W, (b) 7300 W. E X A M P L E 1 2 . 1 4 The two-wattmeter method produces wattmeter readings P1 = 1560 W and P2 = 2100 W when connected to a delta-connected load. If the line voltage is 220 V, calculate: (a) the per-phase average power, (b) the per- phase reactive power, (c) the power factor, and (d) the phase impedance. Solution: We can apply the given results to the delta-connected load. (a) The total real or average power is PT = P1 + P2 = 1560 + 2100 = 3660 W The per-phase average power is then Pp = 1 3 PT = 1220 W (b) The total reactive power is QT = √ 3(P2 − P1) = √ 3(2100 − 1560) = 935.3 VAR so that the per-phase reactive power is Qp = 1 3 QT = 311.77 VAR (c) The power angle is θ = tan−1 QT PT = tan−1 935.3 3660 = 14.33◦ Hence, the power factor is cos θ = 0.9689 (leading) It is a leading pf because QT is positive or P2 P1.
  • 515. CHAPTER 12 Three-Phase Circuits 513 (c) The phase impedance is Zp = Zp θ. We know that θ is the same as the pf angle; that is, θ = 14.57◦ . Zp = Vp Ip We recall that for a delta-connected load, Vp = VL = 220 V. From Eq. (12.46), Pp = VpIp cos θ ⇒ Ip = 1220 220 × 0.9689 = 5.723 A Hence, Zp = Vp Ip = 220 5.723 = 38.44 and Zp = 38.44 14.33◦ P R A C T I C E P R O B L E M 1 2 . 1 4 Let the line voltage VL = 208 V and the wattmeter readings of the bal- anced system in Fig. 12.35 be P1 = −560 W and P2 = 800 W. Deter- mine: (a) the total average power (b) the total reactive power (c) the power factor (d) the phase impedance Is the impedance inductive or capacitive? Answer: (a) 240 W, (b) 2355.6 VAR, (c) 0.1014, (d) 18.25 84.18◦ , inductive. E X A M P L E 1 2 . 1 5 The three-phase balanced load in Fig. 12.35 has impedance per phase of ZY = 8 + j6 . If the load is connected to 208-V lines, predict the read- ings of the wattmeters W1 and W2. Find PT and QT . Solution: The impedance per phase is ZY = 8 + j6 = 10 36.87◦ so that the pf angle is 36.87◦ . Since the line voltage VL = 208 V, the line current is IL = Vp |ZY | = 208/ √ 3 10 = 12 A
  • 516. 514 PART 2 AC Circuits Then P1 = VLIL cos(θ + 30◦ ) = 208 × 12 × cos(36.87◦ + 30◦ ) = 980.48 W P2 = VLIL cos(θ − 30◦ ) = 208 × 12 × cos(36.87◦ − 30◦ ) = 2478.1 W Thus, wattmeter 1 reads 980.48 W, while wattmeter 2 reads 2478.1 W. Since P2 P1, the load is inductive. This is evident from the load ZY itself. Next, PT = P1 + P2 = 3.4586 kW and QT = √ 3(P2 − P1) = √ 3(1497.6) VAR = 2.594 kVAR P R A C T I C E P R O B L E M 1 2 . 1 5 If the load in Fig. 12.35 is delta-connected with impedance per phase of Zp = 30−j40 and VL = 440 V, predict the readings of the wattmeters W1 and W2. Calculate PT and QT . Answer: 6.166 kW, 0.8021 kW, 6.968 kW, −9.291 kVAR. 12.10.2 Residential Wiring In the United States, most household lighting and appliances operate on 120-V, 60-Hz, single-phase alternating current. (The electricity may also be supplied at 110, 115, or 117 V, depending on the location.) The local power company supplies the house with a three-wire ac system. Typically, as in Fig. 12.37, the line voltage of, say, 12,000 V is stepped down to 120/240 V with a transformer (more details on transformers ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; Light pole Grounded metal stake Ground Wall of house Circuit # 1 120 V Circuit # 2 120 V Circuit # 3 120 V Fuse Fuse Fuses Switch Watt-hour meter Step-down transformer Figure 12.37 A 120/240 household power system. (Source: A. Marcus and C. M. Thomson, Electricity for Technicians, 2nd ed. [Englewood Cliffs, NJ: Prentice Hall, 1975], p. 324.)
  • 517. CHAPTER 12 Three-Phase Circuits 515 in the next chapter). The three wires coming from the transformer are typically colored red (hot), black (hot), and white (neutral). As shown in Fig. 12.38, the two 120-V voltages are opposite in phase and hence add up to zero. That is, VW = 0 0◦ , VB = 120 0◦ , VR = 120 180◦ = −VB. VBR = VB − VR = VB − (−VB) = 2VB = 240 0◦ (12.72) Since most appliances are designed to operate with 120 V, the lighting and appliances are connected to the 120-V lines, as illustrated in Fig. 12.39 for a room. Notice in Fig. 12.37 that all appliances are connected in parallel. Heavy appliances that consume large currents, such as air conditioners, dishwashers, ovens, and laundry machines, are connected to the 240-V power line. 120 V lights 120 V appliance 120 V lights 120 V appliance 120 V 120 V − + + − 240 V appliance Black (hot) B White (neutral) Red (hot) W R Ground To other houses Transformer House Figure12.38 Single-phase three-wire residential wiring. ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; Lamp sockets Base outlets 120 volts Ungrounded conductor Switch Neutral Figure 12.39 A typical wiring diagram of a room. (Source: A. Marcus and C. M. Thomson, ElectricityforTech- nicians, 2nd ed. [Englewood Cliffs, NJ:PrenticeHall, 1975], p. 325.) Because of the dangers of electricity, house wiring is carefully reg- ulated by a code drawn by local ordinances and by the National Electrical Code (NEC). To avoid trouble, insulation, grounding, fuses, and circuit breakers are used. Modern wiring codes require a third wire for a sep- arate ground. The ground wire does not carry power like the neutral wire but enables appliances to have a separate ground connection. Figure 12.40 shows the connection of the receptacle to a 120-V rms line and to the ground. As shown in the figure, the neutral line is connected to the ground (the earth) at many critical locations. Although the ground line + − Fuse or circuit breaker 120 V rms Hot wire Receptacle To other appliances Neutral wire Power system ground Service panel ground Ground wire Figure12.40 Connection of a receptacle to the hot line and to the ground.
  • 518. 516 PART 2 AC Circuits seems redundant, grounding is important for many reasons. First, it is re- quired by NEC. Second, grounding provides a convenient path to ground for lightning that strikes the power line. Third, grounds minimize the risk of electric shock. What causes shock is the passage of current from one part of the body to another. The human body is like a big resistor R. If V is the potential difference between the body and the ground, the current through the body is determined by Ohm’s law as I = V R (12.73) The value of R varies from person to person and depends on whether the body is wet or dry. How great or how deadly the shock is depends on the amount of current, the pathway of the current through the body, and the length of time the body is exposed to the current. Currents less than 1 mA may not be harmful to the body, but currents greater than 10 mA can cause severe shock. A modern safety device is the ground-fault circuit interrupter (GFCI), used in outdoor circuits and in bathrooms, where the risk of electric shock is greatest. It is essentially a circuit breaker that opens when the sum of the currents iR, iW , and iB through the red, white, and the black lines is not equal to zero, or iR + iW + iB = 0. The best way to avoid electric shock is to follow safety guidelines concerning electrical systems and appliances. Here are some of them: • Never assume that an electrical circuit is dead. Always check to be sure. • Use safety devices when necessary, and wear suitable clothing (insulated shoes, gloves, etc.). • Never use two hands when testing high-voltage circuits, since the current through one hand to the other hand has a direct path through your chest and heart. • Do not touch an electrical appliance when you are wet. Remember that water conducts electricity. • Be extremely careful when working with electronic appliances such as radio and TV because these appliances have large capacitors in them. The capacitors take time to discharge after the power is disconnected. • Always have another person present when working on a wiring system, just in case of an accident. 12.11 SUMMARY 1. The phase sequence is the order in which the phase voltages of a three-phase generator occur with respect to time. In an abc sequence of balanced source voltages, Van leads Vbn by 120◦ , which in turn leads Vcn by 120◦ . In an acb sequence of balanced voltages, Van leads Vcn by 120◦ , which in turn leads Vbn by 120◦ . 2. A balanced wye- or delta-connected load is one in which the three- phase impedances are equal. 3. The easiest way to analyze a balanced three-phase circuit is to transform both the source and the load to a Y-Y system and then
  • 519. CHAPTER 12 Three-Phase Circuits 517 analyze the single-phase equivalent circuit. Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four possible configurations. 4. The line current IL is the current flowing from the generator to the load in each transmission line in a three-phase system. The line voltage VL is the voltage between each pair of lines, excluding the neutral line if it exists. The phase current Ip is the current flowing through each phase in a three-phase load. The phase voltage Vp is the voltage of each phase. For a wye-connected load, VL = √ 3Vp and IL = Ip For a delta-connected load, VL = Vp and IL = √ 3Ip 5. The total instantaneous power in a balanced three-phase system is constant and equal to the average power. 6. The total complex power absorbed by a balanced three-phase Y-connected or -connected load is S = P + jQ = √ 3VLIL θ where θ is the angle of the load impedances. 7. An unbalanced three-phase system can be analyzed using nodal or mesh analysis. 8. PSpice is used to analyze three-phase circuits in the same way as it is used for analyzing single-phase circuits. 9. The total real power is measured in three-phase systems using either the three-wattmeter method or the two-wattmeter method. 10. Residential wiring uses a 120/240-V, single-phase, three-wire system. REVIEW QUESTIONS 12.1 What is the phase sequence of a three-phase motor for which VAN = 220 − 100◦ V and VBN = 220 140◦ V? (a) abc (b) acb 12.2 If in an acb phase sequence, Van = 100 − 20◦ , then Vcn is: (a) 100 − 140◦ (b) 100 100◦ (c) 100 − 50◦ (d) 100 10◦ 12.3 Which of these is not a required condition for a balanced system: (a) |Van| = |Vbn| = |Vcn| (b) Ia + Ib + Ic = 0 (c) Van + Vbn + Vcn = 0 (d) Source voltages are 120◦ out of phase with each other. (e) Load impedances for the three phases are equal. 12.4 In a Y-connected load, the line current and phase current are equal. (a) True (b) False 12.5 In a -connected load, the line current and phase current are equal. (a) True (b) False 12.6 In a Y-Y system, a line voltage of 220 V produces a phase voltage of: (a) 381 V (b) 311 V (c) 220 V (d) 156 V (e) 127 V
  • 520. 518 PART 2 AC Circuits 12.7 In a - system, a phase voltage of 100 V produces a line voltage of: (a) 58 V (b) 71 V (c) 100 V (d) 173 V (e) 141 V 12.8 When a Y-connected load is supplied by voltages in abc phase sequence, the line voltages lag the corresponding phase voltages by 30◦ . (a) True (b) False 12.9 In a balanced three-phase circuit, the total instantaneous power is equal to the average power. (a) True (b) False 12.10 The total power supplied to a balanced -load is found in the same way as for a balanced Y-load. (a) True (b) False Answers: 12.1a, 12.2a, 12.3c, 12.4a, 12.5b, 12.6e, 12.7c, 12.8b, 12.9a, 12.10a. PROBLEMS1 Section 12.2 Balanced Three-Phase Voltages 12.1 If Vab = 400 V in a balanced Y-connected three-phase generator, find the phase voltages, assuming the phase sequence is: (a) abc (b) acb 12.2 What is the phase sequence of a balanced three-phase circuit for which Van = 160 30◦ V and Vcn = 160 − 90◦ V? Find Vbn. 12.3 Determine the phase sequence of a balanced three-phase circuit in which Vbn = 208 130◦ V and Vcn = 208 10◦ V. Obtain Van. 12.4 Assuming the abc sequence, if Vca = 208 20◦ V in a balanced three-phase circuit, find Vab, Vbc, Van, and Vbn. 12.5 Given that the line voltages of a three-phase circuit are Vab = 420 0◦ , Vbc = 420 − 120◦ Vac = 420 120◦ V find the phase voltages Van, Vbn, and Vcn. Section 12.3 Balanced Wye-Wye Connection 12.6 For the Y-Y circuit of Fig. 12.41, find the line currents, the line voltages, and the load voltages. a A b B c C n N − + − + − + 220 0° V 220 −120° V 220 120° V 10 Ω j5 Ω 10 Ω j5 Ω 10 Ω j5 Ω Figure 12.41 For Prob. 12.6. 12.7 Obtain the line currents in the three-phase circuit of Fig. 12.42 below. + − A N n a 440 0° V 440 120° V 440 −120° V + − − + 6 − j8 Ω 6 − j8 Ω 6 − j8 Ω Ia Ib Ic Figure12.42 For Prob. 12.7. 1Remember that unless stated otherwise, all given voltages and currents are rms values.
  • 521. CHAPTER 12 Three-Phase Circuits 519 12.8 A balanced Y-connected load with a phase impedance of 16 + j9 is connected to a balanced three-phase source with a line voltage of 220 V. Calculate the line current IL. 12.9 A balanced Y-Y four-wire system has phase voltages Van = 120 0◦ , Vbn = 120 − 120◦ Vcn = 120 120◦ V The load impedance per phase is 19 + j13 , and the line impedance per phase is 1 + j2 . Solve for the line currents and neutral current. 12.10 For the circuit in Fig. 12.43, determine the current in the neutral line. + − + − − + 2 Ω 2 Ω 20 Ω 2 Ω 220 −120° V 220 120° V 10 + j5 Ω 25 − j10 Ω 220 0° V Figure 12.43 For Prob. 12.10. Section 12.4 Balanced Wye-Delta Connection 12.11 For the three-phase circuit of Fig. 12.44, IbB = 30 60◦ A and VBC = 220 10◦ V. Find Van, VAB , IAC, and Z. + − + − − + C B A Z Z Z Vcn V an V bn n b a c Figure 12.44 For Prob. 12.11. 12.12 Solve for the line currents in the Y- circuit of Fig. 12.45. Take Z = 60 45◦ . Z∆ A C c B a + − + − Ia Ib Ic + − n b Z∆ Z∆ 110 0° V 110 −120° V 110 120° V Figure 12.45 For Prob. 12.12. 12.13 The circuit in Fig. 12.46 is excited by a balanced three-phase source with a line voltage of 210 V. If Z = 1 + j1 , Z = 24 − j30 , and ZY = 12 + j5 , determine the magnitude of the line current of the combined loads. a b c Zl Zl Zl Z∆ Z∆ Z∆ ZY ZY ZY Figure 12.46 For Prob. 12.13. 12.14 A balanced delta-connected load has a phase current IAC = 10 − 30◦ A. (a) Determine the three line currents assuming that the circuit operates in the positive phase sequence. (b) Calculate the load impedance if the line voltage is VAB = 110 0◦ V. 12.15 In a wye-delta three-phase circuit, the source is a balanced, positive phase sequence with Van = 120 0◦ V. It feeds a balanced load with Z = 9 + j12 per phase through a balanced line with Z = 1 + j0.5 per phase. Calculate the phase voltages and currents in the load.
  • 522. 520 PART 2 AC Circuits 12.16 If Van = 440 60◦ V in the network of Fig. 12.47, find the load phase currents IAB , IBC, and ICA. Three-phase, Y-connected generator (+) phase sequence 12 Ω j9 Ω j9 Ω 12 Ω 12 Ω j9 Ω a b c A B C Figure 12.47 For Prob. 12.16. Section 12.5 Balanced Delta-Delta Connection 12.17 For the - circuit of Fig. 12.48, calculate the phase and line currents. + − + − + − 30 Ω 30 Ω 173 −120° V 173 0° V j10 Ω j10 Ω 30 Ω j10 Ω A B a b c C 173 120° V Figure 12.48 For Prob. 12.17. 12.18 Refer to the - circuit in Fig. 12.49. Find the line and phase currents. Assume that the load impedance is 12 + j9 per phase. A C B Ia Ib Ic + − + − − + 210 0° V 210 −120° V 210 120° V IBC IAB ICA ZL ZL ZL Figure 12.49 For Prob. 12.18. 12.19 Find the line currents Ia, Ib, and Ic in the three-phase network of Fig. 12.50 below. Take Z = 12 − j15 , ZY = 4 + j6 , and Z = 2 . 12.20 A balanced delta-connected source has phase voltage Vab = 416 30◦ V and a positive phase sequence. If this is connected to a balanced delta-connected load, find the line and phase currents. Take the load impedance per phase as 60 30◦ and line impedance per phase as 1 + j1 . A C B Ia Ib Ic + − + − − + 208 0° V 208 −120° V 208 120° V Zl Zl Zl ZY ZY Z∆ Z∆ Z∆ ZY Figure12.50 For Prob. 12.19.
  • 523. CHAPTER 12 Three-Phase Circuits 521 Section 12.6 Balanced Delta-Wye Connection 12.21 In the circuit of Fig. 12.51, if Vab = 440 10◦ , Vbc = 440 250◦ , Vca = 440 130◦ V, find the line currents. b c a 3 + j2 Ω 3 + j2 Ω 10 − j8 Ω 10 − j8 Ω 10 − j8 Ω 3 + j2 Ω + − + − + − Vca Vab Vbc Ib Ic Ia Figure 12.51 For Prob. 12.21. 12.22 For the balanced circuit in Fig. 12.52, Vab = 125 0◦ V. Find the line currents IaA, IbB , and IcC. N IbB IcC IaA 24 Ω 24 Ω 24 Ω −j15 Ω −j15 Ω −j15 Ω a b c A C B Three-phase, ∆-connected generator (+) phase sequence Figure 12.52 For Prob. 12.22. 12.23 In a balanced three-phase -Y circuit, the source is connected in the positive sequence, with Vab = 220 20◦ V and ZY = 20 + j15 . Find the line currents. 12.24 A delta-connected generator supplies a balanced wye-connected load with an impedance of 30 − 60◦ . If the line voltages of the generator have a magnitude of 400 V and are in the positive phase sequence, find the line current IL and phase voltage Vp at the load. Section 12.7 Power in a Balanced System 12.25 A balanced wye-connected load absorbs a total power of 5 kW at a leading power factor of 0.6 when connected to a line voltage of 240 V. Find the impedance of each phase and the total complex power of the load. 12.26 A balanced wye-connected load absorbs 50 kVA at a 0.6 lagging power factor when the line voltage is 440 V. Find the line current and the phase impedance. 12.27 A three-phase source delivers 4800 VA to a wye-connected load with a phase voltage of 208 V and a power factor of 0.9 lagging. Calculate the source line current and the source line voltage. 12.28 A balanced wye-connected load with a phase impedance of 10 − j16 is connected to a balanced three-phase generator with a line voltage of 220 V. Determine the line current and the complex power absorbed by the load. 12.29 The total power measured in a three-phase system feeding a balanced wye-connected load is 12 kW at a power factor of 0.6 leading. If the line voltage is 208 V, calculate the line current IL and the load impedance ZY . 12.30 Given the circuit in Fig. 12.53 below, find the total complex power absorbed by the load. + − + − − + 1 Ω 9 Ω 9 Ω 9 Ω 110 120° V j2 Ω 1 Ω j2 Ω 1 Ω j2 Ω 1 Ω j2 Ω j12 Ω j12 Ω j12 Ω 110 240° V 110 0° V Figure12.53 For Prob. 12.30.
  • 524. 522 PART 2 AC Circuits 12.31 Find the real power absorbed by the load in Fig. 12.54. A −j6 Ω j3 Ω C b B c a 100 −120° V − + + − + − 100 120° V 100 0° V 5 Ω 5 Ω 5 Ω 8 Ω 4 Ω 10 Ω Figure 12.54 For Prob. 12.31. 12.32 For the three-phase circuit in Fig. 12.55, find the average power absorbed by the delta-connected load with Z = 21 + j24 . 1 Ω − + 1 Ω 1 Ω j0.5 Ω j0.5 Ω j0.5 Ω 100 0° V rms 100 −120° V rms 100 120° V rms − + − + Z∆ Z∆ Z∆ Figure 12.55 For Prob. 12.32. 12.33 A balanced delta-connected load draws 5 kW at a power factor of 0.8 lagging. If the three-phase system has an effective line voltage of 400 V, find the line current. 12.34 A balanced three-phase generator delivers 7.2 kW to a wye-connected load with impedance 30 − j40 per phase. Find the line current IL and the line voltage VL. 12.35 Refer to Fig. 12.46. Obtain the complex power absorbed by the combined loads. 12.36 A three-phase line has an impedance of 1 + j3 per phase. The line feeds a balanced delta-connected load, which absorbs a total complex power of 12 + j5 kVA. If the line voltage at the load end has a magnitude of 240 V, calculate the magnitude of the line voltage at the source end and the source power factor. 12.37 A balanced wye-connected load is connected to the generator by a balanced transmission line with an impedance of 0.5 + j2 per phase. If the load is rated at 450 kW, 0.708 power factor lagging, 440-V line voltage, find the line voltage at the generator. 12.38 A three-phase load consists of three 100- resistors that can be wye- or delta-connected. Determine which connection will absorb the most average power from a three-phase source with a line voltage of 110 V. Assume zero line impedance. 12.39 The following three parallel-connected three-phase loads are fed by a balanced three-phase source. Load 1: 250 kVA, 0.8 pf lagging Load 2: 300 kVA, 0.95 pf leading Load 3: 450 kVA, unity pf If the line voltage is 13.8 kV, calculate the line current and the power factor of the source. Assume that the line impedance is zero. Section 12.8 Unbalanced Three-Phase Systems 12.40 For the circuit in Fig. 12.56, Za = 6 − j8 , Zb = 12 + j9 , and Zc = 15 . Find the line currents Ia, Ib, and Ic. Ib Ic Ia 150 120° V 150 0° V 150 −120° V + − + − + − Zb Za Zc Figure 12.56 For Prob. 12.40. 12.41 A four-wire wye-wye circuit has Van = 120 120◦ , Vbn = 120 0◦ Vcn = 120 − 120◦ V If the impedances are ZAN = 20 60◦ , ZBN = 30 0◦ Zcn = 40 30◦ find the current in the neutral line. 12.42 For the wye-connected load of Fig. 12.57, the line voltages all have a magnitude of 250 V and are in a positive phase sequence. Calculate the line currents and the neutral current. 40 60° Ω Ia 60 −45° Ω Ib 20 0° Ω Ic In Figure 12.57 For Prob. 12.42.
  • 525. CHAPTER 12 Three-Phase Circuits 523 12.43 A delta-connected load whose phase impedances are ZAB = 50 , ZBC = −j50 , and ZCA = j50 is fed by a balanced wye-connected three-phase source with Vp = 100 V. Find the phase currents. 12.44 A balanced three-phase wye-connected generator with Vp = 220 V supplies an unbalanced wye-connected load with ZAN = 60 + j80 , ZBN = 100 − j120 , and ZCN = 30 + j40 . Find the total complex power absorbed by the load. 12.45 Refer to the unbalanced circuit of Fig. 12.58. Calculate: (a) the line currents (b) the real power absorbed by the load (c) the total complex power supplied by the source 440 0° V 440 120° V 440 −120° V + − + − − + a b B c A C 20 Ω j10 Ω −j5 Ω Figure 12.58 For Prob. 12.45. Section 12.9 PSpice for Three-Phase Circuits 12.46 Solve Prob. 12.10 using PSpice. 12.47 The source in Fig. 12.59 is balanced and exhibits a positive phase sequence. If f = 60 Hz, use PSpice to find VAN , VBN , and VCN . 100 0° V + − + − − + a b B c A C 40 Ω n N 0.2 mF 10 mF Figure 12.59 For Prob. 12.47. 12.48 Use PSpice to determine Io in the single-phase, three-wire circuit of Fig. 12.60. Let Z1 = 15 − j10 , Z2 = 30 + j20 , and Z3 = 12 + j5 . Z1 Z2 Z3 4 Ω 4 Ω 4 Ω + − + − 220 0° V 220 0° V Io Figure 12.60 For Prob. 12.48. 12.49 Given the circuit in Fig. 12.61, use PSpice to determine currents IaA and voltage VBN . a n A C N 4 Ω − + b 4 Ω c 4 Ω 10 Ω B j3 Ω j15 Ω j15 Ω j3 Ω j3 Ω −j36 Ω −j36 Ω 240 0° V 240 −120° V 240 120° V −j36 Ω − + 10 Ω j15 Ω 10 Ω − + Figure 12.61 For Prob. 12.49. 12.50 The circuit in Fig. 12.62 operates at 60 Hz. Use PSpice to find the source current Iab and the line current IbB . A N b c a 1 Ω 16 Ω 2 mH 2 mH 27 mH 2 mH 133 mF 1 Ω 1 Ω + − + − + − B C 110 120° V 110 −120° V 110 0° V Figure 12.62 For Prob. 12.50. 12.51 For the circuit in Fig. 12.54, use PSpice to find the line currents and the phase currents. 12.52 A balanced three-phase circuit is shown in Fig. 12.63 on the next page. Use PSpice to find the line currents IaA, IbB , and IcC. Section 12.10 Applications 12.53 A three-phase, four-wire system operating with a 208-V line voltage is shown in Fig. 12.64. The source voltages are balanced. The power absorbed by the resistive wye-connected load is measured by the three-wattmeter method. Calculate: (a) the voltage to neutral (b) the currents I1, I2, I3, and In (c) the readings of the wattmeters (d) the total power absorbed by the load
  • 526. 524 PART 2 AC Circuits A b c a 0.6 Ω 0.2 Ω 0.2 Ω 30 Ω j0.5 Ω j1 Ω j1 Ω −j20 Ω j0.5 Ω j0.5 Ω 0.2 Ω j1 Ω 0.6 Ω 0.6 Ω 30 Ω −j20 Ω + − + − + − B C 240 130° V 240 −110° V 240 10° V 30 Ω −j20 Ω Figure12.63 For Prob. 12.52. n I2 In I3 I1 40 Ω 48 Ω 60 Ω W1 W2 W3 Figure 12.64 For Prob. 12.53. 12.54 ∗ As shown in Fig. 12.65, a three-phase four-wire line with a phase voltage of 120 V supplies a balanced motor load at 260 kVA at 0.85 pf lagging. The motor load is connected to the three main lines marked a, b, and c. In addition, incandescent lamps (unity pf) are connected as follows: 24 kW from line a to the neutral, 15 kW from line b to the neutral, and 9 kW from line a to the neutral. (a) If three wattmeters are arranged to measure the power in each line, calculate the reading of each meter. (b) Find the current in the neutral line. *An asterisk indicates a challenging problem. a b c d 24 kW 15 kW 9 kW Motor load 260 kVA, 0.85 pf, lagging Lighting loads Figure 12.65 For Prob. 12.54. 12.55 Meter readings for a three-phase wye-connected alternator supplying power to a motor indicate that the line voltages are 330 V, the line currents are 8.4 A, and the total line power is 4.5 kW. Find: (a) the load in VA (b) the load pf (c) the phase current (d) the phase voltage 12.56 The two-wattmeter method gives P1 = 1200 W and P2 = −400 W for a three-phase motor running on a 240-V line. Assume that the motor load is wye- connected and that it draws a line current of 6 A. Calculate the pf of the motor and its phase impedance.
  • 527. CHAPTER 12 Three-Phase Circuits 525 12.57 In Fig. 12.66, two wattmeters are properly connected to the unbalanced load supplied by a balanced source such that Vab = 208 0◦ V with positive phase sequence. (a) Determine the reading of each wattmeter. (b) Calculate the total apparent power absorbed by the load. a b 0 B c A C 10 Ω 12 Ω 20 Ω j5 Ω −j10 Ω W1 W2 Figure 12.66 For Prob. 12.57. 12.58 If wattmeters W1 and W2 are properly connected respectively between lines a and b and lines b and c to measure the power absorbed by the delta-connected load in Fig. 12.44, predict their readings. 12.59 For the circuit displayed in Fig. 12.67, find the wattmeter readings. Z Z Z = 10 + j30 Ω + − + − 240 −60° V 240 −120° V ± ± ± ± W1 W2 Figure 12.67 For Prob. 12.59. 12.60 Predict the wattmeter readings for the circuit in Fig. 12.68. Z Z Z = 60 − j30 Ω + − + − 208 0° V 208 −60° V ± ± ± ± W1 W2 Figure 12.68 For Prob. 12.60. 12.61 A man has a body resistance of 600 . How much current flows through his ungrounded body: (a) when he touches the terminals of a 12-V autobattery? (b) when he sticks his finger into a 120-V light socket? 12.62 Show that the I2 R losses will be higher for a 120-V appliance than for a 240-V appliance if both have the same power rating. COMPREHENSIVE PROBLEMS 12.63 A three-phase generator supplied 3.6 kVA at a power factor of 0.85 lagging. If 2500 W are delivered to the load and line losses are 80 W per phase, what are the losses in the generator? 12.64 A three-phase 440-V, 51-kW, 60-kVA inductive load operates at 60 Hz and is wye-connected. It is desired to correct the power factor to 0.95 lagging. What value of capacitor should be placed in parallel with each load impedance? 12.65 A balanced three-phase generator has an abc phase sequence with phase voltage Van = 255 0◦ V. The generator feeds an induction motor which may be represented by a balanced Y-connected load with an impedance of 12 + j5 per phase. Find the line currents and the load voltages. Assume a line impedance of 2 per phase.
  • 528. 526 PART 2 AC Circuits 12.66 Three balanced loads are connected to a distribution line as depicted in Fig. 12.69. The loads are Transformer: 12 kVA at 0.6 pf lagging Motor: 16 kVA at 0.8 pf lagging Unknown load: − − −− If the line voltage is 220 V, the line current is 120 A, and the power factor of the combined load is 0.95 lagging, determine the unknown load. Transformer Motor Unknown Load Figure 12.69 For Prob. 12.66. 12.67 A professional center is supplied by a balanced three-phase source. The center has four plants, each a balanced three-phase load as follows: Load 1: 150 kVA at 0.8 pf leading Load 2: 100 kW at unity pf Load 3: 200 kVA at 0.6 pf lagging Load 4: 80 kW and 95 kVAR (inductive) If the line impedance is 0.02 + j0.05 per phase and the line voltage at the loads is 480 V, find the magnitude of the line voltage at the source. 12.68 ∗ Figure 12.70 displays a three-phase delta-connected motor load which is connected to a line voltage of 440 V and draws 4 kVA at a power factor of 72 percent lagging. In addition, a single 1.8 kVAR capacitor is connected between lines a and b, while a 800-W lighting load is connected between line c and neutral. Assuming the abc sequence and taking Van = Vp 0◦ , find the magnitude and phase angle of currents Ia, Ib, Ic, and In. a b c d Motor load 4 kVA, pf = 72%, lagging 800 W lighting load Ia Ib Ic In 1.8 kVAR Figure 12.70 For Prob. 12.68. 12.69 Design a three-phase heater with suitable symmetric loads using wye-connected pure resistance. Assume that the heater is supplied by a 240-V line voltage and is to give 27 kW of heat. 12.70 For the single-phase three-wire system in Fig. 12.71, find currents IaA, IbB , and InN . 24 − j2 Ω 15 + j4 Ω 1 Ω 1 Ω 1 Ω + − + − 120 0° V rms 120 0° V rms a A n b B N Figure 12.71 For Prob. 12.70. 12.71 Consider the single-phase three-wire system shown in Fig. 12.72. Find the current in the neutral wire and the complex power supplied by each source. Take Vs as a 115 0◦ -V, 60-Hz source. 1 Ω 2 Ω 20 Ω 15 Ω 30 Ω 50 mH 1 Ω + − + − V s Vs Figure 12.72 For Prob. 12.71.
  • 529. 527 C H A P T E R MAGNETICALLY COUPLED CIRCUITS 1 3 People want success but keep running away from problems, and yet it is only in tackling problems that success is achieved. — Josiah J. Bonire Enhancing Your Career Career in Electromagnetics Electromagnetics is the branch of electrical engineering (or physics) that deals with the analysis and application of electric and magnetic fields. In electromagnetics, electric circuit analysis is applied at low frequencies. The principles of electromagnetics (EM) are applied in various allied disciplines, such as electric machines, electromechanical energy conversion, radar meteorology, remote sensing, satellite communications, bioelectromag- netics, electromagnetic interference and compatibility, plas- mas, and fiber optics. EM devices include electric motors and generators, transformers, electromagnets, magnetic lev- itation, antennas, radars, microwave ovens, microwave dishes, superconductors, and electrocardiograms. The de- sign of these devices requires a thorough knowledge of the laws and principles of EM. EM is regarded as one of the more difficult disci- plines in electrical engineering. One reason is that EM phenomena are rather abstract. But if one enjoys working with mathematics and can visualize the invisible, one should consider being a specialist in EM, since few electrical engineers specialize in this area. Electrical engineers who specialize in EM are needed in microwave industries, radio/TV broadcasting stations, electromagnetic research laboratories, and several communications industries. Telemetry receiving station for space satellites. Source: T. J. Mal- oney, Modern Industrial Electronics, 3rd ed. Englewood Cliffs, NJ: Prentice Hall, 1996, p. 718.
  • 530. 528 PART 2 AC Circuits 13.1 INTRODUCTION The circuits we have considered so far may be regarded as conductively coupled, because one loop affects the neighboring loop through current conduction. When two loops with or without contacts between them affect each other through the magnetic field generated by one of them, they are said to be magnetically coupled. The transformer is an electrical device designed on the basis of the concept of magnetic coupling. It uses magnetically coupled coils to transfer energy from one circuit to another. Transformers are key circuit elements. They are used in power systems for stepping up or stepping down ac voltages or currents. They are used in electronic circuits such as radio and television receivers for such purposes as impedance matching, isolating one part of a circuit from another, and again for stepping up or down ac voltages and currents. We will begin with the concept of mutual inductance and introduce the dot convention used for determining the voltage polarities of induc- tively coupled components. Based on the notion of mutual inductance, we then introduce the circuit element known as the transformer. We will consider the linear transformer, the ideal transformer, the ideal autotrans- former, and the three-phase transformer. Finally, among their important applications, we look at transformers as isolating and matching devices and their use in power distribution. 13.2 MUTUAL INDUCTANCE When two inductors (or coils) are in a close proximity to each other, the magnetic flux caused by current in one coil links with the other coil, thereby inducing voltage in the latter. This phenomenon is known as mutual inductance. i(t) v + − f Figure13.1 Magnetic flux produced by a single coil with N turns. Let us first consider a single inductor, a coil with N turns. When current i flows through the coil, a magnetic flux φ is produced around it (Fig. 13.1). According to Faraday’s law, the voltage v induced in the coil is proportional to the number of turns N and the time rate of change of the magnetic flux φ; that is, v = N dφ dt (13.1) But the flux φ is produced by current i so that any change in φ is caused by a change in the current. Hence, Eq. (13.1) can be written as v = N dφ di di dt (13.2) or v = L di dt (13.3) which is the voltage-current relationship for the inductor. From Eqs. (13.2) and (13.3), the inductance L of the inductor is thus given by L = N dφ di (13.4)
  • 531. CHAPTER 13 Magnetically Coupled Circuits 529 This inductance is commonly called self-inductance, because it relates the voltage induced in a coil by a time-varying current in the same coil. Now consider two coils with self-inductances L1 and L2 that are in close proximity with each other (Fig. 13.2). Coil 1 has N1 turns, while coil 2 has N2 turns. For the sake of simplicity, assume that the second inductor carries no current. The magnetic flux φ1 emanating from coil 1 has two components: one component φ11 links only coil 1, and another component φ12 links both coils. Hence, φ1 = φ11 + φ12 (13.5) Although the two coils are physically separated, they are said to be mag- netically coupled. Since the entire flux φ1 links coil 1, the voltage induced in coil 1 is v1 = N1 dφ1 dt (13.6) Only flux φ12 links coil 2, so the voltage induced in coil 2 is v2 = N2 dφ12 dt (13.7) Again, as the fluxes are caused by the current i1 flowing in coil 1, Eq. (13.6) can be written as v1 = N1 dφ1 di1 di1 dt = L1 di1 dt (13.8) where L1 = N1 dφ1/di1 is the self-inductance of coil 1. Similarly, Eq. (13.7) can be written as v2 = N2 dφ12 di1 di1 dt = M21 di1 dt (13.9) where M21 = N2 dφ12 di1 (13.10) M21 is known as the mutual inductance of coil 2 with respect to coil 1. Subscript 21 indicates that the inductance M21 relates the voltage induced in coil 2 to the current in coil 1. Thus, the open-circuit mutual voltage (or induced voltage) across coil 2 is v2 = M21 di1 dt (13.11) i1(t) v1 + − v2 + − f11 f12 L1 L2 N1 turns N2 turns Figure 13.2 Mutual inductance M21 of coil 2 with respect to coil 1. v1 + − v2 + − i2(t) f22 f21 L1 L2 N1 turns N2 turns Figure 13.3 Mutual inductance M12 of coil 1 with respect to coil 2. Suppose we now let current i2 flow in coil 2, while coil 1 carries no current(Fig.13.3). Themagneticfluxφ2 emanatingfromcoil2comprises flux φ22 that links only coil 2 and flux φ21 that links both coils. Hence, φ2 = φ21 + φ22 (13.12) The entire flux φ2 links coil 2, so the voltage induced in coil 2 is v2 = N2 dφ2 dt = N2 dφ2 di2 di2 dt = L2 di2 dt (13.13)
  • 532. 530 PART 2 AC Circuits where L2 = N2 dφ2/di2 is the self-inductance of coil 2. Since only flux φ21 links coil 1, the voltage induced in coil 1 is v1 = N1 dφ21 dt = N1 dφ21 di2 di2 dt = M12 di2 dt (13.14) where M12 = N1 dφ21 di2 (13.15) which is the mutual inductance of coil 1 with respect to coil 2. Thus, the open-circuit mutual voltage across coil 1 is v1 = M12 di2 dt (13.16) We will see in the next section that M12 and M21 are equal, that is, M12 = M21 = M (13.17) and we refer to M as the mutual inductance between the two coils. Like self-inductance L, mutual inductance M is measured in henrys (H). Keep in mind that mutual coupling only exists when the inductors or coils are in close proximity, and the circuits are driven by time-varying sources. We recall that inductors act like short circuits to dc. From the two cases in Figs. 13.2 and 13.3, we conclude that mutual inductance results if a voltage is induced by a time-varying current in another circuit. It is the property of an inductor to produce a voltage in reaction to a time-varying current in another inductor near it. Thus, Mutual inductance is the ability of one inductor to induce a voltage across a neighboring inductor, measured in henrys (H). Although mutual inductance M is always a positive quantity, the mutual voltage M di/dt may be negative or positive, just like the self- induced voltage L di/dt. However, unlike the self-induced L di/dt, whose polarity is determined by the reference direction of the current and the reference polarity of the voltage (according to the passive sign con- vention), the polarity of mutual voltage M di/dt is not easy to determine, because four terminals are involved. The choice of the correct polarity for M di/dt is made by examining the orientation or particular way in which both coils are physically wound and applying Lenz’s law in conjunction with the right-hand rule. Since it is inconvenient to show the construction details of coils on a circuit schematic, we apply the dot convention in cir- cuit analysis. By this convention, a dot is placed in the circuit at one end of each of the two magnetically coupled coils to indicate the direction of the magnetic flux if current enters that dotted terminal of the coil. This is illustrated in Fig. 13.4. Given a circuit, the dots are already placed beside the coils so that we need not bother about how to place them. The dots are used along with the dot convention to determine the polarity of the mutual voltage. The dot convention is stated as follows:
  • 533. CHAPTER 13 Magnetically Coupled Circuits 531 i1 f21 f11 f22 f12 v1 + − i2 Coil 1 Coil 2 v2 + − Figure13.4 Illustration of the dot convention. If a current enters the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is positive at the dotted terminal of the second coil. Alternatively, If a current leaves the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is negative at the dotted terminal of the second coil. + − M i1 v2 = M di1 dt (a) + − M i1 v2 = –M di1 dt v1 = –M di2 dt (b) + − M (c) (d) i2 v1 = M di2 dt + − M i2 Figure13.5 Examples illustrating how to apply the dot convention. Thus, the reference polarity of the mutual voltage depends on the refer- ence direction of the inducing current and the dots on the coupled coils. Application of the dot convention is illustrated in the four pairs of mu- tually coupled coils in Fig. 13.5. For the coupled coils in Fig. 13.5(a), the sign of the mutual voltage v2 is determined by the reference polarity for v2 and the direction of i1. Since i1 enters the dotted terminal of coil 1 and v2 is positive at the dotted terminal of coil 2, the mutual voltage is +M di1/dt. For the coils in Fig. 13.5(b), the current i1 enters the dot- ted terminal of coil 1 and v2 is negative at the dotted terminal of coil 2. Hence, the mutual voltage is −M di1/dt. The same reasoning applies to the coils in Fig. 13.5(c) and 13.5(d). Figure 13.6 shows the dot conven- tion for coupled coils in series. For the coils in Fig. 13.6(a), the total inductance is L = L1 + L2 + 2M (Series-aiding connection) (13.18) For the coil in Fig. 13.6(b), L = L1 + L2 − 2M (Series-opposing connection) (13.19) Now that we know how to determine the polarity of the mutual voltage, we are prepared to analyze circuits involving mutual inductance.
  • 534. 532 PART 2 AC Circuits i i L1 L2 M (+) (a) i i L1 L2 M (−) (b) Figure 13.6 Dot convention for coils in series; the sign indicates the polarity of the mutual voltage: (a) series-aiding connection, (b) series-opposing connection. As the first example, consider the circuit in Fig. 13.7. Applying KVL to coil 1 gives v1 = i1R1 + L1 di1 dt + M di2 dt (13.20a) For coil 2, KVL gives v2 = i2R2 + L2 di2 dt + M di1 dt (13.20b) We can write Eq. (13.20) in the frequency domain as V1 = (R1 + jωL1)I1 + jωMI2 (13.21a) V2 = jωMI1 + (R2 + jωL2)I2 (13.21b) As a second example, consider the circuit in Fig. 13.8. We analyze this in the frequency domain. Applying KVL to coil 1, we get V = (Z1 + jωL1)I1 − jωMI2 (13.22a) For coil 2, KVL yields 0 = −jωMI1 + (ZL + jωL2)I2 (13.22b) Equations (13.21) and (13.22) are solved in the usual manner to determine the currents. v1 v2 R1 R2 + − L1 L2 i1 i2 M + − Figure 13.7 Time-domain analysis of a circuit containing coupled coils. V ZL Z1 + − jvL1 jvL2 I1 I2 jvM Figure 13.8 Frequency-domain analysis of a circuit containing coupled coils. At this introductory level we are not concerned with the determi- nation of the mutual inductances of the coils and their dot placements. Like R, L, and C, calculation of M would involve applying the theory of electromagnetics to the actual physical properties of the coils. In this text, we assume that the mutual inductance and the dots placement are the “givens” of the circuit problem, like the circuit components R, L, and C.
  • 535. CHAPTER 13 Magnetically Coupled Circuits 533 E X A M P L E 1 3 . 1 Calculate the phasor currents I1 and I2 in the circuit of Fig. 13.9. V 12 Ω −j4 Ω + − j5 Ω j6 Ω I1 I2 j3 Ω 12 0° Figure13.9 For Example 13.1. Solution: For coil 1, KVL gives −12 + (−j4 + j5)I1 − j3I2 = 0 or jI1 − j3I2 = 12 (13.1.1) For coil 2, KVL gives −j3I1 + (12 + j6)I2 = 0 or I1 = (12 + j6)I2 j3 = (2 − j4)I2 (13.1.2) Substituting this in Eq. (13.1.1), we get (j2 + 4 − j3)I2 = (4 − j)I2 = 12 or I2 = 12 4 − j = 2.91 14.04◦ A (13.1.3) From Eqs. (13.1.2) and (13.1.3), I1 = (2 − j4)I2 = (4.472 − 63.43◦ )(2.91 14.04◦ ) = 13.01 − 49.39◦ A P R A C T I C E P R O B L E M 1 3 . 1 Determine the voltage Vo in the circuit of Fig. 13.10. V 10 Ω 4 Ω + − j8 Ω j5 Ω I1 I2 j1 Ω Vo + − 6 90° Figure13.10 For Practice Prob. 13.1. Answer: 0.6 − 90◦ V.
  • 536. 534 PART 2 AC Circuits E X A M P L E 1 3 . 2 Calculate the mesh currents in the circuit of Fig. 13.11. V 5 Ω 4 Ω j8 Ω + − j6 Ω j2 Ω I1 I2 −j3 Ω 100 0° Figure13.11 For Example 13.2. Solution: The key to analyzing a magnetically coupled circuit is knowing the po- larity of the mutual voltage. We need to apply the dot rule. In Fig. 13.11, suppose coil 1 is the one whose reactance is 6 , and coil 2 is the one whose reactance is 8 . To figure out the polarity of the mutual voltage in coil 1 due to current I2, we observe that I2 leaves the dotted terminal of coil 2. Since we are applying KVL in the clockwise direction, it implies that the mutual voltage is negative, that is, −j2I2. + − j2 I2 V1 (a) V1 = –2jI2 I1 j6 Ω j8 Ω Coil 1 Coil 2 − + j2 Ω I1 V2 (b) V2 = –2jI1 I2 j6 Ω j8 Ω Coil 1 Coil 2 Figure13.12 For Example 13.2; redrawing the relevant portion of the circuit in Fig. 13.11 to find mutual voltages by the dot convention. Alternatively, it might be best to figure out the mutual voltage by redrawing the relevant portion of the circuit, as shown in Fig. 13.12(a), where it becomes clear that the mutual voltage is V1 = −2jI2. Thus, for mesh 1 in Fig. 13.11, KVL gives −100 + I1(4 − j3 + j6) − j6I2 − j2I2 = 0 or 100 = (4 + j3)I1 − j8I2 (13.2.1) Similarly, to figure out the mutual voltage in coil 2 due to current I1, consider the relevant portion of the circuit, as shown in Fig. 13.12(b). Applying the dot convention gives the mutual voltage as V2 = −2jI1. Also, current I2 sees the two coupled coils in series in Fig. 13.11; since it leaves the dotted terminals in both coils, Eq. (13.18) applies. Therefore, for mesh 2, KVL gives 0 = −2jI1 − j6I1 + (j6 + j8 + j2 × 2 + 5)I2 or 0 = −j8I1 + (5 + j18)I2 (13.2.2) Putting Eqs. (13.2.1) and (13.2.2) in matrix form, we get 100 0 = 4 + j3 −j8 −j8 5 + j18 I1 I2
  • 537. CHAPTER 13 Magnetically Coupled Circuits 535 The determinants are = 4 + j3 −j8 −j8 5 + j18 = 30 + j87 1 = 100 −j8 0 5 + j18 = 100(5 + j18) 2 = 4 + j3 100 −j8 0 = j800 Thus, we obtain the mesh currents as I1 = 1 = 100(5 + j18) 30 + j87 = 1868.2 74.5◦ 92.03 71◦ = 20.3 3.5◦ A I2 = 2 = j800 30 + j87 = 800 90◦ 92.03 71◦ = 8.693 19◦ A P R A C T I C E P R O B L E M 1 3 . 2 Determine the phasor currents I1 and I2 in the circuit of Fig. 13.13. V 5 Ω j2 Ω + − j6 Ω j3 Ω I1 I2 −j4 Ω 12 60° Figure13.13 For Practice Prob. 13.2. Answer: 2.15 86.56◦ , 3.23 86.56◦ A. 13.3 ENERGY IN A COUPLED CIRCUIT In Chapter 6, we saw that the energy stored in an inductor is given by w = 1 2 Li2 (13.23) We now want to determine the energy stored in magnetically coupled coils. + − M i1 v1 + − v2 i2 L1 L2 Figure13.14 The circuit for deriving energy stored in a coupled circuit. Consider the circuit in Fig. 13.14. We assume that currents i1 and i2 are zero initially, so that the energy stored in the coils is zero. If we let i1 increase from zero to I1 while maintaining i2 = 0, the power in coil 1 is p1(t) = v1i1 = i1L1 di1 dt (13.24) and the energy stored in the circuit is w1 = p1 dt = L1 I1 0 i1 di1 = 1 2 L1I2 1 (13.25)
  • 538. 536 PART 2 AC Circuits If we now maintain i1 = I1 and increase i2 from zero to I2, the mutual voltage induced in coil 1 is M12 di2/dt, while the mutual voltage induced in coil 2 is zero, since i1 does not change. The power in the coils is now p2(t) = i1M12 di2 dt + i2v2 = I1M12 di2 dt + i2L2 di2 dt (13.26) and the energy stored in the circuit is w2 = p2 dt = M12I1 I2 0 di2 + L2 I2 0 i2 di2 = M12I1I2 + 1 2 L2I2 2 (13.27) The total energy stored in the coils when both i1 and i2 have reached constant values is w = w1 + w2 = 1 2 L1I2 1 + 1 2 L2I2 2 + M12I1I2 (13.28) If we reverse the order by which the currents reach their final values, that is, if we first increase i2 from zero to I2 and later increase i1 from zero to I1, the total energy stored in the coils is w = 1 2 L1I2 1 + 1 2 L2I2 2 + M21I1I2 (13.29) Since the total energy stored should be the same regardless of how we reach the final conditions, comparing Eqs. (13.28) and (13.29) leads us to conclude that M12 = M21 = M (13.30a) and w = 1 2 L1I2 1 + 1 2 L2I2 2 + MI1I2 (13.30b) This equation was derived based on the assumption that the coil currents bothenteredthedottedterminals. Ifonecurrententersonedottedterminal while the other current leaves the other dotted terminal, the mutual voltage is negative, so that the mutual energy MI1I2 is also negative. In that case, w = 1 2 L1I2 1 + 1 2 L2I2 2 − MI1I2 (13.31) Also, since I1 and I2 are arbitrary values, they may be replaced by i1 and i2, which gives the instantaneous energy stored in the circuit the general expression w = 1 2 L1i2 1 + 1 2 L2i2 2 ± Mi1i2 (13.32) The positive sign is selected for the mutual term if both currents enter or leave the dotted terminals of the coils; the negative sign is selected otherwise. We will now establish an upper limit for the mutual inductance M. The energy stored in the circuit cannot be negative because the circuit is
  • 539. CHAPTER 13 Magnetically Coupled Circuits 537 passive. This means that the quantity 1/2L1i2 1 + 1/2L2i2 2 − Mi1i2 must be greater than or equal to zero, 1 2 L1i2 1 + 1 2 L2i2 2 − Mi1i2 ≥ 0 (13.33) To complete the square, we both add and subtract the term i1i2 √ L1L2 on the right-hand side of Eq. (13.33) and obtain 1 2 (i1 L1 − i2 L2)2 + i1i2( L1L2 − M) ≥ 0 (13.34) The squared term is never negative; at its least it is zero. Therefore, the second term on the right-hand side of Eq. (13.34) must be greater than zero; that is, L1L2 − M ≥ 0 or M ≤ L1L2 (13.35) Thus, the mutual inductance cannot be greater than the geometric mean of the self-inductances of the coils. The extent to which the mutual inductance M approaches the upper limit is specified by the coefficient of coupling k, given by k = M √ L1L2 (13.36) or M = k L1L2 (13.37) where 0 ≤ k ≤ 1 or equivalently 0 ≤ M ≤ √ L1L2. The coupling coefficient is the fraction of the total flux emanating from one coil that links the other coil. For example, in Fig. 13.2, k = φ12 φ1 = φ12 φ11 + φ12 (13.38) and in Fig. 13.3, k = φ21 φ2 = φ21 φ21 + φ22 (13.39) If the entire flux produced by one coil links another coil, then k = 1 and we have 100 percent coupling, or the coils are said to be perfectly coupled. Thus, The coupling coefficient k is a measure of the magnetic coupling between two coils; 0 ≤ k ≤ 1. For k 0.5, coils are said to be loosely coupled; and for k 0.5, they are said to be tightly coupled. (a) (b) Air or ferrite core Figure13.15 Windings: (a) loosely coupled, (b) tightly coupled; cutaway view demonstrates both windings. We expect k to depend on the closeness of the two coils, their core, their orientation, and their windings. Figure 13.15 shows loosely coupled
  • 540. 538 PART 2 AC Circuits windings and tightly coupled windings. The air-core transformers used in radio frequency circuits are loosely coupled, whereas iron-core trans- formers used in power systems are tightly coupled. The linear transform- ers discussed in Section 3.4 are mostly air-core; the ideal transformers discussed in Sections 13.5 and 13.6 are principally iron-core. E X A M P L E 1 3 . 3 Consider the circuit in Fig. 13.16. Determine the coupling coefficient. Calculate the energy stored in the coupled inductors at time t = 1 s if v = 60 cos(4t + 30◦ ) V. v 10 Ω + − 5 H 4 H 2.5 H F 1 16 Figure13.16 For Example 13.3. Solution: The coupling coefficient is k = M √ L1L2 = 2.5 √ 20 = 0.56 indicating that the inductors are tightly coupled. To find the energy stored, we need to obtain the frequency-domain equivalent of the circuit. 60 cos(4t + 30◦ ) ⇒ 60 30◦ , ω = 4 rad/s 5 H ⇒ jωL1 = j20 2.5 H ⇒ jωM = j10 4 H ⇒ jωL2 = j16 1 16 F ⇒ 1 jωC = −j4 The frequency-domain equivalent is shown in Fig. 13.17. We now apply mesh analysis. For mesh 1, (10 + j20)I1 + j10I2 = 60 30◦ (13.3.1) For mesh 2, j10I1 + (j16 − j4)I2 = 0 or I1 = −1.2I2 (13.3.2) Substituting this into Eq. (13.3.1) yields I2(−12 − j14) = 60 30◦ ⇒ I2 = 3.254 − 160.6◦ A and I1 = −1.2I2 = 3.905 − 19.4◦ A In the time-domain, i1 = 3.905 cos(4t − 19.4◦ ), i2 = 3.254 cos(4t − 199.4◦ ) At time t = 1 s, 4t = 4 rad = 229.2◦ , and i1 = 3.905 cos(229.2◦ − 19.4◦ ) = −3.389 A i2 = 3.254 cos(229.2◦ + 160.6◦ ) = 2.824 A
  • 541. CHAPTER 13 Magnetically Coupled Circuits 539 The total energy stored in the coupled inductors is w = 1 2 L1i2 1 + 1 2 L2i2 2 + Mi1i2 = 1 2 (5)(−3.389)2 + 1 2 (4)(2.824)2 + 2.5(−3.389)(2.824) = 20.73 J V 10 Ω + − j20 Ω j16 Ω I1 I2 j10 −j4 Ω 60 30° Figure13.17 Frequency-domain equivalent of the circuit in Fig. 13.16. P R A C T I C E P R O B L E M 1 3 . 3 For the circuit in Fig. 13.18, determine the coupling coefficient and the energy stored in the coupled inductors at t = 1.5 s. 20 cos 2t V 4 Ω + − 2 H 1 H 1 H 2 Ω F 1 8 Figure13.18 For Practice Prob. 13.3. Answer: 0.7071, 9.85 J. 13.4 LINEAR TRANSFORMERS Hereweintroducethetransformerasanewcircuitelement. Atransformer is a magnetic device that takes advantage of the phenomenon of mutual inductance. A transformer is generally a four-terminal device comprising two (or more) magnetically coupled coils. As shown in Fig. 13.19, the coil that is directly connected to the voltage source is called the primary winding. The coil connected to the load is called the secondary winding. The resistances R1 and R2 are included to account for the losses (power dissipation) in the coils. The trans- former is said to be linear if the coils are wound on a magnetically linear
  • 542. 540 PART 2 AC Circuits material—a material for which the magnetic permeability is constant. Such materials include air, plastic, Bakelite, and wood. In fact, most ma- terials are magnetically linear. Linear transformers are sometimes called air-core transformers, although not all of them are necessarily air-core. They are used in radio and TV sets. Figure 13.20 portrays different types of transformers. Alineartransformermayalsoberegardedasone whose flux is proportional to the currents in its windings. V ZL + − L1 L2 I1 I2 M R1 R2 Primary coil Secondary coil Figure13.19 A linear transformer. (b) (a) Figure 13.20 Different types of transformers: (a) copper wound dry power transformer, (b) audio transformers. (Courtesy of: (a) Electric Service Co., (b) Jensen Transformers.) We would like to obtain the input impedance Zin as seen from the source, because Zin governs the behavior of the primary circuit. Applying KVL to the two meshes in Fig. 13.19 gives V = (R1 + jωL1)I1 − jωMI2 (13.40a) 0 = −jωMI1 + (R2 + jωL2 + ZL)I2 (13.40b)
  • 543. CHAPTER 13 Magnetically Coupled Circuits 541 In Eq. (13.40b), we express I2 in terms of I1 and substitute it into Eq. (13.40a). We get the input impedance as Zin = V I1 = R1 + jωL1 + ω2 M2 R2 + jωL2 + ZL (13.41) Notice that the input impedance comprises two terms. The first term, (R1 + jωL1), is the primary impedance. The second term is due to the coupling between the primary and secondary windings. It is as though this impedance is reflected to the primary. Thus, it is known as the reflected impedance ZR, and ZR = ω2 M2 R2 + jωL2 + ZL (13.42) It should be noted that the result in Eq. (13.41) or (13.42) is not affected by the location of the dots on the transformer, because the same result is produced when M is replaced by −M. Some authors call this the coupled impedance. The little bit of experience gained in Sections 13.2 and 13.3 in analyzingmagneticallycoupledcircuitsisenoughtoconvinceanyonethat analyzing these circuits is not as easy as circuits in previous chapters. For this reason, it is sometimes convenient to replace a magnetically coupled circuit by an equivalent circuit with no magnetic coupling. We want to replace the linear transformer in Fig. 13.19 by an equivalent T or circuit, a circuit that would have no mutual inductance. Ignore the resistances of the coils and assume that the coils have a common ground as shown in Fig. 13.21. The assumption of a common ground for the two coils is a major restriction of the equivalent circuits. A common ground is imposed on the linear transformer in Fig. 13.21 in view of the necessity of having a common ground in the equivalent T or circuit; see Figs. 13.22 and 13.23. + − M I1 V1 + − V2 I2 L1 L2 Figure13.21 Determining the equivalent circuit of a linear transformer. + − I1 V1 + − V2 I2 Lc La Lb Figure 13.22 An equivalent T circuit. + − I1 V1 + − V2 I2 LB LA LC Figure 13.23 An equivalent circuit. The voltage-current relationships for the primary and secondary coils give the matrix equation V1 V2 = jωL1 jωM jωM jωL2 I1 I2 (13.43) By matrix inversion, this can be written as I1 I2 =       L2 jω(L1L2 − M2) −M jω(L1L2 − M2) −M jω(L1L2 − M2) L1 jω(L1L2 − M2)       V1 V2 (13.44) Our goal is to match Eqs. (13.43) and (13.44) with the corresponding equations for the T and networks. For the T (or Y) network of Fig. 13.22, mesh analysis provides the terminal equations as V1 V2 = jω(La + Lc) jωLc jωLc jω(Lb + Lc) I1 I2 (13.45) If the circuits in Figs. 13.21 and 13.22 are equivalents, Eqs. (13.43) and (13.45) must be identical. Equating terms in the impedance matrices of
  • 544. 542 PART 2 AC Circuits Eqs. (13.43) and (13.45) leads to La = L1 − M, Lb = L2 − M, Lc = M (13.46) For the (or ) network in Fig. 13.23, nodal analysis gives the terminal equations as I1 I2 =       1 jωLA + 1 jωLC − 1 jωLC − 1 jωLC 1 jωLB + 1 jωLC       V1 V2 (13.47) Equating terms in admittance matrices of Eqs. (13.44) and (13.47), we obtain LA = L1L2 − M2 L2 − M , LB = L1L2 − M2 L1 − M LC = L1L2 − M2 M (13.48) Note that in Figs. 13.23 and 13.24, the inductors are not magnetically coupled. Also note that changing the locations of the dots in Fig. 13.21 can cause M to become −M. As Example 13.6 illustrates, a negative value of M is physically unrealizable but the equivalent model is still mathematically valid. ZL + − j20 Ω j40 Ω I1 I2 j5 Ω Z1 Z2 V 50 60° Figure13.24 For Example 13.4. E X A M P L E 1 3 . 4 In the circuit of Fig. 13.24, calculate the input impedance and current I1. Take Z1 = 60 − j100 , Z2 = 30 + j40 , and ZL = 80 + j60 . Solution: From Eq. (13.41), Zin = Z1 + j20 + (5)2 j40 + Z2 + ZL = 60 − j100 + j20 + 25 110 + j140 = 60 − j80 + 0.14 − 51.84◦ = 60.09 − j80.11 = 100.14 − 53.1◦
  • 545. CHAPTER 13 Magnetically Coupled Circuits 543 Thus, I1 = V Zin = 50 60◦ 100.14 − 53.1◦ = 0.5 113.1◦ A P R A C T I C E P R O B L E M 1 3 . 4 Find the input impedance of the circuit of Fig. 13.25 and the current from the voltage source. 4 Ω + − j8 Ω j10 Ω j3 Ω j4 Ω 6 Ω −j6 Ω V 10 0° Figure13.25 For Practice Prob. 13.4. Answer: 8.58 58.05◦ , 1.165 − 58.05◦ A. E X A M P L E 1 3 . 5 DeterminetheT-equivalentcircuitofthelineartransformerinFig.13.26(a). 2 H (a) 10 H 4 H a b c d (b) a b c d 2 H 8 H 2 H Figure 13.26 For Example 13.5: (a) a linear transformer, (b) its T-equivalent circuit. Solution: Given that L1 = 10, L2 = 4, and M = 2, the T equivalent network has the following parameters: La = L1 − M = 10 − 2 = 8 H Lb = L2 − M = 4 − 2 = 2 H, Lc = M = 2 H The T-equivalent circuit is shown in Fig. 13.26(b). We have assumed that reference directions for currents and voltage polarities in the primary and secondary windings conform to those in Fig. 13.21. Otherwise, we may need to replace M with −M, as Example 13.6 illustrates.
  • 546. 544 PART 2 AC Circuits P R A C T I C E P R O B L E M 1 3 . 5 For the linear transformer in Fig. 13.26 (a), find the equivalent network. Answer: LA = 18 H, LB = 4.5 H, LC = 18 H. E X A M P L E 1 3 . 6 Solve for I1, I2, and Vo in Fig. 13.27 (the same circuit as for Practice Prob. 13.1) using the T-equivalent circuit for the linear transformer. + − j8 Ω j5 Ω I1 I2 j1 Ω 4 Ω V 60 90° 10 Ω + − Vo Figure13.27 For Example 13.6. Solution: Notice that the circuit in Fig. 13.27 is the same as that in Fig. 13.10 except that the reference direction for current I2 has been reversed, just to make the reference directions for the currents for the magnetically coupled coils conform with those in Fig. 13.21. + − j1 Ω (a) (b) V1 + − V2 j8 Ω j5 Ω j9 Ω j6 Ω −j1 Ω I1 I2 Figure13.28 For Example 13.6: (a) circuit for coupled coils of Fig. 13.27, (b) T-equivalent circuit. We need to replace the magnetically coupled coils with the T- equivalent circuit. The relevant portion of the circuit in Fig. 13.27 is shown in Fig. 13.28(a). Comparing Fig. 13.28(a) with Fig. 13.21 shows that there are two differences. First, due to the current reference direc- tions and voltage polarities, we need to replace M by −M to make Fig. 13.28(a) conform with Fig. 13.21. Second, the circuit in Fig. 13.21 is in the time-domain, whereas the circuit in Fig. 13.28(a) is in the frequency- domain. The difference is the factor jω; that is, L in Fig. 13.21 has been replaced with jωL and M with jωM. Since ω is not specified, we can assume ω = 1 or any other value; it really does not matter. With these two differences in mind, La = L1 − (−M) = 8 + 1 = 9 H Lb = L2 − (−M) = 5 + 1 = 6 H, Lc = −M = −1 H Thus, the T-equivalent circuit for the coupled coils is as shown in Fig. 13.28(b). Inserting the T-equivalent circuit in Fig. 13.28(b) to replace the two coils in Fig. 13.27 gives the equivalent circuit in Fig. 13.29, which can be solved using nodal or mesh analysis. Applying mesh analysis, we obtain j6 = I1(4 + j9 − j1) + I2(−j1) (13.6.1) and 0 = I1(−j1) + I2(10 + j6 − j1) (13.6.2) From Eq. (13.6.2), I1 = (10 + j5) j I2 = (5 − j10)I2 (13.6.3)
  • 547. CHAPTER 13 Magnetically Coupled Circuits 545 Substituting Eq. (13.6.3) into Eq. (13.6.1) gives j6 = (4 + j8)(5 − j10)I2 − jI2 = (100 − j)I2 100I2 Since 100 is very large compared to 1, the imaginary part of (100 − j) can be ignored so that 100 − j 100. Hence, I2 = j6 100 = j0.06 = 0.06 90◦ A From Eq. (13.6.3), I1 = (5 − j10)j0.06 = 0.6 + j0.3 A and Vo = −10I2 = −j0.6 = 0.6 − 90◦ V This agrees with the answer to Practice Prob. 13.1. Of course, the direc- tion of I2 in Fig. 13.10 is opposite to that in Fig. 13.27. This will not affect Vo, but the value of I2 in this example is the negative of that of I2 in Practice Prob. 13.1. The advantage of using the T-equivalent model for the magnetically coupled coils is that in Fig. 13.29 we do not need to bother with the dot on the coupled coils. j6 V 4 Ω j9 Ω + − −j1 Ω I1 I2 j6 Ω 10 Ω I1 I2 + − V o Figure13.29 For Example 13.6. P R A C T I C E P R O B L E M 1 3 . 6 Solve the problem in Example 13.1 (see Fig. 13.9) using the T-equivalent model for the magnetically coupled coils. Answer: 13 − 49.4◦ A, 2.91 14.04◦ A. 13.5 IDEAL TRANSFORMERS An ideal transformer is one with perfect coupling (k = 1). It consists of two (or more) coils with a large number of turns wound on a common core of high permeability. Because of this high permeability of the core, the flux links all the turns of both coils, thereby resulting in a perfect coupling. To see how an ideal transformer is the limiting case of two cou- pled inductors where the inductances approach infinity and the coupling is perfect, let us reexamine the circuit in Fig. 13.14. In the frequency domain, V1 = jωL1I1 + jωMI2 (13.49a) V2 = jωMI1 + jωL2I2 (13.49b)
  • 548. 546 PART 2 AC Circuits From Eq. (13.49a), I1 = (V1 − jωMI2)/jωL1. Substituting this in Eq. (13.49b) gives V2 = jωL2I2 + MV1 L1 − jωM2 I2 L1 But M = √ L1L2 for perfect coupling (k = 1). Hence, V2 = jωL2I2 + √ L1L2V1 L1 − jωL1L2I2 L1 = L2 L1 V1 = nV1 where n = √ L2/L1 and is called the turns ratio. As L1, L2, M → ∞ such that n remains the same, the coupled coils become an ideal trans- former. A transformer is said to be ideal if it has the following properties: 1. Coils have very large reactances (L1, L2, M → ∞). 2. Coupling coefficient is equal to unity (k = 1). 3. Primary and secondary coils are lossless (R1 = 0 = R2). An ideal transformer is a unity-coupled, lossless transformer in which the primary and secondary coils have infinite self-inductances. Iron-core transformers are close approximations to ideal transformers. These are used in power systems and electronics. Figure 13.30(a) shows a typical ideal transformer; the circuit sym- bol is in Fig. 13.30(b). The vertical lines between the coils indicate an iron core as distinct from the air core used in linear transformers. The primary winding has N1 turns; the secondary winding has N2 turns. N1 N2 (a) (b) N1 N2 Figure13.30 (a) Ideal transformer, (b) circuit symbol for ideal transformers. ZL + − V1 V2 1:n V + − + − I1 I2 Figure13.31 Relating primary and secondary quantities in an ideal transformer. When a sinusoidal voltage is applied to the primary winding as shown in Fig. 13.31, the same magnetic flux φ goes through both wind- ings. According to Faraday’s law, the voltage across the primary winding is v1 = N1 dφ dt (13.50a) while that across the secondary winding is v2 = N2 dφ dt (13.50b) Dividing Eq. (13.50b) by Eq. (13.50a), we get v2 v1 = N2 N1 = n (13.51) where n is, again, the turns ratio or transformation ratio. We can use the phasor voltages V1 and V2 rather than the instantaneous values v1 and v2. Thus, Eq. (13.51) may be written as V2 V1 = N2 N1 = n (13.52)
  • 549. CHAPTER 13 Magnetically Coupled Circuits 547 For the reason of power conservation, the energy supplied to the primary must equal the energy absorbed by the secondary, since there are no losses in an ideal transformer. This implies that v1i1 = v2i2 (13.53) In phasor form, Eq. (13.53) in conjunction with Eq. (13.52) becomes I1 I2 = V2 V1 = n (13.54) showing that the primary and secondary currents are related to the turns ratio in the inverse manner as the voltages. Thus, I2 I1 = N1 N2 = 1 n (13.55) When n = 1, we generally call the transformer an isolation transformer. The reason will become obvious in Section 13.9.1. If n 1, we have a step-up transformer, as the voltage is increased from primary to sec- ondary (V2 V1). On the other hand, if n 1, the transformer is a step-down transformer, since the voltage is decreased from primary to secondary (V2 V1). A step-down transformer is one whose secondary voltage is less than its primary voltage. A step-up transformer is one whose secondary voltage is greater than its primary voltage. The ratings of transformers are usually specified as V1/V2. A transformer with rating 2400/120 V should have 2400 V on the primary and 120 in the secondary (i.e., a step-down transformer). Keep in mind that the voltage ratings are in rms. Power companies often generate at some convenient voltage and use a step-up transformer to increase the voltage so that the power can be transmitted at very high voltage and low current over transmission lines, resulting in significant cost savings. Near residential consumer premises, step-down transformers are used to bring the voltage down to 120 V. Section 13.9.3 will elaborate on this. It is important that we know how to get the proper polarity of the voltages and the direction of the currents for the transformer in Fig. 13.31. If the polarity of V1 or V2 or the direction of I1 or I2 is changed, n in Eqs. (13.51) to (13.55) may need to be replaced by −n. The two simple rules to follow are: 1. If V1 and V2 are both positive or both negative at the dotted terminals, use +n in Eq. (13.52). Otherwise, use −n. 2. If I1 and I2 both enter into or both leave the dotted terminals, use −n in Eq. (13.55). Otherwise, use +n. The rules are demonstrated with the four circuits in Fig. 13.32. V1 V2 N1:N2 + − + − I1 I2 V2 V1 N2 N1 = I2 I1 N1 N2 = (a) V1 V2 N1:N2 + − + − I1 I2 V2 V1 N2 N1 = I2 I1 N1 N2 = − (b) V1 V2 N1:N2 + − + − I1 I2 V2 V1 N2 N1 = − V2 V1 N2 N1 = − I2 I1 N1 N2 = (c) V1 V2 N1:N2 + − + − I1 I2 I2 I1 N1 N2 = − (d) Figure13.32 Typical circuits illustrating proper voltage polarities and current directions in an ideal transformer.
  • 550. 548 PART 2 AC Circuits Using Eqs. (13.52) and (13.55), we can always express V1 in terms of V2 and I1 in terms of I2, or vice versa: V1 = V2 n or V2 = nV1 (13.56) I1 = nI2 or I2 = I1 n (13.57) The complex power in the primary winding is S1 = V1I∗ 1 = V2 n (nI2)∗ = V2I∗ 2 = S2 (13.58) showing that the complex power supplied to the primary is delivered to the secondary without loss. The transformer absorbs no power. Of course, we should expect this, since the ideal transformer is lossless. The input impedance as seen by the source in Fig. 13.31 is found from Eqs. (13.56) and (13.57) as Zin = V1 I1 = 1 n2 V2 I2 (13.59) It is evident from Fig. 13.31 that V2/I2 = ZL, so that Zin = ZL n2 (13.60) The input impedance is also called the reflected impedance, since it ap- pears as if the load impedance is reflected to the primary side. This ability of the transformer to transform a given impedance into another impedance provides us a means of impedance matching to ensure maximum power transfer. The idea of impedance matching is very useful in practice and will be discussed more in Section 13.9.2. Notice that an ideal transformer reflects an im- pedance as the square of the turns ratio. In analyzing a circuit containing an ideal transformer, it is common practicetoeliminatethetransformerbyreflectingimpedancesandsources from one side of the transformer to the other. In the circuit of Fig. 13.33, suppose we want to reflect the secondary side of the circuit to the primary side. We find the Thevenin equivalent of the circuit to the right of the terminals a-b. We obtain VTh as the open-circuit voltage at terminals a-b, as shown in Fig. 13.34(a). Since terminals a-b are open, I1 = 0 = I2 so that V2 = Vs2. Hence, from Eq. (13.56), VTh = V1 = V2 n = Vs2 n (13.61) + − Z1 Z2 Vs1 + − Vs2 I1 I2 a b c d V1 V2 + − + − 1:n Figure 13.33 Ideal transformer circuit whose equivalent circuits are to be found.
  • 551. CHAPTER 13 Magnetically Coupled Circuits 549 Z2 + − Vs2 I1 I2 a b a b V1 VTh V2 + − + − 1:n (a) + − I1 I2 V1 V2 + − + − 1:n (b) + − Z2 V 1 0° Figure13.34 (a) Obtaining VTh for the circuit in Fig. 13.33, (b) obtaining ZTh for the circuit in Fig. 13.33. To get ZTh, we remove the voltage source in the secondary winding and insert a unit source at terminals a-b, as in Fig. 13.34(b). From Eqs. (13.56) and (13.57), I1 = nI2 and V1 = V2/n, so that ZTh = V1 I1 = V2/n nI2 = Z2 n2 , V2 = Z2I2 (13.62) which is what we should have expected from Eq. (13.60). Once we have VTh and ZTh, we add the Thevenin equivalent to the part of the circuit in Fig. 13.33 to the left of terminals a-b. Figure 13.35 shows the result. + − Z1 Vs1 + − Vs2 n a b V1 + − n2 Z2 Figure13.35 Equivalent circuit for Fig. 13.33 obtained by reflecting the secondary circuit to the primary side. The general rule for eliminating the transformer and reflecting the secondary circuit to the primary side is: divide the secondary impedance by n2 , divide the secondary voltage by n, and multiply the secondary current by n. We can also reflect the primary side of the circuit in Fig. 13.33 to the secondary side. Figure 13.36 shows the equivalent circuit. The rule for eliminating the transformer and reflecting the primary circuit to the secondary side is: multiply the primary impedance by n2 , multiply the primary voltage by n, and divide the primary current by n. According to Eq. (13.58), the power remains the same, whether calculated on the primary or the secondary side. But realize that this reflection approach only applies if there are no external connections between the primary and secondary windings. When we have external connections between the primary and secondary windings, we simply use regular mesh and nodal analysis. Examples of circuits where there are external connections between the primary and secondary windings are in Figs. 13.39 and 13.40. Also note that if the locations of the dots in Fig. 13.33 are changed, we might have to replace n by −n in order to obey the dot rule, illustrated in Fig. 13.32. + − n2 Z1 Z2 nVs1 Vs2 + − c d V2 + − Figure13.36 Equivalent circuit for Fig. 13.33 obtained by reflecting the primary circuit to the secondary side. E X A M P L E 1 3 . 7 An ideal transformer is rated at 2400/120 V, 9.6 kVA, and has 50 turns on the secondary side. Calculate: (a) the turns ratio, (b) the number of
  • 552. 550 PART 2 AC Circuits turns on the primary side, and (c) the current ratings for the primary and secondary windings. Solution: (a) This is a step-down transformer, since V1 = 2400 V V2 = 120 V. n = V2 V1 = 120 2400 = 0.05 (b) n = N2 N1 ⇒ 0.05 = 50 N1 or N1 = 50 0.05 = 1000 turns (c) S = V1I1 = V2I2 = 9.6 kVA. Hence, I1 = 9600 V1 = 9600 2400 = 4 A I2 = 9600 V2 = 9600 120 = 80 A or I2 = I1 n = 4 0.05 = 80 A P R A C T I C E P R O B L E M 1 3 . 7 The primary current to an ideal transformer rated at 3300/110 V is 3 A. Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current. Answer: (a) 1/30, (b) 9.9 kVA, (c) 90 A. E X A M P L E 1 3 . 8 For the ideal transformer circuit of Fig. 13.37, find: (a) the source current I1, (b) the output voltage Vo, and (c) the complex power supplied by the source. 4 Ω + − 20 Ω −j6 Ω V1 V2 Vo 1:2 + − + − I1 I2 V rms 120 0° + − Figure13.37 For Example 13.8. Solution: (a) The 20- impedance can be reflected to the primary side and we get ZR = 20 n2 = 20 4 = 5 Thus, Zin = 4 − j6 + ZR = 9 − j6 = 10.82 − 33.69◦ I1 = 120 0◦ Zin = 120 0◦ 10.82 − 33.69◦ = 11.09 33.69◦ A
  • 553. CHAPTER 13 Magnetically Coupled Circuits 551 (b) Since both I1 and I2 leave the dotted terminals, I2 = − 1 n I1 = −5.545 33.69◦ A Vo = 20I2 = 110.9 213.69◦ V (c) The complex power supplied is S = VsI∗ 1 = (120 0◦ )(11.09 − 33.69◦ ) = 1330.8 − 33.69◦ VA P R A C T I C E P R O B L E M 1 3 . 8 In the ideal transformer circuit of Fig. 13.38, find Vo and the complex power supplied by the source. 2 Ω + − 16 Ω V1 V2 Vo 1:4 + − + − I1 I2 V rms 100 0° + − −j24 Ω Figure13.38 For Practice Prob. 13.8. Answer: 178.9 116.56◦ V, 2981.5 − 26.56◦ VA. E X A M P L E 1 3 . 9 Calculate the power supplied to the 10- resistor in the ideal transformer circuit of Fig. 13.39. 20 Ω 10 Ω V1 V2 2:1 + − + − V rms 120 0° 30 Ω + − I1 I2 Figure13.39 For Example 13.9. Solution: Reflection to the secondary or primary side cannot be done with this circuit: there is direct connection between the primary and secondary sides due to the 30- resistor. We apply mesh analysis. For mesh 1, −120 + (20 + 30)I1 − 30I2 + V1 = 0 or 50I1 − 30I2 + V1 = 120 (13.9.1)
  • 554. 552 PART 2 AC Circuits For mesh 2, −V2 + (10 + 30)I2 − 30I1 = 0 or −30I1 + 40I2 − V2 = 0 (13.9.2) At the transformer terminals, V2 = − 1 2 V1 (13.9.3) I2 = −2I1 (13.9.4) (Note that n = 1/2.) We now have four equations and four unknowns, but our goal is to get I2. So we substitute for V1 and I1 in terms of V2 and I2 in Eqs. (13.9.1) and (13.9.2). Equation (13.9.1) becomes −55I2 − 2V2 = 120 (13.9.5) and Eq. (13.9.2) becomes 15I2 + 40I2 − V2 = 0 ⇒ V2 = 55I2 (13.9.6) Substituting Eq. (13.9.6) in Eq. (13.9.5), −165I2 = 120 ⇒ I2 = − 120 165 = −0.7272 A The power absorbed by the 10- resistor is P = (−0.7272)2 (10) = 5.3 W P R A C T I C E P R O B L E M 1 3 . 9 Find Vo in the circuit in Fig. 13.40. 4 Ω 8 Ω 1:2 V 60 0° + − 2 Ω 8 Ω + − Vo Figure13.40 For Practice Prob. 13.9. Answer: 24 V. 13.6 IDEAL AUTOTRANSFORMERS Unlike the conventional two-winding transformer we have considered so far, an autotransformer has a single continuous winding with a connection point called a tap between the primary and secondary sides. The tap is
  • 555. CHAPTER 13 Magnetically Coupled Circuits 553 often adjustable so as to provide the desired turns ratio for stepping up or stepping down the voltage. This way, a variable voltage is provided to the load connected to the autotransformer. An autotransformer is a transformer in which both the primary and the secondary are in a single winding. Figure 13.41 A typical autotransformer. (Courtesy of Todd Systems, Inc.) Figure 13.41 shows a typical autotransformer. As shown in Fig. 13.42, the autotransformer can operate in the step-down or step-up mode. The autotransformer is a type of power transformer. Its major advantage over the two-winding transformer is its ability to transfer larger apparent power. Example 13.10 will demonstrate this. Another advantage is that an autotransformer is smaller and lighter than an equivalent two-winding transformer. However, since both the primary and secondary windings are one winding, electrical isolation (no direct electrical connection) is lost. (We will see how the property of electrical isolation in the conven- tional transformer is practically employed in Section 13.9.1.) The lack of electrical isolation between the primary and secondary windings is a major disadvantage of the autotransformer. Some of the formulas we derived for ideal transformers apply to ideal autotransformers as well. For the step-down autotransformer circuit of Fig. 13.42(a), Eq. (13.52) gives V1 V2 = N1 + N2 N2 = 1 + N1 N2 (13.63) As an ideal autotransformer, there are no losses, so the complex power remains the same in the primary and secondary windings: S1 = V1I∗ 1 = S2 = V2I∗ 2 (13.64) Equation (13.64) can also be expressed with rms values as V1I1 = V2I2 or V2 V1 = I1 I2 (13.65) Thus, the current relationship is I1 I2 = N2 N1 + N2 (13.66) + − I1 I2 V1 N1 N2 + − V2 + − (a) V + − I1 I2 V1 N1 N2 + − V2 + − (b) ZL ZL V Figure 13.42 (a) Step-down autotransformer, (b) step-up autotransformer. For the step-up autotransformer circuit of Fig. 13.42(b), V1 N1 = V2 N1 + N2 or V1 V2 = N1 N1 + N2 (13.67)
  • 556. 554 PART 2 AC Circuits The complex power given by Eq. (13.64) also applies to the step-up auto- transformer so that Eq. (13.65) again applies. Hence, the current relation- ship is I1 I2 = N1 + N2 N1 = 1 + N2 N1 (13.68) A major difference between conventional transformers and auto- transformers is that the primary and secondary sides of the autotrans- former are not only coupled magnetically but also coupled conductively. The autotransformer can be used in place of a conventional transformer when electrical isolation is not required. E X A M P L E 1 3 . 1 0 Compare the power ratings of the two-winding transformer in Fig. 13.43(a) and the autotransformer in Fig. 13.43(b). + − (a) (b) 240 V + − 12 V Vs Vp 0.2 A 4 A 4 A + − Vs = 12 V + − Vp = 240 V + − + − + − 240 V + − 252 V 4 A 0.2 A Figure13.43 For Example 13.10. Solution: Although the primary and secondary windings of the autotransformer are together as a continuous winding, they are separated in Fig. 13.43(b) for clarity. We note that the current and voltage of each winding of the autotransformer in Fig. 13.43(b) are the same as those for the two-winding transformer in Fig. 13.43(a). This is the basis of comparing their power ratings. For the two-winding transformer, the power rating is S1 = 0.2(240) = 48 VA or S2 = 4(12) = 48 VA For the autotransformer, the power rating is S1 = 4.2(240) = 1008 VA or S2 = 4(252) = 1008 VA which is 21 times the power rating of the two-winding transformer. P R A C T I C E P R O B L E M 1 3 . 1 0 Refer to Fig. 13.43. If the two-winding transformer is a 60-VA, 120 V/10 V transformer, what is the power rating of the autotransformer? Answer: 780 VA.
  • 557. CHAPTER 13 Magnetically Coupled Circuits 555 E X A M P L E 1 3 . 1 1 Refer to the autotransformer circuit in Fig. 13.44. Calculate: (a) I1, I2, and Io if ZL = 8+j6 , and (b) the complex power supplied to the load. + − I1 I2 V1 120 turns 80 turns + − V2 + − ZL Io 120 30° V rms Figure13.44 For Example 13.11. Solution: (a) This is a step-up autotransformer with N1 = 80, N2 = 120, V1 = 120 30◦ , so Eq. (13.67) can be used to find V2 by V1 V2 = N1 N1 + N2 = 80 200 or V2 = 200 80 V1 = 200 80 (120 30◦ ) = 300 30◦ V I2 = V2 ZL = 300 30◦ 8 + j6 = 300 30◦ 10 36.87◦ = 30 − 6.87◦ A But I1 I2 = N1 + N2 N1 = 200 80 or I1 = 200 80 I2 = 200 80 (30 − 6.87◦ ) = 75 − 6.87◦ A At the tap, KCL gives I1 + Io = I2 or Io = I2 − I1 = 30 − 6.87◦ − 75 − 6.87◦ = 45 173.13◦ A (b) The complex power supplied to the load is S2 = V2I∗ 2 = |I2|2 ZL = (30)2 (10 36.87◦ ) = 9 36.87◦ kVA
  • 558. 556 PART 2 AC Circuits P R A C T I C E P R O B L E M 1 3 . 1 1 In the autotransformer circuit in Fig. 13.45, find currents I1, I2, and Io. Take V1 = 1250 V, V2 = 800 V. I1 I2 V2 + − V1 + − Io 16 kVA load Figure13.45 For Practice Prob. 13.11. Answer: 12.8 A, 20 A, 7.2 A. †13.7 THREE-PHASE TRANSFORMERS To meet the demand for three-phase power transmission, transformer connections compatible with three-phase operations are needed. We can achieve the transformer connections in two ways: by connecting three single-phase transformers, thereby forming a so-called transformer bank, or by using a special three-phase transformer. For the same kVA rat- ing, a three-phase transformer is always smaller and cheaper than three single-phase transformers. When single-phase transformers are used, one must ensure that they have the same turns ratio n to achieve a balanced three-phase system. There are four standard ways of connecting three single-phase transformers or a three-phase transformer for three-phase operations: Y-Y, -, Y-, and -Y. For any of the four connections, the total apparent power ST , real power PT , and reactive power QT are obtained as ST = √ 3VLIL (13.69a) PT = ST cos θ = √ 3VLIL cos θ (13.69b) QT = ST sin θ = √ 3VLIL sin θ (13.69c) where VL and IL are, respectively, equal to the line voltage VLp and the line current ILp for the primary side, or the line voltage VLs and the line current ILs for the secondary side. Notice from Eq. (13.69) that for each of the four connections, VLsILs = VLpILp, since power must be conserved in an ideal transformer. For the Y-Y connection (Fig. 13.46), the line voltage VLp at the primary side, the line voltage VLs on the secondary side, the line current ILp on the primary side, and the line current ILs on the secondary side are related to the transformer per phase turns ratio n according to Eqs. (13.52) and (13.55) as VLs = nVLp (13.70a) ILs = ILp n (13.70b)
  • 559. CHAPTER 13 Magnetically Coupled Circuits 557 For the - connection (Fig. 13.47), Eq. (13.70) also applies for the line voltages and line currents. This connection is unique in the sense that if one of the transformers is removed for repair or maintenance, the other two form an open delta, which can provide three-phase voltages at a reduced level of the original three-phase transformer. + − VLp + − VLs = nVLp ILp ILs = ILp n 1:n Figure13.46 Y-Y three-phase transformer connection. + − VLp + − VLs = nVLp ILp ILs = ILp n 1:n Figure13.47 - three-phase transformer connection. For the Y- connection (Fig. 13.48), there is a factor of √ 3 arising from the line-phase values in addition to the transformer per phase turns ratio n. Thus, VLs = nVLp √ 3 (13.71a) ILs = √ 3ILp n (13.71b) Similarly, for the -Y connection (Fig. 13.49), VLs = n √ 3VLp (13.72a) ILs = ILp n √ 3 (13.72b) + − VLp + − ILp 1:n ILs = 3ILp n VLs = nVLp 3 Figure13.48 Y- three-phase transformer connection. + − VLs = n VLp ILp 1:n ILs = ILp 3 n 3 + − VLp Figure13.49 -Y three-phase transformer connection.
  • 560. 558 PART 2 AC Circuits E X A M P L E 1 3 . 1 2 The 42-kVA balanced load depicted in Fig. 13.50 is supplied by a three- phase transformer. (a) Determine the type of transformer connections. (b) Find the line voltage and current on the primary side. (c) Determine the kVA rating of each transformer used in the transformer bank. Assume that the transformers are ideal. a b c A B C 42 kVA Three-phase load 240 V 1:5 Figure13.50 For Example 13.12. Solution: (a) A careful observation of Fig. 13.50 shows that the primary side is Y-connected, while the secondary side is -connected. Thus, the three- phase transformer is Y-, similar to the one shown in Fig. 13.48. (b) Given a load with total apparent power ST = 42 kVA, the turns ra- tio n = 5, and the secondary line voltage VLs = 240 V, we can find the secondary line current using Eq. (13.69a), by ILs = ST √ 3VLs = 42,000 √ 3(240) = 101 A From Eq. (13.71), ILp = n √ 3 ILs = 5 × 101 √ 3 = 292 A VLp = √ 3 n VLs = √ 3 × 240 5 = 83.14 V (c) Because the load is balanced, each transformer equally shares the total load and since there are no losses (assuming ideal transformers), the kVA rating of each transformer is S = ST /3 = 14 kVA. Alternatively, the transformer rating can be determined by the product of the phase current and phase voltage of the primary or secondary side. For the primary side, for example, we have a delta connection, so that the phase voltage is the same as the line voltage of 240 V, while the phase current is ILp/ √ 3 = 58.34 A. Hence, S = 240 × 58.34 = 14 kVA. P R A C T I C E P R O B L E M 1 3 . 1 2 A three-phase - transformer is used to step down a line voltage of 625 kV, to supply a plant operating at a line voltage of 12.5 kV. The plant
  • 561. CHAPTER 13 Magnetically Coupled Circuits 559 draws 40 MW with a lagging power factor of 85 percent. Find: (a) the currentdrawnbytheplant, (b)theturnsratio, (c)thecurrentontheprimary side of the transformer, and (d) the load carried by each transformer. Answer: (a) 2.1736 kA, (b) 0.02, (c) 43.47 A, (d) 15.69 MVA. 13.8 PSPICE ANALYSIS OF MAGNETICALLY COUPLED CIRCUITS PSpice analyzes magnetically coupled circuits just like inductor circuits except that the dot convention must be followed. In PSpice Schematic, the dot (not shown) is always next to pin 1, which is the left-hand terminal of the inductor when the inductor with part name L is placed (horizontally) without rotation on a schematic. Thus, the dot or pin 1 will be at the bottom after one 90◦ counterclockwise rotation, since rotation is always about pin 1. Once the magnetically coupled inductors are arranged with the dot convention in mind and their value attributes are set in henries, we use the coupling symbol K−LINEAR to define the coupling. For each pair of coupled inductors, take the following steps: 1. Select Draw/Get New Part and type K−LINEAR. 2. Hit enter or click OK and place the K−LINEAR symbol on the schematic, as shown in Fig. 13.51. (Notice that K−LINEAR is not a component and therefore has no pins.) 3. DCLICKL on COUPLING and set the value of the coupling coefficient k. 4. DCLICKL on the boxed K (the coupling symbol) and enter the reference designator names for the coupled inductors as values of Li, i = 1, 2, . . . , 6. For example, if inductors L20 and L23 are coupled, we set L1 = L20 and L2 = L23. L1 and at least one other Li must be assigned values; other Li’s may be left blank. In step 4, up to six coupled inductors with equal coupling can be specified. For the air-core transformer, the partname is XFRM−LINEAR. It can be inserted in a circuit by selecting Draw/Get Part Name and then typing in the part name or by selecting the part name from the analog.slb library. As shown typically in Fig. 13.51, the main attributes of the linear transformer are the coupling coefficient k and the inductance values L1 and L2 in henries. If the mutual inductance M is specified, its value must be used along with L1 and L2 to calculate k. Keep in mind that the value of k should lie between 0 and 1. TX2 COUPLING=0.5 L1_VALUE=1mH L2_VALUE=25mH Figure13.51 Linear trans- former XFRM LINEAR. TX4 COUPLING=0.5 L1_TURNS=500 L2_TURNS=1000 kbreak Figure13.52 Ideal trans- former XFRM NONLINEAR. For the ideal transformer, the part name is XFRM−NONLINEAR and is located in the breakout.slb library. Select it by clicking Draw/Get Part Name and then typing in the part name. Its attributes are the cou- pling coefficient and the numbers of turns associated with L1 and L2, as illustrated typically in Fig. 13.52. The value of the coefficient of mutual coupling must lie between 0 and 1.
  • 562. 560 PART 2 AC Circuits PSpice has some additional transformer configurations that we will not discuss here. E X A M P L E 1 3 . 1 3 Use PSpice to find i1, i2, and i3 in the circuit displayed in Fig. 13.53. 40 cos 12pt V 60 cos (12pt – 10°) V 4 H 2 H 1.5 H 270 mF 3 H 3 H 1 H 2 H + − + − 70 Ω 100 Ω i2 i1 i3 Figure13.53 For Example 13.13. Solution: The coupling coefficients of the three coupled inductors are determined as follows. k12 = M12 √ L1L2 = 1 √ 3 × 3 = 0.3333 k13 = M13 √ L1L3 = 1.5 √ 3 × 4 = 0.433 k23 = M23 √ L2L3 = 2 √ 3 × 4 = 0.5774 The operating frequency f is obtained from Fig. 13.53 as ω = 12π = 2πf → f = 6 Hz. The right-hand values are the reference designa- tors of the inductors on the schematic. The schematic of the circuit is portrayed in Fig. 13.54. Notice how the dot convention is adhered to. For L2, the dot (not shown) is on pin 1 (the left-hand terminal) and is therefore placed without rotation. For L1, in order for the dot to be on the right-hand side of the inductor, the inductor must be rotated through 180◦ . For L3, the inductor must be rotated through 90◦ so that the dot will be at the bottom. Note that the 2-H inductor (L4) is not coupled. To handle the three coupled inductors, we use three K−LINEAR parts provided in the analog library and set the following attributes (by double-clicking on the symbol K in the box): K1 - K_LINEAR L1 = L1 L2 = L2 COUPLING = 0.3333 K2 - K_LINEAR L1 = L2 L2 = L3 COUPLING = 0.433
  • 563. CHAPTER 13 Magnetically Coupled Circuits 561 K3 - K_LINEAR L1 = L1 L2 = L3 COUPLING = 0.5774 ThreeIPRINTpseudocomponentsareinsertedintheappropriatebranches to obtain the required currents i1, i2, and i3. As an AC single-frequency analysis, we select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq = 6, and Final Freq = 6. After saving the schematic, we select Analysis/Simulate to simulate it. The output file includes: FREQ IM(V_PRINT2) IP(V_PRINT2) 6.000E+00 2.114E-01 -7.575E+01 FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E+00 4.654E-01 -7.025E+01 FREQ IM(V_PRINT3) IP(V_PRINT3) 6.000E+00 1.095E-01 1.715E+01 From this we obtain I1 = 0.4654 − 70.25◦ I2 = 0.2114 − 75.75◦ , I3 = 0.1095 17.15◦ Thus, i1 = 0.4654 cos(12πt − 70.25◦ ) A i2 = 0.2114 cos(12πt − 75.75◦ ) A i3 = 0.1095 cos(12πt + 17.15◦ ) A V2 4H L3 270u 0 L2 3H 3H L1 L4 R1 2H 70 R2 100 − + ACMAG=40V ACPHASE=0 V1 − + ACMAG=60V ACPHASE=-10 MAG=ok AC=ok PHASE=ok K K1 K_Linear COUPLING=0.3333 L1=L1 L2=L2 K K2 K_Linear COUPLING=0.433 L1=L2 L2=L3 K K3 K_Linear COUPLING=0.5774 L1=L1 L2=L3 IPRINT IPRINT IPRINT C1 Figure13.54 Schematic of the circuit of Fig. 13.53. P R A C T I C E P R O B L E M 1 3 . 1 3 Find io in the circuit of Fig. 13.55.
  • 564. 562 PART 2 AC Circuits 20 Ω 6 H 5 H 4 H 8 cos (4t + 50°) V + − 10 Ω 8 Ω 12 Ω k = 0.4 25 mF io Figure13.55 For Practice Prob. 13.13. Answer: 0.1006 cos(4t + 68.52◦ ) A. E X A M P L E 1 3 . 1 4 Find V1 and V2 in the ideal transformer circuit of Fig. 13.56 using PSpice. 80 Ω 6 Ω V1 V2 4:1 + − + − V 20 Ω + − 120 30° −j40 Ω j10 Ω Figure13.56 For Example 13.14. Solution: As usual, we assume ω = 1 and find the corresponding values of capac- itance and inductance of the elements: j10 = jωL ⇒ L = 10 H −j40 = 1 jωC ⇒ C = 25 mF Figure 13.57 shows the schematic. For the ideal transformer, we set the coupling factor to 0.999 and the numbers of turns to 400,000 and 100,000. The two VPRINT2 pseudocomponents are connected across the transformer terminals to obtain V1 and V2. As a single-frequency analysis, we select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq = 0.1592, and Final Freq = 0.1592. After saving the schematic, we select Analysis/Simulate to simulate it. The output file includes: Reminder: For an ideal transformer, the induc- tances of both the primary and secondary wind- ings are infinitely large. FREQ VM(C,A) VP(C,A) 1.592E-01 1.212E+02 -1.435E+02
  • 565. CHAPTER 13 Magnetically Coupled Circuits 563 FREQ VM(B,C) VP(B,C) 1.592E-01 2.775E+02 2.789E+01 From this we obtain V1 = −V(C,A) = 121.1 36.5◦ V V2 = V(B,C) = 27.75 27.89◦ V R1 V1 80 6 R3 10H L1 TX1 ACMAG=120V ACPHASE=30 AC=ok MAG=ok PHASE=ok AC=ok MAG=ok PHASE=ok L1_TURNS=400000 L2_TURNS=100000 COUPLING=0.999 20 R2 C1 A B C 25m − + 0 kbreak Figure13.57 The schematic for the circuit in Fig. 13.56. P R A C T I C E P R O B L E M 1 3 . 1 4 Obtain V1 and V2 in the circuit of Fig. 13.58 using PSpice. 10 Ω 2:3 30 Ω 20 Ω V1 V2 + − + − V + − 100 20° −j16 Ω j15 Ω Figure13.58 For Practice Prob. 13.14. Answer: 63.1 28.65◦ V, 94.64 − 151.4◦ V. †13.9 APPLICATIONS Transformers are the largest, the heaviest, and often the costliest of circuit components. Nevertheless, they are indispensable passive devices in
  • 566. 564 PART 2 AC Circuits electric circuits. They are among the most efficient machines, 95 percent efficiency being common and 99 percent being achievable. They have numerous applications. For example, transformers are used: • To step up or step down voltage and current, making them useful for power transmission and distribution. • To isolate one portion of a circuit from another (i.e., to transfer power without any electrical connection). • As an impedance-matching device for maximum power transfer. • In frequency-selective circuits whose operation depends on the response of inductances. Because of these diverse uses, there are many special designs for transformers (only some of which are discussed in this chapter): voltage transformers, currenttransformers, powertransformers, distributiontrans- formers, impedance-matching transformers, audio transformers, single- phase transformers, three-phase transformers, rectifier transformers, inverter transformers, and more. In this section, we consider three im- portant applications: transformer as an isolation device, transformer as a matching device, and power distribution system. Formoreinformationonthemanykindsoftrans- formers, a good text is W. M. Flanagan, Hand- book of Transformer Design and Applications, 2nd ed. (New York: McGraw-Hill, 1993). 13.9.1 Transformer as an Isolation Device Electrical isolation is said to exist between two devices when there is no physical connection between them. In a transformer, energy is transferred by magnetic coupling, without electrical connection between the primary circuit and secondary circuit. We now consider three simple practical examples of how we take advantage of this property. First, consider the circuit in Fig. 13.59. A rectifier is an electronic circuit that converts an ac supply to a dc supply. A transformer is often used to couple the ac supply to the rectifier. The transformer serves two purposes. First, it steps up or steps down the voltage. Second, it provides electrical isolation between the ac power supply and the rectifier, thereby reducing the risk of shock hazard in handling the electronic device. va + − 1:n Fuse Rectifier Isolation transformer Figure 13.59 A transformer used to isolate an ac supply from a rectifier. As a second example, a transformer is often used to couple two stages of an amplifier, to prevent any dc voltage in one stage from affecting the dc bias of the next stage. Biasing is the application of a dc voltage to a transistor amplifier or any other electronic device in order to produce a desired mode of operation. Each amplifier stage is biased separately to operate in a particular mode; the desired mode of operation will be compromised without a transformer providing dc isolation. As shown in Fig. 13.60, only the ac signal is coupled through the transformer from one stage to the next. We recall that magnetic coupling does not exist with a dc voltage source. Transformers are used in radio and TV receivers to couple stages of high-frequency amplifiers. When the sole purpose of a transformer is to provide isolation, its turns ratio n is made unity. Thus, an isolation transformer has n = 1. As a third example, consider measuring the voltage across 13.2-kV lines. It is obviously not safe to connect a voltmeter directly to such high-voltage lines. A transformer can be used both to electrically isolate the line power from the voltmeter and to step down the voltage to a safe level, as shown in Fig. 13.61. Once the voltmeter is used to measure the
  • 567. CHAPTER 13 Magnetically Coupled Circuits 565 secondary voltage, the turns ratio is used to determine the line voltage on the primary side. 1:1 Amplifier stage 2 Amplifier stage 1 Isolation transformer ac only ac + dc Figure13.60 A transformer providing dc isolation between two amplifier stages. 1:n V 120 V – + 13,200 V – + Power lines Voltmeter Figure13.61 A transformer providing isolation between the power lines and the voltmeter. E X A M P L E 1 3 . 1 5 Determine the voltage across the load in Fig. 13.62. Solution: We can apply the superposition principle to find the load voltage. Let vL = vL1 + vL2, where vL1 is due to the dc source and vL2 is due to the ac source. We consider the dc and ac sources separately, as shown in Fig. 13.63. The load voltage due to the dc source is zero, because a time- varying voltage is necessary in the primary circuit to induce a voltage in the secondary circuit. Thus, vL1 = 0. For the ac source, V2 V1 = V2 120 = 1 3 or V2 = 120 3 = 40 V Hence, VL2 = 40 V ac or vL2 = 40 cos ωt; that is, only the ac voltage is passed to the load by the transformer. This example shows how the transformer provides dc isolation. 3:1 + − 120 V ac 12 V dc RL = 5 kΩ Figure13.62 For Example 13.15. 3:1 6 V dc RL V2 = 0 3:1 + − 120 V ac RL + − + − V2 V1 + − (a) (b) Figure13.63 For Example 13.15: (a) dc source, (b) ac source. P R A C T I C E P R O B L E M 1 3 . 1 5 Refer to Fig. 13.61. Calculate the turns ratio required to step down the 13.2-kV line voltage to a safe level of 120 V. Answer: 1/110.
  • 568. 566 PART 2 AC Circuits 13.9.2 Transformer as a Matching Device Werecallthatformaximumpowertransfer, theloadresistanceRL mustbe matched with the source resistance Rs. In most cases, the two resistances are not matched; both are fixed and cannot be altered. However, an iron- core transformer can be used to match the load resistance to the source resistance. This is called impedance matching. For example, to connect a loudspeaker to an audio power amplifier requires a transformer, because the speaker’s resistance is only a few ohms while the internal resistance of the amplifier is several thousand ohms. ConsiderthecircuitshowninFig.13.64. WerecallfromEq.(13.60) that the ideal transformer reflects its load back to the primary with a scaling factor of n2 . To match this reflected load RL/n2 with the source resistance Rs, we set them equal, Rs = RL n2 (13.73) Equation (13.73) can be satisfied by proper selection of the turns ratio n. From Eq. (13.73), we notice that a step-down transformer (n 1) is needed as the matching device when Rs RL, and a step-up (n 1) is required when Rs RL. vs + − 1:n Matching transformer Load Source Rs RL Figure 13.64 Transformer used as a matching device. E X A M P L E 1 3 . 1 6 The ideal transformer in Fig. 13.65 is used to match the amplifier circuit to the loudspeaker to achieve maximum power transfer. The Thevenin (or output) impedance of the amplifier is 192 , and the internal impedance of the speaker is 12 . Determine the required turns ratio. Amplifier circuit 1:n Speaker Figure13.65 Using an ideal transformer to match the speaker to the amplifier; for Example 13.16. Solution: We replace the amplifier circuit with the Thevenin equivalent and reflect the impedance ZL = 12 of the speaker to the primary side of the ideal transformer. Figure 13.66 shows the result. For maximum power transfer, ZTh = ZL n2 or n2 = ZL ZTh = 12 192 = 1 16 Thus, the turns ratio is n = 1/4 = 0.25. VTh ZL n2 ZTh + − Figure13.66 Equivalent circuit of the circuit in Fig. 13.65, for Example 13.16. Using P = I2 R, we can show that indeed the power delivered to the speaker is much larger than without the ideal transformer. Without the ideal transformer, the amplifier is directly connected to the speaker. The power delivered to the speaker is PL = VTh ZTh + ZL 2 ZL = 288 V2 Th µW With the transformer in place, the primary and secondary currents are Ip = VTh ZTh + ZL/n2 , Is = Ip n
  • 569. CHAPTER 13 Magnetically Coupled Circuits 567 Hence, PL = I2 s ZL = VTh/n ZTh + ZL/n2 2 ZL = nVTh n2ZTh + ZL 2 ZL = 1302V2 Th µW confirming what was said earlier. P R A C T I C E P R O B L E M 1 3 . 1 6 Calculate the turns ratio of an ideal transformer required to match a 100- load to a source with internal impedance of 2.5 k. Find the load voltage when the source voltage is 30 V. Answer: 0.2, 3 V. One may ask, How would increasing the voltage not increase the current, thereby increasing I2 R losses? Keep in mind that I = V-/R, where V- is the potential difference between the sending and receiving ends of the line. The voltage that is stepped up is the sending end voltage V, not V-. If the receiving end is VR, then V- = V − VR. Since V and VR are close to each other, V- is small even when V is stepped up. 13.9.3 Power Distribution A power system basically consists of three components:generation, trans- mission, and distribution. The local electric company operates a plant that generates several hundreds of megavolt-amperes (MVA), typically at about 18 kV. As Fig. 13.67 illustrates, three-phase step-up transformers are used to feed the generated power to the transmission line. Why do we need the transformer? Suppose we need to transmit 100,000 VA over a distance of 50 km. Since S = V I, using a line voltage of 1000 V implies that the transmission line must carry 100 A and this requires a transmission line of a large diameter. If, on the other hand, we use a line voltage of 10,000 V, the current is only 10 A. The smaller current reduces the required conductor size, producing considerable savings as well as minimizing transmission line I2 R losses. To minimize losses requires a step-up transformer. Without the transformer, the majority of the power generated would be lost on the transmission line. The ability 3f 345,000 V 3f 60 Hz ac 18,000 V Generator 3f Step-up transformer 3f Step-down transformer Neutral 3f 60 Hz ac 208 V 345,000 V 345,000 V Neutral Neutral Neutral Tower Tower Insulators Figure 13.67 A typical power distribution system. (Source: A. Marcus and C. M. Thomson, Electricity for Technicians, 2nd ed. [Englewood Cliffs, NJ: Prentice Hall, 1975], p. 337.)
  • 570. 568 PART 2 AC Circuits of the transformer to step up or step down voltage and distribute power economically is one of the major reasons for generating ac rather than dc. Thus, for a given power, the larger the voltage, the better. Today, 1 MV is the largest voltage in use; the level may increase as a result of research and experiments. Beyond the generation plant, the power is transmitted for hundreds of miles through an electric network called the power grid. The three- phase power in the power grid is conveyed by transmission lines hung overhead from steel towers which come in a variety of sizes and shapes. The (aluminum-conductor, steel-reinforced) lines typically have overall diameters up to about 40 mm and can carry current of up to 1380 A. At the substations, distribution transformers are used to step down the voltage. The step-down process is usually carried out in stages. Power may be distributed throughout a locality by means of either overhead or underground cables. The substations distribute the power to residential, commercial, and industrial customers. At the receiving end, a residen- tial customer is eventually supplied with 120/240 V, while industrial or commercial customers are fed with higher voltages such as 460/208 V. Residential customers are usually supplied by distribution transformers often mounted on the poles of the electric utility company. When direct current is needed, the alternating current is converted to dc electronically. E X A M P L E 1 3 . 1 7 A distribution transformer is used to supply a household as in Fig. 13.68. The load consists of eight 100-W bulbs, a 350-W TV, and a 15-kW kitchen range. If the secondary side of the transformer has 72 turns, calculate: (a) the number of turns of the primary winding, and (b) the current Ip in the primary winding. Ip 2400 V 120 V + − + − 120 V – + TV Kitchen range 8 bulbs Figure13.68 For Example 13.17. Solution: (a) The dot locations on the winding are not important, since we are only interested in the magnitudes of the variables involved. Since Np Ns = Vp Vs
  • 571. CHAPTER 13 Magnetically Coupled Circuits 569 we get Np = Ns Vp Vs = 72 2400 240 = 720 turns (b) The total power absorbed by the load is S = 8 × 100 + 350 + 15,000 = 16.15 kW But S = VpIp = VsIs, so that Ip = S Vp = 16,150 2400 = 6.729 A P R A C T I C E P R O B L E M 1 3 . 1 7 In Example 13.17, if the eight 100-W bulbs are replaced by twelve 60-W bulbs and the kitchen range is replaced by a 4.5-kW air-conditioner, find: (a) the total power supplied, (b) the current Ip in the primary winding. Answer: (a) 5.57 kW, (b) 2.321 A. 13.10 SUMMARY 1. Two coils are said to be mutually coupled if the magnetic flux φ emanating from one passes through the other. The mutual induc- tance between the two coils is given by M = k L1L2 where k is the coupling coefficient, 0 k 1. 2. If v1 and i1 are the voltage and current in coil 1, while v2 and i2 are the voltage and current in coil 2, then v1 = L1 di1 dt + M di2 dt and v2 = L2 di2 dt + M di1 dt Thus, the voltage induced in a coupled coil consists of self-induced voltage and mutual voltage. 3. The polarity of the mutually induced voltage is expressed in the schematic by the dot convention. 4. The energy stored in two coupled coils is 1 2 L1i2 1 + 1 2 L2i2 2 ± Mi1i2 5. A transformer is a four-terminal device containing two or more magnetically coupled coils. It is used in changing the current, volt- age, or impedance level in a circuit. 6. A linear (or loosely coupled) transformer has its coils wound on a magnetically linear material. It can be replaced by an equivalent T or network for the purposes of analysis. 7. An ideal (or iron-core) transformer is a lossless (R1 = R2 = 0) transformer with unity coupling coefficient (k = 1) and infinite inductances (L1, L2, M → ∞).
  • 572. 570 PART 2 AC Circuits 8. For an ideal transformer, V2 = nV1, I2 = I1 n , S1 = S2, ZR = ZL n2 where n = N2/N1 is the turns ratio. N1 is the number of turns of the primary winding and N2 is the number of turns of the secondary winding. The transformer steps up the primary voltage when n 1, steps it down when n 1, or serves as a matching device when n = 1. 9. An autotransformer is a transformer with a single winding common to both the primary and the secondary circuits. 10. PSpice is a useful tool for analyzing magnetically coupled circuits. 11. Transformers are necessary in all stages of power distribution systems. Three-phase voltages may be stepped up or down by three-phase transformers. 12. Important uses of transformers in electronics applications are as electrical isolation devices and impedance-matching devices. REVIEW QUESTIONS 13.1 Refer to the two magnetically coupled coils of Fig. 13.69(a). The polarity of the mutual voltage is: (a) Positive (b) Negative M i1 (b) i2 M i1 (a) i2 Figure 13.69 For Review Questions 13.1 and 13.2. 13.2 For the two magnetically coupled coils of Fig. 13.69(b), the polarity of the mutual voltage is: (a) Positive (b) Negative 13.3 The coefficient of coupling for two coils having L1 = 2 H, L2 = 8 H, M = 3 H is: (a) 0.1875 (b) 0.75 (c) 1.333 (d) 5.333 13.4 A transformer is used in stepping down or stepping up: (a) dc voltages (b) ac voltages (c) both dc and ac voltages 13.5 The ideal transformer in Fig. 13.70(a) has N2/N1 = 10. The ratio V2/V1 is: (a) 10 (b) 0.1 (c) −0.1 (d) −10 N1:N2 I1 (a) I2 + − + − N1:N2 I1 (b) I2 V1 V2 Figure 13.70 For Review Questions 13.5 and 13.6. 13.6 For the ideal transformer in Fig. 13.70(b), N2/N1 = 10. The ratio I2/I1 is: (a) 10 (b) 0.1 (c) −0.1 (d) −10 13.7 A three-winding transformer is connected as portrayed in Fig. 13.71(a). The value of the output voltage Vo is: (a) 10 (b) 6 (c) −6 (d) −10 50 V + − Vo 2 V 8 V + − (a) 50 V + − Vo 2 V 8 V + − (b) Figure 13.71 For Review Questions 13.7 and 13.8. 13.8 If the three-winding transformer is connected as in Fig. 13.71(b), the value of the output voltage Vo is: (a) 10 (b) 6 (c) −6 (d) −10
  • 573. CHAPTER 13 Magnetically Coupled Circuits 571 13.9 In order to match a source with internal impedance of 500 to a 15- load, what is needed is: (a) step-up linear transformer (b) step-down linear transformer (c) step-up ideal transformer (d) step-down ideal transformer (e) autotransformer 13.10 Which of these transformers can be used as an isolation device? (a) linear transformer (b) ideal transformer (c) autotransformer (d) all of the above Answers: 13.1b, 13.2a, 13.3b, 13.4b, 13.5d, 13.6b, 13.7c, 13.8a, 13.9d, 13.10b. PROBLEMS Section 13.2 Mutual Inductance 13.1 For the three coupled coils in Fig. 13.72, calculate the total inductance. 6 H 10 H 2 H 8 H 4 H 5 H Figure 13.72 For Prob. 13.1. 13.2 Determine the inductance of the three series- connected inductors of Fig. 13.73. 10 H 8 H 4 H 12 H 6 H 6 H Figure 13.73 For Prob. 13.2. 13.3 Two coils connected in series-aiding fashion have a total inductance of 250 mH. When connected in a series-opposing configuration, the coils have a total inductance of 150 mH. If the inductance of one coil (L1) is three times the other, find L1, L2, and M. What is the coupling coefficient? 13.4 (a) For the coupled coils in Fig. 13.74(a), show that Leq = L1 + L2 + 2M (b) For the coupled coils in Fig. 13.74(b), show that Leq = L1L2 − M2 L1L2 − 2M2 M L2 L1 L1 Leq (b) M L2 Leq (a) Figure 13.74 For Prob. 13.4. 13.5 Determine V1 and V2 in terms of I1 and I2 in the circuit in Fig. 13.75. + − jvM I1 V1 + − V2 I2 jvL1 jvL2 R1 R2 Figure 13.75 For Prob. 13.5. 13.6 Find Vo in the circuit of Fig. 13.76. V –j6 Ω j8 Ω + − j12 Ω 10 Ω j4 Ω 20 30° + − Vo Figure 13.76 For Prob. 13.6.
  • 574. 572 PART 2 AC Circuits 13.7 Obtain Vo in the circuit of Fig. 13.77. 1 Ω + − j6 Ω j4 Ω −j3 Ω 4 Ω j2 Ω Vo + − V 10 0° Figure 13.77 For Prob. 13.7. 13.8 Find Vx in the network shown in Fig. 13.78. 2 Ω + − j4 Ω j4 Ω −j1 Ω 2 Ω j1 Ω V 8 30° + − Vx A 2 0° Figure 13.78 For Prob. 13.8. 13.9 Find Io in the circuit of Fig. 13.79. Im cos vt L C L R io k = 1 Figure 13.79 For Prob. 13.9. 13.10 Obtain the mesh equations for the circuit in Fig. 13.80. V1 V2 R1 I1 I2 I3 R2 jvL1 jvL2 jvM + − + − jvC 1 Figure 13.80 For Prob. 13.10. 13.11 Obtain the Thevenin equivalent circuit for the circuit in Fig. 13.81 at terminals a-b. j6 Ω b a j8 Ω 2 Ω j2 Ω + − V 10 90° A 4 0° 5 Ω −j3 Ω Figure 13.81 For Prob. 13.11. 13.12 Find the Norton equivalent for the circuit in Fig. 13.82 at terminals a-b. b a j20 Ω j5 Ω j10 Ω + − V 60 30° 20 Ω Figure 13.82 For Prob. 13.12. Section 13.3 Energy in a Coupled Circuit 13.13 Determine currents I1, I2, and I3 in the circuit of Fig. 13.83. Find the energy stored in the coupled coils at t = 2 ms. Take ω = 1000 rad/s. j10 Ω j10 Ω 4 Ω k = 0.5 + − A 3 90° V 20 0° 8 Ω −j5 Ω I2 I3 I1 Figure 13.83 For Prob. 13.13.
  • 575. CHAPTER 13 Magnetically Coupled Circuits 573 13.14 Find I1 and I2 in the circuit of Fig. 13.84. Calculate the power absorbed by the 4- resistor. 5 Ω 4 Ω j6 Ω j3 Ω 2 Ω + − I1 I2 V 36 30° j1 Ω –j4 Ω Figure 13.84 For Prob. 13.14. 13.15 ∗ Find current Io in the circuit of Fig. 13.85. j80 Ω j30 Ω j10 Ω j60 Ω j20 Ω + − Io V 50 0° 100 Ω j40 Ω –j50 Ω Figure 13.85 For Prob. 13.15. 13.16 If M = 0.2 H and vs = 12 cos 10t V in the circuit of Fig. 13.86, find i1 and i2. Calculate the energy stored in the coupled coils at t = 15 ms. 0.5 H 25 mF 1 H 5 Ω M i1 i2 + − vs Figure 13.86 For Prob. 13.16. 13.17 In the circuit of Fig. 13.87, (a) find the coupling coefficient, (b) calculate vo, (c) determine the energy stored in the coupled inductors at t = 2 s. 2 Ω + − 4 H 2 H 1 Ω vo 1 H 12 cos 4t V + − F 1 4 Figure 13.87 For Prob. 13.17. 13.18 For the network in Fig. 13.88, find Zab and Io. 4 Ω 0.5 F 1 Ω 2 Ω io + − 12 sin 2t V 1 H 1 H 2 H k = 0.5 3 Ω a b Figure 13.88 For Prob. 13.18. 13.19 Find Io in the circuit of Fig. 13.89. Switch the dot on the winding on the right and calculate Io again. −j30 Ω j20 Ω j40 Ω 10 Ω 50 Ω k = 0.601 Io A 4 60° Figure 13.89 For Prob. 13.19. 13.20 Rework Example 13.1 using the concept of reflected impedance. Section 13.4 Linear Transformers 13.21 In the circuit of Fig. 13.90, find the value of the coupling coefficient k that will make the 10- resistor dissipate 320 W. For this value of k, find the energy stored in the coupled coils at t = 1.5 s. ∗An asterisk indicates a challenging problem.
  • 576. 574 PART 2 AC Circuits 10 Ω + − 30 mH 50 mH 20 Ω k 165 cos 103 t V Figure 13.90 For Prob. 13.21. 13.22 (a) Find the input impedance of the circuit in Fig. 13.91 using the concept of reflected impedance. (b) Obtain the input impedance by replacing the linear transformer by its T equivalent. j30 Ω j20 Ω −j6 Ω j10 Ω 8 Ω 25 Ω j40 Ω Zin Figure 13.91 For Prob. 13.22. 13.23 For the circuit in Fig. 13.92, find: (a) the T -equivalent circuit, (b) the -equivalent circuit. 15 H 20 H 5 H Figure 13.92 For Prob. 13.23. 13.24 ∗ Two linear transformers are cascaded as shown in Fig. 13.93. Show that Zin = ω2 R(L2 a + LaLb − M2 a ) +jω3 (L2 aLb + LaL2 b − LaM2 b − LbM2 a ) ω2(LaLb + L2 b − M2 b ) − jωR(La + Lb) La La Ma Lb Lb Mb R Zin Figure 13.93 For Prob. 13.24. 13.25 Determine the input impedance of the air-core transformer circuit of Fig. 13.94. j12 Ω j40 Ω −j5 Ω j15 Ω 20 Ω 10 Ω Zin Figure 13.94 For Prob. 13.25. Section 13.5 Ideal Transformers 13.26 As done in Fig. 13.32, obtain the relationships between terminal voltages and currents for each of the ideal transformers in Fig. 13.95. V1 V2 1:n + − + − I1 I2 (a) V1 V2 1:n + − + − I1 I2 (b) V1 V2 1:n + − + − I1 I2 (d) V1 V2 1:n + − + − I1 I2 (c) Figure 13.95 For Prob. 13.26. 13.27 A 4-kVA, 2300/230-V rms transformer has an equivalent impedance of 2 10◦ on the primary side. If the transformer is connected to a load with 0.6 power factor leading, calculate the input impedance. 13.28 A 1200/240-V rms transformer has impedance 60 − 30◦ on the high-voltage side. If the transformer is connected to a 0.8 10◦ - load on the low-voltage side, determine the primary and secondary currents. 13.29 Determine I1 and I2 in the circuit of Fig. 13.96. 2 Ω 10 Ω 3:1 I1 I2 + − V 14 0° Figure 13.96 For Prob. 13.29.
  • 577. CHAPTER 13 Magnetically Coupled Circuits 575 13.30 Obtain V1 and V2 in the ideal transformer circuit of Fig. 13.97. 10 Ω A 12 Ω 2 0° A 1 0° V1 V2 + − + − 1:4 Figure 13.97 For Prob. 13.30. 13.31 In the ideal transformer circuit of Fig. 13.98, find i1(t) and i2(t). R 1:n i1(t) i2(t) + − Vm cos vt Vo dc Figure 13.98 For Prob. 13.31. 13.32 (a) Find I1 and I2 in the circuit of Fig. 13.99 below. (b) Switch the dot on one of the windings. Find I1 and I2 again. 13.33 For the circuit in Fig. 13.100, find Vo. Switch the dot on the secondary side and find Vo again. 10 Ω 2 Ω 3:1 20 mF 10 cos 5t V + − + − Vo Figure 13.100 For Prob. 13.33. 13.34 Calculate the input impedance for the network in Fig. 13.101 below. 13.35 Use the concept of reflected impedance to find the input impedance and current I1 in Fig. 13.102 below. 12 Ω 10 Ω j16 Ω + − V 16 60° 10 30° I1 I2 1:2 + − –j8 Ω Figure13.99 For Prob. 13.32. Zin 1:5 4:1 a b 8 Ω 24 Ω 6 Ω j12 Ω −j10 Ω Figure13.101 For Prob. 13.34. 1:2 1:3 5 Ω 8 Ω 36 Ω j18 Ω + − I1 –j2 Ω V 24 0° Figure13.102 For Prob. 13.35.
  • 578. 576 PART 2 AC Circuits 13.36 For the circuit in Fig. 13.103, determine the turns ratio n that will cause maximum average power transfer to the load. Calculate that maximum average power. 40 Ω 1:n + − 10 Ω V rms 120 0° Figure 13.103 For Prob. 13.36. 13.37 Refer to the network in Fig. 13.104. (a) Find n for maximum power supplied to the 200- load. (b) Determine the power in the 200- load if n = 10. 3 Ω 1:n 200 Ω 5 Ω A rms 4 0° Figure 13.104 For Prob. 13.37. 13.38 A transformer is used to match an amplifier with an 8- load as shown in Fig. 13.105. The Thevenin equivalent of the amplifier is: VTh = 10 V, ZTh = 128 . (a) Find the required turns ratio for maximum energy power transfer. (b) Determine the primary and secondary currents. (c) Calculate the primary and secondary voltages. 1:n 8 Ω Amplifier circuit Figure 13.105 For Prob. 13.38. 13.39 In Fig. 13.106 below, determine the average power delivered to Zs. 13.40 Find the power absorbed by the 10- resistor in the ideal transformer circuit of Fig. 13.107. 2 Ω 10 Ω 1:2 V 46 0° 5 Ω + − Figure 13.107 For Prob. 13.40. 13.41 For the ideal transformer circuit of Fig. 13.108 below, find: (a) I1 and I2, (b) V1, V2, and Vo, (c) the complex power supplied by the source. 1:10 + − V rms 120 0° Zs = 500 – j200 Ω Zp = 3 + j4 Ω Figure13.106 For Prob. 13.39. 12 Ω j3 Ω V1 V2 Vo − − 1:2 + − V rms 60 90° I1 I2 2 Ω + + + − −j6 Ω Figure13.108 For Prob. 13.41.
  • 579. CHAPTER 13 Magnetically Coupled Circuits 577 13.42 Determine the average power absorbed by each resistor in the circuit of Fig. 13.109. 20 Ω 100 Ω 1:5 80 cos 4t V + − 20 Ω Figure 13.109 For Prob. 13.42. 13.43 Find the average power delivered to each resistor in the circuit of Fig. 13.110. 8 Ω 4 Ω 2:1 V 20 0° + − 2 Ω Figure 13.110 For Prob. 13.43. 13.44 Refer to the circuit in Fig. 13.111 below. (a) Find currents I1, I2, and I3. (b) Find the power dissipated in the 40- resistor. 13.45 ∗ For the circuit in Fig. 13.112 below, find I1, I2, and Vo. 13.46 For the network in Fig. 13.113 below, find (a) the complex power supplied by the source, (b) the average power delivered to the 18- resistor. 1:4 1:2 4 Ω 5 Ω + − I1 I2 I3 V 120 0° 40 Ω 10 Ω Figure13.111 For Prob. 13.44. 1:5 3:4 2 Ω 14 Ω + − I1 I2 V 24 0° 160 Ω 60 Ω Vo + − Figure13.112 For Prob. 13.45. 2:5 1:3 6 Ω 8 Ω + − V 40 0° 18 Ω –j20 Ω j4 Ω j45 Ω Figure13.113 For Prob. 13.46.
  • 580. 578 PART 2 AC Circuits 13.47 Find the mesh currents in the circuit of Fig. 13.114 below. Section 13.6 Ideal Autotransformers 13.48 An ideal autotransformer with a 1:4 step-up turns ratio has its secondary connected to a 120- load and the primary to a 420-V source. Determine the primary current. 13.49 In the ideal autotransformer of Fig. 13.115, calculate I1, I2, and Io. Find the average power delivered to the load. + − I1 I2 10 + j40 Ω 2 – j6 Ω Io 20 30° V rms 80 turns 200 turns Figure 13.115 For Prob. 13.49. 13.50 ∗ In the circuit of Fig. 13.116, ZL is adjusted until maximum average power is delivered to ZL. Find ZL and the maximum average power transferred to it. Take N1 = 600 turns and N2 = 200 turns. + − V rms N1 N2 75 Ω j125 Ω ZL 120 0° Figure 13.116 For Prob. 13.50. 13.51 In the ideal transformer circuit shown in Fig. 13.117, determine the average power delivered to the load. + − V rms 20 – j40 Ω 120 0° 30 + j12 Ω 1000 turns 200 turns Figure 13.117 For Prob. 13.51. 13.52 In the autotransformer circuit in Fig. 13.118, show that Zin = 1 + N1 N2 2 ZL ZL Zin Figure 13.118 For Prob. 13.52. Section 13.7 Three-Phase Transformers 13.53 In order to meet an emergency, three single-phase transformers with 12,470/7200 V rms are connected in -Y to form a three-phase transformer which is fed by a 12,470-V transmission line. If the transformer supplies 60 MVA to a load, find: (a) the turns ratio for each transformer, (b) the currents in the primary and secondary windings of the transformer, (c) the incoming and outgoing transmission line currents. 1:2 1:3 1 Ω 9 Ω 7 Ω + − V 12 0° –j6 Ω j18 Ω I1 I1 I2 Figure13.114 For Prob. 13.47.
  • 581. CHAPTER 13 Magnetically Coupled Circuits 579 13.54 Figure 13.119 below shows a three-phase transformer that supplies a Y-connected load. (a) Identify the transformer connection. (b) Calculate currents I2 and Ic. (c) Find the average power absorbed by the load. 13.55 Consider the three-phase transformer shown in Fig. 13.120. The primary is fed by a three-phase source with line voltage of 2.4 kV rms, while the secondary supplies a three-phase 120-kW balanced load at pf of 0.8. Determine: (a) the type of transformer connections, (b) the values of ILS and IP S, (c) the values of ILP and IP P , (d) the kVA rating of each phase of the transformer. 13.56 A balanced three-phase transformer bank with the -Y connection depicted in Fig. 13.121 below is used to step down line voltages from 4500 V rms to 900 V rms. If the transformer feeds a 120-kVA load, find: (a) the turns ratio for the transformer, (b) the line currents at the primary and secondary sides. Load 120 kW pf = 0.8 4:1 2.4 kV ILP ILS IPS IPP Figure 13.120 For Prob. 13.55. 3:1 V 450 0° I1 I2 I3 Ic Ib Ia 450 120° V 450 –120° V 8 Ω −j6 Ω 8 Ω −j6 Ω 8 Ω −j6 Ω Figure13.119 For Prob. 13.54. 1:n 4500 V 900 V 42 kVA Three-phase load Figure13.121 For Prob. 13.56.
  • 582. 580 PART 2 AC Circuits 13.57 A Y- three-phase transformer is connected to a 60-kVA load with 0.85 power factor (leading) through a feeder whose impedance is 0.05 + j0.1 per phase, as shown in Fig. 13.122 below. Find the magnitude of: (a) the line current at the load, (b) the line voltage at the secondary side of the transformer, (c) the line current at the primary side of the transformer. 13.58 The three-phase system of a town distributes power with a line voltage of 13.2 kV. A pole transformer connected to single wire and ground steps down the high-voltage wire to 120 V rms and serves a house as shown in Fig. 13.123. (a) Calculate the turns ratio of the pole transformer to get 120 V. (b) Determine how much current a 100-W lamp connected to the 120-V hot line draws from the high-voltage line. 13.2 kV 120 V ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; Figure 13.123 For Prob. 13.58. Section 13.8 PSpice Analysis of Magnetically Coupled Circuits 13.59 Rework Prob. 13.14 using PSpice. 13.60 Use PSpice to find I1, I2, and I3 in the circuit of Fig. 13.124. j15 Ω j80 Ω j0 Ω j100 Ω j10 Ω + − V 60 0° j50 Ω –j20 Ω 20 90° + − V I1 I3 I2 40 Ω 80 Ω Figure 13.124 For Prob. 13.60. 13.61 Rework Prob. 13.15 using PSpice. 13.62 Use PSpice to find I1, I2, and I3 in the circuit of Fig. 13.125. I1 I2 I3 100 Ω 8 H 2 H 1 H 4 H 3 H 2 H 60 mF 200 Ω 70 Ω 50 mF + − V 120 0° f = 100 Figure 13.125 For Prob. 13.62. 1:n 0.05 Ω j0.1 Ω 0.05 Ω j0.1 Ω 0.05 Ω j0.1 Ω Balanced load 60 kVA 0.85 pf leading 2640 V 240 V Figure13.122 For Prob. 13.57.
  • 583. CHAPTER 13 Magnetically Coupled Circuits 581 13.63 Use PSpice to find V1, V2, and Io in the circuit of Fig. 13.126. 2 Ω 1:2 + − 20 Ω 16 Ω V1 V2 + − + − V 40 60° + − V 30 0° –j12 Ω –j4 Ω j8 Ω Io Figure 13.126 For Prob. 13.63. 13.64 Find Ix and Vx in the circuit of Fig. 13.127 below using PSpice. 13.65 Determine I1, I2, and I3 in the ideal transformer circuit of Fig. 13.128 using PSpice. j80 Ω −j30 Ω 50 Ω + − I1 I2 1:2 1:3 V 440 0° 40 Ω j50 Ω I3 60 Ω Figure 13.128 For Prob. 13.65. Section 13.9 Applications 13.66 A stereo amplifier circuit with an output impedance of 7.2 k is to be matched to a speaker with an input impedance of 8 by a transformer whose primary side has 3000 turns. Calculate the number of turns required on the secondary side. 13.67 A transformer having 2400 turns on the primary and 48 turns on the secondary is used as an impedance-matching device. What is the reflected value of a 3- load connected to the secondary? 13.68 A radio receiver has an input resistance of 300 . When it is connected directly to an antenna system with a characteristic impedance of 75 , an impedance mismatch occurs. By inserting an impedance-matching transformer ahead of the receiver, maximum power can be realized. Calculate the required turns ratio. 13.69 A step-down power transformer with a turns ratio of n = 0.1 supplies 12.6 V rms to a resistive load. If the primary current is 2.5 A rms, how much power is delivered to the load? 13.70 A 240/120-V rms power transformer is rated at 10 kVA. Determine the turns ratio, the primary current, and the secondary current. 13.71 A 4-kVA, 2400/240-V rms transformer has 250 turns on the primary side. Calculate: (a) the turns ratio, (b) the number of turns on the secondary side, (c) the primary and secondary currents. 13.72 A 25,000/240-V rms distribution transformer has a primary current rating of 75 A. (a) Find the transformer kVA rating. (b) Calculate the secondary current. 13.73 A 4800-V rms transmission line feeds a distribution transformer with 1200 turns on the primary and 28 turns on the secondary. When a 10- load is connected across the secondary, find: (a) the secondary voltage, (b) the primary and secondary currents, (c) the power supplied to the load. 1:2 2:1 1 Ω 6 Ω 2Vx 8 Ω j2 Ω + − Ix –j10 Ω V 6 0° + − Vx 4 Ω + − Vo + − Figure13.127 For Prob. 13.64.
  • 584. 582 PART 2 AC Circuits COMPREHENSIVE PROBLEMS 13.74 A four-winding transformer (Fig. 13.129) is often used in equipment (e.g., PCs, VCRs) that may be operated from either 110 V or 220 V. This makes the equipment suitable for both domestic and foreign use. Show which connections are necessary to provide: (a) an output of 12 V with an input of 110 V, (b) an output of 50 V with an input of 220 V. a b c d e f g h 32 V 18 V 110 V 110 V Figure 13.129 For Prob. 13.74. 13.75 ∗ A 440/110-V ideal transformer can be connected to become a 550/440-V ideal autotransformer. There are four possible connections, two of which are wrong. Find the output voltage of: (a) a wrong connection, (b) the right connection. 13.76 Ten bulbs in parallel are supplied by a 7200/120-V transformer as shown in Fig. 13.130, where the bulbs are modeled by the 144- resistors. Find: (a) the turns ratio n, (b) the current through the primary winding. 1:n 144 Ω 7200 V 120 V 144 Ω Figure 13.130 For Prob. 13.76.
  • 585. 583 C H A P T E R FREQUENCY RESPONSE 1 4 One machine can do the work of fifty ordinary men. No machine can do the work of one extraordinary man. — Elbert G. Hubbard Enhancing Your Career Career in Control Systems Control systems are another area of electrical engineering where circuit analysis is used. A control system is designed to regulate the behavior of one or more variables in some desired manner. Control systems play major roles in our everyday life. Household appliances such as heating and air-conditioning systems, switch-controlled thermostats, washers and dryers, cruise controllers in automobiles, elevators, traffic lights, manu- facturing plants, navigation systems—all utilize control sys- tems. In the aerospace field, precision guidance of space probes, the wide range of operational modes of the space shuttle, and the ability to maneuver space vehicles remotely from earth all require knowledge of control systems. In the manufacturing sector, repetitive production line opera- tions are increasingly performed by robots, which are pro- grammable control systems designed to operate for many hours without fatigue. Control engineering integrates circuit theory and communication theory. It is not limited to any specific engi- neering discipline but may involve environmental, chemical, aeronautical, mechanical, civil, and electrical engineering. For example, a typical task for a control system engineer might be to design a speed regulator for a disk drive head. A thorough understanding of control systems tech- niques is essential to the electrical engineer and is of great value for designing control systems to perform the desired task. A welding robot. (Courtesy of Shela Terry/Science Photo Library.)
  • 586. 584 PART 2 AC Circuits 14.1 INTRODUCTION Inoursinusoidalcircuitanalysis, wehavelearnedhowtofindvoltagesand currents in a circuit with a constant frequency source. If we let the ampli- tude of the sinusoidal source remain constant and vary the frequency, we obtain the circuit’s frequency response. The frequency response may be regarded as a complete description of the sinusoidal steady-state behavior of a circuit as a function of frequency. The frequency response of a circuit is the variation in its behavior with change in signal frequency. The frequency response of a circuit may also be considered as the variation of the gain and phase with frequency. The sinusoidal steady-state frequency responses of circuits are of significance in many applications, especially in communications and con- trol systems. A specific application is in electric filters that block out or eliminate signals with unwanted frequencies and pass signals of the de- sired frequencies. Filters are used in radio, TV, and telephone systems to separate one broadcast frequency from another. We begin this chapter by considering the frequency response of sim- ple circuits using their transfer functions. We then consider Bode plots, which are the industry-standard way of presenting frequency response. We also consider series and parallel resonant circuits and encounter im- portant concepts such as resonance, quality factor, cutoff frequency, and bandwidth. We discuss different kinds of filters and network scaling. In the last section, we consider one practical application of resonant circuits and two applications of filters. 14.2 TRANSFER FUNCTION The transfer function H(ω) (also called the network function) is a useful analytical tool for finding the frequency response of a circuit. In fact, the frequency response of a circuit is the plot of the circuit’s transfer function H(ω) versus ω, with ω varying from ω = 0 to ω = ∞. A transfer function is the frequency-dependent ratio of a forced function to a forcing function (or of an output to an input). The idea of a transfer function was implicit when we used the concepts of impedance and admittance to relate voltage and current. In general, a linear network can be represented by the block diagram shown in Fig. 14.1. Input Output Linear network H(v) Y(v) X(v) Figure 14.1 A block diagram representation of a linear network. The transfer function H(ω) of a circuit is the frequency-dependent ratio of a phasor output Y(ω) (an element voltage or current) to a phasor input X(ω) (source voltage or current). In this context, X(ω) and Y(ω) denote the input andoutputphasorsofanetwork;theyshouldnot be confused with the same symbolism used for reactance and admittance. The multiple usage of symbols is conventionally permissible due to lack of enough letters in the English language to express all circuit variables distinctly. Thus, H(ω) = Y(ω) X(ω) (14.1)
  • 587. CHAPTER 14 Frequency Response 585 assuming zero initial conditions. Since the input and output can be either voltageorcurrentatanyplaceinthecircuit, therearefourpossibletransfer functions: H(ω) = Voltage gain = Vo(ω) Vi(ω) (14.2a) H(ω) = Current gain = Io(ω) Ii(ω) (14.2b) H(ω) = Transfer Impedance = Vo(ω) Ii(ω) (14.2c) H(ω) = Transfer Admittance = Io(ω) Vi(ω) (14.2d) where subscripts i and o denote input and output values. Being a complex quantity, H(ω) has a magnitude H(ω) and a phase φ; that is, H(ω) = H(ω) φ. To obtain the transfer function using Eq. (14.2), we first obtain the frequency-domain equivalent of the circuit by replacing resistors, inductors, and capacitors with their impedances R, jωL, and 1/jωC. We then use any circuit technique(s) to obtain the appropriate quantity in Eq. (14.2). We can obtain the frequency response of the circuit by plotting the magnitude and phase of the transfer function as the frequency varies. A computer is a real time-saver for plotting the transfer function. Some authors use H(jω) for transfer instead of H(ω), since ω and j are an inseparable pair. The transfer function H(ω) can be expressed in terms of its numer- ator polynomial N(ω) and denominator polynomial D(ω) as H(ω) = N(ω) D(ω) (14.3) where N(ω) and D(ω) are not necessarily the same expressions for the input and output functions, respectively. The representation of H(ω) in Eq. (14.3) assumes that common numerator and denominator factors in H(ω) have canceled, reducing the ratio to lowest terms. The roots of N(ω) = 0 are called the zeros of H(ω) and are usually represented as jω = z1, z2, . . . . Similarly, the roots of D(ω) = 0 are the poles of H(ω) and are represented as jω = p1, p2, . . . . A zero, as a root of the numerator polynomial, is a value that results in a zero value of the function. A pole, as a root of the denominator polynomial, is a value for which the function is infinite. A zero may also be regarded as the value of s = jω that makes H(s) zero, and a pole as the value of s = jω that makes H(s) infinite. To avoid complex algebra, it is expedient to replace jω temporarily with s when working with H(ω) and replace s with jω at the end. E X A M P L E 1 4 . 1 For the RC circuit in Fig. 14.2(a), obtain the transfer function Vo/Vs and its frequency response. Let vs = Vm cos ωt.
  • 588. 586 PART 2 AC Circuits Solution: The frequency-domain equivalent of the circuit is in Fig. 14.2(b). By voltage division, the transfer function is given by H(ω) = Vo Vs = 1/jωC R + 1/jωC = 1 1 + jωRC Comparing this with Eq. (9.18e), we obtain the magnitude and phase of H(ω) as H = 1 1 + (ω/ω0)2 , φ = − tan−1 ω ω0 where ω0 = 1/RC. To plot H and φ for 0 ω ∞, we obtain their values at some critical points and then sketch. vs(t) vo(t) R (a) (b) C + − Vs Vo R jvC 1 + − + − + − Figure 14.2 For Example 14.1: (a) time-domain RC circuit, (b) frequency-domain RC circuit. At ω = 0, H = 1 and φ = 0. At ω = ∞, H = 0 and φ = −90◦ . Also, at ω = ω0, H = 1/ √ 2 and φ = −45◦ . With these and a few more points as shown in Table 14.1, we find that the frequency response is as shown in Fig. 14.3. Additional features of the frequency response in Fig. 14.3 will be explained in Section 14.6.1 on lowpass filters. TABLE 14.1 For Example 14.1. ω/ω0 H φ ω/ω0 H φ 0 1 0 10 0.1 −84◦ 1 0.71 −45◦ 20 0.05 −87◦ 2 0.45 −63◦ 100 0.01 −89◦ 3 0.32 −72◦ ∞ 0 −90◦ 0 0.707 H 1 v0 = 1 RC v0 = 1 RC v 0 v −90° −45° (a) (b) f Figure14.3 Frequency response of the RC circuit: (a) amplitude response, (b) phase response. P R A C T I C E P R O B L E M 1 4 . 1 Obtain the transfer function Vo/Vs of the RL circuit in Fig. 14.4, assuming vs = Vm cos ωt. Sketch its frequency response. Answer: jωL/(R + jωL); see Fig. 14.5 for the response. vs vo R L + − + − Figure14.4 RL circuit for Practice Prob. 14.1.
  • 589. CHAPTER 14 Frequency Response 587 1 H 0.707 0 v0 = R L v0 = R L v (a) (b) 90° 45° f 0 v Figure14.5 Frequency response of the RL circuit in Fig. 14.4. E X A M P L E 1 4 . 2 For the circuit in Fig. 14.6, calculate the gain Io(ω)/Ii(ω) and its poles and zeros. ii (t) io(t) 0.5 F 2 H 4 Ω Figure14.6 For Example 14.2. Solution: By current division, Io(ω) = 4 + j2ω 4 + j2ω + 1/j0.5ω Ii(ω) or Io(ω) Ii(ω) = j0.5ω(4 + j2ω) 1 + j2ω + (jω)2 = s(s + 2) s2 + 2s + 1 , s = jω The zeros are at s(s + 2) = 0 ⇒ z1 = 0, z2 = −2 The poles are at s2 + 2s + 1 = (s + 1)2 = 0 Thus, there is a repeated pole (or double pole) at p = −1. P R A C T I C E P R O B L E M 1 4 . 2 Find the transfer function Vo(ω)/Ii(ω) for the circuit of Fig. 14.7. Obtain its poles and zeros. vo(t) ii(t) 0.1 F 2 H 3 Ω 5 Ω + − Figure14.7 For Practice Prob. 14.2. Answer: 5(s + 2)(s + 1.5) s2 + 4s + 5 , s = jω; poles: −2, −1.5; zeros: −2±j.
  • 590. 588 PART 2 AC Circuits †14.3 THE DECIBEL SCALE It is not always easy to get a quick plot of the magnitude and phase of the transfer function as we did above. A more systematic way of obtaining the frequency response is to use Bode plots. Before we begin to construct Bode plots, we should take care of two important issues: the use of logarithms and decibels in expressing gain. Since Bode plots are based on logarithms, it is important that we keep the following properties of logarithms in mind: 1. log P1P2 = log P1 + log P2 2. log P1/P2 = log P1 − log P2 3. log Pn = n log P 4. log 1 = 0 In communications systems, gain is measured in bels. Historically, the bel is used to measure the ratio of two levels of power or power gain G; that is, G = Number of bels = log10 P2 P1 (14.4) The decibel (dB) provides us with a unit of less magnitude. It is 1/10th of a bel and is given by GdB = 10 log10 P2 P1 (14.5) When P1 = P2, there is no change in power and the gain is 0 dB. If P2 = 2P1, the gain is GdB = 10 log10 2 = 3 dB (14.6) and when P2 = 0.5P1, the gain is GdB = 10 log10 0.5 = −3 dB (14.7) Equations (14.6) and (14.7) show another reason why logarithms are greatly used: The logarithm of the reciprocal of a quantity is simply negative the logarithm of that quantity. Historical note: The bel is named after Alexander Graham Bell, the inventor of the telephone. V2 − + V1 R2 Network I1 I2 P1 P2 R1 + − Figure 14.8 Voltage-current relationships for a four-terminal network. Alternatively, the gain G can be expressed in terms of voltage or current ratio. To do so, consider the network shown in Fig. 14.8. If P1 is the input power, P2 is the output (load) power, R1 is the input resistance, and R2 is the load resistance, then P1 = 0.5V 2 1 /R1 and P2 = 0.5V 2 2 /R2, and Eq. (14.5) becomes GdB = 10 log10 P2 P1 = 10 log10 V 2 2 /R2 V 2 1 /R1 = 10 log10 V2 V1 2 + 10 log10 R1 R2 (14.8) GdB = 20 log10 V2 V1 − 10 log10 R2 R1 (14.9)
  • 591. CHAPTER 14 Frequency Response 589 For the case when R2 = R1, a condition that is often assumed when comparing voltage levels, Eq. (14.9) becomes GdB = 20 log10 V2 V1 (14.10) Instead, if P1 = I2 1 R1 and P2 = I2 2 R2, for R1 = R2, we obtain GdB = 20 log10 I2 I1 (14.11) Two things are important to note from Eqs. (14.5), (14.10), and (14.11): 1. That 10 log is used for power, while 20 log is used for voltage or current, because of the square relationship between them (P = V 2 /R = I2 R). 2. That the dB value is a logarithmic measurement of the ratio of one variable to another of the same type. Therefore, it applies in expressing the transfer function H in Eqs. (14.2a) and (14.2b), which are dimensionless quantities, but not in expressing H in Eqs. (14.2c) and (14.2d). With this in mind, we now apply the concepts of logarithms and decibels to construct Bode plots. 14.4 BODE PLOTS Obtaining the frequency response from the transfer function as we did in Section 14.2 is an uphill task. The frequency range required in frequency response is often so wide that it is inconvenient to use a linear scale for the frequency axis. Also, there is a more systematic way of locating the important features of the magnitude and phase plots of the transfer function. For these reasons, it has become standard practice to use a logarithmic scale for the frequency axis and a linear scale in each of the separate plots of magnitude and phase. Such semilogarithmic plots of the transfer function—known as Bode plots—have become the industry standard. Historical note: Named after Hendrik W. Bode (1905–1982),anengineerwiththeBellTelephone Laboratories,forhispioneeringworkinthe1930s and 1940s. Bode plots are semilog plots of the magnitude (in decibels) and phase (in degrees) of a transfer function versus frequency. Bode plots contain the same information as the nonlogarithmic plots dis- cussed in the previous section, but they are much easier to construct, as we shall see shortly. The transfer function can be written as H = H φ = Hejφ (14.12) Taking the natural logarithm of both sides, ln H = ln H + ln ejφ = ln H + jφ (14.13)
  • 592. 590 PART 2 AC Circuits Thus, the real part of ln H is a function of the magnitude while the imag- inary part is the phase. In a Bode magnitude plot, the gain HdB = 20 log10 H (14.14) is plotted in decibels (dB) versus frequency. Table 14.2 provides a few values of H with the corresponding values in decibels. In a Bode phase plot, φ is plotted in degrees versus frequency. Both magnitude and phase plots are made on semilog graph paper. TABLE 14.2 Specific gains and their decibel values. Magnitude H 20 log10 H (dB) 0.001 −60 0.01 −40 0.1 −20 0.5 −6 1/ √ 2 −3 1 0 √ 2 3 2 6 10 20 20 26 100 40 1000 60 A transfer function in the form of Eq. (14.3) may be written in terms of factors that have real and imaginary parts. One such representation might be H(ω) = K(jω)±1 (1 + jω/z1)[1 + j2ζ1ω/ωk + (jω/ωk)2 ] · · · (1 + jω/p1)[1 + j2ζ2ω/ωn + (jω/ωn)2] · · · (14.15) which is obtained by dividing out the poles and zeros in H(ω). The representation of H(ω) as in Eq. (14.15) is called the standard form. In this particular case, H(ω) has seven different factors that can appear in various combinations in a transfer function. These are: The origin is where ω = 1 or log ω = 0 and the gain is zero. 1. A gain K 2. A pole (jω)−1 or zero (jω) at the origin 3. A simple pole 1/(1 + jω/p1) or zero (1 + jω/z1) 4. A quadratic pole 1/[1 + j2ζ2ω/ωn + (jω/ωn)2 ] or zero [1 + j2ζ1ω/ωk + (jω/ωk)2 ] In constructing a Bode plot, we plot each factor separately and then com- bine them graphically. The factors can be considered one at a time and then combined additively because of the logarithms involved. It is this mathematical convenience of the logarithm that makes Bode plots a pow- erful engineering tool. We will now make straight-line plots of the factors listed above. We shall find that these straight-line plots known as Bode plots approximate the actual plots to a surprising degree of accuracy. Constant term: For the gain K, the magnitude is 20 log10 K and the phase is 0◦ ; both are constant with frequency. Thus the magnitude and phase plots of the gain are shown in Fig. 14.9. If K is negative, the magnitude remains 20 log10 |K| but the phase is ±180◦ . (a) 0.1 1 10 100 v 20 log10K H (b) 0.1 1 10 100 v 0 f Figure14.9 Bode plots for gain K: (a) magnitude plot, (b) phase plot.
  • 593. CHAPTER 14 Frequency Response 591 Pole/zero at the origin: For the zero (jω) at the origin, the magnitude is 20 log10 ω and the phase is 90◦ . These are plotted in Fig. 14.10, where we notice that the slope of the magnitude plot is 20 dB/decade, while the phase is constant with frequency. A decade is an interval between two frequen- cies with a ratio of 10; e.g., between ω0 and 10ω0, or between 10 and 100 Hz. Thus, 20 dB/decade means that the magnitude changes 20 dB whenever the frequency changes tenfold or one decade. The special case of dc (ω = 0) does not appear on Bode plots because log 0 = −∞, implying that zero frequency is infinitely far to the left of the origin of Bode plots. The Bode plots for the pole (jω)−1 are similar except that the slope of the magnitude plot is −20 dB/decade while the phase is −90◦ . In general, for (jω)N , where N is an integer, the magnitude plot will have a slope of 20N dB/decade, while the phase is 90N degrees. Simple pole/zero: For the simple zero (1 + jω/z1), the magnitude is 20 log10 |1 + jω/z1| and the phase is tan−1 ω/z1. We notice that HdB = 20 log10 1 + jω z1 ⇒ 20 log10 1 = 0 as ω → 0 (14.16) HdB = 20 log10 1 + jω z1 ⇒ 20 log10 ω z1 as ω → ∞ (14.17) showing that we can approximate the magnitude as zero (a straight line with zero slope) for small values of ω and by a straight line with slope 20 dB/decade for large values of ω. The frequency ω = z1 where the two asymptotic lines meet is called the corner frequency or break frequency. Thus the approximate magnitude plot is shown in Fig. 14.11(a), where the actual plot is also shown. Notice that the approximate plot is close to the actual plot except at the break frequency, where ω = z1 and the deviation is 20 log10 |(1 + j1)| = 20 log10 √ 2 = 3 dB. (a) (b) 0.1 1.0 Slope = 20 dB/decade 10 v 0 20 –20 H 0.1 1.0 10 v 90° 0° f Figure14.10 Bode plot for a zero (jω) at the origin: (a) magnitude plot, (b) phase plot. The phase tan−1 (ω/z1) can be expressed as φ = tan−1 ω z1 =    0, ω = 0 45◦ , ω = z1 90◦ , ω → ∞ (14.18) As a straight-line approximation, we let φ 0 for ω ≤ z1/10, φ 45◦ for ω = z1, and φ 90◦ for ω ≥ 10z1. As shown in Fig. 14.11(b) along with the actual plot, the straight-line plot has a slope of 45◦ per decade. The Bode plots for the pole 1/(1 + jω/p1) are similar to those in Fig. 14.11 except that the corner frequency is at ω = p1, the magnitude (a) Approximate Exact 3 dB 0.1z1 10z1 z1 v 20 H (b) Approximate Exact 45°/decade 0.1z1 10z1 z1 v 45° 0° 90° f Figure14.11 Bode plots of zero (1 + jω/z1): (a) magnitude plot, (b) phase plot.
  • 594. 592 PART 2 AC Circuits has a slope of −20 dB/decade, and the phase has a slope of −45◦ per decade. Quadratic pole/zero: The magnitude of the quadratic pole 1/[1 + j2ζ2ω/ωn + (jω/ωn)2 ] is −20 log10 |1 + j2ζ2ω/ωn + (jω/ωn)2 | and the phase is − tan−1 (2ζ2ω/ωn)/(1 − ω/ω2 n). But HdB = −20 log10 1 + j2ζ2ω ωn + jω ωn 2 ⇒ 0 as ω → 0 (14.19) and HdB = −20 log10 1 + j2ζ2ω ωn + jω ωn 2 ⇒ −40 log10 ω ωn as ω → ∞ (14.20) Thus, the amplitude plot consists of two straight asymptotic lines: one with zero slope for ω ωn and the other with slope −40 dB/decade for ω ωn, with ωn as the corner frequency. Figure 14.12(a) shows the approximate and actual amplitude plots. Note that the actual plot depends on the damping factor ζ2 as well as the corner frequency ωn. The significant peaking in the neighborhood of the corner frequency should be added to the straight-line approximation if a high level of accuracy is desired. However, we will use the straight-line approximation for the sake of simplicity. (a) 0.01vn 100vn 10vn 0.1vn z2 = 0.05 z2 = 0.2 z2 = 0.4 z2 = 0.707 z2 = 1.5 vn v 20 0 –20 –40 H –40 dB/dec (b) 0.01vn 100vn 10vn 0.1vn z2 = 0.4 z2 = 1.5 z2 = 0.2 z2 = 0.05 vn v 0° –90° –180° f –90°/dec z2 = 0.707 Figure14.12 Bode plots of quadratic pole [1 + j2ζω/ωn − ω2/ω2 n]−1: (a) magnitude plot, (b) phase plot. The phase can be expressed as φ = − tan−1 2ζ2ω/ωn 1 − ω2/ω2 n =    0, ω = 0 −90◦ , ω = ωn −180◦ , ω → ∞ (14.21) The phase plot is a straight line with a slope of 90◦ per decade starting at ωn/10 and ending at 10ωn, as shown in Fig. 14.12(b). We see again that the difference between the actual plot and the straight-line plot is due to the damping factor. Notice that the straight-line approximations for both magnitude and phase plots for the quadratic pole are the same
  • 595. CHAPTER 14 Frequency Response 593 as those for a double pole, i.e. (1 + jω/ωn)−2 . We should expect this because the double pole (1 + jω/ωn)−2 equals the quadratic pole 1/[1 + j2ζ2ω/ωn + (jω/ωn)2 ] when ζ2 = 1. Thus, the quadratic pole can be treated as a double pole as far as straight-line approximation is concerned. For the quadratic zero [1+j2ζ1ω/ωk +(jω/ωk)2 ], the plots in Fig. 14.12 are inverted because the magnitude plot has a slope of 40 dB/decade while the phase plot has a slope of 90◦ per decade. Table 14.3 presents a summary of Bode plots for the seven factors. To sketch the Bode plots for a function H(ω) in the form of Eq. (14.15), for example, we first record the corner frequencies on the semilog graph pa- per, sketch the factors one at a time as discussed above, and then combine additively the graphs of the factors. The combined graph is often drawn from left to right, changing slopes appropriately each time a corner fre- quency is encountered. The following examples illustrate this procedure. There is another procedure for obtaining Bode plots that is faster and perhaps more efficient than the one we have just discussed. It consists in realizing that zeros cause an increase in slope, while poles cause a decrease. By starting with the low-frequency asymptote of the Bode plot, moving along the frequency axis, and increasing ordecreasingtheslopeateachcornerfrequency, one can sketch the Bode plot immediately from thetransferfunctionwithouttheeffortofmaking individualplotsandaddingthem. Thisprocedure can be used once you become proficient in the one discussed here. Digital computers have rendered the pro- cedure discussed here almost obsolete. Several software packages such as PSpice, Matlab, Math- cad, and Micro-Cap can be used to generate fre- quency response plots. We will discuss PSpice later in the chapter. TABLE 14.3 Summary of Bode straight-line magnitude and phase plots. Factor Magnitude Phase K v 20 log10 K v 0° (jω)N v 20N dB⁄decade 1 v 90N° 1 (jω)N v 1 −20N dB⁄decade v −90N° 1 + jω z N v z 20N dB⁄decade v 90N° 0° z 10 z 10z 1 (1 + jω/p)N v p −20N dB⁄decade v −90N° 0° p 10 p 10p
  • 596. 594 PART 2 AC Circuits TABLE 14.3 (continued) Factor Magnitude Phase 1 + 2jωζ ωn + jω ωn 2 N v vn 40N dB⁄decade v 180N° 0° vn vn 10vn 10 1 [1 + 2jωζ/ωk + (jω/ωk)2]N v vk −40N dB⁄decade v −180N° 0° vk 10vk vk 10 E X A M P L E 1 4 . 3 Construct the Bode plots for the transfer function H(ω) = 200jω (jω + 2)(jω + 10) Solution: We first put H(ω) in the standard form by dividing out the poles and zeros. Thus, H(ω) = 10jω (1 + jω/2)(1 + jω/10) = 10|jω| |1 + jω/2| |1 + jω/10| 90◦ − tan−1 ω/2 − tan−1 ω/10 Hence the magnitude and phase are HdB = 20 log10 10 + 20 log10 |jω| − 20 log10 1 + jω 2 − 20 log10 1 + jω 10 φ = 90◦ − tan−1 ω 2 − tan−1 ω 10 We notice that there are two corner frequencies at ω = 2, 10. For both the magnitude and phase plots, we sketch each term as shown by the dotted lines in Fig. 14.13. We add them up graphically to obtain the overall plots shown by the solid curves.
  • 597. CHAPTER 14 Frequency Response 595 (a) 1  1 + jv/2  1 2 10 100 20 log1010 20 log10 20 log10 20 log10 jv 200 v 0 20 H (dB) 0.1 20 1  1 + jv/10  (b) 0.2 0.2 100 200 v 90° 90° 0° –90° f 0.1 20 1 2 10 –tan–1 v 2 –tan–1 v 10 Figure14.13 For Example 14.3: (a) magnitude plot, (b) phase plot. P R A C T I C E P R O B L E M 1 4 . 3 Draw the Bode plots for the transfer function H(ω) = 5(jω + 2) jω(jω + 10) Answer: See Fig. 14.14. (a) 20 log10 1 + 20 log10 20 log101 v 20 0 –20 H (dB) 100 1 1 2 10 jv  20 log10 1  1+ jv/10  (b) 90° −90° 0° –90° f v 100 1 0.2 2 10 20 0.1 tan–1 v 2 –tan–1 v 10 0.1 jv 2 Figure 14.14 For Practice Prob. 14.3: (a) magnitude plot, (b) phase plot.
  • 598. 596 PART 2 AC Circuits E X A M P L E 1 4 . 4 Obtain the Bode plots for H(ω) = jω + 10 jω(jω + 5)2 Solution: Putting H(ω) in the standard form, we get H(ω) = 0.4 (1 + jω/10) jω (1 + jω/5)2 From this, we obtain the magnitude and phase as HdB = 20 log10 0.4 + 20 log10 1 + jω 10 − 20 log10 |jω| − 40 log10 1 + jω 5 φ = 0◦ + tan−1 ω 10 − 90◦ − 2 tan−1 ω 5 There are two corner frequencies at ω = 5, 10 rad/s. For the pole with cor- ner frequency at ω = 5, the slope of the magnitude plot is −40 dB/decade and that of the phase plot is −90◦ per decade due to the power of 2. The magnitude and the phase plots for the individual terms (in dotted lines) and the entire H(jω) (in solid lines) are in Fig. 14.15. (a) 20 0 –20 –8 –40 H (dB) v 100 50 1 0.5 10 –20 dB/decade –60 dB/decade –40 dB/decade 5 20 log10 20 log100.4 0.1 1 jv  40 log10 1  1 + jv/5  20 log10 1 + jv 10 (b) 90° 0° –90° –180° f v 100 50 –90° 1 0.5 10 –90°/decade –45°/decade 45°/decade 5 0.1 tan–1 10 v –2 tan–1 v 5 Figure 14.15 Bode plots for Example 14.4: (a) magnitude plot, (b) phase plot.
  • 599. CHAPTER 14 Frequency Response 597 P R A C T I C E P R O B L E M 1 4 . 4 Sketch the Bode plots for H(ω) = 50jω (jω + 4)(jω + 10)2 Answer: See Fig. 14.16. 20 –20 –40 H (dB) 100 40 1 10 4 (a) 20 log10 jv  0.1 –20 log108 20 log10 1  1 + jv/4  40 log10 1  1 + jv/10  v 0 90° –90° –180° f v 100 90° 40 1 0.4 10 4 (b) 0.1 – tan–1 4 v –2 tan–1 v 10 0° Figure 14.16 For Practice Prob. 14.4: (a) magnitude plot, (b) phase plot. E X A M P L E 1 4 . 5 Draw the Bode plots for H(s) = s + 1 s2 + 60s + 100 Solution: We express H(s) as H(ω) = 1/100(1 + jω) 1 + jω6/10 + (jω/10)2 For the quadratic pole, ωn = 10 rad/s, which serves as the corner fre-
  • 600. 598 PART 2 AC Circuits quency. The magnitude and phase are HdB = −20 log10 100 + 20 log10 |1 + jω| − 20 log10 1 + jω6 10 − ω2 100 φ = 0◦ + tan−1 ω − tan−1 ω6/10 1 − ω2/100 Figure 14.17 shows the Bode plots. Notice that the quadratic pole is treated as a repeated pole at ωk, that is, (1 + jω/ωk)2 , which is an ap- proximation. 20 0 –20 –40 H (dB) v 100 1 10 (a) 20 log10  1 + jv  0.1 20 log10 –20 log10 100 1  1 + j6v/10 – v2 /100  90° 0° –90° –180° f v 100 1 6v/10 1 – v2 /100 10 (b) 0.1 –tan–1 tan–1 v Figure 14.17 Bode plots for Example 14.5: (a) magnitude plot, (b) phase plot. P R A C T I C E P R O B L E M 1 4 . 5 Construct the Bode plots for H(s) = 10 s(s2 + 80s + 400) Answer: See Fig. 14.18.
  • 601. CHAPTER 14 Frequency Response 599 20 0 –20 –40 –32 H (dB) v 100 200 1 20 10 (a) 0.1 20 log10 –20 log10 40 –20 dB/decade –60 dB/decade 1  1 + jv0.2 – v2 /400  20 log10 1 jv  –90° –90° 0° –180° –270° f v 1 2 20 10 (b) 0.1 –tan–1 v  1 – v2 /400 2 100 200 Figure 14.18 For Practice Prob. 14.5: (a) magnitude plot, (b) phase plot. E X A M P L E 1 4 . 6 Given the Bode plot in Fig. 14.19, obtain the transfer function H(ω). 0.1 1 5 10 20 100 –20 dB/decade v 40 dB 0 H +20 dB/decade –40 dB/decade Figure14.19 For Example 14.6. Solution: To obtain H(ω) from the Bode plot, we keep in mind that a zero always causes an upward turn at a corner frequency, while a pole causes a down- ward turn. We notice from Fig. 14.19 that there is a zero jω at the origin which should have intersected the frequency axis at ω = 1. This is indi- cated by the straight line with slope +20 dB/decade. The fact that this straight line is shifted by 40 dB indicates that there is a 40-dB gain; that is, 40 = 20 log10 K ⇒ log10 K = 2
  • 602. 600 PART 2 AC Circuits or K = 102 = 100 In addition to the zero jω at the origin, we notice that there are three factors with corner frequencies at ω = 1, 5, and 20 rad/s. Thus, we have: 1. A pole at p = 1 with slope −20 dB/decade to cause a down- ward turn and counteract the pole at the origin. The pole at z = 1 is determined as 1/(1 + jω/1). 2. Another pole at p = 5 with slope −20 dB/decade causing a downward turn. The pole is 1/(1 + jω/5). 3. A third pole at p = 20 with slope −20 dB/decade causing a further downward turn. The pole is 1/(1 + jω/20). Putting all these together gives the corresponding transfer function as H(ω) = 100jω (1 + jω/1)(1 + jω/5)(1 + jω/20) = jω104 (jω + 1)(jω + 5)(jω + 20) or H(s) = 104 s (s + 1)(s + 5)(s + 20) , s = jω P R A C T I C E P R O B L E M 1 4 . 6 Obtain the transfer function H(ω) corresponding to the Bode plot in Fig. 14.20. 0.1 1 0.5 10 100 –40 dB/decade v 0 dB 0 H +20 dB/decade Figure14.20 For Practice Prob. 14.6. Answer: H(ω) = 200(s + 0.5) (s + 1)(s + 10)2 . 14.5 SERIES RESONANCE The most prominent feature of the frequency response of a circuit may be the sharp peak (or resonant peak) exhibited in its amplitude characteristic. The concept of resonance applies in several areas of science and engi- neering. Resonance occurs in any system that has a complex conjugate pair of poles; it is the cause of oscillations of stored energy from one form to another. It is the phenomenon that allows frequency discrimination in communications networks. Resonance occurs in any circuit that has at least one inductor and one capacitor.
  • 603. CHAPTER 14 Frequency Response 601 Resonance is a condition in an RLC circuit in which the capacitive and inductive reactances are equal in magnitude, thereby resulting in a purely resistive impedance. Resonant circuits (series or parallel) are useful for constructing filters, as their transfer functions can be highly frequency selective. They are used in many applications such as selecting the desired stations in radio and TV receivers. R jvL jvC 1 I + − Vs = Vm u Figure14.21 The series resonant circuit. Consider the series RLC circuit shown in Fig. 14.21 in the fre- quency domain. The input impedance is Z = H(ω) = Vs I = R + jωL + 1 jωC (14.22) or Z = R + j ωL − 1 ωC (14.23) Resonance results when the imaginary part of the transfer function is zero, or Im(Z) = ωL − 1 ωC = 0 (14.24) The value of ω that satisfies this condition is called the resonant frequency ω0. Thus, the resonance condition is ω0L = 1 ω0C (14.25) or ω0 = 1 √ LC rad/s (14.26) Since ω0 = 2πf0, f0 = 1 2π √ LC Hz (14.27) Note that at resonance: Note No. 4 becomes evident from the fact that |VL| = Vm R ω0L = QVm |VC| = Vm R 1 ω0C = QVm whereQisthequalityfactor,definedinEq.(14.38). 1. The impedance is purely resistive, thus, Z = R. In other words, the LC series combination acts like a short circuit, and the entire voltage is across R. 2. The voltage Vs and the current I are in phase, so that the power factor is unity. 3. The magnitude of the transfer function H(ω) = Z(ω) is minimum. 4. The inductor voltage and capacitor voltage can be much more than the source voltage.
  • 604. 602 PART 2 AC Circuits The frequency response of the circuit’s current magnitude I = |I| = Vm R2 + (ωL − 1/ωC)2 (14.28) is shown in Fig. 14.22; the plot only shows the symmetry illustrated in this graph when the frequency axis is a logarithm. The average power dissipated by the RLC circuit is P(ω) = 1 2 I2 R (14.29) The highest power dissipated occurs at resonance, when I = Vm/R, so that P(ω0) = 1 2 V 2 m R (14.30) At certain frequencies ω = ω1, ω2, the dissipated power is half the maximum value; that is, P(ω1) = P(ω2) = (Vm/ √ 2)2 2R = V 2 m 4R (14.31) Hence, ω1 and ω2 are called the half-power frequencies. 0 Bandwidth B v v1 v0 v2 I Vm/R 0.707Vm/R Figure14.22 The current amplitude versus frequency for the series resonant circuit of Fig. 14.21. The half-power frequencies are obtained by setting Z equal to √ 2R, and writing R2 + ωL − 1 ωC 2 = √ 2R (14.32) Solving for ω, we obtain ω1 = − R 2L + R 2L 2 + 1 LC ω2 = R 2L + R 2L 2 + 1 LC (14.33) We can relate the half-power frequencies with the resonant frequency. From Eqs. (14.26) and (14.33), ω0 = √ ω1ω2 (14.34) showing that the resonant frequency is the geometric mean of the half- power frequencies. Notice that ω1 and ω2 are in general not symmetrical around the resonant frequency ω0, because the frequency response is not generally symmetrical. However, as will be explained shortly, symmetry of the half-power frequencies around the resonant frequency is often a reasonable approximation. Although the height of the curve in Fig. 14.22 is determined by R, the width of the curve depends on other factors. The width of the response curve depends on the bandwidth B, which is defined as the difference between the two half-power frequencies, B = ω2 − ω1 (14.35)
  • 605. CHAPTER 14 Frequency Response 603 This definition of bandwidth is just one of several that are commonly used. Strictly speaking, B in Eq. (14.35) is a half-power bandwidth, because it is the width of the frequency band between the half-power frequencies. The “sharpness” of the resonance in a resonant circuit is measured quantitatively by the quality factor Q. At resonance, the reactive energy inthecircuitoscillatesbetweentheinductorandthecapacitor. Thequality factor relates the maximum or peak energy stored to the energy dissipated in the circuit per cycle of oscillation: Q = 2π Peak energy stored in the circuit Energy dissipated by the circuit in one period at resonance (14.36) It is also regarded as a measure of the energy storage property of a circuit in relation to its energy dissipation property. In the series RLC circuit, the peak energy stored is 1 2 LI2 , while the energy dissipated in one period is 1 2 (I2 R)(1/f ). Hence, Q = 2π 1 2 LI2 1 2 I2R(1/f ) = 2πf L R (14.37) or Q = ω0L R = 1 ω0CR (14.38) Notice that the quality factor is dimensionless. The relationship between the bandwidth B and the quality factor Q is obtained by substituting Eq. (14.33) into Eq. (14.35) and utilizing Eq. (14.38). B = R L = ω0 Q (14.39) or B = ω2 0CR. Thus AlthoughthesamesymbolQisusedforthereac- tivepower, thetwoarenotequalandshouldnot be confused. Q here is dimensionless, whereas reactive power Q is in VAR. This may help distin- guish between the two. The quality factor of a resonant circuit is the ratio of its resonant frequency to its bandwidth. Keep in mind that Eqs. (14.26), (14.33), (14.38), and (14.39) only apply to a series RLC circuit. As illustrated in Fig. 14.23, the higher the value of Q, the more selective the circuit is but the smaller the bandwidth. The selectivity of an RLC circuit is the ability of the circuit to respond to a certain frequency and discriminate against all other frequencies. If the band of frequencies to be selected or rejected is narrow, the quality factor of the resonant circuit must be high. If the band of frequencies is wide, the quality factor must be low. The quality factor is a measure of the selectivity (or “sharpness” of resonance) of the circuit. B3 Q3 (greatest selectivity) Q2 (medium selectivity) Q1 (least selectivity) B2 B1 v Amplitude Figure 14.23 The higher the circuit Q, the smaller the bandwidth. A resonant circuit is designed to operate at or near its resonant frequency. It is said to be a high-Q circuit when its quality factor is
  • 606. 604 PART 2 AC Circuits equal to or greater than 10. For high-Q circuits (Q ≥ 10), the half- power frequencies are, for all practical purposes, symmetrical around the resonant frequency and can be approximated as ω1 ω0 − B 2 , ω2 ω0 + B 2 (14.40) High-Q circuits are used often in communications networks. We see that a resonant circuit is characterized by five related param- eters: the two half-power frequencies ω1 and ω2, the resonant frequency ω0, the bandwidth B, and the quality factor Q. E X A M P L E 1 4 . 7 In the circuit in Fig. 14.24, R = 2 , L = 1 mH, and C = 0.4 µF. (a) Find the resonant frequency and the half-power frequencies. (b) Cal- culate the quality factor and bandwidth. (c) Determine the amplitude of the current at ω0, ω1, and ω2. 20 sin vt R L C + − Figure14.24 For Example 14.7. Solution: (a) The resonant frequency is ω0 = 1 √ LC = 1 √ 10−3 × 0.4 × 10−6 = 50 krad/s METHOD 1 The lower half-power frequency is ω1 = − R 2L + R 2L 2 + 1 LC = − 2 2 × 10−3 + (103)2 + (50 × 103)2 = −1 + √ 1 + 2500 krad/s = 49 krad/s Similarly, the upper half-power frequency is ω2 = 1 + √ 1 + 2500 krad/s = 51 krad/s (b) The bandwidth is B = ω2 − ω1 = 2 krad/s or B = R L = 2 10−3 = 2 krad/s The quality factor is Q = ω0 B = 50 2 = 25 METHOD 2 Alternatively, we could find Q = ω0L R = 50 × 103 × 10−3 2 = 25 From Q, we find B = ω0 Q = 50 × 103 25 = 2 krad/s
  • 607. CHAPTER 14 Frequency Response 605 Since Q 10, this is a high-Q circuit and we can obtain the half-power frequencies as ω1 = ω0 − B 2 = 50 − 1 = 49 krad/s ω2 = ω0 + B 2 = 50 + 1 = 51 krad/s as obtained earlier. (c) At ω = ω0, I = Vm R = 20 2 = 10 A At ω = ω1, ω2, I = Vm √ 2R = 10 √ 2 = 7.071 A P R A C T I C E P R O B L E M 1 4 . 7 A series-connected circuit has R = 4 and L = 25 mH. (a) Calculate the value of C that will produce a quality factor of 50. (b) Find ω1, ω2, and B. (c) Determine the average power dissipated at ω = ω0, ω1, ω2. Take Vm = 100 V. Answer: (a) 0.625 µF, (b) 7920 rad/s, 8080 rad/s, 160 rad/s, (c) 1.25 kW, 0.625 kW, 0.625 kW. 14.6 PARALLEL RESONANCE 1 jvC jvL R V + − I = Im u Figure14.25 The parallel resonant circuit. 0 Bandwidth B v v1 v0 v2 V  ImR 0.707 ImR Figure14.26 The current amplitude versus frequency for the series resonant circuit of Fig. 14.25. The parallel RLC circuit in Fig. 14.25 is the dual of the series RLC circuit. So we will avoid needless repetition. The admittance is Y = H(ω) = I V = 1 R + jωC + 1 jωL (14.41) or Y = 1 R + j ωC − 1 ωL (14.42) Resonance occurs when the imaginary part of Y is zero, ωC − 1 ωL = 0 (14.43) or ω0 = 1 √ LC rad/s (14.44) which is the same as Eq. (14.26) for the series resonant circuit. The voltage |V| is sketched in Fig. 14.26 as a function of frequency. Notice that at resonance, the parallel LC combination acts like an open circuit, so
  • 608. 606 PART 2 AC Circuits that the entire currents flows through R. Also, the inductor and capacitor current can be much more than the source current at resonance. We can see this from the fact that |IL| = ImR ω0L = QIm |IC| = ω0CImR = QIm whereQisthequalityfactor,definedinEq.(14.47). We exploit the duality between Figs. 14.21 and 14.25 by comparing Eq. (14.42) with Eq. (14.23). By replacing R, L, and C in the expressions for the series circuit with 1/R, 1/C, and 1/L respectively, we obtain for the parallel circuit ω1 = − 1 2RC + 1 2RC 2 + 1 LC ω2 = 1 2RC + 1 2RC 2 + 1 LC (14.45) B = ω2 − ω1 = 1 RC (14.46) Q = ω0 B = ω0RC = R ω0L (14.47) Using Eqs. (14.45) and (14.47), we can express the half-power frequen- cies in terms of the quality factor. The result is ω1 = ω0 1 + 1 2Q 2 − ω0 2Q , ω2 = ω0 1 + 1 2Q 2 + ω0 2Q (14.48) Again, for high-Q circuits (Q ≥ 10) ω1 ω0 − B 2 , ω2 ω0 + B 2 (14.49) Table 14.4 presents a summary of the characteristics of the series and parallel resonant circuits. Besides the series and parallel RLC considered here, other resonant circuits exist. Example 14.9 treats a typical example. TABLE 14.4 Summary of the characteristics of resonant RLC circuits. Characteristic Series circuit Parallel circuit Resonant frequency, ω0 1 √ LC 1 √ LC Quality factor, Q ω0L R or 1 ω0RC R ω0L or ω0RC Bandwidth, B ω0 Q ω0 Q Half-power frequencies, ω1, ω2 ω0 1 + 1 2Q 2 ± ω0 2Q ω0 1 + 1 2Q 2 ± ω0 2Q For Q ≥ 10, ω1, ω2 ω0 ± B 2 ω0 ± B 2
  • 609. CHAPTER 14 Frequency Response 607 E X A M P L E 1 4 . 8 In the parallel RLC circuit in Fig. 14.27, let R = 8 k, L = 0.2 mH, and C = 8 µF. (a) Calculate ω0, Q, and B. (b) Find ω1 and ω2. (c) Deter- mine the power dissipated at ω0, ω1, and ω2. 10 sin vt C L R io + − Figure14.27 For Example 14.8. Solution: (a) ω0 = 1 √ LC = 1 √ 0.2 × 10−3 × 8 × 10−6 = 105 4 = 25 krad/s Q = R ω0L = 8 × 103 25 × 103 × 0.2 × 10−3 = 1600 B = ω0 Q = 15.625 rad/s (b) Due to the high value of Q, we can regard this as a high-Q circuit. Hence, ω1 = ω0 − B 2 = 25,000 − 7.812 = 24,992 rad/s ω2 = ω0 + B 2 = 25,000 + 7.8125 = 25,008 rad/s (c) At ω = ω0, Y = 1/R or Z = R = 8 k. Then Io = V Z = 10 − 90◦ 8000 = 1.25 − 90◦ mA Since the entire current flows through R at resonance, the average power dissipated at ω = ω0 is P = 1 2 |Io|2 R = 1 2 (1.25 × 10−3 )2 (8 × 103 ) = 6.25 mW or P = V 2 m 2R = 100 2 × 8 × 103 = 6.25 mW At ω = ω1, ω2, P = V 2 m 4R = 3.125 mW P R A C T I C E P R O B L E M 1 4 . 8 A parallel resonant circuit has R = 100 k, L = 20 mH, and C = 5 nF. Calculate ω0, ω1, ω2, Q, and B. Answer: 100 krad/s, 99 krad/s, 101 krad/s, 50, 2 krad/s. E X A M P L E 1 4 . 9 Determine the resonant frequency of the circuit in Fig. 14.28.
  • 610. 608 PART 2 AC Circuits Solution: The input admittance is Y = jω0.1 + 1 10 + 1 2 + jω2 = 0.1 + jω0.1 + 2 − jω2 4 + 4ω2 At resonance, Im(Y) = 0 and ω00.1 − 2ω0 4 + 4ω2 0 = 0 ⇒ ω0 = 2 rad/s Im cos vt 0.1 F 10 Ω 2 H 2 Ω Figure14.28 For Example 14.9. P R A C T I C E P R O B L E M 1 4 . 9 Calculate the resonant frequency of the circuit in Fig. 14.29. Vm cos vt 10 Ω 0.2 F 1 H + − Figure14.29 For Practice Prob. 14.9. Answer: 2.179 rad/s. 14.7 PASSIVE FILTERS The concept of filters has been an integral part of the evolution of electri- cal engineering from the beginning. Several technological achievements would not have been possible without electrical filters. Because of this prominent role of filters, much effort has been expended on the theory, design, and construction of filters and many articles and books have been written on them. Our discussion in this chapter should be considered introductory. A filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others. As a frequency-selective device, a filter can be used to limit the frequency spectrum of a signal to some specified band of frequencies. Filters are the circuits used in radio and TV receivers to allow us to select one desired signal out of a multitude of broadcast signals in the environment. A filter is a passive filter if it consists of only passive elements R, L, and C. It is said to be an active filter if it consists of active elements (such as transistors and op amps) in addition to passive elements R, L, and C. We consider passive filters in this section and active filters in the next section. Besides the filters we study in these sections, there are other kinds of filters—such as digital filters, electromechanical filters, and microwave filters—which are beyond the level of the text. As shown in Fig. 14.30, there are four types of filters whether passive or active:
  • 611. CHAPTER 14 Frequency Response 609 1. A lowpass filter passes low frequencies and stops high frequencies, as shown ideally in Fig. 14.30(a). 2. A highpass filter passes high frequencies and rejects low frequencies, as shown ideally in Fig. 14.30(b). 3. A bandpass filter passes frequencies within a frequency band and blocks or attenuates frequencies outside the band, as shown ideally in Fig. 14.30(c). 4. A bandstop filter passes frequencies outside a frequency band and blocks or attenuates frequencies within the band, as shown ideally in Fig. 14.30(d). 0 (b) v vc H(v) 1 0 (a) v vc H(v) 1 0 (c) v v1 v2 H(v) 1 0 (d) v v1 v2 H(v) 1 Figure14.30 Ideal frequency response of four types of filter: (a) lowpass filter, (b) highpass filter, (c) bandpass filter, (d) bandstop filter. Table 14.5 presents a summary of the characteristics of these filters. Be aware that the characteristics in Table 14.5 are only valid for first- or second-order filters—but one should not have the impression that only these kinds of filter exist. We now consider typical circuits for realizing the filters shown in Table 14.5. TABLE 14.5 Summary of the characteristics of filters. Type of Filter H(0) H(∞) H(ωc) or H(ω0) Lowpass 1 0 1/ √ 2 Highpass 0 1 1/ √ 2 Bandpass 0 0 1 Bandstop 1 1 0 ωc is the cutoff frequency for lowpass and highpass filters; ω0 is the center frequency for bandpass and bandstop filters. vi(t) R C + − vo(t) + − Figure14.31 A lowpass filter. 14.7.1 Lowpass Filter A typical lowpass filter is formed when the output of an RC circuit is taken off the capacitor as shown in Fig. 14.31. The transfer function (see also Example 14.1) is H(ω) = Vo Vi = 1/jωC R + 1/jωC H(ω) = 1 1 + jωRC (14.50) Note that H(0) = 1, H(∞) = 0. Figure 14.32 shows the plot of |H(ω)|, along with the ideal characteristic. The half-power frequency, which is equivalent to the corner frequency on the Bode plots but in the context of filters is usually known as the cutoff frequency ωc, is obtained by setting the magnitude of H(ω) equal to 1/ √ 2, thus H(ωc) = 1 1 + ω2 c R2C2 = 1 √ 2 or ωc = 1 RC (14.51)
  • 612. 610 PART 2 AC Circuits vc v 0.707 Ideal Actual 1 0 H(v) Figure14.32 Ideal and actual fre- quency response of a lowpass filter. The cutoff frequency is also called the rolloff frequency. The cutoff frequency is the frequency at which the transfer function H drops in magnitude to 70.71% of its maximum value. It is also regarded as the frequency at which the power dissipated in a circuit is half of its maximum value. A lowpass filter is designed to pass only frequencies from dc up to the cutoff frequency ωc. A lowpass filter can also be formed when the output of an RL circuit is taken off the resistor. Of course, there are many other circuits for lowpass filters. vi(t) R C + − vo(t) + − Figure14.33 A highpass filter. 14.7.2 Highpass Filter A highpass filter is formed when the output of an RC circuit is taken off the resistor as shown in Fig. 14.33. The transfer function is H(ω) = Vo Vi = R R + 1/jωC H(ω) = jωRC 1 + jωRC (14.52) Note that H(0) = 0, H(∞) = 1. Figure 14.34 shows the plot of |H(ω)|. Again, the corner or cutoff frequency is ωc = 1 RC (14.53) vc v 0.707 Ideal Actual 1 0 H(v) Figure14.34 Ideal and actual fre- quency response of a highpass filter. A highpass filter is designed to pass all frequencies above its cutoff frequency ωc. vi(t) R C + − vo(t) L + − Figure14.35 A bandpass filter. A highpass filter can also be formed when the output of an RL circuit is taken off the inductor. 14.7.3 Bandpass Filter The RLC series resonant circuit provides a bandpass filter when the out- put is taken off the resistor as shown in Fig. 14.35. The transfer function is H(ω) = Vo Vi = R R + j(ωL − 1/ωC) (14.54)
  • 613. CHAPTER 14 Frequency Response 611 We observe that H(0) = 0, H(∞) = 0. Figure 14.36 shows the plot of |H(ω)|. The bandpass filter passes a band of frequencies (ω1 ω ω2) centered on ω0, the center frequency, which is given by ω0 = 1 √ LC (14.55) A bandpass filter is designed to pass all frequencies within a band of frequencies, ω1 ω ω2. Since the bandpass filter in Fig. 14.35 is a series resonant circuit, the half- power frequencies, the bandwidth, and the quality factor are determined as in Section 14.5. A bandpass filter can also be formed by cascading the lowpass filter (where ω2 = ωc) in Fig. 14.31 with the highpass filter (where ω1 = ωc) in Fig. 14.33. v0 v1 v2 v 0.707 Ideal Actual 1 0  H(v) Figure 14.36 Ideal and actual frequency response of a bandpass filter. 14.7.4 Bandstop Filter A filter that prevents a band of frequencies between two designated values (ω1 and ω2) from passing is variably known as a bandstop, bandreject, or notch filter. A bandstop filter is formed when the output RLC series resonant circuit is taken off the LC series combination as shown in Fig. 14.37. The transfer function is H(ω) = Vo Vi = j(ωL − 1/ωC) R + j(ωL − 1/ωC) (14.56) Notice that H(0) = 1, H(∞) = 1. Figure 14.38 shows the plot of |H(ω)|. Again, the center frequency is given by ω0 = 1 √ LC (14.57) whilethehalf-powerfrequencies, thebandwidth, andthequalityfactorare calculated using the formulas in Section 14.5 for a series resonant circuit. Here, ω0 is called the frequency of rejection, while the corresponding bandwidth (B = ω2 − ω1) is known as the bandwidth of rejection. Thus, vi(t) R C + − – + vo(t) L Figure14.37 A bandstop filter. v0 v1 v2 v 0.707 Ideal Actual 1 0  H(v) Figure 14.38 Ideal and actual frequency response of a bandstop filter. A bandstop filter is designed to stop or eliminate all frequencies within a band of frequencies, ω1 ω ω2. Notice that adding the transfer functions of the bandpass and the bandstop gives unity at any frequency for the same values of R, L, and C. Of course, this is not true in general but true for the circuits treated here. This is due to the fact that the characteristic of one is the inverse of the other. In concluding this section, we should note that: 1. From Eqs. (14.50), (14.52), (14.54), and (14.56), the maximum gain of a passive filter is unity. To generate a gain greater than unity, one should use an active filter as the next section shows.
  • 614. 612 PART 2 AC Circuits 2. There are other ways to get the types of filters treated in this section. 3. The filters treated here are the simple types. Many other filters have sharper and complex frequency responses. E X A M P L E 1 4 . 1 0 Determine what type of filter is shown in Fig. 14.39. Calculate the corner or cutoff frequency. Take R = 2 k, L = 2 H, and C = 2 µF. vi(t) C R + − vo(t) L + − Figure14.39 For Example 14.10. Solution: The transfer function is H(s) = Vo Vi = R 1/sC sL + R 1/sC , s = jω (14.10.1) But R 1 sC = R/sC R + 1/sC = R 1 + sRC Substituting this into Eq. (14.10.1) gives H(s) = R/(1 + sRC) sL + R/(1 + sRC) = R s2RLC + sL + R , s = jω or H(ω) = R −ω2RLC + jωL + R (14.10.2) Since H(0) = 1 and H(∞) = 0, we conclude from Table 14.5 that the circuit in Fig. 14.39 is a second-order lowpass filter. The magnitude of H is H = R (R − ω2RLC)2 + ω2L2 (14.10.3) The corner frequency is the same as the half-power frequency, i.e., where H is reduced by a factor of 1 √ 2. Since the dc value of H(ω) is 1, at the corner frequency, Eq. (14.10.3) becomes after squaring H2 = 1 2 = R2 (R − ω2 c RLC)2 + ω2 c L2 or 2 = (1 − ω2 c LC)2 + ωcL R 2 Substituting the values of R, L, and C, we obtain 2 = 1 − ω2 c 4 × 10−6 2 + (ωc 10−3 )2 Assuming that ωc is in krad/s, 2 = (1 − 4ωc)2 + ω2 c or 16ω4 c − 7ω2 c − 1 = 0 Solving the quadratic equation in ω2 c , we get ω2 c = 0.5509, or ωc = 0.742 krad/s = 742 rad/s
  • 615. CHAPTER 14 Frequency Response 613 P R A C T I C E P R O B L E M 1 4 . 1 0 For the circuit in Fig. 14.40, obtain the transfer function Vo(ω)/Vi(ω). Identify the type of filter the circuit represents and determine the corner frequency. Take R1 = 100 = R2, L = 2 mH. vi(t) R1 R2 + − vo(t) L + − Figure14.40 For Practice Prob. 14.10. Answer: Highpass filter, R2 R1 + R2 jω jω + ωc , ωc = R1R2 (R1 + R2)L = 25 krad/s. E X A M P L E 1 4 . 1 1 IfthebandstopfilterinFig.14.37istorejecta200-Hzsinusoidwhilepass- ing other frequencies, calculate the values of L and C. Take R = 150 and the bandwidth as 100 Hz. Solution: We use the formulas for a series resonant circuit in Section 14.5. B = 2π(100) = 200π rad/s But B = R L ⇒ L = R B = 150 200π = 0.2387 H Rejection of the 200-Hz sinusoid means that f0 is 200 Hz, so that ω0 in Fig. 14.38 is ω0 = 2πf0 = 2π(200) = 400π Since ω0 = 1/ √ LC, C = 1 ω2 0L = 1 (400π)2(0.2387) = 2.66 µF P R A C T I C E P R O B L E M 1 4 . 1 1 Design a bandpass filter of the form in Fig. 14.35 with a lower cutoff fre- quency of 20.1 kHz and an upper cutoff frequency of 20.3 kHz. Take R = 20 k. Calculate L, C, and Q. Answer: 7.96 H, 3.9 pF, 101. 14.8 ACTIVE FILTERS Therearethreemajorlimitstothepassivefiltersconsideredintheprevious section. First, they cannot generate gain greater than 1; passive elements cannot add energy to the network. Second, they may require bulky and expensive inductors. Third, they perform poorly at frequencies below the audio frequency range (300 Hz f 3000 Hz). Nevertheless, passive filters are useful at high frequencies.
  • 616. 614 PART 2 AC Circuits Active filters consist of combinations of resistors, capacitors, and op amps. They offer some advantages over passive RLC filters. First, they are often smaller and less expensive, because they do not require inductors. This makes feasible the integrated circuit realizations of fil- ters. Second, they can provide amplifier gain in addition to providing the same frequency response as RLC filters. Third, active filters can be combined with buffer amplifiers (voltage followers) to isolate each stage of the filter from source and load impedance effects. This isolation allows designing the stages independently and then cascading them to realize the desired transfer function. (Bode plots, being logarithmic, may be added when transfer functions are cascaded.) However, active filters are less reliable and less stable. The practical limit of most active filters is about 100 kHz—most active filters operate well below that frequency. Filters are often classified according to their order (or number of poles) or their specific design type. 14.8.1 First-Order Lowpass Filter One type of first-order filter is shown in Fig. 14.41. The components selected for Zi and Zf determine whether the filter is lowpass or highpass, but one of the components must be reactive. + − − + V o + – V i Zi Zf Figure 14.41 A general first- order active filter. Figure 14.42 shows a typical active low-pass filter. For this filter, the transfer function is H(ω) = Vo Vi = − Zf Zi (14.58) where Zi = Ri and Zf = Rf 1 jωCf = Rf /jωCf Rf + 1/jωCf = Rf 1 + jωCf Rf (14.59) Therefore, H(ω) = − Rf Ri 1 1 + jωCf Rf (14.60) We notice that Eq. (14.60) is similar to Eq. (14.50), except that there is a low frequency (ω → 0) gain or dc gain of −Rf /Ri. Also, the corner frequency is ωc = 1 Rf Cf (14.61) which does not depend on Ri. This means that several inputs with dif- ferent Ri could be summed if required, and the corner frequency would remain the same for each input. + − + – V o + – V i Ri Rf Cf Figure 14.42 Active first-order lowpass filter. 14.8.2 First-Order Highpass Filter + − + – V o + – V i Ri Ci Rf Figure 14.43 Active first-order highpass filter. Figure 14.43 shows a typical highpass filter. As before, H(ω) = Vo Vi = − Zf Zi (14.62) where Zi = Ri + 1/jωCi and Zf = Rf so that H(ω) = − Rf Ri + 1/jωCi = − jωCiRf 1 + jωCiRi (14.63)
  • 617. CHAPTER 14 Frequency Response 615 This is similar to Eq. (14.52), except that at very high frequencies (ω → ∞), the gain tends to −Rf /Ri. The corner frequency is ωc = 1 RiCi (14.64) 14.8.3 Bandpass Filter The circuit in Fig. 14.42 may be combined with that in Fig. 14.43 to form a bandpass filter that will have a gain K over the required range of fre- quencies. By cascading a unity-gain lowpass filter, a unity-gain highpass filter, and an inverter with gain −Rf /Ri, as shown in the block diagram of Fig. 14.44(a), we can construct a bandpass filter whose frequency re- sponse is that in Fig. 14.44(b). The actual construction of the bandpass filter is shown in Fig. 14.45. This way of creating a bandpass filter, not neces- sarily the best, is perhaps the easiest to under- stand. v0 v1 v2 v 0.707 K K B 0 (a) (b) Low-pass filter vi vo H High-pass filter Inverter Figure14.44 Active bandpass filter: (a) block diagram, (b) frequency response. + − + – vi R R C1 C2 + − R Stage 1 Low-pass filter sets v2 value Stage 2 High-pass filter sets v1 value Stage 3 An inverter provides gain R + − + – vo Ri Rf Figure14.45 Active bandpass filter. The analysis of the bandpass filter is relatively simple. Its transfer function is obtained by multiplying Eqs. (14.60) and (14.63) with the gain of the inverter; that is H(ω) = Vo Vi = − 1 1 + jωC1R − jωC2R 1 + jωC2R − Rf Ri = − Rf Ri 1 1 + jωC1R jωC2R 1 + jωC2R (14.65)
  • 618. 616 PART 2 AC Circuits The lowpass section sets the upper corner frequency as ω2 = 1 RC1 (14.66) while the highpass section sets the lower corner frequency as ω1 = 1 RC2 (14.67) With these values of ω1 and ω2, the center frequency, bandwidth, and quality factor are found as follows: ω0 = √ ω1ω2 (14.68) B = ω2 − ω1 (14.69) Q = ω0 B (14.70) To find the passband gain K, we write Eq. (14.65) in the standard form of Eq. (14.15), H(ω) = −Kjω/ω1 (1 + jω/ω1)(1 + jω/ω2) = −Kjωω2 (ω1 + jω)(ω2 + jω) (14.71) At the center frequency ω0 = √ ω1ω2, the magnitude of the transfer function is H(ω0) = −Kjω0ω2 (ω1 + jω0)(ω2 + jω0) = Kω2 ω1 + ω2 (14.72) We set this equal to the gain of the inverting amplifier, as Kω2 ω1 + ω2 = Rf Ri (14.73) from which the gain K can be determined. 14.8.4 Bandreject (or Notch) Filter A bandreject filter may be constructed by parallel combination of a low- pass filter and a highpass filter and a summing amplifier, as shown in the block diagram of Fig. 14.46(a). The circuit is designed such that the v0 v1 v2 v 0.707 K K B (b) (a) 0 H vi vo = v1 + v2 v1 v2 Low-pass filter sets v1 High-pass filter sets v2 v1 Summing amplifier Figure14.46 Active bandreject filter: (a) block diagram, (b) frequency response.
  • 619. CHAPTER 14 Frequency Response 617 lower cutoff frequency ω1 is set by the lowpass filter while the upper cut- off frequency ω2 is set by the highpass filter. The gap between ω1 and ω2 is the bandwidth of the filter. As shown in Fig. 14.46(b), the filter passes frequencies below ω1 and above ω2. The block diagram in Fig. 14.46(a) is actually constructed as shown in Fig. 14.47. The transfer function is H(ω) = Vo Vi = − Rf Ri − 1 1 + jωC1R − jωC2R 1 + jωC2R (14.74) The formulas for calculating the values of ω1, ω2, the center frequency, bandwidth, and quality factor are the same as in Eqs. (14.66) to (14.70). + − + – vi + – vo R R C1 Rf C2 + − + − R R Ri Ri Figure14.47 Active bandreject filter. To determine the passband gain K of the filter, we can write Eq. (14.74) in terms of the upper and lower corner frequencies as H(ω) = Rf Ri 1 1 + jω/ω2 + jω/ω1 1 + jω/ω1 = Rf Ri (1 + j2ω/ω1 + (jω)2 /ω1ω1) (1 + jω/ω2)(1 + jω/ω1) (14.75) Comparing this with the standard form in Eq. (14.15) indicates that in the two passbands (ω → 0 and ω → ∞) the gain is K = Rf Ri (14.76) We can also find the gain at the center frequency by finding the magnitude of the transfer function at ω0 = √ ω1ω2, writing H(ω0) = Rf Ri (1 + j2ω0/ω1 + (jω0)2 /ω1ω1) (1 + jω0/ω2)(1 + jω0/ω1) = Rf Ri 2ω1 ω1 + ω2 (14.77) Again, the filters treated in this section are only typical. There are many other active filters that are more complex.
  • 620. 618 PART 2 AC Circuits E X A M P L E 1 4 . 1 2 Design a low-pass active filter with a dc gain of 4 and a corner frequency of 500 Hz. Solution: From Eq. (14.61), we find ωc = 2πfc = 2π(500) = 1 Rf Cf (14.12.1) The dc gain is H(0) = − Rf Ri = −4 (14.12.2) We have two equations and three unknowns. If we select Cf = 0.2 µF, then Rf = 1 2π(500)0.2 × 10−6 = 1.59 k and Ri = Rf 4 = 397.5 We use a 1.6-k resistor for Rf and a 400- resistor for Ri. Figure 14.42 shows the filter. P R A C T I C E P R O B L E M 1 4 . 1 2 Design a highpass filter with a high-frequency gain of 5 and a corner fre- quency of 2 kHz. Use a 0.1-µF capacitor in your design. Answer: Ri = 800 and Rf = 4 k. E X A M P L E 1 4 . 1 3 Design a bandpass filter in the form of Fig. 14.45 to pass frequencies be- tween 250 Hz and 3000 Hz and with K = 10. Select R = 20 k. Solution: Since ω1 = 1/RC2, we obtain C2 = 1 Rω1 = 1 2πf1R = 1 2π × 250 × 20 × 103 = 31.83 nF Similarly, since ω2 = 1/RC1, C1 = 1 Rω2 = 1 2πf2R = 1 2π × 3000 × 20 × 103 = 2.65 nF From Eq. (14.73), Rf Ri = Kω2 ω1 + ω2 = Kf2 f1 + f2 = 10 3000 3250 = 9.223 If we select Ri = 10 k, then Rf = 9.223Ri 92 k.
  • 621. CHAPTER 14 Frequency Response 619 P R A C T I C E P R O B L E M 1 4 . 1 3 Design a notch filter based on Fig. 14.47 for ω0 = 20 krad/s, K = 5, and Q = 10. Use R = Ri = 10 k. Answer: C1 = 47.62 nF, C2 = 52.63 nF, and Rf = 50 k. †14.9 SCALING In designing and analyzing filters and resonant circuits or in circuit anal- ysis in general, it is sometimes convenient to work with element values of 1 , 1 H, or 1 F, and then transform the values to realistic values by scaling. We have taken advantage of this idea by not using realistic el- ement values in most of our examples and problems; mastering circuit analysis is made easy by using convenient component values. We have thus eased calculations, knowing that we could use scaling to then make the values realistic. There are two ways of scaling a circuit: magnitude or impedance scaling, and frequency scaling. Both are useful in scaling responses and circuit elements to values within the practical ranges. While magnitude scaling leaves the frequency response of a circuit unaltered, frequency scaling shifts the frequency response up or down the frequency spectrum. 14.9.1 Magnitude Scaling Magnitude scaling is the process of increasing all impedance in a network by a factor, the frequency response remaining unchanged. Recall that impedances of individual elements R, L, and C are given by ZR = R, ZL = jωL, ZC = 1 jωC (14.78) In magnitude scaling, we multiply the impedance of each circuit element by a factor Km and let the frequency remain constant. This gives the new impedances as Z R = KmZR = KmR, Z L = KmZL = jωKmL Z C = KmZC = 1 jωC/Km (14.79) Comparing Eq. (14.79) with Eq. (14.78), we notice the following changes in the element values: R → KmR, L → KmL, and C → C/Km. Thus, in magnitude scaling, the new values of the elements and frequency are R = KmR, L = KmL C = C Km , ω = ω (14.80)
  • 622. 620 PART 2 AC Circuits The primed variables are the new values and the unprimed variables are the old values. Consider the series or parallel RLC circuit. We now have ω 0 = 1 √ LC = 1 √ KmLC/Km = 1 √ LC = ω0 (14.81) showing that the resonant frequency, as expected, has not changed. Sim- ilarly, the quality factor and the bandwidth are not affected by magnitude scaling. Also, magnitude scaling does not affect transfer functions in the forms of Eqs. (14.2a) and (14.2b), which are dimensionless quantities. 14.9.2 Frequency Scaling Frequency scaling is the process of shifting the frequency response of a network up or down the frequency axis while leaving the impedance the same. We achieve frequency scaling by multiplying the frequency by a factor Kf while keeping the impedance the same. Frequency scaling is equivalent to relabeling the frequency axis of a frequency response plot. It is needed when translating such frequencies such as a resonant frequency, a corner frequency, a bandwidth, etc., to a realistic level. It can be used to bring capacitance and inductance values into a range that is convenient to work with. From Eq. (14.78), we see that the impedances of L and C are frequency-dependent. If we apply frequency scaling to ZL(ω) and ZC(ω) in Eq. (14.78), we obtain ZL = j(ωKf )L = jωL ⇒ L = L Kf (14.82a) ZC = 1 j(ωKf )C = 1 jωC ⇒ C = C Kf (14.82b) since the impedance of the inductor and capacitor must remain the same after frequency scaling. We notice the following changes in the element values: L → L/Kf and C → C/Kf . The value of R is not affected, since its impedance does not depend on frequency. Thus, in frequency scaling, the new values of the elements and frequency are R = R, L = L Kf C = C Kf , ω = Kf ω (14.83) Again, if we consider the series or parallel RLC circuit, for the resonant frequency ω 0 = 1 √ LC = 1 (L/Kf )(C/Kf ) = Kf √ LC = Kf ω0 (14.84) and for the bandwidth B = Kf B (14.85) but the quality factor remains the same (Q = Q).
  • 623. CHAPTER 14 Frequency Response 621 14.9.3 Magnitude and Frequency Scaling If a circuit is scaled in magnitude and frequency at the same time, then R = KmR, L = Km Kf L C = 1 KmKf C, ω = Kf ω (14.86) These are more general formulas than those in Eqs. (14.80) and (14.83). We set Km = 1 in Eq. (14.86) when there is no magnitude scaling or Kf = 1 when there is no frequency scaling. E X A M P L E 1 4 . 1 4 A fourth-order Butterworth lowpass filter is shown in Fig. 14.48(a). The filter is designed such that the cutoff frequency ωc = 1 rad/s. Scale the circuit for a cutoff frequency of 50 kHz using 10-k resistors. 1 Ω 1 Ω (a) + − vo vs + − 1.848 F 0.765 F 1.848 H 0.765 H 10 kΩ 10 kΩ (b) + − vo vs + − 588.2 pF 243.5 pF 58.82 mH 24.35 H Figure14.48 For Example 14.14: (a) Normalized Butterworth lowpass filter, (b) scaled version of the same lowpass filter. Solution: If the cutoff frequency is to shift from ωc = 1 rad/s to ω c = 2π(50) krad/s, then the frequency scale factor is Kf = ω c ωc = 100π × 103 1 = π × 105 Also, if each 1- resistor is to be replaced by a 10-k resistor, then the magnitude scale factor must be Km = R R = 10 × 103 1 = 104 Using Eq. (14.86), L 1 = Km Kf L1 = 104 π × 105 (1.848) = 58.82 mH L 2 = Km Kf L2 = 104 π × 105 (0.765) = 24.35 mH C 1 = C1 KmKf = 0.765 π × 109 = 243.5 pF C 2 = C2 KmKf = 1.848 π × 109 = 588.2 pF
  • 624. 622 PART 2 AC Circuits The scaled circuit is as shown in Fig. 14.48(b). This circuit uses practical values and will provide the same transfer function as the prototype in Fig. 14.48(a), but shifted in frequency. P R A C T I C E P R O B L E M 1 4 . 1 4 A third-order Butterworth filter normalized to ωc = 1 rad/s is shown in Fig. 14.49. Scale the circuit to a cutoff frequency of 10 kHz. Use 15-nF capacitors. 1 Ω 1 Ω + − vo vs + − 1 F 1 F 2 H Figure14.49 For Practice Prob. 14.14. Answer: R 1 = R 2 = 1.061 k, C 1 = C 2 =15 nF,