Electrical engineering principles lecture slides. circuit analysis by Maina

ELECTRICAL ENGINEERING
PRINCIPLES
Course Lecturer: Prof C Maina Muriithi, PhD Electrical Engineering

COURSE CONTENT
• Current, voltage and resistance. Power sources. Electric
power Joule’s law series and parallel circuits. Ohm and
Kirchoff’s laws.
• Direct current (D.C) circuit analysis. Electric fields and
capacitance. Magnetic circuits. Self-inductance. Mutual
inductance.
• Transient analysis of D.C. circuits with inductance (L),
resistance(R), capacitance (c).
• Alternating voltage and current. Expressions of sinusoidal
current and voltage using vector and complex numbers.
Impedance and admittance. Power in Alternating current
(A.C) circuits. Analysis of three-phase A.C. circuits.
• Electrical measurements and measuring instruments.
Course Lecturer: Prof C Maina Muriithi, PhD
Electrical Engineering

• Expected Learning outcomes:
• At the end of this course, the learner should be able to:
• Describe circuit elements and their characteristics:
resistors, capacitors, inductors, voltage and current sources.
• Solve problems applying Ohms Law, Kirchhoff’s Voltage
Law, Kirchhoff’s Current Law, linearity, and the maximum
power theorem.
• Solve circuit problems using resistance and DC source
combinations.
• Solve problems using basic techniques such as current
division, voltage division, node-voltage analysis, mesh
analysis, and superposition.

Assessment & Grading
Final Exam 70%
CAT 1,2,3 25%
Homework Assignments 1,2,3 5%
TOTAL 100%
Textbooks (Reading list):
1. B.L. Theraja, A.K. Theraja: Electrical Technology, S. Chand & Co. Ltd.
2. Edward Hughes: Electrical Technology, ELBS
3. H. Cotton: Electrical Technology, CBS

BASIC INTRODUCTION

1.1 Basic Concepts and Electric Circuits
 Electrical power conversion and transmission

1.1 Basic Concepts and Electric Circuits
Question: What is the current
through the bulb?
Concept of Abstraction
Solution:
In order to calculate the current, we can replace the bulb with a resistor.
R is the only subject of interest, which serves as an abstraction of the bulb.

1.2 Basic Quantities
Units
• Standard SI Prefixes
– 10-12 pico (p)
– 10-9 nano (n)
– 10-6 micro ()
– 10-3 milli (m)
– 103 kilo (k)
– 106 mega (M)
– 109 giga (G)
– 1012 tera (T)
• Electric charge (q)
– in Coulombs (C)
• Current (I)
– in Amperes (A)
• Voltage (V)
– in Volts (V)
• Energy (W)
– in Joules (J)
• Power (P)
– in Watts (W)
 I t q 
V
I
R

IR V
W qV Pt V I t    
P VI

Current
• Time rate of change of charge t
q
I Constant current tIq 
dttdqti /)()( Time varying current 

t
dxxitq )()(
Unit mAA 3
101 
 AmA 3
101 
 (1 A = 1 C/s)
• Notation: Current flow represents the flow of positive charge
• Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCAC
Time – varying current Steady current
• A mount of electric charges flowing through the surface per unit time.

Current
Positive versus negative current
2 A -2 A

P1.1, In the wire electrons moving left to right to create a current of 1 mA.
Determine I1 and I2.
Ans: I1 = -1 mA; I2 = +1 mA.
Current is always associated with arrows (directions)
Negative charge of -2C/s moving
Positive charge of 2C/s moving
or
Negative charge of -2C/s moving
Positive charge of 2C/s moving
or

Voltage(Potential)
baab VVV 

 


b
a
b
a
ab ldE
q
ldF
q
W
V


Voltage Units: 1 V = 1 J/C
Positive versus negative voltage

+
–
–
+
2 V -2 V
• Energy per unit charge.
• It is an electrical force drives an electric current.
+/- of voltage (V) tell the actual polarity of a certain point .DN
Two “Do Not (DN)”
+/- of current (I) tell the actual direction of particle’s movement .DN

Voltage (Potential)


a
b
 VVab 5 a、b, which point’s potential is higher？
b
a
 V6aV  V4bV Vab = ?
a b
 +Q from point b to point a get energy ，Point a is
Positive? or negative ?
Example

Voltage (Potential)
ab
c´
c d
d´
 
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
,
0
rRrR
EE
I
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a














Example
I

Voltage (Potential)
K Open
K Close
Va=?
)V(52.1
)V(1.8


a
a
V
V
Example
I
I

I
1
1 2
a
E
v E R
R R
  

Example
I
1 2
1 1
1 2
a
E E
v E R
R R

  

1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a a
a
v E v E v E v E R R R E R R R E R R R
v
R R R R R R R R R R R R R R R R
     
    
  Course Lecturer: Prof C Maina Muriithi, PhD

Power
• One joules of energy is expanded per second.
• Rate of change of energy
P = W/t )()()(/)()( titV
dt
dq
tVdttdwtp abab 
• Used to determine the electrical power is being absorbed or supplied
– if P is positive (+), power is absorbed
– if P is negative (–), power is supplied
+
–
v(t)
i(t)
p(t) = v(t) i(t)
v(t) is defined as the voltage with
positive reference at the same
terminal that the current i(t) is entering.

Power
Example
2A+
–
-5V 5 2 10WP      Power is supplied. delivered power
to external element.
+
–
5V
2A
5 2 10WP    Power is absorbed. Power delivered
to
Note :
+
–
+5V
+
–
-5V
2A
-2A
Power absorbed .

Power
• Power absorbed by a resistor:
)()()( titvtp 
)(2
tiR
Rtv /)(2

)(2
tvG
Gti /)(2


Power
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P1.5 Find the power absorbed by each element in the circuit.
A21 I A12 I
A13 I
 V35 V
 V41 V
 V82 V  V43 V
 V74 V
;3
;16
;7
;4
;8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP





Supply energy : element 1、3、4 .
Absorb energy : element 2、5

Open Circuit R=
I=0, V=E , P=0
E
R0
Short Circuit R=0
E
R0
R＝0
0R
E
I  00  IREV 0
2
RIPE 

RR
E
I
o 
 0IREIRV 
0
2
RIEIVI 
Loaded Circuit
E
R0 R
I
0PPP E 

1.3 Circuit Elements
Key Words:
Resistors, Capacitors, Inductors, voltage source,
current source

• Passive elements (cannot generate energy)
– e.g., resistors, capacitors, inductors, etc.
• Active elements (capable of generating energy)
– batteries, generators, etc.
• Important active elements
– Independent voltage source
– Independent current source
– Dependent voltage source
• voltage dependent and current dependent
– Dependent current source
• voltage dependent and current dependent

Resistors
 Dissipation Elements
S
l
R  v=iR P=vi=Ri2=v2/R >0 ，
 v-i relationship
v
i
 Resistors connected in series:
– Equivalent Resistance is found by Req= R1
+ R2 + R3 + …
R1 R2 R3
 Resistors connected in parallel 1/Req=1/R1 +
1/R2 + 1/R3 + …
R1 R2 R3

Independent voltage source
+
VS
RS＝0
v
i
VS
Ideal
sS
sS
IRVV
IRV

practical

Independent current source
I
v
iIS
RS＝ ∞
Ideal
SS
SS
RVII
RVI
/
/

practical



n
k
SkS VV
1
Voltage source connected in series:


n
k
SkS RR
1
Voltage source connected in parallel:


n
k
SkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
//////
21
21





Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv 
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q: What are the units for  and r?

Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
Sv Sgvi 
Si Sii 
Q: What are the units for  and g?

Independent source
dependent source
Can provide power to the circuit;
Excitation to circuit ;
Output is not controlled by external.
Can provide power to the circuit;
No excitation to circuit;
Output is controlled by external.

• So far, we have talked about two kinds of circuit elements:
– Sources (independent and dependent)
• active, can provide power to the circuit.
– Resistors
• passive, can only dissipate power.
Review
The energy supplied by the active
elements is equivalent to the energy
absorbed by the passive elements!

DC CIRCUIT ANALYSIS

Nodes, Branches, Loops, mesh
Node: point where two or more elements are joined (e.g., big node 1)
Loop: A closed path that never goes twice over a node (e.g., the blue line)
Branch: Component connected between two nodes (e.g., component R4)
The red path is NOT a loop
Mesh: A loop that does not contain any other loops in it.
Kirchhoff's Current and Voltage Laws

Nodes, Branches, Loops, mesh
• A circuit containing three nodes and five branches.
• Node 1 is redrawn to look like two nodes; it is still one nodes.
P1.8
1.4 Kirchhoff's Current and Voltage Laws

• sum of all currents entering a node is zero
• sum of currents entering node is equal to sum of currents leaving node
KCL
KCL Mathematically
i1(t)
i2(t)
i4(t)
i5(t)
i3(t)


n
j
j ti
1
0)( 

n
j
jI
1
0

• sum of all currents entering a node is zero
• sum of currents entering node is equal to sum of currents
leaving node
KCL
P1.9
DCBA iiii 
In
Out
0A B C O
I
I
i i i i
 
 
    

KCL
+
-120V
50* 1W Bulbs
Is
P1.10
• Find currents through each light bulb:
IB = 1W/120V = 8.3mA
• Apply KCL to the top node:
IS - 50IB = 0
• Solve for IS: IS = 50 IB = 417mA
KCL-Christmas Lights

KCL
P1.11 We can make supernodes by aggregting node.
0:
0:
7542
461


iiii
iii
3Leaving
2Leaving
076521  iiiii:3&2Adding

KCL
Current divider


N V
G1
G2
I
+
-
I1 I2
I
GG
G
G
G
I
VGI
21
1
111

 I
GG
G
VGI
21
2
22


I
G
G
I n
k
k
k
k



1
121
21
11
1
11
RRR
RR
I
R
RI
R
V
I 

 I
RR
R
I
21
1
2


In case of parallel : 1 2
1 2
1 1 1
, , V=
I I
G G G
R R R R G
    

 sum of voltages around any loop in a circuit is zero.
KVL
• A voltage encountered + to - is positive.
• A voltage encountered - to + is negative.
KVL Mathematically 0)(
1

n
j
j tv 0
1

n
j
jV

 KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with
higher voltage and releases electrical energy if it moves to a point
with lower voltage

AV
B
BV)( AB VVqW 
q
 abV
a b

q
abqVW LOSES
 cdV
c d

q
cdqVW GAINS 
AV
B BV
q
CV


AB
V

BCV
 CAV
If the charge comes back to the same Initial point the net energy gain Must be zero.
0)(  CABCAB VVVq

KVL
P1.13 Determine the voltages Vae and Vec.
10 24 0aeV   
16 12 4 6 0aeV    
4 + 6 + Vec = 0

KVL
Voltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
1
11
RR
R
VIRV


21
2
22
RR
R
VIRV


Important voltage Divider
equations
N
V
R
R
V n
k
k
k
k


1

KVL
Voltage divider
 kR 151
Volume control?
P1.14 Example: Vs = 9V, R1 = 90kΩ, R2 = 30kΩ

R3=80
R2=0.4
+
_
VS=14V E2=12V
R1=0.5
2.2 Basic Nodal and Mesh Analysis
Branch Analysis
I2
I3
I1
P2.2 How do we find I1 and I2, I3?
KVL
Mesh 1:
Mesh 2:
1 214 0.5 0.4 12 0I I    
2
1 2
2 0.4
4 0.8
0.5
I
I I

  
2 312 0.4 80 0I I   
2
3 2
12 0.4
0.15 0.005
80
I
I I

  
2 24 0.8 0.15 0.005I I   
2 2.13AI  1 2.29AI  3 0.16AI KCL 1 2 3I I I 
2 2 2 2
1
1
2 0.4
0.5
sV E I R I
I
R
  
  2 2 2 2
3
3
12 0.4
80
E I R I
I
R
 
 
2 2
2
2 0.4 12 0.4
0.5 80
I I
I
 
  2 2.14AI  1 2.29AI  3 0.14AI 
Ch2 Basic Analysis Methods to Circuits

Branch Analysis
 write KCL equation for each independent node.
——(n-1) KCL equations
 write KVL equation for each independent mesh/loop
——m-(n-1) KVL equations
Suppose m branches, n nodals

Mesh (Loop) Analysis
_USI1
I1
I1
I1
I1 I1 I1 I1
 


R1
R3
5
5
5 5
+
+
_
VS1
IS

E


R2
R6
R5 R4
VS2
VS3 VS4
+
+
+_ _
_



I1
I2
I3
I5
I6
I4
33333232131
22323222121
11313212111
Smmm
Smmm
Smmm
VIRIRIR
VIRIRIR
VIRIRIR



1 2 6 2 6 1 1 2
2 2 3 5 5 2 2 3
6 5 4 5 6 3 4
m S S
m S S
m S
R R R R R I V V
R R R R R I V V
R R R R R I V
       
           
          
   
   
    0
0
0
443613532
2212532332
2212631111



Smmmmm
mmSmmSm
mmSmmSm
VRIRIIRII
RIIVRIIVRI
RIIVRIIVRIMesh 1:
Mesh 2:
Mesh 3:
436542516
3235253212
2136221621
)(
)(
)(
Smmm
SSmmm
SSmmm
VIRRRIRIR
VVIRIRRRIR
VVIRIRIRRR



Im1 Mesh 1
Im2
Mesh 2
Im3
Mesh 3

Nodal Analysis
1) Choose a reference node
The reference node is called the ground node.
+
-
V 500
500
1k
500
500
I1 I2

Ⅱ:
Nodal Analysis
1) Choose a reference node
0
+
-
V 500
500
1k
500
500
I1 I2
I4
I5
I6
I7
I8
Ⅰ Ⅱ
KCL Ⅰ:
Ⅲ :
Ⅲ

Nodal Analysis
2) Assign node voltages to the other nodes
V1, V2, and V3 are unknowns for which we solve using KCL.
500
500
1k
500
500
I1 I2
1 2 3
V1 V2 V3I4
I5
I6
I7
I8
0

Nodal Analysis
3) Apply KCL to each node other than the reference-express
currents in terms of node voltages.
500
500
1k
500
500
I1 I2
1 2 3
V1 V2 V3I4
I5
I6
I7
I8
0



500
21
4
VV
I


500
1
5
V
I
2
6
1K
V
I 

3 2
7
500
V V
I



3
8
500
V
I 

, ,
, ,
0
500500
121
1 





VVV
I
0
500k1500
32212







 VVVVV
0
500500
2
323





I
VVV
Node 1:
Node 2:
Node 3:
• Node 1:
• Node 2:
• Node 3:
4 5 1 0I I I  
6 4 7 0I I I  
8 7 2 0I I I  
KCL

Nodal Analysis
4) Solve the resulting system of linear equations.
• Node 1:
• Node 2:
• Node 3:
1
2
1
500500
1
500
1
I
V
V 










0
500500
1
k1
1
500
1
500
3
2
1
















V
V
V
23
2
500
1
500
1
500
IV
V












• The left hand side of the equation:
– The node voltage is multiplied by the sum of conductances of all
resistors connected to the node.
– The neighbourly node voltages are multiplied by the conductance of the
resistor(s) connecting to the two nodes and to be subtracted.
• The right hand side of the equation:
– The right side of the equation is the sum of currents from sources
entering the node.
500
500
1k
500
500
I1
I2
1 2 3
V1 V2 V3
0

Nodal Analysis
• Node 1:
• Node 2:
• Node 3:
1
2
1
500500
1
500
1
I
V
V 










0
500500
1
k1
1
500
1
500
3
2
1
















V
V
V
23
2
500
1
500
1
500
IV
V












500
500
1k
500
500
I1
I2
1 2 3
V1 V2 V3






















































2
1
3
2
1
0
500
1
500
1
500
1
0
500
1
500
1
k1
1
500
1
500
1
0
500
1
500
1
500
1
I
I
V
V
V
Matrix Notation(Symmetric)

Nodal Analysis
• Node 1:
• Node 2:
• Node 3:
1
2
1
500500
1
500
1
I
V
V 










0
500500
1
k1
1
500
1
500
3
2
1
















V
V
V
23
2
500
1
500
1
500
IV
V












500
500
1k
500
500
I1
I2
1 2 3
V1 V2 V3
G11V1+G12V2 +G13V3 =I11
G21V2+G22V2 +G23V3 =I22
G31V1+G32V2+G33V3=I33

Nodal Analysis
What if there are dependent sources?
1k5mA 100Ib
+
-
Vo
50
Ib
1k
Example:
21
V2V1



50
21 VV
Ib
0
k150
100
50
22112







 VVVVV



































0
mA5
k1
1
50
100
50
1
50
100
50
1
50
1
50
1
k1
1
2
1
V
V
Matrix is not symmetric due to the dependent source.
mA5
50k1
211





VVV
Node ①： 0
k1
100
50
212




 V
I
VV
bNode ② ：

Basic Circuits
Mesh Analysis: Example 7.2
+
_
6 
10 
9 
11 
3 
4 
20V 10V
8V
12V
I1 I2
I3
+
+
__
_
_
+
+
_
Mesh 1: 6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 + 10
Mesh 2: 4(I2 – I1) + 11(I2 – I3) + 3I2 = - 10 - 8
Mesh 3: 9I3 + 11(I3 – I2) + 10(I3 – I1) = 12 + 8
Eq (7.13)
Eq (7.14)
Eq (7.15)

Circuit Theorems
• Linear Circuits and Superposition
• Thevenin's Theorem
• Norton's Theorem
• Maximum Power Transfer

Superposition Principle
Because the circuit is linear we can find the response of the circuit to each source
acting alone, and then add them up to find the response of the circuit to all sources
acting together. This is known as the superposition principle.
The superposition principle states that the voltage across (or the 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.

Turning sources off
a
b
si
si i
Current source:
We replace it by a current source
where
0si 
An open-circuit
Voltage source:
DC
+
-
sv vsv
We replace it by a voltage source
where
0sv 
An short-circuit
i

Steps in Applying the Superposition Principle
1. Turn off all independent sources except
one. Find the output (voltage or current)
due to the active source.
2. Repeat step 1 for each of the other
independent sources.
3. Find the total output by adding
algebraically all of the results found in
steps 1 & 2 above.
In some cases, but certainly not all, superposition can simplify the analysis.

DC
12V
 


3A
i
DC
24V
Example: In the circuit below, find the current i by superposition
Turn off the two voltage sources (replace by
short circuits).
12V
 


3A
1i
1v 2v

12V
 


3A
1i
1v 2v
1
2
1 4 1 3 1 4 1 4 0
1 4 1 4 1 8 3
v
v
       
         
1 2
5 1
0
6 4
v v 
1 2
1 3
3
4 8
v v  
2 1
10
3
v v 1
10 2
3
8 8
v
 
  
 
1 3v 1 1i 

DC
12V
 


3A
i
DC
24V
Turn off the 24V & 3A sources:
2i
DC
12V
 


2i
1i
O.C.

12 4
3
16


2
12
2
6
i  
DC
12V
 


2i
O.C.
DC
12V



2i
O.C.
DC
12V


2i
O.C.

DC
12V
 


3A
i
DC
24V
Turn off the 3A & 12V sources:
3i
 


3i
2i
O.C.
DC
24V

2
3
4 8 4 4 24
4 4 3 0
i
i
       
         
2 316 4 24i i  
2 34 7 0i i  
2 3
7
4
i i  3 28 4 24i   
3 1i  
3i
 


3i
2i
O.C.
DC
24V

DC
12V
 


3A
i
DC
24V
12V
 


3A
1i
1v 2v
1 1i  2 2i 
2i
DC
12V
 


2i
1i
O.C.
3 1i  
3i
 


3i
2i
O.C.
DC
24V
1 2 3 1A 2A 1A 2Ai i i i      

Thevenin's Theorem
Linear
Circuit
+
-
V
Variable
R
b
a
In many applications we want to find the response to a particular element which may, at
least at the design stage, be variable.
Each time the variable element changes we
have to re-analyze the entire circuit. To avoid
this we would like to have a technique that
replaces the linear circuit by something simple
that facilitates the analysis.
A good approach would be to have a simple equivalent circuit to replace everything in the
circuit except for the variable part (the load).

Thevenin's Theorem
Thevenin’s theorem states that a linear two-terminal resistive 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 all turned off.
Linear
Circuit
b
a
inR
LR
i
DC
b
a
inR
LR
i
ThR
ThV

Thevenin's Theorem
Linear
Circuit
b
a
inR
LR
i
DC
b
a
inR
LR
i
ThR
ThV
Thevenin’s theorem states that the two circuits given below are equivalent as seen from
the load RL that is the same in both cases.
VTh = Thevenin’s voltage = Vab with RL disconnected (= ) = the open-circuit voltage = VOC

Thevenin's Theorem
Linear
Circuit
b
a
inR
LR
i
DC
b
a
inR
LR
i
ThR
ThV
RTh = Thevenin’s resistance = the input resistance with all independent sources turned off
(voltage sources replaced by short circuits and current sources replaced by open
circuits). This is the resistance seen at the terminals ab when all independent sources
are turned off.

Example
DC
10V
 

a
b
2
10V 5V
2 2
OC Thv V  

DC
10V
 

a
b
10 2 10
2.5A
2 3 42
3
SCi   

5
2
2.5
Th
Th
SC
V
R
i
   
DC
a
b
5VThV 
2ThR  
 

a
b
2 2
1 2
2 2
ThR

   


Norton's Theorem
Norton’s equivalent circuit can be found by transforming the Thevenin equivalent into a
current source in parallel with the Thevenin resistance. Thus, the Norton equivalent circuit is
given below.
Formally, Norton’s Theorem states that a linear two terminal resistive 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 all independent sources are all turned off.
b
a
LR
i
Th
N
Th
V
I
R
 N ThR R

THEVENIN & NORTON
THEVENIN’S THEOREM:
Consider the following:
Network
1
Network
2
•
•
A
B
Figure 10.1: Coupled networks.
For purposes of discussion, at this point, we consider
that both networks are composed of resistors and
independent voltage and current sources
1 Course Lecturer: Prof C Maina Muriithi, PhD

THEVENIN & NORTON
Suppose Network 2 is detached from Network 1 and
we focus temporarily only on Network 1.
Network
1
•
•
A
B
Figure 10.2: Network 1, open-circuited.
Network 1 can be as complicated in structure as one
can imagine. Maybe 45 meshes, 387 resistors, 91
voltage sources and 39 current sources.

Network
1
•
•
A
B
THEVENIN & NORTON
Now place a voltmeter across terminals A-B and
read the voltage. We call this the open-circuit voltage.
No matter how complicated Network 1 is, we read one
voltage. It is either positive at A, (with respect to B)
or negative at A.
We call this voltage Vos and we also call it VTHEVENIN = VTH

THEVENIN & NORTON
• We now deactivate all sources of Network 1.
• To deactivate a voltage source, we remove
the source and replace it with a short circuit.
• To deactivate a current source, we remove
the source.

THEVENIN & NORTON
Consider the following circuit.
+_
+
+
_ _
A
B
V1
I2
V2
I1
V3
R1
R2
R3
R4
Figure 10.3: A typical circuit with independent sources
How do we deactivate the sources of this circuit?

THEVENIN & NORTON
When the sources are deactivated the circuit appears
as in Figure 10.4.
R1
R2
R3
R4
A
B
Figure 10.4: Circuit of Figure 10.3 with sources deactivated
Now place an ohmmeter across A-B and read the resistance.
If R1= R2 = R4= 20  and R3=10  then the meter reads 10 .
6

THEVENIN & NORTON
We call the ohmmeter reading, under these conditions,
RTHEVENIN and shorten this to RTH. Therefore, the
important results are that we can replace Network 1
with the following network.
VTH
RTH
A
B
+_


Figure 10.5: The Thevenin equivalent structure.

THEVENIN & NORTON
We can now tie (reconnect) Network 2 back to
terminals A-B.
A
B
Network
2
VTH
RTH
+
_


Figure 10.6: System of Figure 10.1 with Network 1
replaced by the Thevenin equivalent circuit.
We can now make any calculations we desire within
Network 2 and they will give the same results as if we
still had Network 1 connected.

THEVENIN & NORTON
It follows that we could also replace Network 2 with a
Thevenin voltage and Thevenin resistance. The results
would be as shown in Figure 10.7.
A
B
+ +_ _
RTH 1 RTH 2
VTH 1 VTH 2


Figure 10.7: The network system of Figure 10.1
replaced by Thevenin voltages and resistances.

THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.1.
Find VX by first finding VTH and RTH to the left of A-B.
12  4 
6  2  VX30 V +_
+
_
A
B


Figure 10.8: Circuit for Example 10.1.
10
First remove everything to the right of A-B.

THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.1. continued
12  4 
6 30 V +_
A
B


Figure 10.9: Circuit for finding VTH for Example 10.1.
(30)(6)
10
6 12
ABV V 

Notice that there is no current flowing in the 4  resistor
(A-B) is open. Thus there can be no voltage across the
resistor.

THEVENIN & NORTON
We now deactivate the sources to the left of A-B and find
the resistance seen looking in these terminals.
12  4 
6 
A
B


RTH
Figure 10.10: Circuit for find RTH for Example 10.10.
We see,
RTH = 12||6 + 4 = 8 

THEVENIN & NORTON
After having found the Thevenin circuit, we connect this
to the load in order to find VX.
8 
10 VVTH
RTH
2  VX
+
_
+
_
A
B


Figure 10.11: Circuit of Ex 10.1 after connecting Thevenin
circuit.
10 2
2
2 8
 

( )( )
XV V

THEVENIN & NORTON
In some cases it may become tedious to find RTH by reducing
the resistive network with the sources deactivated. Consider
the following:
VTH
RTH
+
_
A
B


ISS
Figure 10.12: A Thevenin circuit with the output shorted.
We see;
TH
TH
SS
V
R
I

14
Eq 10.1

THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.2.
For the circuit in Figure 10.13, find RTH by using Eq 10.1.
12  4 
6 30 V +_
A
B


ISS


C
D
Figure 10.13: Given circuit with load shorted
The task now is to find ISS. One way to do this is to replace
the circuit to the left of C-D with a Thevenin voltage and
Thevenin resistance.

THEVENIN & NORTON
Applying Thevenin’s theorem to the left of terminals C-D
and reconnecting to the load gives,
4  4 
10 V +_
A
B


ISS


C
D
Figure 10.14: Thevenin reduction for Example 10.2.
10
8
10
8
TH
TH
SS
V
R
I
   

THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.3
For the circuit below, find VAB by first finding the Thevenin
circuit to the left of terminals A-B.
+_20 V
5 
20 
10 
17 
1.5 A
A
B


We first find VTH with the 17  resistor removed.
Next we find RTH by looking into terminals A-B
with the sources deactivated.

THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.3 continued
+_20 V
5 
20 
10 
1.5 A
A
B


Figure 10.16: Circuit for finding VOC for Example 10.3.
20(20)
(1.5)(10)
(20 5)
31
OS AB TH
TH
V V V
V V
   

 

THEVENIN & NORTON
5 
20 
10 
A
B


Figure 10.17: Circuit for find RTH for Example 10.3.
5(20)
10 14
(5 20)
THR    


THEVENIN & NORTON
14 
31 VVTH
RTH
17  VAB
+
_
+
_
A
B


Figure 10.18: Thevenin reduced circuit for Example 10.3.
We can easily find that,
17ABV V

THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.4: Working
with a mix of independent and dependent sources.
Find the voltage across the 100  load resistor by first finding
the Thevenin circuit to the left of terminals A-B.
+_ 86 V
50 
30 
40 
100 
6 IS
IS


A
B
Figure 10.19: Circuit for Example 10.4

THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.4: continued
First remove the 100  load resistor and find VAB = VTH to
the left of terminals A-B.
+_ 86 V
50 
30 
40 
6 IS
IS


A
B
Figure 10.20: Circuit for find VTH, Example 10.4.
86 80 6 0 1
6 30 36
S S S
AB S S
I I I A
V I I V
     
   

THEVENIN & NORTON
To find RTH we deactivate all independent sources but retain
all dependent sources as shown in Figure 10.21.
50 
30 
40 
6 IS
IS


A
B
RTH
Figure 10.21: Example 10.4, independent sources deactivated.
We cannot find RTH of the above circuit, as it stands. We
must apply either a voltage or current source at the load
and calculate the ratio of this voltage to current to find RTH.

THEVENIN & NORTON
50 
30 
40 
6 IS
IS
1 A
1 A
IS + 1 V
Figure 10.22: Circuit for find RTH, Example 10.4.
Around the loop at the left we write the following equation:
50 30( 1) 6 0S S SI I I   
From which
15
43
SI A



THEVENIN & NORTON
50 
30 
40 
6 IS
IS
1 A = I
1 A
IS + 1 V
Using the outer loop, going in the cw direction, using drops;
15
50 1(40) 0
43
V
 
   
 
or 57.4V volts
25
57.4
1
TH
V V
R
I
   

THEVENIN & NORTON
The Thevenin equivalent circuit tied to the 100  load
resistor is shown below.
+_
RTH
VTH
57.4 
36 V 100 
Figure 10.24: Thevenin circuit tied to load, Example 10.4.
100
36 100
22.9
57.4 100
x
V V 


THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.5: Finding
the Thevenin circuit when only resistors and dependent
sources are present. Consider the circuit below. Find Vxy
by first finding the Thevenin circuit to the left of x-y.
+_
x
y


10Ix
20 
50  60 
50 
100 V
IX
For this circuit, it would probably be easier to use mesh or nodal analysis
to find Vxy. However, the purpose is to illustrate Thevenin’s theorem.

THEVENIN & NORTON
We first reconcile that the Thevenin voltage for this circuit
must be zero. There is no “juice” in the circuit so there cannot
be any open circuit voltage except zero. This is always true
when the circuit is made up of only dependent sources and
resistors.
To find RTH we apply a 1 A source and determine V for
the circuit below.
20 
50  60 
20 
V
1 A
IX1 - IX
10IX
Figure 10.26: Circuit for find RTH, Example 10.5.Course Lecturer: Prof C Maina Muriithi, PhD

THEVENIN & NORTON
20 
50  60 
20 
V
1 A
IX1 - IX
10IX
m
Write KVL around the loop at the left, starting at “m”, going
cw, using drops:
29
060)1(2010)1(50  XXXX IIII
AIX 5.0Course Lecturer: Prof C Maina Muriithi, PhD

THEVENIN & NORTON
20 
50  60 
20 
V
1 A
IX1 - IX
10IX
m
n
Figure 10.28: Determining RTH for Example 10.5.
We write KVL for the loop to the right, starting at n, using
drops and find;
or
50V volts
0201)5.0(60  Vx

THEVENIN & NORTON
We know that, ,TH
V
R
I
 where V = 50 and I = 1.
Thus, RTH = 50 . The Thevenin circuit tied to the
load is given below.
+_
50 
50 
x
y


100 V
Figure 10.29: Thevenin circuit tied to the load, Example 10.5.
Obviously, VXY = 50 V

THEVENIN & NORTON
NORTON’S THEOREM:
Assume that the network enclosed below is composed
of independent sources and resistors.
Network
Norton’s Theorem states that this network can be
replaced by a current source shunted by a resistance R.
I R

ISS RN = RTH
THEVENIN & NORTON
NORTON’S THEOREM:
In the Norton circuit, the current source is the short circuit
current of the network, that is, the current obtained by
shorting the output of the network. The resistance is the
resistance seen looking into the network with all sources
deactivated. This is the same as RTH.

THEVENIN & NORTON
NORTON’S THEOREM:
We recall the following from source transformations.
+
_
R
RV I =
V
R
In view of the above, if we have the Thevenin equivalent
circuit of a network, we can obtain the Norton equivalent
by using source transformation.
However, this is not how we normally go about finding
the Norton equivalent circuit.

THEVENIN & NORTON
NORTON’S THEOREM: Example 10.6.
Find the Norton equivalent circuit to the left of terminals A-B
for the network shown below. Connect the Norton equivalent
circuit to the load and find the current in the 50  resistor.
+_
20 
60 
40 
50 
10 A
50 V


A
B

THEVENIN & NORTON
NORTON’S THEOREM: Example 10.6. continued
+_
20 
60 
40 
10 A
50 V
ISS
Figure 10.31: Circuit for find INORTON.
It can be shown by standard circuit analysis that
10.7SSI A
36

THEVENIN & NORTON
It can also be shown that by deactivating the sources,
We find the resistance looking into terminals A-B is
55NR  
RN and RTH will always be the same value for a given circuit.
The Norton equivalent circuit tied to the load is shown below.
10.7 A 55  50 
Figure 10.32: Final circuit for Example 10.6.

THEVENIN & NORTON
NORTON’S THEOREM: Example 10.7. This example
illustrates how one might use Norton’s Theorem in electronics.
the following circuit comes close to representing the model of a
transistor.
For the circuit shown below, find the Norton equivalent circuit
to the left of terminals A-B.
+_5 V
1 k
3 VX 25 IS
+
_
VX
A
B
IS
40 

THEVENIN & NORTON
+_5 V
1 k
3 VX 25 IS
+
_
VX
A
B
IS
40 
We first find;
SS
OS
N
I
V
R 
We first find VOS:
SSXOS IIVV 1000)40)(25( 

THEVENIN & NORTON
+_5 V
1 k
3 VX 25 IS
+
_
VX
A
B
IS
40  ISS
Figure 10.34: Circuit for find ISS, Example 10.7.
We note that ISS = - 25IS. Thus,



 40
25
1000
S
S
SS
OS
N
I
I
I
V
R
40

THEVENIN & NORTON
+_5 V
1 k
3 VX 25 IS
+
_
VX
A
B
IS
40 
Figure 10.35: Circuit for find VOS, Example 10.7.
From the mesh on the left we have;
0)1000(310005  SS II
From which,
mAIS 5.2

THEVENIN & NORTON
We saw earlier that,
SSS II 25
Therefore;
mAISS 5.62
The Norton equivalent circuit is shown below.
IN = 62.5 mA RN = 40 
A
B
42
Norton Circuit for Example 10.7

THEVENIN & NORTON
Extension of Example 10.7:
Using source transformations we know that the
Thevenin equivalent circuit is as follows:
+
_ 2.5 V
40 
Figure 10.36: Thevenin equivalent for Example 10.7.

Maximum Power Transfer
In all practical cases, energy sources have non-zero internal resistance. Thus, there are
losses inherent in any real source. Also, in most cases the aim of an energy source is to
provide power to a load. Given a circuit with a known internal resistance, what is the
resistance of the load that will result in the maximum power being delivered to the load?
Consider the source to be modeled by its Thevenin equivalent.
DC
b
a
LR
i
ThR
ThV

Maximum Power
Transfer
• For any power source, the maximum power
transferred from the power source to the load
is when the resistance of the load RL is equal
to the equivalent or input resistance of the
power source (Rin = RTh or RN).
– The process used to make RL = Rin is called
impedance matching.

Power Transfer
Calculation
RTh 50
VTh
1V
L
Th
ThL
L
L
LLL
R
V
RR
R
P
RVP
2
2










DC
b
a
LR
i
ThR
ThV
The power delivered to the load (absorbed by RL) is
 
22
L Th Th L Lp i R V R R R    
This power is maximum when
   2 32
2 0Th Th L L Th L
L
p
V R R R R R
R
       
 
0Lp R  

   2 32
2 0Th Th L L Th L
L
dp
V R R R R R
dR
 
     
 
2Th L LR R R 
L ThR R
 
2
max L Th
Th Th L L R R
p V R R R 
   
 
2 2
max 2 4Th Th Th Th Thp V R R V R   
Thus, maximum power transfer takes place when the resistance of the load equals the
Thevenin resistance RTh. Note also that
Thus, at best, one-half of the power is dissipated in the internal resistance and one-half
in the load.

Application
• When developing new circuits for a known
application, optimize the power transfer by
designing the circuit to have an input
resistance close to the load resistance.
• When selecting a source to power a circuit,
one of the selection criteria is to match the
input impedance to the load resistance.

Summary
• Maximum power transfer theorem is used
frequently to insure that the greatest power
can be transferred from a power source to a
load.

DC TRANSIENT ANALYSIS

SUB - TOPICS
• NATURAL RESPONSE OF RL CIRCUIT
• NATURAL RESPONSE OF RC CIRCUIT
• STEP RESPONSE OF RL CIRCUIT
• STEP RESPONSE OF RC CIRCUIT

OBJECTIVES
• To investigate the behavior of currents and
voltages when energy is either released or
acquired by inductors and capacitors when
there is an abrupt change in dc current or
voltage source.
• To do an analysis of natural response and step
response of RL and RC circuit.

AC CIRCUIT ANALYSIS

Aims of AC circuit
analysis
• To use phasors to describe sinusoidally varying
quantities
• To use reactance to describe voltage in a circuit
• To analyze an L-R-C series circuit
• To determine power in ac circuits
• To see how an L-R-C circuit responds to
frequency

GRAPHICAL
PRESENTATION OF
SINUSOIDS
0 1 2 3 4 5 6 7
-10
-8
-6
-4
-2
0
2
4
6
8
10
1 5sin( )e t
t=0:pi/100:2*pi;
e1=10*sin(t);
e2=5*sin(t);
e3=5*sin(t+pi/4);
e=e2+e3;
plot(t,e1,'w',t,e2)
plot(t,e1,'w',t,e2,t,e3)
plot(t,e1,'w',t,e2,t,e3,t,e)
grid on

0 1 2 3 4 5 6 7
-10
-8
-6
-4
-2
0
2
4
6
8
10
2 5sin( )
4
e t

 
1 5sin( )e t

0 1 2 3 4 5 6 7
-10
-8
-6
-4
-2
0
2
4
6
8
10
1 2totale e e 

Phasors
The phasor spins around the
complex plane as a function
of time.
Phasors of the same frequency
can be added.
A phasor, or phase vector, is a representation
of a sinusoidal wave whose amplitude ,
phase , and frequency are time-
invariant.
This is an animation
But it’s a known factCourse Lecturer: Prof C Maina Muriithi, PhD

Converting Between Forms
To convert from the Cartesian form to polar form,
note:

What’s the Difference between i and j ?
Engineering Physics
j i 
Can go back and forth between physics and engineering literature
If we adopt the convention
 
0
i t kx
E E e  
 
0
j t kx
E E e  

Clipart images are in the public domain.

Power Factor Correction

ELECTRICAL MEASUREMENTS

Electrical engineering principles lecture slides. circuit analysis by Maina

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