Circuit Analysis – DC Circuits

Circuit analysis: DC Circuits (3 cr)
Fall 2009 / Class AS09

Vesa Linja-aho

Metropolia

October 8, 2010

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Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 1 / 125

Table of Contents

Click the lecture name to jump onto the ﬁrst slide of the lecture.

1 1. lecture 7 7. lecture


1. lecture

About the Course

Lecturer: M.Sc. Vesa Linja-aho
Lectures on Mon 11:00-14:00 and Thu 14:00-16:30, room P113
To pass the course: Home assignments and ﬁnal exam. The exam is
on Monday 12th October 2009 at 11:00-14:00.
All changes to the schedule are announced in the Tuubi-portal.


1. lecture

The Home Assignments
There are 12 home assignments.
Each assignment is graded with 0, 0,5 or 1 points.
To pass the course, the student must have at least 4 points from the
assignments.
Each point exceeding the minimum of 4 points will give you 0,5 extra
points in the exam.
In the exam, there are 5 assignments, with maximum of 6 points each.
To pass the exam, you need to get 15 points from the exam.
All other grade limits (for grades 2-5) are ﬂexible.

Example
The student has 8 points from the home assignments. He gets 13 points
from the exam. He will pass the exam, because he gets extra points from
the home assignments and his total score is (8 − 4) · 0,5 + 13 = 15 points.

However, is one gets 8 of 12 points from the home assignments, he usually
gets more than 13 points from the exam :-).

1. lecture

The Course Objectives

From the curriculum:
Learning outcomes of the course unit
Basic concepts and basic laws of electrical engineering. Analysis of direct current
(DC) circuits.

Course contents
Basic concepts and basic laws of electrical engineering, analysis methods,
controlled sources. Examples and exercises.


1. lecture

The Course Schedule
1 The basic quantities and units. Voltage source and resistance.
Kirchhoﬀ’s laws and Ohm’s law.
2 Conductance. Electric power. Series and parallel circuits. Node.
Ground.
3 Current source. Applying the Kirchhoﬀ’s laws to solve the circuit.
Node-voltage analysis.
4 Exercises on node-voltage analysis.
5 Source transformation.
6 Th´venin equivalent and Norton equivalent.
e
7 Superposition principle.
8 Voltage divider and current divider.
9 Inductance and capacitance in DC circuits.
10 Controlled sources.
11 Recap.
12 Recap.

1. lecture

The Course is Solid Ground for Further Studies in
Electronics

The basic knowledge on DC circuits is needed on the courses Circuit
Analysis: Basic AC-Theory, Measuring Technology, Automotive
Electronics 1, Automotive Electrical Engineering Labs, . . .
Important!
By studying this course well, studying the upcoming courses will be
easier!
The basics of DC circuits are vital for automotive electronics engineer, just
like the basics of accounting are vital for an auditor, and basics of strength
of materials are vital for a bridge-building engineer etc.


1. lecture

What is Not Covered on This Course

The basic physical characteristics of electricity is not covered on this
course. Questions like ”What is electricity?” are covered on the course
Rotational motion and electromagnetism.


1. lecture

Studying in Our School

You have an opportunity to learn on the lectures. I can not force you
to learn.
You have more responsibility on your learning than you had in
vocational school or senior high school.
1 cr ≈ 26,7 hours of work. 3 cr = 80 hours of work. You will spend
39 hours on the lectures.
Which means that you should use about 40 hours of your own time
for studying!
If I proceed too fast or too slow, please interject me (or tell me by
email).
Do not hesitate to ask. Ask also the ”stupid questions”.


1. lecture

What Is Easy and What Is Hard?

Diﬀerent things are hard for diﬀerent people. But my own experience
shows that
DC analysis is easy, because the math involved is very basic.
DC analysis is hard, because the circuits are not as intuitive as, for
example, mechanical systems are.
Studying your math courses well is important for the upcoming courses on
circuit analysis. For example, in AC circuits analysis you have to use
complex arithmetics.


1. lecture

Now, Let’s Get into Business

Any questions on the practical arrangements of the course?


1. lecture

Electric Current

Electric current is a ﬂow of electric charge.
The unit for electric current is the ampere (A).
The abbreviation for the quantity is I .
One may compare the electric current with water ﬂowing in a pipe (so
called hydraulic analogy).
The current always circulates in a loop: current does not compress
nor vanish.
The current in a wire is denoted like this:

-
I = 2 mA


1. lecture

Kirchhoff’s Current Law
As mentioned on the previous slide, the current can not vanish
anywhere.

Kirchhoff’s Current Law (or: Kirchhoff’s First Law)
At any area in an electrical circuit, the sum of currents flowing into that
area is equal to the sum of currents flowing out of that area.

I3 = 1 mA
6
- -
I1 = 3 mA I2 = 2 mA
If you draw a circle in any place in the circuit, you can observe that there
is as the same amount of current flowing into the circle and out from the
circle!

1. lecture

Be Careful with Signs

One can say: ”The balance of my account -50 euros” or equally ”I
owe 50 euros to my bank”.
One can say: ”The proﬁt of the company was -500000 euros” or
equally ”The loss of the company was 500000 euros”.
If you measure a current with an ammeter and it reads −15 mA, by
reversing the wires of the ammeter it will show 15 mA.
The sign of the current shows the direction of the current. The two
circuits below are exactly identical.

I3 = 1 mA I3 = 1 mA
6 6
- -
I1 = 3 mA I2 = 2 mA Ia = −3 mA Ib = −2 mA


1. lecture

Voltage

The potential difference between two points is called voltage.
The abbreviation for the quantity is U.
In circuit theory, it is insignificant how the potential difference is
generated (chemically, by induction etc.).
The unit of voltage is the volt (V).
One may compare the voltage with a pressure difference in hydraulic
system, or to a difference in altitude.
Voltage is denoted with an arrow between two points.

+
12 V U = 12 V

− c


1. lecture

Kirchhoff’s voltage law

The voltage between two points is the same, regardless of the path
chosen.
This is easy to understand by using the analogy of differences in
altitude. If you leave your home, go somewhere and return to your
home, you have traveled uphill as much as you have traveled downhill.

Kirchhoff’s Voltage Law (or: Kirchhoff’s Second Law)
The directed sum of the voltages around any closed circuit is zero.
r'
4,5 V r

− − −
+ + +
1,5 V 1,5 V 1,5 V

1. lecture

Ohm’s law

Resistance is a measure of the degree to which an object opposes an
electric current through it.
The larger the current, the larger the voltage – and vice versa.
The abbreviation of the quantity is R and the unit is ( Ω) (ohm).
The deﬁnition of resistance is the ratio of the voltage over the
element divided with the current through the element. R = U/I

U = RI

U E
-
I R


1. lecture

Definitions

Electric circuit A system consisting of compontents, in where electric
current flows.
Direct current (DC) The electrical quantities (voltage and current) are
constant (or nearly constant) over time.
Direct current circuit An electric circuit, where voltages and currents are
constant over time.

Example
In a flashlight, there is a direct current circuit consisting of a
battery/batteries, a switch and a bulb. In a bicycle there is an alternating
current circuit (dynamo and bulb).


1. lecture

An alternate definition for direct current

One may define also that direct current means a current, which does not
change its direction (sign), but the magnitude of the current can vary over
time. For example, a simple lead acid battery charger outputs a pulsating
voltage, which varies between 0 V ... ≈ 18 V. This can be also called DC
voltage.
Agreement
On this course, we define DC to mean constant voltage and current. The
magnitude and sign are constant over time.


1. lecture

A simple DC circuit

A light bulb is wired to a battery. The resistance of the ﬁlament is
10 Ω.
-
I =?

+
12 V

d

d
−


1. lecture

A simple DC circuit

10 Ω.
-

+ I =?
12 V 10 Ω

−


1. lecture

A simple DC circuit

10 Ω.
-

+ I =?
12 V 10 Ω 12 V

− c


1. lecture

A simple DC circuit

10 Ω.
-
+ I = 1,2 A

12 V 10 Ω 12 V

− c

U = RI
I =U =
R
12 V
10 Ω = 1,2 A


1. lecture

Homework 1 (released 31st Aug, to be returned 3rd Sep)

The homework are to be returned at the beginning of the next lecture.
Remember to include your name and student number.

Homework 1
Find the current I .

+
1,5 V

−
R = 20 Ω

+
1,5 V I
?

−


2. lecture

Homework 1 - Model solution
Homework 1

+
1,5 V

−
R = 20 Ω

+
1,5 V I
?

−


2. lecture

Homework 1

+
1,5 V U2

− c
UR R = 20 Ω
c
+
1,5 V U1 I
?

− c

U1 + U2 − UR = 0 ⇔ UR = U1 + U2


2. lecture

Homework 1

+
1,5 V U2

− c
UR R = 20 Ω
c
+
1,5 V U1 I
?

− c

U1 + U2 − UR = 0 ⇔ UR = U1 + U2
UR U1 + U2 1,5 V + 1,5 V
U = RI ⇒ UR = RI ⇒ I = = = = 150 mA
R R 20 Ω


2. lecture

Conductance

Resistance is a measure of the degree to which an object opposes an
electric current through it.
The inverse of resistance is conductance. The symbol for
conductance is G and the unit is Siemens (S).
Conductance measures how easily electricity ﬂows along certain
element.
For example, if resistance R = 10 Ω then conductance G = 0,1 S.
1
G= R U = RI ⇔ GU = I

U E
-
IG = 1
R


2. lecture

Electric Power

In physics, power is the rate at which work is performed.
The symbol for power is P and the unit is the Watt (W).
The DC power consumed by an electric element is P = UI

U E
-
I
If the formula outputs a positive power, the element is consuming
power from the circuit. If the formula outputs a negative power, the
element is delivering power to the circuit.


2. lecture

Electric Power

Energy can not be created nor destroyed
The power consumed by the elements in the circuit = the power delivered
by the elements in the circuit.
U
I =
I I R

?
+6
U2
E R PR = UI = U U =
R R
−
2
PE = U · (−I ) = U −U = − U
R R

The power delivered by the voltage source is consumed by the resistor.


2. lecture

Series and Parallel Circuits

Definition: series circuit
The elements are in series, if they are connected so that the same current
flows through the elements.

Definition: parallel circuit
The elements are in parallel, if they are connected so that there is the
same voltage across them.


2. lecture

Series and Parallel Circuits

Series circuit
- -
I I

Parallel Circuit
U E

U E


2. lecture

Resistors in series and in parallel

In series
⇐⇒
R1 R2 R = R1 + R2

In parallel

R2
⇐⇒
1
R= 1
+R1
R1 2
R1

Or, by using conductances: G = G1 + G2 .


2. lecture

Resistors in series and in parallel

The formulae on the previous slide can be applied to an arbitrary
number of resistors. For instance, the total resistance of ﬁve resistors
in series is R = R1 + R2 + R3 + R4 + R5 .


2. lecture

Voltage Sources in Series

The voltages can be summed like resistances, but be careful with
correct signs.
Voltage sources in parallel are inadmissible in circuit theory. There can

not be two diﬀerent voltages between two nodes at the same time.
r r
− + −
+ − +
E1 E2 E3
⇐⇒

r r
−
+
E = E 1 − E2 + E3


2. lecture

What Series and Parallel Circuits are NOT
Just the fact that two components seem to be one after the other,
does not mean that they are in series.
Just the fact that two components seem to be side by side, does not
mean that they are in parallel.
In the ﬁgure below, which of the resistors are in parallel and which are
in series with each other?

+ R1 R2 +
E1 R3 E2

− −


2. lecture

What Series and Parallel Circuits are NOT
Just the fact that two components seem to be one after the other,
does not mean that they are in series.
Just the fact that two components seem to be side by side, does not
mean that they are in parallel.
In the ﬁgure below, which of the resistors are in parallel and which are
in series with each other?

+ R1 R2 +
E1 R3 E2

− −

Solution
None! E1 ja R1 are in series and E2 ja R2 are in series. Both of these serial circuits are in
parallel with R3 . But no two resistors are in parallel nor in series.


2. lecture

Terminal and Gate

A point which provides a point of connection to external circuits is
called a terminal (or pole).
Two terminals form a gate.
An easy example: a car battery with internal resistance.
˜

+ RS
E

−
˜


2. lecture

Node

A node means an area in the circuit where there are no potential
diﬀerences, or alternatively a place where two or more circuit
elements meet.
A ”for dummies” –way to ﬁnd nodes in the circuit: put your pen on a
wire in the circuit. Start coloring the wire, and backtrack when your
pen meets a circuit element. The area you colored is one node.
How many nodes are there in the circuit below?
-
+ I R1
R3 R5
E R2 R4 R6

−


2. lecture

Ground

One of the nodes in the circuit can be appointed the ground node.
By selecting one of the nodes to be the ground node, the circuit
diagram usually appear cleaner.
The car battery is connected to the chassis of the car. Therefore it is
convenient to handle the chassis as the ground node.
When we say ”the voltage of this node is 12 volts” it means that the
voltage between that node and the ground node is 12 volts.
-
+ I R1
R3 R5
E R2 R4 R6

−
r


2. lecture

Ground

The ground node can be connected to the chassis of the device or it
can be leave not connected to the chassis.
Therefore, the existence of the ground node does not mean that the
device is ”grounded”.
The circuit on the previous slide can be presented also like this:
-
+ I R1
R3 R5
E R2 R4 R6

−


2. lecture

Homework 2 (released 3rd Sep, to be returned 7th Sep)

Homework 2
-
+ I R1
R3 R5
E R2 R4 R6

−

R1 = R2 = R3 = R4 = R5 = R6 = 1 Ω E = 9V


3. lecture


Homework 2
-
+ I R1
R3 R5
E R2 R4 R6

−

R1 = R2 = R3 = R4 = R5 = R6 = 1 Ω E = 9V

R5 ja R6 are in series. The total resistance of the serial connection is
R5 + R6 = 2 Ω.
Furthermore, the serial connection is in parallel with R4 . The
resistance of this parallel circuit is 1 + 1 Ω = 2 Ω.
1
3
1 2


3. lecture

Solution continues

R3 is in series with the parallel circuit calculated on the previous slide.
The resistance for this circuit is R3 + 2 Ω = 3 Ω.
3
5

And the serial connection is in parallel with R2 . The resistance for the
parallel circuit is ( 5 )−1 + 1 = 5 Ω.
1
8
3 1
Lastly, R1 is in series with the resistance computed in the previous
step. Therefore, the total resistance seen by voltage source E is
5 13
8 Ω + R1 = 8 Ω.
E 72
The current I is computed from Ohm’s law I = 13
Ω
= 13 A ≈ 5,5 A.
8


3. lecture

The Current Source

The current source is a circuit element which delivers a certain
current throught it, just like the voltage source keeps a certain
voltage between its nodes.
The current can be constant or it can vary by some rule.

6
J R


3. lecture

The Current Source

If there is a current source in a wire, you know the current of that
wire.
-
A
6
I =1
J = 1A R1 R2


3. lecture

Applying Kirchhoff’s Laws Systematically to the Circuit
When solving a circuit, it is highly recommended to use a systematic
mehtod to find the voltages and/or currents. Otherwise it is easy to end
up with writing a bunch of equations which can not be solved. One
systematic method is called the nodal analysis:
1 Name each current in the circuit.

2 Select one node as the ground node. Assign a variable for each

voltage between each node and ground node.
3 Write an equation based on Kirchhoff’s current law for each node

(except the ground node).
4 State the voltage of each resistor by using the node voltage variables

in step 2. Draw the voltage arrows at the same direction you used for
the current arrows (this makes it easier to avoid sign mistakes).
5 State every current by using the voltages and substitute them into the

current equations in step 2.
6 Solve the set of equations to find the voltage(s) asked.

7 If desired, solve the currents by using the voltages you solved.


3. lecture

Example

R1 R2

+
+
E1 R3 E2

− −
I
?


3. lecture

Example

I2
R1 - R2

+ I1
+
E1 R3 E2

− −
?I


3. lecture

Example

I2
R1 - R2

+ I1
+
E1 R3 U3 E2

− c −
?I


3. lecture

Example

I2
R1 - R2

+ I1
+
E1 R3 U3 E2 I = I1 + I2

− c −
?I


3. lecture

Example

E1 − U3 '2 − U3
E I2
E
R1 - R2

+ I1
+
E1 R3 U3 E2 I = I1 + I2

− c −
?I


3. lecture

Example

E1 − U3 '2 − U3
E I2
E
R1 - R2

+ I1
+
E1 R3 U3 E2 I = I1 + I2

− c −
?I
U3 E1 − U3 E2 − U3
= +
R3 R1 R2


3. lecture

Example

E1 − U3 '2 − U3
E I2
E
R1 - R2

+ I1
+
E1 R3 U3 E2 I = I1 + I2

− c −
?I
U3 E1 − U3 E2 − U3 R2 E1 + R1 E2
= + =⇒ U3 = R3
R3 R1 R2 R1 R2 + R2 R3 + R1 R3


3. lecture

Example

E1 − U3 '2 − U3
E I2
E
R1 - R2

+ I1
+
E1 R3 U3 E2 I = I1 + I2

− c −
?I
U3 E1 − U3 E2 − U3 R2 E1 + R1 E2
= + =⇒ U3 = R3
R3 R1 R2 R1 R2 + R2 R3 + R1 R3
U3 R2 E1 + R1 E2
I = =
R3 R1 R2 + R2 R3 + R1 R3


3. lecture

Some Remarks

There are many methods for writing the circuit equations, and there
is no such thing as ”right” method.
The only requirement is that you follow Kirchhoﬀ’s laws and Ohm’s
law1 and you have an equal number of equations and unknowns.
If there is a current source in the circuit, it will (usually) make the
circuit easier to solve, as you then have one unknown less to solve.
By using conductances instead of resistances, the equations look a
little cleaner.

1
Ohm’s law can only be utilized for resistors. If you have other elements, you
must know their current-voltage equation.

3. lecture

Another Example

-R1 - R2 - R5
+ I1

I2 I5
+
E1 R3 U3 R4 U4 E2 I1 = I2 + I3

− c c − I2 = I4 + I5
I
?3 I
?4

E1 − U3 U3 − U4 U3 U3 − U4 U4 U4 − E2
= + ja = +
R1 R2 R3 R2 R4 R5

G1 (E1 − U3 ) = G2 (U3 − U4 ) + G3 U3 ja G2 (U3 − U4 ) = G4 U4 + G5 (U4 − E2 )
Two equations, two unknowns → can be solved. Use conductances!


3. lecture

Some Remarks

There are many other methods available too: mesh analysis, modiﬁed
nodal analysis, branch current method . . .
If there are ideal voltage sources in the circuit (=voltage sources
which are connected to a node without a series resistance), you need
one more unknown (the current of the voltage source) and one more
equation (the voltage source will determine the voltage between the
nodes it is connected to).


3. lecture

Homework 3 (released 7th Sep, to be returned 10th Sep)

Homework 3a)
Find the current I4 .

Homework 3b)
Verify your solution by writing down all the voltages and currents to the
circuit diagram and checking that the solution does not contradict Ohm’s
and Kirchhoﬀ’s laws.

− +
6 R1 R4
R2 ER R5
J 3

I
?4
R1 = R2 = R3 = R4 = R5 = 1 Ω E = 9V J = 1A


4. lecture

Homework 3 - Model Solution

Homework 3a)
Find current I4 .

Homework 3b)
Verify your solution by writing down all the voltages and currents to the
circuit diagram and checking that the solution does not contradict Ohm’s
and Kirchhoﬀ’s laws.

− +
6 R1 R4
R2 ER R5
J 3

I
?4
R1 = R2 = R3 = R4 = R5 = 1 Ω E = 9V J = 1A


4. lecture

Solution
-
−
+I
6 R1 R4
J R2 U2ER3 U3 R5

c c
I
?4
R1 = R2 = R3 = R4 = R5 = 1 Ω E = 9V J = 1A

First we write two current equations and one voltage equation. The
conductance of the series circuit formed by R4 ja R5 is denoted with G45 .
J = U2 G2 + I
I = U3 G3 + U3 G45
U2 + E = U3
By substituting I from the second equation to the ﬁrst equation and then
substituting U2 from the third equation, we get
J = (U3 − E )G2 + U3 (G3 + G45 )

4. lecture

By substituting the component values, we get

U3 = 4 V

Therefore the current I is 4 V · 1 S = 4 A.

From the voltage equation U2 + E = U3 we can solve U2 = −5 V,
therefore the current through R2 is 5 A upwards.

The current I is therefore 1 A + 5 A = 6 A, of which 4 A goes through
R3 :n and the remaining 2 A goes through R4 ja R5 .

There is no contradiction with Kirchhoﬀ’s laws and therefore we can
be certain that our solution is correct.

4. lecture

Example 1
Find I and U.

− +
E3

+ I R1 +

?
6
E1 E2 R2 U J

− − c


4. lecture

Example 1
Find I and U.

I3 − +
E3

+ I R1 +

?
6
E1 E2 R2 U J

− − c

J = UG2 + I3
I3 = I + (E1 − E2 )G1
U = E1 + E 3


4. lecture

Example 2
Find U2 and I1 .

R
' 2
U

-

+
+ +
J1 J2
E1 E2 E3

− − −

I1


4. lecture

Example 2
Find U2 and I1 .

R
' 2
U

-

+
+ +
J1 J2
E1 E2 E3

− − −

I1

I1 = (E1 − E3 )G + J1
E2 + U2 = E3


4. lecture

How To Get Extra Exercise?

There are plenty of problems with solutions available at
http://users.tkk.fi/~ksilvone/Lisamateriaali/
lisamateriaali.htm
For example, you can find 175 DC circuit problems at http:
//users.tkk.fi/~ksilvone/Lisamateriaali/teht100.pdf
At the end of the pdf file you can find the model solutions, so you can
check your solution.
If you are enthusiastic, you can install and learn to use a circuit
simulator:
http://www.linear.com/designtools/software/ltspice.jsp


4. lecture

Example 3
Find U4 .

+ R1 R3
E R2 R4 U4

− c


4. lecture

Example 3
Find U4 .

+ R1 R3
E R2 U2 R4 U4

− c c

(E − U2 )G1 = U2 G2 + (U2 − U4 )G3
(U2 − U4 )G3 = G4 U4


4. lecture


Homework 4
Find the voltage U1 . All resistors have the value 10 Ω, E = 10 V ja
J = 1 A.

6 R2
+
J R1 U1 R3 E

c −

R4


5. lecture

Homework 4
Find the voltage U1 . All resistors have the value 10 Ω, E = 10 V ja
J = 1 A. U1 − U2
E
-
R2 I
+
6 U
2
J R1 U R3
1 E

U2 − U3−
©

c c

' R4
U3

J = U1 G1 + (U1 − U2 )G2
(U1 − U2 )G2 = (U2 − U3 )G3 + I
G3 (U2 − U3 ) + I = U3 G4
U2 − U3 = E


5. lecture

Solution

J = U1 G1 + (U1 − U2 )G2
(U1 − U2 )G2 = EG3 + I
G3 E + I = U3 G4
U2 − U3 = E

I is solved from the third equation and substituted into the second
equation, then U3 is solved from the equation and substituted.

J = U1 G1 + (U1 − U2 )G2
(U1 − U2 )G2 = EG3 + (U2 − E )G4 − G3 E

1 = 0,2U1 − 0,1U2
0,1U1 − 0,1U2 = 0,1U2 − 1


5. lecture

Solution

1 = 0,2U1 − 0,1U2
0,1U1 − 0,1U2 = 0,1U2 − 1

Which is solved

U1 = 10
U2 = 10

Therefore the voltage U1 is 10 Volts. This is easy to verify with a circuit
simulator.


5. lecture

Circuit Transformation

1 An operation which transforms a part of the circuit into an internally
diﬀerent, but externally equally acting circuit, is called a circuit
transformation.
2 For example, combining series resistors or parallel resistors into one
resistor, is a circuit transformation. Combining series voltage sources
into one voltage source is a circuit transformation too.
3 On this lecture, we learn dealing with parallel current sources and the
source transformation, with which we can transform a voltage source
with series resistance into a current source with parallel resistance.


5. lecture

An Example of a Circuit Transformation
Two (or more) resistors are combined to a single resistor, which acts just
like the original circuit of resistors.
Resistors in series
⇐⇒
R1 R2 R = R1 + R2

Resistors in parallel

R2
⇐⇒
1
R= 1
+R1
R1 2
R1

Or, by using conductance: G = G1 + G2 .


5. lecture

Current Sources in Parallel

One or more current sources are transformed into single current source,
which acts just like the original parallel circuit of current sources.
Current sources in parallel
˜ ˜

6 6
6
J1 J2 J3 ⇐⇒ J = J1 + J2 − J3

˜ ˜
?

Just like connecting two or more voltage sources in parallel, connecting
current sources in series is an undeﬁned (read: forbidden) operation in
circuit theory, just like divide by zero is undeﬁned in mathematics. There
can not be two currents in one wire!


5. lecture

The Source Transformation

A voltage source with series resistance acts just like current source
with parallel resistance.
The source transformation
˜ ˜

+ R
6
E ⇐⇒ J R E = RJ

−
˜ ˜


5. lecture

Important

Note that an ideal voltage or current source can not be transformed
like in previous slide. The voltage source to be transformed must have
series resistance and the current source must have parallel resistance.
The resistance remains the same, and the value for the source is
found from formula E = RJ, which is based on Ohm’s law.
The source transform is not just a curiosity. It can save from many
lines of manual calculations, for example when analyzing a transistor
ampliﬁer.


5. lecture

Rationale for the Source Transformation

The source transformation
- -

+ R I I
6
E
E U R U
R
− c c

In the ﬁgure left:
E −U
I = U = E − RI
R
In the ﬁgure right:
E U E −U E
I = − = U=( − I )R = E − RI
R R R R
Both the circuits function equally.


5. lecture

Example
Solve U.

+ R1 R2 +
E1 R3 U E

− c −

The circuit is transformed

6
6
J1 R1 R2 R3 J2

And we get the result:
J1 + J2
U=
G1 + G2 + G3


5. lecture

A Very Important Notice!

The value of the resistance remains the same, but the resistor is not
the same resistor! For instance, in the previous example the current
through the original resistor is not same as current through the
transformed resistor!


5. lecture


Homework 5
Find current I by using source transformation. J1 = 10 A, J2 = 1 A,
R1 = 100 Ω, R2 = 200 Ω ja R3 = 300 Ω.
-
I R2
6 6
J1 R1 R3 J2

This is easy and fast assignment. If you ﬁnd yourself writing many lines of
equations, you have done something wrong.


6. lecture

Homework 5
Find current I by using source transformation. J1 = 10 A, J2 = 1 A,
R1 = 100 Ω, R2 = 200 Ω ja R3 = 300 Ω.
-
I R2
6 6
J1 R1 R3 J2

-

+ R1 I R2 R3 +
R1 J1 R3 J2

− −

R1 J1 − R3 J2 1000 V − 300 V 7
I = = = A ≈ 1,17 A.
R1 + R2 + R3 600 Ω 6

6. lecture

Th´venin’s Theorem and Norton’s Theorem
e

So far we have learned the following circuit transformations: voltage
sources in series, current sources in parallel, resistances in parallel and
in series and the source transformation.
Th´venin’s theorem and Norton’s theorem relate to circuit
e
transformations too.
By Th´venin’s and Norton’s theorems an arbitrary circuit constisting
e
of voltage sources, current sources and resistances can be transformed
into a single voltage source with series resistance (Th´venin’s
e
equivalent) or a single current source with parallel resistance
(Norton’s equivalent).


6. lecture

Th´venin’s Theorem and Norton’s Theorem
e

Th´venin’s Theorem
e
An arbitrary linear circuit with two terminals is electrically equivalent to a
single voltage source and a single series resistor, called Th´venin’s
e
equivalent.

Norton’s Theorem
An arbitrary linear circuit with two terminals is electrically equivalent to a
single current source and a single parallel resistor, called Norton’s
equivalent.


6. lecture

Calculating the Th´venin Equivalent
e
˜ ˜

+ R1
+ RT
E R2 ⇐⇒ ET

− −
˜ ˜

The voltage ET in the Th´venin equivalent is solved simply by calculating
e
the voltage between the terminals. For solving RT , there are two ways:
By turning oﬀ all independent (= non-controlled) sources in the
circuit, and calculating the resistance between the terminals.
By calculating the short circuit current of the port and applying
Ohm’s law.
Independent source is a source, whose value does not depend on any other
voltage or current in the circuit. All sources we have dealt with for now,
have been independent. Controlled sources covered later in this course.

6. lecture

e

˜ ˜

+ R1
+ RT
E R2 ⇐⇒ ET

− −
˜ ˜

The voltage at the port is found by calculating the current through the
resistors and multiplying it with R2 . The voltage at the port, called also
the idle voltage of the port, is equal to ET .
E
ET = R2
R1 + R2


6. lecture

e
There are two ways to solve RT . Way 1: turn oﬀ all the (independent)
sources, and calculate the voltage at the port. A turned-oﬀ voltage source
is a voltage source, whose voltage is zero, which is same as just a wire:
˜ ˜
R1 RT
R2 ⇐⇒

˜ ˜

Now it is easy to solve the resistance between the nodes of the port: R1 ja
R2 are in parallel, and therefore the resistance is
1 R1 R2
RT = = .
G1 + G2 R1 + R2
This method is usually simpler than the other way with short circuit
current!

6. lecture

Solving RT by Using the Short-circuit Current
There are two ways to solve RT . Way 2: short-circuit the port, and
calculate the current through the short-circuit wire. This current is called
the short-circuit current:
˜ ˜

+ R1
+ RT
E R2 I
?K ⇐⇒ ET I
?K

− −
˜ ˜

The value for the short-circuit current is
E
IK =
R1
and the resistance RT is (by applying Ohm’s law to the ﬁgure on the right):
E
ET ET R1 +R2 R2 R1 R2
RT = = E = E
=
IK R R1
R1 + R2
1


6. lecture

Norton Equivalent

The Norton equivalent is simply a Th´venin equivalent, which has been
e
source transformed into a current source and parallel resistance (or vice
versa). The resistance has the same value in both equivalents. The value
of the current source is the same as the short-circuit current of the port.
˜

6
JN RN

˜


6. lecture

Example 1

Calculate the Th´venin equivalent.
e All component values = 1.
˜
− +
J1
6
R1 ER
2

˜


6. lecture

Example 2

Calculate the Th´venin equivalent. All component values = 1.
e
˜

+ R1 R3
6
E R2 J

−
˜


6. lecture


Homework 6
Calculate the Th´venin equivalent. All the component values are 1. (Every
e
resistance is 1 Ω and the current source is J1 = 1 A.)
˜

6 R2
J1 R1 R3

˜


7. lecture

Homework 6
Calculate the Th´venin equivalent. All the component values are 1. (Every
e
resistance is 1 Ω and the current source is J1 = 1 A.)
˜

6 R2
J1 R1 R3

˜

First we solve the voltage ET . This can be done by using applying source
transformation:
˜

+ R1 R2
J1 R1 1
J1 R1 R3 ET = R1 +R2 +R3 R3 = V
3
− c ˜

7. lecture

Solution
Next we solve the resistance RT of the Th´venin equivalent. The easiest
e
way to do it is to turn off all the sources and calculate the resistance
between the output port. (The other way is to find out the short-circuit
current.) The resistance can be solver either from the original or the
transformed circuit. Let’s use the transformed circuit and turn off the
voltage source:
˜
R1 R2 1 2
R3 RT = 1
+R1 = 3 Ω
R1 +R2 3

˜

Now the resistors R1 ja R2 are in series, and the series circuit is in parallel
with R3 . Now we know both ET ja RT and we can draw the Th´venin e
equivalent (on the next slide).


7. lecture

The Final Circuit

˜
+ RT = 2
3 Ω
1
ET = V
3
−
˜


7. lecture

Superposition Principle

A circuit consisting of resistances and constant-valued current and
voltage sources is linear.
If a circuit is linear, all the voltages and currents can be solved by
calculating the eﬀect of each source one at the time.
This principle is called the method of superposition.


7. lecture

Superposition Principle

The method of superposition is applied as follows
The current(s) and/or voltage(s) caused by each source is calculated
one at a time so that all other sources are turned off.
A turned-off voltage source = short circuit (a wire), a turned-off
current source = open circuit (no wire).
Finally, all results are summed together.


7. lecture

Superposition Principle: an Example
Find current I3 by using the superposition principle.

+ R1 R2 +
E1 R3 E2

− −
I
?3
First, we turn oﬀ the rightmost voltage source:

+ R1 R2 E1 1
E1 R3 I31 = G
R1 + G +G G2 +G3 3
1
2 3
−
I
?31
Next, we turn oﬀ the leftmost voltage source:

R1 R2 +
E2 1
I32 = G
R2 + G +G G1 +G3 3
1 R3 E2
1 3
−
I
?31

7. lecture

Superposition Principle: an Example

The current I3 is obtained by summing the partial currents I31 and I32 .
E1 1 E2 1
I3 = I31 + I32 = 1
G3 + 1
G3
R1 + G2 +G3 G2 + G3 R2 + G1 +G3 G1 + G3


7. lecture

When Is It Handy to Use the Superposition Principle?

If one doesn’t like solving equations but likes ﬁddling with the circuit.
If there are many sources and few resistors, the method of
superposition is usually fast.
If there are sources with diﬀerent frequencies (as we learn on the AC
Circuits course), the analysis of such a circuit is based on the
superposition principle.


7. lecture

Linearity and the Justification for the Superposition
Principle

The method of superposition is based on the linearity of the circuit,
which means that every source affects every voltage and current with
a constant factor.
This means that if there are sources E1 , E2 , E3 , J1 , J2 in the circuit,
then every voltage and current is of form
k1 E1 + k2 E2 + k3 E3 + k4 J1 + k5 J2 , where constants kn are real
numbers.
If all the sources are turned off (= 0), then all currents and voltages
in the circuit are zero. Therefore, by nullifying all sources except one,
we can find out the multiplier for the source in question.


7. lecture

Homework 7 (released 21st Sep, to be returned 24th Sep)

Homework 7

J1 = 1 A R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω E1 = 5 V
-

I2 R
+
6 2
J1 R1 R3 E1

−


8. lecture


Homework 7

J1 = 1 A R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω E1 = 5 V
-

I2 R
+
6 2
J1 R1 R3 E1

−


8. lecture

Ratkaisu

First, we ﬁnd the eﬀect of the current source:
-

I21 R
6 2
J1 R1 R3

The voltage over R1 and R2 is the same (they are in parallel) and R2 is
twice as large as R1 and therefore the current through R2 is half of the
current of R1 . Because the total current through the resistors is J1 = 1 A,
the current through R1 :n is 2/3 A and the current through R2
isI21 = 1/3 A.


8. lecture

Ratkaisu
Next, we ﬁnd out the eﬀect of the voltage source:
-
I22 R
+
2
R1 R3 E1

−

The resistors R1 and R2 are now in series and the total voltage over them
is E = 5 V, and therefore
E 5V 1
I22 = − =− = − V.
R1 + R2 10 Ω + 20 Ω 6
The minus sign comes from the fact that the direction of the current I22 is
upwards and the direction of the voltage E is downwards.
Finally, we sum the partial results:
1 1 1
I2 = I21 + I22 = A − A = A.
3 6 6

8. lecture

Voltage Divider
U1 E

R1
U R2 U2
c c

U1 = U R1R1 2 ja U2 = U R1R2 2
+R +R
It is quite common in electronic circuit design, that we need a
reference voltage formed from another voltage in the circuit.
The formula is valid also for multiple resistors in series. The
denominator is formed by summing all the resistances and the resistor
whose voltage is to be solved is in the numerator.


8. lecture

Current Divider

-
I I
?1 I
?2
R1 R2

I1 = I G1G1 2 ja I2 = I G1G2 2
+G +G
The formula is valid also for multiple resistors in parallel.
The formula for current divider is not used as frequently as the
voltage divider, but it is natural to discuss it in this concept.


8. lecture

Example 1

I
?1 I
?2 I
?3
6 R4
J R1 R2 R3


8. lecture

Example 1

I
?1 I
?2 I
?3
6 R4
J R1 R2 R3

G1
I1 = J G1 +G2 +G3


8. lecture

Example 1

I
?1 I
?2 I
?3
6 R4
J R1 R2 R3

G1 G
I1 = J G1 +G2 +G3 I2 = J G1 +G2 +G3
2


8. lecture

Example 1

I
?1 I
?2 I
?3
6 R4
J R1 R2 R3

G1 G G
I1 = J G1 +G2 +G3 I2 = J G1 +G2 +G3
2
I3 = J G1 +G3 +G3
2


8. lecture

Example 2

U1 U2 U3

‡ ‡ ‡

+ R1 R2 R3
E R4 U4

− W


8. lecture

Example 2

U1 U2 U3

‡ ‡ ‡

+ R1 R2 R3
E R4 U4

− W

U1 = E R1 +R2R1 3 +R4
+R


8. lecture

Example 2

U1 U2 U3

‡ ‡ ‡

+ R1 R2 R3
E R4 U4

− W

U1 = E R1 +R2R1 3 +R4
+R U2 = E R1 +R2R2 3 +R4
+R


8. lecture

Example 2

U1 U2 U3

‡ ‡ ‡

+ R1 R2 R3
E R4 U4

− W

U1 = E R1 +R2R1 3 +R4
+R U2 = E R1 +R2R2 3 +R4
+R U3 = E R1 +R2R3 3 +R4
+R


8. lecture

Example 2

U1 U2 U3

‡ ‡ ‡

+ R1 R2 R3
E R4 U4

− W

U1 = E R1 +R2R1 3 +R4
+R U2 = E R1 +R2R2 3 +R4
+R U3 = E R1 +R2R3 3 +R4
+R
U4 = E R1 +R2R4 3 +R4
+R


8. lecture


Homework 8
Find the voltage U by applying the voltage divider formula.

E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω
R4 = 40 Ω R5 = 50 Ω E2 = 15 V
U E

+ R1 R4 R5 +
E1 R2 R3 E2

− −


9. lecture

Homework 8
Find the voltage U by applying the voltage divider formula.

E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω
R4 = 40 Ω R5 = 50 Ω E2 = 15 V
U E

+ R1 R4 R5 +
E1 R2 U2 R3 U3 E2

− c c −

U2 = E1 R1R2 2 = 10 V 10 Ω+20 Ω = 6 2 V
+R
20 Ω
3
R
U3 = E2 R3 +R3 +R5 = 15 V 30 Ω+40 Ω
4
30
Ω+50 Ω = 3,75 V
11
U = U2 − U3 = 2 12 V ≈ 2,92 V.

9. lecture

Inductors and Capacitors

u u u

- §¤¤¤
§§ ‡
‡ ‡
- -
i R i L i
C

di
u = Ri u = L dt i = C du
dt


9. lecture

Inductors and Capacitors in DC Circuit

u u u

- ‡ - §¤¤¤
§§ ‡ - ‡
i R i L i
C

di
u = Ri u = L dt i = C du
dt

DC voltage and current remain constant as function of time or the time
derivatives of the voltage and current is zero. Therefore the voltage of an
inductor and the current of a capacitor is zero in a DC circuit.


9. lecture

Exception 1

The capacitor is fed with DC current so that the current has no other
route.

6
J C

i = C du ⇒ J = C du ⇒
dt dt
du
dt = J
C. The voltage of the capacitor rises at
constant speed.


9. lecture

Exception 2

A constant voltage source is connected to the terminals of an inductor.

+ ¤
¥
¤
E L ¤
¥
¥
−

di di di E
u = L dt ⇒ E = L dt ⇒ dt = L. The current of the inductor rises at
constant speed.


9. lecture

Dealing with all other cases involving inductors and
capacitors in DC circuits

Inductors are replaced with short circuits and capacitors are replaced with
open circuits.


9. lecture

Homework 9 (released 28th Sep, to be returned 1st Oct)

Homework 9
Solve the voltage U from this DC circuit.

E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω
R4 = 40 Ω L = 500 mH C = 2 F E2 = 15 V
§¤¤¤
§§

+ R1 L R4 +
C
E1 R2 R3 U E2

− W −


10. lecture


Homework 9
Solve the voltage U from this DC circuit.

E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω
R4 = 40 Ω L = 500 mH C = 2 F E2 = 15 V
§¤¤¤
§§

+ R1 L R4 +
C
E1 R2 R3 U E2

− W −


10. lecture


Because there are no parallel connections of inductors and voltage sources
and no serial connections of capacitors and current sources and the circuit
is a DC circuit (= constant voltages and currents), we can replace the
inductors with short circuits and the capacitors with open circuits.

+ R1 R4 +
E1 R2 R3 U E2

− W −

in which case we obtain U easily by applying the voltage divider formula:
R3 3
U = E2 = 6 V ≈ 6,4 V
R3 + R4 7


10. lecture

Controlled Sources

So far, all of our sources have been constant valued.
If the value of a source does not depend on any of the voltages or
currents in the circuit, the source is an independent source. For
example, constant valued sources and sources varying as function of
time (only) are independent sources.
If the value of a source is a function of a voltage and/or current in
the circuit, the source is a controlled source.


10. lecture

Voltage Controlled Voltage Source (VCVS)

r

+
u e = Au

−
r
c

The voltage e of VCVS is dependent of some voltage u.
The multiplier A is called voltage gain.
A real-world example: an audio ampliﬁer.


10. lecture

Current Controlled Voltage Source (CCVS)

r

+
i
? e = ri

−
r

The voltage e of CCVS is dependent of some current i.
The multiplier r is called transresistance.
There is no good everyday example of this source available (of course
we can construct this kind of source by using an operational
ampliﬁer).


10. lecture

Voltage Controlled Current Source (VCCS)

r

6
u j = gu

r
c

The current j of VCCS is dependent of some voltage u.
The multiplier g is called transconductance.
A real-world example: a ﬁeld-eﬀect transistor (JFET or MOSFET).


10. lecture

Current Controlled Current Source (CCCS)

r

6
i
? j = βi

r

The current j of CCCS is dependent of some current i.
The multiplier β is called current gain.
A real-world example: a (bipolar junction) transistor.


10. lecture

Homework 10 (released 1st Oct, to be returned 5th Oct)

Homework 10
Find the voltage U.

E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω
R4 = 40 Ω r = 2Ω

-

+ R1 i R2 R4 +
E1 R3 U e2 = ri

− W −

Note that the source on the right is a controlled source.


11. lecture


Homework 10
Find the voltage U.

E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω
R4 = 40 Ω r = 2Ω

-

+ R1 i R2 R4 +
E1 R3 U e2 = ri

− W −

Note that the source on the right is a controlled source.


11. lecture

-

+ R1 i R2 R4 +
E1 R3 U e2 = ri

− W −

Let’s denote the total resistance of R1 :n ja R2 with symbol R12 and write a
nodal equation:
UG3 = (E1 − U)G12 + (ri − U)G4
There are two unknowns in the circuit and therefore we need another
equation with the same unknowns:

i = (E1 − U)G12

Then we substitute i to the ﬁrst equation:

E1 G12 − UG12 + rG4 G12 E1 − rG4 G12 U − UG4 = UG3


11. lecture

E1 G12 − UG12 + rG4 G12 E1 − rG4 G12 U − UG4 = UG3
from which we get

G12 E1 (1 + rG4 ) = U(G3 + G12 + G4 + rG4 G12 ).

Then we substitute the component values and solve U:
10 2
30 (1 + 40 )
U= 1 1 1 2
= 3,75 V
30 + 30 + 40 + 40·30


11. lecture

Recapitulation

On this lesson, we solve some refresher assignments. If you have solved all
the circuits, solve the home assignment.


11. lecture

Recap assignment 1

Recap assignment 1
Find U and I ﬁrst by using the method of superposition and then by some
other method of your choice.

R1 = 1 Ω R2 = 2 Ω J = 1A E = 3V
-

+ I R2
E R1 J U

− ?c

Vastaus: I = 4 A ja U = 1 V.


11. lecture

Recap assignment 2
Recap assignment 2
Form a Th´venin equivalent of the circuit on the left. Then, compute the
e
current IX , when the switches are closed and RX is a) 0 Ω, b) 8 Ω ja c)
12 Ω.

R1 = 5 Ω R2 = 3 Ω R3 = 8 Ω R4 = 4 Ω E = 16 V
˜
¨ ˜
¨
R2 R4
R3

RX

+
R1 E

− I
?X
˜
¨ ˜
¨

: Vastaus: RT = 8 Ω, ET = 8 V. a) 1 A b) 0,5 A c)0,4 A.

11. lecture

Recap assignment 3
Recap assignment 3
Find U3 .

G1 = 1 S G2 = 2 S G3 = 3 S G4 = 4 S G5 = 5 S g = 6 S J = 3 A

gU1

r

6 G4 G5
J U1 G1 G2 G3 U3

c c

48
U3 = − 115 V ≈ −417 mV


11. lecture

Homework 11 (released 5th Oct, to be returned 8th Oct)

Homework 11
We are given a fact that the current I3 = 0 A. Find E1 .

R1 = 5 Ω R2 = 4 Ω R3 = 2 Ω R4 = 5 Ω R5 = 6 Ω E2 = 30 V
-

+ R1 R3 I3 R2 +
E1 R4 R5 E2

− −


12. lecture


Homework 11

R1 = 5 Ω R2 = 4 Ω R3 = 2 Ω R4 = 5 Ω R5 = 6 Ω E2 = 30 V
-

+ R1 R3 I3 R2 +
E1 R4 R5 E2

− −


12. lecture


Homework 11

R1 = 5 Ω R2 = 4 Ω R3 = 2 Ω R4 = 5 Ω R5 = 6 Ω E2 = 30 V
-

+ R1 R3 I3 R2 +
E1 R4 U4 R5 U5 E2

− c c −


12. lecture

Because I3 = 0 A, the current through R1 equals the current through R4
and the current through R2 equals the current through R5 . Therefore, the
resistors are in series2 and we may use the voltage divider formula to ﬁnd
voltages over R4 and R5 . The voltage over R5 is U5 = E2 R2R5 5 = 18 V.
+R
Therefore the voltage over R4 is 18 V too. Now, by the voltage divider
rule:
R4 5Ω
U4 = E1 ⇒ 18 V = E1
R1 + R4 5Ω + 5Ω
from which we can solve E1 = 36 V. Note: it is completely correct to write
nodal equations for the circuit and solve E1 from them, too.

2
Because and only because we know that I3 is zero.

12. lecture

Recapitulation

On this lesson, we solve some refresher assignments. I can also
demonstrate some examples on the blackboard or to your booklets, too.


12. lecture

Recap assignment 4

R2 = 5 Ω E1 = 3 V E2 = 2 V

-
I1 I2 +
R1 E2

+ −
E1

−
R2

a) How should we choose R1 , if we want I2 to be 0 A?
b) How large is I1 then?
a) 10 Ω ja b) 0,2 A.


12. lecture

Recap assignment 5

R1 = 100 Ω R2 = 500 Ω R3 = 1,5 kΩ R4 = 1 kΩ E1 = 5 V

J1 = 100 mA J2 = 150 mA
-

J2

r

6 R2 +R3
J1 R1 E1 R4 U4

− c

Find U4 .
U4 = 92 V


12. lecture

Recap assignment 6

R1 = 12 Ω R2 = 25 Ω J = 1A E1 = 1 V E2 = 27 V

−
+ R1 +
6 E2
E1 J U R2

− c

Find voltage U.
1
Solution: 37 V ≈ 27 mV


12. lecture

Homework 12 (released 8th Oct, to be returned 12th Oct)

Write a short essay on following subjects:
What did you learn on the course?
Did the course suck or was it worthwhile?
What could the lecturer do better?
How should this course be improved?
The essay will not aﬀect the grading of the exam — please give honest
feedback3 .
How to return this homework: Write the essay as a plain text email (no
attachments) and send it to me no later than the exam day at 18:00. The
subject of the email message must be ’DC Circuits course feedback
2009 Firstname Surname’.

3
I am really interested in how I could make the course better.

12. lecture

Final Notices on these Slides

The slides are licensed with CC By 1.0 4 . In short: you can use and
modify the slides freely as long as you mention my name (= Vesa
Linja-aho) somewhere.
Single examples and circuits can be of course used without any name
mentioning, because they are not an object of copyright (legal term:
”Threshold of originality”).
The origin of these slides is the DC Circuits course in Metropolia
polytechnic in Helsinki, Finland.
If you ﬁnd typos, misspellings or errors in facts, please give me
feedback.

4
http://creativecommons.org/licenses/by/1.0/

Circuit Analysis – DC Circuits

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