ET 162 Circuit Analysis
Methods of Analysis
Electrical and Telecommunication
Engineering Technology
Professor Jang
Acknowledgement
I want to express my gratitude to Prentice Hall giving me the permission
to use instructor’s material for developing this module. I would like to
thank the Department of Electrical and Telecommunications Engineering
Technology of NYCCT for giving me support to commence and complete
this module. I hope this module is helpful to enhance our students’
academic performance.
OUTLINES
 Introduction to Network Theorems
 Superposition
 Thevenin’s Theorem
 Norton’s Theorem
 Maximum Power Transfer Theorem
Key Words: Network Theorem, Superposition, Thevenin, Norton, Maximum Power
ET162 Circuit Analysis – Network Theorems
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Introduction to Network Theorems
This chapter will introduce the important fundamental
theorems of network analysis. Included are the superposition,
Thevenin’s, Norton’s, and maximum power transfer theorems.
We will consider a number of areas of application for each. A
through understanding of each theorems will be applied
repeatedly in the material to follow.
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Superposition Theorem
The superposition theorem can be used to find the solution to networks
with two or more sources that are not in series or parallel. The most
advantage of this method is that it does not require the use of a
mathematical technique such as determinants to find the required
voltages or currents.
The current through, or voltage across, an element in a linear bilateral
network is equal to the algebraic sum of the current or voltages
produced independently by each source.
Number of networks
to be analyzed
ET162 Circuit Analysis – Network Theorems
=
Number of
independent sources
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Figure 9.1 reviews the various
substitutions required when
removing an ideal source, and
Figure 9.2 reviews the
substitutions with practical
sources that have an internal
resistance.
FIGURE 9.1 Removing the effects of practical sources
FIGURE 9.2 Removing the effects of ideal sources
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Ex. 9-1 Determine I1 for the network of Fig. 9.3.
I 1'  0 A
FIGURE 9.3
E 30V
I 

 5A
R1
6
''
1
I 1  I 1'  I 1''
 0A5A
 5A
ET162 Circuit Analysis – Network Theorems
FIGURE 9.4
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Ex. 9-2 Using superposition, determine the current through the 4-Ω resistor of
Fig. 9.5. Note that this is a two-source network of the type considered in chapter 8.
RT  R1  R2 / / R3
 24   12  / / 4 
 24   3   27 
E1 54V
I

2A
RT 27 
12 A2 A

R2 I
I 

 15
. A
R2  R3 12   4 
'
3
FIGURE 9.5
FIGURE 9.6
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RT  R3  R1 / / R2  4   24  / / 12   4   8   12 
FIGURE 9.7
E 2 48V
I 

4A
RT 12 
''
3
I 3  I 3''  I 3'  4 A  15
. A  2.5 A
(direction of I 3'' )
FIGURE 9.8
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Ex. 9-3 a. Using superposition, find the current through the 6-Ω resistor of Fig.
9.9. b. Determine that superposition is not applicable to power levels.
FIGURE 9.9
FIGURE 9.10
a. considering that the effect of the 36V source ( Fig. 9.10):
I 2' 
36V
E
E


2A
RT R1  R2 12   6 
considering that the effect of the 9 A source ( Fig. 9.11):
I 2'' 
ET162 Circuit Analysis – Network Theorems
FIGURE 9.11
R1 I
(12 )(9 A)

6A
R1  R2 12   6 
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The total current through the
6  resistor ( Fig 9.12) is
I 2  I 2'  I 2''  2 A  6 A  8 A
b.
The power to the 6  resistor is
Power  I 2 R  (8 A) 2 (6 )  384 W
FIGURE 9.12
The calculated power to the 6  resistor
due to each source, misu sin g the principle
of sup erposition, is
P1  ( I 2' ) 2 R  (2 A) 2 (6 )  24 W
P2  ( I 2'' ) 2 R  (6 A) 2 (6 )  216 W
P1  P2  240 W  384 W
because (2 A)  (6 A) 2  (8 A) 2
2
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Ex. 9-4 Using the principle of superposition, find the current through the 12-kΩ
resistor of Fig. 9.13.
R1 I
(6 k)(6 mA)
I 

 2 mA
R1  R2 6 k  12 k
'
2
FIGURE 9.13
FIGURE 9.14
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FIGURE 9.15
considering that the effect of the 9V voltage source ( Fig. 9.15):
9V
E
I 

 0.5 mA
R1  R2 6 k  12 k
''
2
Since I 2' and I 2'' have the same direction through R2 ,
the desired current is the sum of the two:
I 2  I 2'  I 2''  2 mA  0.5 mA  2.5 mA
12
Ex. 9-5 Find the current through the 2-Ω resistor of the network of Fig. 9.16. The
presence of three sources will result in three different networks to be analyzed.
FIGURE 9.16
FIGURE
ET162 Circuit Analysis – Network Theorems
FIGURE 9.17
9.18
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FIGURE 9.19
13
considering the effect of the 12 V source ( Fig. 9.17):
I 1' 
E1
12 V

2A
R1  R2
2  4
considering that the effect of the 6V source ( Fig. 9.18):
I 1'' 
E2
6V

 1A
R1  R2
2  4
considering the effect of the 3 A source ( Fig. 9.19):
I 1''' 
R2 I
(4 )(3 A)

2A
R1  R2
2  4
The total current through the 2  resistor
FIGURE 9.20
appears in Fig.9.20, and
''
'''
'
IET162

I

I

I
AAnalysis
 2 A 2 A  1A
Circuit
Analysis
–
Methods
1
1
1
1  1of
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Thevenin’s Theorem
Any two-terminal, linear bilateral
dc network can be replaced by an
equivalent circuit consisting of a
voltage source and a series resistor,
as shown in Fig. 9.21.
FIGURE 9.21 Thevenin equivalent circuit
FIGURE 9.22 The effect of
applying Thevenin’s theorem.
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FIGURE 9.23 Substituting the Thevenin equivalent circuit for a complex network.
1.
Remove that portion of the network across which the Thevenin equivalent
circuit is to be found. In Fig. 9.23(a), this requires that the road resistor RL be
temporary removed from the network.
2.
Make the terminals of the remaining two-terminal network.
3.
Calculate RTH by first setting all sources to zero (voltage sources are replaced
by short circuits, and current sources by open circuit) and then finding the
resultant resistance between the two marked terminals.
4.
Calculate ETH by first returning all sources to their original position and
finding the open-circuit voltage between the marked terminals.
5.
Draw the Thevenin equivalent circuit with the portion of the circuit previously
removed
replaced
the equivalent circuit.
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Circuit Analysis
– Methodsbetween
of Analysis the terminals ofBoylestad
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Ex. 9-6 Find the Thevenin equivalent circuit for the network in the shaded area of
the network of Fig. 9.24. Then find the current through RL for values of 2Ω, 10Ω,
and 100Ω.
FIGURE 9.24
FIGURE 9.25 Identifying the terminals of particular
importance when applying Thevenin’s theorem.
RTH  R1 / / R2
(3 )(6 )

3   6
 2
FIGURE 9.26 Determining RTH for the network of Fig. 9.25.
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FIGURE 9.28
FIGURE 9.27
RTH 
IL 
FIGURE 9.29 Substituting the Thevenin equivalent
circuit for the network external to RL in Fig. 9.23.
R2 E1
(6 )(9V )

 6V
R2  R1
6   3
E TH
RTH  R L
6V
 15
. A
2  2
R L  2 :
IL 
R L  10 :
6V
IL 
 0.5 A
2   10 
R L  100 : I L 
6V
 0.059 A
2   100 
18
Ex. 9-7 Find the Thevenin equivalent circuit for the network in the shaded area of
the network of Fig. 9.30.
FIGURE 9.30
FIGURE 9.31
RTH  R1  R2
 4  2
FIGURE 9.32
ET162 Circuit Analysis – Network Theorems
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 6
19
V2  I 2 R2  (0) R2  0V
E TH  V1  I 1 R1  I R1
 (12 A)(4 )
 48 
FIGURE 9.33
FIGURE 9.34 Substituting the Thevenin
equivalent circuit in the network external
to the resistor R3 of Fig. 9.30.
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Ex. 9-8 Find the Thevenin equivalent circuit for the network in the shaded area of
the network of Fig. 9.35. Note in this example that there is no need for the section of
the network to be at preserved to be at the “end” of the configuration.
FIGURE 9.36
FIGURE 9.35
RTH  R1 / / R2
(6 )(4 )

6  4 
 2.4 
ET162 Circuit Analysis – Methods of Analysis
FIGURE 9.37
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FIGURE 9.38
FIGURE 9.39
E TH
R1 E1

R1  R2

(6 )(8V )
6  4 
 4.8V
FIGURE 9.40 Substituting the Thevenin
equivalent circuit in the network external
to the resistor R4 of Fig. 9.35.
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Ex. 9-9 Find the Thevenin equivalent circuit for the network in the shaded area of
the network of Fig. 9.41.
FIGURE 9.42
FIGURE 9.41
RTH  R1 / / R3  R2 / / R4
 6  / / 3   4  / / 12 
 2   3  5
9.43
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Methods of Analysis
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FIGURE 9.45
FIGURE 9.44
V1 
R1 E
(6 )(72V )

 48V
R1  R3
6   3
R2 E
(12 )(72V )
V2 

 54V
R2  R4
12   4 
V   E
TH
 V1  V2  0
E TH  V2  V1  54V  48V  6V
FIGURE 9.46 Substituting the Thevenin equivalent circuit
in the network external to the resistor RL of Fig. 9.41.
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Ex. 9-10 (Two sources) Find the Thevenin equivalent circuit for the network
within the shaded area of Fig. 9.47.
FIGURE 9.48
FIGURE 9.47
RTH  R4  R1 / / R2 / / R3
 14
. k  0.8 k / / 4 k / / 6 k
 14
. k  0.8 k / / 2.4 k
 14
. k  0.6 k  2 k
FIGURE
ET162 Circuit Analysis
– Methods 9.49
of Analysis
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V4  I 4 R4  (0) R4  0V
'
E TH
 V3
RT'  R2 / / R3  4 k / / 6 k  2.4 k
FIGURE 9.50
RT' E1
(2.4 k)(6V )
V3  '

 4.5V
RT  R1 2.4 k  0.8 k
'
E TH
 V3  4.5V
V4  I 4 R4  (0) R4  0V
FIGURE 9.51
''
E TH
 V3
RT'  R1 / / R3  0.8 k / / 6 k  0.706 k
RT' E 2
(0.706 k)(10V )
V3  '

 15
.V
0.706 k  4 k
RT  R2
'
''
''
E TH
 V3  15
. V E TH  E TH  E TH
 4.5V  15
.V
FIGURE 9.52 Substituting the Thevenin equivalent circuit
Circuit
Analysis to
– Methods
of Analysis
in theET162
network
external
the resistor
RL of Fig. 9.47.
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'
 3V ( polarity of E TH
26 )
Experimental Procedures
FIGURE 9.53
FIGURE 9.54
FIGURE 9.55
ET162 Circuit Analysis – Network Theorems
I SC 
E TH
RTH
RTH 
E TH
I SC
RTH 
VOC
I SC
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27
Norton’s Theorem
Any two-terminal, linear bilateral dc network can be replaced by an
equivalent circuit consisting of a current source and a parallel
resistor, as shown in Fig. 9.56.
FIGURE 9.56 Norton equivalent circuit
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1.
Remove that portion of the network across which the Thevenin equivalent
circuit is found.
2.
Make the terminals of the remaining two-terminal network.
3.
Calculate RN by first setting all sources to zero (voltage sources are replaced by
short circuits, and current sources by open circuit) and then finding the
resultant resistance between the two marked terminals.
4.
Calculate IN by first returning all sources to their original position and finding
the short-circuit current between the marked terminals.
5.
Draw the Norton equivalent circuit with the portion of the circuit previously
removed replaced between the terminals of the equivalent circuit.
FIGURE 9.57 Substituting the Norton equivalent circuit for a complex network.
29
Ex. 9-11 Find the Norton equivalent circuit for the network in the shaded area of
Fig. 9.58.
FIGURE 9.58
FIGURE 9.59 Identifying the terminals of particular
interest for the network of Fig. 9.58.
R N  R1 / / R2
 3  / / 6
(3 )(6 )

3   6
 2
FIGURE 9.60 Determining RN for the network of Fig. 9.59.
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V2  I 2 R2  (0) 6   0 V
IN
E 9V


3 A
R1 3 
FIGURE 9.61 Determining RN for the network of Fig. 9.59.
FIGURE 9.62 Substituting the Norton
equivalent circuit for the network
external to the resistor RL of Fig. 9.58.
FIGURE 9.63 Converting the Norton equivalent circuit of
Fig. 9.62 to a Thevenin’s equivalent circuit.
Ex. 9-12 Find the Norton equivalent circuit for the network external to the 9-Ω
resistor in Fig. 9.64.
FIGURE 9.64
FIGURE 9.65
RN = R 1 + R 2
=5Ω+4Ω
=9Ω
FIGURE 9.66
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IN
R1 I

R1  R2
(5 )(10 )

5  4 
 5556
.
A
FIGURE 9.67 Determining IN for the network of Fig. 9.65.
ET162 Circuit Analysis – Network Theorems
FIGURE 9.68 Substituting the Norton
equivalent circuit for the network
external to the resistor RL of Fig. 9.64.
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Ex. 9-13 (Two sources) Find the Norton equivalent circuit for the portion of the
network external to the left of a-b in Fig. 9.69.
FIGURE 9.70
FIGURE 9.69
R N  R1 / / R2
 4  / / 6
(4 )(6 )

4   6
 2.4 
FIGURE 9.71
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FIGURE 9.72
FIGURE 9.73
I
'
N
E1 7 V


 175
. A
R1 4 
I N''  I  8 A
I N  I N''  I N'
 8 A  175
. A  6.25 A
FIGURE 9.74 Substituting the Norton equivalent circuit
for the network to the left of terminals a-b in Fig. 9.69.
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Maximum Power Transfer Theorem
A load will receive maximum power from a linear bilateral dc network
when its total resistive value is exactly equal to the Thevenin resistance
of the network as “seen” by the load.
RL = RTH
RL = RN
FIGURE 9.74 Defining the conditions for maximum FIGURE 9.75 Defining the conditions for maximum
power to a load using the Thevenin equivalent circuit. power to a load using the Norton equivalent circuit.
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