Detailed Notes: Network Theorems in A.C. circuits - 1 | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

 Page 1


Network Theorems (ac)
18.1 INTRODUCTION
Due to the need for developing confidence in the application of the various theorems to net-
works with controlled (dependent) sources, some sections have been divided into two parts:
independent sources and dependent sources.
Theorems to be considered in detail include the superposition theorem, Thévenin’s and
Norton’s theorems, and the maximum power transfer theorem. The substitution and reciproc-
ity theorems and Millman’s theorem are not discussed in detail her                      . e
18.2 SUPERPOSITION THEOREM
You will recall from Chapter 9 that the superposition theorem eliminated the need for solv-
ing simultaneous linear equations by considering the effects of each source independently. To
consider the effects of each source, we had to remove the remaining sources. This was ac-
complished by setting voltage sources to zero (short-circuit representation) and current sources
to zero (open-circuit representation). The current through, or voltage across, a portion of the
network produced by each source was then added algebraically to find the total solution for
the current or voltage.
The only variation in applying this method to ac networks with independent sources is that
we are now working with impedances and phasors instead of just resistors and real numbers.
The superposition theorem is not applicable to power effects in ac networks since we are
still dealing with a nonlinear relationship. It can be applied to networks with sources of dif-
ferent frequencies only if the total response for each frequency is found independently and the
results are expanded in a nonsinusoidal expression, as appearing in Chapter 25.
One of the most frequent applications of the superposition theorem is to electronic systems
in which the dc and ac analyses are treated separately and the total solution is the sum of the
two. It is an important application of the theorem because the impact of the reactive elements
• Be able to apply the superposition theorem to ac
networks with independent and dependent
sources.
• Become proficient in applying Thévenin’s theorem
to ac networks with independent and dependent
sources.
• Be able to apply Norton’s theorem to ac networks
with independent and dependent sources.
• Clearly understand the conditions that must be
met for maximum power transfer to a load in an
ac network with independent or dependent
sources.
Objectives
Network Theorems (ac)
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 787
Page 2


Network Theorems (ac)
18.1 INTRODUCTION
Due to the need for developing confidence in the application of the various theorems to net-
works with controlled (dependent) sources, some sections have been divided into two parts:
independent sources and dependent sources.
Theorems to be considered in detail include the superposition theorem, Thévenin’s and
Norton’s theorems, and the maximum power transfer theorem. The substitution and reciproc-
ity theorems and Millman’s theorem are not discussed in detail her                      . e
18.2 SUPERPOSITION THEOREM
You will recall from Chapter 9 that the superposition theorem eliminated the need for solv-
ing simultaneous linear equations by considering the effects of each source independently. To
consider the effects of each source, we had to remove the remaining sources. This was ac-
complished by setting voltage sources to zero (short-circuit representation) and current sources
to zero (open-circuit representation). The current through, or voltage across, a portion of the
network produced by each source was then added algebraically to find the total solution for
the current or voltage.
The only variation in applying this method to ac networks with independent sources is that
we are now working with impedances and phasors instead of just resistors and real numbers.
The superposition theorem is not applicable to power effects in ac networks since we are
still dealing with a nonlinear relationship. It can be applied to networks with sources of dif-
ferent frequencies only if the total response for each frequency is found independently and the
results are expanded in a nonsinusoidal expression, as appearing in Chapter 25.
One of the most frequent applications of the superposition theorem is to electronic systems
in which the dc and ac analyses are treated separately and the total solution is the sum of the
two. It is an important application of the theorem because the impact of the reactive elements
• Be able to apply the superposition theorem to ac
networks with independent and dependent
sources.
• Become proficient in applying Thévenin’s theorem
to ac networks with independent and dependent
sources.
• Be able to apply Norton’s theorem to ac networks
with independent and dependent sources.
• Clearly understand the conditions that must be
met for maximum power transfer to a load in an
ac network with independent or dependent
sources.
Objectives
Network Theorems (ac)
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 787
X
L
2
4 
–
+
X
C
3 
I
E
2  
=  5 V ? 0° E
1  
=  10 V ? 0°
–
+
X
L
1
4 
FIG. 18.1
Example 18.1.
changes dramatically in response to the two types of independent
sources. In addition, the dc analysis of an electronic system can often de-
fine important parameters for the ac analysis. Example 18.4 demon-
strates the impact of the applied source on the general configuration of
the network.
We first consider networks with only independent sources to provide
a close association with the analysis of Chapter 9.
Independent Sources
EXAMPLE 18.1 Using the superposition theorem, find the current I
through the 4  reactance in Fig. 18.1. 1X
L
2
2
–
+
I
E
2
E
1
–
+
Z
1
Z
2
Z
3
FIG. 18.2
Assigning the subscripted impedances to the network
in Fig. 18.1.
I
E
1
–
+
Z
1
Z
2
Z
3
E
1
–
+
Z
1
Z
23
I
s
1
I
s
1
FIG. 18.3
Determining the effect of the voltage source E
1
on the current I of the network in
Fig. 18.1.
Solution: For the redrawn circuit (Fig. 18.2),
Considering the effects of the voltage source E
1
(Fig. 18.3), we have
  1.25 A 90°
  I
s
1

E
1
Z
2  3
 Z
1

10 V 0°
j 12  j 4 

10 V 0°
8  90°
  12  90°
 Z
2  3

Z
2
Z
3
Z
2
 Z
3

1j 4 21j 3 2
j 4  j 3 

12 
j
j 12 
 Z
3
j X
C
j 3 
 Z
2
j X
L
2
 j 4 
 Z
1
j X
L
1
 j 4 
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 788
Page 3


Network Theorems (ac)
18.1 INTRODUCTION
Due to the need for developing confidence in the application of the various theorems to net-
works with controlled (dependent) sources, some sections have been divided into two parts:
independent sources and dependent sources.
Theorems to be considered in detail include the superposition theorem, Thévenin’s and
Norton’s theorems, and the maximum power transfer theorem. The substitution and reciproc-
ity theorems and Millman’s theorem are not discussed in detail her                      . e
18.2 SUPERPOSITION THEOREM
You will recall from Chapter 9 that the superposition theorem eliminated the need for solv-
ing simultaneous linear equations by considering the effects of each source independently. To
consider the effects of each source, we had to remove the remaining sources. This was ac-
complished by setting voltage sources to zero (short-circuit representation) and current sources
to zero (open-circuit representation). The current through, or voltage across, a portion of the
network produced by each source was then added algebraically to find the total solution for
the current or voltage.
The only variation in applying this method to ac networks with independent sources is that
we are now working with impedances and phasors instead of just resistors and real numbers.
The superposition theorem is not applicable to power effects in ac networks since we are
still dealing with a nonlinear relationship. It can be applied to networks with sources of dif-
ferent frequencies only if the total response for each frequency is found independently and the
results are expanded in a nonsinusoidal expression, as appearing in Chapter 25.
One of the most frequent applications of the superposition theorem is to electronic systems
in which the dc and ac analyses are treated separately and the total solution is the sum of the
two. It is an important application of the theorem because the impact of the reactive elements
• Be able to apply the superposition theorem to ac
networks with independent and dependent
sources.
• Become proficient in applying Thévenin’s theorem
to ac networks with independent and dependent
sources.
• Be able to apply Norton’s theorem to ac networks
with independent and dependent sources.
• Clearly understand the conditions that must be
met for maximum power transfer to a load in an
ac network with independent or dependent
sources.
Objectives
Network Theorems (ac)
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 787
X
L
2
4 
–
+
X
C
3 
I
E
2  
=  5 V ? 0° E
1  
=  10 V ? 0°
–
+
X
L
1
4 
FIG. 18.1
Example 18.1.
changes dramatically in response to the two types of independent
sources. In addition, the dc analysis of an electronic system can often de-
fine important parameters for the ac analysis. Example 18.4 demon-
strates the impact of the applied source on the general configuration of
the network.
We first consider networks with only independent sources to provide
a close association with the analysis of Chapter 9.
Independent Sources
EXAMPLE 18.1 Using the superposition theorem, find the current I
through the 4  reactance in Fig. 18.1. 1X
L
2
2
–
+
I
E
2
E
1
–
+
Z
1
Z
2
Z
3
FIG. 18.2
Assigning the subscripted impedances to the network
in Fig. 18.1.
I
E
1
–
+
Z
1
Z
2
Z
3
E
1
–
+
Z
1
Z
23
I
s
1
I
s
1
FIG. 18.3
Determining the effect of the voltage source E
1
on the current I of the network in
Fig. 18.1.
Solution: For the redrawn circuit (Fig. 18.2),
Considering the effects of the voltage source E
1
(Fig. 18.3), we have
  1.25 A 90°
  I
s
1

E
1
Z
2  3
 Z
1

10 V 0°
j 12  j 4 

10 V 0°
8  90°
  12  90°
 Z
2  3

Z
2
Z
3
Z
2
 Z
3

1j 4 21j 3 2
j 4  j 3 

12 
j
j 12 
 Z
3
j X
C
j 3 
 Z
2
j X
L
2
 j 4 
 Z
1
j X
L
1
 j 4 
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 788
and
(current divider rule)
Considering the effects of the voltage source E
2
(Fig. 18.4), we have
 
1j 3 21 j 1.25 A2
j 4  j 3 

3.75 A
j 1
 3.75 A 90°
 I 
Z
3
I
s
1
Z
2
 Z
3
4 
I
I'
I?
X
L
2
FIG. 18.5
Determining the resultant current for the network in
Fig. 18.1.
I
E
2
–
+
Z
1
Z
2
Z
3
E
2
–
+
Z
3
Z
12
I
s
2
I
s
2
FIG. 18.4
Determining the effect of the voltage source E
2
on the current I of the network
in Fig. 18.1.
and
The resultant current through the 4 reactance (Fig. 18.5) is
I  I  I
 3.75 A 90°  2.50 A 90°j 3.75 A  j 2.50 A
j 6.25 A
I  6.25 A 90°
EXAMPLE 18.2 Using superposition, find the current I through the 
6  resistor in Fig. 18.6.
X
L
2
I
I
s
2
2
 2.5 A 90°
 I
s
2

E
2
Z
1  2
 Z
3

5 V 0°
j 2  j 3 

5 V 0°
1  90°
 5 A 90°
 Z
1  2

Z
1
N

j 4 
2
 j 2 
X
C  
=  8 
I
E
1  
=  20 V ? 30°
–
+
R
  
=  6  X
L  
=  6 
I
1
2 A ? 0°
FIG. 18.6
Example 18.2.
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 789
Page 4


Network Theorems (ac)
18.1 INTRODUCTION
Due to the need for developing confidence in the application of the various theorems to net-
works with controlled (dependent) sources, some sections have been divided into two parts:
independent sources and dependent sources.
Theorems to be considered in detail include the superposition theorem, Thévenin’s and
Norton’s theorems, and the maximum power transfer theorem. The substitution and reciproc-
ity theorems and Millman’s theorem are not discussed in detail her                      . e
18.2 SUPERPOSITION THEOREM
You will recall from Chapter 9 that the superposition theorem eliminated the need for solv-
ing simultaneous linear equations by considering the effects of each source independently. To
consider the effects of each source, we had to remove the remaining sources. This was ac-
complished by setting voltage sources to zero (short-circuit representation) and current sources
to zero (open-circuit representation). The current through, or voltage across, a portion of the
network produced by each source was then added algebraically to find the total solution for
the current or voltage.
The only variation in applying this method to ac networks with independent sources is that
we are now working with impedances and phasors instead of just resistors and real numbers.
The superposition theorem is not applicable to power effects in ac networks since we are
still dealing with a nonlinear relationship. It can be applied to networks with sources of dif-
ferent frequencies only if the total response for each frequency is found independently and the
results are expanded in a nonsinusoidal expression, as appearing in Chapter 25.
One of the most frequent applications of the superposition theorem is to electronic systems
in which the dc and ac analyses are treated separately and the total solution is the sum of the
two. It is an important application of the theorem because the impact of the reactive elements
• Be able to apply the superposition theorem to ac
networks with independent and dependent
sources.
• Become proficient in applying Thévenin’s theorem
to ac networks with independent and dependent
sources.
• Be able to apply Norton’s theorem to ac networks
with independent and dependent sources.
• Clearly understand the conditions that must be
met for maximum power transfer to a load in an
ac network with independent or dependent
sources.
Objectives
Network Theorems (ac)
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 787
X
L
2
4 
–
+
X
C
3 
I
E
2  
=  5 V ? 0° E
1  
=  10 V ? 0°
–
+
X
L
1
4 
FIG. 18.1
Example 18.1.
changes dramatically in response to the two types of independent
sources. In addition, the dc analysis of an electronic system can often de-
fine important parameters for the ac analysis. Example 18.4 demon-
strates the impact of the applied source on the general configuration of
the network.
We first consider networks with only independent sources to provide
a close association with the analysis of Chapter 9.
Independent Sources
EXAMPLE 18.1 Using the superposition theorem, find the current I
through the 4  reactance in Fig. 18.1. 1X
L
2
2
–
+
I
E
2
E
1
–
+
Z
1
Z
2
Z
3
FIG. 18.2
Assigning the subscripted impedances to the network
in Fig. 18.1.
I
E
1
–
+
Z
1
Z
2
Z
3
E
1
–
+
Z
1
Z
23
I
s
1
I
s
1
FIG. 18.3
Determining the effect of the voltage source E
1
on the current I of the network in
Fig. 18.1.
Solution: For the redrawn circuit (Fig. 18.2),
Considering the effects of the voltage source E
1
(Fig. 18.3), we have
  1.25 A 90°
  I
s
1

E
1
Z
2  3
 Z
1

10 V 0°
j 12  j 4 

10 V 0°
8  90°
  12  90°
 Z
2  3

Z
2
Z
3
Z
2
 Z
3

1j 4 21j 3 2
j 4  j 3 

12 
j
j 12 
 Z
3
j X
C
j 3 
 Z
2
j X
L
2
 j 4 
 Z
1
j X
L
1
 j 4 
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 788
and
(current divider rule)
Considering the effects of the voltage source E
2
(Fig. 18.4), we have
 
1j 3 21 j 1.25 A2
j 4  j 3 

3.75 A
j 1
 3.75 A 90°
 I 
Z
3
I
s
1
Z
2
 Z
3
4 
I
I'
I?
X
L
2
FIG. 18.5
Determining the resultant current for the network in
Fig. 18.1.
I
E
2
–
+
Z
1
Z
2
Z
3
E
2
–
+
Z
3
Z
12
I
s
2
I
s
2
FIG. 18.4
Determining the effect of the voltage source E
2
on the current I of the network
in Fig. 18.1.
and
The resultant current through the 4 reactance (Fig. 18.5) is
I  I  I
 3.75 A 90°  2.50 A 90°j 3.75 A  j 2.50 A
j 6.25 A
I  6.25 A 90°
EXAMPLE 18.2 Using superposition, find the current I through the 
6  resistor in Fig. 18.6.
X
L
2
I
I
s
2
2
 2.5 A 90°
 I
s
2

E
2
Z
1  2
 Z
3

5 V 0°
j 2  j 3 

5 V 0°
1  90°
 5 A 90°
 Z
1  2

Z
1
N

j 4 
2
 j 2 
X
C  
=  8 
I
E
1  
=  20 V ? 30°
–
+
R
  
=  6  X
L  
=  6 
I
1
2 A ? 0°
FIG. 18.6
Example 18.2.
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 789
Solution: For the redrawn circuit (Fig. 18.7),
Z
1
 j 6  Z
2
 6  j 8 
Consider the effects of the current source (Fig. 18.8). Applying the cur-
rent divider rule, we have
Consider the effects of the voltage source (Fig. 18.9). Applying Ohm’s
law gives us
The total current through the 6  resistor (Fig. 18.10) is
I  I  I
 1.9 A 108.43°  3.16 A 48.43°
 (0.60 A  j 1.80 A)  (2.10 A  j 2.36 A)
 1.50 A  j 4.16 A
I  4.42 A 70.2°
EXAMPLE 18.3 Using superposition, find the voltage across the 6 
resistor in Fig. 18.6. Check the results against V
6
 I(6 ), where I is
the current found through the 6  resistor in Example 18.2.
Solution: For the current source,
For the voltage source,
The total voltage across the 6  resistor (Fig. 18.11) is
 V
6
 26.5 V 70.2°
  8.98 V  j 25.0 V
 13.60 V  j 10.82 V2 112.58 V  j 14.18 V2
  11.4 V 108.43°  18.96 V 48.43°
 V
6
 V
6
 V
6
V
6
 I162 13.16 A 48.43°216 2  18.96 V 48.43°
V
6
 I16 2 11.9 A 108.43°216 2  11.4 V 108.43°
  3.16 A 48.43°
 I 
E
1
Z
T

E
1
Z
1
 Z
2

20 V 30°
6.32  18.43°
 I  1.9 A 108.43°
 
12 A 90°
6.32 18.43°
 I 
Z
1
I
1
Z
1
 Z
2

1j 6 212 A2
j 6  6  j 8 

j 12 A
6  j 2
I
Z
1
Z
2
–
E
1
+
FIG. 18.9
Determining the effect of the voltage source E
1
on
the current I of the network in Fig. 18.6.
I
I'
R
6 
I?
FIG. 18.10
Determining the resultant current I for the network
in Fig. 18.6.
R
6 
V?
6
+ –
V'
6
+ –
V
6
+ –
FIG. 18.11
Determining the resultant voltage V
6
for the network in Fig. 18.6.
–
I
Z
1
Z
2
E
1
+
I
1
FIG. 18.7
Assigning the subscripted impedances to the network
in Fig. 18.6.
I
Z
1
Z
2
I
1
FIG. 18.8
Determining the effect of the current source I
1
on the
current I of the network in Fig. 18.6.
boy30444_ch18.qxd  3/24/06  2:54 PM  Page 790
Page 5


Network Theorems (ac)
18.1 INTRODUCTION
Due to the need for developing confidence in the application of the various theorems to net-
works with controlled (dependent) sources, some sections have been divided into two parts:
independent sources and dependent sources.
Theorems to be considered in detail include the superposition theorem, Thévenin’s and
Norton’s theorems, and the maximum power transfer theorem. The substitution and reciproc-
ity theorems and Millman’s theorem are not discussed in detail her                      . e
18.2 SUPERPOSITION THEOREM
You will recall from Chapter 9 that the superposition theorem eliminated the need for solv-
ing simultaneous linear equations by considering the effects of each source independently. To
consider the effects of each source, we had to remove the remaining sources. This was ac-
complished by setting voltage sources to zero (short-circuit representation) and current sources
to zero (open-circuit representation). The current through, or voltage across, a portion of the
network produced by each source was then added algebraically to find the total solution for
the current or voltage.
The only variation in applying this method to ac networks with independent sources is that
we are now working with impedances and phasors instead of just resistors and real numbers.
The superposition theorem is not applicable to power effects in ac networks since we are
still dealing with a nonlinear relationship. It can be applied to networks with sources of dif-
ferent frequencies only if the total response for each frequency is found independently and the
results are expanded in a nonsinusoidal expression, as appearing in Chapter 25.
One of the most frequent applications of the superposition theorem is to electronic systems
in which the dc and ac analyses are treated separately and the total solution is the sum of the
two. It is an important application of the theorem because the impact of the reactive elements
• Be able to apply the superposition theorem to ac
networks with independent and dependent
sources.
• Become proficient in applying Thévenin’s theorem
to ac networks with independent and dependent
sources.
• Be able to apply Norton’s theorem to ac networks
with independent and dependent sources.
• Clearly understand the conditions that must be
met for maximum power transfer to a load in an
ac network with independent or dependent
sources.
Objectives
Network Theorems (ac)
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 787
X
L
2
4 
–
+
X
C
3 
I
E
2  
=  5 V ? 0° E
1  
=  10 V ? 0°
–
+
X
L
1
4 
FIG. 18.1
Example 18.1.
changes dramatically in response to the two types of independent
sources. In addition, the dc analysis of an electronic system can often de-
fine important parameters for the ac analysis. Example 18.4 demon-
strates the impact of the applied source on the general configuration of
the network.
We first consider networks with only independent sources to provide
a close association with the analysis of Chapter 9.
Independent Sources
EXAMPLE 18.1 Using the superposition theorem, find the current I
through the 4  reactance in Fig. 18.1. 1X
L
2
2
–
+
I
E
2
E
1
–
+
Z
1
Z
2
Z
3
FIG. 18.2
Assigning the subscripted impedances to the network
in Fig. 18.1.
I
E
1
–
+
Z
1
Z
2
Z
3
E
1
–
+
Z
1
Z
23
I
s
1
I
s
1
FIG. 18.3
Determining the effect of the voltage source E
1
on the current I of the network in
Fig. 18.1.
Solution: For the redrawn circuit (Fig. 18.2),
Considering the effects of the voltage source E
1
(Fig. 18.3), we have
  1.25 A 90°
  I
s
1

E
1
Z
2  3
 Z
1

10 V 0°
j 12  j 4 

10 V 0°
8  90°
  12  90°
 Z
2  3

Z
2
Z
3
Z
2
 Z
3

1j 4 21j 3 2
j 4  j 3 

12 
j
j 12 
 Z
3
j X
C
j 3 
 Z
2
j X
L
2
 j 4 
 Z
1
j X
L
1
 j 4 
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 788
and
(current divider rule)
Considering the effects of the voltage source E
2
(Fig. 18.4), we have
 
1j 3 21 j 1.25 A2
j 4  j 3 

3.75 A
j 1
 3.75 A 90°
 I 
Z
3
I
s
1
Z
2
 Z
3
4 
I
I'
I?
X
L
2
FIG. 18.5
Determining the resultant current for the network in
Fig. 18.1.
I
E
2
–
+
Z
1
Z
2
Z
3
E
2
–
+
Z
3
Z
12
I
s
2
I
s
2
FIG. 18.4
Determining the effect of the voltage source E
2
on the current I of the network
in Fig. 18.1.
and
The resultant current through the 4 reactance (Fig. 18.5) is
I  I  I
 3.75 A 90°  2.50 A 90°j 3.75 A  j 2.50 A
j 6.25 A
I  6.25 A 90°
EXAMPLE 18.2 Using superposition, find the current I through the 
6  resistor in Fig. 18.6.
X
L
2
I
I
s
2
2
 2.5 A 90°
 I
s
2

E
2
Z
1  2
 Z
3

5 V 0°
j 2  j 3 

5 V 0°
1  90°
 5 A 90°
 Z
1  2

Z
1
N

j 4 
2
 j 2 
X
C  
=  8 
I
E
1  
=  20 V ? 30°
–
+
R
  
=  6  X
L  
=  6 
I
1
2 A ? 0°
FIG. 18.6
Example 18.2.
boy30444_ch18.qxd  3/24/06  2:53 PM  Page 789
Solution: For the redrawn circuit (Fig. 18.7),
Z
1
 j 6  Z
2
 6  j 8 
Consider the effects of the current source (Fig. 18.8). Applying the cur-
rent divider rule, we have
Consider the effects of the voltage source (Fig. 18.9). Applying Ohm’s
law gives us
The total current through the 6  resistor (Fig. 18.10) is
I  I  I
 1.9 A 108.43°  3.16 A 48.43°
 (0.60 A  j 1.80 A)  (2.10 A  j 2.36 A)
 1.50 A  j 4.16 A
I  4.42 A 70.2°
EXAMPLE 18.3 Using superposition, find the voltage across the 6 
resistor in Fig. 18.6. Check the results against V
6
 I(6 ), where I is
the current found through the 6  resistor in Example 18.2.
Solution: For the current source,
For the voltage source,
The total voltage across the 6  resistor (Fig. 18.11) is
 V
6
 26.5 V 70.2°
  8.98 V  j 25.0 V
 13.60 V  j 10.82 V2 112.58 V  j 14.18 V2
  11.4 V 108.43°  18.96 V 48.43°
 V
6
 V
6
 V
6
V
6
 I162 13.16 A 48.43°216 2  18.96 V 48.43°
V
6
 I16 2 11.9 A 108.43°216 2  11.4 V 108.43°
  3.16 A 48.43°
 I 
E
1
Z
T

E
1
Z
1
 Z
2

20 V 30°
6.32  18.43°
 I  1.9 A 108.43°
 
12 A 90°
6.32 18.43°
 I 
Z
1
I
1
Z
1
 Z
2

1j 6 212 A2
j 6  6  j 8 

j 12 A
6  j 2
I
Z
1
Z
2
–
E
1
+
FIG. 18.9
Determining the effect of the voltage source E
1
on
the current I of the network in Fig. 18.6.
I
I'
R
6 
I?
FIG. 18.10
Determining the resultant current I for the network
in Fig. 18.6.
R
6 
V?
6
+ –
V'
6
+ –
V
6
+ –
FIG. 18.11
Determining the resultant voltage V
6
for the network in Fig. 18.6.
–
I
Z
1
Z
2
E
1
+
I
1
FIG. 18.7
Assigning the subscripted impedances to the network
in Fig. 18.6.
I
Z
1
Z
2
I
1
FIG. 18.8
Determining the effect of the current source I
1
on the
current I of the network in Fig. 18.6.
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Checking the result, we have
V
6
 I(6 )  (4.42 A 70.2°)(6 )
 26.5 V 70.2° (checks)
EXAMPLE 18.4 For the network in Fig. 18.12, determine the sinu-
soidal expression for the voltage y
3
using superposition.
–
+
R
2
1 k
R
1
0.5 k
R
3
3 k
V
3
E
1
  =  12 V
FIG. 18.13
Determining the effect of the dc voltage source E
1
on
the voltage y
3
of the network in Fig. 18.12.
–
+
R
2
1 k
R
1
0.5 k
X
L
2 k
R
3
3 k
v
3
X
C
10 k
E
2
  =  4 V ?0°
–
+
E
1
  =  12 V
FIG. 18.12
Example 18.4.
Solution: For the dc analysis, the capacitor can be replaced by an open-
circuit equivalent, and the inductor by a short-circuit equivalent. The re-
sult is the network in Fig. 18.13.
The resistors R
1
and R
3
are then in parallel, and the voltage V
3
can be
determined using the voltage divider rule:
and
For the ac analysis, the dc source is set to zero and the network is re-
drawn, as shown in Fig. 18.14.
  V
3  

  
3.6 V
 
10.429 k2112 V2
0.429 k 1 k

5.148 V
1.429
  V
3

RE
1
R R
2
R R
1
    R
3
 0.5 k    3 k 0.429 k
X
C
  =  10 k
–
+
R
2 
 =  1 k
R
1
0.5 k
R
3 
 =  3 k
V
3
X
L
2 k
E
2
  =  4 V ?0°
–
+
FIG. 18.14
Redrawing the network in Fig. 18.12 to determine the effect of the ac voltage
source E
2
.
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