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Electric Circuit Chap 4
Chapter 4, Problem 16.
Given the circuit in Fig. 4.84, use superposition to get
io.
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Electric Circuit Chap 4
Solution 4.16
Let io = io1 + io2 + io3, where io1, io2, and io3 are due to
the 12-V, 4-A, and 2-A sources. For io1, consider the circuit below.
4Ω
io1
12V
+
3Ω
2Ω
10 Ω
−
5Ω
10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A
4A
For io2, consider the circuit below.
Ω
io2
4Ω
Ω
3
2
10Ω
5Ω
i1
2 + 5 + 4||10 = 7 + 40/14 = 69/7
i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9
For io3, consider the circuit below.
3Ω
io3
2Ω
i2
4Ω
10 Ω
5Ω
2A
3 + 2 + 4||10 = 5 + 20/7 = 55/7
i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9
io = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA
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Electric Circuit Chap 4
Problem 4.19 Use superposition to solve for vx and ix in the circuit of Fig. 4.87.
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Electric Circuit Chap 4
Solution 4.19: Let vx = v1 + v2, where v1 and v2 are due to the 4-A and 6-A sources
respectively.
ix
v1
ix
v2
+
2Ω
8Ω
4A
−
+
2Ω
v1
6A 8Ω
−
+
−
4ix
+
v2
−
4ix
(a)
(b)
To find v1, consider the circuit in Fig. (a).
v1/8 = 4 + (-4ix – v1)/2
But,
-ix = (-4ix – v1)/2 and we have -2ix = v1. Thus,
v1/8 = 4 + (2v1 – v1)/8, which leads to v1 = -32/3
To find v2, consider the circuit shown in Fig. (b).
v2/2 = 6 + (4ix – v2)/8
But i = v /2 and 2i = v . Therefore,
x
2
x
2
v2/2 = 6 + (2v2 – v2)/8 which leads to v2 = -16
Hence,
vx = –(32/3) – 16
Writing a mesh equation around the outside loop we get,
–2ix + vx + 4ix = 0 or 2ix = –vx
Thus, we get,
ix = 13.333 A.
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Electric Circuit Chap 4
Chapter 4, Problem 27.
Apply source transformation to find vx in the circuit of Fig. 4.95.
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Electric Circuit Chap 4
Transforming the voltage sources to current sources gives the circuit in Fig. (a).
10||40 = 8 ohms
Transforming the current sources to voltage sources yields the circuit in Fig. (b).
Applying KVL to the loop,
-40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4
vx = 12i = -48 V
12 Ω
+ vx
5A
10Ω
40Ω
−
8A
20Ω
2A
(a)
8Ω
+
−
12 Ω
+ vx
40V
i
20 Ω
−
+
−
200V
(b)
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Electric Circuit Chap 4
Chapter 4, Problem 29.
Use source transformation to find
vo
in the circuit of Fig. 4.97.
4 kΩ
3vo
2 kΩ
−
+
+
3 mA
vo
1 kΩ
−
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Electric Circuit Chap 4
Chapter 4, Solution 29.
Transform the dependent voltage source to a current source as shown in Fig. (a). 2||4 =
(4/3) k ohms
4 kΩ
2 kΩ
2vo
(4/3) kΩ
−
1.5vo
+
3 mA
1 kΩ
3 mA
i
+
1 kΩ
+
vo
−
vo
−
(a)
(b)
It is clear that i = 3 mA which leads to vo = 1000i = 3 V
If the use of source transformations was not required for this problem, the actual answer
could have been determined by inspection right away since the only current that could
have flowed through the 1 k ohm resistor is 3 mA.
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Electric Circuit Chap 4
Chapter 4, Problem 41.
Find the Thèvenin and Norton equivalents at terminals a-b of the circuit shown in
Fig. 4.108.
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Electric Circuit Chap 4
Chapter 4, Solution 41
To find RTh, consider the circuit below
14 Ω
a
6Ω
5Ω
b
R
=
+
=
Ω= R
N
5 //(14 6) 4
Applying source transformation to the 1-A current source, we obtain the circuit below.
Th
6Ω
- 14V +
14 Ω
VTh
a
+
6V
3A
5Ω
b
At node a,
14 + 6 − VTh
6 + 14
IN =
VTh
RTh
=
3+
VTh

→
5
VTh = −8 V
= ( −8) / 4 = −2 A
Thus,
RTh = R N =
4Ω,
VTh = −8V,
I N = −2 A
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Electric Circuit Chap 4
Chapter 4, Problem 54.
Find the Thèvenin equivalent between terminals a-b of the circuit in Fig. 4.120
+
-
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Electric Circuit Chap 4
Chapter 4, Solution 54
To find VTh =Vx, consider the left loop.
− 3 + 1000io + 2V x =
0
3 = 1000io

→
(1)
+ 2V x
For the right loop,
V x = −50 x 40i o = −2000io
Combining (1) and (2),
3 = 1000io − 4000io
V x = −2000io =
(2)
= −3000io
2

→
VTh =

→
io = −1mA
2
To find RTh, insert a 1-V source at terminals a-b and remove the 3-V independent
source, as shown below.
1 kΩ
ix
.
io
V x = 1,
ix =
io = −
40io
+
Vx
+
2Vx
-
2V x
1000
1
ix
+
Vx
-
50
Ω
+
1V
-
= −2mA
= −80mA +
50
RTh =
40io
1
A
=
-60mA
50
= −1 / 0.060 = − 16.67Ω
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Electric Circuit Chap 4
Chapter 4, Problem 69.
Find the maximum power transferred to resistor
R
in the circuit of Fig. 4.135
0.003vo
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Electric Circuit Chap 4
Chapter 4, Solution 69.
We need the Thevenin equivalent across the resistor R. To find RTh, consider the circuit
below.
22 kΩ
v1
+
10 kΩ
40 kΩ
vo
30 kΩ
0.003vo
1mA
−
Assume that all resistances are in k ohms and all currents are in mA.
10||40 = 8, and 8 + 22 = 30
1 + 3vo = (v1/30) + (v1/30) = (v1/15)
15 + 45vo = v1
But vo = (8/30)v1, hence,
15 + 45x(8v1/30) v1, which leads to v1 = 1.3636
RTh = v1/1 = –1.3636 k ohms
RTh being negative indicates an active circuit and if you now make R equal to 1.3636 k
ohms, then the active circuit will actually try to supply infinite power to the resistor. The
correct answer is therefore:
2
2
VTh


V 
pR = 
 1363.6 =  Th  1363.6 =
 − 1363.6 + 1363.6 
 0 
∞
It may still be instructive to find VTh. Consider the circuit below.
10 kΩ vo 22 kΩ
v1
+
100V
+
−
vo
+
40 kΩ
30 kΩ
0.003vo
−
VTh
−
(100 – vo)/10 = (vo/40) + (vo – v1)/22
(1)
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Electric Circuit Chap 4
[(vo – v1)/22] + 3vo = (v1/30)
(2)
Solving (1) and (2),
v1 = VTh = -243.6 volts
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Electric Circuit Chap 4
Chapter 5, Problem 67
Obtain the output vo in the circuit of Fig. 5.95.
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Electric Circuit Chap 4
Solution 5.67
Voltage follower , inv amplifier , summer
Voltage follower, summer
vo =
=
−
80 
80 
80
 − (0.4) − (0.2)
40  20 
20
3.2 − 0.8 = 2.4V
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Electric Circuit Chap 4
Chapter 5, Problem 72
Find the load voltage vL in the circuit of Fig. 5.98.
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Electric Circuit Chap 4
Since no current flows into the input terminals of ideal op amp, there is no voltage drop
across the 20 kΩ resistor. As a voltage summer, the output of the first op amp is
v01 = 0.4
The second stage is an inverter
v2
250
v 01
100
= −2.5(0.4) = -1V
=−
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Electric Circuit Chap 4
Chapter 5, Problem 73
Determine the load voltage vL in the circuit of Fig. 5.99.
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Electric Circuit Chap 4
Chapter 5, Solution 73.
The first stage is a noninverting circuit. The output is
50 + 10
v 01 =
(1.8) = 10.8V
10
The second stage is a voltage follower,
v2
=
v 01
= 10.8V
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