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Chapter 3 - Alexander`s Fundamentals of Electric circuits
Ch.3 Solution
Fundamentals of Electric circuits (군산대학교)
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Chapter 3, Solution 1
Using Fig. 3.50, design a problem to help other students to better understand
nodal analysis.
R1
R2
Ix
12 V
+

R3
9V
+

Figure 3.50
For Prob. 3.1 and Prob. 3.39.
Solution
Given R 1 = 4 kΩ, R 2 = 2 kΩ, and R 3 = 2 kΩ, determine the value of I x using
nodal analysis.
Let the node voltage in the top middle of the circuit be designated as V x .
[(V x –12)/4k] + [(V x –0)/2k] + [(V x –9)/2k] = 0 or (multiply this by 4 k)
(1+2+2)V x = 12+18 = 30 or V x = 30/5 = 6 volts and
I x = 6/(2k) = 3 mA.
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Chapter 3, Solution 2
At node 1,
v  v2
 v1 v1

 6 1
2
10
5
60 = - 8v 1 + 5v 2
(1)
At node 2,
v2
v  v2
 36 1
4
2
36 = - 2v 1 + 3v 2
Solving (1) and (2),
v 1 = 0 V, v 2 = 12 V
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Chapter 3, Solution 3
Applying KCL to the upper node,
8
v0 vo vo
v


 20  0 = 0 or v 0 = –60 V
60
10 20 30
v0
v
 –6 A , i 2 = 0  –3 A,
10
20
v0
v
i3 =
 –2 A, i 4 = 0  1 A.
30
60
i1 =
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Chapter 3, Solution 4
3A
v1
6A
i1
i2
20 
10 
v2
i3
40 
i4
40 
At node 1,
–6 – 3 + v 1 /(20) + v 1 /(10) = 0 or v 1 = 9(200/30) = 60 V
At node 2,
3 – 2 + v 2 /(10) + v 2 /(5) = 0 or v 2 = –1(1600/80) = –20 V
i 1 = v 1 /(20) = 3 A, i 2 = v 1 /(10) = 6 A,
i 3 = v 2 /(40) = –500 mA, i 4 = v 2 /(40) = –500 mA.
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2A
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Chapter 3, Solution 5
Apply KCL to the top node.
30  v 0 20  v 0 v 0


2k
5k
4k
v 0 = 20 V
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Chapter 3, Solution 6.
Solve for V 1 using nodal analysis.
10 
10 V
–
+
5
10 
+
V1
10 
20 V
+

–
Figure 3.55
For Prob. 3.6.
Step 1.
The first thing to do is to select a reference node and to identify all the
unknown nodes. We select the bottom of the circuit as the reference node. The only
unknown node is the one connecting all the resistors together and we will call that node
V 1 . The other two nodes are at the top of each source. Relative to the reference, the one
at the top of the 10-volt source is –10 V. At the top of the 20-volt source is +20 V.
Step 2.
unknown).
Setup the nodal equation (there is only one since there is only one
Step 3.
Simplify and solve.
or
V 1 = –2 V.
The answer can be checked by calculating all the currents and see if they add up to zero.
The top two currents on the left flow right to left and are 0.8 A and 1.6 A respectively.
The current flowing up through the 10-ohm resistor is 0.2 A. The current flowing right to
left through the 10-ohm resistor is 2.2 A. Summing all the currents flowing out of the
node, V 1 , we get, +0.8+1.6 –0.2–2.2 = 0. The answer checks.
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Chapter 3, Solution 7
2
Vx  0 Vx  0

 0.2Vx  0
10
20
0.35V x = 2 or V x = 5.714 V.
Substituting into the original equation for a check we get,
0.5714 + 0.2857 + 1.1428 = 1.9999 checks!
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Chapter 3, Solution 8
6
i1
v1
i3
20
i2
+
V0
60V
4
+
–
+ 5V 0
–
–
20
i1 + i2 + i3 = 0
But
v1 ( v1  60)  0 v1  5v 0
0


10
20
20
2
v1 so that 2v 1 + v 1 – 60 + v 1 – 2v 1 = 0
5
or v 1 = 60/2 = 30 V, therefore v o = 2v 1 /5 = 12 V.
v0 
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Chapter 3, Solution 9
Let V 1 be the unknown node voltage to the right of the 250-Ω resistor. Let the ground
reference be placed at the bottom of the 50-Ω resistor. This leads to the following nodal
equation:
V1  24 V1  0 V1  60I b  0


0
250
50
150
simplifying we get
3V1  72  15V1  5V1  300I b  0
But I b 
24  V1
. Substituting this into the nodal equation leads to
250
24.2V1  100.8  0 or V 1 = 4.165 V.
Thus,
I b = (24 – 4.165)/250 = 79.34 mA.
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Chapter 3, Solution 10
v1
v2
v3
At node 1.
[(v 1 –0)/8] + [(v 1 –v 3 )/1] + 4 = 0
At node 2.
–4 + [(v 2 –0)/2] + 2i o = 0
At node 3.
–2i o + [(v 3 –0)/4] + [(v 3 –v 1 )/1] = 0
Finally, we need a constraint equation,
i o = v 1 /8
This produces,
1.125v 1 – v 3 = 4
(1)
0.25v 1 + 0.5v 2 = 4
(2)
–1.25v 1 + 1.25v 3 = 0 or v 1 = v 3
(3)
Substituting (3) into (1) we get (1.125–1)v 1 = 4 or v 1 = 4/0.125 = 32 volts. This leads to,
i o = 32/8 = 4 amps.
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Chapter 3, Solution 11
Find V o and the power absorbed by all the resistors in the circuit of Fig. 3.60.
12 
60 V
Vo
+
_
12 
6
–
+
24 V
Figure 3.60
For Prob. 3.11.
Solution
At the top node, KCL produces
Vo  60 Vo  0 Vo  (24)
0


12
12
6
(1/3)V o = 1 or V o = 3 V.
P 12Ω = (3–60)2/1 = 293.9 W (this is for the 12 Ω resistor in series with the 60 V
source)
P 12Ω = (V o )2/12 = 9/12 = 750 mW (this is for the 12 Ω resistor connecting V o to
ground)
P 4Ω = (3–(–24))2/6 = 121.5 W.
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Chapter 3, Solution 12
There are two unknown nodes, as shown in the circuit below.
20 
10 
V1
Vo
Ix
40 V
At node 1,
At node o,
+
_
20 
10 
4 Ix
V1  40 V1  0 V1  Vo
 0 or


20
20
10
(0.05+0.05+.1)V 1 – 0.1V o = 0.2V 1 – 0.1V o = 2
(1)
Vo  V1
V 0
 0 and I x = V 1 /20
 4I x  o
10
10
–0.1V 1 – 0.2V 1 + 0.2V o = –0.3V 1 + 0.2V o = 0 or
(2)
V 1 = (2/3)V o
(3)
Substituting (3) into (1),
0.2(2/3)V o – 0.1V o = 0.03333V o = 2 or
V o = 60 V.
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Chapter 3, Solution 13
Calculate v 1 and v 2 in the circuit of Fig. 3.62 using nodal analysis.
10 V
15 A
Figure 3.62
For Prob. 3.13.
Solution
At node number 2, [((v 2 + 10) – 0)/10] + [(v 2 –0)/4] – 15 = 0 or
(0.1+0.25)v 2 = 0.35v 2 = –1+15 = 14 or
v 2 = 40 volts.
Next, I = [(v 2 + 10) – 0]/10 = (40 + 10)/10 = 5 amps and
v 1 = 8x5 = 40 volts.
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Chapter 3, Solution 14
Using nodal analysis, find v o in the circuit of Fig. 3.63.
12.5 A
8
2
1
+
4
vo
100 V
+
–
50 V
–
+

Figure 3.63
For Prob. 3.14.
12.5 A
Solution
v0
v1
1
2
+
4
vo
100 V
+
–
8
50 V
–
+

At node 1,
[(v 1 –100)/1] + [(v 1 –v o )/2] + 12.5 = 0 or 3v 1 – v o = 200–25 = 175
At node o,
[(v o –v 1 )/2] – 12.5 + [(v o –0)/4] + [(v o +50)/8] = 0 or –4v 1 + 7v o = 50
(2)
Adding 4x(1) to 3x(2) yields,
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4(1) + 3(2) = –4v o + 21v o = 700 + 150 or 17v o = 850 or
v o = 50 V.
Checking, we get v 1 = (175+v o )/3 = 75 V.
At node 1,
[(75–100)/1] + [(75–50)/2] + 12.5 = –25 + 12.5 + 12.5 = 0!
At node o,
[(50–75)/2] + [(50–0)/4] + [(50+50)/8] – 12.5 = –12.5 + 12.5 + 12.5 – 12.5 = 0!
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Chapter 3, Solution 15
5A
v0
v1
1
8
2
4
40 V
20 V
–
+
+
–
Nodes 1 and 2 form a supernode so that v 1 = v 2 + 10
At the supernode, 2 + 6v 1 + 5v 2 = 3 (v 3 - v 2 )
At node 3, 2 + 4 = 3 (v 3 - v 2 )
(1)
2 + 6v 1 + 8v 2 = 3v 3
v3 = v2 + 2
Substituting (1) and (3) into (2),
2 + 6v 2 + 60 + 8v 2 = 3v 2 + 6
v 1 = v 2 + 10 =
v2 =
 56
11
54
11
i 0 = 6v i = 29.45 A
2
P 65 =
v12
 54 
 v12 G    6  144.6 W
R
 11 
2
  56 
P 55 = v G  
 5  129.6 W
 11 
2
2
P 35 = v L  v 3  G  (2) 2 3  12 W
2
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(3)
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Chapter 3, Solution 16
2S
v2
v1
i0
2A
+
1S
v0
4S
8S
v3
13 V
–
+
–
At the supernode,
2 = v 1 + 2 (v 1 - v 3 ) + 8(v 2 – v 3 ) + 4v 2 , which leads to 2 = 3v 1 + 12v 2 - 10v 3 (1)
But
v 1 = v 2 + 2v 0 and v 0 = v 2 .
Hence
v 1 = 3v 2
v 3 = 13V
Substituting (2) and (3) with (1) gives,
v 1 = 18.858 V, v 2 = 6.286 V, v 3 = 13 V
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(3)
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Chapter 3, Solution 17
v1
i0
4
2
10 
v2
60 V
60 V
8
3i 0
+
–
60  v1 v1 v1  v 2


4
8
2
60  v 2 v1  v 2

0
At node 2, 3i 0 +
10
2
At node 1,
But i 0 =
120 = 7v 1 - 4v 2
(1)
60  v1
.
4
Hence
360  v1  60  v 2 v1  v 2


0
4
10
2
1020 = 5v 1 + 12v 2
Solving (1) and (2) gives v 1 = 53.08 V. Hence i 0 =
60  v1
 1.73 A
4
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Chapter 3, Solution 18
–+
v2
v1
2
v3
2
15A
+
+
8
4
30 V
v1
v3
–
–
(b)
(a)
v 2  v1 v 2  v 3

–15 = 0 or –0.5v 1 + v 2 – 0.5v 3 = 15
(1)
2
2
v  v1 v 2  v 3 v1 v 3
At the supernode, 2

  = 0 and (v 1 /4) – 15 + (v 3 /8) = 0 or
2
2
4 8
2v 1 +v 3 = 120
(2)
At node 2, in Fig. (a),
From Fig. (b), – v 1 – 30 + v 3 = 0 or v 3 = v 1 + 30
Solving (1) to (3), we obtain,
v 1 = 30 V, v 2 = 60 V = v 3
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Chapter 3, Solution 19
At node 1,
V1  V3 V1  V2 V1


2
8
4
At node 2,
5  3
V1  V2 V2 V2  V3


8
2
4
At node 3,
12  V3




0  V1  7V2  2V3
V1  V3 V2  V3

0
8
2
4
From (1) to (3),
3

 7  1  4  V1   16 

  

  1 7  2 V2    0 
4
2  7  V3    36 

Using MATLAB,
 10 
V  A 1 B   4.933 
12.267 
16  7V1  V2  4V3


(1)
(2)
 36  4V1  2V2  7V3 (3)


AV  B


V1  10 V, V2  4.933 V, V3  12.267 V
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Chapter 3, Solution 20
Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence
V1 V2 V3


0

 V1  4V2  V3  0
(1)
4
1
4
.
V1
V2
.
4
2
V3
1
Between nodes 1 and 3,
 V1  12  V3  0

 V3  V1  12
Similarly, between nodes 1 and 2,
V1  V2  2i
But i  V3 / 4 . Combining this with (2) and (3) gives
. V2  6  V1 / 2
4
(2)
(3)
(4)
Solving (1), (2), and (4) leads to
V1  3V, V2  4.5V, V3  15V
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Chapter 3, Solution 21
4 k
v1
2 k
3v 0
v3
+
3v 0
v2
+
+
v0
3 mA
–
+
1 k
+
v3
v2
–
–
(b)
(a)
Let v 3 be the voltage between the 2k resistor and the voltage-controlled voltage source.
At node 1,
v  v 2 v1  v 3
3x10 3  1
12 = 3v 1 - v 2 - 2v 3
(1)

4000
2000
At node 2,
v1  v 2 v1  v 3 v 2


4
2
1
3v 1 - 5v 2 - 2v 3 = 0
(2)
Note that v 0 = v 2 . We now apply KVL in Fig. (b)
- v 3 - 3v 2 + v 2 = 0
v 3 = - 2v 2
From (1) to (3),
v 1 = 1 V, v 2 = 3 V
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Chapter 3, Solution 22
At node 1,
12  v 0 v1
v  v0

3 1
2
4
8
At node 2, 3 +
24 = 7v 1 - v 2
(1)
v 1  v 2 v 2  5v 2

8
1
But, v 1 = 12 - v 1
Hence, 24 + v 1 - v 2 = 8 (v 2 + 60 + 5v 1 ) = 4 V
456 = 41v 1 - 9v 2
(2)
Solving (1) and (2),
v 1 = - 10.91 V, v 2 = - 100.36 V
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Chapter 3, Solution 23
We apply nodal analysis to the circuit shown below.
1
30 V
+
_
4
Vo
2
2 Vo
+
+
Vo
V1
–
16 
_
3A
At node o,
Vo  30 Vo  0 Vo  (2Vo  V1 )
 0  1.25Vo  0.25V1  30


1
2
4
(1)
At node 1,
(2Vo  V1 )  Vo V1  0

 3  0  5V1  4Vo  48
4
16
From (1), V 1 = 5V o – 120. Substituting this into (2) yields
29V o = 648 or V o = 22.34 V.
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Chapter 3, Solution 24
Consider the circuit below.
8
+ Vo _
4A
V1
1
4
V2
2A
V3
2
V4
2
V  V4
V1  0
4 1
 0  1.125V1  0.125V4  4
8
1
V  0 V2  V3
 0  0.75V2  0.25V3  4
4 2

2
4
V3  V2 V3  0

 2  0  0.25V2  0.75V3  2
4
2
V  V1 V4  0
2 4

 0  0.125V1  1.125V4  2
8
1
1
(1)
(2)
(3)
(4)
0
0
 0.125
4
 1.125

  4
 0
0.75  0.25
0 

V 
  2
 0
 0.25 0.75
0 

 

0
0
1.125 
2
 0.125
Now we can use MATLAB to solve for the unknown node voltages.
>> Y=[1.125,0,0,-0.125;0,0.75,-0.25,0;0,-0.25,0.75,0;-0.125,0,0,1.125]
Y=
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1.1250
0
0 -0.1250
0 0.7500 -0.2500
0
0 -0.2500 0.7500
0
-0.1250
0
0 1.1250
>> I=[4,-4,-2,2]'
I=
4
-4
-2
2
>> V=inv(Y)*I
V=
3.8000
-7.0000
-5.0000
2.2000
V o = V 1 – V 4 = 3.8 – 2.2 = 1.6 V.
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Chapter 3, Solution 25
Consider the circuit shown below.
20

4


10
1  
1
2

10



3 

30
8
4
20
At node 1.
V  V2 V1  V4
4 1


 80  21V1  20V2  V4
1
20
At node 2,
V1  V2 V2 V2  V3




0  80V1  98V2  8V3
1
8
10
At node 3,
V2  V3 V3 V3  V4



 0  2V2  5V3  2V4
10
20
10
At node 4,
V1  V4 V3  V4 V4



 0  3V1  6V3  11V4
20
10
30
Putting (1) to (4) in matrix form gives:
(1)
(2)
(3)
(4)
1   V1 
 80   21 20 0
  
 
 0    80 98 8 0  V2 
0  0
2 5 2  V3 
  
 
0
6 11 V4 
0  3
B =A V
V = A-1 B
Using MATLAB leads to
V 1 = 25.52 V,
V 2 = 22.05 V, V 3 = 14.842 V, V 4 = 15.055 V
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Chapter 3, Solution 26
At node 1,
V  V3 V1  V2
15  V1
 3 1



 45  7V1  4V2  2V3
20
10
5
At node 2,
V1  V2 4 I o  V2 V2  V3


5
5
5
V1  V3
But I o 
. Hence, (2) becomes
10
0  7V1  15V2  3V3
At node 3,
V  V3  10  V3 V2  V3
3 1


0

 70  3V1  6V2  11V3
10
15
5
Putting (1), (3), and (4) in matrix form produces
 7  4  2  V1    45 

  

 7  15 3  V2    0 
  3  6 11  V   70 

 3  

Using MATLAB leads to
  7.19 


V  A 1B    2.78 
 2.89 




AV  B
Thus,
V 1 = –7.19V; V 2 = –2.78V; V 3 = 2.89V.
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(1)
(2)
(3)
(4)
lOMoARcPSD|30544978
Chapter 3, Solution 27
At node 1,
2 = 2v 1 + v 1 – v 2 + (v 1 – v 3 )4 + 3i 0 , i 0 = 4v 2 . Hence,
2 = 7v 1 + 11v 2 – 4v 3
(1)
At node 2,
v 1 – v 2 = 4v 2 + v 2 – v 3
0 = – v 1 + 6v 2 – v 3
(2)
At node 3,
2v 3 = 4 + v 2 – v 3 + 12v 2 + 4(v 1 – v 3 )
or
– 4 = 4v 1 + 13v 2 – 7v 3
(3)
In matrix form,
7 11  4  v 1   2 
1  6 1   v    0 
 2   

4 13  7  v 3    4
7
2
11
4
1  176,  1  0
6
1  110
13
7
4
13
7
2
4
7
11
2
11
  1 6
4
7
4
1  66,
2  1 0
4 4 7
v1 =
 3  1  6 0  286
4 13  4
 1 110

66

 0.625V, v 2 = 2 
 0.375V

176
176

v3 =
3
286

 1.625V.
176

v 1 = 625 mV, v 2 = 375 mV, v 3 = 1.625 V.
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Chapter 3, Solution 28
At node c,
V d  V c V c  Vb V c



 0  5Vb  11Vc  2Vd
(1)
10
4
5
At node b,
Va  90  Vb Vc  Vb Vb




 90  Va  4Vb  2Vc (2)
8
4
8
At node a,
Va  60  Vd Va Va  90  Vb
60  7Va  2Vb  4Vd (3)


0


4
16
8
At node d,
Va  60  Vd Vd Vd  Vc



 300  5Va  2Vc  8Vd (4)
4
20
10
In matrix form, (1) to (4) become
 0  5 11  2  Va   0 

  

 1  4 2 0  Vb    90 
AV  B


 7  2 0  4  V    60 
c

  

 5 0 2  8  V   300 

 d  

We use MATLAB to invert A and obtain
  10.56 


 20.56 
1
VA B
1.389 


  43.75 


Thus,
V a = –10.56 V; V b = 20.56 V; V c = 1.389 V; VC d = –43.75 V.
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Chapter 3, Solution 29
At node 1,
5  V1  V4  2V1  V1  V2  0


 5  4V1  V2  V4
At node 2,
V1  V2  2V2  4(V2  V3 )  0

 0  V1  7V2  4V3
At node 3,
6  4(V2  V3 )  V3  V4

 6  4V2  5V3  V4
At node 4,
2  V3  V4  V1  V4  3V4

 2  V1  V3  5V4
In matrix form, (1) to (4) become
 4  1 0  1 V1    5 
   

  1 7  4 0 V2   0 
AV  B


 0  4 5  1 V    6 
3
   

  1 0  1 5 V   2 
 4   

Using MATLAB,
(1)
(2)
(3)
(4)
  0.7708 


 1.209 
1
V  A B
2.309 


 0.7076 


i.e.
V1  0.7708 V, V 2  1.209 V, V 3  2.309 V, V 4  0.7076 V
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Chapter 3, Solution 30
i0
+
80 V
96 V
v0
1
2
4v 0
–
–+
20 
v1
10 
40 
+
–
2i 0
80 
At node 1,
[(v 1 –80)/10]+[(v 1 –4v o )/20]+[(v 1 –(v o –96))/40] = 0 or
(0.1+0.05+0.025)v 1 – (0.2+0.025)v o =
0.175v 1 – 0.225v o = 8–2.4 = 5.6
(1)
At node 2,
–2i o + [((v o –96)–v 1 )/40] + [(v o –0)/80] = 0 and i o = [(v 1 –(v o –96))/40]
–2[(v 1 –(v o –96))/40] + [((v o –96)–v 1 )/40] + [(v o –0)/80] = 0
–3[(v 1 –(v o –96))/40] + [(v o –0)/80] = 0 or
–0.0.075v 1 + (0.075+0.0125)v o = 7.2 =
–0.075v 1 + 0.0875v o = 7.2
(2)
Using (1) and (2) we get,
 0.175  0.225  v1  5.6
 0.075 0.0875  v   7.2 or
 o   

0.0875 0.225
0.0875 0.225


 v1 
0.075 0.175 5.6  0.075 0.175 5.6

v   0.0153125  0.016875 7.2   0.0015625 7.2
 
 
 o
v 1 = –313.6–1036.8 = –1350.4
v o = –268.8–806.4 = –1.0752 kV
and i o = [(v 1 –(v o –96))/40] = [(–1350.4–(–1075.2–96))/40] = –4.48 amps.
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lOMoARcPSD|30544978
Chapter 3, Solution 31
1
+ v0 –
v2
v1
2v 0
v3
2
i0
1A
4
1
10 V
4
–
At the supernode,
1 + 2v 0 =
v1 v 2 v1  v 3


4
1
1
(1)
But v o = v 1 – v 3 . Hence (1) becomes,
4 = -3v 1 + 4v 2 +4v 3
(2)
At node 3,
2v o +
or
v3
10  v 3
 v1  v 3 
4
2
20 = 4v 1 + 0v 2 – v 3
At the supernode, v 2 = v 1 + 4i o . But i o =
(3)
v3
. Hence,
4
v2 = v1 + v3
Solving (2) to (4) leads to,
v 1 = 4.97V, v 2 = 4.85V, v 3 = –0.12V.
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+
(4)
lOMoARcPSD|30544978
Chapter 3, Solution 32
5 k
v1
v3
v2
+
10 k
v1
10 V
20 V
–+
+–
12 V
–
4 mA
+
loop 1
+
loop 2
–
v3
–
(b)
(a)
We have a supernode as shown in figure (a). It is evident that v 2 = 12 V, Applying KVL
to loops 1and 2 in figure (b), we obtain,
-v 1 – 10 + 12 = 0 or v 1 = 2 and -12 + 20 + v 3 = 0 or v 3 = -8 V
Thus,
v 1 = 2 V, v 2 = 12 V, v 3 = -8V.
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Chapter 3, Solution 33
(a) This is a planar circuit. It can be redrawn as shown below.
5
3
1
2
4
6
2A
(b) This is a planar circuit. It can be redrawn as shown below.
4
3
5
12 V
+
2
–
1
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Chapter 3, Solution 34
(a)
This is a planar circuit because it can be redrawn as shown below,
7
2
1
3
6
10 V
5
+
–
4
(b)
This is a non-planar circuit.
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lOMoARcPSD|30544978
Chapter 3, Solution 35
30 V
20 V
+
–
+
–
i1
2 k
+
i2
v0
4 k
–
5 k
Assume that i 1 and i 2 are in mA. We apply mesh analysis. For mesh 1,
-30 + 20 + 7i 1 – 5i 2 = 0 or 7i 1 – 5i 2 = 10
(1)
For mesh 2,
-20 + 9i 2 – 5i 1 = 0 or -5i 1 + 9i 2 = 20
Solving (1) and (2), we obtain, i 2 = 5.
v 0 = 4i 2 = 20 volts.
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(2)
lOMoARcPSD|30544978
Chapter 3, Solution 36
10 V
4
+–
i1
12 V
i2
+
I1
6
–
i3
I2
2
Applying mesh analysis gives,
10I 1 – 6I 2 = 12 and –6I 1 + 8I 2 = –10
or
4 3


 6   5  3  I1   6   I1  3 5  6 
 5   3 4  I    5 or I   11  5
 
 2     2 
  
I 1 = (24–15)/11 = 0.8182 and I 2 = (18–25)/11 = –0.6364
i 1 = –I 1 = –818.2 mA; i 2 = I 1 – I 2 = 0.8182+0.6364 = 1.4546 A; and
i 3 = I 2 = –636.4 mA.
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Chapter 3, Solution 37
6
60 V +
+
v0
20
–
i1
4
i2
–
5v 0
+
–
20
Applying mesh analysis to loops 1 and 2, we get,
30i 1 – 20i 2 + 60 = 0 which leads to i 2 = 1.5i 1 + 3
(1)
–20i 1 + 40i 2 – 60 + 5v 0 = 0
(2)
But, v 0 = –4i 1
(3)
Using (1), (2), and (3) we get –20i 1 + 60i 1 + 120 – 60 – 20i 1 = 0 or
20i 1 = –60 or i 1 = –3 amps and i 2 = 7.5 amps.
Therefore, we get,
v 0 = –4i 1 = 12 volts.
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lOMoARcPSD|30544978
Chapter 3, Solution 38
Consider the circuit below with the mesh currents.
4
+
_
60 V
3
I3
I4
1
10 A
2
2
Io
1
1
I1
I2
+
_
22.5V
4
5A
I 1 = –5 A
(1)
1(I 2 –I 1 ) + 2(I 2 –I 4 ) + 22.5 + 4I 2 = 0
7I 2 – I 4 = –27.5
(2)
–60 + 4I 3 + 3I 4 + 1I 4 + 2(I 4 –I 2 ) + 2(I 3 – I 1 ) = 0 (super mesh)
–2I 2 + 6 I 3 + 6I 4 = +60 – 10 = 50
(3)
But, we need one more equation, so we use the constraint equation –I 3 + I 4 = 10. This
now gives us three equations with three unknowns.
0  1 I 2   27.5
7
 2 6 6   I    50 

 3  

 0  1 1  I 4   10 
We can now use MATLAB to solve the problem.
>> Z=[7,0,-1;-2,6,6;0,-1,0]
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lOMoARcPSD|30544978
Z=
7 0 -1
-2 6 6
0 -1 0
>> V=[–27.5,50,10]'
V=
–27.5
50
10
>> I=inv(Z)*V
I=
–1.3750
–10.0000
17.8750
I o = I 1 – I 2 = –5 – 1.375 = –6.375 A.
Check using the super mesh (equation (3)):
–2I 2 + 6 I 3 + 6I 4 = 2.75 – 60 + 107.25 = 50!
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lOMoARcPSD|30544978
Chapter 3, Solution 39
Using Fig. 3.50 from Prob. 3.1, design a problem to help other students to better
understand mesh analysis.
Solution
Given R 1 = 4 kΩ, R 2 = 2 kΩ, and R 3 = 2 kΩ, determine the value of I x using
mesh analysis.
R1
R2
Ix
12 V
+

I2
I1
R3
9V
+

Figure 3.50
For Prob. 3.1 and 3.39.
For loop 1 we get –12 +4kI 1 + 2k(I 1 –I 2 ) = 0 or 6I 1 – 2I 2 = 0.012 and at
loop 2 we get 2k(I 2 –I 1 ) + 2kI 2 + 9 = 0 or –2I 1 + 4I 2 = –0.009.
Now 6I 1 – 2I 2 = 0.012 + 3[–2I 1 + 4I 2 = –0.009] leads to,
10I 2 = 0.012 – 0.027 = –0.015 or I 2 = –1.5 mA and I 1 = (–0.003+0.012)/6 = 1.5
mA.
Thus,
I x = I 1 –I 2 = (1.5+1.5) mA = 3 mA.
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lOMoARcPSD|30544978
Chapter 3, Solution 40
2 k
6 k
6 k
56V
+
–
i2
2 k
i1
i3
4 k
4 k
Assume all currents are in mA and apply mesh analysis for mesh 1.
–56 + 12i 1 – 6i 2 – 4i 3 = 0 or 6i 1 – 3i 2 – 2i 3 = 28
(1)
for mesh 2,
–6i 1 + 14i 2 – 2i 3 = 0 or –3i 1 + 7i 2 – i 3
=0
(2)
=0
(3)
for mesh 3,
–4i 1 – 2i 2 + 10i 3 = 0 or –2i 1 – i 2 + 5i 3
Solving (1), (2), and (3) using MATLAB, we obtain,
i o = i 1 = 8 mA.
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Chapter 3, Solution 41
10 
i1
6V
2
+–
1
i2
4
5
i3
8V
+
–
i
i2
i3
0
For loop 1,
6 = 12i 1 – 2i 2
3 = 6i 1 – i 2
(1)
For loop 2,
-8 = – 2i 1 +7i 2 – i 3
(2)
For loop 3,
-8 + 6 + 6i 3 – i 2 = 0
2 = – i 2 + 6i 3
We put (1), (2), and (3) in matrix form,
6  1 0  i1   3
 2  7 1  i    8 
 2   

0  1 6 i 3  2
6
1 0
6 3 0
  2  7 1  234,  2  2 8 1  240
0
1 6
0 2 6
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(3)
lOMoARcPSD|30544978
6
1 3
 3  2  7 8  38
0 1 2
At node 0, i + i 2 = i 3 or i = i 3 – i 2 =
3  2
 38  240

= 1.188 A

 234
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lOMoARcPSD|30544978
Chapter 3, Solution 42
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Determine the mesh currents in the circuit of Fig. 3.88.
Figure 3.88
Solution
For mesh 1,
 12  50I 1  30I 2  0

 12  50I1  30I 2
For mesh 2,
 8  100 I 2  30 I 1  40 I 3  0


8  30 I 1  100 I 2  40 I 3
For mesh 3,
 6  50 I 3  40 I 2  0


6  40 I 2  50 I 3
Putting eqs. (1) to (3) in matrix form, we get
0  I 1  12 
 50  30
   

  30 100  40  I 2    8 
 0
 40 50  I 3   6 

Using Matlab,


AI  B
 0.48 


I  A B   0.40 
 0.44 


1
i.e. I 1 = 480 mA, I 2 = 400 mA, I 3 = 440 mA
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(1)
(2)
(3)
lOMoARcPSD|30544978
Chapter 3, Solution 43
20 
a
80 V
+
i1
–
30 
+
30 
i3
20 
80 V
+
i2
–
30 
20 
V ab
–
b
For loop 1,
80 = 70i 1 – 20i 2 – 30i 3
8 = 7i 1 – 2i 2 – 3i 3
(1)
80 = 70i 2 – 20i 1 – 30i 3
8 = -2i 1 + 7i 2 – 3i 3
(2)
0 = -30i 1 – 30i 2 + 90i 3
0 = i 1 + i 2 – 3i 3
For loop 2,
For loop 3,
(3)
Solving (1) to (3), we obtain i 3 = 16/9
I o = i 3 = 16/9 = 1.7778 A
V ab = 30i 3 = 53.33 V.
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lOMoARcPSD|30544978
Chapter 3, Solution 44
90 V
+
2
4
i3
i2
1
180V +
–
5
i1
45 A
i1
i2
Loop 1 and 2 form a supermesh. For the supermesh,
6i 1 + 4i 2 – 5i 3 + 180 = 0
(1)
For loop 3,
–i 1 – 4i 2 + 7i 3 + 90 = 0
(2)
Also,
i 2 = 45 + i 1
(3)
Solving (1) to (3), i 1 = –46, i 3 = –20;
i o = i 1 – i 3 = –26 A
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lOMoARcPSD|30544978
Chapter 3, Solution 45
30V
+
–
4
8
i3
i4
2
6
i1
3
i2
1
For loop 1,
30 = 5i 1 – 3i 2 – 2i 3
(1)
For loop 2,
10i 2 - 3i 1 – 6i 4 = 0
(2)
For the supermesh,
6i 3 + 14i 4 – 2i 1 – 6i 2 = 0
(3)
But
i 4 – i 3 = 4 which leads to i 4 = i 3 + 4
Solving (1) to (4) by elimination gives i = i 1 = 8.561 A.
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(4)
lOMoARcPSD|30544978
Chapter 3, Solution 46
For loop 1,
 12  3i1  8(i1  i 2 )  12  11i1  8i 2  0
For loop 2,
8(i 2  i1 )  6i 2  2v o  8i1  14i 2  2v o  0
But vo  3i1 ,


11i1  8i 2  12
 8i1  14i2  6i1  0

 i1  7i2
Substituting (2) into (1),
77i2  8i2  12

 i 2  0.1739 A and i1  7i2  1.217 A
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(1)
(2)
lOMoARcPSD|30544978
Chapter 3, Solution 47
First, transform the current sources as shown below.
- 6V +
2
8
V1
4
V2
I3
V3
4
2
I1
8
I2
+
20V
-
+
12V
-
For mesh 1,
 20  14 I 1  2 I 2  8I 3  0
For mesh 2,
10  7 I 1  I 2  4 I 3
(1)
 6   I1  7 I 2  2I 3
(2)
 6  14 I 3  4 I 2  8I 1  0

 3  4 I 1  2 I 2  7 I 3
Putting (1) to (3) in matrix form, we obtain
(3)
12  14 I 2  2 I 1  4 I 3  0
For mesh 3,




 7  1  4  I 1   10 
   

  1 7  2  I 2     6 
  4  2 7  I   3 
 3   

Using MATLAB,
 2 
I  A 1 B  0.0333
1.8667 
But


AI  B

 I 1  2.5, I 2  0.0333, I 3  1.8667
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lOMoARcPSD|30544978
I1 
20  V
4


V1  20  4 I 1  10 V
V 2  2( I 1  I 2 )  4.933 V
Also,
I2 
V3  12
8


V3  12  8I 2  12.267 V.
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lOMoARcPSD|30544978
Chapter 3, Solution 48
We apply mesh analysis and let the mesh currents be in mA.
3k 
I4
4k 
2k 
Io
1k 
I1
+
6V
-
5k 
I3
I2
+
4V
-
10k 
For mesh 1,
 6  8  5I1  I 2  4I 4  0

 2  5I1  I 2  4I 4
For mesh 2,
 4  13I 2  I1  10I 3  2I 4  0

 4  I1  13I 2  10I 3  2I 4
For mesh 3,
 3  15I 3  10I 2  5I 4  0

 3  10I 2  15I 3  5I 4
For mesh 4,
 4 I 1  2 I 2  5I 3  14 I 4  0
Putting (1) to (4) in matrix form gives
1
 4  I1   2 
0
 5

   
  1 13  10  2  I 2   4 

 AI B
 0  10 15  5  I    3 
3

   
  4  2  5 14  I   0 

 4   
Using MATLAB,
 3.608 


 4.044 
1
IA B
0.148
3.896 


 3 


The current through the 10k  resistor is I o = I 2 – I 3 = 148 mA.
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3V
+
(1)
(2)
(3)
(4)
lOMoARcPSD|30544978
Chapter 3, Solution 49
3
i3
2
1
2
i1
27 V
i2
+
–
2i 0
i1
i2
0
(a)
2
1
2
i1
+
+
v0
or
v0
–
i2
27V
+
–
(b)
For the supermesh in figure (a),
3i 1 + 2i 2 – 3i 3 + 27 = 0
(1)
At node 0,
i 2 – i 1 = 2i 0 and i 0 = –i 1 which leads to i 2 = –i 1
(2)
For loop 3,
–i 1 –2i 2 + 6i 3 = 0 which leads to 6i 3 = –i 1
(3)
Solving (1) to (3), i 1 = (–54/3)A, i 2 = (54/3)A, i 3 = (27/9)A
i 0 = –i 1 = 18 A, from fig. (b), v 0 = i 3 –3i 1 = (27/9) + 54 = 57 V.
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lOMoARcPSD|30544978
Chapter 3, Solution 50
i1
4
2
i3
10 
8
35 V
+
–
i2
3i 0
i2
For loop 1,
i3
16i 1 – 10i 2 – 2i 3 = 0 which leads to 8i 1 – 5i 2 – i 3 = 0
(1)
For the supermesh, –35 + 10i 2 – 10i 1 + 10i 3 – 2i 1 = 0
or
–6i 1 + 5i 2 + 5i 3 = 17.5
Also, 3i 0 = i 3 – i 2 and i 0 = i 1 which leads to 3i 1 = i 3 – i 2
Solving (1), (2), and (3), we obtain i 1 = 1.0098 and
i 0 = i 1 = 1.0098 A
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(2)
(3)
lOMoARcPSD|30544978
Chapter 3, Solution 51
5A
i1
8
2
i3
1
+
i2
40 V
4
+
v0
20V
–
+

–
For loop 1,
i 1 = 5A
(1)
For loop 2,
-40 + 7i 2 – 2i 1 – 4i 3 = 0 which leads to 50 = 7i 2 – 4i 3
(2)
For loop 3,
-20 + 12i 3 – 4i 2 = 0 which leads to 5 = - i 2 + 3 i 3
(3)
Solving with (2) and (3),
And,
i 2 = 10 A, i 3 = 5 A
v 0 = 4(i 2 – i 3 ) = 4(10 – 5) = 20 V.
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lOMoARcPSD|30544978
Chapter 3, Solution 52
+
v0 2 
i2
–
VS
+
–
8
3A
i2
i1
i3
4
i3
+
–
2V 0
For mesh 1,
2(i 1 – i 2 ) + 4(i 1 – i 3 ) – 12 = 0 which leads to 3i 1 – i 2 – 2i 3 = 6
(1)
For the supermesh, 2(i 2 – i 1 ) + 8i 2 + 2v 0 + 4(i 3 – i 1 ) = 0
But v 0 = 2(i 1 – i 2 ) which leads to -i 1 + 3i 2 + 2i 3 = 0
(2)
For the independent current source, i 3 = 3 + i 2
Solving (1), (2), and (3), we obtain,
i 1 = 3.5 A, i 2 = -0.5 A, i 3 = 2.5 A.
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(3)
lOMoARcPSD|30544978
Chapter 3, Solution 53
Applying mesh analysis leads to;
–12 + 4kI 1 – 3kI 2 – 1kI 3 = 0
(1)
–3kI 1 + 7kI 2 – 4kI 4 = 0
–3kI 1 + 7kI 2 = –12
(2)
–1kI 1 + 15kI 3 – 8kI 4 – 6kI 5 = 0
–1kI 1 + 15kI 3 – 6k = –24
(3)
I 4 = –3mA
(4)
–6kI 3 – 8kI 4 + 16kI 5 = 0
–6kI 3 + 16kI 5 = –24
(5)
Putting these in matrix form (having substituted I 4 = 3mA in the above),
 4  3  1 0   I1   12 
 3 7
0
0  I 2    12 


k
  1 0 15  6  I 3   24

   

0  6 16   I 5   24
0
ZI = V
Using MATLAB,
>> Z = [4,-3,-1,0;-3,7,0,0;-1,0,15,-6;0,0,-6,16]
Z=
4 -3 -1 0
-3 7 0 0
-1 0 15 -6
0 0 -6 16
>> V = [12,-12,-24,-24]'
V=
12
-12
-24
-24
We obtain,
>> I = inv(Z)*V
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lOMoARcPSD|30544978
I=
1.6196 mA
–1.0202 mA
–2.461 mA
3 mA
–2.423 mA
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lOMoARcPSD|30544978
Chapter 3, Solution 54
Let the mesh currents be in mA. For mesh 1,
 12  10  2 I 1  I 2  0


2  2I1  I 2
For mesh 2,
 10  3I 2  I 1  I 3  0

 10   I 1  3I 2  I 3
For mesh 3,
 12  2 I 3  I 2  0

 12   I 2  2 I 3
Putting (1) to (3) in matrix form leads to
 2  1 0  I 1   2 
   

  1 3  1 I 2   10 
 0  1 2  I  12 
 3   

Using MATLAB,
 5.25 
I  A B   8.5 
10.25
1


AI  B

 I 1  5.25 mA, I 2  8.5 mA, I 3  10.25 mA
I 1 = 5.25 mA, I 2 = 8.5 mA, and I 3 = 10.25 mA.
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(1)
(2)
(3)
lOMoARcPSD|30544978
Chapter 3, Solution 55
10 V
I2
b
i1
4A
c
+
1A
I2
6
1A
2
i2
i3
I4
4A
I3
d
I1
12 
4
+–
a
I3
0
I4
8V
It is evident that I 1 = 4
For mesh 4,
12(I 4 – I 1 ) + 4(I 4 – I 3 ) – 8 = 0
For the supermesh
At node c,
(1)
6(I 2 – I 1 ) + 10 + 2I 3 + 4(I 3 – I 4 ) = 0
or -3I 1 + 3I 2 + 3I 3 – 2I 4 = -5
I2 = I3 + 1
(2)
(3)
(4)
Solving (1), (2), (3), and (4) yields, I 1 = 4A, I 2 = 3A, I 3 = 2A, and I 4 = 4A
At node b,
i 1 = I 2 – I 1 = -1A
At node a,
i 2 = 4 – I 4 = 0A
At node 0,
i 3 = I 4 – I 3 = 2A
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lOMoARcPSD|30544978
Chapter 3, Solution 56
+ v1 –
2
i2
2
2
2
12 V
+
i1
–
2
i3
+
v2
–
For loop 1, 12 = 4i 1 – 2i 2 – 2i 3 which leads to 6 = 2i 1 – i 2 – i 3
(1)
For loop 2, 0 = 6i 2 –2i 1 – 2 i 3 which leads to 0 = -i 1 + 3i 2 – i 3
(2)
For loop 3, 0 = 6i 3 – 2i 1 – 2i 2 which leads to 0 = -i 1 – i 2 + 3i 3
(3)
In matrix form (1), (2), and (3) become,
 2  1  1  i1  6
  1 3  1 i   0
 2   

  1  1 3  i 3  0
1 1
2
 = 1
3
1 1
2
2
6 1
 1  8,  2 =  1 3  1  24
3
1 0
3
1 6
 3 =  1 3 0  24 , therefore i 2 = i 3 = 24/8 = 3A,
1 1 0
v 1 = 2i 2 = 6 volts, v = 2i 3 = 6 volts
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lOMoARcPSD|30544978
Chapter 3, Solution 57
Assume R is in kilo-ohms.
V2  4kx15mA  60V,
V1  90  V2  90  60  30V
Current through R is
3
3

 30 
(15)R
iR 
i o,
V1  i R R
3 R
3 R
This leads to R = 90/15 = 6 kΩ.
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lOMoARcPSD|30544978
Chapter 3, Solution 58
30 
i2
30 
10 
i1
10 
30 
i3
+
–
120 V
For loop 1, 120 + 40i 1 – 10i 2 = 0, which leads to -12 = 4i 1 – i 2
(1)
For loop 2, 50i 2 – 10i 1 – 10i 3 = 0, which leads to -i 1 + 5i 2 – i 3 = 0
For loop 3, -120 – 10i 2 + 40i 3 = 0, which leads to 12 = -i 2 + 4i 3
Solving (1), (2), and (3), we get, i 1 = -3A, i 2 = 0, and i 3 = 3A
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(2)
(3)
lOMoARcPSD|30544978
Chapter 3, Solution 59
40 
–+
I0
96 V
i2
10 
20 
+
80V
i1
+
4v 0
–
i3
+
–
80 
v0
–
2I 0
i2
i3
For loop 1, -80 + 30i 1 – 20i 2 + 4v 0 = 0, where v 0 = 80i 3
or 4 = 1.5i 1 – i 2 + 16i 3
(1)
For the supermesh, 60i 2 – 20i 1 – 96 + 80i 3 – 4 v 0 = 0, where v 0 = 80i 3
or 4.8 = -i 1 + 3i 2 – 12i 3
(2)
Also, 2I 0 = i 3 – i 2 and I 0 = i 2 , hence, 3i 2 = i 3
(3)
From (1), (2), and (3),
 3  2 32   i1   8 
  1 3  12 i   4.8

  2  
 1  i3   0 
3
 0
3
2
32
3
2
8
 = 1
3
 12  5,  2 =  1 4.8  12  22.4,  3 =  1
3
4.8  67.2
0
3
1
3
0
3
0
8
0
32
1
0
I 0 = i 2 =  2 / = -28/5 = –4.48 A
v 0 = 8i 3 = (-84/5)80 = –1.0752 kvolts
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lOMoARcPSD|30544978
Chapter 3, Solution 60
0.5i 0
4
56 V
8
v1
v2
1
56 V
+
2
–
i0
At node 1, [(v 1 –0)/1] + [(v 1 –56)/4] + 0.5[(v 1 –0)/1] = 0 or 1.75v 1 = 14 or v 1 = 8 V
At node 2, [(v 2 –56)/8] – 0.5[8/1] + [(v 2 –0)/2] = 0 or 0.625v 2 = 11 or v 2 = 17.6 V
P 1 = (v 1 )2/1 = 64 watts, P 2 = (v 2 )2/2 = 154.88 watts,
P 4 = (56 – v 1 )2/4 = 576 watts, P 8 = (56 – v 2 )2/8 = 1.84.32 watts.
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lOMoARcPSD|30544978
Chapter 3, Solution 61
v1
20 
v2
10 
i0
is
+
v0
–
30 
–
+ 5v 0
At node 1, i s = (v 1 /30) + ((v 1 – v 2 )/20) which leads to 60i s = 5v 1 – 3v 2
But v 2 = -5v 0 and v 0 = v 1 which leads to v 2 = -5v 1
Hence, 60i s = 5v 1 + 15v 1 = 20v 1 which leads to v 1 = 3i s , v 2 = -15i s
i 0 = v 2 /50 = -15i s /50 which leads to i 0 /i s = -15/50 = –0.3
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40 
(1)
lOMoARcPSD|30544978
Chapter 3, Solution 62
4 k
100V +
–
A
8 k
i2
i1
B
2 k
i3
+
–
40 V
We have a supermesh. Let all R be in k, i in mA, and v in volts.
For the supermesh, -100 +4i 1 + 8i 2 + 2i 3 + 40 = 0 or 30 = 2i 1 + 4i 2 + i 3 (1)
At node A,
i1 + 4 = i2
(2)
At node B,
i 2 = 2i 1 + i 3
(3)
Solving (1), (2), and (3), we get i 1 = 2 mA, i 2 = 6 mA, and i 3 = 2 mA.
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lOMoARcPSD|30544978
Chapter 3, Solution 63
10 
A
5
50 V
+
–
i1
i2
+
–
4i x
For the supermesh, -50 + 10i 1 + 5i 2 + 4i x = 0, but i x = i 1 . Hence,
50 = 14i 1 + 5i 2
At node A, i 1 + 3 + (v x /4) = i 2 , but v x = 2(i 1 – i 2 ), hence, i 1 + 2 = i 2
Solving (1) and (2) gives i 1 = 2.105 A and i 2 = 4.105 A
v x = 2(i 1 – i 2 ) = –4 volts and i x = i 2 – 2 = 2.105 amp
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(1)
(2)
lOMoARcPSD|30544978
Chapter 3, Solution 64
i1
50 
A
i0
i1
i 2 10 
+

v0
10 
i2
+
–
4i 0
i3
40 
250V +
–
5A
0.2V 0
B
i1
For mesh 2,
i3
20i 2 – 10i 1 + 4i 0 = 0
(1)
But at node A, i o = i 1 – i 2 so that (1) becomes i 1 = (16/6)i 2
(2)
For the supermesh, –250 + 50i 1 + 10(i 1 – i 2 ) – 4i 0 + 40i 3 = 0
or
28i 1 – 3i 2 + 20i 3 = 125
(3)
At node B,
But,
i 3 + 0.2v 0 = 2 + i 1
v 0 = 10i 2 so that (4) becomes i 3 = 5 + (2/3)i 2
Solving (1) to (5), i 2 = 0.2941 A,
v 0 = 10i 2 = 2.941 volts, i 0 = i 1 – i 2 = (5/3)i 2 = 490.2mA.
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(4)
(5)
lOMoARcPSD|30544978
Chapter 3, Solution 65
For mesh 1,
–12 + 12I 1 – 6I 2 – I 4 = 0 or
12  12 I 1  6 I 2  I 4
(1)
–6I 1 + 16I 2 – 8I 3 – I 4 – I 5 = 0
(2)
–8I 2 + 15I 3 – I 5 – 9 = 0 or
9 = –8I 2 + 15I 3 – I 5
(3)
For mesh 2,
For mesh 3,
For mesh 4,
–I 1 – I 2 + 7I 4 – 2I 5 – 6 = 0 or
6 = –I 1 – I 2 + 7I 4 – 2I 5
(4)
For mesh 5,
–I 2 – I 3 – 2I 4 + 8I 5 – 10 = 0 or
10   I 2  I 3  2 I 4  8I 5
Casting (1) to (5) in matrix form gives
1
0  I1  12 
 12  6 0

   
  6 16  8  1  1  I 2   0 
 0  8 15 0  1  I    9 



 3   
7  2  I 4   6 
 1 1 0
 0  1  1  2 8  I  10 

 5   
(5)
AI  B
Using MATLAB we input:
Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]
and V=[12;0;9;6;10]
This leads to
>> Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]
Z=
12
-6
0
-1
0
-6 0 -1 0
16 -8 -1 -1
-8 15 0 -1
-1 0 7 -2
-1 -1 -2 8
>> V=[12;0;9;6;10]
V=
12
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lOMoARcPSD|30544978
0
9
6
10
>> I=inv(Z)*V
I=
2.1701
1.9912
1.8119
2.0942
2.2489
Thus,
I = [2.17, 1.9912, 1.8119, 2.094, 2.249] A.
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lOMoARcPSD|30544978
Chapter 3, Solution 66
The mesh equations are obtained as follows.
12  24  30I1  4I2  6I3  2I4  0
or
30I 1 – 4I 2 – 6I 3 – 2I 4 = –12
24  40  4I1  30I2  2I4  6I5  0
or
–4I 1 + 30I 2 – 2I 4 – 6I 5 = –16
(1)
(2)
–6I 1 + 18I 3 – 4I 4 = 30
(3)
–2I 1 – 2I 2 – 4I 3 + 12I 4 –4I 5 = 0
(4)
–6I 2 – 4I 4 + 18I 5 = –32
(5)
Putting (1) to (5) in matrix form
 30  4  6  2 0    12 
 4 30 0  2  6   16 

 

  6 0 18  4 0  I   30 

 

 2  2  4 12  4  0 
 0  6 0  4 18   32
ZI = V
Using MATLAB,
>> Z = [30,-4,-6,-2,0;
-4,30,0,-2,-6;
-6,0,18,-4,0;
-2,-2,-4,12,-4;
0,-6,0,-4,18]
Z=
30 -4 -6 -2 0
-4 30 0 -2 -6
-6 0 18 -4 0
-2 -2 -4 12 -4
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lOMoARcPSD|30544978
0
-6
0
-4
18
>> V = [-12,-16,30,0,-32]'
V=
-12
-16
30
0
-32
>> I = inv(Z)*V
I=
-0.2779 A
-1.0488 A
1.4682 A
-0.4761 A
-2.2332 A
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lOMoARcPSD|30544978
Chapter 3, Solution 67
Consider the circuit below.
5A
V1
4
V2
2
V3
+ Vo 3 Vo
10 
5
10 A
0 
 5  3Vo 
 0.35  0.25

 0.25 0.95  0.5 V  
0




 15 
 0
 0 .5
0.5 
Since we actually have four unknowns and only three equations, we need a constraint
equation.
Vo = V2 – V3
Substituting this back into the matrix equation, the first equation becomes,
0.35V 1 – 3.25V 2 + 3V 3 = –5
This now results in the following matrix equation,
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lOMoARcPSD|30544978
3 
 5
 0.35  3.25
 0.25 0.95  0.5 V   0 
 


 15 
 0
0.5 
 0.5
Now we can use MATLAB to solve for V.
>> Y=[0.35,-3.25,3;-0.25,0.95,-0.5;0,-0.5,0.5]
Y=
0.3500 -3.2500 3.0000
-0.2500 0.9500 -0.5000
0 -0.5000 0.5000
>> I=[–5,0,15]'
I=
–5
0
15
>> V=inv(Y)*I
V=
–410.5262
–194.7368
–164.7368
V o = V 2 – V 3 = –77.89 + 65.89 = –30 V.
Let us now do a quick check at node 1.
–3(–30) + 0.1(–410.5) + 0.25(–410.5+194.74) + 5 =
90–41.05–102.62+48.68+5 = 0.01; essentially zero considering the
accuracy we are using. The answer checks.
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lOMoARcPSD|30544978
Chapter 3, Solution 68
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Find the voltage V o in the circuit of Fig. 3.112.
3A
10 
25 
+
4A
20 
Vo
40 
+
_
24 V
+
_
24 V
_
Figure 3.112
For Prob. 3.68.
Solution
Consider the circuit below. There are two non-reference nodes.
3A
V1
10 
25 
Vo
+
4A
40 
Vo
20 
_
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lOMoARcPSD|30544978
0.125  0.1
 43   7 
V

  0.1 0.19 
 3  24 / 25   2.04

 



Using MATLAB, we get,
>> Y=[0.125,-0.1;-0.1,0.19]
Y=
0.1250 -0.1000
-0.1000 0.1900
>> I=[7,-2.04]'
I=
7.0000
-2.0400
>> V=inv(Y)*I
V=
81.8909
32.3636
Thus, V o = 32.36 V.
We can perform a simple check at node V o ,
3 + 0.1(32.36–81.89) + 0.05(32.36) + 0.04(32.36–24) =
3 – 4.953 + 1.618 + 0.3344 = – 0.0004; answer checks!
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lOMoARcPSD|30544978
Chapter 3, Solution 69
Assume that all conductances are in mS, all currents are in mA, and all voltages are in
volts.
G 11 = (1/2) + (1/4) + (1/1) = 1.75, G 22 = (1/4) + (1/4) + (1/2) = 1,
G 33 = (1/1) + (1/4) = 1.25, G 12 = -1/4 = -0.25, G 13 = -1/1 = -1,
G 21 = -0.25, G 23 = -1/4 = -0.25, G 31 = -1, G 32 = -0.25
i 1 = 20, i 2 = 5, and i 3 = 10 – 5 = 5
The node-voltage equations are:
 1   v 1   20
 1.75  0.25
  0.25
1
 0.25 v 2    5 

  1
 0.25 1.25  v 3   5 
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Chapter 3, Solution 70
 4I x  20 
3 0
0 5  V    4 I  7 


x


With two equations and three unknowns, we need a constraint equation,
I x = 2V 1 , thus the matrix equation becomes,
 20 
  5 0
 8 5 V    7 


 
This results in V 1 = 20/(–5) = –4 V and
V 2 = [–8(–4) – 7]/5 = [32 – 7]/5 = 5 V.
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lOMoARcPSD|30544978
Chapter 3, Solution 71
 9  4  5  30 
 4 7  1 I   15

 

  5  1 9   0 
We can now use MATLAB solve for our currents.
>> R=[9,–4,–5;–4,7,–1;–5,–1,9]
R=
9 –4 –5
–4 7 –1
–5 –1 9
>> V=[30,–15,0]'
V=
30
–15
0
>> I=inv(R)*V
I=
6.255 A
1.9599 A
3.694 A
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lOMoARcPSD|30544978
Chapter 3, Solution 72
R 11 = 5 + 2 = 7, R 22 = 2 + 4 = 6, R 33 = 1 + 4 = 5, R 44 = 1 + 4 = 5,
R 12 = -2, R 13 = 0 = R 14 , R 21 = -2, R 23 = -4, R 24 = 0, R 31 = 0,
R 32 = -4, R 34 = -1, R 41 = 0 = R 42 , R 43 = -1, we note that R ij = R ji for
all i not equal to j.
v 1 = 8, v 2 = 4, v 3 = -10, and v 4 = -4
Hence the mesh-current equations are:
0   i1   8 
 7 2 0
 2 6  4 0  i   4 

 2   

 0  4 5  1  i 3    10

  

0  1 5  i4    4 
 0
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lOMoARcPSD|30544978
Chapter 3, Solution 73
R 11 = 2 + 3 +4 = 9, R 22 = 3 + 5 = 8, R 33 = 1+1 + 4 = 6, R 44 = 1 + 1 = 2,
R 12 = -3, R 13 = -4, R 14 = 0, R 23 = 0, R 24 = 0, R 34 = -1
v 1 = 6, v 2 = 4, v 3 = 2, and v 4 = -3
Hence,
 9  3  4 0   i1   6 
 3 8
0
0   i 2   4 


 4 0
6  1  i 3   2 

   
0  1 2   i 4    3
 0
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lOMoARcPSD|30544978
Chapter 3, Solution 74
R 11 = R 1 + R 4 + R 6 , R 22 = R 2 + R 4 + R 5 , R 33 = R 6 + R 7 + R 8 ,
R 44 = R 3 + R 5 + R 8 , R 12 = -R 4 , R 13 = -R 6 , R 14 = 0, R 23 = 0,
R 24 = -R 5 , R 34 = -R 8 , again, we note that R ij = R ji for all i not equal to j.
 V1 
 V 
2
The input voltage vector is = 
 V3 


  V4 
 R 1  R4  R6

 R4


 R6

0

 R4
R2  R4  R5
0
 R5
 R6
0
R6  R7  R8
 R8
  i1   V1 
 i   V 
 2    2 
  i 3   V3 
  

R3  R5  R8   i 4    V4 
0
 R5
 R8
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lOMoARcPSD|30544978
Chapter 3, Solution 75
* Schematics Netlist *
R_R4
R_R2
R_R1
R_R3
R_R5
V_V4
v_V3
v_V2
v_V1
$N_0002 $N_0001 30
$N_0001 $N_0003 10
$N_0005 $N_0004 30
$N_0003 $N_0004 10
$N_0006 $N_0004 30
$N_0003 0 120V
$N_0005 $N_0001 0
0 $N_0006 0
0 $N_0002 0
i3
i1
i2
Clearly, i 1 = –3 amps, i 2 = 0 amps, and i 3 = 3 amps, which agrees with the answers in
Problem 3.44.
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lOMoARcPSD|30544978
Chapter 3, Solution 76
* Schematics Netlist *
I_I2
R_R1
R_R3
R_R2
F_F1
VF_F1
R_R4
R_R6
I_I1
R_R5
0 $N_0001 DC 4A
$N_0002 $N_0001 0.25
$N_0003 $N_0001 1
$N_0002 $N_0003 1
$N_0002 $N_0001 VF_F1 3
$N_0003 $N_0004 0V
0 $N_0002 0.5
0 $N_0001 0.5
0 $N_0002 DC 2A
0 $N_0004 0.25
Clearly, v 1 = 625 mVolts, v 2 = 375 mVolts, and v 3 = 1.625 volts, which agrees with
the solution obtained in Problem 3.27.
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lOMoARcPSD|30544978
1
Chapter 3, Solution 77
As a check we can write the nodal equations,
 1.7  0.2
5
 1.2 1.2  V   2


 
Solving this leads to V 1 = 3.111 V and V 2 = 1.4444 V. The answer checks!
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lOMoARcPSD|30544978
Chapter 3, Solution 78
The schematic is shown below. When the circuit is saved and simulated the node
voltages are displayed on the pseudocomponents as shown. Thus,
V1  3V, V2  4.5V, V3  15V,
.
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lOMoARcPSD|30544978
Chapter 3, Solution 79
The schematic is shown below. When the circuit is saved and simulated, we obtain the
node voltages as displayed. Thus,
V a = –10.556 volts; V b = 20.56 volts; V c = 1.3889 volts; and V d = –43.75 volts.
1.3889 V
R3
10
–43.75 V
R6
4
R4
5
R5
4
R1
R2
20
8
20.56 V
R7
16
R8
8
V1
V2
60Vdc
90Vdc
–10.556 V
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lOMoARcPSD|30544978
Chapter 3, Solution 80
* Schematics Netlist *
H_H1
VH_H1
I_I1
V_V1
R_R4
R_R1
R_R2
R_R5
R_R3
$N_0002 $N_0003 VH_H1 6
0 $N_0001 0V
$N_0004 $N_0005 DC 8A
$N_0002 0 20V
0 $N_0003 4
$N_0005 $N_0003 10
$N_0003 $N_0002 12
0 $N_0004 1
$N_0004 $N_0001 2
Clearly, v 1 = 84 volts, v 2 = 4 volts, v 3 = 20 volts, and v 4 = -5.333 volts
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lOMoARcPSD|30544978
Chapter 3, Solution 81
Clearly, v 1 = 26.67 volts, v 2 = 6.667 volts, v 3 = 173.33 volts, and v 4 = –46.67 volts
which agrees with the results of Example 3.4.
This is the netlist for this circuit.
* Schematics Netlist *
R_R1
R_R2
R_R3
R_R4
R_R5
I_I1
V_V1
E_E1
0 $N_0001 2
$N_0003 $N_0002 6
0 $N_0002 4
0 $N_0004 1
$N_0001 $N_0004 3
0 $N_0003 DC 10A
$N_0001 $N_0003 20V
$N_0002 $N_0004 $N_0001 $N_0004 3
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lOMoARcPSD|30544978
Chapter 3, Solution 82
2i 0
+ v0 –
3 k
1
2 k
3v 0
2
3
6 k
4
+
4A
8 k
4 k
100V +
–
0
This network corresponds to the Netlist.
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lOMoARcPSD|30544978
Chapter 3, Solution 83
The circuit is shown below.
1
20 V
+
20 
70 
2
50 
2A
3
30 
–
0
When the circuit is saved and simulated, we obtain v 2 = –12.5 volts
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lOMoARcPSD|30544978
Chapter 3, Solution 84
From the output loop, v 0 = 50i 0 x20x103 = 106i 0
(1)
From the input loop, 15x10-3 + 4000i 0 – v 0 /100 = 0
(2)
From (1) and (2) we get, i 0 = 2.5 A and v 0 = 2.5 volt.
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lOMoARcPSD|30544978
Chapter 3, Solution 85
The amplifier acts as a source.
Rs
+
Vs
-
RL
For maximum power transfer,
R L  R s  9
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lOMoARcPSD|30544978
Chapter 3, Solution 86
Let v 1 be the potential across the 2 k-ohm resistor with plus being on top. Then,
Since i = [(0.047–v 1 )/1k]
[(v 1 –0.047)/1k] – 400[(0.047–v 1 )/1k] + [(v 1 –0)/2k] = 0 or
401[(v 1 –0.047)] + 0.5v 1 = 0 or 401.5v 1 = 401x0.047 or
v 1 = 0.04694 volts and i = (0.047–0.04694)/1k = 60 nA
Thus,
v 0 = –5000x400x60x10-9 = –120 mV.
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lOMoARcPSD|30544978
Chapter 3, Solution 87
v 1 = 500(v s )/(500 + 2000) = v s /5
v 0 = -400(60v 1 )/(400 + 2000) = -40v 1 = -40(v s /5) = -8v s ,
Therefore, v 0 /v s = –8
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lOMoARcPSD|30544978
Chapter 3, Solution 88
Let v 1 be the potential at the top end of the 100-ohm resistor.
(v s – v 1 )/200 = v 1 /100 + (v 1 – 10-3v 0 )/2000
(1)
For the right loop, v 0 = -40i 0 (10,000) = -40(v 1 – 10-3)10,000/2000,
or, v 0 = -200v 1 + 0.2v 0 = -4x10-3v 0
(2)
Substituting (2) into (1) gives, (v s + 0.004v 1 )/2 = -0.004v 0 + (-0.004v 1 – 0.001v 0 )/20
This leads to 0.125v 0 = 10v s or (v 0 /v s ) = 10/0.125 = –80
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lOMoARcPSD|30544978
Chapter 3, Solution 89
Consider the circuit below.
C
_
15 V
|
0.7 V
+
100 k
|
+ IC
V CE
_
2.25 V
1 k
+
_
E
For the left loop, applying KVL gives
–2.25 – 0.7 + 105I B + V BE = 0 but V BE = 0.7 V means 105I B = 2.25 or
I B = 22.5 µA.
For the right loop, –V CE + 15 – I C x103 = 0. Addition ally, I C = βI B = 100x22.5x10–6 =
2.25 mA. Therefore,
V CE = 15–2.25x10–3x103 = 12.75 V.
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lOMoARcPSD|30544978
Chapter 3, Solution 90
1 k
IB
10 k
vs
+
-
i1
i2
+
V CE
+
V BE –
–
18V
500 
+
IE
–
+
-
V0
For loop 1, -v s + 10k(I B ) + V BE + I E (500) = 0 = -v s + 0.7 + 10,000I B + 500(1 + )I B
which leads to v s + 0.7 = 10,000I B + 500(151)I B = 85,500I B
But, v 0 = 500I E = 500x151I B = 4 which leads to I B = 5.298x10-5
Therefore, v s = 0.7 + 85,500I B = 5.23 volts
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lOMoARcPSD|30544978
Chapter 3, Solution 91
We first determine the Thevenin equivalent for the input circuit.
R Th = 6||2 = 6x2/8 = 1.5 k and V Th = 2(3)/(2+6) = 0.75 volts
5 k
IC
1.5 k
+
0.75 V
-
i1
IB
i2
+
V CE
+
V BE –
–
9V
400 
+
IE
–
+
-
V0
For loop 1, -0.75 + 1.5kI B + V BE + 400I E = 0 = -0.75 + 0.7 + 1500I B + 400(1 + )I B
I B = 0.05/81,900 = 0.61 A
v 0 = 400I E = 400(1 + )I B = 49 mV
For loop 2, -400I E – V CE – 5kI C + 9 = 0, but, I C = I B and I E = (1 + )I B
V CE = 9 – 5kI B – 400(1 + )I B = 9 – 0.659 = 8.641 volts
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lOMoARcPSD|30544978
Chapter 3, Solution 92
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find I B and V C for the circuit in Fig. 3.128. Let  = 100, V BE = 0.7V.
Figure 3.128
Solution
I1
5 k
10 k
VC
IB
IC
+
V CE
+
V BE –
–
4 k
IE
12V
+
V0
–
I 1 = I B + I C = (1 + )I B and I E = I B + I C = I 1
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+
-
lOMoARcPSD|30544978
Applying KVL around the outer loop,
4kI E + V BE + 10kI B + 5kI 1 = 12
12 – 0.7 = 5k(1 + )I B + 10kI B + 4k(1 + )I B = 919kI B
I B = 11.3/919k = 12.296 A
Also, 12 = 5kI 1 + V C which leads to V C = 12 – 5k(101)I B = 5.791 volts
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lOMoARcPSD|30544978
Chapter 3, Solution 93
1
4
v1
i1
24V
+
3v 0
i
2
2
2
v2 i3
+
8
i
i2
–
3v 0
4
+
+
+
v0
v1
v2
–
–
–
(a)
+
(b)
From (b), -v 1 + 2i – 3v 0 + v 2 = 0 which leads to i = (v 1 + 3v 0 – v 2 )/2
At node 1 in (a), ((24 – v 1 )/4) = (v 1 /2) + ((v 1 +3v 0 – v 2 )/2) + ((v 1 – v 2 )/1), where v 0 = v 2
or 24 = 9v 1 which leads to v 1 = 2.667 volts
At node 2, ((v 1 – v 2 )/1) + ((v 1 + 3v 0 – v 2 )/2) = (v 2 /8) + v 2 /4, v 0 = v 2
v 2 = 4v 1 = 10.66 volts
Now we can solve for the currents, i 1 = v 1 /2 = 1.333 A, i 2 = 1.333 A, and
i 3 = 2.6667 A.
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