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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku Chapter 2, Problem 7. Determine the number of branc
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku Chapter 2, Problem 7. Determine the number of branches and nodes in the circuit in Fig. 2.71.
Figure 2.71
Chapter 2, Solution 7 7 elements or 7 branches and 4 nodes, as indicated. 1
2A
2
20 Ω
30 Ω
30 V
60 Ω
++++ -
3
40 Ω
10 Ω
4
Chapter 2, Problem 8. Use KCL to obtain currents i1, i2, and i3 in the circuit shown in Fig. 2.72. Copyright ©2004 The McGraw-Hill Companies Inc.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Figure 2.72
Chapter 2, Solution 8
At node a, At node c, At node d,
8 = 12 + i1 9 = 8 + i2 9 = 12 + i3
i1 = - 4A i2 = 1A i3 = -3A
Chapter 2, Problem 9. Find i1, i2, and i3 in the circuit in Fig. 2.73.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Figure 2.73
Chapter 2, Solution 9 Applying KCL, i1 + 1 = 10 + 2 1 + i2 = 2 + 3 i2 = i3 + 3
i1 = 11A i2 = 4A i3 = 1A
Chapter 2, Problem 11. Determine v1 through v4 in the circuit in Fig. 2.75.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Figure 2.75
Chapter 2, Solution 11 Applying KVL to each loop gives -8 + v1 + 12 = 0 -12 - v2 + 6 = 0 10 - 6 - v3 = 0 -v4 + 8 - 10 = 0
v1 = 4v v2 = -6v v3 = 4v v4 = -2v
Chapter 2, Problem 13. For the circuit in Fig. 2.77, use KCL to find the branch currents I1 to I4.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku 2A
I2
I4
7A
3A
I1
I3
4A
Figure 2.77
Chapter 2, Solution 13 2A
1
I2
2
7A
I1
3 3A
I4
4 I3
At node 2, 3 + 7 + I2 = 0
→
I 2 = −10 A
At node 1, I1 + I 2 = 2
→
I 1 = 2 − I 2 = 12 A
At node 4, 2 = I4 + 4
→
I 4 = 2 − 4 = −2 A
At node 3, 7 + I4 = I3
→
I3 = 7 − 2 = 5 A
Hence, I 1 = 12 A,
I 2 = −10 A,
I 3 = 5 A,
I 4 = −2 A
Copyright ©2004 The McGraw-Hill Companies Inc.
4A
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 2, Problem 15. Find v1, v2, and v3 in the circuit in Fig. 2.79.
+
+
+
12 V –
– –
v1 –
v2
+ v3
8V + – + 10 V
–
Figure 2.79
Chapter 2, Solution 15
+
12V -
v1 -
3
+
v3
+
1 - 8V + 2
(a)
+
v2 -
10V +
For loop 1, 8 − 12 + v2 = 0
→
v2 = 4V
For loop 2, −v3 − 8 − 10 = 0
→
v3 = −18V
→
v1 = −6V
For loop 3, −v1 + 12 + v3 = 0
Thus, v1 = −6V ,
v2 = 4V ,
v3 = −18V
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 2, Problem 17. Obtain v1 through v3 in the circuit in Fig. 2.81.
Figure 2.81
Chapter 2, Solution 17
12V It is evident that v3 = 10V Applying KVL to loop 2,
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
v2 + v3 + 12 = 0
v2 = -22V
Applying KVL to loop 1, -24 + v1 - v2 = 0
v1 = 2V
Thus, v1 = 2V, v2 = -22V, v3 = 10V
Chapter 2, Problem 20. Determine io in the circuit of Fig. 2.84.
Figure 2.84
Chapter 2, Solution 20 Applying KVL around the loop, -36 + 4i0 + 5i0 = 0
i0 = 4A
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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Chapter 2, Problem 21. Calculate the power dissipated in the 5-Ω resistor in the circuit of Fig. 2.85.
Figure 2.85
Chapter 2, Solution 21 Apply KVL to obtain -45 + 10i - 3V0 + 5i = 0 But v0 = 10i,
-
-45 + 15i - 30i = 0
i = -3A
P3 = i2R = 9 x 5 = 45W
Chapter 2, Problem 22.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku Find Vo in the circuit in Fig. 2.86 and the power dissipated by the controlled source.
Figure 2.86
Chapter 2, Solution 22 4Ω
At the node, KCL requires that v0 + 10 + 2 v 0 = 0 4
v0 = –4.444V
The current through the controlled source is i = 2V0 = -8.888A and the voltage across it is v = (6 + 4) i0 = 10
v0 = −11.111 4
Hence, p2 vi = (-8.888)(-11.111) = 98.75 W
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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Chapter 2, Problem 23. In the circuit shown in Fig. 2.87, determine vx and the power absorbed by the 12Ω resistor. 1Ω
1.2 Ω
+v – x
4Ω 8Ω
2Ω
6A
12 Ω
6Ω
3Ω
Figure 2.87
Chapter 2, Solution 23 8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3 The circuit is reduced to that shown below. ix
1Ω +
6A
vx
2Ω
3Ω
Applying current division, ix =
2 (6 A) = 2 A, 2 + 1+ 3
v x = 1i x = 2V
The current through the 1.2- Ω resistor is 0.5ix = 1A. The voltage across the 12- Ω resistor is 1 x 4.8 = 4.8 V. Hence the power is p=
v 2 4.8 2 = = 1.92W R 12
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 2, Problem 28. Find v1, v2, and v3 in the circuit in Fig. 2.92.
Figure 2.92.
Chapter 2, Solution 28 We first combine the two resistors in parallel
15 10 = 6 Ω We now apply voltage division, v1 =
14 (40) = 20 V 14 + 6
v2 = v3 = Hence,
6 (40) = 12 V 14 + 6
v1 = 28 V, v2 = 12 V, vs = 12 V
Chapter 2, Problem 31. In the circuit in Fig. 2.95, find v, i, and the power absorbed by the 4-Ω resistor. Copyright ©2004 The McGraw-Hill Companies Inc.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Figure 2.95
Chapter 2, Solution 31 The 5 Ω resistor is in series with the combination of 10 (4 + 6) = 5Ω . Hence by the voltage division principle, v=
5 (20V ) = 10 V 5+5
by ohm's law,
i=
v 10 = = 1A 4 + 6 4+ 6
pp = i2R = (1)2(4) = 4 W
Chapter 2, Problem 34. Determine i1, i2, v1, and v2 in the ladder network in Fig. 2.98. Calculate the power dissipated in the 2-Ω resistor.
Figure 2.98
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 2, Solution 34 By parallel and series combinations, the circuit is reduced to the one below:
8Ω
10 x15 10 ( 2 + 13 ) = = 6Ω 25 15 x15 15 (4 + 6) = = 6Ω 25 12 (6 + 6) = 6Ω Thus
i1 =
6Ω
-
28 = 2 A and v1 = 6i1 = 12 V 8+6
We now work backward to get i2 and v2. 1A 1A
+ 6V -
+ 12V i1 = 2A
8Ω
6Ω
1A
0.6A
4Ω
1A
28V
Thus, v2 =
+ 12V -
+
-
12 Ω
+ 6V -
15 Ω
+
3.6V
v 13 (3 ⋅ 6) = 3 ⋅ 12, i2 = 2 = 0.24 15 13
p2 = i2R = (0.24)2 (2) = 0.1152 W i1 = 2 A, i2 = 0.24 A, v1 = 12 V, v2 = 3.12 V, p2 = 0.1152 W
Chapter 2, Problem 35. Calculate Vo and Io in the circuit of Fig. 2.99. Copyright ©2004 The McGraw-Hill Companies Inc.
-
6Ω
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Figure 2.99
Chapter 2, Solution 35 + V1 -
50V
20 Ω
I0 +
i2
V0 5 Ω -
Combining the versions in parallel,
70 30 =
i=
70 x 30 = 21Ω , 100
20 15 =
20 x 5 =4 Ω 25
50 =2 A 21 + 4
vi = 21i = 42 V, v0 = 4i = 8 V v v i1 = 1 = 0.6 A, i2 = 2 = 0.4 A 70 20 At node a, KCL must be satisfied i1 = i2 + I0
0.6 = 0.4 + I0
I0 = 0.2 A
Hence v0 = 8 V and I0 = 0.2A
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
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Chapter 2, Problem 37. In the circuit of Fig. 2.101, find R if Vo = 4 V.
Figure 2.101
Chapter 2, Solution 37 Let I = current through the 16Ω resistor. If 4 V is the voltage drop across the 6 R combination, then 20 - 4 = 16 V in the voltage drop across the 16Ω resistor. 16 Hence, I = = 1 A. 16 20 6R But I = =1 4= 6R= R = 12 Ω 6+R 16 + 6 R
Chapter 2, Problem 39. Find the equivalent resistance at terminals a-b for each of the networks in Fig. 2.103.
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Figure 2.103
Chapter 2, Solution 39 (a)
Req = R 0 = 0
R R + = R 2 2 Req = (R + R ) (R + R ) = 2R 2R = R
Req = R R + R R =
1 Req = 3R (R + R R ) = 3R (R + R ) 2 3 3Rx R 2 =R = 3 3R + R 2 R ⋅ 2R Req = R 2R 3R = 3R 3R
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
= 3R
2 R= 3
2 3Rx R 3 = 6R 2 11 3R + R 3
Chapter 2, Problem 41. If Req = 50 Ω in the circuit in Fig. 2.105, find R.
Figure 2.105
Chapter 2, Solution 41 Let R0 = combination of three 12Ω resistors in parallel 1 1 1 1 = + + R o 12 12 12
Ro = 4
R eq = 30 + 60 (10 + R 0 + R ) = 30 + 60 (14 + R ) 50 = 30 +
60(14 + R ) 74 + R
74 + R = 42 + 3R
or R = 16 Ω
Chapter 2, Problem 43. Calculate the equivalent resistance Rab at terminals a-b for each of the circuits in Copyright ©2004 The McGraw-Hill Companies Inc.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku Fig. 2.107.
Figure 2.107
Chapter 2, Solution 43 (a)
Rab = 5 20 + 10 40 =
60 20 30 =
5x 20 400 + = 4 + 8 = 12 Ω 25 50
1 1 1 + + 60 20 30
Rab = 80 (10 + 10) =
−1
=
60 = 10Ω 6
80 + 20 = 16 Ω 100
Chapter 2, Problem 45. Find the equivalent resistance at terminals a-b of each circuit in Fig. 2.109.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku 10 Ω 40 Ω 20 Ω a 30 Ω
5Ω
50 Ω b (a)
30 Ω
12 Ω 5Ω
20 Ω
60 Ω
25 Ω 10 Ω
15 Ω (b)
Figure 2.109
Chapter 2, Solution 45 (b) 10//40 = 8, 20//30 = 12, 8//12 = 4.8
Rab = 5 + 50 + 4.8 = 59.8 Ω (c) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus Rab = 5 + 12.8 + 15 = 32.5Ω
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 2, Problem 46. Find Req at terminals a-b for each of the circuits in Fig. 2.110.
Figure 2.110
Chapter 2, Solution 46 Rab = 30 70 + 40 + 60 20 =
30 x 70 60 + 20 + 40 + 100 80
= 21 + 40 + 15 = 76 Ω Copyright ©2004 The McGraw-Hill Companies Inc.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
22
The 10-Ω, 50-Ω, 70-Ω, and 80-Ω resistors are shorted.
20 30 =
20 x 30 = 12Ω 50
40 60 =
40 x 60 = 24 100
Rab = 8 + 12 + 24 + 6 + 0 + 4 = 54 Ω
Chapter 2, Problem 56. Determine V in the circuit of Fig. 1.120.
Figure 2.120
Chapter 2, Solution 56 We need to find Req and apply voltage division. We first tranform the Y network to ∆ . 30 Ω 16 Ω
30 Ω 16 Ω
10 Ω
+ 100 V
+ 100 V
-
a
35 Ω
-
Req Rab =
15x10 + 10 x12 + 12 x15 450 = = 37.5Ω 12 12
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37.5 Ω 30 Ω 45 Ω
c
b 20 Ω
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku Rac = 450/(10) = 45Ω, Rbc = 450/(15) = 30Ω Combining the resistors in parallel, 30||20 = (600/50) = 12 Ω, 37.5||30 = (37.5x30/67.5) = 16.667 Ω 35||45 = (35x45/80) = 19.688 Ω Req = 19.688||(12 + 16.667) = 11.672Ω By voltage division, v =
11.672 100 = 42.18 V 11.672 + 16
Chapter 2, Problem 58. The lightbulb in Fig. 2.122 is rated 120 V, 0.75 A. Calculate Vs to make the lightbulb operate at the rated conditions.
Figure 2.122
Chapter 2, Solution 58 The resistor of the bulb is 120/(0.75) = 160Ω 40 Ω
2.25 A
+ 90 V - 0.75 A VS
+
-
160 Ω
1.5 A
+
120 V
-
80 Ω
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Once the 160Ω and 80Ω resistors are in parallel, they have the same voltage 120V. Hence the current through the 40Ω resistor is 40(0.75 + 1.5) = 2.25 x 40 = 90 Thus vs = 90 + 120 = 210 V
Chapter 2, Problem 68. Find the current I in the circuit of Fig. 2.128(a). An ammeter with an internal resistance of 1 Ω is inserted in the network to measure I’ as shown in Fig. 2.128 (b). What is I’? Calculate the percent error introduced by the meter as I −I' ×100% I
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku Figure 2.128
Chapter 2, Solution 68 (a)
40 = 24 60Ω 4 = 0.1 A 16 + 24 4 i' = = 0.09756 A 16 + 1 + 24 0.1 − 0.09756 % error = x100% = 2.44% 0 .1
i=
Chapter 2, Problem 69. A voltmeter is used to measure Vo in the circuit in Fig. 2.129. The voltmeter model consists of an ideal voltmeter in parallel with a 100-kΩ resistor. Let Vs = 40 V, Rs = 10 kΩ, and R1 = 20 kΩ. Calculate Vo with and without the voltmeter when (a) R2 = 1 kΩ (b) R2 = 10 kΩ (c) R2 = 100 kΩ
Figure 2.129
Chapter 2, Solution 69
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku With the voltmeter in place, R2 Rm V0 = VS R1 + R S + R 2 R m where Rm = 100 kΩ without the voltmeter, R2 V0 = VS R1 + R 2 + R S When R2 = 1 kΩ, R m R 2 =
100 kΩ 101
100 V0 = 101 (40) = 1.278 V (with) 100 101 + 30 1 V0 = (40) = 1.29 V (without) 1 + 30 1000 When R2 = 10 kΩ, R 2 R m = = 9.091kΩ 110 9.091 V0 = (40) = 9.30 V (with) 9.091 + 30 10 V0 = (40) = 10 V (without) 10 + 30 When R2 = 100 kΩ, R 2 R m = 50kΩ 50 (40) = 25 V (with) 50 + 30 100 V0 = (40) = 30.77 V (without) 100 + 30 V0 =
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