Chapter 4, Solution 1.
1Ω i 5Ω io
+
1V 8Ω 3Ω
−
1 1
8 (5 + 3) = 4Ω , i = =
1+ 4 5
1 1
io = i= = 0.1A
2 10
Chapter 4, Solution 2.
1
6 (4 + 2) = 3Ω, i1 = i 2 = A
2
1 1
io = i1 = , v o = 2i o = 0.5V
2 4
5Ω i1 4Ω io
i2
1A 8Ω 6Ω 2Ω
If is = 1µA, then vo = 0.5µV
Chapter 4, Solution 3.
R
3R
io
3R 3R +
+ 3R 1.5R
+ R 1V
Vs vo −
−
−
(b)
(a)
(a) We transform the Y sub-circuit to the equivalent ∆ .
3R 2 3 3 3 3
R 3R = = R, R + R = R
4R 4 4 4 2
vs
vo = independent of R
2
io = vo/(R)
When vs = 1V, vo = 0.5V, io = 0.5A
(b) When vs = 10V, vo = 5V, io = 5A
(c) When vs = 10V and R = 10Ω,
vo = 5V, io = 10/(10) = 500mA
Chapter 4, Solution 4.
If Io = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω
resistor is 2A.
2A 2Ω 2Ω
1A 3A 3A i1
+
3Ω 6Ω 4Ω Is 2Ω 4Ω Is
v1
−
(a) (b)
vo
3 6 = 2Ω , vo = 3(4) = 12V, i1 = = 3A.
4
Hence Is = 3 + 3 = 6A
If Is = 6A Io = 1
Is = 9A Io = 6/(9) = 0.6667A
Chapter 4, Solution 5.
2Ω v1 3Ω vo
+
Vs 6Ω 6Ω 6Ω
−
1
If vo = 1V, V1 = + 1 = 2V
3
2 10
Vs = 2 + v1 =
3 3
10
If vs = vo = 1
3
3
Then vs = 15 vo = x15 = 4.5V
10
Chapter 4, Solution 6
R2 R3 RT
Let RT = R2 // R3 = , then Vo = Vs
R2 + R3 RT +R1
R2 R3
V RT R2 + R3 R2 R3
k= o = = =
Vs RT + R1 R2 R3 R1 R2 + R2 R3 + R3 R1
+ R1
R2 + R3
Chapter 4, Solution 7
We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the
circuit below.
3Vx
5Ω 5Ω
+
+
4V 15 Ω VTh
- 6Ω
-
+ Vx -
From the figure,
15
V x = 0, VTh = (4) = 3V
15 + 5
To find RTh, consider the circuit below:
3Vx
5Ω 5Ω
V1 V2
+
4V 15 Ω 1A
- 6Ω
+ Vx -
At node 1,
4 − V1 V V − V2
= 3V x + 1 + 1 , V x = 6 x1 = 6
→ 258 = 3V2 − 7V1 (1)
5 15 5
At node 2,
V1 − V2
1 + 3V x + =0 → V1 = V2 − 95 (2)
5
Solving (1) and (2) leads to V2 = 101.75 V
2
V V 9
RTh = 2 = 101.75Ω, p max = Th = = 22.11 mW
1 4 RTh 4 x101.75
Chapter 4, Solution 8.
Let i = i1 + i2,
where i1 and iL are due to current and voltage sources respectively.
6Ω
i2
i1
6Ω 4Ω 5A + 4Ω
20V
−
(a) (b)
6 20
i1 = (5) = 3A, i 2 = = 2A
6+4 6+4
Thus i = i1 + i2 = 3 + 2 = 5A
Chapter 4, Solution 9.
Let i x = i x1 + i x 2
where i x1 is due to 15V source and i x 2 is due to 4A source,
12 Ω i
ix1 ix2 -4A
+ 10 Ω 40Ω 12Ω 10Ω 40Ω
15V
−
(a) (b)
For ix1, consider Fig. (a).
10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75
ix1 = [40/(40 + 10)]i = (4/5)0.75 = 0.6
For ix2, consider Fig. (b).
12||40 = 480/52 = 120/13
ix2 = [(120/13)/((120/13) + 10)](-4) = -1.92
ix = 0.6 – 1.92 = -1.32 A
p = vix = ix2R = (-1.32)210 = 17.43 watts
Chapter 4, Solution 10.
Let vab = vab1 + vab2 where vab1 and vab2 are due to the 4-V and the 2-A sources
respectively.
3vab1 3vab2
10 Ω 10 Ω
+− +−
+ +
+
4V vab1 2A vab2
−
− −
(a) (b)
For vab1, consider Fig. (a). Applying KVL gives,
- vab1 – 3 vab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V
For vab2, consider Fig. (b). Applying KVL gives,
- vab2 – 3vab2 + 10x2 = 0, which leads to vab2 = 5
vab = 1 + 5 = 6 V
Chapter 4, Solution 11.
Let i = i1 + i2, where i1 is due to the 12-V source and i2 is due to the 4-A source.
6Ω io
i1
+ 2Ω 3Ω
12V
−
(a)
i2 4A ix2 4A
6Ω 2Ω 3Ω 2Ω 2Ω
(b)
For i1, consider Fig. (a).
2||3 = 2x3/5 = 6/5, io = 12/(6 + 6/5) = 10/6
i1 = [3/(2 + 3)]io = (3/5)x(10/6) = 1 A
For i2, consider Fig. (b), 6||3 = 2 ohm, i2 = 4/2 = 2 A
i = 1+2 = 3A
Chapter 4, Solution 12.
Let vo = vo1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V, and 19-V
sources respectively. For vo1, consider the circuit below.
2A
2A
5Ω 4Ω
io 5 Ω
+ vo1 −
+ vo1 −
6Ω 3Ω 12 Ω
5Ω
6||3 = 2 ohms, 4||12 = 3 ohms. Hence,
io = 2/2 = 1, vo1 = 5io = 5 V
For vo2, consider the circuit below.
6Ω 5Ω 4Ω 6Ω 5Ω
+ vo2 − + + vo2 −
+ 3Ω 12 Ω + 3Ω 3Ω
12V 12V v1
− −
−
3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5
vo2 = (5/8)v1 = (5/8)(16/5) = 2 V
For vo3, consider the circuit shown below.
5Ω 4Ω 5Ω 4Ω
+ vo3 − + vo3 − +
+ 2Ω +
6Ω 12 Ω 19V 12 Ω v2
3Ω − − 19V
−
7||12 = (84/19) ohms, v2 = [(84/19)/(4 + 84/19)]19 = 9.975
v = (-5/7)v2 = -7.125
vo = 5 + 2 – 7.125 = -125 mV
Chapter 4, Solution 13
Let
io = i1 + i2 + i3 ,
where i1, i2, and i3 are the contributions to io due to 30-V, 15-V, and 6-mA sources
respectively. For i1, consider the circuit below.
1 kΩ 2 kΩ 3 kΩ
+ i1
30V
- 4 kΩ 5 kΩ
3//5 = 15/8 = 1.875 kohm, 2 + 3//5 = 3.875 kohm, 1//3.875 = 3.875/4.875 = 0.7949
kohm. After combining the resistors except the 4-kohm resistor and transforming the
voltage source, we obtain the circuit below.
i1
30 mA
4 kΩ 0.7949 k Ω
Using current division,
0.7949
i1 = (30mA) = 4.973 mA
4.7949
For i2, consider the circuit below.
1 kΩ 2 kΩ 3 kΩ
i2 -
15V
4 kΩ 5 kΩ +
After successive source transformation and resistance combinations, we obtain the circuit
below:
2.42mA
i2
4 kΩ 0.7949 k Ω
Using current division,
0.7949
i2 = − (2.42mA) = −0.4012 mA
4.7949
For i3, consider the circuit below.
6mA
1 kΩ 2 kΩ 3 kΩ
i3
4 kΩ 5 kΩ
After successive source transformation and resistance combinations, we obtain the circuit
below:
3.097mA
i3
4 kΩ 0.7949 k Ω
0.7949
i3 = − (3.097mA) = −0.5134 mA
4.7949
Thus,
io = i1 + i2 + i3 = 4.058 mA
Chapter 4, Solution 14.
Let vo = vo1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20-V, 1-A, and 2-A
sources respectively. For vo1, consider the circuit below.
6Ω
4Ω 2Ω
+
+
vo1 3Ω
− 20V
−
6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V
For vo2, consider the circuit below.
6Ω 6Ω
4V
4Ω 2Ω 4Ω 2Ω
−+
+ +
1A vo2 3Ω vo2 3Ω
− −
3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V
For vo3, consider the circuit below.
6Ω
2A 2A
4Ω 2Ω 3Ω
+
vo3 3Ω 3Ω
− − vo3 +
6||(4 + 2) = 3, vo3 = (-1)3 = -3
vo = 10 + 1 – 3 = 8 V
Chapter 4, Solution 15.
Let i = i1 + i2 + i3, where i1 , i2 , and i3 are due to the 20-V, 2-A, and 16-V sources. For
i1, consider the circuit below.
io
+
20V 1Ω
−
i1 4Ω
2Ω 3Ω
4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 A
For i3, consider the circuit below.
+
1Ω 4Ω
2Ω i3
vo ’
− 16V
3Ω +
−
2||(1 + 3) = 4/3, vo’ = [(4/3)/((4/3) + 4)](-16) = -4
i3 = vo’/4 = -1
For i2, consider the circuit below.
1Ω 2A 1Ω 2A
2Ω (4/3)Ω
i2 4Ω i2
3Ω 3Ω
2||4 = 4/3, 3 + 4/3 = 13/3
Using the current division principle.
i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375
i = 2.5 + 0.375 - 1 = 1.875 A
p = i2R = (1.875)23 = 10.55 watts
Chapter 4, Solution 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.
io1 4Ω 3Ω 2Ω
+ 10 Ω 5Ω
12V
−
10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A
For io2, consider the circuit below. 4A
io2 3Ω 2Ω
4Ω 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.
io3 3Ω 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
Chapter 4, Solution 17.
Let vx = vx1 + vx2 + vx3, where vx1,vx2, and vx3 are due to the 90-V, 6-A, and 40-V
sources. For vx1, consider the circuit below.
30 Ω 10 Ω 20 Ω
+ −
vx1
+ 60 Ω 30 Ω
90V
−
io 10 Ω
+ −
vx1
3A 20 Ω 12 Ω
20||30 = 12 ohms, 60||30 = 20 ohms
By using current division,
io = [20/(22 + 20)]3 = 60/42, vx1 = 10io = 600/42 = 14.286 V
For vx2, consider the circuit below.
10 Ω i ’ 10 Ω i ’
o o
+ vx2 − + vx2 −
30 Ω 60 Ω 6A 30 Ω 20 Ω 20 Ω 6A 12 Ω
io’ = [12/(12 + 30)]6 = 72/42, vx2 = -10io’ = -17.143 V
For vx3, consider the circuit below.
10 Ω 10 Ω 10 Ω io”
+ − + −
vx3 vx3
30 Ω 60 Ω 30 Ω + 20 Ω 7.5Ω
40V 4A
−
io” = [12/(12 + 30)]2 = 24/42, vx3 = -10io” = -5.714
vx = 14.286 – 17.143 – 5.714 = -8.571 V
Chapter 4, Solution 18.
Let ix = i1 + i2, where i1 and i2 are due to the 10-V and 2-A sources respectively. To
obtain i1, consider the circuit below.
2Ω
10i1
i1 1Ω i1 1Ω 2Ω
+−
+ 5i1 4Ω + 4Ω
10V 10V
− −
-10 + 10i1 + 7i1 = 0, therefore i1 = (10/17) A
For i2, consider the circuit below.
10i2
i2 1Ω 2Ω i io 1Ω 2Ω
o
+− +−
10i2 4Ω + 4Ω
2A 2V
−
-2 + 10i2 + 7io = 0, but i2 + 2 = io. Hence,
-2 + 10i2 +7i2 + 14 = 0, or i2 = (-12/17) A
vx = 1xix = 1(i1 + i2) = (10/17) – (12/17) = -2/17 = -117.6 mA
Chapter 4, Solution 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 4A 8Ω v1 2Ω 6A 8Ω v2
−+ − −+ −
4ix 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 ix = v2/2 and 2ix = v2. Therefore,
v2/2 = 6 + (2v2 – v2)/8 which leads to v2 = -16
Hence, vx = –(32/3) – 16 = -26.67 V
Chapter 4, Solution 20.
Transform the voltage sources and obtain the circuit in Fig. (a). Combining the 6-ohm
and 3-ohm resistors produces a 2-ohm resistor (6||3 = 2). Combining the 2-A and 4-A
sources gives a 6-A source. This leads to the circuit shown in Fig. (b).
i
i
2A 6Ω 2Ω 3Ω 4A 2Ω 2Ω 6A
(a) (b)
From Fig. (b), i = 6/2 = 3 A
Chapter 4, Solution 21.
To get io, transform the current sources as shown in Fig. (a).
io 6Ω 3Ω
i +
+ +
6Ω 3Ω vo 2 A
− 12V − 6V 2 A
−
(a) (b)
From Fig. (a), -12 + 9io + 6 = 0, therefore io = 666.7 mA
To get vo, transform the voltage sources as shown in Fig. (b).
i = [6/(3 + 6)](2 + 2) = 8/3
vo = 3i = 8 V
Chapter 4, Solution 22.
We transform the two sources to get the circuit shown in Fig. (a).
5Ω 5Ω
−
4Ω 10Ω
+ 10V 2A
(a)
i
1A 10Ω 4Ω 10Ω 2A
(b)
We now transform only the voltage source to obtain the circuit in Fig. (b).
10||10 = 5 ohms, i = [5/(5 + 4)](2 – 1) = 5/9 = 555.5 mA
Chapter 4, Solution 23
If we transform the voltage source, we obtain the circuit below.
8Ω
10 Ω 6Ω 3Ω 5A
3A
3//6 = 2-ohm. Convert the current sources to voltages sources as shown below.
10 Ω 8Ω 2Ω
+ +
10V
30V -
-
Applying KVL to the loop gives
− 30 + 10 + I (10 + 8 + 2) = 0
→ I = 1A
p = VI = I 2 R = 8 W
Chapter 4, Solution 24
Convert the current source to voltage source.
16 Ω 1Ω
4Ω
+ 5Ω +
48 V
10 Ω Vo
- + -
12 V
-
Combine the 16-ohm and 4-ohm resistors and convert both voltages sources to current
Sources. We obtain the circuit below.
1Ω
2.4A 20 Ω 5Ω 2.4A 10 Ω
Combine the resistors and current sources.
20//5 = (20x5)/25 = 4 Ω , 2.4 + 2.4 = 4.8 A
Convert the current source to voltage source. We obtain the circuit below.
4Ω 1Ω
+ +
19.2V Vo 10 Ω
- -
Using voltage division,
10
Vo = (19.2) = 12.8 V
10 + 4 + 1
Chapter 4, Solution 25.
Transforming only the current source gives the circuit below.
9Ω 18 V
−+
+ 5Ω
12V
−
i
−
4Ω 30 V
vo +
+ −
+−
2Ω 30 V
Applying KVL to the loop gives,
(4 + 9 + 5 + 2)i – 12 – 18 – 30 – 30 = 0
20i = 90 which leads to i = 4.5
vo = 2i = 9 V
Chapter 4, Solution 26.
Transform the voltage sources to current sources. The result is shown in Fig. (a),
30||60 = 20 ohms, 30||20 = 12 ohms
10 Ω
+ vx −
3A 30Ω 60Ω 6A 30Ω 20Ω 2A
(a)
20 Ω 10 Ω 12 Ω
+ vx −
+ +
− 60V i − 96V
(b)
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