Solution manual fundamentals of electric circuits 3rd edition chapter03

Chapter 3, Problem 1.
Determine Ix in the circuit shown in Fig. 3.50 using nodal analysis.
1 kΩ

4 kΩ
Ix

9V

+
_

2 kΩ

+
_

6V

Figure 3.50 For Prob. 3.1.

Chapter 3, Solution 1
Let Vx be the voltage at the node between 1-kΩ and 4-kΩ resistors.

9 − Vx 6 − Vx Vk
+
=
1k
4k
2k
Vx

Ix =
= 3 mA
2k

⎯⎯
→ Vx = 6

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Chapter 3, Problem 2.
For the circuit in Fig. 3.51, obtain v1 and v2.

Figure 3.51

Chapter 3, Solution 2
At node 1,

− v1 v1
v − v2
−
= 6+ 1
10
5
2

60 = - 8v1 + 5v2

(1)

At node 2,

v2
v − v2
= 3+ 6+ 1
4
2

36 = - 2v1 + 3v2

(2)

Solving (1) and (2),
v1 = 0 V, v2 = 12 V

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Chapter 3, Problem 3.
Find the currents i1 through i4 and the voltage vo in the circuit in Fig. 3.52.

Figure 3.52

Chapter 3, Solution 3
Applying KCL to the upper node,
10 =

v0 vo vo
v
+
+
+2+ 0
10 20 30
60

i1 =

v0
v
v
v
= 4 A , i2 = 0 = 2 A, i3 = 0 = 1.3333 A, i4 = 0 = 666.7 mA
10
20
30
60

v0 = 40 V

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Chapter 3, Problem 4.
Given the circuit in Fig. 3.53, calculate the currents i1 through i4.

Figure 3.53

Chapter 3, Solution 4
2A

v1
i1
4A

5Ω

i2

v2
i3

10 Ω

10 Ω

i4
5Ω

5A

At node 1,
4 + 2 = v1/(5) + v1/(10)

v1 = 20

At node 2,
5 - 2 = v2/(10) + v2/(5)

v2 = 10

i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A

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Chapter 3, Problem 5.
Obtain v0 in the circuit of Fig. 3.54.

Figure 3.54
Chapter 3, Solution 5
Apply KCL to the top node.

30 − v 0 20 − v 0 v 0
+

=
2k
5k
4k

v0 = 20 V

Chapter 3, Problem 6.
Use nodal analysis to obtain v0 in the circuit in Fig. 3.55.

Figure 3.55
Chapter 3, Solution 6
i1 + i2 + i3 = 0

v 2 − 12 v 0 v 0 − 10
+
+
=0
4
6
2

or v0 = 8.727 V
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Chapter 3, Problem 7.
Apply nodal analysis to solve for Vx in the circuit in Fig. 3.56.

+
2A

10 Ω

Vx

20 Ω

_

0.2 Vx

Figure 3.56 For Prob. 3.7.

Chapter 3, Solution 7

V − 0 Vx − 0
−2+ x
+
+ 0.2Vx = 0
10
20
0.35Vx = 2 or Vx = 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, Problem 8.
Using nodal analysis, find v0 in the circuit in Fig. 3.57.

Figure 3.57

Chapter 3, Solution 8
3Ω

i1

v1

i3

5Ω

i2
+
V0

3V
2Ω

+

–

+ 4V0
–

–
1Ω

v1 v1 − 3 v1 − 4 v 0
+
+
=0
5
1
5
2
8
v 0 = v1 so that v1 + 5v1 - 15 + v1 - v1 = 0
5
5
or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V
i1 + i2 + i3 = 0

But

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Chapter 3, Problem 9.
Determine Ib in the circuit in Fig. 3.58 using nodal analysis.
60 Ib
Ib
250 Ω
+ –

24 V

+
_

50 Ω

150 Ω

Figure 3.58 For Prob. 3.9.

Chapter 3, Solution 9
Let V1 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 V1 = 4.165 V.
Thus,

Ib = (24 – 4.165)/250 = 79.34 mA.

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Chapter 3, Problem 10.

Find i0 in the circuit in Fig. 3.59.

Figure 3.59
Chapter 3, Solution 10

3Ω

i1

v1

6Ω

i3

+ v0 –
i2
12V

+

–

+
v1

8Ω

+
–

–

2v0

At the non-reference node,

12 − v1 v1 v1 − 2v 0
=
+
3
8
6

(1)

But
-12 + v0 + v1 = 0

v0 = 12 - v1

(2)

Substituting (2) into (1),

12 − v1 v1 3v1 − 24
=
+
3
8
6

v0 = 3.652 V

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Chapter 3, Problem 11.

Find Vo and the power dissipated in all the resistors in the circuit of Fig. 3.60.
1Ω

36 V

Vo

+
_

4Ω

2Ω

–
+

12 V

Figure 3.60 For Prob. 3.11.

Chapter 3, Solution 11

At the top node, KVL gives

Vo − 36 Vo − 0 Vo − (−12)
+
+
=0
1
2
4
1.75Vo = 33 or Vo = 18.857V

P1Ω = (36–18.857)2/1 = 293.9 W
P2Ω = (Vo)2/2 = (18.857)2/2 = 177.79 W
P4Ω = (18.857+12)2/4 = 238 W.

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Chapter 3, Problem 12.

Using nodal analysis, determine Vo in the circuit in Fig. 3.61.
10 Ω

1Ω
Ix

30 V

+
_

2Ω

5Ω
4 Ix

+
Vo
_

Figure 3.61 For Prob. 3.12.

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Chapter 3, Solution 12

There are two unknown nodes, as shown in the circuit below.

10 Ω

30 V

At node 1,

+
_

1Ω

V1

Vo

2Ω

V1 − 30 V1 − 0 V1 − Vo
=0
+
+
10
2
1
16V1 − 10Vo = 30

4 Ix

5Ω

(1)

At node o,

Vo − V1
V −0
=0
− 4I x + o
1
5
− 5V1 + 6Vo − 20I x = 0
But Ix = V1/2. Substituting this in (2) leads to
–15V1 + 6Vo = 0 or V1 = 0.4Vo

(2)

(3)

Substituting (3) into 1,
16(0.4Vo) – 10Vo = 30 or Vo = –8.333 V.

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Chapter 3, Problem 13.

Calculate v1 and v2 in the circuit of Fig. 3.62 using nodal analysis.

Figure 3.62

Chapter 3, Solution 13

At node number 2, [(v2 + 2) – 0]/10 + v2/4 = 3 or v2 = 8 volts
But, I = [(v2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts

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Chapter 3, Problem 14.

Using nodal analysis, find vo in the circuit of Fig. 3.63.

Figure 3.63

Chapter 3, Solution 14

5A

v0

v1
1Ω

8Ω

2Ω
4Ω

40 V

20 V

–

+

+

–

At node 1,

v1 − v 0
40 − v 0
+5=
2
1

At node 0,

v1 − v 0
v
v + 20
+5= 0 + 0

2
4
8

v1 + v0 = 70

4v1 - 7v0 = -20

(1)

(2)

Solving (1) and (2), v0 = 27.27 V

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Chapter 3, Problem 15.

Apply nodal analysis to find io and the power dissipated in each resistor in the circuit of
Fig. 3.64.

Figure 3.64

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Chapter 3, Solution 15
5A

8Ω

v0

v1
1Ω

2Ω
4Ω

40 V

20 V

–

+

+

–

Nodes 1 and 2 form a supernode so that v1 = v2 + 10
At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2)
At node 3, 2 + 4 = 3 (v3 - v2)

(1)
2 + 6v1 + 8v2 = 3v3

v3 = v2 + 2

(2)
(3)

Substituting (1) and (3) into (2),
2 + 6v2 + 60 + 8v2 = 3v2 + 6
v1 = v2 + 10 =

v2 =

− 56
11

54
11

i0 = 6vi = 29.45 A
2

v12
⎛ 54 ⎞
P65 =

= v12 G = ⎜ ⎟ 6 = 144.6 W
R
⎝ 11 ⎠
2

⎛ − 56 ⎞
P55 = v G = ⎜
⎟ 5 = 129.6 W
⎝ 11 ⎠
2
2

P35 = (v L − v 3 ) G = (2) 2 3 = 12 W
2

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Chapter 3, Problem 16.

Determine voltages v1 through v3 in the circuit of Fig. 3.65 using nodal analysis.

Figure 3.65
Chapter 3, Solution 16
2S

v2

v1
i0
2A

+

1S

v0

4S

8S

v3
13 V

–

+

–

At the supernode,
2 = v1 + 2 (v1 - v3) + 8(v2 – v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3

(1)

But
v1 = v2 + 2v0 and v0 = v2.
Hence
v1 = 3v2
v3 = 13V

(2)
(3)

Substituting (2) and (3) with (1) gives,
v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V
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Chapter 3, Problem 17.

Using nodal analysis, find current io in the circuit of Fig. 3.66.

Figure 3.66

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Chapter 3, Solution 17
v1
i0
4Ω

2Ω
10 Ω

v2

60 V
60 V

+

8Ω

3i0

–

60 − v1 v1 v1 − v 2
=
+
4
8
2
60 − v 2 v1 − v 2
+

=0
At node 2, 3i0 +
10
2

At node 1,

But i0 =

120 = 7v1 - 4v2

(1)

60 − v1
.
4

Hence

3(60 − v1 ) 60 − v 2 v1 − v 2
+
+
=0
4
10
2

1020 = 5v1 + 12v2

Solving (1) and (2) gives v1 = 53.08 V. Hence i0 =

60 − v1
= 1.73 A
4

(2)

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Chapter 3, Problem 18.

Determine the node voltages in the circuit in Fig. 3.67 using nodal analysis.

Figure 3.67
Chapter 3, Solution 18

–+
v2

v1
2Ω

5A

v3

2Ω
8Ω

4Ω

10 V

+

+

v1

v3

–

(a)

At node 2, in Fig. (a), 5 =

At the supernode,

–

(b)

v 2 − v1 v 2 − v3
+
2

2

10 = - v1 + 2v2 - v3

v 2 − v1 v 2 − v 3 v1 v 3
+
=
+
2
2
4
8

From Fig. (b), - v1 - 10 + v3 = 0

v3 = v1 + 10

40 = 2v1 + v3

(1)

(2)
(3)

Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3

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Chapter 3, Problem 19.
Use nodal analysis to find v1, v2, and v3 in the circuit in Fig. 3.68.

Figure 3.68
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 ⎟⎜ V ⎟ ⎜ − 36 ⎟
⎝
⎠⎝ 3 ⎠ ⎝
⎠
Using MATLAB,
⎡ 10 ⎤
−1

V = A 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, Problem 20.
For the circuit in Fig. 3.69, find v1, v2, and v3 using nodal analysis.

Figure 3.69
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

.

4Ω

Between nodes 1 and 3,
− V1 + 12 + V3 = 0
⎯

⎯→

2Ω

V3

1Ω

V3 = V1 − 12

Similarly, between nodes 1 and 2,
V1 = V2 + 2i

4Ω

(2)

(3)

But i = V3 / 4 . Combining this with (2) and (3) gives

V2 = 6 + V1 / 2

(4)

Solving (1), (2), and (4) leads to
V1 = −3V, V2 = 4.5V, V3 = −15V
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Chapter 3, Problem 21.

For the circuit in Fig. 3.70, find v1 and v2 using nodal analysis.

Figure 3.70

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Chapter 3, Solution 21
4 kΩ

v1

2 kΩ

v3

3v0
+

3v0

v2

+
v0

3 mA

–

1 kΩ

+

+

+

v3

v2

–

–

(b)

(a)

Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source.

At node 1,
v − v 2 v1 − v 3
3x10 −3 = 1
+
12 = 3v1 - v2 - 2v3
(1)
4000
2000
At node 2,
v1 − v 2 v1 − v 3 v 2
+
=
4
2
1

3v1 - 5v2 - 2v3 = 0

(2)

Note that v0 = v2. We now apply KVL in Fig. (b)
- v3 - 3v2 + v2 = 0

v3 = - 2v2

(3)

From (1) to (3),
v1 = 1 V, v2 = 3 V

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Chapter 3, Problem 22.

Determine v1 and v2 in the circuit in Fig. 3.71.

Figure 3.71

Chapter 3, Solution 22

At node 1,

12 − v 0 v1
v − v0
=
+3+ 1
2
4
8

At node 2, 3 +

24 = 7v1 - v2

(1)

v 1 − v 2 v 2 + 5v 2
=
8
1

But, v1 = 12 - v1
Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V
456 = 41v1 - 9v2

(2)

Solving (1) and (2),
v1 = - 10.91 V, v2 = - 100.36 V

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