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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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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