Chapter 2, Solution 2 p = v2/R → R = v2/p = 14400/60 = 240 ohms Chapter 2, Solution 4

(a) i = 15/100 = 150 mA (b) i = 15/150 = 100 mA

Solution 7 (a) 1 Ω 2 Ω 4 V 5 Ω 3 Ω 10 Ω 6 branches and 5 nodes. (b) 12V 2 Ω 1 A 4 A 1Ω 3 Ω 5 Ω 7 branches and 5 nodes.

+ _

- +

Chapter 2, Problem 9. Find 1 2 3, i , and ii in Fig. 2.73.

Figure 2.73 For Prob. 2.9. Use KCL at node A to find i1: i1 – 2 – 12 = 0; i1 - 14 = 0; i1 = 14A Use KCL at node B to find i2: -i2 + 12 – 14 = 0; -i2 - 2 = 0; i2 = -2A Use KCL at node C to find i1: -i3 – 4 +14 = 0; -i3 +10 = 0; i3 = 10A Chapter 2, Problem 12. In the circuit in Fig. 2.76, obtain v1, v2, and v3.

Use KVL to find v1: -20 -25 + 10 + v1 = 0; -35 + v1 = 0; v1 = 35V Use KVL to find v2: -10 + 15 – v2 = 0; 5 – v2 = 0; v2 = 5V Use KVL to find v3: -20 -25 + 15 + v3 = 0; -30 + v3 = 0; v3 = 30V

4 A

8 A

i3 A

12 A

14 A

B

C

i1

10 A

2 A i2

A

B C

Chapter 2, Problem 15. Calculate v and ix in the circuit of Fig. 2.79.

Figure 2.79 For Prob. 2.15. Solution: Use Ohms Law to find v: -12 + v + 2 = 0, v = 10V Use KVL to find ix: -2 +8 + 3ix = 0; 6 + 3ix = 0; 3ix = -6; ix = -2 A Chapter 2, Problem 18. Find I and Vab in the circuit of Fig. 2.79.

Solution: Use KVL to Find I: -30 + V1 -10 + V2 + 8 = 0; V1 = 3I; V2 = 5I; -30 + 3I -10 + 5I + 8 = 0; -32 + 8I = 0; 8I = 32; I = 4A

Use KVL to find Vab: -Vab + V2 + 8 = 0; -Vab + 5I + 8 = 0; -Vab + 5*4 + 8 = 0; -Vab + 20 + 8 = 0; -Vab + 28 = 0; Vab = 28V;

+

12 V 3 ix

+ 2 V

_

+

v –

8 V – 12 Ω

+ _

ix

+ _

V0 i1

+V1- +V2-

Chapter 2, Solution 20

Applying KVL around the loop,

-36 + 4i0 + 5i0 = 0 i0 = 4A Chapter 2, Problem 21. Find Vx in the circuit of Fig. 2.85.

Figure 2.85 For Prob. 2.21. Solution: Use KVL to find I: -15 + 1*I + 2Vx + Vx + 2*I = 0; -15 + I + 3Vx + 2I = 0; -15 + 3I + 3Vx = 0; (Eqn 1) By Ohms Law: Vx = 5*I; Substitute for Vx in eqn 1: -15 + 3I + 3(5*I) = 0; -15 + 3I + 15I = 0; 18I = 15;

1 Ω

2 Ω

+ –

5 Ω + _ 15 V Vx

2 Vx

_

+

I

I = 15/18 = 5/6; Use Ohms Law to find Vx: Vx = 5*I = 5*(5/6) = 25/6V; Vx = 25/6V Problem 22: Chapter 2, Solution 22 Assign I0. I0=v0/4. At the node, KCL requires that

00 v210

4v

++ = 0 v0 = –4.444V

The current through the controlled source is i = 2V0 = -8.888A and the voltage across it is

v = (6 + 4) i0 (where i0 = v0/4) = 10 111.114

v0 −=

Hence, p2 = v i = (-8.888)(-11.111) = 98.75 W Chapter 2, Problem 25:

6 Ω 2v0

+ v0 -

10A

4 Ω I0

Solution:

V0 = 5 x 10-3 x 10 x 103 = 50V

Using current division,

I20 5 (0.01 50)

5 20= × =

+ 0.1 A

V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW Chapter 2, Problem 26. For the circuit in Fig. 2.90, io =2 A. Calculate ix and the total power dissipated by the circuit.

Figure 2.90 For Prob. 2.26. Solution: By Ohms Law: V = 2*i0 = 2 * 16 = 32V By Ohms Law: i1 = V/8 = 32/8 = 4A i2 = V/4 = 32/4 = 8A i3 = V/2 = 32/2 = 16A By KCL: ix = i0 + i1 + i2 + i3; ix = 2 + 4 + 8 + 16; ix = 30A Total Power: Pt = V*ix = 32 * 30 = 960W

2 Ω 4 Ω 8 Ω

io

16 Ω

ix

i3 i2 i1

+ V -