Chapter 4, Problem 32. Use source transformation to find ix in the circuit of Fig. 4.100.

Figure 4.100

Chapter 4, Solution 32. As shown in Fig. (a), we transform the dependent current source to a voltage source,

(a)

15 Ω

40 Ω

5ix

60V

+ − 50 Ω

− + 10 Ω

ix − +

(c)

15 Ω

2.5ix 60V

+ −

25 Ωix

60V

+ −

(b)

15 Ω

50 Ω 0.1ix 50 Ω

In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c), -60 + 40ix – 2.5ix = 0, or ix = 1.6 A Chapter 4, Problem 33. Determine RTh and VTh at terminals 1-2 of each of the circuits of Fig. 4.101.

Figure 4.101

Chapter 4, Solution 33. (a) RTh = 10||40 = 400/50 = 8 ohms

VTh = (40/(40 + 10))20 = 16 V (b) RTh = 30||60 = 1800/90 = 20 ohms

2 + (30 – v1)/60 = v1/30, and v1 = VTh

120 + 30 – v1 = 2v1, or v1 = 50 V

VTh = 50 V

Chapter 4, Problem 39. Obtain the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.106.

Figure 4.106 For Prob. 4.39.

Chapter 4, Solution 39. We obtain RTh using the circuit below.

20 516 20//5 16 20 25Th

xR = + = + = Ω

5 Ω 8 V

16 Ω 10 Ω

1 A

+ _

10 Ω

a

b

10 Ω 16

5 Ω 10 Ω RTh

To find VTh, we use the circuit below.

At node 1, [(V1–8)/10] – 1 + [(V1–V2)/10] = 0 or 2V1 – V2 = 18 (1) At node 2, [(V2–V1)/10] + [(V2–0)/5] + 1 = 0 or –V1 + 3V2 = –10 (2) Adding 3(1) to (2) gives 5V1 = 44 or V1 = 8.8 V Using (2) we get 3V2 = 8.8 – 10 = –1.2 or V2 = –400 mV. Finally, VTh = V2 + (–1)(16) = –0.4 – 16 = –16.4 V \

5 8V

16 10

1 A

+ _

10 Ω

_

V2

+

V2 V1

_

+

VTh

Chapter 4, Problem 44. For the circuit in Fig. 4.111, obtain the Thevenin equivalent as seen from terminals

(a) a-b (b) b-c

Figure 4.111

Chapter 4, Solution 44. (a) For RTh, consider the circuit in Fig. (a).

RTh = 1 + 4||(3 + 2 + 5) = 3.857 ohms For VTh, consider the circuit in Fig. (b). Applying KVL gives,

10 – 24 + i(3 + 4 + 5 + 2), or i = 1

VTh = 4i = 4 V (b)

24V

+ −

VTh

+

4 Ω

1Ω3Ω

5 Ω2 Ω

10V

+ −

i

b

a

b

a

(a)

RTh 4 Ω

1Ω 3Ω

5 Ω

2 Ω

(b) For RTh, consider the circuit in Fig. (c).

RTh = 5||(2 + 3 + 4) = 3.214 ohms To get VTh, consider the circuit in Fig. (d). At the node, KCL gives,

[(24 – vo)/9] + 2 = vo/5, or vo = 15

VTh = vo = 15 V Chapter 4, Problem 47.

Obtain the Thèvenin and Norton equivalent circuits of the circuit in Fig. 4.114 with respect to terminals a and b.

Figure 4.114

b

c

(c)

RTh

4 Ω

1Ω 3Ω

5 Ω

2 Ω vo

(d)

VTh

+

4 Ω

1Ω3Ω

5 Ω

b

c

2A

2 Ω

24V

+ −

50 V

Chapter 4, Solution 47 Since VTh = Vab = Vx, we apply KCL at the node a and obtain

V 9841.1126/250VV260

V12

V50ThTh

ThTh ==⎯→⎯+=−

To find RTh, consider the circuit below.

12Ω Vx a 2Vx 60Ω

1A

At node a, KCL gives

4762.0126/601260

21 ==⎯→⎯++= xxx

x VVV

V

A167.44762.0/9841.1RV

I,4762.01

VR

Th

ThN

xTh ===Ω==

Thus, VTh = 1.9841 V, Req = RTh = RN = 476.2 mΩ, IN = 4.167 A

Chapter 4, Problem 48. Determine the Norton equivalent at terminals a-b for the circuit in Fig. 4.115.

Figure 4.115

Chapter 4, Solution 48.

4 A

4

8

To get RTh, consider the circuit in Fig. (a). From Fig. (a), Io = 1, 4 – 10 +8 – V = 0, or V = 2

RN = RTh = 2/1 = 2 ohms To get VTh, consider the circuit in Fig. (b),

Io = 4, VTh = -10Io + 8Io = –8 V

IN = VTh/RTh = -8/2 = -4A Chapter 4, Problem 51. Given the circuit in Fig. 4.117, obtain the Norton equivalent as viewed from terminals

(a) a-b (b) c-d

Figure 4.117 Chapter 4, Solution 51. (a) From the circuit in Fig. (a),

RN = 4||(2 + 6||3) = 4||4 = 2 ohms

(b)

2 Ω

4 Ω

Io

4A

+ − +

VTh

−

10Io

Io

(a)

2 Ω

4 Ω 1A

+ − +

V

−

10Io

8 8

4

4

For IN or VTh, consider the circuit in Fig. (b). After some source transformations, the circuit becomes that shown in Fig. (c). Applying KVL to the circuit in Fig. (c),

-40 + 8i + 12 = 0 which gives i = 7/2

VTh = 4i = 14 therefore IN = VTh/RN = 14/2 = 7 A

6 Ω

2 Ω

(a)

3 Ω

4 Ω

RTh

6 Ω

2 Ω

(b)

3 Ω

4 Ω

VTh

6A

+

120V

+ −

2 Ω

(c)

4 Ω

VTh+

40V

+ − 12V

+ −

2 Ω

i

(b) To get RN, consider the circuit in Fig. (d).

RN = 2||(4 + 6||3) = 2||6 = 1.5 ohms To get IN, the circuit in Fig. (c) applies except that it needs slight modification as in Fig. (e).

i = 7/2, VTh = 12 + 2i = 19, IN = VTh/RN = 19/1.5 = 12.667 A Chapter 4, Problem 53. Find the Norton equivalent at terminals a-b of the circuit in Fig. 4.119.

Figure 4.119 Chapter 4, Solution 53. To get RTh, consider the circuit in Fig. (a).

6 Ω

2 Ω

(d)

3 Ω

4 Ω

RN

(e)

VTh

+

12V

+ −

2 Ωi

From Fig. (b),

vo = 2x1 = 2V, -vab + 2x(1/2) +vo = 0

vab = 3V

RN = vab/1 = 3 ohms To get IN, consider the circuit in Fig. (c).

[(18 – vo)/6] + 0.25vo = (vo/2) + (vo/3) or vo = 4V

But, (vo/2) = 0.25vo + IN, which leads to IN = 1 A Chapter 4, Problem 59. Determine the Thevenin and Norton equivalents at terminals a-b of the circuit in Fig. 4.125.

(a)

3 Ω

+

vo

−

1A

2 Ω

6 Ω

0.25vo

b

a

0.25vo

(b)

2 Ω

+

vo

−

1A

2 Ω

b

a

+

vab

−

1/2

1/2

18V

+ −

(c)

3 Ω

+

vo

−

Isc = IN

2 Ω

0.25vo

b

a 6 Ω

Figure 4.125 Chapter 4, Solution 59. RTh = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms To find VTh, consider the circuit below.

i1 = i2 = 8/2 = 4, 10i1 + VTh – 20i2 = 0, or VTh = 20i2 –10i1 = 10i1 = 10x4

VTh = 40V, and IN = VTh/RTh = 40/22.5 = 1.7778 A Chapter 4, Problem 66. Find the maximum power that can be delivered to the resistor R in the circuit in Fig. 4.132.

Figure 4.132 Chapter 4, Solution 66.

40 Ω

i1

8A

VTh +

20 Ω

50 Ω

10 Ωi2

We first find the Thevenin equivalent at terminals a and b. We find RTh using the circuit in Fig. (a).

RTh = 2||(3 + 5) = 2||8 = 1.6 ohms By performing source transformation on the given circuit, we obatin the circuit in (b). We now use this to find VTh.

10i + 30 + 20 + 10 = 0, or i = –6

VTh + 10 + 2i = 0, or VTh = 2 V

p = VTh2/(4RTh) = (2)2/[4(1.6)] = 625 m watts

3 Ω

5 Ω

(a)

b a

RTh

2 Ω

i 20V

+ −

− +

3 Ω

5 Ω

(b)

b a

VTh

2 Ω− +

+

30V

10V

Chapter 4, Problem 73.

Determine the maximum power that can be delivered to the variable resistor R in the circuit of Fig. 4.139.

Figure 4.139

Chapter 4, Solution 73 Find the Thevenin’s equivalent circuit across the terminals of R.

10 Ω 25Ω RTh 20Ω 5Ω

Ω==+= 833.1030/3255//2520//10ThR

10 Ω 25Ω + + VTh - 60 V + + - Va Vb 20Ω 5Ω - -

10)60(305,40)60(

3020

==== ba VV

V 3010400 =−=−=⎯→⎯=++− baThbTha VVVVVV

W77.20833.104

304

22

max ===xR

Vp

Th

Th