0195136047.pdf

68 CIRCUIT ANALYSIS TECHNIQUES

a

b

24 Ω

48 Ω

6 A

(a)

96 V

I =?

+

−

R = 16 Ω

a

Voc

b

IL

48 Ω 24 Ω

(b)

96 V 144 V

+

−

+

+

−

−

a

b

48 Ω 24 Ω

(c)

a

b

16 Ω

(d)

R = 16 Ω

I

128 V

+

−

Isc

a

b

48 Ω 24 Ω

(e)

2A 6A

+

−

a

b

48 Ω 24 Ω

(f)

Figure E2.1.1

a

b

I

8 A 16 Ω R = 16 Ω

(g)

Solution

The 6-A source with 24in parallel can be replaced by a voltage source of 6× 24 =144 V with

24 in series. Thus, by using source transformation, in terms of voltage sources, the equivalent

circuit to the left of terminalsa–bis shown in Figure E2.1.1(b).