0195136047.pdf

70 CIRCUIT ANALYSIS TECHNIQUES

a

b

I

200 Ω 9 I

2000 Ω

+

−

(a)

I 1

10 V

Figure E2.1.2

a

b

Isc

I

200 Ω 9 I

2000 Ω

+

−

(b)

I 1

10 V

a

b

RTh = 10 Ω

+

−

+

−

(c)

0.5 V

Solution

First, the open-circuit voltage at terminalsa–bis to be found.

KCL at nodea: I+ 9 I=I 1 ,orI 1 = 10 I

KVL for the left-hand mesh: 2000 I+ 200 I 1 = 10 ,or 4000I= 10 ,orI= 1 /400 A

Voc= 200 I 1 = 200 ( 1 / 400 )= 0 .5V

Because of the presence of a dependent source, in order to findRTh, one needs to determineIsc

after shorting terminalsa–b,as shown in Figure E2.1.2(b).

Note thatI 1 =0, sinceVab=0.

KCL at nodea: Isc= 9 I+I= 10 I

KVL for the outer loop: 2000 I= 10 ,orI= 1 /200 A

Isc= 10 ( 1 / 200 )= 1 /20 A

Hence the equivalent Thévenin resistanceRThviewed from terminalsa–bis

RTh=

Voc

Isc

=

0. 5

1 / 20

= 10 

Thus, the Thévenin equivalent is given in Figure E2.1.2(c).

The preceding examples illustrate how a complex network could be reduced to a simple

representation at an output port. The effect of load on the terminal behavior or the effect of an

output load on the network can easily be evaluated. Thévenin and Norton equivalent circuits

help us in matching, for example, the speakers to the amplifier output in a stereo system. Such

equivalent circuit concepts permit us to represent the entire system (generation and distribution)