Introduction to Electric Circuits

3.7 Norton'stheorem 53

and that

V~ = I~ r (3.13)

Figure 3.23

,so I

ISC

1

Rsc

A

lvL

To calculate Isc we short circuit the load resistor r as shown in Fig. 3.24 and

determine the current through this short circuit. We can make use of the

Figure 3.24

Vl (~V2

~R1 ~R2

R3

F ] ~,A

principle of superposition to do this. Replace the battery 1/2 by a short circuit to

give the circuit of Fig. 3.25.

Figure 3.25

R3

I I

R2

A

ILl

--B

From Fig. 3.25 we note that the resistor R~ is in series with the parallel

combination of the resistors R2 and R3 so that

I~- V~/[R~ + (R2R3/R2 + R3)I = 20/[5 + (100/20)1- 20/10 = 2 A

By current division, since R2 - R3, IL~ - 1 A.

Now reconnect the battery V~ and replace the battery V2 by a short circuit to

give the circuit of Fig. 3.26 overleaf. From Fig. 3.26 shown overleaf we see that

the resistor R2 is in series with the parallel combination of the resistors R~ and

R 3 so that

12 = V2/[R2 + (R1R3/(R1 + R3) ] = 10/[10 + 50/151 = (10/13.33)A