CHAP. 4]

ANALYSIS METHODS

Req ¼ 47 þ

45

ð27Þð4 þ 23Þ ¼ 60:5  54

200 ¼ 3:31 A 60:5   27 ð3:31Þ ¼ 1:65 A ¼ 54

IT ¼ 0 I23

When the 20-A source acts alone, the 200-V source is replaced by a short circuit, Fig. 4-11(c). resistance to the left of the source is

The equivalent

ð27Þð47Þ ¼ 21:15  Req ¼ 4 þ 74   21:15 00 ð20Þ ¼ 9:58 A ¼ I23 21:15 þ 23

Then The total current in the 23- resistor is

0 00 I23 ¼ I23 þ I23 ¼ 11:23 A

4.9

THE´VENIN’S AND NORTON’S THEOREMS

A linear, active, resistive network which contains one or more voltage or current sources can be replaced by a single voltage source and a series resistance (The´venin’s theorem), or by a single current source and a parallel resistance (Norton’s theorem). The voltage is called the The´venin equivalent voltage, V 0 , and the current the Norton equivalent current, I 0 . The two resistances are the same, R 0 . When terminals ab in Fig. 4-12(a) are open-circuited, a voltage will appear between them.

Fig. 4-12

From Fig. 4-12(b) it is evident that this must be the voltage V 0 of the The´venin equivalent circuit. If a short circuit is applied to the terminals, as suggested by the dashed line in Fig. 4-12(a), a current will result. From Fig. 4-12(c) it is evident that this current must be I 0 of the Norton equivalent circuit. Now, if the circuits in (b) and (c) are equivalents of the same active network, they are equivalent to each other. It follows that I 0 ¼ V 0 =R 0 . If both V 0 and I 0 have been determined from the active network, then R 0 ¼ V 0 =I 0 . EXAMPLE 4.8 Obtain the The´venin and Norton equivalent circuits for the active network in Fig. 4-13(a). With terminals ab open, the two sources drive a clockwise current through the 3- and 6- resistors [Fig. 4-13(b)]. I¼

20 þ 10 30 ¼ A 3þ6 9

Since no current passes through the upper right 3- resistor, the The´venin voltage can be taken from either active branch:

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An