Page 66 - Schaum's Outline of Theory and Problems of Electric Circuits

ANALYSIS METHODS
CHAP. 4]

Fig. 4-29 55

Fig. 4-30

54 27 I 1 ¼ 460
27 74 I 2 ¼ 200

From the 460-V source,

ð 460Þð74Þ
0
0
I 1 ¼ I ¼ ¼ 10:42 A
3267
and for the 200-V source

00 00 ð200Þð 27Þ
I 1 ¼ I ¼ ¼ 1:65 A
3267
00
0
Then, I ¼ I þ I ¼ 10:42 þ 1:65 ¼ 8:77 A

Fig. 4-31(a)

4.15 Obtain the current in each resistor in Fig. 4-31(a), using network reduction methods.
As a first step, two-resistor parallel combinations are converted to their equivalents. For the 6
and
3
, R eq ¼ð6Þð3Þ=ð6 þ 3Þ¼ 2
. For the two 4-
resistors, R eq ¼ 2
. The circuit is redrawn with series
resistors added [Fig. 4-31(b)]. Now the two 6-
resistors in parallel have the equivalent R eq ¼ 3
, and this
is in series with the 2
. Hence, R T ¼ 5
, as shown in Fig. 4-31(c). The resulting total current is