Page 54 - Schaum's Outline of Theory and Problems of Electric Circuits

43
ANALYSIS METHODS
CHAP. 4]
the network provides a very clear picture of the overall functioning of the network in terms of voltages,
currents, and power. The reduction begins with a scan of the network to pick out series and parallel
combinations of resistors.

EXAMPLE 4.6 Obtain the total power supplied by the 60-V source and the power absorbed in each resistor in the
network of Fig. 4-8.
R ab ¼ 7 þ 5 ¼ 12
ð12Þð6Þ
R cd ¼ ¼ 4
12 þ 6

Fig. 4-8
These two equivalents are in parallel (Fig. 4-9), giving
ð4Þð12Þ
R ef ¼ ¼ 3
4 þ 12
Then this 3-
equivalent is in series with the 7-
resistor (Fig. 4-10), so that for the entire circuit,
R eq ¼ 7 þ 3 ¼ 10

Fig. 4-9
Fig. 4-10

The total power absorbed, which equals the total power supplied by the source, can now be calculated as
V 2 ð60Þ 2
P T ¼ ¼ ¼ 360 W
R eq 10
This power is divided between R ge and R ef as follows:

7 3
P ge ¼ P 7
¼ ð360Þ¼ 252 W P ef ¼ ð360Þ¼ 108 W
7 þ 3 7 þ 3
Power P ef is further divided between R cd and R ab as follows:

12 4
P cd ¼ ð108Þ¼ 81 W P ab ¼ ð108Þ¼ 27 W
4 þ 12 4 þ 12