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CHAP. 4]

43

ANALYSIS METHODS

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. Rab ¼ 7 þ 5 ¼ 12  ð12Þð6Þ Rcd ¼ ¼ 4 12 þ 6

Fig. 4-8 These two equivalents are in parallel (Fig. 4-9), giving Ref ¼

ð4Þð12Þ 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, Req 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 PT ¼

V2 ð60Þ2 ¼ 360 W ¼ Req 10

This power is divided between Rge and Ref as follows: Pge ¼ P7 ¼

7 ð360Þ ¼ 252 W 7þ3

Pef ¼

3 ð360Þ ¼ 108 W 7þ3

Power Pef is further divided between Rcd and Rab as follows: Pcd ¼

12 ð108Þ ¼ 81 W 4 þ 12

Pab ¼

4 ð108Þ ¼ 27 W 4 þ 12

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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