Electric Circuits 8th Edition 143

Electric Circuits 8th Edition 143 - 4 .10 T h-;venin N...

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4.10 Th-;venin and Nofton Equivatents 119 souce, posilive at the upper terminal of the We now use a source transformation to replace the 250V source and 25 O resistor with a 10A source in parallel with the 25 O rcsistor, as shoqn in Fig.4.4).We cdn noq simpliJy lhe cir- cuit showr in Fig. 4.42 by using Kircbhoff's cur- rent law to combine the parallel curent sources into a single source. The pamllel resistors com- bine into a single resistor Figure 4.43 shows the Iesull.Hence r), = 20 V. b) The current supplied by the 250 V source equals the current in the 125 O resistor plus the current in the 25 O resistor-Thus 250 250 20 .-^ '' 125 25 "' - Therefore the power developed by the vottage /25ov(developed) : (250)01.2) = 2800 W. source, we fi$t find the voltage across the source. It we let o" represent the voltage aqoss the DJ + 8(10) = ro = 20. ot .N.: and the polver developed by the 8 A source
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This note was uploaded on 07/05/2010 for the course EE 100 taught by Professor Boser during the Spring '07 term at University of California, Berkeley.

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