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 a10A source in parallel with the 25 O rcsistor, asshoqn 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 sourcesinto a single source. The pamllel resistors com-bine into a single resistor Figure 4.43 shows theIesull.Hence r), = 20 V.b) The current supplied by the 250 V source equals the current in the 125 O resistor plus the currentin 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 theDJ + 8(10) = ro = 20. ot .N.: and the polver developed by the 8 A source
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