Electric Circuits 8th Edition 99

Electric Circuits 8th Edition 99 - se es-parallel...

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Applying a Delta-to-Wye Transform Find the current and powei supplied by the 40 V souce in the circuit shown in Fig.3.32. Figure 3.32 A lhe circuit for Exampte 3.7. Solution We are interested oDly in the current and power drain on the 40 V source, so the problem has been solved once we obtain the equivalent rcsistance across ttre terminals of the source. We can find this equivalenl resistance easily after replacing eilher the upper A (100, 125,25 O) or the lower A (40, 25, 37.5 O) with its equivalent Y We choose to replace the upper A- We then compute the three Y resistances, defined in Fig. 3.33, from Eqs. 3.44 to 3.46.Thus, 3.7 Detta-to-Wye (Pi-ro-TeF) Equivdlert (ncrirs 75
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Unformatted text preview: se es-parallel simplifications: (501450) R," : 55 + -_- = 80 ().-' I ll(, The final step is to note that the circuit reduces to an 80 f) resistor across a 40 V souice, as shown ln Fig. 3.35, from which it is apparent t]lat the 40 V source delivers 0-5 A and 20 W to the circuit. Figure 3.33 A The equivalentY rcsistor. 37.5 100 x 125 =50r), 1 2 5 2 5 - ^ _ ^ 250 100 25 .^ ^ 250 Flgure 3.34 t A trandomd version ofthe cncuit shown in Fjg. 3.32. Flgure 3.35 ^ Thefinatstep jn the simptification of the cjrdrit shown jn Fig.3.32. 250 Substituting the Y-rcsistoN into t]le circuit shown iD Fig. l.J2 produce( the ciJcuit sholn in Fig.3.34- From Fig.3.34,we can easily calculate the ,J'...
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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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