Electric Circuits 8th Edition 91

Electric Circuits 8th Edition 91 - the branch con taining...

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equivalent rcsislance of the parallel-connecled ser ofresistors to the siDgle resrstance. or Rcq/&. Note that the constanl of proportionaliry in thc cul' rent division equation is rhe inverse of rhc constant of proportionalily il1 the voltage division equationl Examplc 3.4 uses voltage divislon and current division to solve for voltages and culrenls in a circuit. Using Vottage Division and Current Division to Solve a Circuit 3.4 Vottage Divisjon and Current Division Figur€ 3.20 ,", The circuitfor Examph 3.4. ,, = l!,** u, : ,av. 6l [Jse currenr division to find thc cullett i, and use voltage division to find the voltage r,, for the circuit ]n Fig.3.20. Solution We san use Eq. 3.32 if we can find the tcsislance of the four parallel branches rcsislors. Symbolically, Req = (36 + 44) l0 (40 + 10 + 30) 24 =80 108024- equivalcnl containing =60. 1 * ro * Applyhg Eq.3.32. 6 i, = t(8A) = 2A. We can usc Ohm's law to find thc \,ottage drop across the 24 () resistor:
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Unformatted text preview: the branch con taining the 40 Q, thc 10 O, and the 30 0 resistors in .<ri(s We c.n rhcl U.c \ ollxg< di\ ision ro de.cr1riir the voltage drop ,. across the 30 () resistor gi\,cn llJL $< knoq the roltrgi orup rcfo. . rl'e icric\ connected resistors, using Eq. 3.30. To do this, wc recognize that the cquivalenl resisrance of the series-connected rcsislors is 40 + 10 + 30 = 80 O: 1 t s 0 - t I lJ0 obiective 3-ge abte to use vottage and current division to sotve simple circuits 3.4 a) Use voltage division to determine the voltagc ?), across the 40 O resistor in lhe circuit shown. b) Use o,, from part (a) to determinc rhe cur-rent through the 40 J) resistor, and use this currcnt and currelt division to calcu]ate the current in the 30 O resistor. c) How much power is absorbed by lhe 50 O NOTE: Alxo ttr Chapter Problens 3.22 and 3.23. + ',o,,1 rrl ,0,, 7 0 o 4 0 0 6 0 v Answer: (a) 20 V; (b) 166.67 I1lA; (c) 347.22mw. 50(:}...
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