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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This note was uploaded on 07/05/2010 for the course EE 100 taught by Professor Boser during the Spring '07 term at Berkeley.
- Spring '07