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Unformatted text preview: ofthe cjrcuit shown jn Fjg.3.9. +\ 12A v figuru 3.r1 A lhe circuit of Fig. 3.10(b) showins the numedcaL 61 3() 4 0 t20 v 9f) 1\ = (12)(6)  '72v. (3.16) But t 1 is the voltage drop fmm node x to node y, so we can return to the circuit shown in Fig. 3.10(a) and again use Ohm's law to calculate tr and ir. Thus, (3.17) (3.18) 120 V 18 72 1 8 : 4 A ' : 8 A . 4f) 1)1 :72 9 9 6() We have found t}le thrce specified currents by using seriesparallel reductions in combination with Ohm's law' Belore leaving Example 3.1, we suggest that you take the tlme to show that the solution satisfies Kirchloffs curent law at every node and Kirchloff's voltage law around every closed path. (Note that there are thee closed paths that can be tested.) Showing t]Iat the power delivered by the voltage souce equals the total power dissipated in the rcsistors also is informative. fsee Prcblems 3.3 and 3.4.)...
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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.
 Spring '07
 Boser

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