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Unformatted text preview: voltage drop across R 2 . 30V = I 2 R 2 + I 3 R 3 30V = I 2 (300 )+(0.129A)(100 ) 30V12.9V=17.1V= I 2 (300 ) I 2 =17.1V/300 =57mA Now determine the power: P=I 2 R=(57mA) 2 (300 )=0.974W c) Now consider R 3 to be shorted with a wire. How much power is now dissipated in R 2 ? I 2 =30V/300 =100mA P=I 2 R=(100mA) 2 (300 )=3.0W R V R E V / I IV P IR2 2...
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This note was uploaded on 04/17/2011 for the course PHY 7b taught by Professor Taylor during the Winter '08 term at UC Davis.
 Winter '08
 Taylor
 Current

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