Q3_7B_W10_Solution

Q3_7B_W10_Solution - voltage drop across R 2 . 30V = I 2 R...

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Name: , DL Section: Last First Office Use Quiz 3 Physics 7B Winter 2010 Jan. 27, 2010 1. Consider the circuit shown in the diagram above. a) The switch is initially open. What is the magnitude and direction of the current flowing through resistor 2? 30V – 20V = I(300 ) + I(100 ) 10V = I(400 ) I = 10V/400 I = 25 mA b) The switch is closed, and a current of 129 mA is measured in the ammeter. How much power is dissipated in resistor 2? First we need to determine the current in resistor 2. To do that, let’s first find the
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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 ) 30V-12.9V=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 IR-2 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.

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