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Capacitor in circuit

  1. Feb 18, 2008 #1
    1. The problem statement, all variables and given/known data
    V = 115 V, R = 30 , and the switch has been closed for a very long time.


    2. Relevant equations


    3. The attempt at a solution

    I thought it would be a good idea to use the Loop Laws I get the following equations (If Loop 1 is the larger loop with the battery, and Loop 2 is the smaller one):

    L1: 115-60I1-30I3=0
    L2: 30I3-10I2-Q/2.0microF=0

    Where I1 is the current through the battery and 60 ohm resistor; I2 is the current through the 10 ohm resistor and capacitor; I3 is the current through the middle resistor.

    However, with this there are 3 equations and 4 variables, so something is wrong, or I'm approaching it wrong. Thanks for any help.
  2. jcsd
  3. Feb 18, 2008 #2


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    Homework Helper

    First of all you should note that if the switch has been closed for an infinite period of time and you're using a DC source, then the current flowing through the right loop containing the 10 ohm resistor and the capacitor is 0A. You need to solve it for the case where t<0 and t>0 separately.
  4. Feb 18, 2008 #3
    Sorry, I forgot one part of the problem. I'm supposed to solve for what the charge on the capacitor is after the switch has been closed for a very long time...and then solve for how much time it takes for the capacitor to lose 10% of it's charge when the switch is opened.
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