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Solving an RC Circuit Using Differential Equations

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data
    hwAvh.png

    3. The attempt at a solution
    I tried two different methods and came to an odd conclusion:
    qEbub.png
     
  2. jcsd
  3. Nov 4, 2012 #2
    The situation begins with the switch closed.
    Before the switch is opened, the capacitor will charge up to some voltage, driven by both power supplies. You can calculate what that is.

    When the switch opens, it removes one of the voltage sources so that the capacitor begins to discharge into the resistor network. A dropping current will start to flow and you are asked to find what that current is 0.1 seconds after the switch is opened.
     
  4. Nov 4, 2012 #3
    In method 1, Voc is not zero. No current flows so Voc = 12. If Voc were zero, you would be claiming the open circuit behaved like a short circuit and there would be a path for current to flow through Voc to the resistors.

    In method 2, you are mixing up the conditions at t=0- and t=0+. At t=0-, the switch is closed and the capacitor is open (no current flows through it). At t=0+, the switch is open but the capacitor is in there because current will flow through it. The connection between t=0- and t=0+ is that currents and voltages cannot instantly change so the initial condition prior to the switch opening will be the same as after it opens.
     
    Last edited: Nov 4, 2012
  5. Nov 4, 2012 #4
    Wait, why is Voc = 12? If there's no current through the three resistors on the left loop then the voltage would be zero wouldn't it?
     
  6. Nov 4, 2012 #5
    No, there is no current in the three resistors so there is no IR drop either so the voltage at the -ve terminal of the battery is the same as at the -ve side of Voc. You can make the same argument for R5 and come to Voc= 12.
     
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