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Capacitor voltage at time = infinity

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data

    I am trying to understand how to get the voltages Va and Vc in the following circuit.
    It is assumed the circuit has been like this for a very long time.


    2. Relevant equations

    Kirchoff's voltage law

    3. The attempt at a solution

    So I know that the capacitor acts like an open circuit to DC. The current through the resistor is obviously 0, so the voltage across the resistor is also 0.

    I redraw it like this ( easier to visualize for me ).


    Now, I have the answer from my book to be that Va is 0 and Vc is 80. This allows KVL to work.

    How do you know that Va is not 80 and Vc is 0? To find the voltage between nodes "a" and "b" I usually do Voltage @ a - Voltage @ b (using correct signs). Doing this between the top of the 80V source, and the top of the resistor gives: Va = 80 - 0 = 80. If I do this at the bottom, I get Vc = 0 - (-80) = 80. Obviously this is wrong because KVL won't hold.

    Thanks for any help
  2. jcsd
  3. Dec 1, 2009 #2


    User Avatar

    Staff: Mentor

    There must be a switch opening or closing in this problem, or it makes no sense. Was a switch closed at time t=0 across that gap or something?
  4. Dec 1, 2009 #3
    Ya, thanks I think I figured it out. I tried to simplify the problem by leaving the switch out, but that lets you figure out the voltage.

    What I didn't tell you was that the switch causes Vc to be 40V before and after 0.

    Then, I just think that no current can flow through the capacitor, so 0=I=C dv/dt means dv/dt = 0. Therefore, Vc does not change as t goes to infinity. Thus it remains at 40V.

    If there is another way of thinking about it please let me know.

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