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Homework Help: Another RC circuit problem, don't understand why my answer is wrong

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data
    In the figure below, R1 = 13.0 kΩ, R2 = 18.0 kΩ, C = 0.600 µF, and the ideal battery has emf = 20.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 4.00 ms?


    2. Relevant equations

    V = IR

    [itex]V_{Capacitor}(t) = V_{0}(1-e^{-t/\tau})[/itex]

    3. The attempt at a solution

    Okay so i put a decent amount into this one but the answer I'm coming up with is wrong and I can't figure out why. Here's what I've done:

    When the capacitor is fully charged, the current through the battery is [itex]\frac{\xi}{R_{1} + R_{2}}[/itex] which is [itex]\frac{20}{13 * 10^{3} + 18 * 10^{3}}[/itex] = 6.45e-4

    The current through R2 would be the same as this, 6.45e-4, so the potential difference over R2 = IR = 6.45e-4 * 18e3 = 11.61 V

    When the switch is first opened, the potential difference across the capacitor would be the same as this, 11.61.

    The time constant for the capacitor would be RC = 18e3 * .6e-6 = 0.0108

    so to calculate the potential difference at t = 4 ms, you would use the following equation:
    [itex]V_{capacitor}(0.004) = 11.61 * (1 - e^{-0.004 / 0.0108})[/itex] which comes out to 3.59

    This is the same as the potential difference in R2, and so current through R2 should be 3.59 / 18e3 = 1.99e-4, but that is not right. Can someone let me know where I went wrong?
  2. jcsd
  3. Oct 17, 2011 #2


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    Staff: Mentor

    The potential across the capacitor will be falling, not rising. Check your equation for the potential on the capacitor!
  4. Oct 17, 2011 #3


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    Staff Emeritus
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    This equation

    [itex]\displaystyle V_\text{Capacitor}(t) = V_{0}(1-e^{-t/\tau})[/itex]

    is for the case of charging a capacitor which initially has a charge (and thus potential) of zero. How can I tell ? Look at V when t = 0 .

    What's the correct equation for discharging a capacitor which initially has a potential of V0 across its plates?
  5. Oct 17, 2011 #4
    is it just the same equation except without the "1 -" part? so V0(e^-t/TAU)?
  6. Oct 17, 2011 #5
    got it, 0.000445358666. Thanks for putting me on the right track guys
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