1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook