1. Not finding help here? Sign up for a free 30min 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!

Maximum current IN the resistor

  1. Dec 4, 2006 #1
    Here's what I know:
    A 1.89 nF capacitor with an initial charge of 5.26 mC is discharged through a 1.34 kW resistor.

    I'm supposed to find the maximum current in the resistor.

    I know the maximum current through the resistor is -2.077 A from the equation I(t) = -(Q/RC)*e^(-t/RC) and I know how to find the terminal voltage - E - Ir, but I'm at a loss on how to find the maximum current in the resistor. Any help would be awesome.

    Thanks a lot.
     
  2. jcsd
  3. Dec 4, 2006 #2
    The maximum current is right in front of your eyes!

    You either solved or were given the equation that I(t) = -(Q/RC)*e^(-t/RC) for the capacitor. Think about how this solution will look qualitatively - that is, graph it. I didn't plug the numbers in to verify, but it looks like you figured the capacitors max current. What does Kirchoff's current law tell you (you'll have to figure out the passive sign convention for this i.e. what way the cap is oriented)?
     
  4. Dec 4, 2006 #3
    When t = 0 it is the highest (or lowest since it's negative) because e^0=1 so it's just -Q/RC

    As t approaches infinity e^(-t/RC) will equal 0, so the extremes will be 0 and -Q/RC, right?

    Is there a difference between that current and the current that's actually in the resistor?

    Thanks.
     
  5. Dec 4, 2006 #4
    Right, the max current value will be the initial value.

    If the capacitor and resistor are in series, which they would have to be for that equation, then the magnitude of the current is the same because Kirchoff's current law will tell you that the current coming out of the cap must be the current into the resistor.
     
  6. Dec 4, 2006 #5
    Great.

    Yeah, it was the magnitude not the actual current.

    Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?