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Discharging Capacitor answer check

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data

    I have finished this and gotten an answer, but I have no solutions on hand and would like to see if anyone else gets the same answer :]

    A 2-nF capacitor with an initial charge of 5.1 microCoulombs is discharged through a 1.3- kiloOhm resistor.

    (a) Calculatue the current in the resistor 9 microseconds after the resistor is connected acorss the terminals of the capacitor.
    (b) What charge remains on the capacitor after 8 microseconds?




    2. Relevant equations

    I(t) = - Q/RC(e^(-t/RC))
    q(t) = Qe^-t/RC


    3. The attempt at a solution

    (A) Find the current through the resistor. So I differentiated equation one to get

    I = Q (e^(-t/RC)
    = 0.16 microAmperes...

    (B) Differentiate so
    q = (-Q)(-RC) (e^(-t/RC))
    = 6.11 E -11 C

    My answers seem way too small to me. So my question is, was I supposed to differentiate?? Forgive my ignorance, but what does I(t) mean? Is it the same as dI/dt? Also, if I do not differentiate the question seems exceedingly simple (it's worth 10 marks.) Is there something in my process that I am over-simplifying?

    Thanks so much to everyone in advance.
     
  2. jcsd
  3. Oct 10, 2009 #2
    I(t) stands for the current I at t seconds. and Q/RC is the initial current I(o).If you ln your equation you get
    lnI=lnI(o)-t/RC
     
  4. Oct 10, 2009 #3

    ideasrule

    User Avatar
    Homework Helper

    I don't get why it's necessary to differentiate, ln, use the logarithm rules, or do anything else. You have a formula for I:

    I = Q (e^(-t/RC)

    and the question asks for I. Just plug in the numbers. Similarly, you have a formula for q:

    q(t) = Qe^-t/RC

    and the question asks for q. Plug in the numbers.
     
  5. Oct 10, 2009 #4
    True, just plug in the numbers.
     
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