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Capacitor circuit, time for current to drop

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    the capacitor is originally charged to 100 volts. the switch is closed at t=0. how long does it take for the current to drop to 0.068 amps?

    circuit drawing attached

    2. Relevant equations

    capacitance C = Q/V where Q is charge, V is electric potential

    V = IR where I is current, R is resistance

    I = dW/dt where W is work, t is time

    I = epsilon/ R_1 + R_2 where epsilon is battery emf, R is resistance

    QV/t = I^2R where t is time ---> t = QV/I^2R
    3. The attempt at a solution

    right now, i am trying to figure out the equation to use, i have a feeling the capacitance formula does not come into play. but there is a capacitance value given in the diagram of 100*10^-6 farads, should i use C = Q/V to find charge Q?

    Q = CV = 100*10^-6 (100) = 0.01 coulombs

    the variables i do have are R and V, so i can find I using I = V/R = 100/200 = 0.5 ampere

    t = QV/I^2R ---> should I be the change in I (0.5 - 0.068 = 0.432)?

    t = (0.01*100)/(0.068^2(200)) = 1/0.9248 = 1.08 seconds

    did i do it correctly?
     

    Attached Files:

    Last edited: Jun 11, 2008
  2. jcsd
  3. Jun 11, 2008 #2

    rl.bhat

    User Avatar
    Homework Helper

    In RC circuit current increases exponentially. The instantaneous current is gives by
    I = Io*e^-t/RC where Io = V/R
     
  4. Jun 11, 2008 #3
    okay so use that equation i got this:

    0.068 = (100/200)e^(-t/(200(100*10^-6)))
    0.136 = e^(-t/0.02)
    ln(0.136) = -t/0.02
    -t = -0.039 = 0.04 seconds

    does that make sense?
     
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