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Discharging a capacitor through a resistor

  1. Mar 7, 2009 #1
    In the circuit below, the time required (s) for the charge on the capacitor to drop to 47% of its value after the switch is closed is 2.83 s. If R = 1340 ohms, what is C in μF (microfarad)?

    (dq/dt) + q/(RC) = 0

    (.47q)/(t) = q/(RC)
    t/(.47R)= C
    Convert C from F to uF by multiplying by 10^6

    I am getting 4993 uF, but that is the wrong answer and I'm not sure where I'm going wrong. Thanks for any help.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Mar 7, 2009 #2


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    Homework Helper

    Welcome to PF.

    Maybe try starting with an equation that describes the discharging of a capacitor?

    Q = C*Vo*e-t/rc

    Plugging the values and dividing by Q = CVo at t=0

    .47 = e-2.83/1340*C
  4. Mar 7, 2009 #3
    Thank you, the equation worked!
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