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!

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

    LowlyPion

    User Avatar
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?