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Charging and Discharging capacitor problem .

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data
    The question says
    A capacitor of capacitance C is given a charge Q . at time t = 0 This capacitor is then connected to another identical capacitor which was initially uncharged through a resistance R .
    Find the charge on the second capacitor as a function of time .
    ( I am assuming that the second capacitor is the uncharged capacitor .)

    2. Relevant equations

    Charging a capacitor through resistance R and Potential difference ε :
    Q(t) = εC(1 -e^(-t/RC) )
    where e is the constant ( e = 2.78... )

    Discharging a capacitor through a resistance r
    Q(t) = Q(initial) * e^(-t/RC)
    where e is the constant ( e = 2.78... )

    The Answer to the question is :
    Q(t) = Q/2(1-e^(-2t/RC))

    3. The attempt at a solution

    Alright so the Capacitor 1 initially had charge Q so when it will be connected to another uncharged capacitor it will discharge whereas the other capacitor which was initially uncharged will get charged and the total charge will remain constant at any time t.
    So if the first capacitor is getting Discharged then
    Q(t) = Q(* e^(-t/RC) )

    And the second one is getting discharged so
    Q(t) = εC(1 -e^(-t/RC) ) -------- (i)

    Since the Potential difference ε itself is a function of time so ,
    I went to the charge on first capacitor and used
    Q = Cε
    ε = Q/C
    Since C is constant so the ε across first capacitor will also be changing as (and the 1 capcitor is discharging also)
    ε(t) = [Q * e^(-t/RC) ]/C

    so now I substitute this in (i)
    Q(t) = [Q * e^(-t/RC) ]/C * C(1 -e^(-t/RC) )
    so the charge at t on the second will be
    = [Q * e^(-t/RC) ] * (1 -e^(-t/RC) )

    but the answer is
    Q(t) = Q/2(1-e^(-2t/RC))

    Where am i wrong ?
     
    Last edited: Jun 7, 2012
  2. jcsd
  3. Jun 7, 2012 #2

    ehild

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    Homework Helper
    Gold Member

    The method you tried is not correct. You need to solve the differential equation of the circuit. Write up the voltages in terms of q, the charge on the initially empty capacitor. Their sum is zero. Solve the equation with the initial condition that q(0)=0.

    ehild
     
  4. Jun 8, 2012 #3
    I did this and got the answer but I am not sure whether its right
    so first I found the equivalent capacitance which would be for two Capacitors connected in series would be C/2
    so now the time constant (RC) would be RC/2
    also for charging the second capacitor since it was initially uncharged the final charge on it would be half of the charge on the first since they both are identical
    so final charge on second is Q/2
    Now the equation
    Q(t) = εC(1-e^-t/RC)

    Using Q = CV
    V = Q/C
    I am thinking that
    εC would be the final charge on the capacitor
    that is Q/2 so using the relation
    I found
    ε = Q/2C

    I substituted it in the Q(t) equation
    THis gives
    Q(t) = Q/2 (1-e^-2t/RC)
    Thats the answer
    is the method right ? any problems with it ?
     
  5. Jun 8, 2012 #4

    ehild

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    Homework Helper
    Gold Member

    Correct :smile:
    Right!

    You meant Q(t) = εC(1-e^(-t/RC')) where C'=C/2...And it is better to denote the charge on the second capacitor by q, as Q is used for the initial charge on the first one. q(t)=εC(1-e^(-2t/RC)). Also do not forget the parentheses!

    It is all right...

    ehild
     
    Last edited: Jun 8, 2012
  6. Jun 8, 2012 #5
    Yeah its C' . Thanks .
     
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