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Discharging a capacitor - how long does it take

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data

    A capacitor stores 1.09 C of charge.
    The capacitor has a p.d of 16V and a capacitance of 68 000 µF.
    The maximum leakage current is 0.003 µA per µF per V therefore for this capacitor the maximum leakage current = 3260 µA.
    How long does it take for the capacitor to discharge by leaking?

    2. Relevant equations



    3. The attempt at a solution

    So Q=It
    therefore t = Q/I = 334 seconds
    BUT apparently this is the time constant, why?

    (now I realise my method is wrong because discharging current will change but I have no idea how to do this Q so just thought I would use Q=It)
     
    Last edited: Jun 5, 2012
  2. jcsd
  3. Jun 5, 2012 #2

    NascentOxygen

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    At any time t, the current flowing from the capacitor is given by the differential equation:
    1/C * dVc(t)/dt = ......
     
  4. Jun 5, 2012 #3
    how does this help me fidn the time constant and then (5RC) time to discharge?
     
  5. Jun 5, 2012 #4

    NascentOxygen

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    The dots ..... indicate where you fill in using the data you are given.
     
  6. Jun 5, 2012 #5
    but I don't have dVc(t)/dt
     
  7. Jun 5, 2012 #6

    NascentOxygen

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    Expressing it in symbols rather than words:
    Ic(t) = 1/C * dVc(t)/dt = ......
     
  8. Jun 5, 2012 #7
    sorry...I am very much confused....

    What i still don't see is why the t I found using Q/I was time constant...I realise it is not discharge time but cannot see why it is time constant...
     
  9. Jun 5, 2012 #8

    NascentOxygen

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    You are told what the capacitor's leakage current is, here:
     
  10. Jun 5, 2012 #9
    yes so leakage current changes BECAUSE p.d will change...but this also means dV/dt changes....
     
  11. Jun 5, 2012 #10

    NascentOxygen

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    It means V changes. So you express leakage current in terms of V, and this leads to a differential equation to solve.
     
  12. Jun 5, 2012 #11

    gneill

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    You may gather some insight though unit analysis. Take a look at the units of the leakage specification. If capacitance remains constant, how does the leakage current vary with the capacitor voltage? What do the units remind you of?
     
  13. Jun 5, 2012 #12
    this is A2 physics...we don't need to solve differential equations....note the question was only worth 2 marks so clearly this cannot be the solution

    This is what the MS wrote:

    Estimate of time for capacitor to discharge with reasoning:
    Q/I0 = 334 s [This is time constant] (1) - why??
    Q=1.09

    Numerical example such as: for less than 0.7% remaining t = 5τ =
    1670 s OR well-reasoned estimate showing t >> 300 s (1)
    (I know this bit BUT NOT the first bit??)
     
  14. Jun 5, 2012 #13
    Leakge: uA/uF/V
    If capacitance is constant then we get uA/V
    this = 1/R
     
  15. Jun 5, 2012 #14

    NascentOxygen

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    Avoiding the D.E. makes it too easy! :tongue:
     
  16. Jun 5, 2012 #15
    yes but thats all I need (no D.E but working as per mark scheme)....I am still struggling to get why the mark scheme is correct
     
  17. Jun 5, 2012 #16

    gneill

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    Given the voltage V on capacitor C, what's the value of the charge Q on the capacitor?
     
  18. Jun 5, 2012 #17
    q=cv
    =v*680000*10-6
     
  19. Jun 5, 2012 #18

    gneill

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    Yes, but the numerical value is not important for the moment. If you divide both sides by I what do you get?
     
  20. Jun 5, 2012 #19
    Q/I = RC....

    ok...so the next question is why must I use the initial current....I presume this is because I am also using the initial charge, right?
     
  21. Jun 5, 2012 #20

    gneill

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    You can use any set of Q,I, and V which hold at a given instant; It just so happens that you know those particular values. And the R=V/I is constant... that's the leakage resistance you worked out earlier.
     
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