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Capacitance Problem

  1. May 6, 2014 #1
    Hi all,

    I have been looking at this problem for a while now and can not seem to be able to reach the correct answer, hopefully someone out there can help me!

    A capacitor of capacitance 180pF is charged to a P.D of 12V and then discharged through a sensitive meter. This sequence of operations is repeated 250 times per second, what is the average current registered by the meter?


    The way I see it, the maximum charge on the capacitor is is 12*(180*10^-12) = 2.16*10^-9. The switch charges and discharged 250 times per second, so in a second the current goes from 0 to 2.16*10^-9 and then back down again to 0. I would therefore assume that the average current is the halfway point, so (2.16*10^-9)/2. Apparently this is not correct and I can not see how to find a valid answer.

    Any help would be much appreciated.
     
  2. jcsd
  3. May 6, 2014 #2

    BvU

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    Point is the voltage doesn't drop linearly with time during discharge, so you can't take a simple average.

    The way to approach this one is to start from what you know and see if you can use it someway. In the template, we have a provision for that: relevant equations. Please use the template. Can you give me (yourself) a relevant equation in this context where C and V are known ?

    You have to go back to the "definition" of current.
     
  4. May 6, 2014 #3
    Thank you for your reply.

    The equations know are C=Q/V, I think that is about the only thing I have covered yet with regards to capacitance. The question is what I have posted above and included no additional details. Current is the rate of flow of charge, I am aware that current decays and increases exponentially but I don't believe the question is asking for that much depth, and is possibly just asking for the average current over the whole time period, to be honest I am not to sure what you are getting at with regards to current?
     
  5. May 6, 2014 #4

    BvU

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    I think you are getting it. You have a charge of 2.16 nC per discharge, so 250 * 2.16 = 0.54 μC per second. The exercise explicitly asks for the average current: 0.54 μC/s !

    I missed the relevant equation upon first glancing over, hence the question and the request to use the template. But you did have it.
     
  6. May 6, 2014 #5
    Ahh I see now,

    I was getting charge and current confused, and the average charge on the capacitor would be half of maximum charge, but the current would be the rate of flow of this charge which stays constant (or is for the purpose of finding an average).

    Thank you very much for your help.
     
  7. May 7, 2014 #6

    BvU

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    Rate of flow is of course anything but constant: it is proportional to the (rapidly dropping) voltage.
    Point is that for the average current you need something like ##\int I \, dt = \int {dQ\over dt}\, dt = Q -Q_0##
    So knowing Q=CV and Q0=0 you have the average for one discharge.
     
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