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Finding Charge, Voltage, and Stored Energy on Capacitor given only a graph of current

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    The current shown below is applied to a 0.25uF capacitor. The initial voltage on the capacitor is zero.
    (a) Find the charge on the capacitor at t = 30us.
    (b) Find the voltage on the capacitor at t = 50us.
    (c) How much energy is stored in the capacitor by this current?

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    Thank you very much for any help you offer. This has had me bogged down for a couple hours now and I just want to understand how they got those answers in the back of the book.
     

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    Last edited: Nov 13, 2008
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  3. Nov 13, 2008 #2

    cepheid

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    Re: Finding Charge, Voltage, and Stored Energy on Capacitor given only a graph of cur

    There is no need to "work backwards."

    The total charge accumulated at time t is the integral of the current up to time t. So find the area underneath the curve up to t = 30 us in order to find q(30 us). You should get 1.25 uC.
     
  4. Nov 13, 2008 #3

    cepheid

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    Re: Finding Charge, Voltage, and Stored Energy on Capacitor given only a graph of cur

    Finding the voltage requires exactly the same procedure as I described above for the charge. (After all, the voltage is simply proporational to the charge accumulated). You can see that from the equation you've been given. You just have to compute the integral up until t = 50 us, and then divide the answer by C.
     
  5. Nov 13, 2008 #4

    cepheid

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    Re: Finding Charge, Voltage, and Stored Energy on Capacitor given only a graph of cur

    I interpret part c to be asking how much energy is stored in the process of charging up this capacitor using this current (i.e., at the end of it all, how much energy is stored)? Therefore, all you have to do is calculate the energy based on the capacitor's final voltage.
     
  6. Nov 14, 2008 #5
    Re: Finding Charge, Voltage, and Stored Energy on Capacitor given only a graph of cur

    Thank you very much for your help. It all works out now.
     
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