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Capacitor Voltage

  1. Feb 9, 2012 #1
    The current through an initially uncharged 4uF capacitor is as attached. Find the voltage across the capacitor for 0<t<3.

    I have following calculation:

    i = 40mA 0 <t < 1
    i = 0 1 < t < 2
    i = -40mA 2 < t < 3

    since the voltage across capacitor is given by:

    v = (1/c) ∫ i dt + v(t0)

    v (@40mA) = (1/c) ∫40m dt + 0 = 10t KV
    v(@0mA) = (1/c) ∫0m dt + 10KV = 10KV
    v(@ -40mA) = (1/c) ∫-40m + 10KV = -10t + 10 KV (Correct answer is -10t + 30 KV)

    What I understood is the voltage across capacitor is

    ∫ idt + v(t0), where i is current and v(t0) is initial voltage on the capacitor.

    In this case between time span 2 to 3 sec the current is constant -40mA and hence the voltage will linearly increase. Looking back at t=2 sec the initial voltage is 10KV due to steady capacitor voltage and which was not discharge. And hence I take 10KV at v(t0). I am unable to understand where 30 KV come from as I assume the v(t0) is 10KV.

    I understand I am missing something but do not know where exactly and how.
  2. jcsd
  3. Feb 9, 2012 #2


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    I'm no mathematician but if 40m in you equation is for 40mA then you have an error in magnitude value.
    You should be converting it to Amps so it should be 0.04A
    you should always use SI units in equations

    I dont even begin to follow your math but try that see how it works :)

  4. Feb 9, 2012 #3
    I would say that for the first second a current of 40mA means that the capacitor is charging up and a charge of 40 mC is placed on the capacitor.
    This means the voltage = Q/C = 10kV
    For the next second the current is zero so the voltage remains at 10kV
    For the last second a current of -40 A means that the capacitor is discharging and loses 40 mC of charge so ends up with zero charge and therefore V = 0
    The answer given as -10t + 30kV is correct if you put t=3
  5. Feb 9, 2012 #4

    jim hardy

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    """(Correct answer is -10t + 30 KV)""

    not sure i believe that.

    you add 40 milliamp-seconds to capacitor in first second,
    leave them trapped there in next second,
    then remove them in third second.

    result is zero.

    surely calculus will give same result...
    does (-10t + 30) somehow resolve to zero?

    keep it simple.
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