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Integrating Current-Voltage Relationship for a Capacitor

  1. Jan 18, 2007 #1
    1. The problem statement, all variables and given/known data

    To find the voltage-current relationship of a capcitor, integrate both sides of
    i = C(dv/dt)

    3. The attempt at a solution

    In the book they say, v = (1/C)(The Integral from -tve infitity to t) of i dt.


    v = (1/C)(The Integral from tnot to t) of i dt + v(tnot)

    I am trying to understand why they integrated from negative infinity. Why not start the integration at 0. I also do not undersstand in the second equation where the + v(tnot) came from.

    Thanks in advance
  2. jcsd
  3. Jan 19, 2007 #2


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    Science Advisor
    Gold Member

    An ideal capacitor will store all the charge that's ever applied to it. What you call 0 time is arbitrary (it could be a certain date and time, or when you start taking data, or anything else)--but the capacitor voltage also reflects currents that flowed before that time, all the way back.

    BTW, the subscript is "nought" which means zero, not "not".

    V_t0 is the integration constant (look back at your calculus book for an indefinite integral). The capacitor might have started with a voltage on it before it was hooked into your circuit.
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