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Determining current drawn from a capacitor

  1. Apr 7, 2012 #1
    Hello.

    I want to know what current i am getting out of a supercapacitor, but i have nothing to directly measure the currents in the given range with.
    So i tried to determine it from the voltage drop on the cap, and i want to know if i did it right.

    Given is 3000F capacitor, at initial voltage of V1=1.100V.
    After being connected to the load in question for 3.5 seconds, the voltage is V2=1.045V.

    Thus, the energy in the capacitor is 0.5*C*V*V.
    E1=1815J
    E2=1638J
    Which gives spent energy of 177J.

    W=J/s, so for the 3.5 seconds it was giving out 50.5W of power.
    Finally, A=W/V, and average V was 1.0725V, which gives 47.1A of current.

    Is that a correct way of current estimation?
     
  2. jcsd
  3. Apr 7, 2012 #2
    You can also use a capacitor equation.

    C = Q/V = (I*t)/V ---> I = C * V/t = 3000F * (1.1V - 1.045V)/3.5s = 3000F * 0.055V/3.5s = 47.14A
     
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