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Discharging a capacitor

  1. Aug 16, 2010 #1
    Hello, I'm just doing a lab on the discharging of a capacitor.

    The data collected shows that at first the current is large because the voltage is large, so charge is lost quickly and the voltage decreases rapidly. As charge is lost the voltage is reduced making the current smaller so the rate of discharging becomes progressively slower.

    I'm just wondering how the fit equation A*exp(−Cx)+B relates to the equation for a discharging capacitor V = Vo e^-t/RC ?
     
  2. jcsd
  3. Aug 17, 2010 #2
    Substitute in the values for A, C, and B; while the variable x in your case is t, since you're viewing V as a function of time. so A=Vo, B = 0, and C = 1/RC.
     
  4. Aug 17, 2010 #3
    y = A*exp(−Cx)+B is an approximation to V = Vo e^-t/RC, in the xy plane, in cartesian coordinates.
    When you plot your results (V against t) you need to fiddle with the A,B and C constants till you get the best fit.
    Theoretically there would be a perfect combination of these three numbers, that fits your data perfectly, but this is rarely seen in practise - say why.

    A would be Vo. B moves the fit up/down the y axis (its probably 0).
    Then C is then 1/RC, sometimes called the time constant, tau.
     
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