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Deriving current from decharging capacitor and homogenous DE

  1. Nov 30, 2014 #1
    Hello there,

    I want learn to do homogenous differential equations and I cannot work out what is wrong in this example. Hope you'll be able to tell me why my DE is unsatisfactory.

    Assume that a capacitor is has charge Q and that the terminal with the high voltage is connected to a switch in line with a resistor which in turn is connected to the low terminal on the cap.

    When the switch is closed, I write ohms law:
    ##i = \frac{v}{R}##
    The current in the circuit will be proportional to the voltage across the capacitor plates.

    The from there I rewrite ohms law using the definition of capacitance and differentiate to get the DE:

    i=\frac{v}{R} = \frac{q}{RC} \implies \frac{d i}{dt} = \frac{i}{RC}

    I solve the DE and find:

    i = C e^{t/(RC)},
    Where ##C## is some constant. The problem is that the current diverges as a function of time.

    I wanted expected something along the lines of:

    ##i = Ce^{-t/(RC)},##

    Where the current converge to zero of as time pass.

    What is unsatisfactory with the DE I started with? I think it looks correct, and cannot find anything wrong with it. Perhaps you can help. :)

    Thank you for your time.

    Kind regards,
    Last edited: Nov 30, 2014
  2. jcsd
  3. Dec 1, 2014 #2


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    You simply made a sign mistake in your derivation of the DE. Here you integrate the electric along the circuit from the positively charged plate to the negatively charged plate, this gives you [itex]-U=Q/C[/itex]. Further, since [itex]\vec{j} = \sigma \vec{E}[/itex] along the wire, you have [itex]i[/itex] positive in this direction. Thus the right equation is
    [tex]i=-\frac{Q}{R C},[/tex]
    and the rest of the calculation is analogous to the one you've presented.
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