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RC circuit - Capacitor discharge

  1. May 3, 2010 #1
    When a capacitor discharges, the potential difference between its plates and the resistor's terminals is the same. Hence,

    [tex] \frac{Q}{C}=R \frac{dq}{dt}[/tex]
    Solving this equation we get

    [tex] q(t)= Q_{0} exp \frac{t}{RC} [/tex]

    Obviously, this isn't the solution. It is actually [tex] q(t)= Q_0 exp \frac{-t}{RC} [/tex]. So, I'm missing a minus sign on the original equation.

    But why should it be there?
  2. jcsd
  3. May 3, 2010 #2
    According to Kirchoff's Second Law, the sum of the voltages is equal to zero. R(dQ/dt) + (Q/C) = 0, therefore you need -R(dQ/dt) = (Q/C). I apologize for the lack of proper script.
  4. May 4, 2010 #3
    Thank you.
    Didn't know such laws.
  5. May 5, 2010 #4
    If the capacitor discharges, then dq/dt is negative, so according toyour originial equation, Q/C wich is positive equals something negative
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