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Posted by on Tuesday, August 14, 2012 at 11:56pm.

Show that during the charging of the capacitor in an RC circuit, the charge of the capacitor, as a function of time, is: q(t) = C(1 – e-t/RC)

  • Physics 2 - , Tuesday, August 14, 2012 at 11:56pm

    the character that doesn't show up is the emf (E)

  • Physics 2 - , Wednesday, August 15, 2012 at 12:34am

    The current is
    i = dq/dt = C*E*(1/RC)*e^(-t/RC)
    = (E/R)*e^(-t/RC)

    iR + q/C = E*e^(-t/RC)+ E*(1 – e^-t/RC)
    = E

    Therefore the sum of the voltage drop across the capacitor and the resistor equals E at all times, as required.

  • Physics 2 - , Wednesday, August 15, 2012 at 12:47am

    thank you very much drwls, this is a question from a sample exam, getting ready for my finals tomorrow!

  • Physics 2 - , Wednesday, August 15, 2012 at 12:50am

    Good luck with finals! I took a lazy approach to your question, showing that what you wrote is the solution, rather than solving the general differential equation.

  • Physics 2 - , Wednesday, August 15, 2012 at 12:52am

    thanks. It's all good, I understand your approach and I really appreciate it!

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