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 - Robert, Tuesday, August 14, 2012 at 11:56pm
the character that doesn't show up is the emf (E)
Physics 2 - drwls, Wednesday, August 15, 2012 at 12:34am
The current is
i = dq/dt = C*E*(1/RC)*e^(-t/RC)
iR + q/C = E*e^(-t/RC)+ E*(1 – e^-t/RC)
Therefore the sum of the voltage drop across the capacitor and the resistor equals E at all times, as required.
Physics 2 - Robert, 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 - drwls, 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 - Robert, Wednesday, August 15, 2012 at 12:52am
thanks. It's all good, I understand your approach and I really appreciate it!