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September 1, 2014

September 1, 2014

Posted by **drwales** on Sunday, March 21, 2010 at 10:34pm.

the charge of the capacitor is given by

Q(t)=CV[1− e ^ (−t/RC)]

Find an expression for the current passing through the resistor as a function of time, and make a graph clearly indicating the initial values. how long will it take for the current passing through the resistor to drop to 37% of its max value?

- Physics -
**drwls**, Sunday, March 21, 2010 at 11:29pmCurrent is the derivative of the capacitor charge.

I(t) = dQ/dt = CV/(RC) exp(-t/RC)

= (V/R) exp(-t/RC)

Current is 1/e = 37% of the initial value when t/RC = 1. That is called the time constant.

You will have to draw you own graph.

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