A capacitor of capacitance C= 10mF, is charged using a battery of potential V, through a resistor of resistance R = 250 ohms.
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 11:29pm
Current 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.