A 20.0 \mu F capacitor initially charged to 20.0 \mu C is discharged through a 1.90 k\Omega resistor. How long does it take to reduce the capacitor's charge to 5.00 \mu C?

1 answer

The charge of the discharging capacitor, as a function of t, is:

Q(t) = Qo* e^(-t/RC)

Qo is the initial charge, RC is the time constant, which in this case is 1900*20*10^-6 = 0.038 seconds

So, solve

5*10^-6 = 20*10^-6*e^(-t/0.038)

ln(1/4) = -t/0.038

t = ?