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 = ?
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
1 answer