The Discharge Time of a Capacitor in an RLC Circuit
The capacitor in an RLC circuit is discharged with a time constant of 50.4 ms
The formula for calculating the discharge of a capacitor is expressed as
Q = Qo * e-t/b
where
Q is the final charge after time t
Qo is the initial charge
t is the time
b is the time constant
We want to find when Q = Qo/2
Substituting Q = Qo/2 into the formula, it becomes
Qo/2 = Qo * e-t/b
Dividing both sides by Qo, it becomes
1/2 = e-t/b
Taking the natural log of both sides,
ln(1/2) = ln(et/b) = -t/b * ln(e)
Recall, ln(e) = 1
ln(1/2) = -t/b
By cross multiplying,
t = -b * ln(1/2)
From the information given,
b = 50.4
t = -50.4 * ln(1/2)
t = 34.93 ms
The charge is reduced to half its initial value after 34.93 ms
What is the time, in microseconds, after the discharge begins is the charge on the capacitor reduced to half its initial value? The charge on the capacitor is reduced to half its initial value after 34.93 ms.