The formula for calculating the discharge of a capacitor is expesed as
Q = Qoe^-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 = Qoe^-t/b
Dividing both sides by Qo, it becomes
1/2 = e^-t/b
Taking the natural og of both sides,
ln (1/2) = ln e^t/b = -t/b * lne
Recall, lne = 1
ln (1/2) = -t/b
By cross multiplying,
t = - bln (1/2)
From the information given,
b = 50.4
t = - 50.4ln(1/2)
t = 34.93 ms
The charge is reduced to half its initial value after 34.93 ms
