A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is then discharged by connecting it directly across a 100-k\Omega100−kΩ resistor. At what point in time after the resistor is connected will the capacitor have 13.5~\mu C13.5 μC of charge remaining on it?
Question
Explanation:
The given data is as follows.
C = [tex]20 \times 10^{-6} F[/tex]
R = [tex]100 \times 10^{3}[/tex] ohm
[tex]Q_{o} = 100 \times 10^{-6}[/tex] C
Q = [tex]13.5 \times 10^{-6} C[/tex]
Formula to calculate the time is as follows.
[tex]Q_{t} = Q_{o} [e^{\frac{-t}{\tau}][/tex]
[tex]13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}][/tex]
0.135 = [tex]e^{\frac{-t}{2}}[/tex]
[tex]e^{\frac{t}{2}} = \frac{1}{0.135}[/tex]
= 7.407
[tex]\frac{t}{2} = ln (7.407)[/tex]
t = 4.00 s
Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.