Question

A capacitor is charged and then made to discharge through a resistance. The time constant is Ï„. In what time will the potential difference across the capacitor decrease by 10%?

a. t ln 0.1
b. t in 0.9
c. t ln 10/9
d. t in 11/10

Answer 1

Q = Q0 e^-t/RC = Q0 e^-t/T

Charge left on capacitor after time t where T is the time constant of the capacitor - R * C

Q / Q0 = .9 charge decreases by 10%

.9 = e^-t/T

ln .9 = -t / T = .10

Or t = -T * ln .9

When elapsed time t = -ln .9 * time constant