A capacitor with initial charge q0 is discharged through a r

A capacitor with initial charge q_0 is discharged through a resistor. What multiple of the time constant r gives the time the capacitor takes to lose (a) the first 1/6-th of its charge and (b)5/6-th of its charge?

Solution

Q = C*V

tau = RC

voltage on a capacitor while discharging:

V = V0*e^(-t/RC)

Q = Q0*e^(-t/tau)

A. Q = Q0 - Q0/6 = 5Q0/6

5Q0/6 = Q0*e^(-t/tau)

e^(-t/tau) = 5/6

-t/tau = ln(5/6)

t/tau = ln (6/5)

t = 0.182*tau

B. Q = Q0 - 5Q0/6 = Q0/6

Q0/6 = Q0*e^(-t/tau)

e^(-t/tau) = 1/6

-t/tau = ln(1/6)

t/tau = ln (6)

t = tau*ln (6) = 1.79*tau

Let me know if you have any doubt.