A capacitor with initial charge q0 is discharged through a resistor. What multiple of the time constant τ gives the time the capacitor takes to lose

(a) the first 1/8-th of its charge

(b) 7/8-th of its charge?

Answer: a) 0.13*τ; b) 2.08*τ

Explanation: In order to explain the discharg of a capacitor through a resistor, we have to consider the following:

Q (t) = Qo * exp (-t/τ) for a lose of 1/8-th of its charge

in this case, Q (t) = 7/8*Qo=7/8*exp (-t/τ)

ln (7/8) * τ=-t

then, t = - ln (7/8) * τ = 0.13

For a lose of 7/8 th of its charge, we have

Q (t) = 1/7*Qo*exp (-t/τ)

t=-ln (1/8) * τ=2.08