Answer
During the charging of the capacitor, its charge as a function of time is :-
$q=q_{0}(1-e^{-\frac{t}{T}})$, $T$ being the time constant
After 1 time constant, T, has elapsed the charge is
$q=q_{0}(1-e^{-1})\approx 0.632q_{0}$, so time it takes to charge up to $0.25q_{0}$ is less than 1 time constant.
When discharging, the capacitor's charge is $q=q_{0}e^{-\frac{t}{T}}$. After 1 time constant, the charge is $q=q_{0}e^{-1}=0.368q_{0}$, so the time to discharge to $0.25q_{0}$ is more than 1 time constant.
Work Step by Step
During the charging of the capacitor, its charge as a function of time is :-
$q=q_{0}(1-e^{-\frac{t}{T}})$, $T$ being the time constant
After 1 time constant, T, has elapsed the charge is
$q=q_{0}(1-e^{-1})\approx 0.632q_{0}$, so time it takes to charge up to $0.25q_{0}$ is less than 1 time constant.
When discharging, the capacitor's charge is $q=q_{0}e^{-\frac{t}{T}}$. After 1 time constant, the charge is $q=q_{0}e^{-1}=0.368q_{0}$, so the time to discharge to $0.25q_{0}$ is more than 1 time constant.
