Answer
(a) $0.693RC$
(b) $2.303RC$
Work Step by Step
(a) We know that
$q_2=C\epsilon(1-e^{\frac{-t}{RC}})$
$\implies 0.5q_1=C\epsilon(1-e^{-1/RC})$
As $q_1=C\epsilon$ and $0.5C\epsilon=C\epsilon(1-e^{-t/RC})$
$\implies 0.5=1-e^{-t/RC}$
$\implies t=-RC\ln(0.5)$
$t=0.693RC$
(b) We know that the current flowing through the circuit is given as
$I=\frac{\epsilon}{R}e^{-t/RC}$
$\implies 0.1I_{\circ}=\frac{\epsilon}{R}e^{-t/RC}$
$\implies t=-RC\ln (0.1)$
$\implies t=2.303RC$
Thus, this is the time required to drop the current to 10% of its initial condition.
