Question

The capacitor of capacitance $C$ in the circuit shown is fully charged initially. Resistance is $R$.
After the switch $S$ is closed, the time taken to reduce the stored energy in the capacitor to half its initial value is

• A. $\frac{RC}{2}$
• B. $2RC\ln 2$
• C. $RC\ln 2$
• D. $\frac{RC\ln 2}{2}$

Answer 1

The correct option is D $\frac{RC\ln 2}{2}$

we know that $q=q_o{e}^{\frac{-t}{RC}}$, Now we know that energy is directly proportional to $q^2$, So final charge should be $\frac{q_o}{\sqrt{2}}$

on putting the value of $q_o$ we get $t=\frac{RCln2}{2}$