JEE Advanced 2021 Paper 2 Physics: Paragraph 2

Replace the battery, $R_1$ and $R_2$ seen by the capacitor by a Thevenin equivalent: $V_{th}=12\times\dfrac{R_2}{R_1+R_2}=6\ \mathrm{V}$ and $R_{th}=R_1\parallel R_2=1\ \mathrm{k\Omega}$. The time constant is $\tau=R_{th}C=1000\times100\times10^{-6}=0.1\ \mathrm{s}$. The capacitor charges as $q=CV_{th}\left(1-e^{-t/\tau}\right)$ with final charge $600\ \mu\mathrm{C}$. Setting $q=450\ \mu\mathrm{C}$ gives $1-e^{-t/\tau}=0.75$, so $t=\tau\ln4=0.1\times1.386=0.1386\ \mathrm{s}=138.6\ \mathrm{ms}$. The answer $69.3$ ms corresponds to $\tau\ln2$ (half-charge) and $100$ ms is a single time constant, which would give only $q=379\ \mu\mathrm{C}$. So option (C) is the correct choice, and the exponential law must be used rather than a linear charging estimate.

At that instant $q=450\ \mu\mathrm{C}$, so the capacitor voltage, which is also the voltage across $R_2$, is $V_C=q/C=4.5\ \mathrm{V}$. The current through $R_1$ is the battery current, and it equals $I=\dfrac{12-V_C}{R_1}=\dfrac{12-4.5}{2000}=3.75\ \mathrm{mA}$. As a check on the sign of the other branches, the current in $R_2$ is $V_C/R_2=2.25\ \mathrm{mA}$ and the capacitor branch carries $3.75-2.25=1.5\ \mathrm{mA}$, which matches $dq/dt=(q_\infty-q)/\tau=(600-450)\ \mu\mathrm{C}/0.1\ \mathrm{s}=1.5\ \mathrm{mA}$. The battery current falls from $6$ mA at $t=0$ to $3$ mA at steady state, so $3.75$ mA lies between these values.