JEE Advanced 2017 Paper 2 Physics: Paragraph 2

After a long time the capacitor carries no current, so the circuit is simply $R_1$ and $R_2$ in series across the $12\ \mathrm{V}$ battery. The steady current is $I=\dfrac{12}{2+4}=2\ \mathrm{mA}$. The capacitor is in parallel with $R_2$, so its final voltage is $V_C=IR_2=2\ \mathrm{mA}\times4\ \mathrm{k\Omega}=8\ \mathrm{V}$ (equivalently the voltage divider $12\times\tfrac46$). The charge is $Q=CV_C=3\ \mu\mathrm{F}\times8\ \mathrm{V}=24\ \mu\mathrm{C}$. The value $36\ \mu\mathrm{C}$ would result from taking the whole $12\ \mathrm{V}$ across the capacitor. The same result follows from charge $Q=CV$ with $V$ found by the divider; it does not depend on $C$ except through the final multiplication.

Replace the battery, $R_1$ and $R_2$ by their Thevenin equivalent seen by the capacitor: $V_{th}=12\times\tfrac{4}{6}=8\ \mathrm{V}$ and $R_{th}=R_1\parallel R_2=\dfrac{2\times4}{6}=\dfrac43\ \mathrm{k\Omega}$. The time constant is $\tau=R_{th}C=\tfrac43\times10^3\times3\times10^{-6}=4\ \mathrm{ms}$. The capacitor voltage is $V(t)=8(1-e^{-t/\tau})$. Setting $V=4\ \mathrm{V}$ gives $e^{-t/\tau}=\tfrac12$, so $t=\tau\ln2=4\times0.693=2.77\ \mathrm{ms}$. Using $\tau=R_1C=6\ \mathrm{ms}$ would give $4.16\ \mathrm{ms}$, and the value $4.00$ ignores the logarithm. At the first instant the capacitor behaves like a short circuit, so the initial current from the battery is $12/2=6\ \mathrm{mA}$, and it decays to $2\ \mathrm{mA}$ as the capacitor charges. The Thevenin method is justified because for the capacitor the rest of the network is a single source with a single series resistance.