Capacitors in DC Circuits & RC question from JEE Main 2021 (27 Aug, Shift 1)

Answer: (a)

1. Steady state: current only through the $4\ \Omega$: $I = \dfrac{5}{4+1} = 1\text{ A}$, terminal voltage $= 4\text{ V}$.
2. Capacitor branch: $4\ \mu\text{F}$ in series with $(2+2) = 4\ \mu\text{F}$, so $C_{eq} = 2\ \mu\text{F}$ across 4 V.
3. Series charge $Q = 2\times4 = 8\ \mu\text{C}$, same on the $4\ \mu\text{F}$.
4. So the answer is (a) $8\ \mu\text{C}$.