Q 12-02-092JEE MainJEE Main 2023 (13 Apr, Shift 2)Medium
In the network shown below, the charge accumulated in the capacitor in steady state will be
Answer: (D) $7.2\ \mu\text{C}$
In the steady state no current flows in the capacitor branch. The cell drives current through $4\ \Omega$ and the bottom $6\ \Omega$:
$$I=\frac{3}{4+6}=0.3\ \text{A}$$
The capacitor branch is in parallel with the bottom $6\ \Omega$, so $V_C=0.3\times6=1.8\ \text{V}$ and $Q=4\times1.8=7.2\ \mu\text{C}$.
Solution by Sreeraj P, M.Sc Physics