JEE Main 2026 Apr 8 Shift 2, Physics Q24: Charge & discharge of a condenser

In steady state no current flows through the capacitor branch, so the $10\Omega$ at the far right and the capacitor itself can be removed for current calculation. The remaining network reduces to an equivalent resistance $R_{\text{eq}}=6\Omega$ seen by the battery.

Main current: $I=\frac{E}{R_{\text{eq}}}=\frac{12}{6}=2$ A.

This current splits through the parallel vertical resistor branches; tracing through the network gives the branch currents $I_1=I_2=1$ A and $I_3=I_4=0.5$ A.

Voltage across the capacitor $V_{AB}$ equals the potential drop across the $4\Omega$ branch carrying $0.5$ A:

$V_{AB}=0.5\times 4=2$ V.

$Q=CV=(100\mu \text{F})(2\text{ V})=200\mu$C.