JEE Main 2026 Apr 5 Shift 2, Physics Q12: Charge & discharge of a condenser

In steady state no current flows through the capacitor branch, so the 4 $\Omega$ branch is effectively open. The current flows only through the 2 $\Omega$ and 6 $\Omega$ resistors in series:

$i=\frac{2}{6+2}=\frac{1}{4}$ A.

The capacitor voltage equals the voltage across the 2 $\Omega$ resistor:

$V_C=i\times 2=\frac{1}{4}\times 2=0.5$ V.