JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Current Electricity

In the steady state no current flows in the branch containing the capacitor, so the $4\Omega$ resistor in series with it carries no current and has no potential drop. The $2\text{V}$ source drives current around the loop containing the $2\Omega$ and $6\Omega$ resistors in series, giving $i=\frac{2}{2+6}=\frac{1}{4}\text{A}$. The capacitor branch spans the same two nodes as the $2\Omega$ resistor, and with no drop across the $4\Omega$ resistor the capacitor voltage equals the drop across the $2\Omega$ resistor, $V_C=iR=\frac{1}{4}\times 2=0.5\text{V}$. The potential difference across the capacitor is $0.5\text{V}$.