JEE Main 2026 — Current Electricity Question with Solution

In steady state, the capacitor acts as an open circuit, meaning no current flows through the middle branch.

The circuit simplifies to a single loop containing the $2\text{ V}$ battery, the $6\Omega$ resistor, and the $2\Omega$ resistor.

The total resistance of this loop is $R_{\text{eq}}=2+6=8\Omega$.

The current flowing through the loop is $I=\frac{2}{8}=0.25\text{ A}$.

The potential difference across the parallel branches is equal to the voltage drop across the $2\Omega$ resistor:
$V=I\times 2=0.25\times 2=0.5\text{ V}$

Since there is no current in the middle branch, the voltage drop across the $4\Omega$ resistor is zero. Thus, the entire potential difference of $0.5\text{ V}$ appears across the capacitor.

Answer: $0.5$