JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Current Electricity

In the steady state no current passes through the capacitor, so the branch holding the final $10\Omega$ and the capacitor carries none. In the conducting network the $5\Omega$ in series with the middle section (the $10\Omega$ vertical in parallel with the $4+4+2\Omega$ path, which is $5\Omega$) in series with the lower $2\Omega$ totals $5+5+2=12\Omega$, so the $12\text{V}$ source drives a branch current of $1\text{A}$. The potentials across the middle section are then $7\text{V}$ at the top and $2\text{V}$ at the bottom. The $0.5\text{A}$ flowing through the $4\Omega$-$4\Omega$-$2\Omega$ path makes the potential at the capacitor's top plate $5\text{V}$ and at its bottom plate $3\text{V}$, a difference of $2\text{V}$. The stored charge is $Q=CV=100\mu \text{F}\times 2\text{V}=200\mu \text{C}$.