JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Current Electricity

In the steady state the capacitor is fully charged and its branch carries no current. For the polarity the $2.5\mathrm{V}$ cell sets up, the diode in the $3\Omega$ branch is reverse biased and blocks, while the diode in the $4\Omega$ branch is forward biased and conducts. The active loop is therefore the cell with its $1\Omega$ in series with the $4\Omega$ resistor, giving $i=\frac{2.5}{1+4}=0.5\mathrm{A}$. Since no current flows through the $2\Omega$ in series with the capacitor, there is no drop across it and the capacitor voltage equals the voltage across the $4\Omega$ branch, $V_C=0.5\times 4=2\mathrm{V}$. The stored charge is $Q=C{V}_C=5\mu \mathrm{F}\times 2\mathrm{V}=10\mu \mathrm{C}$.