JEE Main 2024 — 1 February, Shift 2 — Question 56

In steady state there will be no current in branch of capacitor, so no voltage drop across $\mathrm{R}_{2}=5 \Omega$ $\mathrm{I}_{2}=0$ $\mathrm{I}_{1}=\mathrm{I}_{3}=\frac{10}{4+6}=1 \mathrm{~A}$ $\mathrm{V}_{\mathrm{R}_{3}}=\mathrm{V}_{\mathrm{c}}+\mathrm{V}_{\mathrm{R}_{2}} \quad \mathrm{~V}_{\mathrm{R}_{2}}=0$ $\mathrm{I}_{3} \mathrm{R}_{3}=\mathrm{V}_{\mathrm{c}}$ $\mathrm{V}_{\mathrm{c}}=1 \times 6=6$ volt $\mathrm{q}_{\mathrm{c}}=\mathrm{CV}_{\mathrm{c}}=10 \times 6=60 \mu \mathrm{C}$