JEE Main 2024 — 4 April, Shift 1 — Question 53

From symmetry, current through e-b & g-d $=0$ $\therefore \mathrm{R}_{\mathrm{eq}}=\frac{3}{4} \times \mathrm{R}=\frac{3}{2} \Omega$

$\therefore$ Current through battery $=\frac{6 \times 2}{3}=4 \mathrm{~A}$

$\mathrm{i}_{2}=\frac{4}{8} \times 2=1 \mathrm{~A}$

$\therefore \Delta \mathrm{V}$ across e- $\mathrm{f}=\frac{\mathrm{i}_{2}}{2} \times \mathrm{R}=\frac{1}{2} \times 2=1 \mathrm{~V}$