a
From symmetry, current through \(e-b \& g-d = 0\)

\(\therefore \mathrm{R}_{\mathrm{eq}}=\frac{3}{4} \times \mathrm{R}=\frac{3}{2} \Omega\)

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

\(\mathrm{i}_2=\frac{4}{8} \times 2=1 \mathrm{~A}\)

\(\therefore \Delta \mathrm{V} \text { across e-f }=\frac{\mathrm{i}_2}{2} \times \mathrm{R}=\frac{1}{2} \times 2=1 \mathrm{~V}\)