
From symmetry, current through e-b & g-d =0 $\begin{aligned}
& \therefore \mathrm{R}_{\text {eq }}=\frac{3}{4} \times \mathrm{R}=\frac{3}{2} \Omega \
& \therefore \text { Current through battery }=\frac{6 \times 2}{3}=4 \mathrm{A} \
& \mathrm{i}_2=\frac{4}{8} \times 2=1 \mathrm{A} \
& \therefore \Delta \mathrm{V} \text { across e-f }=\frac{\mathrm{i}_2}{2} \times \mathrm{R}=\frac{1}{2} \times 2=1 \mathrm{~V}
\end{aligned}$
