Question 107 of 380

Twelve wires each having resistance $2 \Omega$ are joined to form a cube. A battery of $6 \mathrm{~V}$ emf is joined across point $a$ and $c$. The voltage difference between $e$ and $f$ is ______V.

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Answer: 1

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