Twelve wires each having resistance 2Ω are joined to form a cube. A battery of 6 V emf is joined across point a and c . The voltage difference between e and f is ______V.

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From symmetry, current through e-b & g-d =0 $∴ R_{eq} = \frac{3}{4} \times R = \frac{3}{2} Ω ∴ Current through battery = \frac{6 \times 2}{3} = 4 A i_{2} = \frac{4}{8} \times 2 = 1 A ∴ Δ V across e-f = \frac{i _{2}}{2} \times R = \frac{1}{2} \times 2 = 1 V$

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