Twelve wires each having resistance $2 \Omega$ are joined to form a cube. A battery of $6 \mathrm{~V}$ emf is joined across point $\mathrm{a}$ and $\mathrm{c}$. The voltage difference between $e$ and $f$ is.______.V.
JEE MAIN 2024, Diffcult
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From symmetry, current through $e-b \& g-d = 0$
$\therefore \mathrm{R}_{\mathrm{eq}}=\frac{3}{4} \times \mathrm{R}=\frac{3}{2} \Omega$
$\therefore \text { Current through battery }=\frac{6 \times 2}{3}=4 \mathrm{~A}$
$\mathrm{i}_2=\frac{4}{8} \times 2=1 \mathrm{~A}$
$\therefore \Delta \mathrm{V} \text { across e-f }=\frac{\mathrm{i}_2}{2} \times \mathrm{R}=\frac{1}{2} \times 2=1 \mathrm{~V}$
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