Twelve wires each having resistance are joined to form a cube. A battery of emf is joined across points and . The voltage difference between and is

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Solution

Q.
Twelve wires each having resistance $2 Ω$ are joined to form a cube. A battery of $6 V$ emf is joined across points $a$ and $c$ . The voltage difference between $e$ and $f$ is ____ $V$ . [2024]

Ans.
(1)

$∴$ $R_{eq} = \frac{R \times 3 R}{R + 3 R} = \frac{3}{4} R \Rightarrow R_{eq} = \frac{3}{4} \times 2 = \frac{3}{2} Ω$

$∴$ Current through battery $= \frac{6 \times 2}{3} = 4 A$

The current through $a - h, i_{a h} = \frac{4}{8} \times 2 = 1 A$

$∴$ $∆ V$ across $e - f = \frac{i_{a h}}{2} \times R = \frac{1}{2} \times 2 = 1 V$

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