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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2024
  • Hard
  • Numerical

12 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.

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Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

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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