JEE Mains · Physics · STD 12 - 3. current electricity
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.
- A \(1\)
- B \(5\)
- C \(8\)
- D \(10\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
From symmetry, current through \(e-b \& g-d = 0\) \(\therefore \mathrm{R}_{\mathrm{eq}}=\frac{3}{4} \times \mathrm{R}=\frac{3}{2} \Omega\) \(\therefore \text { Current through battery }=\frac{6 \times 2}{3}=4 \mathrm{~A}\) \(\mathrm{i}_2=\frac{4}{8} \times 2=1 \mathrm{~A}\)…
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