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Question

Twelve wires, each having equal resistance of 2 Ω, are joined to form a cube. What is the equivalent resistance between the ends of a face diagonal?


A

32Ω

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B

23Ω

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C

52Ω

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D

25Ω

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Solution

The correct option is A

32Ω



Im going to take a current I flowing in through A due to an externally connected cell V. Let the current going into AC be i. That must be the same current coming through EB into node B because those two arms are symmetric. I’m just calling a junction a node. Now, Currents through arms AD and AH must be the same. Let it be (Ii)2. Now currents must divide at junction C symmetrically as i2 and i2. There must be no currents through the arms HF and DG as the currents are the same through AD and DB; and hence CF and FE.

Thus if there is a cell connected between A and B with voltage V,
Taking loop ABEA,
2(Ii)2+2(Ii)2V=0V=2(Ii)(1)Now take loop ACFEBA,2i+2i2+2i2+2iV=0V=6ii=V6Plugging this back into (1),V=2IV34V3=2IVI=32Ω
The resistance between points A and B is 32Ω.


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