Question

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

A
1.5Ω
B
0.4Ω
C
0.66Ω
D
1.25Ω

Solution

## The correct option is A 1.5Ω 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 (I−i)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(I−i)2+2(I−i)2−V=0V=2(I−i)→(1)Now take loop ACFEBA,2i+2i2+2i2+2i−V=0V=6ii=V6Plugging this back into (1),V=2I−V34V3=2IVI=32Ω The resistance between points A and B is 32Ω.

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