Twelve wires each having resistance 2 Ω are joined to form a cube. A battery of 6 V emf is joined across point a and c. The voltage difference between e and f is ________ V.

Solution

Physics sir · 5 steps

Use the cube's symmetry to fix most of the potentials, then one junction equation gives the rest.

1. Set potentials. Take Vₐ = 6 V and V_c = 0. The edges are a-b, b-c, c-d, d-a on top, h-e, e-f, f-g, g-h at the bottom, and a-h, b-e, c-f, d-g as the verticals.
2. Use symmetry. Swap a ↔ c and h ↔ f while leaving b, d, e and g in place. This maps the cube onto itself and turns every potential V into 6 - V. So the unmoved points b, d, e and g must all sit at 3 V. If Vₕ = x, then V_f = 6 - x.
3. Junction at h. All resistances are equal, so at h the net current is zero when Vₕ is the average of its neighbours a, e and g.

   (6 - x)/2 + (3 - x)/2 + (3 - x)/2 = 0 ⇒ x = 4 V

4. Check. So V_f = 6 - 4 = 2 V. At e the neighbours are b, h and f, at 3, 4 and 2 V. Their average is 3 V, which matches Vₑ. The current leaving a is 1.5 + 1.5 + 1 = 4 A, so R_(eq) = 1.5 Ω = 3R/4, the known face-diagonal result.
5. So. Vₑ - V_f = 3 - 2 = 1 V.

The idea to keep. In a symmetric resistor network, any point the symmetry leaves fixed sits at the average of the two terminal potentials. Fix those first, and only a few junction equations remain.

Answer from the NTA final answer key. Solution by PrepWiser.