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Think of the cube as a wire framewith each edge a resistor.
S.L has got it right.

I set it as a 'stretch and challenge' problem.
There is also a slightly different way of doing it. Given that if two points are at the same potential then no current will flow between them, we can connect any points of the same potential on the cube with wires since it will make no difference.

By using symmetry arguments you can do that, and create a circuit diagram that is pretty easy to solve.

This question came up when I did the AEA back in 2006, I think I solved it by teacher col's method, but thought of the one above after the exam.

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