 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016-resit/6/solution

A = (0, 0, 0)

and

B = (1, 0, 0)

at potentials

V

and zero. The two corners

(0, 1, 0), (0, 0, 1)

have a common potential

x

; their neighbours

(1, 1, 0), (1, 0, 1)

have

V - x

. Write the remaining two potentials as

z, V - z

At a corner of the first pair and at

(0, 1, 1)

3 x = V + (V - x) + z, 3 z = 2 x + (V - z) .

Thus

4 x - z = 2 V

and

4 z - 2 x = V

, yielding

x = 9 V /14

z = 4 V /7

. The supplied current at

A

I = \frac{V}{R} + 2 \frac{V - x}{R} = \frac{12 V}{7 R}, R_{eq} = \frac{V}{I} = \frac{7 R}{12} .

The direct edge between the terminals is one of three current paths leaving

A

; treating the other eleven resistors as a single obvious series chain would miss the branching.

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