Twelve wires, each having resistance r. are joined to form a cube as shown in figure. Find the equivalent resistance between the ends of a face diagonal such as a and c.

Explanation

Suppose a potential difference V is applied between the points a and c so that a current `i` enters at a and the same current leaves at c.
The current distribution is shown in figure

By symmetry, the paths ad and ab are equivalent and hence will carry the same current `i_(1)`. The path ah will carry the remaining current `i-2i_(1)` (using Kirchoff.s junction law).
Similarly at junction c, currents coming from dc and bc will be `i_(1)` each and from fc will be `i-2i_(1)`.
Kirchoff.s junction law at b and d shows that currents through be and dg will be zero and hence may be igonored for further analysis. Omitting these two wires, the circuit is redrawn in figure.
The wire hef and joined in parallel and have equivalent resistance
`((2r)(2r))/((2r)+(2r))=r`
between h and f. This is joined in series with ah and fc giving equivalent resistance
`r+r+r=3r`
This `3r` is joined in parallel with adc (2r) and abc (2r) between a and c.
The equivalent reistance R between a and c is, therefore, given by
`(1)/(R)=(1)/(3r)+(1)/(2r)+(1)/(2r)`
`therefore" Equivalent resistance, R"=(3)/(4)r.`