Find the equivalent resistance of the circuit of the previous problen between the ends of an edge such as a and b in figure.

Explanation

Suppose a current I enters the circuit at the point a and the same current leaves the circuit at the point b.The cirrent distribution is shown in figure. The paths through ad an ah are equivalent and carry equal current `i_(1)`.The current through ab is then `i-2i_(1).`
the same distribution holds at the junction b. Currents in eb and cb are `i_(1)`in he and `i_(1)-I_(2)`in hg. Similar is the division of current `i_(1)`in ad into dc and dg. The rest of the currents may be written easily using Kirchhoff's junction law.
The potential difference V between a and d may be written from the paths ab, aheb and ahgfcb as
`V=(i-2i_(1))r`
`V=(i_(1)+i_(2)+i_(1))r`
`and V=[i_(1)+(i_(1)-i_(2))+2(i_(1)-i_(2))+(i_(1)-i_(2))+i_(1)]r.`
which may be written as
V=(i-2i_(1))r`
`V=(2i_(1)+i_(2))r`
and`V= (6i_(1)-4i_(2))r.`
Eliminating` i_(1) and i_(2)`form these equations,
`V/i=(7/12)r.`
which is the equivalent resistances.