Twelve identical resistances arranged on all edges of a cube. The resisors are all the same. Then find the equivalent resistance between the edges A and B as shown in.

Explanation

Let un number the corners as 1,2,3,…. As
shown in fig 5.46 Let us identify various paths from A to B
`a(1-8 - 7 -6), b(1-2-7-6),c(1-8-5-6),
d(1-2-3-6),e(1-4-5-6), f(1-4-3-6)`
Now for each of these paths
we have identical resistances. Let
the current entering at point A be
I. The current will be distributed
equally among all the three paths.
therefore, we can say that current
is `I_(12) = I_(23) = (1)/(2) ((I)/(3)) = (I)/(6)`
Similarly, we can say that `I_(43) = I_(45) = I_(85) = I_(87) = I//6.` Therefore, to
calculate the potential difference across the paht `(1rarr2rarr7rarr6),`
we have
`V_1 -(I)/(3)R - (I)/(6)R -(I)/(3) R = V_6`
or `V_1V_6 = I[(R )/(3) + (R )/(6) +(R )/(3)] = (5)/(6)IR`
or`((V_1 - V_6))/(I) = (5)/(6)R`
Therefore, the equivalent resistance is `5R//6`.