Twelve equal wires each of resistance r are joined to form a skeleton cube. Find the equivalent resistance between two corners on the same edge of the cube.

Explanation

Method-I
Connect a source between points 1 and 2. Let current `i` enter through point 1 into the network . The network is symmetrical about dotted lines. The currents above and below dotted line are symmetrically distributed as shown in fig.
By junction rule at 1, we have `i= x+ 2y`
`therefore R_(12) = V/i = V/(x+2y) " "....(i)`

In close loop 1-2-9-10-1 , we have
`-rx + V=0 ` or `x= (V)/r " "....(ii)`
In close loop 1-4-3-2-1
`-rz-rz-ry+rx=0` (or) `x-2y - z =0" "....(iii)`
In close loop `4-8-7-3-4 `
`-r(y-z) - r xx 2(y-z) + rz=0`
or `-4(y-z) + z =0` or `4y+ 5z=0" "....(iv)`
From equations (iii) and (iv) , we get
`-4y + 5(x-2y) = 0 ` or `5x=14y`
Since `x = V/r therefore y= 5/(14) xx V/r`
Now `R_(12) = V/(x+2y) = (V)/(V/r + 2 xx(5V)/(14r)) =(7r)/(12`
Method II:
Out previous knowledge reveals that points 3 and 6 must be at the same potential . So must be 4 and 5 . If points of equal potential are joint by a wire, the currents in the circuit do not change. The given network of resistors can be reduced successively as shown in figure.

`R_(12) =(rxx(7r)/(5))/(r+(7r)/(5)) = 7/(12) r`