A bettery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance `1Omega` Determine the equivalent resistance of the network and the current along each edge of the cube.

Explanation

The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.
The paths AA..AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners A.,B and D, the incming current I must split equially into the two outgoing branches. In this manner, the current in all the 12 deges of the cube are esaily written down in terms of I, using kirchhoffs first rule and the symmetry in the problem.
Next take a closed loop, say. ABCCEA, and apply kirchhoffs second rule:
` epsi = 5/2 IR`
The equivalent rsistance `R _(eq)` of the network is
`R _(eq) = (epsi )/(3I)=5/6 R`
For `R =1 Omega, R _(eq) (5//6) Omega and ` for `in = 10V,` the total current `(=3I)` in the network is
`3I =10 V (5//6) Omega =12A, i.e., I=4A`
The current following in each edge can now be read off from the