A battery of 15 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance `5Omega` 22. 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 incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff.s first rule and the symmetry in the problem. Next take a closed loop, say, ABCC.EA, and apply Kirchhoff.s second rule:
`-IR - (1//2)IR - IR + E` = 0 where R is the resistance of each edge and E the emf of battery. Thus, E =`(5)/(2)IR`
The equivalent resistance Ras of the network is
`R_(eq)`=`(E)/(3I)`=`(5)/(6)R`
For R = `6Omega`,` R_(eq)` = `5Omega` and for E = 15 V 
the total current = 31 ) in the network is
3I = `15//5 `= 3 A, i.e., I = 1A
The current flowing in each edge can now be read off from the figure.