Battery 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 `1 Omega` (Fig. 3.23). 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+epsilon=0`
where R is the resistance of each edge and `epsilon` the emf of battery. Thus, `epsilon=(5)/(2)IR`
The equivalent resistance `R_(eq)` of the network is
`R_(eq) = (epsilon)/(3I)=(5)/(6)R`
For `R = 1 Omega, R_(eq) = (5//6)Omega` for `epsilon = 10 V`. the total current `(=3I)` in the network is
`3I = 10 V//(5//6)Omega = 12 A,` i.e., `I = 4A`
The current flowing in each edge can now be read off from the fig.