A battery of $10\mathrm{V}$ 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.). Determine the equivalent resistance of the network and the current along each edge of the cube.

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 ${\mathrm{AA}}^{\prime },\mathrm{AD}$ and $\mathrm{AB}$ are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, $I$. Further, at the corners ${\mathrm{A}}^{\prime },\mathrm{B}$ and $\mathrm{D}$, 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, $\epsilon =\frac{5}{2}IR$ The equivalent resistance $R_{eq}$ of the network is $R_{\text{eq }}=\frac{\epsilon }{3I}=\frac{5}{6}R$ For $R=1\Omega ,R_{eq}=(5/6)\Omega$ and for $\epsilon =10\mathrm{V}$, the total current $(=3I)$ in the network is $3I=10\mathrm{V}/(5/6)\Omega =12\mathrm{A}\text{, i.e., }I=4\mathrm{A}$ The current flowing in each edge can now be read off from the Figure.