$A$ 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 $1 \;\Omega$. Determine the equivalent resistance of the network and the current along each edge of the cube.

(N/A) The network is not reducible to a simple series and parallel combination of resistors. However,there is 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 symmetrically placed in the network. Thus,the current in each must be the same,say $I$.
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 $12$ edges of the cube can be written in terms of $I$ using Kirchhoff's first rule and the symmetry of the problem.
Consider a closed loop,such as $ABCC'EA$,and apply Kirchhoff's second rule:
$-IR - (1/2)IR - IR + \varepsilon = 0$
where $R$ is the resistance of each edge and $\varepsilon$ is the emf of the battery.
Thus,$\varepsilon = \frac{5}{2}IR$.
The equivalent resistance $R_{eq}$ of the network is:
$R_{eq} = \frac{\varepsilon}{3I} = \frac{5}{6}R$.
For $R = 1 \;\Omega$,$R_{eq} = \frac{5}{6} \;\Omega$.
For $\varepsilon = 10 \;V$,the total current $I_{total} = 3I = \frac{\varepsilon}{R_{eq}} = \frac{10}{5/6} = 12 \;A$.
Therefore,$I = 4 \;A$.
The current in the edges connected to the battery is $4 \;A$,the current in the middle edges is $2 \;A$,and the current in the edges connected to the opposite corner is $4 \;A$.