Twelve identical wires, each of resistance 12 Ω, are joined to form a cube. A battery of emf 24 V and internal resistance 2 Ω is connected across the…

By symmetry the cube between opposite corners splits into three groups of wires: three wires from the first corner, six middle wires, and three wires into the last corner. The equivalent resistance is $\frac{R}{3}+\frac{R}{6}+\frac{R}{3}=\frac{5R}{6}=10\ \Omega$ for $R=12\ \Omega$. The total circuit resistance including the battery is $10+2=12\ \Omega$, so the current is $I=24/12=2$ A. At the first corner the current divides equally among the three wires, so each carries $2/3$ A. At the next set of vertices each current of $2/3$ A splits equally into two middle wires, so each middle wire carries $I/6=1/3$ A. The potential difference across a middle wire is $\frac13\times12=4$ V.