Twelve wires, each with an equal resistance r , are connected to...

Let the total current entering at junction $a$ be $I$ and leaving at junction $f$ be $I$.

Due to the symmetry of the cube, the current divides equally into three identical paths at junction $a$.

Current in branches $ab$, $ad$, and $ah$ is $\dfrac{I}{3}$.

At junction $b$, the current $\dfrac{I}{3}$ further divides equally into branches $bc$ and $be$ because the paths to $f$ are symmetric.

Current in branches $bc$ and $be$ is $\dfrac{I}{6}$.

Similarly, the current in branches $dc$, $dg$, $he$, and $hg$ is $\dfrac{I}{6}$.

At junction $c$, currents from $bc$ and $dc$ combine. So, the current in branch $cf$ is $\dfrac{I}{6} + \dfrac{I}{6} = \dfrac{I}{3}$.

The potential difference between points $a$ and $f$ along the path $a \rightarrow b \rightarrow c \rightarrow f$ is:

$V_{af} = V_{ab} + V_{bc} + V_{cf}$

$V_{af} = \left(\dfrac{I}{3}\right)r + \left(\dfrac{I}{6}\right)r + \left(\dfrac{I}{3}\right)r$

$V_{af} = Ir \left(\dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{3}\right) = \dfrac{5}{6}Ir$

The equivalent resistance $R_{eq}$ is given by:

$R_{eq} = \dfrac{V_{af}}{I} = \dfrac{5}{6}r$