Past exam of the physics 2 paper of the University of Oxford / 2016 resit / 6 / Solution

The adjacent-corner resistance of a resistor cube follows from Kirchhoff's current law and cube symmetries. Hold adjacent corners $A=(0,0,0)$ and $B=(1,0,0)$ at potentials $V$ and zero. The two corners $(0,1,0),(0,0,1)$ have a common potential $x$; their neighbours $(1,1,0),(1,0,1)$ have $V-x$. Write the remaining two potentials as $z,V-z$.

At a corner of the first pair and at $(0,1,1)$, Kirchhoff's current law givesThus $4x-z=2V$ and $4z-2x=V$, yielding $x=9V/14$, $z=4V/7$. The supplied current at $A$ isThe direct edge between the terminals is one of three current paths leaving $A$; treating the other eleven resistors as a single obvious series chain would miss the branching.