# 013 Can you solve the Resistor’s Cube?

Resistance across the cube Let’s imagine a current I passing across the cube. By symmetry, the current will split equally at each junction. We can thus tell that the current at each of the blue, green and red resistors will be I/3, I/6 and I/3 respectively.

This means that the potential difference across the cube will be

V = I/3 *R + I/6 *R + I/3 R = 5/6 IR.

Since resistance is the ratio of V to I, Reff = 5/6 R

Resistance across a face If the two red resistors were not there, then the effective resistance between points a and b is simply

Rab     = ( 2R || 2R ) || [ R + (2R||2R) + R]

= R || 3R

= ¾ R

Then we notice that points c and d are at the same potential. This means even if the two red resistors were there, no current will pass through them anyway. This means the two red resistors can be removed without affecting the circuit in any way.

Resistance across an edge By symmetry, we can tell that points c and d are at the same potential, so these two points can be shorted together without affecting the circuit in any way. Likewise, points e and f can be shorted together since they are at the same potential. The circuit can now be redrawn as shown above. The effective resistance between point cd and ef can be simplified to

Rcd,ef = [(R||R) + R + (R||R)] || R || R

= 2R || R || R

= 2/5 R

The effective resistance across an edge will thus be

Rab     = [(R||R) + 2/5 R + (R||R)] || R

= 5/7 R || R

= 7/12 R