I know! I know! Pick me.

I'm glad to see this one come back.

M@Man wrote:

The equivalent resistance of the cube is .  I calculated the answer by writing out conservation of energy over loops around the faces of the cube and conservation of charge at the vertices of the cube (i.e. I used Kirchhoff's Rules).  This gave me a system of 14 equations in 12 variables (the currents through each edge).  I solved the system by writing it as a matrix and row-reducing it to lowest form, which gave me the currents over the cube.  Then, since the potential is independent of the path taken to go from A to B, I just chose a particular path and summed the potential differences across it.

This is the “brute force” approach. (Nothing at all wrong with it, but an experienced electrical engineer likely would look for an easier way, based upon the symmetry of the configuration.)

Looking at the answer and the distribution of currents, it  makes perfect sense; because the resistances are all the same, the current just divides by the number of branches at each vertex, so the three currents leaving point A are I/3, the three currents entering point B are I/3, and all the rest are I/6.

You got it! This is what I meant when I referred to “the symmetry of the configuration.”

Here is what the current distribution looks like:




Here is what the cofiguration looks like when you “stretch it out”:



Once you realize that the vertices 1, 2, and 3 are at the same potential, and the vertices 4, 5, and 6 also are at the same potential, you can see that those respective sets of vertices can be connected by “virtual short circuits” (as shown below), transforming the 3-dimensional cube configuration into a more-familiar planar configuration, and the resistance between A and B readily calculated as a parallel combination of 3 R-ohm resistances in series with a parallel combination of 6 R-ohm resistances, in series with a parallel combination of 3 R-ohm resistances:

Since the potential is the same at points 1, 2, and 3 then we can treat the problem as if there was a "wire" connecting those three points directly.

The potential w/r/t A that the "front end" of eah resistor "sees" is the same.

so if they were all 1k ohm resistors what would be the total resistance?