# Equivalent Resistance of a Cube of unequal resistors

I've been completely stumped on this one physics problem. Essentially, it's a cube of resistance, but two things have been changed: there is a resistance along one of the diagonals of the cube, and the cube is made up of resistors with different resistances.

My solving process started out by trying to find points of equal voltage and connecting them, but it just made the problem more cumbersome. Then, I tried making a $2d$ map of the circuit, but again, the endeavour was fruitless.

Now, my thought process is this: Since the entire cube is just a parallel circuit, I should be able to get both the total current and the voltage through a single path by sending through some current in A and using Kirchhoff's laws to determine how the current splits. Then, using $V=IR$, I could calculate the voltage for each resistor in a path. Adding those resistances together, I could use $V=IR$ in the form of $\frac{V}{I}=R$, using total current and voltage of the circuit to get the resistance.

Am I right or completely off track? Also, is there an easy way to map this to a $2D$ plane so I can actually work with it feasibly?

First, a few remarks:

Then, I tried making a 2d2d map of the circuit, but again, the endeavour was fruitless.

In fact, that is an example of a non-planar network, that is, a network that cannot be drawn on a plane without any intersection.

Since the entire cube is just a parallel circuit

No, it is not. There are also "bridge" connections.

If the resistances are all different, there's really not much you can do to simplify the analysis of that circuit.

Use nodal analysis, choosing node b as ground: the system of (7!) equations can be written directly by inspection, and can then be solved either numerically, if the resistance values are given, or with a computer algebra system.

You can't draw it in the plane without showing a branch intersect. I realized I didn't trust my original argument why so I edited this answer with another one. Adding the extra diagonal lets you connect each of the three points in blue to each of the three points in red in the following diagram.

This diagram or graph is called the complete bipartite graph $K_{3,3}$ and is known to be non-planar. In fact every non-planar graph contains either $K_{3,3}$ or $K_5$ as a homomorphic subgraph. This is called Kuratowski's theorem.

Back to physics, you can do what you suggested and define a current in each branch and the voltage across each resistor and set up a messy system of equations using Kirchoff's laws. It doesn't need to be a planar diagram to do that.