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# Solving the resistor cube

It has been offered, in jest I hope, that this question originated sometime in the middle of the last ice age, just after a successful mastodon hunt. A loop analysis or a node analysis would yield the same answer as shown here, but with much more mental effort. Sometimes, if we keep an open eye, there is an easier way.

This question is actually demonstrative of the benefits that symmetry can offer in some circuit analysis cases. To wit, what is the resistance between the opposite corners of a cube made up of twelve equal resistors (Figure 1)?

Figure 1 A resistor cube consisting of 12 equal resistors on each edge where we are prompted to determine the resistance at each corner. Source: John Dunn

You can write the node or the loop equations to solve this problem, but the process will be tedious. You’ll have to forgive me, but using SPICE would be cheating, at least in my opinion. Trying to see what would go in parallel with what could leave you babbling incoherently. However, using some simple observations and recognizing where symmetry applies will lead smoothly to the correct result.

In Figure 1, we choose node A and node H as opposite corners of the cube. We inject a current into those opposite corners and just to make things easy, we let that current be one ampere.

First, we look at current divisions.

Because we have symmetry, we can state that three resistors leading away from node A will each carry one-third of the injected current. Node B, node C and node D will each receive one-third of an ampere from their respective resistors connected back to node A.

Again, due to symmetry, we can further state that the two resistors leading away from node B, node C and node D will each carry one-half of their respective nodes’ arriving currents. Node B will deliver one-sixth of an ampere each to node F and node G, node C will deliver one-sixth of an ampere each to node E and node G and node D will deliver one-sixth of an ampere each to node E and node F.

Next, we look at current summations.

Node E will receive two currents of one-sixth of an ampere each from node C and node D, node F will receive two currents of one-sixth of an ampere each from node B and node D while node G will receive two currents of one-sixth of an ampere each from node B and node C.

Node H will receive three currents of one-third of an ampere each from node E, node F and node G.

To be sure, this whole chain of reasoning is a linguistic mouthful, but a visual examination of Figure 1 will make it all very easy to see.

If we let each resistor be one ohm, the voltage drops across each resistor that carries one-third of an ampere will be one-third of a volt while the voltage drops across each resistor that carries one-sixth of an ampere will be one-sixth of a volt.

If we add up the voltage drops from node H to node A, the result is five-sixths of a volt no matter what path we choose. Therefore, the corner-to-corner resistance of the cube comes to five-sixths of one ohm.

Quod erat demonstrandum.

John Dunn is an electronics consultant, and a graduate of The Polytechnic Institute of Brooklyn (BSEE) and of New York University (MSEE).

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