## The Solution

A few days ago I posted a little puzzle about the resistance measured between two adjacent nodes in an infinite square grid made of of 1Ω resistors. There was a bit of discussion after the post that hinted at the solution but, since a few people have asked me about it, I thought I’d post a fuller answer here.

The quickest way I know to get the answer uses the Principle of Superposition, as illustrated in the following picture.

Consider two copies of the grid, both earthed at infinity. Imagine injecting a current of 1 Amp into the grid through a wire attached to node A as shown in the top left of the picture. The current will run to earth through the grid, but, by symmetry, it is obvious that 1/4 of the current entering the grid through A must travel through each of the 4 wires radiating out from it. Each of the wires leading out from A therefore carries 0.25 Amps in this solution.

Now, in the top right hand picture, forget about A, but attach a wire to B and pull out 1 Amp from the earth (at infinity). By a similar argument to the first diagram, 0.25 Amps must be flowing into B along each of the wires connected to it in the directions shown.

We now have two perfectly good solutions for currents flowing in resistive networks. The principle of superposition says that if we add the two solutions we also get a solution. Adding the two configurations above means that the resistor joining A to B must be carrying 0.5 Amp (0.25 from the first solution and 0.25 from the second, both in the same direction). But this is a 1Ω resistor so the Voltage across AB must be 0.5 V.

Now think of the whole mesh as being a black box in between the input wire and output wire. This black box has a current of 1 Amp flowing through it and the voltage dropped is 0.5 V. It’s resistance is therefore 0.5 Ω.

If anyone has a better solution than this, I’d like to see it!!

February 24, 2010 at 10:29 pm

This analysis generalises immediately to give the resistance between nearest neighbour nodes in other shapes of regular lattice (triangular, hexagonal) and to three dimensions. But it isn’t generalisable to non-nearest-neighbour nodes. (How’s that phrase for alliteration?)

Some pedants have grumbled that the conditions at infinity, where the current leaks away for each of the two solutions that are superposed, cannnot be neglected. In that case the fact that superposition gives the right answer (for experiments have been done) would be mere coincidence. I do see the point, but when you superpose +1A and -1A then the current at infinity clearly falls away much faster, since no net current leaks away to infinity.

Peter, you asked before if there was a comparably sneaky way to get 2/pi as the resistance across a diagonal of any 4-resistor square in the lattice. I guess that the presence of pi means not, but one of the references quoted in the website you linked to (in the previous post) sugests that a further reference does this.

Resistance between arbitrary nodes of a (regular 3D) tetrahedral lattice, anyone?

Anton

February 24, 2010 at 11:06 pm

Anton,

I take the point that it might be a bit glib to let the current out at infinity, but I think it’s clear it cancels out when you add the solutions.

I won’t name names, but one person I mentioned this problem to insisted that the answer was zero rather than half an Ohm. But surely that would also imply that the resistance measured across a continuous sheet of resistive material would also be zero…

Peter

February 24, 2010 at 11:23 pm

Peter,

Interestingly, the resistance between nodes increases indefinitely in 2D with distance between nodes, but in 3D tends to an asymptote – and even a thin sheet would have so many atoms across its thickness as to be 3-dimensional for the purpose of resistance analysis.

What reasons did your anonymous friend give for zero resistance?

Anton

February 24, 2010 at 11:30 pm

It didn’t really make sense, but I think it was something to do with the fact that there are infinitely many paths between A and B.

February 25, 2010 at 12:32 pm

I would like to see your friend doing integrals.

February 25, 2010 at 11:38 pm

Dan,

Did I say it was a friend? (It was a retired observational astronomer …)

Peter

February 28, 2010 at 3:21 am

Beautiful! ^_^