infinite square grid of resistors

Question

Given an infinite, 2-d, square grid of 1 Ohm resistors, what is the resistance between two adjacent nodes? (Something like a very large window screen, where the wires have finite resistance, but no contact resistance.)

I was given this problem in a math competition (I think it was the Putnam Competition) many years ago, so it must have an answer that can be derived in a finite amount of time with a finite amount of intelligence, but I have never been able to figure it out.

My best idea has been to consider the case of a uniform conductive sheet. I couldn't get it to work, but maybe it will give somebody else a better idea.

At large scales, the sheet will behave the same as the grid if the surface resistivity is equal to the resistance, $R$, of the individual resistors (here 1 Ohm). A current, $I$, fed in at one point will disperse radially with current density equal to $I/(2\pi r)$. Therefore the electric field will be IR/(2pir), and the potential difference between two radii will be equal to $\phi_2 - \phi_1 = \frac{IR}{2\pi}\cdot\ln(r_2/r_1)$.

In this approximation, if the nodes are at $(x=i+0.5d, y=j\cdot d)$, then the Cartesian distance from $(i=\pm0.5, j=0)$ to $(i, j)$ is $(i\pm 0.5)^2 + j^2$. If current is fed into node $a$ and the same current is extracted from node $b$, the solution will be a superposition of the two individual solutions.

$\phi(i, j) = \frac{IR}{2\pi}\cdot\ln(r_{2a}/r_{1a}) - \frac{IR}{2\pi}\cdot\ln(r_{2b}/r_{1b}) = \frac{IR}{2\pi}\cdot\ln\left(\frac{r_{2a}\cdot r_{1b}}{r_{1a}\cdot r_{2b}}\right)$

Let the reference point 2 be on the line of symmetry, where the potential must be zero, so that $r_{2a} = r_{2b}$.

$\phi(i, j) = \frac{IR}{2\pi}\cdot\ln(r_{1b}/r_{1a}) = \frac{IR}{2\pi}\cdot\ln\left(\frac{(i+0.5)^2 + j^2}{(i-0.5)^2 + j^2}\right)$

If this is really a solution, then current continuity at each node requires that the potential at the node be the mean of the potential at the 4 adjacent nodes.

$\ln\left(\frac{(i+0.5)^2 + j^2}{(i-0.5)^2 + j^2}\right) = \ln\left(\frac{(i-0.5)^2 + (j-1)^2}{(i-1.5)^2 + (j-1)^2}\right) + \ln\left(\frac{(i-0.5)^2 + (j+1)^2}{(i-1.5)^2 + (j+1)^2}\right) + \ln\left(\frac{(i+1.5)^2 + (j-1)^2}{(i+0.5)^2 + (j-1)^2}\right) + \ln\left(\frac{(i+1.5)^2 + (j+1)^2}{(i+0.5)^2 + (j+1)^2}\right)$

If you haven't noticed yet, this is not going to get me to a solution. Who has a better idea?

Answer 1

This question should be well-known and easy to find on the internet.

related pages are for example;

http://www.mathpages.com/home/kmath668/kmath668.htm

https://physics.stackexchange.com/questions/2072/on-this-infinite-grid-of-resistors-whats-the-equivalent-resistance

Answer 2

OK. Got it, mostly through the help of the mathpages link posted by both almagest and Stijn Cambie. You can ignore all my unformatted equations because that approach is much too complicated for the simple question asked, and doesn't lead to a solution anyway. All you need to know is that current injected into one node will disperse symmetrically into the 4 adjacent nodes, and that solutions can be superimposed. The solutions you want to superimpose are one that injects a current, I, into one node, and one that extracts the same current from an adjacent node. The current flowing through the resister joining the nodes will be I/4 from the injection solution, plus I/4 from the extraction solution, for a total of I/2. Since exactly half the current flows through the resistance, R (1 Ohm), the resistance of the rest of the network connected in parallel must also be equal to R. The parallel combination of R and R is R/2 (0.5 Ohm). That's the answer. Thanks for leading me to the solution.