# Total Resistance of Infinite Resistor Grid?

The problem of the infinite resistor grid is very common. The solution for the resistance between any 2 nodes in an infinite resistor lattice is all over the internet.

My question is somewhat similar but more pragmatic. If we had a grid that was very large but yet finite... Then what would be the average voltage drop across a given grid for a given current density?

For arguments sake, a grid in the region of say 4000 by 4000. Maybe it would be safe to assume an infinite grid(?)

Very interesting Q. Can anyone shed any light?

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I presume you're talking about the problem as described by Randall Munroe (of xkcd) at his talk at Google. – Warrick Nov 15 '11 at 10:15
– user1011182 Nov 15 '11 at 10:29

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@user1011182: How exactly do you define the notion of total resistance, cf. question(v1), if you say that it is not between two points in the grid? Using some limit? – Qmechanic♦ Nov 15 '11 at 11:53

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I do not know what I want, but I want it really! – Georg Nov 15 '11 at 13:44

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It's still not clear to me what you're asking for... you can't talk about the entire grid if the grid is infinite. So is it a finite grid, or are you looking for the resistance between two given nodes? – David Zaslavsky♦ Nov 15 '11 at 20:57
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## 4 Answers

The total resistance of the grid is infinite when the grid is two dimensional and large.

If you place two point probes at location x and y on an infinite 2-d resistor grid, and impose the voltage V(x)=1 and V(y)=0, the potential obeys the discretized Laplace equation: V(up) + V(down) + V(left) + V(right) - 4 V(center) = 0 with the boundary conditions at the two given points and V=0 at infinity (beyond x and y).

In the limit that x and y are far apart, the discrete Laplace equation might as well be the continuous Laplace equation, and the solution goes like C log(|r-x|/|y-x|), so that the potential difference for any finite C diverges with the distance. This means that C has to go to zero in the large |x-y| limit, so the current vanishes. The same is true in 1d, where a line of resistors has a current which vanishes as 1/L, so the total resistance goes as the total length L. In two dimensions, the total resistance blows up as log(L).

For a three dimensional grid and higher, you do have a finite resistance for a block. Whether the limiting resistance is finite or infinite is the same problem as the recurrence/nonrecurrence of a random walk on the grid.

If you make a pseudo two-d grid using N parallel lines of N resistors in series, then the total resistance is N on each path, but there are N parallel paths, so the total resistance is R, independent of the size. This is not the same as the 2-d resistor grid, because in the 2d grid there is resistance to going vertically a long way which is similar to the resistance to going horizontally, so the horizontal resistor paths are not parallel. If you make all the vertical resistors zero, and make the separation between x and y horizontal, and make the vertical width equal to |x-y|, you recover the series/parallel situation.

The series-parallel example gives intuition about why two dimensions is critical for the transition from infinite resistance to finite resistance.

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I do not understand this ignorant downvote. – Ron Maimon Nov 15 '11 at 22:00
No obvious errors, answers the question, +1 to cancel downvote. – Alexander Nov 15 '11 at 22:29
@alexander: thanks. There were a few bloopers, and I fixed those (fingers move too fast, I last worked out this problem 20 years ago) – Ron Maimon Nov 16 '11 at 0:24

You haven't been very specific about how your voltage probes are attached to this infinite grid.

We could imagine the grid is stretched between two vertical conducting surfaces, one at $x=-\infty$ and one at $x=+\infty$. Then the problem is symmetric under discrete translations in the $y$ direction. In this case it is easy to see that you will have constant current flowing through the horizontal nodes, 0 current through the vertical nodes. Let $N_x$ be the number of nodes in the $x$ direction (to be taken to $\infty$, $N_y$ the number of nodes in the $y$ direction, and $I$ the current flowing through any horizontal resistor.

Total voltage drop $V = N_x * I * R$, total current $I_{tot} = I*N_y$, net resistance $R_{tot} = V/I_{tot} = N_x * R / N_y$, so you can get any answer you want depending on how you take the limit.

Now this probably isn't what you had in mind. You probably had in mind putting two point probes in the network, and taking the locations of those two point probes to $+/- \infty$. That's a more interesting problem.

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I don't get why you speicify that the current flows only through horizontal nodes? – user1011182 Nov 16 '11 at 20:21
Because in this situation, the source and the sink are both infinite and translation invariant in y. This gives the total resistance of a long 1-d chain, and the resistance in this case goes linearly in the length. – Ron Maimon Nov 17 '11 at 18:08

I remember an interesting result that the resistance of uniform sheet is independent of the distance between the two points. And so conductive mats are labelled "ohms/square"

Doesn't an infinite grid of resistors approximate to this?

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Here is a way an electrician solves the problem:

To get an analytical approximation let's approximate the large grid by a solid, homogenous metal sheet with thickness $H$.
Let 2 nodes be 2 cylindrical conductors of radius $r$ both. The distance between nodes, let it be $L$.
Also, for simplicity assume that the conductivity of the nodes is much greater than the conductivity of the sheet's material.

Because of the last assumption we can take that the nodes have a constant potential throughout their lengths. So, to determine the electric field we can consider an electrostatic problem:

Let the linear charge density on conductors be $±\lambda$.

Applying Gauss' theorem to one of the nodes, we find that the field strength of the node at a distance $l$ from its axis is equal to

$$E=\frac{\lambda}{2\pi\epsilon_0l}$$

The potential difference between the nodes are obtained by integration of the field:

$$U=\frac{\lambda}{2\pi\epsilon_0}\int_{r}^{L-r}\left ( \frac{1}{l}-\frac{1}{L-l}\right)dl\approx\frac{\lambda}{\pi\epsilon_0}\ln\frac{L}{r};L>>r$$

Assuming that the current density $j=\gamma E$ ($\gamma$ is sheet's conductivity) is constant over the thickness of the sheet, we obtain for the total current flowing out of a cylindrical node:

$$I=2\pi rHj=2\pi rH\gamma E=\frac{H\lambda\gamma}{\epsilon_0}$$

So, the resistance between 2 nodes approximately:

$$R=\frac{U}{I}\approx\frac{1}{\pi\gamma H}\ln\frac{L}{r}$$

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