Infinite Grid of Resistors

mathpages.com

177 points by niklasbuschmann 15 hours ago


1970-01-01 - a minute ago

This is also known as a high pass filter for first year EE students.

bgnn - 3 hours ago

People think this is not relevant to real world problems but it actually is, albeit all the calculations aren't that relevant. Silicon substrate's resistance is basically an infinitely large grid of unut resistances at the distances relevant for a local point of an IC. Note that silicon substrate is often heavily doped (p-type) and all info you get from the fab is it's resistivity (often somewhere between 1 to 100 ohm per cm). For the most advanced tech nodes its often 10 ohm/cm. If you need to develop some intuition about noise coupling via the substrate you have to think that it's a grid instead of just calculating the resustance between point A and B. We need to distribute a grid of substrate contacts to collect the noisy currents too. So the grid shows up again!

neepi - 14 hours ago

I'm a bit mathematician and a bit electrical engineer.

The electrical engineer suggests it's not measurable unless you apply current and also asks "when" after the current is applied referring to the distributed inductive and capacitive element and the speed of field propagation. The mathematician goes to a bar and has a stiff drink after hearing that.

Balgair - 31 minutes ago

Aside: Veritasium had a great video similar to this on the paths that light takes. I'll link to the part where they do the best physics demo I have ever seen:

https://www.youtube.com/watch?v=qJZ1Ez28C-A&t=1500

kayson - 11 hours ago

A much more useful (in the educational sense) question to ask, in my opinion, is the resistance between opposite corners of a cube of 1ohm resistors. There are some neat intuitions it can help build (circuit symmetry, KCL, etc). The infinite grid is too much an obscure math problem that seems like it might be solvable in an introductory circuits class.

at_a_remove - 7 minutes ago

Odd. As an undergrad in physics, we had a project for our team which involved percolation theory and "testing" it. So, we had to make differing grids of conductive ink, with a certain number of "links" (resistors, edges in the graph) as missing. Getting even-flowing conductive ink was hard. I wrote all of the software for the XY plotter, pushing out instructions to make rectangular and triangular grids. Then we would measure the resistance from one side to another.

dogman1050 - 2 hours ago

This is a discrete case of "sheet resistance."[1] The resistance between any two points, nodes in this case, is the same. We covered this in the EE uni curriculum back in the day, but I don't remember the solution derivation anymore.

[1] https://en.m.wikipedia.org/wiki/Sheet_resistance

mmastrac - 14 hours ago

This was the question I hated in my EE degree. The thought exercise was a favourite of the profs.

praptak - 8 hours ago

What I never got about the simple symmetry-based solution is "if we accept the idea that we can treat the current fields for the positive and negative nodes separately".

Why are the currents in the two node solution (not symmetric) a simple sum of the currents of two single node solutions (symmetric)?

Obviously the 2 node solution still has some symmetries but not the original ones that let us infer same current in every direction.

TheOtherHobbes - 4 hours ago

At infinite scale this reduces to the bulk equation R = rl/A for a rectangular block where r is resistivity, l is length, and A is the area of the block.

Both l and A are infinite. So you get infinity/infinity, which is undefined, proving it's a silly problem and you should go do something useful with your time instead.

quibono - 14 hours ago

There's one thing I don't get about the symmetric+superposition explanation. Why are there alpha - beta - alpha on the adjacent nodes, and not alpha-alpha-alpha? I.e. why is one of the directions distinct while the other two are considered the same?

pyman - 11 hours ago

Re: the infinite resistor grid

If you take an endless grid made of identical resistors and try to measure the resistance between two neighbouring points, the answer turns out to be about one-third of a resistor

clbrmbr - 13 hours ago

The finite grid of resistors (or arbitrary impedances) is actually of great practical usefulness.

rwmj - 3 hours ago

Now I'm wondering if anyone has built a very large grid of resistors to try to approximate / curve fit this. Surely there's a youtube video in this ...

nimish - 12 hours ago

In the integral, the h_m(s) are chebyshev polynomials of the first kind

petschge - 15 hours ago

See also https://xkcd.com/356/

sriku - 8 hours ago

This is cool and I have my own take on it after being nerd sniped by XKCD - https://sriku.org/posts/nerdsniped/ - I link to this article at the end but that post specifically solves the xkcd puzzle.

causality0 - 11 hours ago

My math isn't strong enough to follow the whole article, but my intuition as someone who works in electronics is that when a quantized system interacts with an infinity, the infinity is restricted based on the magnitude of the quantized factor. Electric charge is quantized. Less than one electron cannot pass through a node, therefore an infinite grid of resistors is effectively a finite grid of resistors whose size changes based on how much charge is dumped into the system.

shove - 13 hours ago

Word on the street was that my Physics professor at NCSSM (Dr Britton) worked on this problem during his doctorate

- 13 hours ago
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steamrolled - 12 hours ago

I don't get why EE education emphasizes problems of this sort. The infinite grid is an extreme example, but solving weirdly complicated problems involving Kirchoff's laws and Thevenin's theorem was a common way to torture students back in my day...

Here, I don't think it's even useful to look at this problem in electronic terms. It's a pure math puzzle centered around an "infinite grid of linear A=B/C equations". Not the puzzle I ever felt the need to know the answer to, but I certainly don't judge others for geeking out about it.

kevinmhickey - 12 hours ago

In school I would have tried to solve this… now if I want to know I just get out my multimeter and measure. Faster, simpler, and more practical.