by neepi on 6/14/25, 10:46 PM
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.
by bgnn on 6/15/25, 10:14 AM
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!
by kayson on 6/15/25, 2:26 AM
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.
by mmastrac on 6/14/25, 11:13 PM
This was the question I hated in my EE degree. The thought exercise was a favourite of the profs.
by praptak on 6/15/25, 5:25 AM
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.
by quibono on 6/14/25, 11:26 PM
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?
by at_a_remove on 6/15/25, 12:53 PM
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.
by causality0 on 6/15/25, 1:39 AM
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.
by dogman1050 on 6/15/25, 10:42 AM
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
by Balgair on 6/15/25, 12:30 PM
by TheOtherHobbes on 6/15/25, 9:25 AM
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.
by 1970-01-01 on 6/15/25, 12:59 PM
This is also known as a high pass filter for first year EE students.
by pyman on 6/15/25, 2:03 AM
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
by Koshkin on 6/15/25, 1:31 PM
by shove on 6/15/25, 12:05 AM
Word on the street was that my Physics professor at NCSSM (Dr Britton) worked on this problem during his doctorate
by clbrmbr on 6/15/25, 12:26 AM
The finite grid of resistors (or arbitrary impedances) is actually of great practical usefulness.
by bilsbie on 6/15/25, 1:22 PM
Dumb question but why isn’t a vacuum considered an infinite grid of resistors?
by nimish on 6/15/25, 1:16 AM
In the integral, the h_m(s) are chebyshev polynomials of the first kind
by bilsbie on 6/15/25, 1:23 PM
Offshoot question. Why don’t we make resistors by making wire so thin that only a certain current can fit through?
Wouldn’t that be more efficient than converting current to heat?
by petschge on 6/14/25, 10:16 PM
by sriku on 6/15/25, 5:21 AM
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.
by kevinmhickey on 6/15/25, 1:28 AM
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.
by rwmj on 6/15/25, 10:17 AM
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 ...
by steamrolled on 6/15/25, 12:33 AM
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.