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!
ChoGGi 1 hours ago [-]
My vague understanding of photolithography is that it's hard, though I didn't realise it's bad enough to evoke an egyptian goddess.
I'll see myself out.
eternauta3k 5 hours ago [-]
I'd argue the case you're describing is mathematically simpler precisely because it is continuous.
bgnn 48 minutes ago [-]
True, but the continuous solution is just a limit condition of tge discrete one. It doesn't make it any harder or easier, at least from what I know fron calculus. The software tools use numerical methods to solve this type of problems and they tend to divide the continuous substrate into a mesh of discrete elements to model them as lumped circuit elements so that we can represent them in a matrix and simulate the circuit using linear algebra. They often use random walk in their algorithm to find a mesh which introduces a minimum error.
gugagore 1 hours ago [-]
Right, why is it a 4-connected grid instead of 8-connected, or any other topology, like a hex grid.
Den_VR 4 hours ago [-]
You’re practically describing the invention of Calculus.
neepi 16 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.
Taniwha 16 hours ago [-]
Eventually you need to pullin a physicist too who will point out that at an appropriate distance quantum effects will dominate - because eventually at a far enough distance the number of electrons moving per second (ie current flow) will be either 0 or 1 at some nodes
__MatrixMan__ 11 hours ago [-]
I've not studied QED directly, so by all means correct me if I'm wrong, but it seems to me that we'd get a double-slit like scenario where it's as if a partial electron went through either path. We might want to say that surely a whole electron took one path and not the other but we couldn't say which and if we tried to instrument to and find out we'd affect the resistance.
But that's fine because knowing which path the electron took is not part of the problem. Both paths contributed to the resistance even if one was not taken.
We only have to worry about quantum effects if the probabilities are not a decent proxy for the partial-particles that we suspect don't exist. In this case, the Physicist can probably proceed directly to the bar and have a drink with the Mathematician.
dgfl 4 hours ago [-]
Resistance is inherently dissipative, so there is no coherent path the electron can take. No quantum effects here, the electron is always interacting with the resistor lattice.
viraptor 4 hours ago [-]
I don't think you'll ever get 0/1. You get a difference in voltage that influences all electrons to move slightly more in one direction than another in electric current. They'll just drift very very slightly as a group, not measurable when you get far enough. But they're always all affected, rather than individually.
10 hours ago [-]
mjevans 16 hours ago [-]
Intuitively I knew this class of problem was theoretical only BS when it came up in college...
I hadn't considered that sort of strange effect though! Makes me feel not so bad for 'never really getting it' because I just couldn't wrap my mind around the problem description's obvious inanity and the infinite edges.
aydyn 9 hours ago [-]
The question is not pretending to be realistic. No one is thinking it is possible to build an infinite grid of resistors.
It's simply an evaluation of your mathematical ability to manipulate the equations and overall understanding of them, wrapped up in a cute little thought experiment. This evaluation IS relevant to more realistic scenarios and therefore your grade and engineering ability.
Koshkin 2 hours ago [-]
> apply current
Going on something of a tangent: in engineering, it seems unusual to talk about “applying current,” it’s usually voltage (say, across a resistor) or some sort of an “electromotive force.”
gugagore 1 hours ago [-]
In idealizations, there are both voltage sources and current do m sources.
red75prime 9 hours ago [-]
> The electrical engineer suggests it's not measurable unless you apply current and also asks "when"
Just wait infinite time for all the transient responses to die down. The grid to enter steady state and to became true to the schematic.
eternauta3k 5 hours ago [-]
An infinitely large grid never reaches equilibrium, like it says in the article.
red75prime 2 hours ago [-]
That's why you need to wait longer than ever.
Mostly joking, but an idealized schematic requires idealized conditions to observe.
repiret 13 hours ago [-]
I think there are two interpretations of schematics.
One is where the components on the schematic represent physical things, where the resistors have some inductance and some non-linearity, and some capacitance to the ground plane and so on. This is what we mean by schematics when we’re using OrCad or whatever.
There is another interpretation where resistors are ideal ohms law devices, the traces have no inductance or propagation delay or resistance. Where connecting a trace between both ends of a voltage source is akin to division by zero.
Sometimes you translate from the first interpretation to the second, adding explicit resistors and inductors and so on to model the real world behavior of traces etc. if you don’t, then maybe SPICE does for you.
Infinite resistor lattices exist only in the second interpretation.
divbzero 7 hours ago [-]
An infinite grid of resistors is clearly a toy scenario, but the infinite universe is a reality that astrophysicists try to reason about. I wonder if there are blindspots in astrophysics because we lack intuition about the universe at that scale and are forced to approach it from theory.
bravesoul2 16 hours ago [-]
Given an infinite grid of resistors... would you expect planets to form?
corysama 16 hours ago [-]
They say hydrogen is an odorless colorless gas which, in sufficient quantities, given enough time, turns into people. I’m sure the same could be true of resistors.
bravesoul2 16 hours ago [-]
Resistors are made of heavier elements though. And I remember something like everything wants to become iron (fuse if lighter, decay if heavier)
That said there might be enough energy (infinity!) for anything to be possible.
jfengel 14 hours ago [-]
You get stable things heavier than iron, and they're more common than you'd expect. It's possible that they form in neutron star collisions, which are complete anarchy in atomic terms.
inopinatus 15 hours ago [-]
People are resistors too.
neepi 7 hours ago [-]
Just checked. Around 1.5Mohms today.
temp0826 14 hours ago [-]
I resist that statement
taneq 12 hours ago [-]
Gee, I hope nobody ever puts you in charge of a train!
farhaven 7 hours ago [-]
Well they can still be a conductor, even if they're not a resistor. Actually they'd be a pretty good conductor. A super-conductor, if you will.
taneq 3 hours ago [-]
So wait, if they resist the statement that they're a resistor, are they a resistor or not? I think they're a semiconductor. Maybe they work at a railway junction. :D
QuadmasterXLII 15 hours ago [-]
I think it collapses into a black hole. black hole mass scales with radius, grid mass scales with radius squared
jfengel 14 hours ago [-]
It becomes a black hole, but it doesn't necessarily collapse, at least not at first. A supermassive black hole has very low density and a very gentle gravitational gradient.
All of the mass does end up in the singularity, in finite time (at least for any finite subset of the black hole), but it doesn't automatically become super dense just because it's a black hole. It can remain quite ordinary for a very long time.
kqr 8 hours ago [-]
Wait, can you take this a little slower? I was not aware black holes could have sensible density.
jfengel 2 hours ago [-]
In the formula for a Schwarzchild black hole, the mass is proportional to the radius. Since volume goes up with the cube of the radius, the density drops quickly.
The kind of black hole that forms from a collapsing star is super dense. But a supermassive black hole forms differently, as a denser region of gas and stars during galaxy formation. The density can be lower than that of water. You could be inside it without even realizing anything is odd.
There is no known way to form black holes bigger than that. We had been discussing a pure thought exercise. Though it is possible that the universe as a whole has enough density to be a black hole. (Signs point to no, but it's an open question.)
pixl97 14 hours ago [-]
Thats what I'm not sure about here...
If we assume this is an infinite grid in its own universe then nothing can actually move. The gravitational pull should be the same from every direction. If we assume the grid is perfect then there is no nucleation sites to start a collapse. The grid would be in perfect balance.
The same is thought about our universe. If there hadn't been small quantum fluctuations during the inflationary stage it would have taken much longer for what we see in the modern universe to form.
staplung 14 hours ago [-]
In fact, we might have a different problem: dark energy should tear the grid up into (very large) bits. I guess the question is then would the bits then collapse into black holes or not. I assume so since the mass would not longer be perfectly balanced.
pixl97 14 hours ago [-]
Guess we'll have to wait for an actual answer on dark energy and universal expansion.
What it collapses into depends on the time for the bits to coalesce into. If it's a slow collapse you could get a mass large enough to form a neutron star (like thing) instead of black holes.
If those neutron stars crash into each other they can release a large amount of 'recycled' matter from all over the atomic spectrum back into this universe.
raattgift 11 hours ago [-]
In 1+1 dimensions one can analyse the gravitational behaviour of an infinite line of ...-wire-resistor-wire-resistor-... with an adaptation of Bell's spaceship. Throwing away two dimensions eliminates shear and rotation (and all sorts of interesting matter-matter interactions) so we can take a Raychaudhuri approach.
We impose initial conditions so that there is a congruence of motion of the connected resistors, so that we have a flavour of Born rigidity. Unlike in the special-relativistic Bell's spaceship model (in which the inertial motion of each spaceship identical save for a spatial translation), in our general-relativistic approach none of the line-of-connected-reistors elements' worldlines is inertial, and each worldline's proper acceleration points in a different direction but with the same magnitude. This gives us enough symmetry to grind out an expansion scalar similar to Raychaudhuri's, Θ = ∂_a v^a (<https://en.wikipedia.org/wiki/Raychaudhuri_equation#Mathemat...>). As an aid to understanding, we can rewrite this as 1/v \frac{d v}{d \tau}, and again in terms of a Hubble-like constant, 3H_0.
We can then understand Θ as a dark energy, and with Θ > 0 the infinitely connected line of ...-wire-resistor-wire-resistor-... is forced to expand and will eventually fragment. If Θ < 0, the line will collapse gravitationally.
> no nucleation sites
If Θ = 0 initially, we have a Jeans instability problem to solve. Any small perturbation will either break the infinite ...wire-resistor-wire-..., leading to an evolution comparable to Bell's spaceship: the fragments will grow more and more separated; or it will drive the gravitational collapse of the line. The only way around this is through excruciatingly finely balanced initial conditions that capture all the matter-matter interactions that give rise to fluctuations in density or internal pressure. It is those fluctuations which break the initial worldline congruence.
This is essentially the part of cosmology Einstein struggled with when trying to preserve a static universe.
In higher dimensions (2+1d, 3+1d) the evolution of rotation and shear (instead of just pressure and density) becomes important (indeed, we need an expansion tensor and take its trace, rather than use the expansion scalar above). A different sort of fragmentation becomes available, where some parts of an infinite plane or infinite volume of connected resistors can undergo an Oppenheimer-Snyder type of collapse (probably igniting nuclear fusion, so getting metal-rich stars in the process) and other parts separate; the Lemaître-Tolman-Bondi metric becomes interesting, although the formation of very heavy binaries early on probably mitigates against a Swiss-cheese cosmological model: too much gravitational radiation. The issue is that the chemistry is very different from the neutral-hydrogen domination at recombination during the formation of our own cosmic microwave background, but grossly a cosmos full of luminous filaments of quasi-galaxies and dim voids is a plausible outcome. (It'd be a fun cosmology to try to simulate numerically -- I guess it'd be bound to end up being highly multidisciplinary).
rzzzt 16 hours ago [-]
Assume perfectly spherical through-hole resistors soldered on an infinite PCB.
nofunsir 3 hours ago [-]
Does Schrodinger's cat study Fourier transforms?
Koshkin 1 hours ago [-]
But you already know that it does and it doesn’t at the same time.
sandworm101 15 hours ago [-]
And the electrician knows he can get a 99% answer out of a 10x10 grid on a workbench. The engineer is free to then add more resisters to the periphery until either the grant money runs out or the physicist's publishing deadline approaches.
A really difficult question: At each distance, what percentage of soldering errors in the grid can be tolerated before the fluke meter across the center square detects the fault? (That might actually be a thing as I've heard people talk about using changes of local resistance to detect remote cracks in conductive structures ... like maybe in a carbon fiber submarine hull.)
nerdsniper 13 hours ago [-]
For measuring corrosion in conductive surfaces, “eddy current” testing is often used. It uses AC current of some frequency, so it’s technically measuring inductance rather than resistance.
kayson 13 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.
Balgair 3 hours 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:
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?
mort96 2 hours ago [-]
You're describing thin film resistors, and they exist.
They also just convert current to heat though. Some amount of current moving through a material with some amount of resistance always produces a fixed amount of current in accordance with Ohm's law. You can't really get away from that.
bilsbie 2 hours ago [-]
Thanks. So Is there a physical reason resistors have to make heat? Is it theoretically possible to find a material that limits current but produces very little heat?
I guess the explanations always confuse me. Let’s say a short circuit with no resistors has a certain amount of power. Then we add a resistor and the power in the circuit goes down. The resistor isn’t turning the difference in power into heat, right.
mort96 48 minutes ago [-]
There is no such thing as a short circuit with no resistance, because everything (other than superconductors) has a resistance. If you had a circuit with a magical ideal voltage source and no resistance, you'd have infinite current.
But let's talk about short-circuiting lithium batteries for example. They have a roughly 50 milliohm (aka 0.05 ohm) of "equivalent series resistance".
That means, if you short circuit a lithium battery with a superconductive wire (aka with 0 resistance), the circuit has a resistance of 0.05 ohms. We can compute the current with Ohm's law: I=V/R. V is typically 3.6 volts for li-ion batteries, R is 0.05 ohm, so I (aka current) is 3.6/0.05 = 72 amperes. 72 amperes * 3.6 volts is 259 watts. Now in the real world, the battery's chemistry would step in here and limit current in complicated ways, but this means that under the assumption that our battery would work as an ideal voltage source + a 0.05 ohm resistor, and if there was no extra heat coming from the chemical reactions, a shorted battery would produce 259 watts of heat.
We can add a 1 ohm resistor to the circuit, which means our circuit's combined resistance would be 1.05 ohm. Using Ohm's law again, we find that the current would be 3.6/1.05 = approx 3.43 amperes. 3.43 amperes * 3.6 volts is 12.35 watts of heat.
So thanks to our resistor, we're now producing 12.35 watts instead of 259 watts of heat, because the resistor limits the current going through the circuit. With a higher resistance resistor we'd produce even less heat.
A core idea here is that power consumptions equals heat. I don't understand the physics reasons why, but "this thing consumes 10 watts of power" means the same as "this thing produces 10 watts of heat". Higher resistance means less current which means less watts, which means both less heat and less power consumption because those are the same.
59 minutes ago [-]
analog31 1 hours ago [-]
It's useful to look at the units of measure. Voltage is energy per unit charge. As the charge carriers go across the resistors, their energy changes, and that energy has to go somewhere. It's not always lost as heat in all devices. In an LED, some of the energy is "lost" as light. But still, the sum total of heat and light power generated by an LED is equal to the product of the current and the forward voltage.
Another useful heuristic is that heat is generated from what's left after all of of the other ways of converting energy are used up, such as light, chemical potential, and so forth. It's energy's last resort. The usefulness of a resistor lies in its simple voltage-current relationship, which is equivalent to saying that the only thing it generates is heat.
grues-dinner 1 hours ago [-]
Resistance is V/I. You literally cannot have current flowing in a resistor without a voltage across it (either the voltage causes the current to flow, or the resistor in the path of a flowing current has a voltage appear across it).
A voltage drop with a flowing current is power (P=VI).
There is literally nothing you can do to avoid resistors dissipating that power as heat, it's just what they are. If they didn't do it, they wouldn't be resistors.
What you can do is use larger resistances which need less current to see the same voltage (e.g. change a pull up from 10k to 100k or higher, but that's more sensitive to noise), or smaller resistances that drop less power from a given current (e.g. a miiliohm-range current shunt, and then you need a more sensitive input circuit) or find another way to do what you want (e.g. a switched-mode power supply is far more efficient than a voltage divider at stepping down voltage). This is usually much more complex and often requires fiddly active control, but is worth it in power-constrained applications, and with modern integrated technology, there's often a chip that does what you need "magically" for not much money.
This is also known as a high pass filter for first year EE students.
mmastrac 16 hours ago [-]
This was the question I hated in my EE degree. The thought exercise was a favourite of the profs.
dcassett 3 hours ago [-]
I saw this question only once, as the first of 4 problems on the final exam for my very first EE introductory course. The course had covered an infinite ladder of resistors, but at the time it seemed like quite a leap to apply that knowledge to this problem.
dogman1050 5 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.
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.
IronyMan100 7 hours ago [-]
the Maxwell Equations are linear in the electric and magnetic fields, then you can add Up and subtract fields and Potentials from each other. It's the same Argument for why interfernce works or optical gratings
bilsbie 2 hours ago [-]
Dumb question but why isn’t a vacuum considered an infinite grid of resistors?
TheOtherHobbes 6 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 16 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?
magicalhippo 15 hours ago [-]
Start by assuming they could potentially be all different, so denote the currents i_1 to i_12.
However note the problem is symmetrical about the vertical axis, so flip the figure. The current passing through the flipped paths should be the same as before the flip, so note down which i's equate to each other due to this.
Note that the problem is symmetrical about the horizontal axis, and do the same there. Note that the problem is symmetric when rotated 90 degrees, so do that. And so on.
In the end you'll have a bunch of i's that are equal, and you can group those into two distinct groups. Call those groups alpha and beta.
edit: Another way to look at it is that you can't use the available symmetry operations to take you from any of the alphas to a beta. This is unlike alpha to alpha, or beta to beta.
pyman 13 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
at_a_remove 2 hours 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.
clbrmbr 15 hours ago [-]
The finite grid of resistors (or arbitrary impedances) is actually of great practical usefulness.
Kirr 15 hours ago [-]
This may be as good time as any to plug my calculator for finite resistor networks (including grids) [1]. It works by eliminating non-terminal nodes one by one with the Star-Mesh transform, while keeping the exact rational resistances at each point.
Why mathematicians are three points? I think it is easier to disable a mathematician. Look at this discussion, for example. EE engineers and physicists are dismissing the problem outright, while mathematicians have no issues thinking about it.
quinndexter 7 hours ago [-]
-Why mathematicians are three points?
Possibly based on this ranking. Everything sub-mathematician is 2 points? Maybe there's subdivision of points.
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 ...
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 14 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.
eternauta3k 5 hours ago [-]
That only matters if you're measuring in the time domain and seeing the noise due to individual carriers. Often you just care about averages over some time and space (e.g. the macroscopic flow of water behaves quite different from the speeds of the individual molecules).
morepedantic 10 hours ago [-]
That was my initial thought, but on further reflection it feels wrong. The electron is also a wave, and that wave can spread across the entire grid.
Another interesting aspect is that in an infinite grid, a spontaneous high voltage is going to exist somewhere at all times. It is probably very far away from you, but it's still weird.
yusina 9 hours ago [-]
Funny to put "intuition" and "infinity" into the same sentence.
The only type of person for whom intuition about infinity to form is not entirely unlikely are mathematicians.
shove 15 hours ago [-]
Word on the street was that my Physics professor at NCSSM (Dr Britton) worked on this problem during his doctorate
16 hours ago [-]
steamrolled 15 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.
goochphd 14 hours ago [-]
I was about to say "they still torture students this way" but stopped myself when I remembered I took Circuits 1 and 2 back in 2007. So maybe my knowledge is dated too...
It's a weird butterfly effect moment in my career though. I had an awesome professor for circuits 1, and ended up switching majors to EE after that. Then got two more degrees on top of the bachelor's
jesuslop 14 hours ago [-]
If all you mind is the EE curriculum then ok. Or else there is an interesting work of Gerald Westendorp on the web [1] on how allowing other classical passive components (Ls & Cs) you can get discretizations (and hence alternative views) of a very wide class of iconic Physics partial differential equations (to the point that the question is more what cannot be fit to this technique). G. W. is alive and kicking in mathstodon.
There are two parts to education. One is to impart knowledge, the other is to filter the students.
Nevermark 13 hours ago [-]
The third is to challenge students. With unusual concepts, preferably.
How else to create students capable of solving problems we cannot anticipate today?
Not to mention, that understanding strange problems is a very efficient way to broaden horizons.
dwattttt 13 hours ago [-]
You're missing general problem solving. If all people do is encounter problems they've already seen before, well, we have lookup tables for that kind of thing.
colechristensen 14 hours ago [-]
Not entirely wrong but it's a little too easy to use that argument for squashing any criticism for education content.
ohxh 6 hours ago [-]
> Here, I don't think it's even useful to look at this problem in electronic terms
I always thought this problem was a funny choice for the comic, because it’s not that esoteric! It’s equivalent to asking about a 2d simple random walk on a lattice, which is fairly common. And in general the electrical network <-> random walk correspondence is a useful perspective too
bobmcnamara 13 hours ago [-]
> solving weirdly complicated problems involving Kirchoff's laws and Thevenin's theorem was a common way to torture students back in my day...
Hey now, those actually come up sometimes.
Workaccount2 13 hours ago [-]
One of my core grips with STEM education, mainly the math heavy part of it, which frankly is most of it, is that it is taught primarily by people who love math.
The people who loved application and practical solutions went to industry, the people who got off spending a weekend grinding a theoretical infinite resistor grid solution went into academia.
steamrolled 13 hours ago [-]
Loving math is not necessarily a problem. But if you want others to love it too, you have to explain it in a way that makes them see the light.
A lot of STEM education is more along the lines of "take the rapid-fire calculus class, memorize a bunch of formulas, and then use them to find the transfer function of this weird circuit". It's not entirely useless, but it doesn't make you love the theory.
esafak 13 hours ago [-]
In school I would have relished solving this problem. Now I wonder if it has any application.
bsder 13 hours ago [-]
> I don't get why EE education emphasizes problems of this sort
Last I checked, they don't. I certainly never hit an "infinite grid of resistors" in general circuits and systems except as some weird "bonus" problem in the textbook.
Occasionally, I would hit something like this when we would be talking about "transmission lines" to make life easier, not harder. ("Why can we approximate an infinite grid of inductors and capacitors to look like a resistor?")
It's possible that infinite grid/infinite cube might have some pedagogical context when talking about fields and antennas, but I don't remember any.
9 hours ago [-]
12 hours ago [-]
kevinmhickey 14 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.
personjerry 14 hours ago [-]
Where are you going to find an infinite grid of resistors in real life to measure?
BenjiWiebe 11 hours ago [-]
Measure a couple of different sizes of grids and fit a curve to your results?
terminalbraid 5 hours ago [-]
How many would you need to try to get an acceptable result?
I'll see myself out.
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.
But that's fine because knowing which path the electron took is not part of the problem. Both paths contributed to the resistance even if one was not taken.
We only have to worry about quantum effects if the probabilities are not a decent proxy for the partial-particles that we suspect don't exist. In this case, the Physicist can probably proceed directly to the bar and have a drink with the Mathematician.
I hadn't considered that sort of strange effect though! Makes me feel not so bad for 'never really getting it' because I just couldn't wrap my mind around the problem description's obvious inanity and the infinite edges.
It's simply an evaluation of your mathematical ability to manipulate the equations and overall understanding of them, wrapped up in a cute little thought experiment. This evaluation IS relevant to more realistic scenarios and therefore your grade and engineering ability.
Going on something of a tangent: in engineering, it seems unusual to talk about “applying current,” it’s usually voltage (say, across a resistor) or some sort of an “electromotive force.”
Just wait infinite time for all the transient responses to die down. The grid to enter steady state and to became true to the schematic.
Mostly joking, but an idealized schematic requires idealized conditions to observe.
One is where the components on the schematic represent physical things, where the resistors have some inductance and some non-linearity, and some capacitance to the ground plane and so on. This is what we mean by schematics when we’re using OrCad or whatever.
There is another interpretation where resistors are ideal ohms law devices, the traces have no inductance or propagation delay or resistance. Where connecting a trace between both ends of a voltage source is akin to division by zero.
Sometimes you translate from the first interpretation to the second, adding explicit resistors and inductors and so on to model the real world behavior of traces etc. if you don’t, then maybe SPICE does for you.
Infinite resistor lattices exist only in the second interpretation.
That said there might be enough energy (infinity!) for anything to be possible.
All of the mass does end up in the singularity, in finite time (at least for any finite subset of the black hole), but it doesn't automatically become super dense just because it's a black hole. It can remain quite ordinary for a very long time.
The kind of black hole that forms from a collapsing star is super dense. But a supermassive black hole forms differently, as a denser region of gas and stars during galaxy formation. The density can be lower than that of water. You could be inside it without even realizing anything is odd.
There is no known way to form black holes bigger than that. We had been discussing a pure thought exercise. Though it is possible that the universe as a whole has enough density to be a black hole. (Signs point to no, but it's an open question.)
If we assume this is an infinite grid in its own universe then nothing can actually move. The gravitational pull should be the same from every direction. If we assume the grid is perfect then there is no nucleation sites to start a collapse. The grid would be in perfect balance.
The same is thought about our universe. If there hadn't been small quantum fluctuations during the inflationary stage it would have taken much longer for what we see in the modern universe to form.
What it collapses into depends on the time for the bits to coalesce into. If it's a slow collapse you could get a mass large enough to form a neutron star (like thing) instead of black holes.
If those neutron stars crash into each other they can release a large amount of 'recycled' matter from all over the atomic spectrum back into this universe.
We impose initial conditions so that there is a congruence of motion of the connected resistors, so that we have a flavour of Born rigidity. Unlike in the special-relativistic Bell's spaceship model (in which the inertial motion of each spaceship identical save for a spatial translation), in our general-relativistic approach none of the line-of-connected-reistors elements' worldlines is inertial, and each worldline's proper acceleration points in a different direction but with the same magnitude. This gives us enough symmetry to grind out an expansion scalar similar to Raychaudhuri's, Θ = ∂_a v^a (<https://en.wikipedia.org/wiki/Raychaudhuri_equation#Mathemat...>). As an aid to understanding, we can rewrite this as 1/v \frac{d v}{d \tau}, and again in terms of a Hubble-like constant, 3H_0.
We can then understand Θ as a dark energy, and with Θ > 0 the infinitely connected line of ...-wire-resistor-wire-resistor-... is forced to expand and will eventually fragment. If Θ < 0, the line will collapse gravitationally.
> no nucleation sites
If Θ = 0 initially, we have a Jeans instability problem to solve. Any small perturbation will either break the infinite ...wire-resistor-wire-..., leading to an evolution comparable to Bell's spaceship: the fragments will grow more and more separated; or it will drive the gravitational collapse of the line. The only way around this is through excruciatingly finely balanced initial conditions that capture all the matter-matter interactions that give rise to fluctuations in density or internal pressure. It is those fluctuations which break the initial worldline congruence.
This is essentially the part of cosmology Einstein struggled with when trying to preserve a static universe.
In higher dimensions (2+1d, 3+1d) the evolution of rotation and shear (instead of just pressure and density) becomes important (indeed, we need an expansion tensor and take its trace, rather than use the expansion scalar above). A different sort of fragmentation becomes available, where some parts of an infinite plane or infinite volume of connected resistors can undergo an Oppenheimer-Snyder type of collapse (probably igniting nuclear fusion, so getting metal-rich stars in the process) and other parts separate; the Lemaître-Tolman-Bondi metric becomes interesting, although the formation of very heavy binaries early on probably mitigates against a Swiss-cheese cosmological model: too much gravitational radiation. The issue is that the chemistry is very different from the neutral-hydrogen domination at recombination during the formation of our own cosmic microwave background, but grossly a cosmos full of luminous filaments of quasi-galaxies and dim voids is a plausible outcome. (It'd be a fun cosmology to try to simulate numerically -- I guess it'd be bound to end up being highly multidisciplinary).
A really difficult question: At each distance, what percentage of soldering errors in the grid can be tolerated before the fluke meter across the center square detects the fault? (That might actually be a thing as I've heard people talk about using changes of local resistance to detect remote cracks in conductive structures ... like maybe in a carbon fiber submarine hull.)
https://www.youtube.com/watch?v=qJZ1Ez28C-A&t=1500
Wouldn’t that be more efficient than converting current to heat?
They also just convert current to heat though. Some amount of current moving through a material with some amount of resistance always produces a fixed amount of current in accordance with Ohm's law. You can't really get away from that.
I guess the explanations always confuse me. Let’s say a short circuit with no resistors has a certain amount of power. Then we add a resistor and the power in the circuit goes down. The resistor isn’t turning the difference in power into heat, right.
But let's talk about short-circuiting lithium batteries for example. They have a roughly 50 milliohm (aka 0.05 ohm) of "equivalent series resistance".
That means, if you short circuit a lithium battery with a superconductive wire (aka with 0 resistance), the circuit has a resistance of 0.05 ohms. We can compute the current with Ohm's law: I=V/R. V is typically 3.6 volts for li-ion batteries, R is 0.05 ohm, so I (aka current) is 3.6/0.05 = 72 amperes. 72 amperes * 3.6 volts is 259 watts. Now in the real world, the battery's chemistry would step in here and limit current in complicated ways, but this means that under the assumption that our battery would work as an ideal voltage source + a 0.05 ohm resistor, and if there was no extra heat coming from the chemical reactions, a shorted battery would produce 259 watts of heat.
We can add a 1 ohm resistor to the circuit, which means our circuit's combined resistance would be 1.05 ohm. Using Ohm's law again, we find that the current would be 3.6/1.05 = approx 3.43 amperes. 3.43 amperes * 3.6 volts is 12.35 watts of heat.
So thanks to our resistor, we're now producing 12.35 watts instead of 259 watts of heat, because the resistor limits the current going through the circuit. With a higher resistance resistor we'd produce even less heat.
A core idea here is that power consumptions equals heat. I don't understand the physics reasons why, but "this thing consumes 10 watts of power" means the same as "this thing produces 10 watts of heat". Higher resistance means less current which means less watts, which means both less heat and less power consumption because those are the same.
Another useful heuristic is that heat is generated from what's left after all of of the other ways of converting energy are used up, such as light, chemical potential, and so forth. It's energy's last resort. The usefulness of a resistor lies in its simple voltage-current relationship, which is equivalent to saying that the only thing it generates is heat.
A voltage drop with a flowing current is power (P=VI).
There is literally nothing you can do to avoid resistors dissipating that power as heat, it's just what they are. If they didn't do it, they wouldn't be resistors.
What you can do is use larger resistances which need less current to see the same voltage (e.g. change a pull up from 10k to 100k or higher, but that's more sensitive to noise), or smaller resistances that drop less power from a given current (e.g. a miiliohm-range current shunt, and then you need a more sensitive input circuit) or find another way to do what you want (e.g. a switched-mode power supply is far more efficient than a voltage divider at stepping down voltage). This is usually much more complex and often requires fiddly active control, but is worth it in power-constrained applications, and with modern integrated technology, there's often a chip that does what you need "magically" for not much money.
[1] https://en.m.wikipedia.org/wiki/Sheet_resistance
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.
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.
However note the problem is symmetrical about the vertical axis, so flip the figure. The current passing through the flipped paths should be the same as before the flip, so note down which i's equate to each other due to this.
Note that the problem is symmetrical about the horizontal axis, and do the same there. Note that the problem is symmetric when rotated 90 degrees, so do that. And so on.
In the end you'll have a bunch of i's that are equal, and you can group those into two distinct groups. Call those groups alpha and beta.
edit: Another way to look at it is that you can't use the available symmetry operations to take you from any of the alphas to a beta. This is unlike alpha to alpha, or beta to beta.
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
[1] https://kirill-kryukov.com/electronics/resistor-network-solv...
Possibly based on this ranking. Everything sub-mathematician is 2 points? Maybe there's subdivision of points.
https://xkcd.com/435/
Another interesting aspect is that in an infinite grid, a spontaneous high voltage is going to exist somewhere at all times. It is probably very far away from you, but it's still weird.
The only type of person for whom intuition about infinity to form is not entirely unlikely are mathematicians.
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.
It's a weird butterfly effect moment in my career though. I had an awesome professor for circuits 1, and ended up switching majors to EE after that. Then got two more degrees on top of the bachelor's
[1] https://westy31.nl/Electric.html
How else to create students capable of solving problems we cannot anticipate today?
Not to mention, that understanding strange problems is a very efficient way to broaden horizons.
I always thought this problem was a funny choice for the comic, because it’s not that esoteric! It’s equivalent to asking about a 2d simple random walk on a lattice, which is fairly common. And in general the electrical network <-> random walk correspondence is a useful perspective too
Hey now, those actually come up sometimes.
The people who loved application and practical solutions went to industry, the people who got off spending a weekend grinding a theoretical infinite resistor grid solution went into academia.
A lot of STEM education is more along the lines of "take the rapid-fire calculus class, memorize a bunch of formulas, and then use them to find the transfer function of this weird circuit". It's not entirely useless, but it doesn't make you love the theory.
Last I checked, they don't. I certainly never hit an "infinite grid of resistors" in general circuits and systems except as some weird "bonus" problem in the textbook.
Occasionally, I would hit something like this when we would be talking about "transmission lines" to make life easier, not harder. ("Why can we approximate an infinite grid of inductors and capacitors to look like a resistor?")
It's possible that infinite grid/infinite cube might have some pedagogical context when talking about fields and antennas, but I don't remember any.