41 ms·
Infinite Grid of Resistors
- rwmj 1y agoNow 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 ...
- Kirr 1y agoWell there is this: https://www.youtube.com/watch?v=v1YrANSmOGY https://www.youtube.com/watch?v=v1YrANSmOGY - Not sure if it counts as large, but it's a start.
- BizarroLand 1y agoI imagine that a 100x100 grid of resistors all terminated in a massive resistor would be the same as long as you kept to the middle of the grid for testing.
- bilsbie 1y agoOffshoot 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 1y agoYou'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 1y agoThanks. 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.
- analog31 1y agoIt'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.
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- mort96 1y agoThere 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.
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- grues-dinner 1y agoResistance 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.
- petschge 1y agoSee also https://xkcd.com/356/ https://xkcd.com/356/
- ordu 1y agoWhy 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 1y ago-Why mathematicians are three points? Possibly based on this ranking. Everything sub-mathematician is 2 points? Maybe there's subdivision of points. https://xkcd.com/435/ https://xkcd.com/435/
- Mawr 1y agoSee alsoer https://youtu.be/zJOS0sV2a24?t=932 https://youtu.be/zJOS0sV2a24?t=932
- neepi 1y agoI'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.
- bravesoul2 1y agoGiven an infinite grid of resistors... would you expect planets to form?
- corysama 1y agoThey 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 1y agoResistors 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 1y agoYou 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 1y agoPeople are resistors too.
- temp0826 1y agoI resist that statement
- mmastrac 1y agoThis was the question I hated in my EE degree. The thought exercise was a favourite of the profs.
- dcassett 1y agoI 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.
- quibono 1y agoThere'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 1y agoStart 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.
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- shove 1y agoWord on the street was that my Physics professor at NCSSM (Dr Britton) worked on this problem during his doctorate
- clbrmbr 1y agoThe finite grid of resistors (or arbitrary impedances) is actually of great practical usefulness.
- Kirr 1y agoThis 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. [1] https://kirill-kryukov.com/electronics/resistor-network-solver/ https://kirill-kryukov.com/electronics/resistor-network-solv...
- steamrolled 1y agoI 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.
- choonway 1y agoThere are two parts to education. One is to impart knowledge, the other is to filter the students.
- colechristensen 1y agoNot entirely wrong but it's a little too easy to use that argument for squashing any criticism for education content.
- dwattttt 1y agoYou'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.
- Nevermark 1y agoThe 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.
- goochphd 1y agoI 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
- nimish 1y agoIn the integral, the h_m(s) are chebyshev polynomials of the first kind
- kevinmhickey 1y agoIn 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 1y agoWhere are you going to find an infinite grid of resistors in real life to measure?
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- BenjiWiebe 1y agoMeasure a couple of different sizes of grids and fit a curve to your results?
- terminalbraid 1y agoHow many would you need to try to get an acceptable result?
- causality0 1y agoMy 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.
- morepedantic 1y agoThat 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 1y agoFunny 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.
- eternauta3k 1y agoThat 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).
- pyman 1y agoRe: 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
- kayson 1y agoA 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.
- sriku 1y agoThis is cool and I have my own take on it after being nerd sniped by XKCD - https://sriku.org/posts/nerdsniped/ https://sriku.org/posts/nerdsniped/ - I link to this article at the end but that post specifically solves the xkcd puzzle.
- praptak 1y agoWhat 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 1y agothe 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
- TheOtherHobbes 1y agoAt 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.
- bgnn 1y agoPeople 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!
- eternauta3k 1y agoI'd argue the case you're describing is mathematically simpler precisely because it is continuous.
- Den_VR 1y agoYou’re practically describing the invention of Calculus.
- gugagore 1y agoRight, why is it a 4-connected grid instead of 8-connected, or any other topology, like a hex grid.
- bgnn 1y agoTrue, 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.
- ChoGGi 1y agoMy 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.
- dogman1050 1y agoThis 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 https://en.m.wikipedia.org/wiki/Sheet_resistance
- Balgair 1y agoAside: 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 https://www.youtube.com/watch?v=qJZ1Ez28C-A&t=1500
- nullc 1y agosadly the demo is not so impressive as they make it out to be: The extra-path light is 'caused' by exit diffraction of the light source. Now, the same underlying theory also explains why there will always be diffraction from any finite boundary resulting in a reality which is indistinguishable from one where light actually takes all possible paths. But to argue that it actually does is arguably more meta-physics than physics. The demonstration is further compromised by the fact that the laser's diffraction performance is also presumably far from the physical limit. So a cynic seeing that demo would say "Isn't that just due to some of the light from the laser being off-axis" -- and it absolutely is. The physics means that some of the light will always be off-axis, but the demonstration does nothing to establish that.
- Balgair 1y agoI'm sorry what? What do you mean by diffraction here? Are you talking about the bokeh? The laser isn't interacting with any other elements than the air before it makes the gradient sheet light up. And the permisivity of the air is about the same as vacuum here. Like, the grating is going to show a diffraction pattern similar to any pinhole aperture source [0] because of the 'half cancellations' they kinda explain, but not all that in depth. But since there are no elements in the path yet, the only conclusion we can make is that of Feynman's - that the light is in fact taking all possible paths and then cancelling out to make the laser light we normally experience. What am I missing? To me this demo is like mind boggling as it shows that the wave model is the correct one even with a laser. [0] https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fwww.researchgate.net%2Fpublication%2F226250595%2Ffigure%2Ffig4%2FAS%3A670015766405134%401536755615316%2Fa-Digital-diffraction-pattern-obtained-by-fast-Fourier-transform-of-corresponding.jpg&f=1&nofb=1&ipt=0c6db3e071879707da65fc8261c7b7fb94af4e11d7fe2c0be4ba27f892b34452 https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2F... Something like (a) above, though I'm not confident that this is really showing that exactly.
- at_a_remove 1y agoOdd. 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.
- 1970-01-01 1y agoThis is also known as a high pass filter for first year EE students.
- bilsbie 1y agoDumb question but why isn’t a vacuum considered an infinite grid of resistors?
- jiggawatts 1y agoBecause it is effectively a superconductor! An electron (or proton) with some velocity will keep going in a straight line forever. This is neglecting the influence of stray background magnetic fields and gravitational fields, but the general notion applies.
- Koshkin 1y agohttps://news.ycombinator.com/item?id=44282191 https://news.ycombinator.com/item?id=44282191