Is it possible to know all voltages/amps in a grid or matrix of resistors? - Page 1
Forum section: electronics / beginners

author: Beamin
author_rank: Super Contributor
timestamp: June 06, 2018, 03:43:00 pm
content: I just thought of this: What if you had a large grid of 1k resistors (can't make an ascii diagram for some reason) but rows or horizontal and vertical resistorseach resistor is connected to three other resistors. And you had a 10X10 or 1000X1000 or infinately big grid. How would you figure out the voltages? Is there an elegant math problem for this that could go up to infinate resistors? What would happen if it was a cube or if say you added 1V@1A to each corner? Or put +1 one one corner -1 on opposite corner and ground in the middle? I don't know where to start figuring this out because there would be so many values that change based on other values as every path is connected. My resistors are in storage so I can't build one to test at the moment so I will have to do this expirament in my head. Is this what a differential equation is in real life when your varibles change based on the other variables ie a+b+c=d if b=d+c? There must be a name for thisproblem so I can have a start point for google.

author: JimRemington
author_rank: Regular Contributor
timestamp: June 06, 2018, 03:48:33 pm
content: These are standard problems and there are elegant ways to solve them. Google "solve resistor grid", for example.

author: MK14
author_rank: Super Contributor
timestamp: June 06, 2018, 03:50:31 pm
content: This video, seems to discuss what you are asking, to a reasonable extent.

author: StillTrying
author_rank: Super Contributor
timestamp: June 06, 2018, 03:59:57 pm
content: Quote from: Beamin on June 06, 2018, 03:43:00 pmI'd be doing a lot of copying and pasting in LTSpice.

author: rstofer
author_rank: Super Contributor
timestamp: June 06, 2018, 04:40:10 pm
content: Here is the solution for a finite grid and it is easily solved with MATLAB among other tools:http://www.wiete.com.au/conferences/2wctee/papers/15-13-Zegarmistrz-P.pdfThe infinite grid is a little more difficult and hand-wavy because spooky stuff happens out at infinity.http://www.mathpages.com/home/kmath668/kmath668.htmNothing new here, the finite grid is discussed in just about every EE101 class.Sometimes the grid is wrapped around, like a sphere or just a cubehttp://physicstasks.eu/285/the-resistance-of-a-wire-cubehttp://www.rfcafe.com/miscellany/factoids/kirts-cogitations-256.htmReally, these are just exercises in finding symmetry and creating the matrix.

author: CopperCone
author_rank: Super Contributor
timestamp: June 06, 2018, 06:51:31 pm
content: how does heisenberg uncertainty deal with averages?if you know exact voltage then you know the exact velocity of the particles? But if you know the exact current that means you know the exact location of the electrons.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 06, 2018, 07:45:59 pm
content: The traditional method is:- For adjacent probe points (i.e., straddling a single resistor), symmetry allows a relatively simple reduction in the network, and, a geometric series gives the answer I think?- For arbitrary probe points, the engineer's best friend comes out: Fourier analysis. Instead of a grid of resistors, consider the grid of nodes as a sequence of samples. You are trying to solve for the impulse response due to currents injected at the probe points. This quickly gives interesting results, like the knights-move resistance carrying a factor of pi.I think a circuit-rules-based reduction method can be used to give the same result, but the process is much more complicated (requires wye-delta transformations). The result is infinite resistors in series, of decreasing values, which converges to the correct value.Tim

author: IanB
author_rank: Super Contributor
timestamp: June 06, 2018, 08:24:46 pm
content: Quote from: Beamin on June 06, 2018, 03:43:00 pmIf your grid is of finite size with a known number of resistors then the problem is simply one of simultaneous linear equations or linear algebra. If you search on solving linear systems of simultaneous equations you will find detailed answers.If your grid is infinite then special methods are required as described by others in this thread.

author: rstofer
author_rank: Super Contributor
timestamp: June 06, 2018, 10:07:07 pm
content: Some folks played with this a few weeks ago at hamster_nz's thread:https://www.eevblog.com/forum/beginners/resistance-across-a-1k-resistor-cube/25/Or you can look at Simon's mesh analysis problem from back in January 2017 for a glimpse of how these mesh equations are derived (Kirchhoff's Laws)https://www.eevblog.com/forum/beginners/mesh-analysis/Or look at reply #150 for my final, final, final pass at the problem for node voltages and mesh currentshttps://www.eevblog.com/forum/beginners/mesh-analysis/150/Getting those equations from the original schematic are the hard part. wxMaxima does all the work and it would be quite similar in MATLAB. The 'matrix' isn't really shown.This kind of stuff comes up all the time. Rarely is it simply a fixed DC voltage. More likely it will be like Simon's problem with AC and impedances rather than DC and resistance.Google terms:Kirchhoff's LawsMesh AnalysisNodal AnalysisSuperpositionAny DC circuits book will cover this stuff in the first couple of chapters.

author: rstofer
author_rank: Super Contributor
timestamp: June 06, 2018, 11:51:16 pm
content: There a nice little book "Electronics and Circuit Analysis Using MATLAB" available at Amazon but a little cheaper at Alibris:https://www.alibris.com/Electronics-and-Circuit-Analysis-Using-MATLAB-Second-Edition-John-Okyere-Attia/book/35084421This would be for the serious student of circuit analysis because most classroom problems rarely go beyond a 4x4 matrix as they are darn hard to solve by hand. Maybe newer texts assume computers and get a little more involved. I notice that even lower division Mechanical Engineering has a required MATLAB course. That doesn't seem to be true for Computer Engineering.The good news about matrices for circuit analysis is that they tend to be sparse - containing a lot of zero elements. As a result, Gauss-Jordan Elimination works well, even by hand.http://pages.pacificcoast.net/~cazelais/251/gauss-jordan.pdfMany calculators will solve reasonably sized matrices. Alas, a slide rule will not...

author: Beamin
author_rank: Super Contributor
timestamp: June 07, 2018, 03:54:31 am
content: Quote from: rstofer on June 06, 2018, 04:40:10 pmThe math goes over my head in those two papers. Need simpler examples With the infinite grid wouldn't the edges not matter at some point because the current would be so small and the resistance so high that effectively there would be no or unmeasurable (say your equipment goes down to 0.001 uV/uA) current flow out at those points? If it was all made of wires would the same thing happen or would it be the opposite with the net effect be a batter and better conductor as more connections were formed?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 07, 2018, 05:16:32 am
content: Yes, in this case the currents will go to zero at infinity, and the series converges.However, proof of this is necessary.Without proof, you may get very peculiar results. An example is the infinite sum: 1+1+1+... = -1/2.If the series of all partial sums converges, then the sum exists. The partial sums of 1+1+1+... clearly do not, so we do not say that sum exists, under ordinary circumstances. (However, the existence of such tricks, and its application to particle physics, makes it too good not to have a name -- in that case, we specify what kind of summation is being used to claim the existence of this relationship.)For a grid of linear and positive-valued resistors, we expect a general 1/n^2 behavior at a distance, which will give a convergent series (e.g., the sum of 1/n^2 for n = 1, 2, ..., is exactly 2). Proof of this fact, is, uh, left as an exercise for the student, yes.Tim

author: IanB
author_rank: Super Contributor
timestamp: June 07, 2018, 05:21:39 am
content: Quote from: Beamin on June 07, 2018, 03:54:31 amBut the examples are what you gave in the first post, and the math is what is required to solve those examples.If the math goes over your head you will have to do what I do, which is to read it with fascination and accept that the details will take some effort to understand.The questions you ask about what happens at infinity are very appropriate and reasonable questions and are discussed at some length in the second paper.

author: Kirr
author_rank: Regular Contributor
timestamp: June 07, 2018, 08:36:26 am
content: When googling, also try "resistor lattice" - it will find a lot of research papers on this topic.For finite grids (and not too large), you can try my solver, e.g.:5x5 grid.The largest grid I solved so far with my solver was 251x250 (with terminals across a single central resistor). It took about 2 hours, in three runs - first simplifying a triangle (1/8 of the grid), then merging two triangles, then finally simplifying the rest. The resistance is ~0.50000868 Ohm (13 kB long number in rational form in hexadecimal).

author: CJay
author_rank: Super Contributor
timestamp: June 07, 2018, 02:42:46 pm
content: That's going to be some weird looking grid if each resistor is only connected to three others...

author: rstofer
author_rank: Super Contributor
timestamp: June 07, 2018, 03:16:49 pm
content: Quote from: Beamin on June 07, 2018, 03:54:31 amNow you are putting limits on things happening at infinity. Infinity isn't a countable number, spooky stuff happens out there and the fact that you can't measure it with a bench DMM just means it is best left for theoreticians. The closest EEs get to infinity is in RF using the Earth as a ground plane. Infinity is the biggest number you can imagine, plus one! And that just takes care of the integers. There are an infinite number of real numbers between any two integers. And there are an infinite number of integers. Kind of like infinity squared! It's just spooky!When current flows, it does not, contrary to popular conjecture, take the path of least resistance. It takes ALL of the paths. Even the high resistance paths will have some current flow. The low resistance paths will just have a high percentage of the current flow. So, yes, more parallel paths means lower resistance.The way to get a handle on this is to start with 1 resistor and solve Kirchoff's Laws using Ohm's Law. Add another resistor in series and solve again. Add a resistor in parallel and solve again. Try 4 resistors in a square, try 12 resistors in a cube and then stop. The process is understood and further examples are just more of the same with uglier arithmetic.But you really need a handle on Kirchhoff's Laws for mesh and nodal analysis. You can't even analyze simple op amp circuits without understanding Kirchhoff's Current Law. It'a a simple law: The sum of all currents entering or leaving a node total to zero. Current can't pile up at a node.At ever more complicated levels, math is required every semester of the 4 year EE undergraduate program. In the upper division, math is masquerading in classes like Field Theory (Maxwell's Equations) or Control Systems (Laplace Transforms) or Signal Processing (Fourier Transforms) and this is AFTER taking Calc I, Calc II, Calc III and Differential Equations in the lower division. Electronics, as a profession, is a math intense way to make a living.The good news about today is the tools we have for doing the math. Getting by with a slide rule in the pre-HP35 calculator days made things even more difficult. I remember when nearly every electronics text book had tables for trigonometric functions and logarithms in the back. We actually had to use tables to perform trig problems!

author: Beamin
author_rank: Super Contributor
timestamp: June 07, 2018, 05:49:51 pm
content: Quote from: Kirr on June 07, 2018, 08:36:26 amWhat was the value of the individual resistors? 1k ohms?When I said each is connected to three resistors I meant per lead 6 resistors total like a carbon latus with single bonds. You could use this math to solve graphene (-C=C- etc etc) and other conductivity of nano sheets. You could model out what the resistance is between the bonds and scale it up into ohms to show a life size graphene sheet and would be a great way to teach organic conductors. You could dope by include N and P atoms with different value resistors and geometry.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 07, 2018, 07:03:44 pm
content: Graphene doesn't work, because any given electron zipping around in the field of atoms experiences dozens of atoms simultaneously. The electron is a wave confined by a boundary condition; that it ever has resistance is an accident brought on by defects and impurities, encountered with low probability over long distances -- in other words, in the thermodynamic limit.In other other words, it doesn't help to reason about quantum systems from the smallest possible unit, stacked up. There are long range order, and effects, that pull in later. You have to run an integral over the ensemble -- quantum statistical mechanics. A notoriously hard problem, because to calculate that integral, you must account for, not just the effect of each differential unit upon itself, but upon its neighbors as well!Think about taking the frequency response of an amplifier, by looking at just one tiny little blip. You get good information about the high frequency response, but the low frequency response is a slow and minuscule 'tail' that is, at best, inconvenient to read in this way. How much long-term information can you infer from a very short blip? Not much. Much better, use a holistic method, like frequency domain (Fourier) analysis, to find the properties overall. (In QM, this is called k-space.)The equivalent to this principle in linear electronics is: if you cascade a bunch of AC filters, you must account for the interactions between them, which in general won't be well-behaved if you pick everything randomly. In this case, we can use the transmission ('T' or 'ABCD') matrix to solve the problem: a two-port filter is only two zero-dimensional connections, so we don't have to worry about waves or distance or phase or any of that, we simply get four complex numbers describing the input, transmission and output properties of the block. There are six standard, equivalent matrix types used to describe these things, but the handy part about the T-matrix is, you simply multiply them together, and that's your total answer.Suppose you could derive the electron wave function's T-matrix for graphene, for one very small differential unit; now suppose you can cascade infinity of them in all directions. Well, it's not going to be a two-port, because if nothing else, there are two directions to integrate over. Maybe a four-port would work? Unsure. It's... complicated.Tim

author: Kirr
author_rank: Regular Contributor
timestamp: June 07, 2018, 11:11:31 pm
content: Quote from: Beamin on June 07, 2018, 05:49:51 pm1 Ohm. (You're free to use any other values that you like, including non-identical).

author: The Electrician
author_rank: Frequent Contributor
timestamp: June 08, 2018, 03:14:43 pm
content: Quote from: T3sl4co1l on June 07, 2018, 05:16:32 amMathematica thinks the sum is not 2:

author: rstofer
author_rank: Super Contributor
timestamp: June 08, 2018, 03:54:02 pm
content: Quote from: The Electrician on June 08, 2018, 03:14:43 pmThat is correct but the theory is well beyond my skills:https://en.wikipedia.org/wiki/Basel_problem

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 08, 2018, 04:44:04 pm
content: I meant to say (1/2)^n. But that too.Tim

author: IanB
author_rank: Super Contributor
timestamp: June 08, 2018, 05:00:54 pm
content: Quote from: rstofer on June 08, 2018, 03:54:02 pmThis may help, if you have not seen it before:https://youtu.be/d-o3eB9sfls

author: Beamin
author_rank: Super Contributor
timestamp: June 08, 2018, 06:30:34 pm
content: Quote from: T3sl4co1l on June 07, 2018, 07:03:44 pmIt always amazes me how you can put concepts that I would never hope to understand like k-space and how graphene is NOT a resistor matrix into things I can understand. From what I know about the quantum world that answer makes perfect sense. Each bond is a probability not an actual path for an electron "particle" to take, like in the sea of electrons in metal where it conducts because you have so many. A single atom isn't a metal or a gas or a solid etc. Like NaCl is part of an ionic solid because you can't have just a NaCl molecule without the others to keep it inline in the cube. The stuff they teach you in chemistry class rarely goes into that: protons are also charged hydrogen/an acid and act nothing like a proton in a nuclear reactor. I remember learning nuclear in chemistry and the students were totally baffled including the professor with a phd) on why beta decay doesn't just result in a full orbital or a -1 atom; like you would expect in an aqueous solution. The beta particle has way too much energy: same with how alpha isn't just hot ionized helium bouncing around like in a neon light.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 08, 2018, 08:26:40 pm
content: Yup, those cases are about equilibrium vs. nonequilibrium conditions, and also energy levels relative to orbits and such.Chemistry isn't very meaningful under nonequilibrium conditions; it can be done, but it takes heroic effort (e.g., studying single-molecule reactions with lasers and a frozen argon matrix).Some beta decays do result in ions directly -- read up on the "electron capture" process.I like to say nothing is a stronger Bronsted-Lowry acid than a proton beam, but given that it's more of a mechanical effect (sheer kinetic energy sputters the target away), it's kind of a silly claim.Conversely, I suppose vacuum tubes have to deal with the strongest possible Lewis acid/base, but electrons are rather easier to deal with than protons, even at modest energies (100s keV). Insulators may not find it very pleasant (see Lichtenberg figures!), but it seems metals have little problem with that!Tim

author: Beamin
author_rank: Super Contributor
timestamp: June 10, 2018, 12:46:57 pm
content: Quote from: T3sl4co1l on June 08, 2018, 08:26:40 pmI knew you were going to say that. Would be interesting if we could mix wet chemistry with bare particles from electric devices. Build atoms like legos. We could have cures to almost all dieses. The strongest acid would be HHe; Helium hydride if you could make it stable pH of -40 or something ridiculous.