Finding Equivalent Resistance

The mesh extends infinitely in all the directions. =A0=A0|=A0=A0=A0=A0=A0|=A0=A0=A0=A0=A0|

--+--R--+--R--+---- =A0=A0|=A0=A0=A0=A0=A0|=A0=A0=A0=A0=A0| =A0=A0=A0=A0R =A0 =A0 R =A0 =A0 R =A0=A0|=A0=A0=A0=A0=A0|=A0=A0=A0=A0=A0|

--+--R--+--R--+---- =A0=A0|=A0=A0=A0=A0=A0|=A0=A0=A0=A0=A0| Extending the question, what could we say about an infinite triangular or hexagonal mesh like this? Thanks in advance,

-Bhushit

~> Thevenin was a Genius. =AEoy

Hint

Inject a 1A current into one node and out of the other. Use symmetry and superposition to calculate the voltage drop between the nodes.

martin

The paper: Infinite Resistive Lattices by Atkinson and van Steenwijk provides somewhat more details on the Fourier Transform method of solving periodic networks. It develops solutions for multi- dimensional, trangular and hexagonal lattices.

John Bailey

damned little chance of that ever becoming a topic of discussion amongst these pathetic old trolls...

FWIW, there was some discussion 13 years ago in sci.physics, here:

The solution approach looks like the one Martin hinted at.

"John Bailey" wrote om...

This is almost never true of today's undergraduates in EE programs and nowhere not ubiquitous among graduate students. I've looked at those old network theory books and it's amazing how 'deep' the area goes!

(Granted, when you look at the types of problems engineers typically work on today, it's hard to argue they should be reading Van Valkenburg rather than whoever writes a good Java or Linux book or something...)

WLOG, take R = 1 and suppose a square lattice with unit spacing. The paper cited by John Bailey doesn't say so (afaics), but inspection of the table of values there suggests that between opposite corners of a square of side n in this network, the resistance is exactly R_nn = (2/PI)*sum(1/(2*k-1),k=1..n) for n=0,1,2,... (Between adjacent lattice points it is 1/2.)

--r.e.s.

That seems right for R_nn. But what about R_nm? As you say, at R_01 it is 1/2. Also, at R_12 for example, the exact answer is 4/PI-1/2.

An iterative algorithm I devised (not the first person to invent the idea, by any stretch, but it is what I came up with when thinking about the problem before) is to arbitrarily place a voltage of 1V at the node under consideration (n,m) and 0V at the node of (0,0). I then arbitrarily set all other nodes of some arbitrarily sized finite grid (extending from -S to S in both dimensions) to an assumed guess of 0.5V to start. I then go to alternate nodes on the finite grid, first those spanning from 1-S to S-1, stepped by 2, and compute the node voltage as the mean of the adjacent node voltages (up, down, left, and right, in the case of a square grid.) Once that is done, the other alternate nodes on the finite grid are then visited, those spanning from S to S, stepped by 2, and compute the node voltages there in the same fashion. (Just be sure not to do any such averaging for the node at (0,0) or at (n,m).) Repeat the process over and over until the voltages settle. At that point, the resistance can be computed by assuming that current flows accurately based on the potential differences between (0,0) and its adjacent squares. (Which, of course, determines the current that the node at (n,m) must inevitably supply.) Since your starting potential is 1V between them and since you know the required current to be supplied by examining the nodes next to (0,0), the resistance is 1/sum.

Jon

| This is basic Electrical Engineering puzzle - hounting me since 1993. I | hope this is a correct place to ask this and such questions. Simple | mathematicle induction fails in this 2D arrangement. | | If '+' denotes a join, find equivalent resistances between any two | adjacent and diagonally opposite "+" in the figure below. The mesh | extends infinitely in all the directions. | | | | | | --+--R--+--R--+---- | | | | | R R R | | | | | --+--R--+--R--+---- | | | | | | | Extending the question, what could we say about an infinite triangular | or hexagonal mesh like this?

For AC, this could be even more interesting based on the lengths between nodes and the AC frequency involved.

The OP asked only about points that were "adjacent and diagonally opposite" (which would reasonably be interpreted as referring only to R_11, or possibly to both R_11 and R_01) -- I mentioned the apparent formula for R_nn (n=0,1,2,...) because of its simplicity.

Are you aware of any formulas of comparable simplicity in the more general cases?

--r.e.s.

Yes, but I don't recollect them, now. I only recall what I wrote.

Jon

Resistance (ideally or "ohmically") is independent of frequency.

Sure and why don't you include self and mutual inductances, and the capacitances, for spice.

Martin Demello has the right approach. However, you have to do it in two steps. Inject current into a node with the return connection a "circle" at infinity. The current splits equally in 4 directions. Now disconnect and draw current from the other node with injection at infinity. Superimpose.

By Superimposing too many of the network traingles you can be vexed infinitely, but, even if you don't have a value for R you can use Thevenize the first network array =3D Rth, and thus so forth., without injecting any imaginary currents and no applied voltage, obtaining an R Value for any node you want to analize.

Dang Dog speak for your Mockin Trollin Self., and I am not going over things twice or thrice., & I don't click mice. =AE

a)You don't try to solve all the nodes in the infinite network. You look at the current distribution from a given node to surrounding nodes up to the second node of interest b)You appear to be looking into the Thevenin resistance between the nodes selected. However, Thevenin (unless you are talking of something else than the usual Thevenin model) is not of use for finding this resistance. After all, Thevenin is simply an application of superposition.

Perhaps one can use severel methods to find R at a given node in the array, but Thevenin is not quite Superimposition because we do not need the following I/E/R resolve after the node to resolve for Rnode at any point with Thevenin.=AEoy From: snipped-for-privacy@peeshaw.ca (Don=A0Kelly) "Roy Q.T." wrote in message news: snipped-for-privacy@storefull-3258.bay.webtv.net... By Superimposing too many of the network traingles you can be vexed infinitely, but, even if you don't have a value for R you can Thevenize the first network array =3D Rth, and thus so forth., without injecting any imaginary currents and no applied voltage, obtaining an R Value for any node you want to analize. a)You don't try to solve all the nodes in the infinite network. You look at the current distribution from a given node to surrounding nodes up to the second node of interest [=AE]OF COURSE[=AE] b)You appear to be looking into the Thevenin resistance between the nodes selected. However, Thevenin (unless you are talking of something else than the usual Thevenin model) is not of use for finding this resistance. After all, Thevenin is simply an application of superposition.

From: snipped-for-privacy@peeshaw.ca (Don Kelly) "Roy Q.T." wrote in message news: snipped-for-privacy@storefull-3258.bay.webtv.net... By Superimposing too many of the network traingles you can be vexed infinitely, but, even if you don't have a value for R you can Thevenize the first network array = Rth, and thus so forth., without injecting any imaginary currents and no applied voltage, obtaining an R Value for any node you want to analize. a)You don't try to solve all the nodes in the infinite network. You look at the current distribution from a given node to surrounding nodes up to the second node of interest [®]OF COURSE[®] b)You appear to be looking into the Thevenin resistance between the nodes selected. However, Thevenin (unless you are talking of something else than the usual Thevenin model) is not of use for finding this resistance. After all, Thevenin is simply an application of superposition.