Infinite resistor network

Reasonbly. The extence of Pi in the answer makes it seem high improbable that there's a simple solution.

Sylvia.

THANK you! Now I can get some sleep! ;-)

Cheers! Rich

No, Sylvia, the exact solution is:

R=(8-pi)/(2*pi)

See Question 10 at:

if you need someone else's word for it.

I can provide three or four different integral equation versions with solutions using differing approaches, if you want.

The problem is that although the voltage between nodes D and C are just what you forced them to be, the currents arriving/exiting from nodes D/C aren't what you imagined in your calculation -- they don't remain equal to I.

It's easy to see a huge improvement to your estimate, that gets much closer to the right answer and points out the flaw in your thinking that leads towards understanding the remaining smaller errors, as well.

Imagine a 0.5V battery between D and E, with D positive. Given the correct solution that there is an effective 1/2 ohm between those points, you would get 1A of current spilling out (conventionally) from point D and arriving back at point E.

Now imagine a 2/PI battery between E and C, with E positive. Given the correct solution for the impedance between those two points (and if the earlier battery were removed), we'd see 1A of current spilling out from point E and arriving back at point C.

Now put the two batteries in place and use superposition. Take a close look at the path between node D and E. The 0.5V battery is spilling a total of 1A outward from node D, as before. But the 2/PI battery is spilling current outward from node E and a substantial amount of that must spill along the path from E to D. In other words, node D must also spill out this additional current being generated by the addition of the 2/PI battery. You aren't accounting for this additional current arriving into node D. That additional current must be added before you compute the effective resistance.

If you just took a gross estimate (and it doesn't take into account all the details) and assumed that a net of 1/4 of the current goes equally in all four directions, you'd have at least another 1/4A to add to the 1A that is already spilling outward from D. So you'd need to correct your initial estimate of the resistance by dividing by 5/4A instead of 1A, or multiplying your calculated resistance by a factor of 4/5ths. Which does get you closer to the real value. Still more corrections are required, though.

Jon

(1/2)R is the only case in the above links that appears to me to have a completed answer. Are you saying that figure is incorrect?

Jon

Hey - you don't need to be so aggressive. It is sufficient to explain why you think I'm wrong.

The point is actually easier to see by considering the voltage at D when E is held at 0 and C at -I * 2/Pi. Presumably the voltage at D is not zero. On my analysis applying that voltage to D would involve a current flow into or out of that node, but that's clearly wrong, since applying to a node a voltage that it already has makes no change, and no current would flow.

Sylvia.

Not aggressive. It just takes so long between responses on the web that it speeds things up to put everything down in one go.

I saw the mental error instantly using superposition. And it said the same thing you just wrote. I was just trying to help out with a powerful theoretical viewpoint with broad application easily deduced to circumstances like this. If you prefer keeping track of details that way to arrive at the same place, I say you should use whatever works for you on whatever schedule you require. Use what works.

Jon

But there is. As I understand it the trick is to solve the current in the resistors along a chosen path between the two for one point sourcing current flowing off to infinity

then solve the same for other node sinking it from infinity

then superimpose the two and apply Ohm's law, and kirchoff's loop law.

(nice troll by the way)

bye.

And that makes the answer obvious because....?

Sylvia.

Because some of us are so inclined. To any that do not see the obvious nature of the answer, you should consider pursuing a different technology.

(resistor lattice problem)

I followed the attribution chain to:

which is apparently one of, if not the, fundamental treatments of the general class of "hypercubic, rectangular, triangular and honeycomb lattices of resistors".

Unfortunately I am not a math geek and there weren't any pictures, so I got lost right off the bat.

What I want to know is where exactly does pi get off sneaking into a kind of problem comprising solely straight lines and their intersections, i. e. without any intrinsic curves?

An obviously (to me) related question; is there an equally "neat" general approach to solving lattices _with_ built-in curvature, frinst a geodesic "dome" where the edges are the resistors, or wrapping your sheet of resistors into a cylinder? Such solutions might make it more obvious where pi gets involved.

Mark L. Fergerson

The physical position of the resistors has nothing to do with it.

It is 100% electrical.

I am surprised as you should know that.

Pi appears in many electrical situations where the voltage and/or current vary harmonically in time. The reason is obvious when the variation is graphed.

There is no similarly obvious reason why pi should appear in a DC problem.

Perhaps you might clarify.

Mark L. Fergerson

Me clarify? YOU tell us what you think a physical arrangement of the resistors around a cylinder would do.

It does not even stand to make it easier to visualize.

The total resistance is the limit of an infinite sum. Pi often crops up in such situations; e.g. 1-1/3+1/5-1/7+... converges to pi/4.

Why do you act like this to others?

Jon

Well, it's easy for you to imply that the answer is obvious to you. Easy for me to doubt it too. So let's see the colour of your money.

Though no doubt you'll find some well-worn pretext for not providing anything concrete.

Sylvia.

Sorry, no. That assumes the 2 nodes can be arbitrarily placed, which gives a false "solution".

Wake up and READ what he said! The new matrix is not a simple plane that has been bent; the new matrix is 3-dimensional and not 2 dimensional.

Well, a cylindrical array or matrix of resistors has a very easy and obvious solution that even you could solve without pencil and paper.