Bwa Ha! Ha!

Dave

Joined Nov 17, 2003
6,969
I recall about 3 years ago a similar question was asked, actually as a homework question not as a thought-exercise - it was as bizarre then as it is now! :D

As a first guess I'm going to say on an infinite grid the resistance would also be infinite since current would need to distribute around an infinite number of parallel branches; I = 0 and R = V/I = \(\infty\).

Just a stab-in-the-dark.

Dave
 

Salgat

Joined Dec 23, 2006
218
Well, couldn't you just do R=1/(1OHM*n) as n approaches infinity :p Anyways, what we do know about the problem is that the maximum resistance is 3 Ohms, and the theoretical minimum is 0.
 

Dave

Joined Nov 17, 2003
6,969
So how do we account for the fact the current at node 1 must be distributed around an infinite number of branches?

Dave
 

bloguetronica

Joined Apr 27, 2007
1,541
Should have near zero resistance. It is an infinite resistor grid, right?

Well, couldn't you just do R=1/(1OHM*n) as n approaches infinity :p Anyways, what we do know about the problem is that the maximum resistance is 3 Ohms, and the theoretical minimum is 0.
If we only accounted with those 7 resistors in the middle, we should have actually less than 3Ohm. There is a potential difference across that resistor in the middle, and that is a problem.
 

bloguetronica

Joined Apr 27, 2007
1,541
Yes. But my above question still stands about current distributed through an infinite number of branches.

Dave
I guess the current will not be equal in all branches if we apply voltage to those nodes. My guess is that the current will be greater between those two points.

But then again, it depends on where you apply the voltage.
 

Dave

Joined Nov 17, 2003
6,969
I guess the current will not be equal in all branches if we apply voltage to those nodes. My guess is that the current will be greater between those two points.

But then again, it depends on where you apply the voltage.
Sure, each branch will have a different resistance depending on your routing. It's unimaginable becuase the grid is infinite.

In fact I've just done a Google search and come up something that will help (I say help tongue in cheek!) with the answer. If we assume each resistor is 1\(\Omega\) (keep it simple) then the resistance diagonally across a box is:

R = \(\frac{2}{\pi}\)

See: http://www.geocities.com/frooha/grid/node2.html (get your calculus books out! :D)

Dave
 

bloguetronica

Joined Apr 27, 2007
1,541
Sure, each branch will have a different resistance depending on your routing. It's unimaginable becuase the grid is infinite.

In fact I've just done a Google search and come up something that will help (I say help tongue in cheek!) with the answer. If we assume each resistor is 1\(\Omega\) (keep it simple) then the resistance diagonally across a box is:

R = \(\frac{2}{\pi}\)

See: http://www.geocities.com/frooha/grid/node2.html (get your calculus books out! :D)

Dave
I though I could use the formula of the distance for that, but diagonally across each square you have 1Ohm (not accounting with the other resistors). Was a wild guess. The solution was in Fourier analysis indeed, as I first guessed.
 

Dave

Joined Nov 17, 2003
6,969
The solution was in Fourier analysis indeed, as I first guessed.
When dealing with solutions tending into infinity it is often the soundest (and many cases, along side Laplacian, the only) approach. The treatise of the solution in the above link is an impressive effort, and I cannot (at the moment) see a flaw in the logic.

Dave
 

bloguetronica

Joined Apr 27, 2007
1,541
When dealing with solutions tending into infinity it is often the soundest (and many cases, along side Laplacian, the only) approach. The treatise of the solution in the above link is an impressive effort, and I cannot (at the moment) see a flaw in the logic.

Dave
Would it be necessary to resort to Fourier analysis if we were to calculate the resistance of a finite grid with 7 resistors arranged the same way between those two points? I can't see how to calculate using other method.

Rich (BB code):
0---0---B
|   |   |  Determine resistance between points A and B, assuming that the resistors are 1Ohm ideal resistors.
A---0---0
 

Dave

Joined Nov 17, 2003
6,969
Would it be necessary to resort to Fourier analysis if we were to calculate the resistance of a finite grid with 7 resistors arranged the same way between those two points? I can't see how to calculate using other method.

Rich (BB code):
0---0---B
|   |   |  Determine resistance between points A and B, assuming that the resistors are 1Ohm ideal resistors.
A---0---0
You need to use the DsFT to allow you to calculate the current and voltages at nodes A and B since m and n (the grid dimensions) are infinite; ref. equations (28) and (30). In your example above the current and voltage calculations at A and B would be different and hence the resistance will be different.

Dave
 

thingmaker3

Joined May 16, 2005
5,083
One: there is no such thing as an "ideal" resistor.

Two: there are a finite number of resistors in the real universe.

Three: Having adroitly invoked the logical fallacy of hypothesis contrary to fact I can quite deftly sidestep the oncoming bus, beat up the guy with the sign, steal the sign, and study the problem at my leisure.
 

Dave

Joined Nov 17, 2003
6,969
One: there is no such thing as an "ideal" resistor.

Two: there are a finite number of resistors in the real universe.
Spoilsport! :p

Look at the fun we're having with this thought-exercise :D

Three: Having adroitly invoked the logical fallacy of hypothesis contrary to fact I can quite deftly sidestep the oncoming bus, beat up the guy with the sign, steal the sign, and study the problem at my leisure.
Lol! Is that secret answer number 4?!

Dave
 

mentaaal

Joined Oct 17, 2005
451
being an infinite array of resistors, couldnt you also associate this problem with that of a cable baving a characteristic impedance?
 
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