# Bwa Ha! Ha!

Discussion in 'General Electronics Chat' started by mrmeval, Dec 14, 2007.

1. ### mrmeval Thread Starter AAC Fanatic!

Jun 30, 2006
833
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http://xkcd.com/356/

Solve THAT!

Heck, sneak that into the teachers pile to be graded and watch'em implode!

2. ### beenthere Retired Moderator

Apr 20, 2004
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It's evil. I love it!

3. ### mrmeval Thread Starter AAC Fanatic!

Jun 30, 2006
833
3
The joy would be finding a teacher who could solve it.
The terror would be that it'll be on a test.

4. ### Dave Retired Moderator

Nov 17, 2003
6,960
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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!

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

5. ### recca02 Senior Member

Apr 2, 2007
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then again, infinite resistance in parallel should tend to zero resistance.

6. ### Dave Retired Moderator

Nov 17, 2003
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Hmmm, the delights of infinities. In fact if I = 0 then this would violate KCL so that can't be right.

Dave

7. ### Salgat Active Member

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

8. ### Dave Retired Moderator

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

Dave

9. ### bloguetronica AAC Fanatic!

Apr 27, 2007
1,359
26
Should have near zero resistance. It is an infinite resistor grid, right?

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.

10. ### Dave Retired Moderator

Nov 17, 2003
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Yes. But my above question still stands about current distributed through an infinite number of branches.

Dave

11. ### beenthere Retired Moderator

Apr 20, 2004
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295
THE BUS IS COMING!!!!!!

12. ### bloguetronica AAC Fanatic!

Apr 27, 2007
1,359
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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.

13. ### Dave Retired Moderator

Nov 17, 2003
6,960
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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! )

Dave

14. ### bloguetronica AAC Fanatic!

Apr 27, 2007
1,359
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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.

15. ### Dave Retired Moderator

Nov 17, 2003
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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

16. ### bloguetronica AAC Fanatic!

Apr 27, 2007
1,359
26
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.

Code ( (Unknown Language)):
1.
2. 0---0---B
3. |   |   |  Determine resistance between points A and B, assuming that the resistors are 1Ohm ideal resistors.
4. A---0---0
5.

17. ### Dave Retired Moderator

Nov 17, 2003
6,960
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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

18. ### thingmaker3 Retired Moderator

May 16, 2005
5,073
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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.

19. ### Dave Retired Moderator

Nov 17, 2003
6,960
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Spoilsport!

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

Lol! Is that secret answer number 4?!

Dave

20. ### mentaaal Senior Member

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