# This is great! xkcd and electronics - infinite 1 ohm resistor grid problem

Discussion in 'Off-Topic' started by thatoneguy, Aug 12, 2009.

1. ### thatoneguy Thread Starter AAC Fanatic!

Feb 19, 2009
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I found this hilarious.

Who has the answer:

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2. ### beenthere Retired Moderator

Apr 20, 2004
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This is great! Belongs in Off Topic.

3. ### ELECTRONERD Senior Member

May 26, 2009
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It's simpler than it seems, you just exclude the surrounding resistors outside the two points and find the total resistance. I'm thinking its going to be 6Ω. On the top, you'll find the original circuit and below you'll find the equivalent circuit.

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4. ### thatoneguy Thread Starter AAC Fanatic!

Feb 19, 2009
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Oh no. This may be another Airplane on a treadmill...

I can only see a max of 1.5Ω using the eliminate all outside, but there is an infinite number in parallel connected, making the minimum near zero...

Sorry for posting this in the wrong area, BeenThere!

When I saw this, I was instantly reminded of you guys and that I haven't been back here in a while.

5. ### beenthere Retired Moderator

Apr 20, 2004
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Not a problem. Gotta be close to 0 with all those paths in parallel.

6. ### gerty AAC Fanatic!

Aug 30, 2007
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I agree with beenthere, too many parallel paths, and they cannot be ignored.

7. ### Ratch New Member

Mar 20, 2007
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ELECTRONERD,

No its not.

No you don't.

beenthere and gerty,

Faulty reasoning. Those outer paths have increasing resistance.

To the Ineffable All,

See this link http://sites.google.com/site/resistorgrid/node2 . This shows the solution for a single diagonal, but the method can be extended to the "knight's move" nodes. As all you members of the Ineffable All can see, the problem is not trivial.

Ratch

8. ### loosewire AAC Fanatic!

Apr 25, 2008
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9 ohms resistist

Last edited: Aug 12, 2009

Aug 8, 2005
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10. ### ELECTRONERD Senior Member

May 26, 2009
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I assumed the same current path, but apparently I have the decimal point in the wrong place. Maybe someone can actually construct the seven array of resistors and measure it. I dont have any 1Ω resistors

11. ### loosewire AAC Fanatic!

Apr 25, 2008
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Use different resistors,different math

12. ### beenthere Retired Moderator

Apr 20, 2004
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By measurement, 1.42 ohms. .95 ohm corner-to-corner for one square. Those infinite alternate paths cancel their effects.

13. ### thatoneguy Thread Starter AAC Fanatic!

Feb 19, 2009
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How long did it take you to solder together an infinite number of resistors, and how did you get so many?

14. ### beenthere Retired Moderator

Apr 20, 2004
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That was just for 7 resistors. The external paths make it like the Rt should tend towards 0, but every parallel infinity has a series infinity counteracting it

15. ### Wendy Moderator

Mar 24, 2008
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I suspect (without knowing) it resembles calculas, in that it approaches a limit. So if you go 3 or 4 layers out and calculate it you will be very close to the answer.

16. ### ELECTRONERD Senior Member

May 26, 2009
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Man, I wish I knew calculus! I'm gonna learn though, once I finish high school.

17. ### Wendy Moderator

Mar 24, 2008
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I remember an engineer telling me about sheet resistors (we were making them for substrates on modules). Basically the measurement on a flat sheet is per square. If you increase the size of the square the resistance stays the same, though the wattage goes up. If you put 2 squares (a rectangle) side by side it's 2X the resistance.