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A resistance question...
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by kudoushinichi
on 2007-12-18 09:36:24 (edited 2007-12-18 09:40:56)
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This has been bugging me for hours already~! The question is "On this infinite grid of ideal one-ohm resistors, what is the equivalent resistance between the two marked nodes?" Here's the picture: http://xkcd.com/356/
Shinjitsu wa itsumo hitotsu!
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Re: A resistance question...
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by
 
on 2007-12-18 09:42:02 |
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*runs kudou over with a truck* I have no clue. I don't even know where the current is coming from, let alone the potential difference and such. I don't think it's possible to figure it out. Wanna share your thoughts?  |
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Re: A resistance question...
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by
on 2007-12-18 10:45:01 |
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The spoiler can be found over in the xkcd forum. It's an insanely difficult problem, by the looks of it. I honestly don't understand the problem, sorry. 
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Re: A resistance question...
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by
on 2007-12-18 13:56:27 (edited 2007-12-18 13:58:13)
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Was thinking about it in the shower... 1? Using the equation for resistance, if the resistance is inf. then it would be like 1/inf + 1/inf + 1/inf + ... Then, that would like translate to (1/inf)(inf) meaning inf/inf = 1 ... But, in calculus, inf/inf is not determinable... I think I'm wrong... |
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Re: A resistance question...
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by kudoushinichi
on 2007-12-19 04:26:19
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*from beneath the truck* Yep... insane indeed. Never thought that Fourier is needed in this one...
Shinjitsu wa itsumo hitotsu!
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Re: A resistance question...
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by
on 2007-12-20 11:26:33 |
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"Lets call the resistance of the grid Req. Since the grid is infinite, you could "break off" a segment, and the resistance of the remaining grid will still be Req. The remaining segment is in parallel with one of the 1 ohm resistors that has been "broken off." The equation for reisistors R1 and R2 in parallel is: Rp = R1*R2/(R1 + R2) . There is still a one ohm resistor from the broken-off section that is in series with Req. The equivalent resistance Req can be calculated using the equation: Req = 1 + (Req * 1)/(1+Req) Then Req^2 - Req - 1 = 0 Then use the quadratic equation to solve. Thus, the answer comes to be 2/pi ohms." That actually made sense to me...wow....engineering IS helping.
Beware the quiet people,
You don't know their intentions
(small signatures are sooo much cooler since they don't annoy people trying to read through posts!) |
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Re: A resistance question...
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by
on 2008-01-25 07:37:59 |
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There was some people in the forums of the comic that had a few problems with it. but it being 2/pi ohms? Where did the pi come from? |
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Re: A resistance question...
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by iamlagging
on 2008-01-25 18:57:47
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Haha, I thought about this problem for some time too. I haven't actually solved it but I developed an approach that MAY or MAY NOT work. I haven't crunched any numbers so I don't guarantee its success. But here goes: First, notice that the shortest path is a sequence of 3 resistors. Each path would therefore have a total resistance of 3 ohms. In this case, there are 3 such unique paths (you can just count them). The next case would be looking at paths of 5 resistors (there are no paths with an even number of resistors). Keeping the same approach in mind, we find that each path has a resistance of 5 ohms. The number of paths is another story. You have to keep in mind that current only travels one way through a resistor. This becomes a counting problem. And you can continue the cases infinitely. This is the part where the approach gets shaky since I'm not too sure of it: my mathematical intuition tells me there SHOULD be some sort of pattern between the length of the path and the number of unique paths. So there should be some sort of proof by mathematical induction here. If the above follows, then the numbers you get can be summed up in an infinite series (which should be converge). |