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 Subject: Infinite grid of resistors Category: Science > Physics Asked by: racecar-ga List Price: \$2.00 Posted: 29 Oct 2002 16:18 PST Expires: 28 Nov 2002 16:18 PST Question ID: 92564
 ```Well known problem: Given an infinite grid of 1 ohm risistors, what is the resistance across a single diagonal? Well known answer: 2/pi ohms. Question: Where can I find a full solution to this problem?```
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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 Use quadratic equation to solve```
 ```To clarify: the grid in question is infinite in 2 dimensions, not just 1. The method mentioned by scodad works fine for an infinite "ladder" of resistors, but since the answer has a pi in it, there is no point in looking for a solution by such a simple algebraic method in the case of an infinite (in 2 dimensions) grid.```
 ```By full solution, do you mean across more than one node? If so, the answer is: R(n,n) = 2/pi * Sum{k=1...n: 1/(2*k-1)} R(1,1) = 2/pi R(2,2) = 2/pi * (1 + 1/3) R(3,3) = 2/pi * (1 + 1/3 + 1/5) the derivation of this can be found among the following http://mathquest.com/discuss/sci.math/a/m/277937/277940 (with embedded links) Other links... http://hilbert.dartmouth.edu/~doyle/docs/net/net/net.html#thomasseninfinite (with embedded links) Comments on a three dimensional grid... http://mathforum.org/discuss/sci.math/m/193340/193357```
 ```A current link showing the integral solution for one 2D case http://www.geocities.com/frooha/grid/node2.html```