Given an infinite number of identical resistors connected to form an infinitely large 2D network (each resistor is connected to 4 other resistors), what is the effective resistance between any 2 adjacent junctions?
When you say resistor is connected to 4 other resistors do you mean the resistors are lay out in a "ladder" type format?
Hmm, sorry about the mistake.. Here's what I mean: Given an infinite number of identical resistors connected to form an infinitely large 2D arrays of squares with a resistor at each side (each resistor is connected to 6 other resistors), what is the effective resistance between any 2 adjacent junctions? --|--|--|--|--|--|-- --|--|--|--|--|--|-- --|--|--|--|--|--|-- --|--|--|--|--|--|-- --|--|--|--|--|--|-- --|--|--|--|--|--|-- --|--|--|--|--|--|--
Right, confusing question! If you are looking at the resistance across a resistor that is in parallel with an infinate number of other resistors (i.e. node to node across a resistor) then the resistance is infinate. Think about it that the curent has to be disributed across an infinate number of resistors, therefore there is effectively no current through the resistors. By looking at the equations for parallel resistors: 1/RTotal = 1/R1 + 1/R2 + 1/R3 ....... + 1/Rn As n tends to infinity so RTotal tends to zero, and the reciprocal of zero (or at least a very very small number) is astronomically big, i.e. tending to infinity. The problem with this question is that you are dealing with infinities , which invariabley cause serious problems mathematically.
Does it really converge to zero? The way I see it, it sounds like you have an effective resistance of: R//(R+effective resistance looking into north node)//(R+effective resistance looking into south node)//(R+effective resistance looking into west node) so that the overall effective resistance approaches R/4?
Hmm, my logic seems to be marginally flawed even if I still agree with the answer <_< (I really shouldn't watch the Rugby and type at the same time!) 1/RTotal = 1/R1 + 1/R2 + 1/R3 ....... + 1/Rn As n tends to infinity so RTotal tends to zero, and the reciprocal of zero (or at least a very very small number) is astronomically big, i.e. tending to infinity. This is the bit I think is logically flawed. What I really meant to say was that for an infinate number of resistors in parallel, the above equation gives an infinate number of fractions, so 1/RTotal tends to infinity also. The current is distributed an infinate number of times so the resistance is zero because effectively there is no current through any individual resistor. I can see what your saying Battousai, however I haven't interpreted the question that way. I'm not 100% certain from Dazed's diagram how the configuration works, buy my interpretation each resistor is connected to 8 other resistors and the basis of my answer is on this idea. Its a good question really!
Hmm..I think your answer is correct. But I don't understand your workings. Can you please explain.. Thanks.
Well if you have two resistors in parallel the overall effective resistance must be lower than the value of each resistor alone, right? This problem is a recursive one... 1/Reffective = 1/R + 3/(R+Reffective) Hmm... this actually comes out to Reffective = R/3 My equation may be wrong but I believe my idea is right, which is: If you want to know the effective resistance between two adjacent nodes A and B, (let's say B is east of A): The effective resistance will be the resistance looking north in parallel with the resistance looking south in parallel with the resistance looking west in parallel with the resistance looking east: 1/Reff = 1/Rnorth + 1/Rsouth + 1/Rwest + 1/Reast Reast = R and Rnorth = Rsouth since they are symmetric. I'm not sure what to think about Rwest actually. In the equation where I calculated Reff=R/3 I assumed Rwest = Rnorth = Rsouth. This probably is not true. Anyway the resistance looking north is just R + (effective resistance of node north of A). The effective resistance of of the node to the north of A is the same as the initial question: you go to the node to the north of A and look at all the effective resistances north, east, west, (we already counted south), which will be the same as Reff itself (I think). Anyway I hope this helps. I haven't thought about this problem enough because I have kind of a headache <_< , but I hope I pointed you in the right direction.
Ok bear with me one last time!! Firstly, my previous answer were based on the fact that I didn't understand exactly the configuration, but now I do (I hope <_< ). Correct me if I'm wrong but the resistors are layed out like a net, with each string on the net being analogus to a branch in the resistor network with one resistor on it. This goes off in all directions infinately. Before I was assuming that the resistor networ was a series layout of infinately long parallel resistors. If I've got this wrong this time I really do give up!! If you are looking at the resistance between to adjacent nodes (junctions) across one resistor you can disregard the resistors aft of the nodes (north and south if that makes it easier), and only concentrate on the resistors between the two nodes (junctions) (i.e. east and west). However the paradox of this question is that the resistors network is infinate in all directions, this doesn't concern us north and south of the nodes but it does to the east and west. Ok now lets breakdown the resistor network. If you took one square on the network you would have four branches (the four sides of the square) each with one resistor on, the resistors are equal in value, therefore the resistor around this square is 4R. Now if you put another square next to that square (they share a common branch) you now have one branch with one resistor on in parallel with two loops either side with three resistors on (the three resistor are from the three branches that make the loops), you would have a sum of 3 terms. To find the resistance between the adjacent nodes in this network would require you to use parallel resistor equations like the ones I posted above or the product-over-sum. If you made the value of each resistor = 10Ω Then you would find you have a total resistance of around 6Ω. Now add another two loops either side of this network, one on the east side and one on the west side, so you will now have a sum of 5 terms. Using the same equations as above, the resistance is now 5.somethingΩ. If you continue to add loops the the east and west sides as you must do to correctly evaluate the resistance between adjaecent nodes then you can see the resistance dropping, if you sum that to infinity eventually you will have zero resistance - don't forget for parallel networks it is the reciprocal of your sum, so for an infinate number of sums x: 1/(x tending to infinity) ~ 0. Sorry about the naff explaination, but please correct me if you think I'm wrong, but thats the way I see it, you can ignore all the garbage I said above I didn't interpret the network correctly so I wasn't answering the question asked, although strangely enough I still have the same answer!! Comments :unsure:
:blink: Actually, thats the aproach I'm taking with this question - circuit reduction. However because the circuit is infinate, it can't be reduced!! So its a case of sums tending to infinity.
When you count multiple loops like that, aren't you double counting somewhere? I kind of see where you're headed. I thought at first that since there are infinite paths from one node to another that the overall resistance must tend to 0. But then again there are only four paths exiting a node and only four paths entering a node. I believe that your answer is correct for an infinite matrix of resistors where each node is connected to an infinite number of resistors = infinite number of paths = infinite number of resistors in parallel = 0 resistance. This is making my head hurt