Can we find net resistance between the body diagonal points of a infinite cubic lattice
The objective is to find net resistance between A & B of the given cube which is part of infinite grid. Let the resistance between any two adjacent vertices is R
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Pranjal Prashant
3 years, 4 months ago
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Top NewestOh my.
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Yes, I know about 2 d version for resistors and fourier series for getting across square diagonal 2r/pi.[you know that Ishan :D ] But this is quite different fom those, much more difficult in getting the approach. You should share this so that someone reaches at the final ansWeR.
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Here is the 2-d version of the problem.
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A simpler version would be two find the equivalent resistance between two diagonally opposite points in an infinite grid of resistors, which is still quite difficult. See the page I have linked Infinite grid of resistors
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It must be R/3
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Solution ?
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I asked for diagonally opposite points. Not adjacent ones
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Total resistance (across points A and B) = ( 5/6 ) * R
I can't post a picture of my solution here
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Incorrect, This answer holds if there was a single cube of resistances rather than an infinite one.
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Yes it is haha, I considered only a 'cube'.
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but infinite series of resistors just push the "single cube value" to an exact number so my idea is that effective resistance is near (5/6)R (like 0.85R or 0.9R max but not less than (5/6)R). Please notify me if there is a mistake in my assumption.
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I could not understand "push the single cube value to an exact number". Also think of 2-D grid of infinite resistors in which we have to find the equivalent resistance between adjacent points which everyone knows to be R/2. What will be your argument in this one? Just because other resistances are present does not mean that the net would be greater than 5R/6 because resistances in parallel decrease the value
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Now what significance pushing values is of, the question is not an mcq, I want to know how it can be done, and I am sure that it is solvable. {although answer would not be beautiful perhaps}.
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got it but I dont know how to solve the way you told and yes the answer sure will be weird
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NO, IT IS NOT SO. In the fourier series , steps of integration are nasty, but answer is simply 2r/ \(\pi\)
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thanks for the info I will surely try to get the steps for your answer.
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