Infinite cubic grid of resisrors

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

#ElectricityAndMagnetism #Resistance #Grids #infinite

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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Oh my.

Josh Silverman Staff - 4 years, 4 months ago

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.

Pranjal Prashant - 4 years, 4 months ago

Here is the 2-d version of the problem.

Calvin Lin Staff - 4 years, 4 months ago

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

Ishan Tarunesh - 4 years, 4 months ago

It must be R/3

Jatin Chauhan - 3 years, 9 months ago

Solution ?

Rajdeep Dhingra - 3 years, 9 months ago

I asked for diagonally opposite points. Not adjacent ones

Pranjal Prashant - 3 years, 9 months ago

Total resistance (across points A and B) = ( 5/6 ) * R
I can't post a picture of my solution here

Vincent Miller Moral - 4 years, 4 months ago

Incorrect, This answer holds if there was a single cube of resistances rather than an infinite one.

Ishan Tarunesh - 4 years, 4 months ago

Yes it is haha, I considered only a 'cube'.

Vincent Miller Moral - 4 years, 4 months ago

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.

Yeshas Bharadwaj - 4 years, 4 months ago

@Yeshas Bharadwaj – 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

Ishan Tarunesh - 4 years, 4 months ago

@Yeshas Bharadwaj – 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}.

Pranjal Prashant - 4 years, 4 months ago

@Pranjal Prashant – got it but I dont know how to solve the way you told and yes the answer sure will be weird

Yeshas Bharadwaj - 4 years, 4 months ago

@Yeshas Bharadwaj – NO, IT IS NOT SO. In the fourier series , steps of integration are nasty, but answer is simply 2r/ $\pi$

Pranjal Prashant - 4 years, 4 months ago

@Pranjal Prashant – thanks for the info I will surely try to get the steps for your answer.

Yeshas Bharadwaj - 4 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$