N is the number of nodes along a length of the square.
So for N=4 (a 4 x 4 grid):
R = 1 + 1/2 + 1/3 = 11/6 ohms.
For N=5:
R = 1 + 1/2 + 1/3 + 1/4 = 25/12 ohms.
I got that by drawing diagonal lines; I think all points on each diagonal line are at the same potential, so all points on a line can be treated as one node.
So from the first corner to the first diagonal line (which links 2 nodes), there are 2 resistors in parallel, so the resistance is 1/2 ohm.
That is in series with the resistance between the 1st diagonal line (2 nodes) and the second diagonal line (3 nodes), so there are 4 paralleled resistors between those 2 diagonal lines, so another 1/4 ohm.
The infinite grid is the one often posed and solutions are to be found on the web, for example
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A finite rather than infinite grid can be worked in by setting limits appropriately. I don't know if that leads to a short-cut - give it a go!
The one we were given at Uni was to take an infinite grid of squares of 1 ohm resistors, remove one resistor and calculate the resistance between the two points to which that resistor would have been connected. Again, the solution can be found on the web, for example
When I posted that this morning, I made the assumption based on intuition.
This evening, I looked at it again, and arrived at another approach, which is less based on guesswork and more rigorous (I think!).
It relies on the symmetry of the grid about the diagonal line AB (referring back to your diagram).
Take the N=3 case (and view in Courier or similar font):
a b c
d e f
g h i
The nodes are at positions a to i, and the 1 ohm resistors link them on a square grid.
It is obvious from symmetry that b,d are at the same potential, and f,h, and c,g.
So we can eliminate half of the resistors and nodes b,c,f and change the remaining resistors from 1-ohm ones to 0.5-ohm ones. This just exploits the symmetry about the line a-e-i above.
That leaves us with:
a
d e
g h i
Now all resistors are 0.5 ohms.
Again, it is obvious that e,g are at the same potential, so we use the same trick again on d,e,g,h and remove node e and the 2 0.5-ohm resistors d-e, and e-h, and make resistors d-g and g-h into 0.25-ohm ones.
So now we have:
a
d
g h i
Resistors a-d, and h-i are 0.5 ohms for a subtotal of 1.0 ohms.
Resistors d-g, and g-h are 0.25 ohms for a subtotal of 0.5 ohms.
Therefore resistance a-d-g-h-i is 1.5 ohms.
This is the same value as that predicted by my earlier summation formula, i.e. 1/1 + 1/2 = 3/2 ohms.
Now because the entire network is symmetrical about line M-J, nodes M,J must be at the same potential.
Then, also, because all 4 resistors in square I-J-N-M are equal, the square can be "folded" along the diagonal I-N, so that nodes J,N end up as 1 node (J), node M can be removed, and the remaining resistors I-J and J-N are again halved in value:
E-|-F | | - - | | I-=-J-|-K | | = - | | N-|-O + 1
Where a double-barred link -=- indicates a 1/4 ohm resistor.
Now the resistance R(E-O) is just 2 of network E-F-J-I, in series, so the resistance of the original grid is:
This, IF CORRECT, does NOT equal the value predicted by my initial summation formula, of:
1/1 + 1/2 + 1/3 = 11/6 ohms.
That in turn would seem to render invalid my assumption that all points on the diagonal lines I mentioned are at the same potential.
However, it is not clear to me why that should be.
Also, I do not now see how to generate a formula for R for arbitrary N.
I did try the same approach for N=5, but ran into a problem, because I managed to reduce the network to a small sub-network that I do not know how to solve for overall resistance.
However, the approach I'm using to reduce the square grid network can be summarised into a few steps:
1) Fold it dagonally along diagonal line AP (in top diagram above) to remove one half of the network.
2) Change all resistors in the remaining triangular grid to half of their previous value
3) Take off the 2 resistors at A and P because they're only connected to E and O, and just note their total series resistance and add it back on later.
4) Now consider all nodes on the middle diagonal going from (say) bottom left to top right (in above example), i.e. the longest diagonal at 90 deg to the original diagonal AB. The network is symmetrical about this new diagonal, so all nodes on it are at the same potential. So they can all be collapsed onto 1 node by removing nodes and resistors and reducing values of existing resistors in that area (at least for N=4, N=5).
5) Now you end up with 2 identical subnetworks, in series (see penultimate diagram above), so you only need to solve 1 of them for R, and then double the result. (This caused me a problem for N=5.)
6) Add back on the series resistance removed in step 3 above.
I'm a bit stuck now, though. And I don't know for sure that my value of 13/7 ohms for N=4 above is even correct.
In my particular case, R1=R4=0.5, R2=R5=0.5, R3=R6=0.25, and R7=0.5.
I wonder whether since the vertical chains are identical, I can move the bottom of R7 to the junction of R2, R3, and it'll still be the same in terms of R(A-B).
That assumption is incorrect for the cases where there are 3 or more resistors (4 or more nodes) on a side. For the cases where n >= 4, some of the diagonal lines have equipotential nodes, and some do not.
Inadequate symmetry. There is the classic resistor cube problem:
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People who have seen that problem and know about the method of solution that exploits symmetry, upon seeing the finite resistor grid of this problem, will likely think (as I did at first) that the same kind of symmetry argument will solve it. I haven't been able to come up with a shortcut method yet, but there may well be one. Here are a few solutions, showing the correct value (determined by a brute force solution of the network), and the (incorrect, if n > 3) value determined by the sum of the reciprocals of integers (the method you initially proposed, and which I also initially thought would work):
The n=21 case (the case with 20 resistors on a side) has an exact rational solution, but it involves a really big numerator and denominator, so I give the decimal version.
No, you can't do that. If you replace R4 and R5 with a single equivalent resistor, and R2 and R3 similarly, you end up with a Wheatstone bridge circuit:
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This can be solved with the Wye-Delta transformation:
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See also:
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I get a value of 25/44 for this subnetwork. If you continue with your analysis using that value, do you get the correct value of 47/22 for the entire n=5 network?
If you have a grid like this and want the resistance between a and p:
a b c d e f g h i j k l m n o p
You can reduce it to a grid like this (by symmetry), where you are getting the resistance between a and x:
a b c x f x
For a grid like this:
a b c d e f g h i j k l m n o p q r s t u v w x y
You can reduce it to the equivalent problem, which is the resistance between a and x in this network:
a b c d x g h x x
This makes the problem much easier. The only issue is the delta-wye conversion that is required for Ra=agh, Rb=ch, Rc=bc. However, working it out, you end up with the 47/22 value one might expect.
I'm sure there is a scheme that depends on adding up the total number of different non-looping paths through the network, with the lengths of those paths factored in somehow. I'm not sure that this is simpler than the brute force method, though.
Well, in the continuum approximation, this is one ohm per square sheet material. You can solve Poisson's equation for the potential at any point in the sheet... it's the square boundary that is the problem in that case, an infinite sheet is.. relatively easy.
It's because some nodes have two connections to the next diagonal and some nodes have only one. in other words some nodes aren't evenly spaced between their adjacent diagonals.
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