# Resistor Grid Voltages Challenge

Discussion in 'Math' started by MrAl, Apr 29, 2017.

1. ### MrAl Thread Starter AAC Fanatic!

Jun 17, 2014
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Hello there,

For those that like solving networks here is a not too difficult one that only involves resistors and a single voltage source. However, there are a lot of resistors
The resistors form a nice symmetrical grid too.

So see what you can come up with, and see if you can figure it out in whole or in part without actually calculating too much, or if you feel you have to calculate a lot then go right ahead and do so. I can assure you it is interesting.

In this simpler example, all the resistors have the same value, 10 Ohms each. And only one voltage source, which is 10 volts DC. No other complications. One end of the 'diamond' is connected to 10v (E1 in the schematic) and the bottom point of the 'diamond' is connected to ground which is zero volts. That's it. The schematic is shown in the attachment, and as drawn two green horizontal lines pass through two nodes where each horizontal set has the same voltage for both nodes (but note the bottom set is not the same voltage as the top set of course).

The challenge is to find the other horizontal voltages or at least explain how they might be found (such as a simplification or something), or say something that seems important about the network as is.

Note the voltages are the node voltages and are labeled by row and column, such as v11, v12, v13, etc., but only the first and last for each row are shown in the diagram.

Here we go:

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2. ### WBahn Moderator

Mar 31, 2012
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The total resistance of the network is R times the harmonic number of the square root of the number of resistors used (which is the number of resistors in each side of the grid).

The harmonic numbers have no closed form rational representation. They also diverge as the number of resistors increases.

CORRECTION: While the number of resistors in each side of the grid is correct (so four in the case of the figure provided), the total number of resistors is NOT the square of this -- I didn't take into account that the resistors bridge in both dimensions.

If there are N resistors on each edge, the total number of resistors is 2N(N+1). So with N=4, the total number of resistors is 40.

Last edited: Apr 29, 2017
3. ### MrAl Thread Starter AAC Fanatic!

Jun 17, 2014
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Hi,

That's very interesting, i was not aware of this. That's a fact that is nice to know.

But let me make sure i understand you right in what you are calculating here.

First, are you measuring the resistance across two diagonal corners as in the diagram given, or somewhere else?

Also, when you say the harmonic number of the square root of the number of resistors it sounds like you mean the harmonic number of 5 because sqrt(25)=5 and that by harmonic number you mean the sum of reciprocals from 1 to 5 inclusive, so we would have as result:
HN=1+1/2+1/3+1/4+1/5

and so the total resistance across your two measuring points (which hopefully are the same as mine which is across two diagonal corners) is:
Rtotal=R*NH

where NH is defined above and R is the resistance of one resistor (and they are all the same value).

Or do you instead mean the harmonic number is calculated as:
HN=(1)^2+(1/2)^2+(1/3)^2+(1/4)^2+(1/5)^2

Lastly, is this an approximation or is it an exact calculation of the total resistance?

Sorry for all the questions but i'd like to nail this down and try to verify the results of my own techniques using a program for example which uses a totally different analysis method then we usually use.

Thanks for the reply. Also, if you care to mention where you learned this technique that might help too, thanks.

4. ### WBahn Moderator

Mar 31, 2012
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Ignore my sqrt(number of resistors) -- that was incorrect (see my correction in the post).

The number that matters is the number of resistors along each edge. In your diagram this is N=4.

So the total resistance is

Req = R(1/1 + 1/2 + 1/3 + 1/4) = (10 Ω)(12 + 6 + 4 + 3)/12 = (250/12) Ω = (125/6) Ω = 20.83 Ω

This is exact (except for the rounding at the end, of course).

I didn't "learn" this technique -- I looked at the symmetry of the network that was given.

Start from one of the corner nodes. Between it and the next horizontal node (as you have drawn the diamond) there are two resistors. Because the horizontal nodes are all equipotential, the resistors between any two layers are effectively in parallel. Each layer closer to the center adds two more resistors. The circuit is symmetric about the center horizontal line, so we have

$
R_{eq} \; = \; 2 $$\frac{R_0}{2} \; + \frac{R_0}{4} \; + \frac{R_0}{6} \; + \frac{R_0}{8}$$
R_{eq} \; = \; R_0 $$\frac{1}{1} \; + \frac{1}{2} \; + \frac{1}{3} \; + \frac{1}{4}$$
$

Generalizing this to any square array using equal-valued resistors

$
R_{eq} \; = \; R_o \sum_{k=1}^N \frac{1}{k}
$

5. ### MrAl Thread Starter AAC Fanatic!

Jun 17, 2014
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Hello again,

Are you sure that is not an approximation and actually an exact expression?

I ask because i used to think that when we rotate the grid by 45 degrees (and thus form a diamond shape rather than a square shape) that ALL the horizontal rows are equipotential, but if we look at this next diagram we see that cant really be true.

In the diagram i show the original grid first on the left, then the rotated grid in the middle showing two equipotential lines in red (which we might believe for now are true equipotential lines), then because of those two equipotential horizontal node sets we can redraw the diagram in a simpler form in order to reason out the other nodes a bit further. Therefore drawn to the right is a sub set of the grid, with the assumed equipotential nodes energized by two sources and to make it simpler one is a positive voltage V+ and the other is simply ground, although in the original grid we would only see ground (0v) there if there was a negative supply connected to the bottom of the diamond in the center drawing.

Now if we look at nodes P and R, we see they are connected directly to V+ through one resistor. So they are connected to the most positive source through ONE resistor which might be 2 ohms. If we look at node Q however, we see that it is connected to the most positive node (V+) directly through TWO resistors, which means that it connects to the main source with a total resistance of 1 ohm, which is 1/2 of what the edge nodes are connected through. This tells me that in a finite grid like this the node Q might have a higher potential than the nodes at P and R.

Similarly, if we go down one horizontal row of nodes and label them from left to right as A,B,C,D, then B and C must be of higher potential than either A or D, although B and C could be the same, and A and D could be the same, but A will not be the same as B or C and D will not be the same as B or C either.

What i think is the basic reasoning for this is that the impedance through an edge is different than through the center somewhere, so it is not as symmetrical as it seems at first. This might not be the case if we had the so called "infinite grid" but i think for a finite grid there is a tendency to have higher voltages near the center vertically.
Interestingly though the actual horizontal center of the diamond grid is 1/2 of Vcc at all 5 nodes.

I've done some reading on the web but they mostly like to talk about the 'infinite grid' where we allow it to become a continuous sheet of solid resistance material.

So maybe there is a small modification to that formula?

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6. ### WBahn Moderator

Mar 31, 2012
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If it turns out not to be exact, then it is might not even be a good approximation; it might simply be wrong.

Yes, there are two resistors tied to Q on the top instead of just one like P and R. But there are also more paths downward toward the center line from Q than there are from P or R. If the effective resistance going upward is 1 Ω instead of 2 Ω, that won't matter as long as the effective resistance going downward from the center node as it is going downward from an edge node. It's only the ratios that matter.

But you ask a very reasonable question, so let's explore it.

Would you agree the symmetry demands that each node long the center line be at the same potential, specifically half way between the potential of the top and bottom nodes? If they aren't then some of those nodes would be closer to the top node's voltage than the bottom node's voltage or vice-versa and there is nothing that justifies that.

All of the current entering the top node must split equally between the two resistors; left-right symmetry requires this. IF the next row's three nodes are all at the same potential, then each of the four resistors must carry half of the current as in the top row's two resistors, or 1/4 of the total current. IF we had a 2x2 square then this would satisfy the requirement that for the next layer as they paths start coming back together. But this is probably just because the second layer has twice as many resistors as the first. But if we have s 3x3 square, we to run into problems because one layer has four resistors and the next has six (and not eight). But if they are between equipotential nodes then they all carry the same current. However the outer nodes would have to split the current from the prior layer in half but the inner nodes have to have the same current in both resistors leaving them as in both resistors coming into them. Clearly all of the resistors in a given layer would NOT be carrying the same current in this case, which argues that the nodes across most horizontal lines are not at the same potential.

So I agree with you. I'll have to think if there is an elegant way to determine the way that the currents split from layer to layer -- it seems like there should be, but that might be a false illusion.

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9. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Plain old nodal analysis yields these voltages, starting at the upper left of the square array and moving to the right for 5 voltages, then down a row, etc.

10. ### MrAl Thread Starter AAC Fanatic!

Jun 17, 2014
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Hello WBahn and Electrician,

Hi,

Yes you are definitely correct about knowing the top 'feed' resistor is not enough we must also know the bottom impedances in order to assume that one node voltage is different than another. What i have managed to do though is find a way around this little dilemma by assuming the solution is equipotential and then trying to prove that is wrong. Once we assume the horizontal rows of the 'diamond' are equipotential, we then only have to test a single row that is between two assumed rows to see what the currents tell us. If the two rows we choose are equipotential and the solution really was all rows are equipotential, then the third intermediate row must also be equipotential. If we can show that just one row is not equipotential even when all other rows are, then we disproved that all rows are equipotential.

To attempt this i first did two wye to delta transformations and ended up with the new diamond shape shown in the attachment. Since the upper and lower rows are assumed equipotential, then we can calculate the currents in the resistors. Also, since after the transformation and again assuming equipotential rows we can completely remove two of the horizontal resistors, we end up with three separate series resistor strings energized by the same two voltages on top and bottom.
What we end up with though is not three identical resistor strings, but two identical (outer two) and one with one resistor 1/2 the resistance as one outer resistor (the center string where two of the same value resistors are in parallel).
It is clear that if the two chosen rows were equipotential as a test, then the two outer intermediate nodes are 1/7 of the voltage and the inner node is 1/4 of the total potential. That makes the inner node a different voltage than the two outer nodes, and so the entire grid can not be be horizontally equipotential. This of course does not prove that NONE of the horizontal rows are equipotential, and the top two and bottom two nodes (in the diamond) are definitely equipotential as that is easier to see.
BTW the resistors with heavy leads are 3 times the resistance of the other resistors due to the wye to delta transformation, but since those resistors are all the same value too it makes it a little easier.

Here's a tabular set of data for the nodes with about 4 significant digits for each node voltage:
[11] 10.00 [12] 7.660 [13] 6.277 [14] 5.426 [15] 5.000
[21] 7.660 [22] 6.702 [23] 5.745 [24] 5.000 [25] 4.574
[31] 6.277 [32] 5.745 [33] 5.000 [34] 4.255 [35] 3.723
[41] 5.426 [42] 5.000 [43] 4.255 [44] 3.298 [45] 2.340
[51] 5.000 [52] 4.574 [53] 3.723 [54] 2.340 [55] 0.000

This agrees with Electrician's brute force nodal

There is some symmetry here but not as much as we would like. For example, 6.702+3.298=10.000 volts.

What was so interesting about WBahn's idea is that something like that, when calculate the following:
sum=Sum of all N from 2 to n+1
sum=1/2^2+1/3^2+1/4^2+1/5^2
Rtotal=R/sum

i got a value that is only about 1 percent different than the true result for this grid. I have not tested other grids though yet but i think this was just a coincidence.

The actual value for the total resistance i got was exactly 235/11 Ohms (all resistors 10 Ohms each).

Electrician:
Thanks for the link. Now we can compare results to the 20x20 matrix as well.

LATER:
The result posted in that older thread was 11.86297756 Ohms which is approx 11.86 Ohms, but i am getting a different result by a small amount but still significant: 11.53 Ohms. I will have to go over this though to see if i made a mistake or not. This is using a 20 x 20 grid with each resistor equal to 3 ohms. UPDATE: A new calculation gave 11.863 to those 5 significant digits so this is now in agreement with the older post calculation. I'll go over it one more time to see if i can get better resolution, but i am reasonably happy with it already. This is BTW using an entirely different method for calculation other than Nodal or another regular network theory. It might still be nice to know how the original was calculated however as to the precision of the numerical math used. Not as big a deal now though
In the mean time however, i would want to inquire as to how this 11.86 Ohms was calculated, possibly what program, and what kind of precision was used for the numerical calculations. For example if the program incorporates Gaussian elimination to calculate the matrix solution and it has limited precision in each calculation, we could see small errors like this pop up because of the numerical truncation after many, many multiplications and subtractions and divisions. A 20 x 20 grid requires a huge number of calculations each depending on the previous result, and each previous result is truncated in some way by some amount, using a straight forward Nodal analysis method. But if we could see how the program handles these tasks we could make a better judgement.
One way to tell might be to see a tabular set of data with each node voltage. We could then verify some of the voltages to see if they add up.

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11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The calculation is made using the nodal method. An admittance matrix for the array is created using integers. The matrix is inverted, multiplied by the resistance of the individual resistors, and the 1,1 element is the total resistance. The matrix is inverted with Mathematica and since it's done with arithmetic on rational numbers, there are no rounding errors. The final result is exact and I show it as a fraction and a 15 digit decimal result.

Here's the calculation for the 4 resistor on a side version:

Here's the result for the 20 resistor on a side version from the older thread (the resistors are 3 ohms each). It took my laptop about 14 seconds to invert the matrix of integers:

12. ### MrAl Thread Starter AAC Fanatic!

Jun 17, 2014
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Hello again,

Ok sounds good. I am using a method similar to solving Laplace's equation numerically. For the 20 x 20 grid It takes less time than the time it takes to say, "Laplacian" The 100 x 100 grid takes longer though.

Ok so i think we beat this horse to death, maybe there are some other interesting things about the grid too?
I tried various formulas involving reciprocals but could not get any of them to work on the 20 x 20 grid. for computing the total resistance across opposite corners. I have one more method i was thinking about but am not sure i will get to do it right away because that Inductance Book just got here in the mail yesterday afternoon so i might go back to that. A quick glance, it looks like they give a lot of already worked out formulas in there.

Oh yeah BTW i dont use Mathematica so i dont understand the language it uses either. You'd have to explain what your post meant in terms of symbols and stuff like that. I like the result though, being the ratio of two integers tells me it should be as accurate as we can get with a method like that.

13. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I made a mistake in one of the Mathematica comments. The admittance matrices are not rsiz by rsiz; they're (rsiz)^2 by (rsiz)^2.

14. ### MrAl Thread Starter AAC Fanatic!

Jun 17, 2014
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Hello again,

Ok thanks but i still dont use Mathematica so i dont understand the symbols being used there.
A little explanation of the symbols and stuff would help understand that part.

I really like the result though, being the ratio of two integers. That probably indicates that there is no simple solution, simple that is like some simple algebraic statement like WBahn and I were looking for earlier.

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Looking at the large expression that generates the admittance matrix, the Abs and Mod functions are the usual. The Boole function evaluates the logical expression inside the brackets [], and returns a 1 if the expression evaluates to true, or a zero if the expression evaluates to false. The bold symbol Λ is logical "and", V is logical "or". The Table function generates the elements of the matrix from left to right, up to down as the variables i (the row) and j (the column) run from 1 to (rsiz)^2.

Knowing all this, you can verify that the large expression uses the standard rules for generating an admittance matrix by inspection:

16. ### MrAl Thread Starter AAC Fanatic!

Jun 17, 2014
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Hello again,

I wanted to do one more thnig with this corner to corner resistance property and that is to calculate the NxN resistance for N=1 to 100 and also for N=200. Here is the graph of those results.

Notes:
All resistors are the same value, each 10 ohms.
For a given N there are 2*N*(N+1) resistors in the grid and (N+1)^2 voltage nodes, so for example for N=200 there are 80400 separate resistors involved and 40401 voltage nodes that are calculated.
The resistance is measured from one corner node to an opposite corner node (diagonally), for example from the upper left hand corner to the lower right hand corner of the grid.
The graph is a plot of all integer N values from N=1 to N=100, but just one point for N=200 so there are no points plotted from N=101 to N=199.

There are many other two point calculations that could be done but i will save that for another time.

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17. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The curve you plotted looks a plot of the log function. That doesn't surprise me given that the sum of reciprocals method for calculating Req is an approximation to the correct value, and the integral of 1/x gives a log function.

If you calculate (and possibly plot) the error in the sum of reciprocals method, you get an interesting result. The error is positive for small values of n, and negative for large values of n, crossing over around n=7 or 8.

The error seems to be continually increasing for increasing n. Since Req is apparently unbounded as n increases, I would guess that the error for the sum of reciprocals method also increases without bound for increasing n.

18. ### MrAl Thread Starter AAC Fanatic!

Jun 17, 2014
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Hi again,

Yes it is somewhat like a log function, a little smoother, like a power function in ln(x):
y=(25*ln(x)^(4/3))/4+10

for example (each R=10).

Might be interesting to try to find a formula, or try to see if there is some asymptotic limit.
It's hard to tell even after doing the 200x200 grid case.

19. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Did you have a look at the error behavior of the sum of reciprocals formula compared to the true Req?

20. ### WBahn Moderator

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I would have expected the reverse. I would have thought that, as the size of the mesh grew, that the approximation to the harmonic function would have gotten better as the edge effects became more diffuse.

But as I think about it, I guess I'm not too confident of that conclusion.

But I still think it should approach the solution for a solid sheet of material. A plot of the equipotential lines in such a sheet might be enlightening. When you think about those, you can start to see that they can't be straight lines parallel to the non-contacted diagonals, otherwise the current would not spread out as the width increases. I would expect them to be convex relative to the contact points. If that's true, then the voltage change more quickly (with distance) along the center line (between the contacted corners) than off -- which makes intuitive sense since the path length is longer. Thus as you go across one of the rows parallel to the non-contacted diagonal, the magnitude of the voltages should be higher near the center (if the diagonal is used as the reference).

This agrees with the numerical results shown earlier.

Last edited: May 5, 2017