# resistance across a mesh of wire?

#### redacejr

Joined Apr 22, 2008
85
ok so if you have this 20 x 20cm of 1cm sq squares mesh of wire. if 1 cm of wire has a 3 ohm resistance.... what would the resistance be across the diagonal ( from a to b in the diagram) i cannot really figure it uot. if you need extra info ask pls .

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#### hgmjr

Joined Jan 28, 2005
9,029
This appears to be a homework problem, if so it belongs in the homework section.

One way to solve the problem is to start with a single square. Determine how you would find the resistance of the square. Then I would increase the size to 4 squares and determine how that configuration would be analyzed.

Once you have succeeded in analyzing a 2 by 2 square you would be able to establish a pattern that would allow you to attack the 20 by 20 mesh.

hgmjr

#### studiot

Joined Nov 9, 2007
4,998
Just think about it for a moment. If you have a single entry and exit connection for the current you will have a controlling subsection of the mesh, no matter how big it is.

#### Ratch

Joined Mar 20, 2007
1,070
redacejr,

At first this problem looks formidable, but it is really a pussy cat. This network is really just a symmetric "mesh bridge". In a ordinary bridge, when balanced, no current exists across the crossover resistor because the voltage is equal on both sides. Same here. There is no current existing in the interior of the mesh. All the charge from the ohmmeter is flowing along the edges of the mesh. So we have two legs of 40 cm each in parallel. That makes 3*40 = 120 ohms on each leg. Two 120 ohm parallel legs makes 60 ohms of resistance total across points A and B.

Ratch

#### The Electrician

Joined Oct 9, 2007
2,751
redacejr,

There is no current existing in the interior of the mesh. All the charge from the ohmmeter is flowing along the edges of the mesh.
Ratch

#### Ratch

Joined Mar 20, 2007
1,070
The Electrician,

Perhaps you are right. Increasing the number of threads in the mesh to infinity, would simulate a solid piece of metal. That surely would not direct current to only the outer edges, would it? Back to another solution.

Ratch

#### studiot

Joined Nov 9, 2007
4,998
Ratch,
if the edges are connected this is a two dimensional version of the infinite ladder of resistors problem - a transmission line.

If the edges are open, then as I said before all the current flows through two lenghts of wire (in and out) in series with the rest of the grid.

#### The Electrician

Joined Oct 9, 2007
2,751
Let each little 1 cm piece of wire be replaced by a 3 ohm resistor so we have a rectangular grid of resistors.

To test my algorithm, I solve a couple of smaller grids first.

For a grid with 2 resistors on each edge, the resistance would be 4.5 ohms.

For a grid with 3 resistors on each edge, the resistance would be 39/7, or 5.57143 ohms.

Seems reasonable; let's go for the big one.

For a grid with 20 resistors on each edge, the resistance would be 11.86297756 ohms.

I can't see any shortcut method so I just used the nodal method and solved the network.

After writing the above, I was looking at the grid of resistors again, and I noticed that there is what appears to be a shortcut method. One could use the same kind of argument that is used on the cube of resistors. Draw a grid with 3 resistors on a side and orient it so it looks like a diamond, with a corner at the top and bottom. Now draw a horizontal line through through the two nodes that are just below the top node; draw another line through the 3 nodes that are just below the previous two; then through 4 nodes, then 3 nodes and finally two nodes just above the bottom corner.

It appears that one could say that because of symmetry, the nodes through which each horizontal line passes are at the same potential, and therefore may be connected without changing the resistance of the circuit from corner to corner. Then, starting from the top, we have 2 resistors in parallel, then 4, then 6, then 6, then 4, then 2. If the resistors are all 1 ohm, it would seem that we would have a total corner-to-corner resistance of 1/2 + 1/4 + 1/6 + 1/6 + 1/4 + 1/2 ohm.

But, this isn't correct because some of the nodes through which the horizontal lines pass are equipotential and some are not!!!!

If they were all equipotential, then the solution of the 20 x 20 grid would be 3 times (because in the problem the resistors, wires actually, are 3 ohms each) the sum of the reciprocals of the integers from 1 to 20, or 10.7932 ohms.

I was surprised by this, so much so that I got 24 100 ohm, 1% resistors, and wired up a 3x3 grid and made some measurements. Sure enough, the horizontal line connected nodes are NOT all equipotential. Some of them are!! But not all.

#### hgmjr

Joined Jan 28, 2005
9,029
If they were all equipotential, then the solution of the 20 x 20 grid would be 3 times (because in the problem the resistors, wires actually, are 3 ohms each) the sum of the reciprocals of the integers from 1 to 20, or 10.7932 ohms.
For the 20 by 20 grid, I got the same answer as the electrician. 10.79322 Ohms.

In my solution, I exploited the symmetry between the lower triangle and the upper triangle.

ADDENDUM: The answer that I have provided above is not the correct one. It was based on the assumption that that all nodes are at equipotential. The Electrician kindly brought the error to my attention. Thanks.

hgmjr

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#### The Electrician

Joined Oct 9, 2007
2,751
For the 20 by 20 grid, I got the same answer as the electrician. 10.79322 Ohms.

In my solution, I exploited the symmetry between the lower triangle and the upper triangle.

hgmjr
You do understand, don't you, that that number is incorrect? It *would* be correct if all those groups of nodes I described were equipotential, but not all of them are.

The correct result is 11.86297756 ohms.

If you examine the grid with 3 resistors on a side, it's small enough to solve by conventional methods. You will find that you get a different result by the method that ends up summing the reciprocals of integers versus a conventional solution. That's what tipped me off to the fact that the assumption about equipotentials was false.

For a 3x3 grid, the method of reciprocals gives a result of 11/2, or 5.5 ohms, while a conventional network solution gives a result of 39/7, or 5.57143 ohms, with the latter being correct.

If you inject 1 amp at the top of the diamond in the 3x3 case and solve for node voltages, you will see what I mean about the groups of nodes not all being equipotential.

It just so happens that the node groups *are* equipotential in the 2x2 case, but that's the only one. For any larger number of nodes, they aren't *all* equipotential, and, of course, that includes the 20x20 case.

#### hgmjr

Joined Jan 28, 2005
9,029
I see your point electrician. Thanks for the correction. My solution assumed equipotential and that is not a valid assumption as you have indicated.

The most glaring location in the matrix that makes it obvious that an equipotential in not present is the two corners.

I'll take a closer look in view of the information you have supplied.

Thanks,
hgmjr

#### The Electrician

Joined Oct 9, 2007
2,751
Consider the 3x3 grid. If it's oriented as a diamond with corners up and down, right and left, we can number the nodes like this:

1
2 3
4 5 6
7 8 9 10
11 12 13
14 15
16

Apply 1 volt to node 1 (which can be done when using the nodal method by injecting a current of 7/39 amps if the resistors are all 3 ohms) and ground node 16; then calculate the voltages at the various nodes. I get this result:

1.000
.7308 .7308
.5769 .6154 .5769
.5000 .5000 .5000 .5000
.4321 .3846 .4321
.2692 .2692
0.000

As can be seen, some of the rows have equipotential nodes, some do not.

This is a tricky puzzle, because someone who was familiar with the resistor cube, and the method of solving it using symmetry, would assume that the same kind of symmetry argument would solve this one.

#### Ratch

Joined Mar 20, 2007
1,070
To the Ineffable All,

This lattice has a lot of symmetry in it, so I tried to apply Bartlett's bisection theorem. http://books.google.com/books?id=KkNkAPc-WcIC&pg=PA267&lpg=PA267&dq=bartlett's+bisection+theorem&source=web&ots=IIIVeEDW_A&sig=Jz0nokfzKp50NaF3LnQ5xl7lDEo&hl=en&sa=X&oi=book_result&resnum=7&ct=result#PPA267,M1 . The answer I got was 60.92 ohms, but who knows if I did it correctly. I was able to fold the circuit 6 times and finally wound up with 4 loop mesh to evaluate. There appears to be a dearth in information on how to apply Bartlett's theorem to a lattice, but I think it has possibilities. It would be nice if the OP could give us the answer.

Ratch

#### The Electrician

Joined Oct 9, 2007
2,751
In post #4, you said:
"...So we have two legs of 40 cm each in parallel. That makes 3*40 = 120 ohms on each leg. Two 120 ohm parallel legs makes 60 ohms of resistance total across points A and B."

Surely, the interior resistances can't make the corner to corner resistance greater than 60 ohms; only smaller.

#### Ratch

Joined Mar 20, 2007
1,070
The Electrician,

I may have doubled the value instead of halving it, which would make its value around 15 ohms. Like I said, implementing Bartlett's theorem for a mesh like this is new territory, but I think it can be done. I will look at my calculations again. It might be a while because I became busy lately. Thanks for pointing out the discrepancy.

Ratch

#### Ratch

Joined Mar 20, 2007
1,070
Electrician,

OK, I played with the 20x20 mesh some more. I used SWCADIII to model the mesh using 1 ohm resisters and a 1 volt source. I got 0.252888 amps which divided into 1 volt according to the resistance formula gives 3.95431 which gets multiplied by 3 for a result of 11.8630. That is the result you declared as correct. How did you get that result? You explained that some of the nodes were not at equipotential, but you never explained the solution. Anyway, I was able to apply Bartlett's theorem three times, and reduced the geometry significantly, but it still is too much to do by pencil and paper. I am enclosing three *.asc files which can be read by SWCADIII to display the geometry. After running the DC analysis, each node's voltage can be displayed by selecting with the cursor. Ask if you have any questions.

Ratch

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#### The Electrician

Joined Oct 9, 2007
2,751
Electrician,

OK, I played with the 20x20 mesh some more. I used SWCADIII to model the mesh using 1 ohm resisters and a 1 volt source. I got 0.252888 amps which divided into 1 volt according to the resistance formula gives 3.95431 which gets multiplied by 3 for a result of 11.8630. That is the result you declared as correct. How did you get that result? You explained that some of the nodes were not at equipotential, but you never explained the solution.

Ratch
I simply did a complete solution by the nodal method. I looked at node voltages on several of the diagonals and saw that they weren't all equipotential.

Have a look at post #12 in this thread.

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#### Ratch

Joined Mar 20, 2007
1,070
Electrician,

I simply did a complete solution by the nodal method.
Are you saying you did a 400 node analysis with a 20x20 matrix and solved it by a math program? Were you able to reduce it further to a simpler mesh? I guess I did the same thing using SWCADIII. The simplest I was able to get is shown in the enclosed file. I also tried solving a 5x5 mesh, but the results are not any integer multiple of a 20x20 mesh.

Ratch

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#### The Electrician

Joined Oct 9, 2007
2,751
Electrician,

Are you saying you did a 400 node analysis with a 20x20 matrix and solved it by a math program? Were you able to reduce it further to a simpler mesh?

Ratch
With 20 resistors on a side, there are 21 nodes on a side, for a total of 441 nodes in the circuit. I solved the system with a 441x441 matrix; it took a few minutes.

I didn't even bother trying to reduce it. By solving the entire circuit as it was, I was more confident of the answer.

I said it was brute force!

#### Ratch

Joined Mar 20, 2007
1,070
Electrician,

Thank you. If I see a way to significantly reduce it, I will let everyone know.

Ratch