Honeycomb electrical circuit

Thread Starter

jscivi

Joined Apr 22, 2017
5
Hello everyone,
I am trying to figure out the equivalent resistance of a honeycomb circuit of resistors (cf image attached, where each resistor has the same value).
In the example on the image, there are 3 nodes on the horizontal axis (Nx=3), and 14 nodes on the vertical axis (Ny=14), but I would like to know if there is an analytical solution for any numbers of nodes along both axes (for any Nx and Ny).

The solution for Nx=1 is quite trivial, it is simply: Req=Ny*R.

But do you think it is possible to find an analytical solution for Req as a function of Nx, Ny and R?
If yes, how would you proceed?

If no, what methods do you suggest for computing this?
I was thinking using a program, in which you input one equation for each node and each loop (Kirchoff's laws) and having a variable for each current in each resistor. And then having a program that could analytically solve the system of equation (using matrices for examples). Can you think of something better/faster/easier?

Thank you in advance for your help and time!

Best,
Julia


 

WBahn

Joined Mar 31, 2012
29,932
Looks like homework? Is it?

What would the voltages be across the horizontal resistors if they were all made very large (even infinite)?

How much current will flow in them as their values are made very small (even zero)?
 

Thread Starter

jscivi

Joined Apr 22, 2017
5
Hello,
Thank you for your response.
The thought of ignoring the horizontal resistors crossed my mind.
That would make things much easier: you just have Nx resistors of value R*Ny in parallel

But I am not looking for approximations (it's not homework, it's actually something I am personally interested in).

Do you think it's way too complicated to get a real analytical solution (no approximations)?

Thanks again,
Julia
 

WBahn

Joined Mar 31, 2012
29,932
Again, imagine that the horizontal resistors are either not there or their resistance is very large. What is the voltage across them? Then ask yourself what happens if the their resistances are reduced back down to R. At that point you have your exact solution.
 

Thread Starter

jscivi

Joined Apr 22, 2017
5
Hello,
Thanks again for the response.
So when you make the assumption that the horizontal resistors are very large (i.e. open circuit), the solution is simply: Req=R*(Ny-1)/Nx (cf. picture on the left below).
But when you make the assumption that the horizontal resistors are very small (i.e. shorts), the solution is not trivial. The most I can simplify the circuit too is what is shown on the right of the figure below.
But to be honest, I am not sure I understand where you are trying to get with this methodolody?

Thanks anyway for the time and help :)

Best,
Julia


 

WBahn

Joined Mar 31, 2012
29,932
You are leading yourself down a rabbit hole with the figure on the far right (even if it were correct).

Ask yourself why the top resistor in the left hand string is 2R while the bottom resistor in the right string is R. Doesn't symmetry demand that they be the same? Neither R nor 2R is correct.

But the key is to look at the diagram on the far left and do what I have suggested twice already. What is the voltage difference across each of the pairs of points where you removed a resistor?
 

Thread Starter

jscivi

Joined Apr 22, 2017
5
Yeah sorry for the silly mistake. The top and bottom resistors are R/2.
I am also sorry but I don't quite understand the question. In the case of the short-circuit, the voltage difference between the 2 points is 0.
For the case where we replace the horizontal resistor with an open-circuit, the difference of voltage is:
Voltage at point on the left node i (where i is the node index from 0 to 13) : Vi=V*R*(13-i)/(13R)
Voltage at point on the right node j (where j is the node index from 0 to 13) : Vj=V*R*(13-j)/(13R)
So the voltage difference is Vj-Vi=0
But these calculations (voltage divider) only hold true if there is no current passing in that horizontal branch.
 

WBahn

Joined Mar 31, 2012
29,932
Yeah sorry for the silly mistake. The top and bottom resistors are R/2.
I am also sorry but I don't quite understand the question. In the case of the short-circuit, the voltage difference between the 2 points is 0.
For the case where we replace the horizontal resistor with an open-circuit, the difference of voltage is:
Voltage at point on the left node i (where i is the node index from 0 to 13) : Vi=V*R*(13-i)/(13R)
Voltage at point on the right node j (where j is the node index from 0 to 13) : Vj=V*R*(13-j)/(13R)
So the voltage difference is Vj-Vi=0
But these calculations (voltage divider) only hold true if there is no current passing in that horizontal branch.
I think you are about at a point to make a mental break through.

I have two nodes, A and B. Without any thing connected between them the voltage difference between the nodes is zero. When I connect a resistor between those two nodes, how much current will flow in that resistor?
 

OBW0549

Joined Mar 2, 2015
3,566
Voltage at point on the left node i (where i is the node index from 0 to 13) : Vi=V*R*(13-i)/(13R)
Voltage at point on the right node j (where j is the node index from 0 to 13) : Vj=V*R*(13-j)/(13R)
So the voltage difference is Vj-Vi=0
But these calculations (voltage divider) only hold true if there is no current passing in that horizontal branch.
If the open-circuit voltage between those nodes is zero, how much current will flow if we insert a resistor between them?

You've found the key to unlock this puzzle, you just don't know it (yet).
 

Thread Starter

jscivi

Joined Apr 22, 2017
5
Ok thanks guy. It was not obvious at first for me!
But yeah if the voltage difference is 0, then the current is 0, and you are basically back to the simplified schematic where you can consider the horizontal resistors as open circuits.

So basically: Req=R*(Ny-1)/Nx, where Ny is the number of vertical nodes and Nx is the number of horizontal nodes.

Thanks a lot!

Best,
Julia
 

MrAl

Joined Jun 17, 2014
11,342
Ok thanks guy. It was not obvious at first for me!
But yeah if the voltage difference is 0, then the current is 0, and you are basically back to the simplified schematic where you can consider the horizontal resistors as open circuits.

So basically: Req=R*(Ny-1)/Nx, where Ny is the number of vertical nodes and Nx is the number of horizontal nodes.

Thanks a lot!

Best,
Julia

Hi,

Here's a partial circuit in text form and the solutions shown in groups of three because the voltages for three horizontal nodes of each row are the same. The nodes are numbered and the voltages follow: v1, v2, are for node 1 and node 2 respectively.

The solutions when the resistors are identical (all resistors equal to R1) are pretty simple, but when the resistors are all a different value the solution is quite large, spanning some 100,000 characters or more although i did not spend too much time looking for simplifications.
 

Attachments

WBahn

Joined Mar 31, 2012
29,932
If the resistors are all different is going to be a nightmare very quickly.

Analytically, the best approach is probably to do delta-wye conversions to collapse the network down. Doing this for the all-identical case would be a good exercise because you should be able to show it for a stage and then show how the stages build. This is probably how you would best proceed with the all-different case, too. The idea would be to set it up as a cascade of multi-port networks.
 

MrAl

Joined Jun 17, 2014
11,342
Hi,

Yes i havent tried that yet. I just did a straight up nodal analysis, and with 10 nodes that's a lot of stuff :)
I'd post the solution but it wont do anybody any good.

I might get around to the transformation thing but in the past when i tried this i just got another set that is also hard to solve. That's was with a resistor 'grid' though where all resistors were connected forming squares of resistors that looked like a piece of graph paper. This particular set is probably easier since it is not as wide.
 

WBahn

Joined Mar 31, 2012
29,932
Yeah, if the resistors values are not highly structured then it is going to be a mess. I think a layer-by-layer cascade of multi-port networks is the best way to go in that case. You have a fixed topology for each layer (the very top and bottom might be different) and each layer has a limited number of resistors so the network transfer matrix can be fairly easily calculated and parameterized. Then it should be a simply matter of cascading them together to come up with an equivalent resistance of the whole thing.
 

Janis59

Joined Aug 21, 2017
1,827
If this is the school project, it is for progressions theory on math much more than for physics. There I am not the best guru. But if for practical use - only for 14x3 resistors.... just press the button SPICE-LT, draw the circuit there and measure with a screen tools what current is flowing and what voltage is over there. LT-Spice freeware is downloadable from Linear Technology homepage.
 

atferrari

Joined Jan 6, 2004
4,763
If this is the school project, it is for progressions theory on math much more than for physics. There I am not the best guru. But if for practical use - only for 14x3 resistors.... just press the button SPICE-LT, draw the circuit there and measure with a screen tools what current is flowing and what voltage is over there. LT-Spice freeware is downloadable from Linear Technology homepage.
The first time I ever heard of the famous cube, was in an old forum, where someone did as you say. It sounded almost like cheating...

That triggered the idea of solving it with delta-wye transformations which took me time but proved to be a pleasure to verify that I worked it right.

To confirm once more I breadboarded the cube and measured current. After that, just Georg Simon.
 
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