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# The Resistor Cube Problem - RF Cafe Forums

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Below are all of the forum threads, including all the responses to the original posts.

 Post subject: The Resistor Cube Problem Posted: Sun Jun 13, 2010 1:32 pm

Joined: Sun Aug 03, 2003 2:02 pm
Posts: 308
Location: Erie, PA
Greetings:

Most of us have run into the Resistor Cube problem where it is required to determine the equivalent resistance between opposing corners. In this month's edition of "Kirt's Cogitations," I show the traditional solution, as well as my own solution which I believe is easier and more intuitive. A circuit simulator proof of the solution is also presented.

This often shows up in job interviews, so it's good to know.

Anybody have a better solution?

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- Kirt Blattenberger
RF Cafe Progenitor & Webmaster

 Post subject: Re: The Resistor Cube Problem Posted: Wed Jun 16, 2010 9:36 am
 Lieutenant

Joined: Wed Jun 16, 2010 9:12 am
Posts: 1
Hi Kirt,
Good point to revisit the old resistor cube network. Your appoach shows that there are more ways to tackle a problem. However, I think that your method is not essentially different than the classic one. The classic method uses the equal voltage at several nodes, your method uses the equal currents in several branches. Both methods are right, because the network is symmetrical.

This reminds me of the very elegant solution to the infinite mesh network, which is the ultimate symmetrical network. The problem is to find the resistance between two adjacent nodes, with all resistors equal.

(by the way, how do you insert a picture?)

I'll be glad to show the solution in a next post

Hugo Kroeze
The Netherlands.

PS.
I recently visited the ISMRM 2010 conference in Stockholm. In MRI there is a lot of RF work done, so the conference had a large educational session on this subject. Several lectures mentioned the RFcafe as a good source of RF information.

 Post subject: Re: The Resistor Cube Problem Posted: Mon Jun 28, 2010 4:35 pm
 Lieutenant

Joined: Wed Jun 02, 2010 2:56 pm
Posts: 2
Location: Arizona
Hello Kirt:
When teaching Intro. E&M, I often assign this cube problem. My assignment guides the students to write formulas like "I=(Vn-Vm)/R", then to write KCL at 6 nodes, then to use a program called "maxima" to invert the resulting 6-by-6 matrix to find the node voltages, then find the total current drawn from the source to find Req. I think it makes the students become aware of how circuit theory works.

Maxima was developed years ago with the name "MACSYMA" and has most of the functionality as Mathematica, but it is available as a free download from www.sourceforge.net.

 Post subject: Re: The Resistor Cube Problem Posted: Tue Jun 29, 2010 7:55 am

Joined: Sun Aug 03, 2003 2:02 pm
Posts: 308
Location: Erie, PA
hkroeze wrote:
Hi Kirt,

(by the way, how do you insert a picture?)

Hugo Kroeze
The Netherlands.

PS.
I recently visited the ISMRM 2010 conference in Stockholm. In MRI there is a lot of RF work done, so the conference had a large educational session on this subject. Several lectures mentioned the RFcafe as a good source of RF information.

Greetings Hugo:

Sorry for the delayed response. A couple people have sent me e-mails with alternate solutions. One of them used delta/star transformations to do it. That one has been added to the Resistor Cube page at the bottom.

Thanks for letting me know about RF Cafe being mentioned in the lectures. I'll take all the promotion I can get!

Thanks for writing.

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- Kirt Blattenberger
RF Cafe Progenitor & Webmaster

 Post subject: Re: The Resistor Cube Problem Posted: Tue Jun 29, 2010 8:02 am

Joined: Sun Aug 03, 2003 2:02 pm
Posts: 308
Location: Erie, PA
johnlawvere wrote:
Hello Kirt:
When teaching Intro. E&M, I often assign this cube problem.
Maxima was developed years ago with the name "MACSYMA" and has most of the functionality as Mathematica, but it is available as a free download from http://www.sourceforge.net.

Greetings johnlawvere:

Yeah, trying to solve a 6x6 matrix without the help of a computer can be daunting. That's the purpose of assigning equal resistances so that the poor sap who sees it during an interview has an opportunity to demonstrate knowledge of fundamental principles.

I posted a link to MACSYMA a while back, but never did download and try it. There is a lot of high quality open source software that replicates most or all of the functionality of the commercial stuff. OpenOffice.org is a good example that has a version of Excel (called Calc) that will import excel files, including macros.

Thanks for writing.

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- Kirt Blattenberger
RF Cafe Progenitor & Webmaster

 Post subject: Re: The Resistor Cube Problem Posted: Sat Dec 18, 2010 4:55 pm
 Captain

Joined: Sat Dec 18, 2010 8:56 am
Posts: 5
Location: Long Island, NY
I have a question concerning this problem. I teach amateur radio license classes here on Long Island and lately I've been doing some more advanced stuff with some of the already licensed hams.

Regarding this cube, which I've seen before, when one does the math with the "equivalent" circuits described by most websites, i.e., 3 parallel resistors into 6 parallel resistors and those into another 3 parallel resistors, as long as all the resistors are the same value all the math works out well. Simulations also work as expected.

You can also work the cube as 3 parallel branches of a series resistor feeding 2 parallel resistors and those into another series resistor. Again, all the math works out fine and so do simulations. However...

If just one of the resistors is changed, say doubling the value, then an imbalance occurs and things get weird.

If the above mentioned "equivalent" circuits are used the mathematically calculated answer does not jibe with simulations of the cube. If you simulate the equivalent circuit it does match the math (the simulation that is).

Running a cube simulation you notice this immediately. All three branches have different currents running through them whereas in the equivalent circuits 2 of the branches have the same current and the third branch with the changed resistor is different.

Can anyone help enlighten me on what's going on? I'm confused enough normally, this problem is making me more so.

Thanks.

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Kevin AB2ZI

 Post subject: Re: The Resistor Cube Problem Posted: Wed Dec 22, 2010 12:06 am

Joined: Sun Aug 03, 2003 2:02 pm
Posts: 308
Location: Erie, PA
Greetings Kevin:

I think you've answered your own question. Altering any resistor value so that it does not equal all the others disturbs the symmetry of the cube, so currents in the branch circuits at each node are not equal.

You might be tempted to think that by doubling one resistor value it will cut the current in that branch in half and therefore the voltage across it will still be what it was when at the original value, but the node is not an ideal voltage (or current) source that maintains the same value regardless of the load; its value is dependent on the resistances in all the other branches.

Symmetry alone allows the assumptions that permit the simplified solutions like the series/parallel equivalent circuits, or my method that depends on Kirchhoff's current law.

Does that answer your question, or did I miss the point?

_________________
- Kirt Blattenberger
RF Cafe Progenitor & Webmaster

 Post subject: Re: The Resistor Cube Problem Posted: Wed Dec 22, 2010 9:14 am
 Captain

Joined: Tue Dec 26, 2006 5:39 pm
Posts: 10
It's Gordian's knot.

I solved the problem by BUILDING it and measuring it with an ohm meter.

And I got a different answer.

 Post subject: Re: The Resistor Cube Problem Posted: Wed Dec 22, 2010 10:12 am
 Captain

Joined: Sat Dec 18, 2010 8:56 am
Posts: 5
Location: Long Island, NY
 Kirt Blattenberger wrote: Greetings Kevin:I think you've answered your own question. Altering any resistor value so that it does not equal all the others disturbs the symmetry of the cube, so currents in the branch circuits at each node are not equal. You might be tempted to think that by doubling one resistor value it will cut the current in that branch in half and therefore the voltage across it will still be what it was when at the original value, but the node is not an ideal voltage (or current) source that maintains the same value regardless of the load; its value is dependent on the resistances in all the other branches.Symmetry alone allows the assumptions that permit the simplified solutions like the series/parallel equivalent circuits, or my method that depends on Kirchhoff's current law.Does that answer your question, or did I miss the point? I'm still a bit confused. Mostly I think it's the "equivalent circuits" that mess up the math. Here is an example using 100 ohm resistors and a 5 v supply. When the cube is modeled you get 82.7 ohms with 60 ma of total current. 20 ma through each of the 3 branches and 10 ma distributing nicely among the middle 6 resistors. When you draw it out as an equivalent circuit of 3 parallel feeding 6 parallel outputting to 3 parallel it looks like this:The circuit on the left behaves like a normal series-parallel circuit and the math and simulations come up with the same answer as the cube simulation. However, if one of the middle resistors is changed to 200 ohms, as is the case on the right, then the math and a simulation of this equivalent circuit work as expected, yielding total resistance of 84.85 ohms with 58.93 ma of current. 19.64 ma in each input and output set of 3 resistors, 10.71 ma in the five 100 ohm parallel resistors and 5.36 ma through the 200 ohm resistor.But... When you model the cube and change one of these middle resistors this is what you see:Total resistance is now 85.3 ohms with 58.62 ma total current. Each of the 3 branch input/output resistors (for lack of a better description) has a different total current. One branch has 19.83 ma, the second 20.69 ma and the last, the one with the 200 ohm resistor, has 18.1 ma.In the middle resistors, the first branch divides to 10.34 and 9.48 ma respectively, the second branch divides up to 9.48 and 11.21 ma, and the third to 11.21 and 6.9 ma (the 6.9 ma flowing through the 200 ohm resistor).The circuit follows Kirchoff's law without a problem. The equivalent circuits are just not equivalent unless the resistors are all the same value.My head hurts!  Kevin _________________ Kevin AB2ZI

Posted  11/12/2012
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