Resistor Cube

I

Inka

Cosmic Princess
Registered Senior Member
Hello - Does Anyone know a SIMPLE solution to the age old resistor cube problem??

For anyone who does not know what this is i have detailed it below:

Imagine a cube made of resistors. so that would be a total of 12 identical resistors. Work out the resistance between the front left bottom corner, and the top right back corner. A diagonal thing.

Each edge is a resistor of value R.
 
Inka said:
Hello - Does Anyone know a SIMPLE solution to the age old resistor cube problem??

For anyone who does not know what this is i have detailed it below:

Imagine a cube made of resistors. so that would be a total of 12 identical resistors. Work out the resistance between the front left bottom corner, and the top right back corner. A diagonal thing.

Each edge is a resistor of value R.
Click to expand...

Piece of cake. It requires using "Kirchhoff's Law". You must draw a schematic of your circuit connections showingthe flow of current through the parallel and series paths and then use:

1 - Parallel: Resistance = 1/R1 + 1/R2 + 1/R3 + 1/Rn

2 - Series: Resistance = R1 + R2 + R3 + Rn

And then Theveninize the cicuit by redrawing it for using the effective or equivelent resistors in each block.

Lunch is over so I can't give you an example calculation until this evening but I will.
 
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MacM said:
Oiece of cake. It requires using "Kirchhoff's Law". You must draw a schematic of your circuit connections showingthe flow of current through the parallel and series paths and then use:

1 - Parallel: Resistance = 1/R1 + 1/R2 + 1/R3 + 1/Rn

2 - Series: Resistance = R1 + R2 + R3 + Rn

And then Thevanize the cicuit by redrawing it for using the effective or equivelent resistors in each block.

Lunch is over so I can't give you an example calculation until this evening but I will.
Click to expand...
  1. the rules you quote are not Kirchhoff's laws (though they can be derived from Kirchhoff's laws).
  2. the first rule you write is incorrect. It should read: 1/R<sub>total</sub>=1/R<sub>1</sub> + ... + 1/R<sub>n</sub> for resistors in parallel. you forgot that the left-hand side of the equation must be inverted as well.
  3. The equations you give are not sufficient to solve the cube in any straightforward way. There are no sub graphs of the cube which are all parallel or all series, which is a requirement to be allowed to use the formulas you quote.
  4. If you want to solve the cube, you may use the fully general Kirchhoff's laws (not the laws about parallel and series resistors, which do not apply here), however there is a much easier argument to solve the cube based on symmetry. which is what makes this an interesting physics question, instead of just an exercise in solving a lot of coupled equations. anyone who wants to answer this question should use symmetry arguments.
  5. Note that if all 12 edges had different resistances, the symmetry argument would fail, and a system of coupled equations would have to be solved. MacM, I would be impressed to see you solve this system of equations.
 
lethe said:
  1. the rules you quote are not Kirchhoff's laws (though they can be derived from Kirchhoff's laws).
  2. the first rule you write is incorrect. It should read: 1/R<sub>total</sub>=1/R<sub>1</sub> + ... + 1/R<sub>n</sub> for resistors in parallel. you forgot that the left-hand side of the equation must be inverted as well.
  3. The equations you give are not sufficient to solve the cube in any straightforward way. There are no sub graphs of the cube which are all parallel or all series, which is a requirement to be allowed to use the formulas you quote.
  4. If you want to solve the cube, you may use the fully general Kirchhoff's laws (not the laws about parallel and series resistors, which do not apply here), however there is a much easier argument to solve the cube based on symmetry. which is what makes this an interesting physics question, instead of just an exercise in solving a lot of coupled equations. anyone who wants to answer this question should use symmetry arguments.
  5. Note that if all 12 edges had different resistances, the symmetry argument would fail, and a system of coupled equations would have to be solved. MacM, I would be impressed to see you solve this system of equations.
Click to expand...

Well, I have started. I have redrawn the circuit but it is not as simple as I first tought. It will have to be Theveninized and since it has been 40 years since I done this sort of thing it may take a while. But I will be working on it.

Just because you want to see me do it. :D
 
Lethe,

OK. It wasn't as bad as it looks. Since all resistors are the same I don't need to use Kirchhoff's rules. I simply Theveninize and get 3 groups of parallel resistors and assuming 100 ohms each reduces to:


O-XX 33.333 Ohms XX--O--XX 16.666 Ohms XX--O--XX 33.333 Ohms XX--O

or

83.333 Ohms.
 
MacM said:
Lethe,

OK. It wasn't as bad as it looks. Since all resistors are the same I don't need to use Kirchhoff's rules. I simply Theveninize and get 3 groups of parallel resistors and assuming 100 ohms each reduces to:


O-XX 33.333 Ohms XX--O--XX 16.666 Ohms XX--O--XX 33.333 Ohms XX--O

or

83.333 Ohms.
Click to expand...
that is, of course, the correct answer to the problem with large symmetry, where you do not need Kirchhoff's laws.

It is an interesting physics problem, and symmetry is a physicist's best friend.

however, if you look at my post again, you will see that what i was really hoping for from you was the general problem with no symmetry, where you do need Kirchoff's laws. That is, assume that all 12 resistors are different. Here is what I wrote:

Note that if all 12 edges had different resistances, the symmetry argument would fail, and a system of coupled equations would have to be solved. MacM, I would be impressed to see you solve this system of equations.
Click to expand...

This is an application of Kirchhoff's laws. You claim that the laws are a piece of cake, and I am dubious of your claims. That's why I challenged you to do it. So how about it? can you solve a set of coupled equations?
 
Lethe,

I'll have to be honest and give you two answers.

1 - I used to do it easily.

2 - I haven't done it in 40 years and I don't know how long it would take me to get my sea legs back but basically you sum current at a node or point and trace the currents throught the resistances calculating voltage drops.

The case of the cube would be quite a challenge at this late date. I frankly don't think I can afford the time to go through it just for sport. If I needed that answer then I would get it. I know for you it is probably easy but that is because you are current in your exercising the rules, current into a node equals current out of a node and voltage drops equal net zero, etc.

On the other hand I think it might be worth while just to prove to you that I can. So if you want to draw up your cube, lable the resistors, and mark their values, draw the external battery terminal points and lable the cube corners, I'll will do it.
 
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MacM said:
Lethe,

I'll have to be honest and give you two answers.

1 - I used to do it easily.

2 - I haven't done it in 40 years and I don't know how long it would take me to get my sea legs back but basically you sum current at a node or point and trace the currents throught the resistances calculating voltage drops.

The case of the cube would be quite a challenge at this late date.
Click to expand...
Yes, of course it would be a challenge. That's why I would be impressed if you could do it, and that's why I was surprised when you said "Piece of cake" before.

I frankly don't think I can afford the time to go through it just for sport. If I needed that answer then I would get it. I know for you it is probably easy but that is because you are current in your exercising the rules, current into a node equals current out of a node and voltage drops equal net zero, etc.

On the other hand I think it might be worth while just to prove to you that I can. So if you want to draw up your cube, lable the resistors, and mark their values, draw the external battery terminal points and lable the cube corners, I'll will do it.
Click to expand...

You know what they say, mac, "be careful what you ask for ..."

<img src=http://www.sit.wisc.edu/~jhannon/cube.jpeg>

calculate the resistance between points A and B. I have just finished the calculation myself
 
lethe said:
Yes, of course it would be a challenge. That's why I would be impressed if you could do it, and that's why I was surprised when you said "Piece of cake" before.

You know what they say, mac, "be careful what you ask for ..."

<img src=http://www.sit.wisc.edu/~jhannon/cube.jpeg>

calculate the resistance between points A and B. I have just finished the calculation myself
Click to expand...

Thanks. :D That is what I get for being cocky. HeHe. Back sometime. I had already drawn a ladder diagram. There are 12 resistor but I think it was about 20 "I" loops.

See if you cut me any slack in your choices of resistors. I'm assuming you didn't make it hard on yourself.
 
MacM said:
See if you cut me any slack in your choices of resistors. I'm assuming you didn't make it hard on yourself.
Click to expand...
I don't know what you mean, the choice of resistors is completely arbitrary
 
MacM said:
lethe,

Here is your circuit with current nodes and resistor paths.

I'll start working of the Kirchhoff equations.

Move cursor onto document and click on the enlarge square when it appears.


http://www.sciforums.com/attachment.php?attachmentid=3273&stc=1
Click to expand...

OK good, so now you have a somewhat planar representation of the cubic graph. I checked a couple of vertices of the graph, and it looks like your graph is indeed isomorphic to mine. This should make it a bit easier to write Kirchhoff's laws, so maybe now you're ready to get started?

By the way, if you want a true planar representation of the cube, there is a very simple form, and perhaps you will prefer it

<img src="http://www.sit.wisc.edu/~jhannon/planar_cube.jpeg">

This graph is also isomorphic to the cube, and it fits in a plane without any self-intersections. It probably won't make much difference, and you've obviously already put some work into your other graph, but for your information, there it is.
 
lethe,

That indeed is a much simpler layout. HeHe. Don't know why I didn't think of that, makes perfect sense. Actually you can lay out another way and still not have the cross overs but it isn't as good as this.
 
Solving 12 simultaneous equations manually really sucks!
Doing matrix row operations in Excel isn't much better.

I need to get me some maths software.
 
Pete said:
Solving 12 simultaneous equations manually really sucks!
Doing matrix row operations in Excel isn't much better.

I need to get me some maths software.
Click to expand...

I would have to agree. HeHe.

The answer to lethe's problem is 12.5 ohms.
 
MacM said:
The answer to lethe's problem is 12.5 ohms.
Click to expand...
Yes, that's what I got too. 12.515, to be exact.

Good job, I'm impressed Mac. I didn't think you could do it.
 
lethe said:
Yes, that's what I got too. 12.515, to be exact.

Good job, I'm impressed Mac. I didn't think you could do it.
Click to expand...

Thanks but don't make more of this than warranted. There is however, a differeance in being ignorant and being out of practice. In spite of the many lovely things some here have asserted I am, the facts of the case are that at one time I was fairly high on the hog (as the saying goes).

Indeed to receive the special assignment that I did we were tested extensively and the basis was that we had to be within the top 5% of brain power available. I further honored myself by graduating 2nd in my class by only 0.1% grade average.

Having said that however, it goes without arguement that that was 40 years ago and like anyother muscle that isn't used, the brain can get weak. :D

But I certainly remember the procedures to do many things that I would no longer claim that I can do (without extensive refreshers and review). But that really isn't the same as being uneducated or incapable.

This has however, reminded me of just how atrophied some muscles can get. :D

I'll be more cautious in future posts about claiming just how easy something might be.

Thanks.
 
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A shortcut of sorts (offered without proof)

If you use the principle of "virtual shorts" you can save a lot of work on this. Look for any two points that have equal voltage on them. (This can usually be done by simple inspection). Now SHORT these points out with a piece of wire. See how this simplifies the problem! (You can "short out" any two equal-voltage points on any network without altering the operation of said network)

Eric
 
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