Physics resistance problem 3

My son came home from school today with a problem that his physics teacher gave to the class. The teacher is fairly cocky that no-one (including parents, internet research, other teachers and aquaintances of students) will find the answer!

I would appreciate people's views on the answer!

Here is the question: If you have a cube with an equal resistor (say 1 ohm) on each perimeter edge of the cube (i.e. 12) and the positive terminal was attached to the top left front corner (sorry hard to explain without a diagram) and the negative to the bottom right back corner, what would the equivalent resistance be leaving the circuit?

I got part of the way thru it BUT not having used physics since high school and the loss of brain matter as a result of aging, I soon became upstuck with the number of in series and parallel circuits etc. X-)

Thanks in anticipation of putting him back in his place....

Rhodian

 

[gwc43]

Electripete - Yep, count me in on the 5/6ths deal. I noticed that this has been sucessfully answered .. I think the kid's dad should go visit the teacher and make him pay up. I've got one that involves a triangle in the middle of a circle that we could send to the teacher. I bet some of y'all remember the one I'm talking about. By the way, (for pete) I'm jburn's co-worker y'all can see that maybe I've found a home - I like this site!!

 

[nbucska]

Hi Rhodian:

Have your son ask the guy my two puzzles. (fuse and xformer) <nbucska@pcperipherals.com>

 

[jbartos]

Comment:
1. Practical solution is by far the fastest. For better measurement, 12 resistors can be purchased at 1kiloohm and 1/25 Watt or so ratings and wire them around a paper cube, e.g. 3&quot;x3&quot;x3&quot;. Then, a regular ohmeter can measure three different or possible resistances, namely:
1.1 Along one edge of cube, it means the shortest distance. There are 12 of them, one across of each resistor of the cube.
1.2 Across the cube side diagonal. There are six cube sides and each side has two diagonals. There are 2 x 6 = 12 readings possible. All should be approximately the same.
1.3 Cube body diagonal resistance measurement (I believe that this one is needed and most difficult to calculate.). There are four cube body diagonals. The resistance reading should be the same.
2. The theoretical approach by calculations. I am still working on it. However, what is very important is to collapse the cube by star to delta transfigurations appropriately.

 

[electricpete]

jbartos - I would like to invite you one more time to take a look at the solution I posted at:

http://www.reliability-magazine.com/pub/cube.jpg

I believe it is a fairly straightforward solution.

For the benefit of some of the other readers who may not have understood the approach (although many came up with same approach independently), I'd like to further explain the one &quot;tricky&quot; part - which was the fact that when we identified nodes of equal potential, we &quot;shorted&quot; them together and treated them like one node. Why is this permissible? Three explanations:
1 - Intuitive explanation - since there is no voltage difference between the nodes there will be no current flow between the nodes and the addition of the shorting jumpers has no effect.
2 - Mathematical explanation - The mathematical properites of a node are: A - It has a voltage and B -(KCL)- The sum of the currents entering a node must equal the sum of the currents leaving a node. If we combine like potential nodes, we have not violated item A. Additionally, since the new combined node has all the inputs of the old nodes and all the outputs of the old node, KCL will hold... i.e. sum of currents entering and leaving the new node will be the same.
3 - Analogy - This is the same technique used for single-phase analysis of a three phase balanced system. Even a neutral may be ungrounded, it has the same potential as all the other neutrals in the system (assuming balanced), and we can analyse the system by drawing imaginary short circuit between neutrals and evaluating the single-phase loope circuit which includes that imaginary neutral short circuit.

 

[jbartos]

Suggestion: Very impressive but missing measurement results and the proof.

 

[nbucska]

J. Bartos:
So does the 1/36 ohm solution. <nbucska@pcperipherals.com>

 

[peterb]

For my ten cents worth, I think that the solution offered by electricpete and nbucsa is simple, elegant and a good example of out-of-the-box thinking.
I too started out with the star-delta transformation approach and got bogged down in the tedium of the arithmetic - I really wasn't prepared to invest the time required to run this one down.
The question of experimental proof doesn't really arise - this was meant to be a mental exercise and the solution is readily apparent from the wealth of talent exhibited above. Of course, anyone who wishes to spend the time & effort on the measurement exercise would probably be applauded by the group (I'm not going to volunteer).

 

[redtrumpet]

This circuit has 8 nodes and 12 branches - proof of the 5/6 result should be obtainable by doing a Kirchhoff loop or nodal analysis and solving the simultaneous equations. In my mind, though, the solution has already been proven by electricpete and others.

 

[jbartos]

I beg your pardon, the 1/36 solution for the cube body diagonal resistance is not correct (typos in many formulae).
I provide correct derived and proved results by measurements, namely:
1. Body diagonal is 5/6 = 0.83333 as posted above. It has been measured on my cube consisting of twelve 1 kohm 1/25 Watt resistors with results fluctuating around 833 ohms. When this is divided by 1000 it gives 0.8333 result coinciding with the above posted value. However, I used the William D. Stevenson, Jr., &quot;Elements of Power System Analysis,&quot; Third Edition, McGraw-Hill, Inc., 1975, Section 7 Network Equation and Solutions. Essentially, transfigurations of stars to deltas, and four-corner stars to meshes will lead to the above results.
2. Side diagonal is 3/4 = 0.75. It is verified by measurements, namely 750 ohms on my cube. It is also derived over the mentioned reference concept.
3. Cube edge resistance is 7/12 = 0.583333. It is verified by measurements, namely ~590 ohms on my cube. It is also arrived via the mentioned reference concept.