Solving The "R" Cube

[bountyhunter]

The famous cube built out of resistors of value "R":

For those who have never done this problem, it's a really good one. Build a cube where each side of the cube is a resistor whose value is "R", and then calculate the resistance:

1) Across diagonal points of the cube.

2) Across opposite corners on any face

3) Across any resistor's ends on any face.

The values should be in terms of "R".

See attached PDF.

Obviously, the answer is on the net but it's a very good problem because it requires recognizing symmetry, spatial visualization, and using equipotential nodes to simplify circuits.

I'll publish the answers later.

 

[Ratchit]

The famous cube built out of resistors of value "R":

For those who have never done this problem, it's a really good one. Build a cube where each side of the cube is a resistor whose value is "R", and then calculate the resistance:

1) Across diagonal points of the cube.

2) Across opposite corners on any face

3) Across any resistor's ends on any face.

The values should be in terms of "R".

See attached PDF.

Obviously, the answer is on the net but it's a very good problem because it requires recognizing symmetry, spatial visualization, and using equipotential nodes to simplify circuits.

I'll publish the answers later.

First of all, I promise that I did not Google for the answer. I just solved for the resistance across the diagonal, because the other 2 pair of points are solved the same way. The cube can be drawn as a 2-dimensional mesh. The mesh looks like a 3-step ladder with a resistor on each side. Next I assumed all resistors were one ohm, and a voltage source of one volt was across the diagonal. Then I calculated all the node voltages. That was easier than expected because of the one ohm assumption, making the solution matrix sparse. After knowing all the node voltages, it was easy to calculate the total current leaving the upper diagonal node. As a check, it also matched all the current entering the lower diagonal node. Using the definition of resistance, R = V/I was calculated.

The answer came out to be 5*R/6.

Ratch

 

[bountyhunter]

That's one of the three. When I post the solutions it will have the breakdown of how to generate the equivalent circuits to get to the answer.

 

[MikeMl]

Just have to do this:

A cube has eight vertices. If I force 1A into vertex A, I can choose one of seven other vertices from which to sink the current. Because of symmetry, there are only three different possible voltages at vertex A, however.

Here is how I solved it seven times, connecting only one ground at a time: R=V(a)/1

The numbers are expressed in terms of R*Ohms

 

[bountyhunter]

Well, I was thinking that if somebody did it only using their brain it might be more enlightening.....

It is true that seeing the application of symmetry is required to solve all three problems.

 

[atferrari]

I did it long time ago using delta-wye conversions. More than knowledge, you need being organized, patient and willing to verify your last step prior moving to the next.

/EDIT Uploaded an .htm file but failed to add images, so I deleted all for the moment. EDIT/

/EDIT bis .htm file converted to .pdf, now uploaded. Flaming for the bad quality expected but not appreciated. Wondering why I did not use Corel Draw at that time, many many years ago...?

I know there is a mistake in a decimal figure somewhere but I lack the energy/interest to correct it and retype this properly. EDIT bis/

 

[bountyhunter]

I'll attach the PDF file which has the solutions and the equivalent circuits used to get them. The key to analyzing and simplifying any circuit is to see that any two points in a circuit which have equal voltage (equipotential nodes) under all conditions can be shorted together. Also: any components that connect two equipotential nodes will never have any current pass through them so they can be deleted.

 

[Ratchit]

I'll attach the PDF file which has the solutions and the equivalent circuits used to get them. The key to analyzing and simplifying any circuit is to see that any two points in a circuit which have equal voltage (equipotential nodes) under all conditions can be shorted together. Also: any components that connect two equipotential nodes will never have any current pass through them so they can be deleted.

That method might solve this particular problem, but the methods used by MikeML and I will work for dissimilar impedance values including coils and capacitors connected between the nodes.

Ratch

 

[bountyhunter]

It's not supposed to represent a problem you're likely to see in the world. The point of the exercise was to pose a classic problem that a young engineer should be able to solve with only his mind if he can visualize and see symmetry. I think it's a perfect measure as to whether graduating students can take the circuit analysis tools they were shown and apply them.

 

[JoeJester]

How many employers use tests for their first round of discrimation between the applicants?

 

[MrAl]

It's not supposed to represent a problem you're likely to see in the world. The point of the exercise was to pose a classic problem that a young engineer should be able to solve with only his mind if he can visualize and see symmetry. I think it's a perfect measure as to whether graduating students can take the circuit analysis tools they were shown and apply them.

Hi there,

A real world problem does stem from this simple construction however, but first let me digress...

How about next a dodecahedron?

To get to the real world problem, take 7 more such cubes, connect them all so they form a larger cube, but where edges parallel just use one resistor not the two in parallel. We can then try to find resistances between any two points, which includes the inner points that came from the extra cubes.

To get one more step closer, stack 7 more of THOSE cubes into a larger cube. Might want to calculate resistances again, might not want to.

To go the distance, repeat that procedure an infinite number of times while allowing the resistance of each R to decrease toward zero.
In the limit, we end up with a solid block of conductive material. Find the resistance between any two points on the surface of the block.
In reality we might let the length of each resistance go toward zero where the material is of known resistivity.

 

[bountyhunter]

How many employers use tests for their first round of discrimation between the applicants?

Everyone I ever interviewed did.

True story: I remember one interview where a guy asked me a complicated problem that had a long math derivation and I did it.

Next guy asked me the exact same question..... and I just wrote the answer down. His jaw dropped and I waited a few seconds before I told him:

"The last guy asked the same question."

 

[Mikebits]

A common interview question I recall engineers asking is to ask the applicant to design a divide by 3 with 50% duty cycle using FF and gates.