Without getting into calculations, a symetrical pattern of pathways is formed, so all indirect currents will cancel out if all resistors are 10ohm..
At A there are 3 possible pathways comprised of 3 10ohm resistors.. At the first vertice in each initial path, there are 2 possible pathways comprised 2 10ohm resistors, and at the next vertice reached there is only one pathway to B comprised of one 10ohm resistor..
So perhaps I'm jumping to a conclusion since I haven't done any math here, but don't the parallel and series configurations of the pathways cancel out making the resistance from A to B 10ohms?

 

Do you realize that the exact answer has been posted? And that your answer isn't it?

 

Wasn't sure if it had been confirmed.. I googled it and found your solution/answer, so I was wrong.. Usually brain teasers like this work out to the simplest solutions
..

 

Thing is, this is the math area, some math required.

 

This is a bit like the Medieval philosophers debating the number of teeth in the horse's mouth. Aside from the mathematical solution, has anyone ever soldered twelve 10 ohm resistors together and examined the cube as a physical device? It might be interesting.

 

This is a bit like the Medieval philosophers debating the number of teeth in the horse's mouth. Aside from the mathematical solution, has anyone ever soldered twelve 10 ohm resistors together and examined the cube as a physical device? It might be interesting.

They sure have: http://www.physics.ucsb.edu/~lecturedemonstrations/digital photos/resistor cube, small.jpg

Ok, so they are 100Ohm resistors, but hey!

Dave

 

Well, putting together a model and then measuring it is certainly a practical way of solving the problem.

hgmjr

 

With a +/- 5% tolerance. Some of the guys in my class had done it. I'll probably go through the steps I originally used before I drop this thread, just to see if the delta transformations would handle the off values I'd suggested.

Wonder if the tesseract would offer anything new puzzlewise?

 

Bill,

You keep mentioning the tesseract, so I did it. I don't think it adds too much puzzlewise, but it kept me busy for about 30mins.

Basically, if you look at the image in my excel sheet (or pdf), you will see that I've color coded the vertices red, blue, and green. Each node of the same color has the same voltage, so you can connect them together, and...
A -> Red = 4 resistors in parallel.
Red -> Green = 12 resistors in parallel.
Green -> Blue = 12 resistors in parallel.
Blue -> B = 4 resistors in parallel.
This gives a total resistance of (2/3)R.

And the excel sheet solves for the generic case.

 

Killjoy. As a parent that's my job.

 

Bill,

You keep mentioning the tesseract, so I did it. I don't think it adds too much puzzlewise, but it kept me busy for about 30mins.

Basically, if you look at the image in my excel sheet (or pdf), you will see that I've color coded the vertices red, blue, and green. Each node of the same color has the same voltage, so you can connect them together, and...
A -> Red = 4 resistors in parallel.
Red -> Green = 12 resistors in parallel.
Green -> Blue = 12 resistors in parallel.
Blue -> B = 4 resistors in parallel.
This gives a total resistance of (2/3)R.

And the excel sheet solves for the generic case.

Nice one Caveman

Dave

 

To the Ineffable All,

http://mathforum.org/library/drmath/view/65234.html

Ratch

 

To the Ineffable All,

http://mathforum.org/library/drmath/view/65234.html

Ratch

These are the two methods proposed by both recca02 and Caveman.

Also, without hijacking the thread, can I ask why "the ineffable all"?

Dave

 

Dave,

Also, without hijacking the thread, can I ask why "the ineffable all"?

My reply was to all and everyone, but I don't know who they are. The word "ineffable" emphasizes that fact. Ratch

 

the way i solved it was i added my own extra wire between A and B and threw a voltage source there, which i chose to be 10 volts to make calculations easy. then i simply did mesh analysis on all 7 meshes; the 7th being an arbitrary direct path in the original cube between A and B. Then i have 7 equations and 7 unknown currents; the 7th of which is what would give me the current through the entire cube. a little linear algebra and ohms law got me the same answer you guys get. it might be a little brute force, but it only took me a little less than 10 minutes (and matlab).

so mesh analysis can work if you modify it a bit. i'm sure you could use the same method for if you want to have all different resistor values, but i'll save that problem for later.

 

Bill,

You keep mentioning the tesseract, so I did it. I don't think it adds too much puzzlewise, but it kept me busy for about 30mins.

Basically, if you look at the image in my excel sheet (or pdf), you will see that I've color coded the vertices red, blue, and green. Each node of the same color has the same voltage, so you can connect them together, and...
A -> Red = 4 resistors in parallel.
Red -> Green = 12 resistors in parallel.
Green -> Blue = 12 resistors in parallel.
Blue -> B = 4 resistors in parallel.
This gives a total resistance of (2/3)R.

For the 5 dimensional hypercube, the corner to corner resistance with all equal resistors is (8/15)R

For the 6 dimensional hypercube, it's (13/30)R

It's interesting to display the numbers in decimal format for the 3rd through 6th dimensional hypercube:

.83333333333333
.66666666666667
.53333333333333
.43333333333333

Can you guess what the next would be, without calculating it?

Is there a discernible pattern?

I haven't calculated it myself, so I would be guessing too.

 

We'll it seems that Dave is true..there are several solutions that would be appropriate in solving this problem..One of which is solving it using power dissipated through each resistor using:

Ptotal = [(Itotal)^2] * Rtotal

where:

Ptotal = Total power dissipated in the circuit
Itotal = Total current that enters or exits at node A or B.
Rtotal = Equivalent Resistance of the circuit

I'll work the worksheet out and then post it after..the answer is ultimately [5/6] * R.

 

This of course, is a classic, classic electronics puzzle. Without going into great detail...here's a GREATLY simplifying trick.....called the principal of "virtual shorts." Once you've done this, you can easily figure out the rest:

By inspection, determine which points on the cube would have an equal voltage on them. Connect a wire between those points. From there, the problem becomes a simple series/parallel problem!

eric

 

They sure have: http://www.physics.ucsb.edu/~lecturedemonstrations/digital photos/resistor cube, small.jpg

Ok, so they are 100Ohm resistors, but hey!

Dave

I've had one of these dangling over my desk for 39 years.

 

Hi

This is a problem I always wanted to solve comprehensively.

Please find the attachment.

Regards
Deon