hi rhodian,
  Ok the challenge has been accepted. With 5 crazy engineers in our group at work it will be attacked tommorrow and our best guess posted.
  Knowing some of the smarties on this site I will be the eighth posting tommorrow.
  I am sure my physics tutor thru this same one at us at Uni but that was a long long long time ago.
  
Regards Don

I took a quick stab at looking for simple way by trying to exploit the symmetry.

Put the one corner of block at (0,0,0) and other corner at (1,1,1). Call these the terminals.  Energize one terminal at at +V and the other at -V.

Look for the symmetric points in the middle where voltage will be zero. There are 12 edges of the cube. 6 of these edges touch one terminal and 6 do not touch either terminal.  I believe there are 6 zero voltage points which lie in the center of each of the 6 edges which do not touch either terminal.  For example (1,0.5,0), (1,1,0.5),(0.5,0,1) (0,0.5,1),(0,1,.5),(.5,1,0).


Since these points are at same voltage we should be able to short them together without changing the problem (call them ground voltage = 0).  So we now have separated the problem into to equal halves.  Each half is connected between a terminal and ground.  Each half will include Three parallel branches.  Each of those parallel branches is one full edge in series with the parallel combination of two half edges.

Parallel combination of 2 half edges is 0.25 ohms.

Put in series with a full edge to give a resistance of 1.25 per parallel branch.

3 parallel branches in parallel give resistance of (5/4)/3 = 5/12.

Each of the two sets of 3 parallel branches is attached to an electrode.  The series combination of these two is 2* 5/12 = 10/12 = 5/6.

That's just a stab.  No guarantees.  

In visualizing the geometric aspect it might be better to orient the cube with one terminal at (x,y,z)=(0,0,+Z0) and one terminal at (0,0,-Z0).

The point where the cube crosses the x-y plane is the point of zero voltage.  The three edges attached to the top terminal will be entirely above the x-y plane.  The three edges attached to bottom terminal will be entirely below the x-y planes.  The 6 remaining edges will penetrate the x-y plane at their midpoints (these are the points where V=0).  

There is still one degree of freedom available to rotate the cube about it's diagonal axis but I don't think that that changes the point of intersection with x-y plane. (if it did then it wouldnt be at 1/2 way point thru edge and my analysis would be incorrect).

A slightly different solution:
Let's have a 3 dimensional coordinate system,x-y-z. The plane of x-y is horizontal

The 000 point is far/left/down. Three sides of the cube
are on each axis, from 0 to +1.

Let x=bit0  y=bit1 and z=bit2. In this case the value of the
corners clockwise  from the corner on the Z axis ( x=y=0 )are:

On the far/lower side: 0, 2, 3, 1 and
on the near/upper side 4, 6, 7, 5

Let's use 7 and 0 as inputs : 1,2, and 4 are one unit distance from 0 while 3,5 and 6 from the 7 i.e. they have the same potential so they can be shorted. Call the first node 124 and the second one 356

We have three parallel (#) combinations in series : R1#R2#R4
and R3#R5#R6 and the six remaining ones between nodes 124 and 356.  This is 1/3 + 1/3 + 1/6 = 5/6
 

<nbucska@pcperipherals.com>

nbucska - It looks like we both took the approach of finding points of common potential and "shorting" them together to simplify the analysis.

I chose the set of 6 points (not corners) located midway between the terminals.  You chose two sets of 3 corners each... the three corners located 1 edge away from each terminal.

I like your method better.  A little more elegant. Doesn't require splitting a resistor in half.  And probably easier to explain.

I'm glad we got the same answer.

Thanks E Pete. Actually the nodes shorted are on two
parallel planes between the 3-6-3 R-s. You used one
plane halfway. This just shows there are many ways to
skin a cat...

On a drawing both would be obvious.

<nbucska@pcperipherals.com>

#1 - You're right, a picture is worth a thousand words. Here's my version of nbucska's solution:

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

#2 - Too much time on my hands?  Not really.  Back to work

Two more comments.

#1 - There's a typo in the last line of the proof.  The intent should be obvious.

#2 - The location that I put the file does not belong to me... no guarantees that it will stay there for long.  I suggest that you download it, Rhodian.

I've got a brainteaser for your physics teacher.

Start with two capacitors of identical capacitance C.
Initial condition is that one is charged to V0 and the other has no charge.  

You can calculate that the charged cap has initial energy E0=(1/2)C*V0^2 and initial charge Q0=C*V0. The uncharged cap has no charge or initial energy.

Now short the two capacitors together at both terminals. By conservation of charge, the charge Q=C*V0 should redistribute among the two caps so that each one carries a new charge of Q = Q0/2=C*V0/2.  The voltage in each capacitor will be V = Q/C = (C*V0/2)/C = V0/2.

The energy in each capacitor is (1/2)C*V^2 = 1/2*C*(V0/2)^2
= 1/2*C*(V0^2)/4

The total energy among both capacitors is twice the above or 1/2*C*(V0^2)/2.  But this is E0/2.  i.e. the final energy of the system is 1/2 of the initial energy.  What happened to conservation of energy?

If he figures out that one, then tell him that:
#1 - I have routinely placed my hand for several seconds on directly on an aluminum conductor (no insulation) carrying approx 35,000 amps.  I had no special electrical protective equipment.  I was standing on grounded structure with normal shoes.  
#2 - The above action was perfectly safe.
#3 - I am not crazy.
#4 - Don't try to recreate this at home until you find out the details.  No-one should play with electricity unless they know what they're doing.

You must be very brave E.Pete -- or an EE.

Want some puzzles?

1.) Make two electrically distinct  transformers. 100% of any active part of one transformer -- i.e. coil or core -- MUST be an active part of the other one.  

2.) You have a blown 9 A fuse. Temporarily you want to fix it by using a piece of wire but the only wire you have blows at 6 A.  12 A would be too much.

<nbucska@pcperipherals.com>

electricpete - I will guess your 35,000 amp bus was on an ungrounded low-voltage DC system, presumably used for electrolysis of water or some such process.

RED T:
retrack "ungrounded "

<nbucska@pcperipherals.com>

Suggestion: Simple transfigurations reveal the resistance across the cube side diagonal points to be 1/36 ohm. The computation is much more involved to find out the resistance across the cube body diagonal.

jbartos - I believe you are solving the wrong problem.  There are 12 resistors, one along each edge of the cube.

nbucska - brave?  no.  crazy?  maybe.  ee? guilty as charged.

The wire puzzle sounds pretty easy if I make some assumptions about the thermal model.

Assume I've got magnet wire where the electrical insulation provides almost no thermal insulation.  To make a 3 amp wire-fuse, take a 20' section of wire and double it over to make two series wires of 10' each. Wrap them tightly together so their combined heat adds together.

Put the 3 amp wire-fuse in parallel with a 10' length of wire (6 amp fuse).  They will share load in proportion to their ratings since the 3 amp wire-fuse has twice the resistance of the 6 amp wire-fuse.

That's probably more of a valid model for a slow-blow fuse where heat transfer to ambient is a factor (I think). Probably not too good a model for fast-acting fuse where thermal capacity plays a bigger role (I think).

You've got me on the transformer one. I don't think it's possible.

redtrumpet - The associated power system is a grounded ac system.  

I think my calculations on the fuse were flawed.  One wire at 3 amps generates 1/4 the heat of one wire at 6 amps. To provide equivalent heat with 3 amps I need to increase the resistance of the wire by 4.  I should wrap FOUR 10' lengths of wire together to form the 3-amp wire fuse. Then put it in parallel with one 10' length of wire (6amp wire fuse).  I think.

The fuse solution I provided is not perfect by any means.  There may be a much higher temperature on the 3 amp fuse which makes the resistance increase more than for the 6 amp fuse. Even without temperature effects, it's shaky to rely on the resistance of wires for load sharing. But I can't think of any other way.... except maybe go sell the darned wire and use the money to buy a $1 fuse.  Unless of course you're on a desert Island.   One of those Islands with 110v ac service of course

nbucska - you have to clarify what you mean by electrically distinct.  My normal assumption would be that the inputs/outputs of one transformer should be electrically isolated from the other.  That's pretty tough if they share the same copper.  Do the two transformers have to carry the same frequency?

Suggestion/comment to the original posting marked ///\\\:
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 am calculating/simplifying the cube with 12 resistors of 1 ohm value in each cube edge. So far, I have gotten to resistance between the top left corner and right rear top corner, and I got solution 1/36 ohm. This could possibly be modelled by 12 resistors of 1 Ohm value and measured by ohmeter between the front top left corner and bottom right back corner without any tedious calculation. However, the calculations get pretty interesting. One may use Stevenson W. D. "Elements of Power System Analysis," 3rd Edition, McGraw-Hill, Inc., 1975, Chapter 7 "Network Equations and Solutions."

jbartos - did you look at the link that I posted?  Can you point out an error.  It looks fairly simple to me when you look at the figure. (But I'm not ruling out that I may be wrong.)

jbartos - there is a reasonability test which contradicts your solution (from my perspective).

If I take 12 1-ohm resistors and arrange them in various combinations, then the lowest possible resistance I could obtain is 1/12, corresponding to to the parallel combination of all 12.  Any other combination would include some series combinations and would give a higher resistance.  1/36 ohm does not seem possible.

Yet another way is solve the cube problem is to let the computer do the thinking.  I couldnt remember how to put it into matrix form, but I was able to do it using Maple.  Here is a cut and paste of the entire program.  I suspect that it can be transported to other math programs with little modification.  Comment lines begin with "#". Maple output is listed in bold.

> restart;

> # 8 nodes identified by three binary coordinates. Apply V_applied at node 111 and 0 voltage at node 000. Each of the 6 remaining nodes has a Kirchoffs current law equation of the form (Vneighbor1+Vneighbor2+Vneighbor3)-3Vnode=0 (assumes resistance of 1 in each branch)

> eq1:=V000=0:
> eq2:=V111=V_applied:

> # kcl at node 001 is as follows:
> eq3:=V101+V011+V000 - 3*V001=0:

> # kcl at node 010 is as follows:
> eq4:=V011+V110+V000 - 3*V010=0:

> # kcl at node 011 is as follows:
> eq5:=V010+V111+V001 - 3*V011=0:

> # kcl at node 100 is as follows:
> eq6:=V101+V110+V000 - 3*V100=0:

> # kcl at node 101 is as follows:
> eq7:=V100+V111+V001 - 3*V101=0:

> # kcl at node 110 is as follows:
> eq8:=V111+V010+V100 - 3*V110=0:

> Vsolution:=solve({eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8},{V000,V001,V010,V011,V100,V101,V110,V111});

Vsolution := {V100 = 2/5 V_applied, V011 = 3/5 V_applied,

    V101 = 3/5 V_applied, V010 = 2/5 V_applied, V001 = 2/5 V_applied,

    V110 = 3/5 V_applied, V000 = 0, V111 = V_applied}

> I111:=3*V111-V110-V101-V011:
> R_effective:=V_applied/I111;

                                      V_applied
              R_effective := ---------------------------
                             3 V111 - V110 - V101 - V011

> # Substitude Voltage solution into above expression for R_effective
> R_effective:=subs(Vsolution,R_effective);

                          R_effective := 5/6

>

I fotgot to mention the reason for using binary indexing of the 6 cube corners (nbucska's idea). It makes it effortless to identify the 3 meighbors of each cube (x,y,z) without a picture. Those neighbors are (x',y,z),(x,y',z),(x,y,z').

Let me try that last message again with some corrections (I must have been typing in my sleep):

I forgot to mention the reason for using binary indexing of the 8 cube corners (nbucska's idea). It makes it effortless to identify the 3 neighbors of each corner (x,y,z) without a picture. Those neighbors are (x',y,z),(x,y',z),(x,y,z').

Hi E.Pete:

The two transformers are independent - they may have different frequencies, -- ideally -- no coupling,
may have different xfer ratios etc.

If you want solution, E-mail.

<nbucska@pcperipherals.com>

Hi E.Pete:
The fuse solution is close, but you don't need to twist
the long one to preserve the heat.

If there is a short, the 2L wire has 3 A the 1L 6 A.
The later blows -- now all 9A blows the 2L wire instantly.

You may want to correct for the increased resistance around
fusing temperature.

<nbucska@pcperipherals.com>

0 Ohms!  There is NO resistance LEAVING the circuit.  Everybody else is calculating the equivalent series resistance.  This sounds like a perspective riddle.

Hi E.Pete:

Tried your <...technologist...> from your WEB -came back.
Can you pse send me your new addr ?

<nbucska@pcperipherals.com>

RE: E.Pete's cap puzzle:

Whem you connect the new cap, I is unlimited - halp of the energy is dissipated as I^2*R heat and EM radiation and perhaps spark.

<nbucska@pcperipherals.com>

The simplest (I think)

Let the longest diameter of the cube 1 ". Stand it on
one input corner so the other input corner is on the
same vertical line.

You have 3 corners at the height of 1/3" and 3 at 2/3"

The corners at the same hight are at the same potetial -
-- due to symetry -- so you may short them. Do it.

From 0 to 1/3 you have three, from 1/3 to 2/3 six, from
2/3 to 1" three R-s.  1/3+1/6+1/3=5/6  -----   Q.E.D.

<nbucska@pcperipherals.com>

Rhodian,

What is the answer?

melone - I'm of the firm opinion that any competent electrical engineer who takes a look at
http://www.reliability-magazine.com/pub/cube.jpg
or who studies nbucska's most recent post would quickly come to the same conclusion (5/6).

There are probably quite a few students that would reach the same conclusion.

If it would help to remove the confusion due to  conflicting answers posted on this thread, perhaps jbartos could comment?

I do not dispute the calculated equivalent series resistance (since it is correct).  However, I strongly disagree with the answer.  Thinking back, I was asked the exact same question in a physics class.  The point of the question was not to test your ability to solve a circuit, but to demonstrate that a thorough understanding of the problem allows for a correct solution.  Sometimes we get so focused on solving the problem, we fail to get all of the necessary information.  By the teacher telling the students that noone could get the solution, we automatically assume that the math will be difficult.  Therefore, when we feel that we have enough information to start working, and stop listening to the problem.  Unfortunately, this problem preys on pride and not technical ability.  I applaud everyones mastery of electrical concepts, but this question had nothing to due with circuitry.

Sorry about my last response melone.  I did not pay close attention to your earlier response. I thought your most recent question concerned the two other answers proposed... namely 5/6 and 1/32.  

I'll grant that you have a point about the original wording of the question.  There is no meaning whatsoever to resistance "leaving" the circuit.  (even zero resistance leaving the circuit doesn't make any more physical sense than 5/6... is there a short circuit involved here?).  But given that two "terminals" were identified and an "equivalent" resistance was requested, I believe my interpretation of the question was a reasonable one.  I'll grant that your interpretation is also a reasonable interpretation if the question was intended to be a "trick" question.  I don't have any insight into the intent of the person asking the question, other than the fact that it was a physics teacher.

Try this:

All Resistor = 5 ohms find REQ
Establish points A & B
Establish  paths V1, V2, & V3
V1 = I/3 R
V2 = I/6 R
V3 = I/3 R
Vab = V1 + V2 + V3 = I REQ
Vab = IR/3 + IR/6 + IR/3 = I REQ
IR (1/3 + 1/6 + 1/3) = I REQ
IR (2/6 + 1/6 + 2/6) = I REQ
5ohms (5/6) = REQ
25/6ohms = REQ
4.17ohms = REQ actually 4.166666666666

Sorry I'm not Jburn but a fellow employee that was directed to your site and I couldn't resist this one.  If this isn't correct then I forgot to bring my brain cells.

jburn's coworker - I'm counting you as another vote in the 5/6 camp (asssuming the resistances on each side were one ohm vs 5 ohms).

jbartos - are you standing by your 1/36?  It was after all a pretty definitive statement that you made including references to circuit analysis textbook.  

5/6 Ohm is correct.
When current flows through the network, the set of 3 nodes closest to the input node all adopt the same voltage, and may be considered "virtually connected". Likewise the set of 3 nodes closest to the output node are equipotential. Replacing these two sets of "virual connections" with real connections has no effect on current flows and thus no effect on network resistance. However it gives a manageable network. This is (3 parallel) series (6 parallel) series (3 parallel), giving (1/3+1/6+1/3)* 1 Ohm = 5/6 Ohm. Or, as I call it nowadays, 0R83.
When my physics teacher set this problem (seems they all do it!) we checked our solution practically, and I can report that the soldering iron & multimeter back up the solution given above!

Hey guys....

I'd like to thank everyone that helped out on this particular problem as my son was able to get his teacher to eat humble pie!!!!!

The teacher confessed to him that he expected him to come back with the correct answer! No $5 to date however!

The answer was 5/6....

Thank you to all!

Regards
Rhodian

So in other words, his cockiness succeeded in motivating the students, their parents, and a bunch of engineers scattered around the world?  And now you're telling us he knew all along we'd get it.

Pretty crafty, them physics teachers.

I remeber doing this problem during my training but remebering the solution is another matter.
But no matter someone has already done it for us
See :-    http://eswww.it.dtu.dk/~c49102/Circuit09.html.

A nice little java solution.

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!!

Hi Rhodian:

Have your son ask the guy my two puzzles. (fuse and xformer)

<nbucska@pcperipherals.com>

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"x3"x3". 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.

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 "tricky" part - which was the fact that when we identified nodes of equal potential, we "shorted" 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.

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

J. Bartos:
So does the 1/36 ohm solution.

<nbucska@pcperipherals.com>

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).

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.

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., "Elements of Power System Analysis," 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.