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

 

[don01]

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

 

[electricpete]

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.

 

[electricpete]

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).

 

[nbucska]

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>

 

[electricpete]

nbucska - It looks like we both took the approach of finding points of common potential and &quot;shorting&quot; 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.

 

[nbucska]

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>

 

[electricpete]

#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 ;-)

 

[electricpete]

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.

 

[electricpete]

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.

 

[nbucska]

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>

 

[redtrumpet]

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.

 

[nbucska]

RED T:
retrack &quot;ungrounded &quot; <nbucska@pcperipherals.com>

 

[jbartos]

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.

 

[electricpete]

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.

 

[electricpete]

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.

 

[electricpete]

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 ;-)

 

[electricpete]

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?

 

[jbartos]

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. &quot;Elements of Power System Analysis,&quot; 3rd Edition, McGraw-Hill, Inc., 1975, Chapter 7 &quot;Network Equations and Solutions.&quot;

 

[electricpete]

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.)