## Physics resistance problem

## Physics resistance problem

**BickVess**(visitor)

(OP)

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 (

I would appreciate people's views on the answer!

Here is the question:

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.

Rhodian

**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.

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

## RE: Physics resistance problem

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

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

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).

## RE: Physics resistance problem

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>

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

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>

## RE: Physics resistance problem

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

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

## RE: Physics resistance problem

#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.

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

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>

## RE: Physics resistance problem

## RE: Physics resistance problem

retrack "ungrounded "

<nbucska@pcperipherals.com>

## RE: Physics resistance problem

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

## RE: Physics resistance problem

## RE: Physics resistance problem

## RE: Physics resistance problem

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."

## RE: Physics resistance problem

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

> 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>

## RE: Physics resistance problem

## RE: Physics resistance problem

I fo

rgot to mention the reason for using binary indexing of the8cube corners (nbucska's idea). It makes it effortless to identify the 3neighbors of eachcorner(x,y,z) without a picture. Those neighbors are (x',y,z),(x,y',z),(x,y,z').## RE: Physics resistance problem

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>

## RE: Physics resistance problem

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>

## RE: Physics resistance problem

## RE: Physics resistance problem

Tried your <...technologist...> from your WEB -came back.

Can you pse send me your new addr ?

<nbucska@pcperipherals.com>

## RE: Physics resistance problem

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>

## RE: Physics resistance problem

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>

## RE: Physics resistance problem

What is the answer?

## RE: Physics resistance problem

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?

## RE: Physics resistance problem

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

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!

## RE: Physics resistance problem

BickVess(visitor)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

## RE: Physics resistance problem

Pretty crafty, them physics teachers.

## RE: Physics resistance problem

BickVess(visitor)But no matter someone has already done it for us

See :- http://eswww.it.dtu.dk/~c49102/Circuit09.html.

A nice little java solution.

## RE: Physics resistance problem

## RE: Physics resistance problem

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

<nbucska@pcperipherals.com>

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

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.

## RE: Physics resistance problem

## RE: Physics resistance problem

So does the 1/36 ohm solution.

<nbucska@pcperipherals.com>

## RE: Physics resistance problem

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).

## RE: Physics resistance problem

## RE: Physics resistance problem

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.