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

 

[electricpete]

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.

 

[electricpete]

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

>

 

[electricpete]

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').

 

[electricpete]

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').

 

[nbucska]

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>

 

[nbucska]

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>

 

[melone]

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

 

[nbucska]

Hi E.Pete:

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

 

[nbucska]

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>

 

[nbucska]

The simplest (I think)

Let the longest diameter of the cube 1 &quot;. 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&quot; and 3 at 2/3&quot;

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&quot; three R-s. 1/3+1/6+1/3=5/6 ----- Q.E.D.
<nbucska@pcperipherals.com>

 

[melone]

Rhodian,

What is the answer?

 

[electricpete]

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?

 

[melone]

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.

 

[electricpete]

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 &quot;leaving&quot; 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 &quot;terminals&quot; were identified and an &quot;equivalent&quot; 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 &quot;trick&quot; 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.

 

[jburn]

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.

 

[electricpete]

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.

 

[Resistor]

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 &quot;virtually connected&quot;. Likewise the set of 3 nodes closest to the output node are equipotential. Replacing these two sets of &quot;virual connections&quot; 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!!!!! LOL

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

The answer was [red]5/6[/red]....

Thank you to all!

Regards
Rhodian

 

[electricpete]

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://es

http://www.it.dtu.dk/~c49102/Circuit09.html.

A nice little java solution.