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Posted By: Kenbo Conundrum - 09/20/06 02:45 PM
Here is a small puzzle I was asked about.

12 resistors of 1 ohm each, are conected to form a cube (my drawing skills are awfull)
At two corners diaginaly opposite each other are marked A and B, (so you have to pass through three resistors to get from A to B)
What is the total resistance of this circuit?
Posted By: Alan Belson Re: Conundrum - 09/20/06 03:41 PM
0.5 ohms?

Alan
Posted By: Bob Re: Conundrum - 09/20/06 04:50 PM
"At two corners diaginaly opposite each other are marked A and B, (so you have to pass through three resistors to get from A to B)"
Maybe
I am reading more into this than needed. If each side has 3 resistors you would have 12
installed in the cube. You would go thru 6 resistors to get from A to B. Are you requiring the circuit to be constructed so that you must go thru 3 resistors from A to B?
Posted By: Kenbo Re: Conundrum - 09/20/06 05:06 PM
Bob
Imagine a wire model of a cube It would need 12 lenths of wire to construct. But each is a 1 ohm resistor. What is the restance at the two furthest away corners of the cube?

I will try to post a pict

Alan 0.5 ohm ? how did you get that (not correct though)
Posted By: Max H Re: Conundrum - 09/20/06 05:37 PM
The result is 5/6 ohms

The circuit can be re-drawn as three 1 ohm resistors in parallel(=1/3 ohm), in series with 6 1 ohm resistors in parallel (=1/6 ohms), and then another 3 1 ohm resistors in parallel (=1/3 ohms).

The reason this can be done will become apparent once you re-draw the circuit, and is based on the fact that all the resistors are identical.
Posted By: Bob Re: Conundrum - 09/20/06 05:40 PM
I understand that. The cube has 4 sides. That would take 3 resistors in each side to use 12. If so the equivalent R = 3.
Posted By: Almost Fried Re: Conundrum - 09/20/06 05:59 PM
I concur with MaxH, it's 5/6 or .75 ohm
Posted By: Trumpy Re: Conundrum - 09/20/06 06:51 PM
When I sat my Trade Cert as an Electrician in 1994, this question came up.
With 3 hours for the whole exam, it was certainly a stumbling block.
I think it is the idea of a cube that throws a lot of people off.
Posted By: walrus Re: Conundrum - 09/20/06 07:31 PM
5 divided by 6 isn't .75 its .83
Posted By: Alan Belson Re: Conundrum - 09/20/06 07:40 PM
Kenbo; just a lucky guess!

Actually, I worked it out while roofing a dormer today and the sun must have addled my brains, ie. 1/3 + 1/3 = 1/2 . DOH!
Still, only 1/6 of an ohm out!!! [Linked Image]

This is how I figured it:

Let each corner be designated a letter,
A,B,C,D,E,F,G,H
Each resistor can then be identified by a pair of letters ie. 'AD'

From point A we get, [from my layout]
AD+AH+AE, in parallel, = 1/3 ohm
Call this the first leg.

PLUS:
HC+CB =2ohm parallel 2 ways = 1 ohm
HG+GB =2ohm

EF+FB =2ohm parallel 2 ways = 1 ohm
EG+GB =2ohm

DC+CB =2ohm parallel 2 ways = 1 ohm
DF+FB =2ohm

But these 3 routes are also in parallel to complete the 2nd leg;

Total for second leg = 1/3 ohm

1/3 + 1/3 = 2/3 ohm total resistance

Alan, Class Dunce of 1948!

ps. I just spotted I got 15 resistors, [some used twice], so I give in!






[This message has been edited by Alan Belson (edited 09-20-2006).]

[This message has been edited by Alan Belson (edited 09-20-2006).]
Posted By: Bob Re: Conundrum - 09/20/06 08:42 PM
Kembo
You said cube and I heard square. I have a picture of this is I knew how to post it.
5/6 R
Posted By: Admin Re: Conundrum - 09/20/06 09:25 PM
from Bob:

[Linked Image]
Posted By: NJwirenut Re: Conundrum - 09/21/06 12:11 AM
OK, for the heck of it, I just tacked together a cube of 100K resistors (a nice round value I had a bunch of in my junkbox). Picture of the cube is being sent for posting.

Each resistor was tested before soldering, and found to be pretty close to marked value. Well within the marked 5% tolerance.

Corner to corner resistance as measured with a Fluke 8060A = 82.94 kOhms. With 1 ohm resistors, .83 ohms sounds like a good answer.

Nothing like a real-world confirmation of the theory! [Linked Image]

[Linked Image]

(edited to add image)

[This message has been edited by Webmaster (edited 09-20-2006).]
Posted By: Alan Belson Re: Conundrum - 09/21/06 06:41 AM
Thank's Bob for an excellent diagram, now I see where I went wrong.
I should have stayed up on the roof.

NJ, you actually built one, from tested components, ran an experiment, photographed it and posted it to ECN???
You crazy person! LOL! [Linked Image] [Linked Image] [Linked Image]

Alan

[ ps. Still like to see the math. proof for that central 6-resistor group; I can't work out why it comes to 1/6 ohm. ]
Posted By: Scott35 Re: Conundrum - 09/21/06 11:43 AM
I drew up a couple of drawings regarding this "Resistor Cube Condundrum", and posted them below.

Let me know what 'ya think! [Linked Image]

Scott 35

[Linked Image]
*** Figure 1: "Cube" View ***

---------------------------------------------
---------------------------------------------

[Linked Image]
*** Figure 2: "1-Line" View ***
Posted By: Kenbo Re: Conundrum - 09/21/06 01:52 PM
What can I say excelent replies guys.

Bob that diagram is spot on.

Alan what is wrong with NJs solution? I ended up building one up and actualy testing it as well [Linked Image]

Scott briliant explination. Would you mind if I used your diagrams to explain this to some of my students?
Posted By: pauluk Re: Conundrum - 09/21/06 04:13 PM
Quote
OK, for the heck of it, I just tacked together a cube of 100K resistors

Is it just the lighting/my monitor, or do the yellow bands on those resistors really look close enough to orange that they appear to be 10K ?
Posted By: NJwirenut Re: Conundrum - 09/21/06 04:28 PM
Not your monitor or the lighting. Just an annoying batch of resistors with yellowish-orange (or orangish-yellow) paint.

That's one reason I tested each one before installing them! [Linked Image]
Posted By: Scott35 Re: Conundrum - 09/21/06 06:51 PM
Kenbo:

Quote

Would you mind if I used your diagrams to explain this to some of my students?



For a small fee of $2,048,000², you may make one copy, for use on only one Student.
Any other Students will require additional Licensed copies (1 Licensed copy per Student), at which a reduced rate may be applied for larger group purchases of Licenses (7% discount for 10 to 25 Seats Licensed, 12.285% discount for 26 to 100 Seats Licensed).

[Linked Image]

Just like a typical Software's disclaimer and Licensing agreement, isn't it???
[Linked Image]

Anyhow, please pay no attention to the Legal-Eze stuff above, as it is 100% Male Bovine Fecal Matter (Bull Poop!)

That text above could be spread across a Lawn, and will really make the Grass grow nice and green!

The People at the Bandini Corporation purchase matter similar to that text, and distribute it in 2.0 cubic foot bags, which are sold at Lawn and Garden Centers
(anyone remember the "Ski Bandini Mountain" Commercial?).

But Seriously

Feel free to use these drawings for discussions and/or examples, in your Classes!
In fact, feel free to use any of the stuff I have posted, for Classroom examples / discussions!

This applies to anyone at ECN who might have a use for the posted techno-electro-mumbo-jumbo I have placed at ECN.

Also, there will be no charge -
except the previously described Licensing fees in the Licensing Agreement above...
[Linked Image]
[Linked Image]

Scott35
Posted By: Scott35 Re: Conundrum - 09/21/06 06:59 PM
Oh, forgot to ask/mention:

If needed, contact me via e-mail, and I can send you drawings/information in:
  • Full Size (not reduced to 640x480 dpi),
  • .GIF, .JPG, .TIF, or .BMP file format (for Raster images),
  • .WMF file format (for Vector images),
  • .DWG, or .DXF file format (for AutoCAD drawings),
  • Excel spreadsheets, or text in some usable format.


e-mail address is (spelled phonetically):

setelectric at pacbell dot net

Scott35
Posted By: Alan Belson Re: Conundrum - 09/21/06 07:09 PM
NJ,/Kenbo: No offence intended, so please accept my apology if my remark was taken the wrong way.

regards

Alan
Posted By: Max H Re: Conundrum - 09/21/06 08:18 PM
Alan,

Think of it like this.

Because the circuit is symmetrical, and the resistors identical the current flowing through each of the resistors (R1, R2, R3) leaving point A will be equal.

Since this is true, the voltage at the far end of each of the three resistors in section 1 will be the same, so for the purposes of analysis you can connect them all together (i.e. in parallel) without affecting the operation of the circuit.

The same can be said with the resistors connected to point B, which means that they can also be connected in parallel.

Now, if the top three resistors are all in parallel, and so are the bottom 3, then by default, the middle 6 are also in parallel.


And for resistors in parallel, their resistance is

(1/((1/1ohm)+(1/1ohm)+(1/1ohm)+(1/1ohm)+(1/1ohm)+(1/1ohm))) = 1/6 ohm
------------------------------------------

Another way you can look at it, is since the middle section resistors have half as much current going through them (per resistor) as the top and bottom section (per resistor), they will have half the voltage drop, so by ohm's law this section must have half the resistance of the top or bottom section.
Posted By: Admin Re: Conundrum - 09/22/06 03:26 AM
Quote
Redrawing the circuit you can see that nodes 1,2 and 3 are at the same potential as are the nodes 4, 5 and 6.
[Linked Image]
Quote
If we short ckt these terminals we get the drawing below.
[Linked Image]
Quote
We now have 3R in parallel in series with 6R in parallel in series with 3R in parallel which is easy to solve.

- Bob
Posted By: Kenbo Re: Conundrum - 09/22/06 10:40 AM
Alan no offence taken I was trying to be funny (smiley at end)

Bob took second look at your drawing before I realised you had the same as Scott.

Well done. I was going to offer prize of a free holiday in Scotland. At one of "Her Majesties" prime holiday camps. HMP Glenochil but I do not want to upset the governor (or get sacked) so the offer has expired [Linked Image]
Posted By: Bob Re: Conundrum - 09/22/06 07:16 PM
The secrete to solving the problem is to recognize that since nodes 1,2 and 3 and 4,5 and 6 are at the same potential. This allows
you to short circuit these terminals without changing the ckt and re-draw the circuit into a simpler form.


[This message has been edited by Bob (edited 09-22-2006).]
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