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#69828 - 09/20/06 07:45 AM Conundrum
Kenbo Offline
Member
Registered: 04/07/06
Posts: 234
Loc: Scotland
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?
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#69829 - 09/20/06 08:41 AM Re: Conundrum
Alan Belson Offline
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Registered: 03/23/05
Posts: 1801
Loc: Mayenne N. France
0.5 ohms?

Alan
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#69830 - 09/20/06 09:50 AM Re: Conundrum
Bob Offline
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Registered: 02/05/02
Posts: 182
Loc: Mobile, AL, USA
"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?
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#69831 - 09/20/06 10:06 AM Re: Conundrum
Kenbo Offline
Member
Registered: 04/07/06
Posts: 234
Loc: Scotland
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)
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#69832 - 09/20/06 10:37 AM Re: Conundrum
Max H Offline
Junior Member
Registered: 09/20/06
Posts: 2
Loc: Edmonton, Alberta, Canada
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.
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#69833 - 09/20/06 10:40 AM Re: Conundrum
Bob Offline
Member
Registered: 02/05/02
Posts: 182
Loc: Mobile, AL, USA
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.
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#69834 - 09/20/06 10:59 AM Re: Conundrum
Almost Fried Offline
Member
Registered: 09/04/06
Posts: 104
Loc: Madison County, Ark. USA
I concur with MaxH, it's 5/6 or .75 ohm
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#69835 - 09/20/06 11:51 AM Re: Conundrum
Trumpy Offline


Member
Registered: 07/05/02
Posts: 8560
Loc: SI,New Zealand
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.
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#69836 - 09/20/06 12:31 PM Re: Conundrum
walrus Offline
Member
Registered: 07/25/02
Posts: 671
Loc: Bangor Me. USA
5 divided by 6 isn't .75 its .83
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#69837 - 09/20/06 12:40 PM Re: Conundrum
Alan Belson Offline
Member
Registered: 03/23/05
Posts: 1801
Loc: Mayenne N. France
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!!!

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).]
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