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#69828 - 09/20/06 10:45 AM Conundrum  
Kenbo  Offline
Member
Joined: Apr 2006
Posts: 233
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?


der Gro├čvater

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#69829 - 09/20/06 11:41 AM Re: Conundrum  
Alan Belson  Offline
Member
Joined: Mar 2005
Posts: 1,803
Mayenne N. France
0.5 ohms?

Alan


Wood work but can't!

#69830 - 09/20/06 12:50 PM Re: Conundrum  
Bob  Offline
Member
Joined: Feb 2002
Posts: 182
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?


#69831 - 09/20/06 01:06 PM Re: Conundrum  
Kenbo  Offline
Member
Joined: Apr 2006
Posts: 233
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)


der Gro├čvater

#69832 - 09/20/06 01:37 PM Re: Conundrum  
Max H  Offline
Junior Member
Joined: Sep 2006
Posts: 2
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.


#69833 - 09/20/06 01:40 PM Re: Conundrum  
Bob  Offline
Member
Joined: Feb 2002
Posts: 182
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.


#69834 - 09/20/06 01:59 PM Re: Conundrum  
Almost Fried  Offline
Member
Joined: Sep 2006
Posts: 98
Madison County, Ark. USA
I concur with MaxH, it's 5/6 or .75 ohm


#69835 - 09/20/06 02:51 PM Re: Conundrum  
Trumpy  Offline


Member
Joined: Jul 2002
Posts: 8,217
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.


Let's face it, these days if you're not young, you're old - Red Green grin

#69836 - 09/20/06 03:31 PM Re: Conundrum  
walrus  Offline
Member
Joined: Jul 2002
Posts: 680
Bangor Me. USA
5 divided by 6 isn't .75 its .83


#69837 - 09/20/06 03:40 PM Re: Conundrum  
Alan Belson  Offline
Member
Joined: Mar 2005
Posts: 1,803
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!!! [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).]


Wood work but can't!

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