a circuit puzzle.......help(wiht pics)

[Semidevil]

so lets say that I have 12 resistors put together like a cube. Each resistor is 1 ohm. Looking at the picture, what is R(ab), the resistance of AB??

do I just apply paralell/series relationship for the resistors and slowly reduce it?? there has to be an easier way right? I see that almost all of them are in paralell....

*confused*


link of pic

http://www.geocities.com/hluu410/circuit.bmp

 

[Demon-Xanth]

If they're all 1k ohm, then it's 1.16k ohms (rounded). This is just a guess since I don't feel like working it all out

Picture it like two points with three resistors, and connecting each of those points two in paralell.

 

[Semidevil]

so how did you start it?

each node has 3 branches sticking out. So can I make 2 of those parellel, and then combine that, and make that in series with the last branch, and combine that. and then just keep combining?

 

[dighn]

should be 5/6

just draw the currents and use symmetry of the cube

 

[Demon-Xanth]

Attempt at ASCII schematics:
Points A and B are your nodes, point C is just a common node, "o" are filler characters
ooo|-R5-C
/-R1-R4-\
|oooooo|--R10-\
|ooo|-R6/ooooo|
A-R2--R7-\oooo|
|ooooooo|-R11-|--B
\-R3-R8---/oooo|
ooo|-R9-C-R12/

 

[Demon-Xanth]

Okay, that was fugly. Maybe this can explain it.

R1-1, R2-1, and R3-1 are connected to node A
R1-2 is connected to R4-1 and R5-1
R2-2 is connected to R6-1 and R7-1
R3-2 is connected to R8-1 and R9-1
R4-2 is connected to R10-1 and R6-2
R5-2 is connected to R9-2 and R12-1 (point C)
R7-2 is connected to R11-1 and R8-2
R10-2, R11-2, and R12-2 are connected to B

 

[dighn]

Originally posted by: Demon-Xanth
Okay, that was fugly. Maybe this can explain it.

R1-1, R2-1, and R3-1 are connected to node A
R1-2 is connected to R4-1 and R5-1
R2-2 is connected to R6-1 and R7-1
R3-2 is connected to R8-1 and R9-1
R4-2 is connected to R10-1 and R6-2
R5-2 is connected to R9-2 and R12-1 (point C)
R7-2 is connected to R11-1 and R8-2
R10-2, R11-2, and R12-2 are connected to B

wow that makes it all so clear

 

[Qacer]

Series / Parallel relationships will not help you to find Rab easily. Also, with a quick glance of your drawing, not one of them is parallel to another. Remember, think of parallel as having the two terminals sharing the same nodes (i.e. terminal a1 shares a node with terminal a2 and terminal b1 shares a node with terminal b2).

a1----resistor----b1
a2----resistor----b2

Right now, the best way I can think of to handle this problem is by doing a delta-wye (or pie-T) transformation.
Look it up in your book or in the library. Also, don't forget to try it out in the lab using your handy digital multimeter.



Resistors Cube

 

[Demon-Xanth]

But it can be viewed as paralell. There are only 4 different voltages on the system: points A and B being the source, and there are two others that are split up among three points. An equivilent can be made of having three 2.5k ohm lines in paralell. (okay, so I screwed up the math in my head earlier, 5/6k is correct). Assuming they are all 1k ohm.

 

[Pepsi90919]

it's parallel not paralelllllllllll

 

[Demon-Xanth]

parralel?

 

[dighn]

praelel?!

 

[RichieZ]

wiht?!

 

[Luagsch]

omg it's a bmp!!!

 

[Qacer]

Originally posted by: Demon-Xanth
But it can be viewed as paralell.

Well, once you have the equivalent resistances, then you can view it as being parallel. But the point is you can't use parallel relationships easily to come up with the Rab resistance. You have to transform sections of the circuit configuration into an equivalent form that will eventually lead to finding out Rab.

The original poster wanted an easy way to solve the problem. The easiest way I can think of is delta-wye (pie-T) transformation.