Author |
Topic: Resistor cube using unique value for each side (Read 5171 times) |
|
octave57
Newbie
Posts: 1
|
 |
Resistor cube using unique value for each side
« on: Apr 27th, 2004, 2:18pm » |
 Quote  Modify
|
My proffesor recently approached the class with the typical resistor cube problem. If all resistors are equal, the symetry is easy to spot, therefore the problem is neatly solved. However, my prof is requesting we devise a formula that would measure resistance between opposite corners that would accomidate any ohmic value for any resistor. The thing is: He has never solved this himself! We are not receiving a grade for this, its simply extracurricular, nevertheless I've been obsessing about it. I know that it has to do partly with delta-wye conversions following which, some parallel and series resistance banks should appear, and the problem becomes one of organization, not necesarily critical thinking. I am fairly new to electronics, and was wondering if anyone has solved this or could offer insight on how to identify parallel and series banks within a complex(to me) schematic. If you made it all the way through this, thank you for your time.
O57S
|
|
 IP Logged |
|
|
|
SWF
Uberpuzzler
Posts: 879
|
|
Re: Resistor cube using unique value for each side
« Reply #1 on: Apr 27th, 2004, 7:28pm » |
Quote Modify
|
There is a problem on this board somewhere about infinite resistor networks that you might be interested in.
It may not be the easiest way, but for this one, I think you can just write a system of 12 linear equations in terms of the 12 values for current in each edge of the cube assuming 1 Ampere enters one corner and leaves the other. The voltage drop from corner to corner with on 1 Ampere of current gives the equivalent resistance. You will only need to solve the equations for 3 of the variables to get the result you need.
To get the 12 equations you can use sum of currents at any vertex equals zero (don't forget to include the applied 1 Ampere), and the sum of the voltage drops around a closed loop is zero. I assume this is not difficult to do such that 12 independent equations are obtained.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print