Another resistive circuit

L D[AR2P#0.0]

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L
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Following on from spifldorf's post, here's another circuit to ponder:

Find an expression (in terms of R) for the resistance between terminals A and B for the following circuit:

circuit001.JPG
 
I used the sum of currents at each node being equal to zero. I numbered the nodes 1 through 9. Nodes 3 and 9 were ommitted because they aren't really nodes. The result is 7 equations and seven unknowns. I solved for the voltage at each of these nodes in terms of A B r=1 and R. I then found the current going from A to v1. The voltage A-B divided by the current from A to v1 yields the resistance.

I never liked the current loop way of solving these problems. Keeping track of whether the current adds or subtracts from the loop voltage seems error prone.

Good brain teaser. I don't think I have done this in the optimal way. It is just good I could do this without reading a book or other material after 30 years.

I don't want to ruin anybody's fun yet.
What the heck
ftp://ftp.deltacompsys.com/public/NG/Mathcad%20-%20Norton's%20Method.pdf
 
Last edited:
L D[AR2 said:
Peter - is that link working ? I just get a white screen
Click to expand...
I did too. I had to wait a few seconds and then it showed.

Hard to be humble, hey Peter?
Click to expand...
Didn't you get the first one? I thought I would try this one because no one else did. I would have expect the new forum new engineer to quickly solve this one. Actually, I didn't have to think too much about this at all but I had two entry mistakes that took time to track down.

BTW, this problem is easy because I have Mathcad but there are other and better symbolic processing math packages.

This is one is free
http://sourceforge.net/projects/maxima
There was a recent update.
 
Peter Nachtwey said:
I would try this one because no one else did. I would have expect the new forum new engineer to quickly solve this one.
Click to expand...
It's only because of the time difference :)
The catch is to do subsequent star-delta transformations. In first step, there will be four (top, bottom, left and right)... Just look at the file.
 
This is my take on the problem. I used Kirchoff's Voltage Law to solve the problem. This was the way we were taught how to do it in Circuit Analysis class.

(Finally get to use some of this "new" knowledge. Woohoo!)

Bob
 
Great effort, but note that not all of the resistances have a value of 1, there is a single resistance of value R in there as well.
 
L D[AR2 said:
Great effort, but note that not all of the resistances have a value of 1, there is a single resistance of value R in there as well.
Click to expand...

DAMN! I completely missed that! Hmmm, back to the drawing board I guess...

Maybe I will give it a go again tomorow morning.
 
L D[AR2 said:
Great effort, but note that not all of the resistances have a value of 1, there is a single resistance of value R in there as well.
Click to expand...
Hahaha, I can't believe, I didn't see it before. I must learn myself to read twice before trying to solve it...
 
Here's a final festive fun circuit:

The circuit shown is connected to a balanced 3-phase supply. The magnitude of current x is larger than the magnitude of current y.
Determine the phase sequence of the supply.

three001.JPG
 
L D[AR2,P#0.0], do you know solution?
According to my calculation, if Ua has phase 0, then Ub has phase -120 deg and Uc has +120 deg. I tried several times and every time I got the same on paper. However, I'd like someone who has some kind of simulation software to check.
 
Since nobody offered step by step solution to the 3 phase problem, here is mine (not complete though...). First step is to write equations. My approach is to define parameter alpha, solve circuit and then analyse solutions if alpha is -1 or +1. Here in attachment you'll find procedure for determining magnitude of Ix which I noted as I1. According to this specific circuit's structure, I think it is enough to find magnitude of I1 to make decision. However, complete procedure must be repetead for I2 and decision would be trivial in that case.
 

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