Understanding derivation of delta-y transform

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TheCanadian
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Hi,

I was just looking over my textbook, and it mentions a ## \Delta##-y and y-## \Delta## transformation that is helpful for dealing with circuits in these configurations. The equations can be found here: https://en.wikipedia.org/wiki/Y-Δ_t...xistence_and_uniqueness_of_the_transformation

After looking through the above link and searching for a proper derivation elsewhere, I simply don't seem to understand how the transform equations were derived. If I'm not mistaken, the whole purpose is to be able to evaluate a ## \Delta ## circuit as a Y circuit, and vice versa. Thus, equivalent resistances must be found. But after looking at the derivation provided in the wikipedia link, I don't quite see how it is known that the impedance at a node is: ## R = \frac {R'R''}{\sum R_{\Delta}} ##. Maybe my understanding of nodal analysis is poor, but how is that expression specifically derived?

Any help would be greatly appreciated!
 
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Given RA, RB, RC, determine R1,R2,R3 so as the equivalent resistance between any two points is the same in both circuits.
For 1,2 it means :
[tex]\frac{1}{\frac{1}{R_C}+\frac{1}{R_A+R_B}}=R_1+R_2[/tex]
Write up the other two equations for 2,3 and 3,1, and solve for R1, R2, R3.Or watch the video
 
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ehild said:
View attachment 89767

Given RA, RB, RC, determine R1,R2,R3 so as the equivalent resistance between any two points is the same in both circuits.
For 1,2 it means :
[tex]\frac{1}{\frac{1}{R_C}+\frac{1}{R_A+R_B}}=R_1+R_2[/tex]
Write up the other two equations for 2,3 and 3,1, and solve for R1, R2, R3.Or watch the video


Thank you! Putting the equation into that form really helped!