Wye-delta transform derivation

Question

I have been reading about the Delta-Wye and Wye-Delta transforms for resistors, and I was curious as to how the equations for the Delta-Wye and Wye-Delta transforms are formed. I managed to isolate the variables for \$R_a\$, \$R_b\$, and \$R_c\$ (for here, in terms of \$R_{ab}\$, \$R_{ac}\$, \$R_{bc}\$, but I am unable isolate \$R_{ab}\$, \$R_{bc}\$, and \$R_{ac}\$ in terms of \$R_a\$, \$R_b\$, and \$R_c\$. I guess my question is how, without looking at the answer, one is able to derive the equations for the Wye-Delta transform. Thanks.

EDIT: I guess this is more of a math question than a EE question, but I was also wondering if I'm looking at the problem incorrectly. For Delta-Wye transforms, I calculated resistance from A to B and formulated the equation

\$R_a+R_b=R_{ab}\dfrac{R_{ac}+R_{bc}}{R_{ab}+R_{bc}+R_{ac}}\$

Similarily I calculated resistance from B to C as

\$R_b+R_c=R_{bc}\dfrac{R_{ab}+R_{ac}}{R_{ab}+R_{bc}+R_{ac}} \$,

and resistance from A to C as

\$R_a+R_c=R_{ac}\dfrac{R_{ab}+R_{bc}}{R_{ab}+R_{bc}+R_{ac}} \$

I then added up all three equations and divided by two. \$R_a+R_b+R_c=\dfrac{R_{ab}R_{bc}+R_{ab}R_{ac}+R_{ac}R_{bc}}{R_{ab}+R_{bc}+R_{ac}} \$

To find the individual \$R_a\$, \$R_b\$, and \$R_c\$, I just subtracted the last equation with the first three one at a time. The problem is I can't easily isolate \$R_{ab}\$, \$R_{bc}\$, and \$R_{ac}\$ - The Wye-Delta transform equations seem a lot harder to derive.

Answer 1

There are two ways to do this. First of all, I would change your terminology. Use R1,R2,R3 for the wye and RA,RB and RC for the delta.

1. Use the principle of duality - this is the simplest approach. Take the corresponding equations for the delta-wye transformation, and everywhere there is an R, swap with G, and swap letters for numbers. If you then want the corresponding resistances, the conversion from conductance is straightforward.

2. Re-draw the wye as an inverted-T network, and the delta as a pi network. Then treat both of these as two-ports. Applying an open-circuit to one of the ports and then determining the input impedance at the other is equivalent to the approach of determining the resistance between two nodes of the corresponding wye and delta circuits. However, if you use a short-circuit termination (again, duality comes into play), you'll get the corresponding equations in G. This approach is a little tedious. It is also possible to configure a three-port, but without a diagram can't really help to explain it.

Answer 2

Unfortunally, as I just created my account here, I cannot comment. So this is not a complete answer.

The way you need to look at it, is that the equivalent resistance from nodes a-b, a-c an b-c must be equal comparing between both designs, so that's where you start your math.

If you want to, you can start algebra from that and you will get the answer.

I'll edit this answer ASAP in order to give you the full explanation including algebra stuff, I'm at work now :)

Hope it helps.

Answer 3

Ok, let me help you out. $${R_{a}=\frac{R_{ab}R_{ac}}{R_{ab}+R_{bc}+R_{ca}}}$$ $${R_{b}=\frac{R_{ab}R_{bc}}{R_{ab}+R_{bc}+R_{ca}}}$$ $${R_{c}=\frac{R_{ca}R_{bc}}{R_{ab}+R_{bc}+R_{ca}}}$$ Finding $${R_{a}R_{b}+R_{b}R_{c}+R_{c}R_{a}}$$ gives $${\frac{R_{ab}^2 R_{ac} R_{bc}+R_{ab}R_{bc}^2R_{ca}+R_{ca}^2 R_{bc}R_{ab}}{ {(R_{ab}+R_{bc}+R_{ca})}^2 }}$$ Factoring $${R_{a}}$$ out gives $${\frac{R_{ab}R_{ac}}{R_{ab}+R_{bc}+R_{ca}} \frac{R_{ab}R_{bc}+R_{bc}^2+R_{ca} R_{bc}}{ {(R_{ab}+R_{bc}+R_{ca})}}}$$

$${R_{a}R_{bc}}$$ $${\frac{R_{a}R_{b}+R_{b}R_{c}+R_{c}R_{a}}{R_{a}}=R_{bc}}$$ Similarly $${\frac{R_{a}R_{b}+R_{b}R_{c}+R_{c}R_{a}}{R_{b}}=R_{ac}}$$ $${\frac{R_{a}R_{b}+R_{b}R_{c}+R_{c}R_{a}}{R_{c}}=R_{ab}}$$ Factor out $${\frac{R_{ab}^2 R_{ac} R_{bc}+R_{ab}R_{bc}^2R_{ca}+R_{ca}^2 R_{bc}R_{ab}}{ {(R_{ab}+R_{bc}+R_{ca})}^2 }}$$ appropriately in terms of the wye resistances to get the delta resistances.