Wye Delta Impedence Transform Proof?

  • Thread starter caprice24
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I am studying the familiar Wye Delta Impedence transform and can easily prove that R1, R2 and R3 in the Wye configuration are equal to Ra, Rb, and Rc in the Delta configuration according to

R1= Ra*Rb/(Ra + Rb +Rc)
R2= Ra*Rc/(Ra + Rb + Rc)
R3= Rb*Rc/(Ra + Rb + Rc)

This proof is fairly simple because through the parallel resistance formula
R1 + R3 = Rb(Ra+Rc)/(Ra+Rb+Rc)
R2 + R3 = Rc(Ra+Rb)/(Ra + Rb+Rc)
R2 + R1 = Ra(Rb+Rc)/(Ra + Rb +Rc)

you can do a back substitution of variables to solve for R1, R2, and R3.

No problem.

However, going the other way, that is----trying to solve for Ra, Rb, and Rc is not so simple, and leads to really messy polynomials.

I read somewhere that this is a topology problem and it is hard to solve using simple algebraic equations.

Can anyone enlighten me on why this is so? Or maybe I am mistaken, and there is a simple proof?

Thanks,
Nick
 

s_a

22
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This proof is fairly simple because through the parallel resistance formula
In using this method, you are treating the wye resistors & delta resistors as a 3 terminal black box, and you're determining the resistance seen at any two of these terminals with the remaining third terminal hanging and equating the resistances. This is useful for determining the wye values (R1-R3) in terms of the delta values (Ra-Rb), but not vice versa. To determine the delta values in terms of the wye values, use a similar method, but instead short circuit the third terminal to any other of the two terminals rather than leaving it hanging, and work with conductances/admittances instead of resistances/impedances.

Ga + Gc = G3*(G1+G2)/(G1+G2+G3)
Ga + Gb = G2*(G1+G3)/(G1+G2+G3)
Gb + Gc = G1*(G2+G3)/(G1+G2+G3)

You can solve the above system of linear equations for Ga-Gc in terms of G1-G3, and then convert all conductances to resistances (if you wish).
 
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Proof

Thank you for your explanation. I tried it and it works. However, I am trying to understand why it works, and am having trouble visualising it. When you short out two terminals of the Delta network so that one of the resistors is in effect 0 Ohms, you have a parallel combination of R resistors which is the same as the sum of two Admittances. Looking at the Wye network, those same two terminals that are shorted out give two admittances in parallel and one in series, which is the result of the formula you gave.

But what is the rational for shorting out the terminals in the Delta configuration? I understand you have zero current flowing through one of the resistors in the Delta, the voltage across the resistor is going to be zero, and therefore this is equivalent to shorting it out. When you look at the same transformation in the Wye, you have the same voltage at the ends of two resistors.

But don't you have to assume, under this assumption, that the voltage at the two of the nodes of the Wye have to be the same? What if they are different? I see how this proof works for the situation where the voltage is the same, but I guess I am missing why it also applies to when they are different.

Nick
 

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