- #1
caprice24
- 3
- 0
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
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