How do you work out the impeadance of a Star to Delta A.C circuit?

Question

I know how to do it if there are just resistors in the Star to Delta circuit. I don't know how to work it out when there are capacitors and inductors involved.

Lets say i have a 5 ohm resistor in series with a 7 ohm (+j7) inductor on one of the star arms. how do i work out the impedance of this arm?

Is it 1/(5+j7) to find y and then use the usual formula or do i have to just change the resistor (1/5) and put this with the inductor? (0.2+j7).

Thanks in advance

Answer 1

Instead of Resistance R use the impedance Z where Z = R + jX for an inductive load or R - jX for a capacitive load. So in the example you give, Z will be 5 +j7. Then just use the Star-Delta transformation formula for each of the Zs.

e.g. Z(12)= Z(10) + Z(20) + {Z(10)x Z(20)}/Z(30)

and similarly for Z(23) and Z(31)

By the way I have used Z(12) as the resulting impedance in the delta leg 1,2. The impedances in the star - line to neutral legs are Z(10), Z(20), Z(30).  You obviously need the impedances in the other two star legs given in the same form as a +/- jb.