# Star Delta transformation of balanced 3 phase load

3. For the balanced three-phase loads shown in FIGURE 3,

ZY = (15 + j15) Î© and ZÎ” = (45 + j45) Î©. Determine:

(a) the equivalent single Î”-connected load,

(b) the equivalent single Y-connected load obtained from the Î”-Y transformation of (a) above,

(c) the equivalent single Y-connected load obtained by transforming the Î” sub-load of FIGURE 3 to a Y and with the star-points of the two Y-sub-circuits connected together,

(d) the total power consumed in case (a) above if the line voltage of the three-phase supply is 415 V at 50 Hz.

https://brainmass.com/physics/ac/star-delta-transformation-balanced-phase-load-628598

#### Solution Preview

Q3 .

Solution:

the equivalent single âˆ†-connected load

In general a Y-connected load (impedances Z_1.Z_2,Z_3) is

transformed to a âˆ†-connected (Z_1^',Z_2^',Z_3^' ) load according to {1}

Z_n^'=(Z_1 Z_2+Z_1 Z_3+Z_2 Z_3)/Z_n {1}

But for a balanced, load as given, Z_1=Z_2=Z_3=Z_Y substituted in

{1} we obtain âˆ†-connected load element impedances Z_âˆ† {2}

Z_âˆ†=(Z_Y Z_Y+Z_Y Z_Y+Z_Y Z_Y)/Z_Y =(3Z_Y^2)/Z_Y

Z_âˆ†=3Z_Y {2}

Since Z_Y=(15+j15)Î© then using {2}

Z_âˆ†=3(15+j15)=(45+j45)Î©

This results in a circuit with two âˆ†-connected loads as shown below

You should be able to see from the topology of the above circuits

that there exist || combination of impedances Z_Î” ||Z_Î” {3} between

lines 1& 2, line 1 & 3, line 2 & 3

Z_Î” ||Z_Î”=(Z_Î” Z_Î”)/(Z_Î”+Z_Î” )=Z_Î”/2 ...

#### Solution Summary

A number of questions concerning transforming delta to star and star to delta loads, determining phase current and voltages and equivalent impedances