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# Star Delta transformation of balanced 3 phase load

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3. For the balanced three-phase loads shown in FIGURE 3,
ZY = (15 + j15) Ω and ZΔ = (45 + j45) Ω. Determine:
(a) the equivalent single Δ-connected load,
(b) the equivalent single Y-connected load obtained from the Δ-Y transformation of (a) above,
(c) the equivalent single Y-connected load obtained by transforming the Δ 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.

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

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Q3 .

Solution:

In general a Y-connected load (impedances Z_1.Z_2,Z_3) is
transformed to a ∆-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 ∆-connected load element impedances Z_∆ {2}

Z_∆=(Z_Y Z_Y+Z_Y Z_Y+Z_Y Z_Y)/Z_Y =(3Z_Y^2)/Z_Y

Z_∆=3Z_Y {2}

Since Z_Y=(15+j15)Ω then using {2}

Z_∆=3(15+j15)=(45+j45)Ω

This results in a circuit with two ∆-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_Δ ||Z_Δ {3} between
lines 1& 2, line 1 & 3, line 2 & 3

Z_Δ ||Z_Δ=(Z_Δ Z_Δ)/(Z_Δ+Z_Δ )=Z_Δ/2 ...

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