For the balanced three-phase loads shown in FIGURE 3,
ZY = (15 + j15) Ω and ZΔ = (45 + j45) Ω. Determine:
Uploaded file C1.png
(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.
The Attempt at a Solution
For (a) P=Q=R=(15+15i)
Star "PQR" --->Delta "ABC" equivelant =
Since the loads QPR are all the same value and same equation form then A=B=C, ((15+15i)*(15+15i)+(15+15i)*(15+15i)+(15+15i)*(15+15i))/(15+15i)=45+i45
Delta equivelant is A=45+i45, B=45+i45, C=45+i45
For(b) The reverse of (a) I assume; Delta ---> Star =
Questions seems deceptively easy for my liking
(c) Is the diagram C2.png how the transformation and two Y sub-circuit connected star points should look like?
I need a hint on how to form the equations for this if it is correct(im sure its obvious but im not sure.)
Any help greatly appreciated.
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