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The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current IY.

(a) (10-j0) A

(b) (10+j0) A

(c) (-10+j0) A

(d) (-10-j0) A

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The correct answer is (c) (-10+j0) A

The best I can explain: The voltage VYB is VYB = 400∠-120⁰V. The impedance Z2 is Z2 = 40∠60⁰Ω => IY = (400∠-120^o)/(40∠60^o)=(-10+j0)A.

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