Test_3_Solutions - Question 1: (See Slide 67) No load or...

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Question 1: (See Slide 67) No load or open-circuit conditions: Given : V P 100 10 3 := P 1000 := I P 10 := VA V P I P := pf P VA := pf 1 10 3 × = θ acos pf () := θ 180 π 89.943 = Magnitude of magnetizing impedance: Z Emag V P I P := Z Emag 11 0 4 × = From Slide 73: X M Z Emag sin acos pf := X M 0 4 × = R C Z Emag pf := R C 0 7 × = The magnitizing branch impedance represents a lagging load since it is inductive, which is always the case for a transformer. Some of you solved the problem this way: Y E I P V P cos θ ( ) j sin θ := R C 1 Re Y E := X M 1 Im Y E := R C 10 10 6 × = X M 0 4 × = ___________________________________________________________________________
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Question 2: (See Slide 77) Given : VA 100 10 3 := V H 100 10 3 := V L 10 10 3 := a V H V L := 1000 R P 5 := X P 25 := V Applied 1000 := Rated current on low voltage side: I L VA V L := Current on high voltage side: I H I L a := I H 1 = pf P V Applied I H := θ acos pf () := θ 180 π 0 = II H e j θ := V Applied R eq jX eq + I := R eq Re V Applied I := X eq Im V Applied I := Clearly a silly answer - my mistake! Z eq Z P a 2 Z S + := R S R eq R P a 2 := R S 9.95 = X S X eq X P a 2 0.25 = := ___________________________________________________________________________
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Test_3_Solutions - Question 1: (See Slide 67) No load or...

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