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exam2solution - EE321 Exam 2 Problem 1 rs := 3.5 2 points...

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EE321 Exam 2 Problem 1 r s 3.5 := 2 points L max 120 10 3 := L min 20 10 3 := L A 1 2 L max L min + () := This includes the leakage inductance Lls L A 70 10 3 × = 3 points (Includes Lls) L B 1 2 L max L min := 4 points L B 50 10 3 × = RT 4 := N 3 := TP 2 π RT := SL TP N := SL 180 π 30 10 0 × = 4 points i2 := T eadv 1 2 RT L B i 2 3 2 := T eadv 346.41 10 3 × = 4 points
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Problem 2 4 O 3 O 5O O 2 Problem 3 T e θ rm i as , () 1 2 16 sin 4 θ rm 8 sin 8 θ rm + i as 2 := Derive - 10 pts i as 5 10 := 5 pts θ rm 0.1 := T e θ rm i as , 1.496 10 0 × = 2pts
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Problem 4 k v 0.1 := r a 0.3 := v a 25 := T e v a k v ω r r a k v = T L ω r 200 = Equating v a k v ω r r a k v ω r 200 = 7 points to derive ω r 200 v a k v 200 k v 2 r a + := ω r 217.391 10 0 × = rad/s 3 points to evaluate i a v a k v ω r r a := i a 10.87 10 0 × = 3 points to evaluate T e k v i a := P out ω r T e := P in v a i a := η P out P in := η 100 86.957 10 0 × = % 4 points to evaluate
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Problem 5 We have v a r a i a L AF i f ⋅ω r + = v f r f i f = v t v a v f = 3 points i f i a = 3 points Thus v t r a i a L AF i a r r f i a + = So i a v t r a r f + L AF ω r = 6 points Since T e L AF i a i f = We have T e L AF v t 2 r a r f + L AF ω r () 2 = 5 points Problem 6 r a 1 := i f 2 := i a 5 := ω r 10 := pi f 0 := pi a 50 :=
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We start with v a r a i a p λ a + = Now λ a 5 4 cos 2 θ r () i a 3 sin
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exam2solution - EE321 Exam 2 Problem 1 rs := 3.5 2 points...

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