exam1asolution

exam1asolution - EE321 Spring 04 Exam 1a Solution Basic...

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Φ 1do2 N 2 i 2 R 1 R 3 R 3 R 1 R 3 + = Flux into winding 1 do to current in winding 2 N 2 100 := N 1 50 := R 3 10 5 := R 2 R 1 := R 1 10 4 := Basic Problem 3: R 994.718 = R l µ r µ 0 A := l 0.05 := µ 0 4 π 10 7 := µ r 2000 := A 0.02 = A Φ B := B 0.5 := Φ 0.01 := Basic Problem 2: By inspection, 500 A. Basic Problem 1: EE321 Spring 04 Exam 1a. Solution
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R 2 R 1 R 3 R 1 R 3 + + R 1 R 3 + L 12 N 1 N 2 R 2 R 1 R 3 R 1 R 3 + + R 3 R 1 R 3 + := L 12 0.238 = Basic Problem 4 λ 5 i 100 1 e 3 i ( ) + 2 x + = x 1 := i 0.1 := Absolute inductance λ 5 i 100 1 e 3 i ( ) + 2 x + 10 3 := λ 8.806 10 3 × = L abs λ i := L abs 0.088 = L inc 5 300 e 3 i + 2 x + 10 3 := L inc 0.076 =
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Basic Problem 5 λ 1 5 i 1 7 2 cos 2 θ rm ( ) ( ) 1 e i 1 2 i 2 + = λ 2 3 i 2 2 7 2 cos 2 θ rm ( ) ( ) 1 e i 1 2 i 2 + = Step 1 - Bring up first current with second held at zero W c1 0 i 1f i 1 5 i 1 7 2 cos 2 θ rm ( ) ( ) 1 e i 1 + d =
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