Exam1_solution - ECE321 Fall 2008 Exam 1 Solution Outline Handy Stuff 2 cm:= 1 10 mm:= 1.0 10 3 7 0:= 4 10 Problem 1 Hy:= 20 A:= 100 10 By:= Hy 0:=

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ECE321. Fall 2008 Exam 1 Solution Outline Handy Stuff cm 110 2 := mm 1.0 10 3 := μ 0 4 π 10 7 := Problem 1 H y 20 := A 100 10 := B y H y μ 0 := Φ B y A := Only the y-component couples the loop λΦ := Single turn λ 0.02513 = Problem 2 w1 c m := d s 2cm := g 0.1 mm := w s 5cm := d5 c m := N 100 := B sat 1 := Point where saturation occurs Now let's compute some reluctances. For flux densities that are not saturated u 1500 μ 0 := R 12 w s w + wd u := R 56 R 12 := R 23 w 2w d u := R 81 R 23 :=
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R 34 g wd ⋅μ 0 := R 78 R 34 := R 45 d s w 2 + u := R 67 R 45 := R eff R 12 R 23 + R 34 + R 45 + R 56 + R 67 + R 78 + R 81 + := R eff 5.09296 10 5 × = OK. Now, the cross sectional area we have is A := The flux where we saturate is Φ sat B sat A := The current to achieve this flux i sat Φ sat N R eff := i sat 2.54648 = Problem 3 L N 2 R eff := L 0.01963 = Inductance of UI Core with standard gap
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Problem 4 W f 1 3 λ 3 5x + = f e 1 3 λ 3 + () 2 = Now i5 := x2 := λ i5 x + := λ 5.91608 = f e 1 3 λ 3 + 2 := One negative for differentiation, one for partial derivative f e 1.40859 =
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This note was uploaded on 09/22/2011 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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Exam1_solution - ECE321 Fall 2008 Exam 1 Solution Outline Handy Stuff 2 cm:= 1 10 mm:= 1.0 10 3 7 0:= 4 10 Problem 1 Hy:= 20 A:= 100 10 By:= Hy 0:=

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