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HW3 Solution

# HW3 Solution - EE321 Homework 3 Problem 1 From minimum cost...

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Unformatted text preview: EE321, Homework 3 Problem 1 From minimum cost solution in class N 260 := d 8.3788 10 2 − ⋅ := g 13.069 10 3 − ⋅ := w 1.8361 10 2 − ⋅ := d s 8.9184 10 2 − ⋅ := w s 8.9184 10 2 − ⋅ := μ 4 π ⋅ 10 7 − ⋅ := i 40 := L des 5 10 3 − ⋅ := μ r 5000 := Recomputing the reluctance R 2g μ w d ⋅ 2 w s 2w + ( ) 2d s + μ r μ w d ⋅ + := R 13.565 10 6 × = a) To find the inductance L N 2 R := L 4.984 10 3 − × = b) To find the flux density in the magnetic material Φ Ni R = BA = so B m N i ⋅ R w ⋅ d ⋅ := B m 498.361 10 3 − × = c) The flux density in the air gap is the same (areas assumed equal) B g B m := B g 498.361 10 3 − × = d) The field intensity in the magnetic material is H m B m μ r μ := H m 79.317 10 × = e) The field intensity in the air gap is H g B g μ := H g 396.583 10 3 × = f) The percentage difference between this calculated inductance and the design value is L L des − L des 100 ⋅ 329.835 10 3 − × = percent Problem 6 Solution (I know this is out of order, but is is a convenient point)...
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HW3 Solution - EE321 Homework 3 Problem 1 From minimum cost...

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