Lecture 5 - UCF Hard Material Circuit Analysis Example 5...

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CF ard Material Circuit Analysis UCF Hard Material Circuit Analysis Example 5 Device line: =(B H H B(B r /H c )(H m +H c ) Note: B g and B m are the same. Load line:
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CF Inductance UCF = Φ dS B B ∫∫ Φ = N λ Flux linkage I N L Φ = = I I NI = Φ NI total total R N 2 total L R
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CF Mutual Inductance UCF n n I L I L I L I L I L I L L L + + = + + = 2 1 1 2 12 1 11 1 λ I λ 1 λ n n n I L I L L L M + + = 1 1 2 2 22 1 21 2 1 I n n in j ij i ii i i I L I L M L + λ 2 i p I L i = = , 0 n nn n n n I L I L I L L + + = 2 2 1 1 elf inductance i p I N I i p i ii = = , 0 2 Self inductance I 2 p i R j p I I L p j i ij = = , 0 Mutual inductance
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CF Example 6 UCF 2 22 1 21 2 2 12 1 11 1 I L I L I L I L + = + = λ
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CF Example 7 UCF 2 12 1 11 1 I L I L + = λ 2 22 1 21 2 I L I L + =
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CF Faraday’s Law (I) UCF ∫∫ = S C dS B dl E dt d Electromotive Force (emf) ------ V l f Total Magnetic Flux ------ Wb (Webers) dl E C emf ∫∫ = S dS B ϕ
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CF Faraday’s Law (II) UCF = dS B dl E t d l ∫∫ S C dt C dl E B • The emf generated around a closed contour C is lated to the time rate of change of the total related to the time rate of change of the total
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This document was uploaded on 11/03/2009.

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Lecture 5 - UCF Hard Material Circuit Analysis Example 5...

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