Mathcad - 04-07 - Problem 4-7: Solution Givens: R := 65 ohm...

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φ B A c = The total mmf will be the mmf of the core plus the mmf of the airgap reluct 4.775 10 5 × amp weber = reluct gap μ 0 A c = H 300 350 410 490 520 600 690 766 843 921 1000 amp m = B .5 .55 .6 .65 .7 .75 .8 .85 .9 .95 1.0 tesla = This is a type-two problem. It can be solved iteratively or by making a table. Start by assuming the flux density will be somewhere between 0.5 and 1.0 T. For values of B, we can read the field intensity, H, from the B-H curve for cast steel. μ 0 4 π 10 7 - henry m = free space permeability gap 0.0012 m = length 1 m = I 4.5 amp = A c 0.002 m 2 = N 260 = R 65 ohm = Givens: Problem 4-7: Solution
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φ 0 0 1 2 3 4 5 6 7 8 9 10 1·10 -3 1.1·10 -3 1.2·10 -3 1.3·10 -3 1.4·10 -3 1.5·10 -3 1.6·10 -3 1.7·10 -3 1.8·10 -3 1.9·10 -3 2·10 -3 weber = For the airgap, we can use the reluctance of the gap to find the airgap mmf The core mmf can be found from the magnetic field intensity that was obtained from the
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University-West Lafayette.

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Mathcad - 04-07 - Problem 4-7: Solution Givens: R := 65 ohm...

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