Mathcad - 04-13 - respective flux densities 0 0.2 0.4 0.6...

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I coil 1.84 A = I coil F tot N := The current is the mmf divided by the number of turns F tot 552 turns A = F tot F cs F ci + := The total mmf is the sum of the two F ci 160 A = F cs 392 A = F ci l ci H ci := F cs l cs H cs := We can find the mmf required for each portion of the core by multipying field intensity times length H ci 800 A turns m := H cs 560 A turns m := From the B-H curves, we can obtain the field intensity for the two materials at their
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Unformatted text preview: respective flux densities 0 0.2 0.4 0.6 0.8 1 1.2 1.4 600 800 1000 Cast Iron Cast Steel Sheet Steel H in At/m 0 200 400 B ci 0.3 T := B cs 0.75 T := N 300 turns := l ci 0.2 m := l cs 0.7 m := Problem 4-13: Solution turns 1...
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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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