HW8_solutions

HW8_solutions - EEE 241 Solution to Assignment #8 P.6-28...

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Unformatted text preview: EEE 241 Solution to Assignment #8 P.6-28 Consider the magnetic circuit in Figure 6-45. A current of 3(A) flows through 200 turns of wire on the center leg. Asssuming the core to have a constant cross-sectional area of 10 3 ( m 2 ) and a relative permeability of 5000: a) Determine the magnetic flux in each leg. b) Determine the magnetic field intensity in each leg of the core and in the air gap. Solution: The equivalent magnetic circuit is as in the figure. R 1 = R 3 = 1 c S c = . 64 5000 o 10 3 = 1 . 0186 10 5 H 1 (1) R 2 = . 24- . 002 5000 o 10 3 = 3 . 788 10 4 H 1 (2) R g = . 002 o 10 3 = 1 . 5915 10 6 H 1 (3) Using the node equation, we have parenleftBigg 1 R 1 + 1 R 3 + 1 R 2 + R g parenrightBigg V AB = E s R 2 + R g (4) where E s = NI = 200 3 A . Thus V AB = E s / ( R 2 + R g ) 1 R 1 + 1 R 3 + 1 R 2 + R g = 600 / 1 . 629 10 6 2 1 . 0186 10 5 + 1 1 . 629 10 6 = 18 . 19( A ) (5) The magnetic flux on arm 1 is a) 1 = V AB R 1 = 18 . 19 1 . 0186 10 5 = 1 . 786 10 4 Wb = 3 (6) 2 = 1 + 3 = 3 . 572...
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This note was uploaded on 10/15/2008 for the course EEE 241 taught by Professor Ayanar during the Spring '07 term at ASU.

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HW8_solutions - EEE 241 Solution to Assignment #8 P.6-28...

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