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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 = ℓ 1 μ c S c = . 64 5000 × μ o × 10 − 3 = 1 . 0186 × 10 5 H − 1 (1) R 2 = . 24- . 002 5000 × μ o × 10 − 3 = 3 . 788 × 10 4 H − 1 (2) R g = . 002 μ 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) φ 1 = V AB R 1 = 18 . 19 1 . 0186 × 10 5 = 1 . 786 × 10 − 4 Wb = φ 3 (6) φ 2 = φ 1 + φ 3 = 3 . 572...
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