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exam4_solution

# exam4_solution - EE321 Exam 4 Solution Outline Problem 1 By...

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EE321 Exam 4 Solution Outline Problem 1 By inspection, the secondary to primary turns ratio is 10. N sp 10 = L m 500 = Referred to secondary r s 2 = From secondary r p 1 N sp 2 = r p 100 10 0 × = From secondary The leakage inductances are zero. Problem 2 L ls 0 = r s 0 = L lr 2.65 10 3 - = L M 20.1 10 3 - = r r 41.3 10 3 - = P 4 = N 3 = T e 100 = L rr L M L lr + = L rr 22.75 10 3 - × = ω s r r L rr = ω s 1.815 10 0 × =

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I s 2 T e r r 2 ω s L rr ( 29 2 + N P ω s L M 2 r r = I s 43.325 10 0 × = ω r 2000 2 π 60 P 2 = ω r 418.879 10 0 × = ω e ω r ω s + = ω e 420.694 10 0 × = S ω e ω r - ω e = S 4.315 10 3 - × = Z r s j ω e L ls + 1 1 j L M ω e 1 r r S j ω e L lr + + + = v as Z I s = v as 3 451.72 10 0 × = This is the required line-to-line voltage (rms) Problem 3 w ar 50 cos 4 ϕ sm ( 29 = w br 50 sin 4 ϕ rm ( 29 = i ar 20 cos 100t ( ) = i br 20 sin 100 t ( ) = Thus F r 1000 cos 4 ϕ rm 100t - ( 29 =
Thus the rotor mmf is moving at 25 rad/s in the counterclockwise direction, relative to the rotor.

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exam4_solution - EE321 Exam 4 Solution Outline Problem 1 By...

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