hw12_solution

hw12_solution - EE321 Spring 2009 Homework #12 Problem 54...

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EE321 Spring 2009 Homework #12 Problem 54 We have: i as 50 sin 200 t ( ) = i bs 50 - cos 200 t ( ) = Now, we will also have w as W cos 2 ϕ s ( 29 = w bs W sin 2 ϕ s ( 29 = Thus, the stator MMF is F s 50 W sin 200 t 2 ϕ s - ( 29 = Whereupon the stator MMF is moving at 100 rad/s in the CCW direction (since it is in the positive direction, and CCW is assumed direction for positive. This, of course, is relative to the stator, but what does the rotor see ? ω rm 500 - 2 π 60 = ω rm 52.36 - = The speed of the stator MMF relative to the rotor in the CCW direction is 100 ω rm - 152.36 =
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Problem 55 L asar r L μ 0 g 2 P 2 N s N r 0 2 π ϕ sm cos P 2 ϕ sm sin P 2 ϕ sm P 2 θ rm - d = L asar r L μ 0 g 2 P 2 N s N r 0 2 π ϕ sm 1 2 sin P ϕ sm θ r - ( 29 1 2 sin θ r - ( 29 + d
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This note was uploaded on 10/07/2010 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue.

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hw12_solution - EE321 Spring 2009 Homework #12 Problem 54...

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