hw10_solution

hw10_solution - EE321 Spring 2008 HW#10 Problem 42 The...

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EE321 Spring 2008 HW#10 Problem 42 The winding functions are w as 50 sin 2 φ s () = w bs 50 cos 2 φ s = but now the currents are i as 5 cos 400t = i bs 4 sin 400t = So the total MMF is given by F 250 cos 400 t sin 2 φ s 200 sin 400 t ( ) cos 2 φ s = F 125 sin 2 φ s 400t + sin 2 φ s 400t + 100 sin 400t 2 φ s + sin 400t 2 φ s + = F 225 sin 2 φ s 400t + 25 sin 2 φ s 400t = The first, larger term of the MMF is moving at a speed of 200 rad/s in the CW direction, while the second, smaller term is moving at the same speed in the CCW direction. Problem 43 Starting with 4.5-16 n as φ sm N s sin P φ sm 2 N s3 sin 3 P φ sm 2 + = w as φ sm 1 2 0 2 π P φ sm n as φ sm d 0 φ sm φ sm n as φ sm d =

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w as φ sm () 1 2 2 N s P cos P φ sm 2 2N s3 3P cos 3 P φ sm 2 φ sm 0 = 2 π P ,
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This note was uploaded on 09/22/2011 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University.

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hw10_solution - EE321 Spring 2008 HW#10 Problem 42 The...

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