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exam3_solution

exam3_solution - ECE 321 Exam 3 Solution Spring 2006...

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ECE 321 Exam 3 SolutionSpring 2006 Problem 1 The winding function expresses how many turns of a particular phase are linking the flux at a specific angle. Problem 2 P 6 = W 1 1 2 j 1to N aslt P = N asj = 1 2 j 1to2 = N asj = 1 2 5 5 + ( ) = 5 = W 2 W 1 N 1 = 0 = W 3 W 2 N 2 = 5 = W 4 W 3 N 3 = 0 = etc. W as 5 0 5 0 5 0 5 0 5 0 5 0 ( ) T = Problem 3 P 8 = Note: Check the integral equation below. Some got full credit since this answer can essentially be arrived at by inspection, even though the equation they used might have had errors in it. w as 1 2 0 2 π P φ sm n as d 0 φ sm φ sm n as d = 25 sin 4 φ sm ( ) 25 6 sin 12 φ sm ( ) + = Problem 4 ω rm 2 P ω r = Machines with more poles rotate more slowly (under the same excitation) than otherwise similar machines with fewer poles (and vice versa). Problem 5 F w as i as w bs i bs + = 50 cos 4 φ sm ( ) cos 30t ( ) 50 sin 4 φ sm ( ) sin 30t ( ) + = F 50 cos 4 φ sm 30t ( ) = The maximum MMF occurs at 4 φ sm 30t 0 = Therefore, the speed is t φ sm d d t 7.5t ( ) d d = 7.5 = rad s CCW (since positive)
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Problem 6 The maximum of the MMF is at 4 φ sm 100t 0 = φ sm 100 t 4 = 100 0.1ms 4 = 0.0025 = rad ( ) B µ 0 F g = 4 π 10 7
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