exam3asolution - EE321 Spring 2004 Exam 3a Solution Problem...

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EE321 Spring 2004 Exam 3a Solution Problem 1 N as 10 5 5 10 5 51 05 5 10 5 5 () T := By inspection N sslt 12 := P4 := Now we have Note: array indices have been changed to start at zero W as 1 1 2 1 N sslt P i N as i = := W as 1 5 = Next R2N sslt .. := W as R W as R1 N as := W as T 123456789 1 0 1 1 1 2 1 5 -5 -10 -5 5 10 5 -5 -10 -5 5 10 =
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500 µ 0 g 0.314 = B 500 µ 0 g sin 20 t 8 φ s + () = F s 500 sin 20 t 8 φ s + = g 0.002 := i bs 10 cos 20 t = i as 10 sin 20 t = w b 50 sin 8 φ s = w as 50 cos 8 φ s = Problem 3 L asbs 2.462 10 3 × = L asbs rL ⋅µ 0 g 0 2 π φ s w as φ s w bs φ s d := w bs φ s 12 cos 4 φ s π 6 := w as φ s 12 cos 4 φ s := g 0.002 := L 0.2 := µ 0 4 π 10 7 := r 0.05 := Problem 2
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Since iq is zero v q r s i q ⋅ω r L ss i d r λ m + = Next i d v d r s := Since iq is zero v d r s i d r L ss i q = Now i q 0 := The load torque is zero. Therefore .
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This note was uploaded on 02/19/2012 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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exam3asolution - EE321 Spring 2004 Exam 3a Solution Problem...

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