exam2_solution

exam2_solution - EE321 Exam 2 Problem 1 3 O 4 O2 Problem 2...

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EE321 Exam 2 Problem 1 3 O 4 O 2 Problem 2 A. No B. Yes C. Yes D. Yes E. No
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Problem 3 r a 10 10 3 := L af 150 10 3 := i fd_mx 10 := i a_mx 200 := v a_mx 400 := K l 0.01 := Assume we are against the armature voltage and armature current limits: i f v a_mx r a i a_mx L af ω r = T e L af i f i a_mx = T e L af i a_mx v a_mx r a i a_mx L af ω r = T L k l ω r = k l ω r L af i a_mx v a_mx r a i a_mx L af ω r = ω r L af i a_mx v a_mx r a i a_mx L af K l := ω r 2.821 10 3 × = i f v a_mx r a i a_mx L af ω r := i f 940.449 10 3 × = Just out of curiosity
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Problem 4 v t1 100 := i t1 45 := ω r1 50 := v t2 200 := i t2 25 := ω r2 200 := v t1 v t2 i t1 i t2 ω r1 i t1 ω r2 i t2 r s L af = r s L af i t1 i t2 ω r1 i t1 ω r2 i t2 1 v t1 v t2 := r s 296.296 10 3 × = L af 38.519 10 3
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exam2_solution - EE321 Exam 2 Problem 1 3 O 4 O2 Problem 2...

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