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hw11_solution

# hw11_solution - ECE321 Spring 2009 Homework 11-3-3-3 Ll1:=...

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ECE321 Spring 2009 Homework 11 L l1 1 10 3 - = L l2 0.5 10 3 - = L m 100 10 3 - = ω e 2 π 60 = r 1 2 = r 2 3 = N 12 15 = Problem 49 Z oc r 1 j ω e L l1 + j ω e L m + = Z oc 38.129 = arg Z oc ( 29 180 π 86.993 = Problem 50 Z sc r 1 j ω e L l1 + 1 1 j ω e L m 1 r 2 j ω e L l2 + + + = Z sc 5.016 = arg Z sc ( 29 180 π 9.158 = Problem 51 Z sc 1 N 12 2 r 2 j ω e L l2 + 1 1 j ω e L m 1 r 1 j ω e L l1 ( 29 + + + = Z sc 0.022 = arg Z sc ( 29 180 π 7.637 =

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Problem 52 R load 15 = R load_prime 15 N 12 2 = Referred voltage V 1 100 e j 0 = Z 1 r 1 j ω e L l1 + = Z 2 r 2 j ω e L l2 + = Impedance looking into magnetizing branch and load Z m 1 1 j ω e L m 1 Z 2 R load_prime + + = Phasor voltage across magnizting branch V m Z m Z 1 Z m + V 1 = V 2prime V m R load_prime R load_prime Z 2 + = Phasor representing voltage across load (referred) V 2 V 2prime 1 N 12 = Phasor representing voltage across secondary load t 5 = v 2int 2 V 2 cos ω e t arg V 2 ( 29 + ( 29 = v 2int 9.296 = Instanteous voltage at t=5 s i 2int v 2int R load = i 2int 0.62 =

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hw11_solution - ECE321 Spring 2009 Homework 11-3-3-3 Ll1:=...

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