hw12_solution

hw12_solution - ECE321 Spring 2008 Homework 12 3 Ll1 := 1...

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ECE321 Spring 2008 Homework 12 L l1 110 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 10 := Problem 51 Z oc r 1 j ω e L l1 + j ω e L m + := Z oc 38.129 = arg Z oc () 180 π 86.993 = Problem 52 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 180 π 9.158 = Problem 53 Z sc 1 N 12 2 r 2 j ω e L l2 + 1 1 j ω e L m 1 r 2 j ω e L l2 + + + := Z sc 0.06 = arg Z sc 180 π 5.85 =
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Problem 54 R load 15 := R load_prime 15 N 12 2 := Referred voltage V 1 100 e j0 := 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 loa t5 := v 2int 2 V 2 cos ω e t arg V 2 () + := v 2int 13.918 = Instanteous voltage at t=5 s i 2int v 2int R load := i 2int 0.928
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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.

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hw12_solution - ECE321 Spring 2008 Homework 12 3 Ll1 := 1...

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