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ECE430_Prob3_9

# ECE430_Prob3_9 - used are Loop i 1 v ab = L 1 d i 1-i 2...

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ECE 430 HW #4 Solutions Fall 2010 Problem 3.9 Figure 1: Problem 3.9 (a) Figure 2: Problem 3.9 (b) (a) Using the loops deﬁned in Figure 1, the following equations can be derived for the left and right current loops: Loop i 1 : v ab = L 1 d ( i 1 - i 2 ) dt + M di 2 dt Loop i 2 : 0 = L 2 di 2 dt + M d ( i 1 - i 2 ) dt - " L 1 d ( i 1 - i 2 ) dt + M di 2 dt # v ab = L 2 di 2 dt + M d ( i 1 - i 2 ) dt By subtracting the second loop equation from the ﬁrst: 0 = ( L 1 - M ) di 1 dt + ( M - L 1 - L 2 + M ) di 2 dt di 2 dt = ( L 1 - M ( L 1 + L 2 - 2 M ) di 1 dt Plugging this into the loop equation for i 2 : v ab = M d 1 dt + ( L 2 - M ) ( L 1 - M ( L 1 + L 2 - 2 M ) di 1 dt = L 1 L 2 - M 2 L 1 + L 2 - 2 M ! di 1 dt = L eq di 1 dt X 1

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(b) Using Figure 2, and applying a similar procedure as part (a), two loop equations that may be
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Unformatted text preview: used are: Loop i 1 : v ab = L 1 d ( i 1-i 2 ) dt-M di 2 dt Loop i 2 : = L 2 di 2 dt-M d ( i 1-i 2 ) dt-" L 1 d ( i 1-i 2 ) dt-M di 2 dt # v ab = L 2 di 2 dt-M d ( i 1-i 2 ) dt By applying the same elimination procedure as before, the equivalent inductance can be de-termined: v ab = L 1 L 2-M 2 L 1 + L 2 + 2 M ! di 1 dt = L eq di 1 dt 2...
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ECE430_Prob3_9 - used are Loop i 1 v ab = L 1 d i 1-i 2...

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