exam 5 solution - ECE321. Spring 2009. Exam 5 Solution...

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ECE321. Spring 2009. Exam 5 Solution Outline Problem 1 w 0.01 = g 0.001 = w s 0.05 = N 100 = d s 0.02 = μ r 1000 = d 0.05 = μ 0 4 π 10 7 - = Step 1: Lets compute how much flux we are talking about (5 pts) B 1.2 = Φ B w d = Φ 6 10 4 - × = Step 2: Lets find the reluctance of the MEC (10 pts) R gap g d w μ 0 = R gap 1.59155 10 6 × = R I w s w 2 + w 2 + d w μ 0 μ r = R I 9.5493 10 4 × = R U 2 d s w 2 + w s + w + d w μ 0 μ r = R U 1.7507 10 5 × = R total R U R I + 2 R gap + = R total 3.45366 10 6 × =
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Step 3: Let's find the current (5 pts) Φ N i R total = i Φ R total N = i 20.72197 = Problem 2 Step 1: Let's find the co-energy. (10 pts) Step 1a: Let's bring up i1 with i2=0 W c1 0 i 1 i 1 2 i 1 5 3 x 2 + 1 e 2 - i 1 - + d = W c1 i 1 2 5 3 x 2 + i 1 1 2 e 2 - i 1 + + = evaluated between 0 and i1 W c1 i 1 2 5 3 x 2 + i 1 1 2 e 2 - i 1 1 - + + = Step 1b: Let's bring up i2 with i1 fixed
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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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exam 5 solution - ECE321. Spring 2009. Exam 5 Solution...

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