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 coenergy. (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
W
c2
0
i
2
i
1
4 i
2
⋅
10
3
x
2
+
1
e
2 i
1
⋅
4 i
2
⋅
+
(
29


⋅
+
⌠
⌡
d
=
W
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 Spring '08
 Staff
 Flux, Cos, Akhenaten, −sin θr, kv⋅ ωr ra

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