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ECE321.
Fall 2008
Exam 1 Solution Outline
Handy Stuff
cm
110
2
−
⋅
:=
mm
1.0 10
3
−
⋅
:=
μ
0
4
π
⋅
10
7
−
⋅
:=
Problem 1
H
y
20
:=
A
100 10
⋅
:=
B
y
H
y
μ
0
⋅
:=
Φ
B
y
A
⋅
:=
Only the ycomponent couples the loop
λΦ
:=
Single turn
λ
0.02513
=
Problem 2
w1
c
m
⋅
:=
d
s
2cm
⋅
:=
g
0.1 mm
⋅
:=
w
s
5cm
⋅
:=
d5
c
m
⋅
:=
N
100
:=
B
sat
1
:=
Point where saturation occurs
Now let's compute some reluctances.
For flux densities that are not saturated
u
1500
μ
0
⋅
:=
R
12
w
s
w
+
wd
⋅
u
⋅
:=
R
56
R
12
:=
R
23
w
2w
⋅
d
⋅
u
⋅
:=
R
81
R
23
:=
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34
g
wd
⋅μ
0
⋅
:=
R
78
R
34
:=
R
45
d
s
w
2
+
⋅
u
⋅
:=
R
67
R
45
:=
R
eff
R
12
R
23
+
R
34
+
R
45
+
R
56
+
R
67
+
R
78
+
R
81
+
:=
R
eff
5.09296
10
5
×
=
OK.
Now,
the cross sectional area we have is
A
⋅
:=
The flux where we saturate is
Φ
sat
B
sat
A
⋅
:=
The current to achieve this flux
i
sat
Φ
sat
N
R
eff
⋅
:=
i
sat
2.54648
=
Problem 3
L
N
2
R
eff
:=
L
0.01963
=
Inductance of UI Core with standard gap
Problem 4
W
f
1
3
λ
3
5x
+
⋅
=
f
e
1
−
3
λ
3
+
()
2
⋅
=
Now
i5
:=
x2
:=
λ
i5 x
+
⋅
:=
λ
5.91608
=
f
e
1
3
λ
3
+
2
⋅
:=
One negative for differentiation, one for partial derivative
f
e
1.40859
=
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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 UniversityWest Lafayette.
 Spring '08
 Staff

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