p33_045.pdf
(
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)
Pobierz
Chapter 33 - 33.45
45. (a) For a given amplitude (
E
)
m
of the generator emf, the current amplitude is given by
I
=
(
E
)
m
Z
=
(
E
)
m
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
.
We find the maximum by setting the derivative with respect to
ω
d
equal to zero:
dI
dω
d
=
−
(
E
)
m
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
−
3
/
2
ω
d
L
−
1
ω
d
C
L
+
1
ω
d
C
.
(1
/ω
d
C
); it does so for
ω
d
=1
/
√
LC
=
ω
.Forthis
The only factor that can equal zero is
ω
d
L
−
circuit,
1
√
LC
=
1
(1
.
00H)(20
.
0
10
−
6
F)
= 224 rad
/
s
.
(b) When
ω
d
=
ω
, the impedance is
Z
=
R
, and the current amplitude is
I
=
(
E
)
m
R
ω
d
=
×
=
30
.
0V
5
.
00Ω
=6
.
00 A
.
(c) We want to find the (positive) values of
ω
d
for which
I
=
(
E
)
m
2
R
:
(
E
)
m
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
=
(
E
)
m
2
R
.
This may be rearranged to yield
ω
d
L
2
1
ω
d
C
−
=3
R
2
.
Taking the square root of both sides (acknowledging the two
±
roots) and multiplying by
ω
d
C
,we
obtain
ω
d
√
3
CR
ω
d
(
LC
)
±
−
1=0
.
Using the quadratic formula, we find the smallest positive solution
ω
2
=
−
√
3
CR
+
√
3
C
2
R
2
+4
LC
2
LC
=
−
√
3(20
.
0
10
−
6
F)(5
.
00Ω)
2(1
.
00H)(20
.
0
×
×
10
−
6
F)
+
3(20
.
0
×
10
−
6
F)
2
(5
.
00Ω)
2
+4(1
.
00H)(20
.
0
×
10
−
6
F)
2(1
.
00H)(20
.
0
×
10
−
6
F)
= 219 rad
/
s
,
and the largest positive solution
ω
1
=
+
√
3
CR
+
√
3
C
2
R
2
+4
LC
2
LC
=
10
−
6
F)(5
.
00Ω)
2(1
.
00H)(20
.
0
×
×
10
−
6
F)
+
3(20
.
0
×
10
−
6
F)
2
(5
.
00Ω)
2
+4(1
.
00H)(20
.
0
×
10
−
6
F)
2(1
.
00H)(20
.
0
×
10
−
6
F)
= 228 rad
/
s
.
(d) The fractional width is
ω
1
−
ω
2
=
228rad
/
s
219rad
/
s
224rad
/
s
−
=0
.
04
.
ω
0
+
√
3(20
.
0
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(59 KB)
p33_011.pdf
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p33_005.pdf
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p33_018.pdf
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p33_017.pdf
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Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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