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45.
(a) For a given amplitude (
E
)
m
of the generator emf, the current amplitude is given by
I
=
(
E
)
m
Z
=
(
E
)
m
p
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
.
We ±nd 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
2
d
C
¸
.
The only factor that can equal zero is
ω
d
L
−
(1
/ω
d
C
); it does so for
ω
d
=1
/
√
LC
=
ω
.F
o
rt
h
i
s
circuit,
ω
d
=
1
√
LC
=
1
p
(1
.
00 H)(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
=
30
.
0V
5
.
00 Ω
=6
.
00 A
.
(c) We want to ±nd the (positive) values of
ω
d
for which
I
=
(
E
)
m
2
R
:
(
E
)
m
p
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
=
(
E
)
m
2
R
.
This may be rearranged to yield
µ
ω
d
L
−
1
ω
d
C
¶
2
=3
R
2
.
Taking the square root of both sides (acknowledging the two
±
roots) and multiplying by
ω
d
C
,we
obtain
ω
2
d
(
LC
)
±
ω
d
³
√
3
CR
´
−
1=0
.
Using the quadratic formula, we ±nd the smallest positive solution
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Current

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