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59. We use the result of problem 54:
P
avg
=
(
E
)
2
m
R
2
Z
2
=
(
E
)
2
m
R
2[
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
]
.
We use the expression
Z
=
p
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
for the impedance in terms of the angular frequency.
(a) Considered as a function of
C
,
P
avg
has its largest value when the factor
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
has
the smallest possible value. This occurs for
ω
d
L
=1
/ω
d
C
,or
C
=
1
ω
2
d
L
=
1
(2
π
)
2
(60
.
0Hz)
2
(60
.
0
×
10
−
3
H)
=1
.
17
×
10
−
4
F
.
The circuit is then at resonance.
(b) In this case, we want
Z
2
to be as large as possible. The impedance becomes large without bound
as
C
becomes very small. Thus, the smallest average power occurs for
C
=0(wh
ichisnotvery
di±erent from a simple open switch).
(c) When
ω
d
L
=1
/ω
d
C
, the expression for the average power becomes
P
avg
=
(
E
)
2
m
2
R
,
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 Fall '08
 SPRUNGER
 Physics

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