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where
R
is the resistance,
L
is the inductance,
C
is the capacitance, and
ω
d
is the angular
frequency. To find the maximum, set the derivative with respect to
ω
d
equal to zero and solve
for
ω
d
. The derivative is
dI
dω
d
=
−E
m
±
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
²
−
3
/
2
³
ω
d
L
−
1
ω
d
C
´³
L
+
1
ω
2
d
C
´
.
The only factor that can equal zero is
ω
d
L
−
(1
/ω
d
C
) and it does for
ω
d
=1
/
√
LC
.F
o
rt
h
e
given circuit,
ω
d
=
1
√
LC
=
1
µ
(1
.
00 H)(20
.
0
×
10
−
6
F)
= 224 rad
/
s
.
(b) For this value of the angular frequency, the impedance is
Z
=
R
and the current amplitude is
I
=
E
m
R
=
30
.
0V
5
.
00
Ω
=6
.
00 A
.
(c) and (d) You want to find the values of
ω
d
for which
I
=
E
m
/
2
R
. This means
E
m
µ
R
2
+(
ω
d
L
−
1
/ω
d
C
)
2
=
E
m
2
R
.
Cancel the factors
E
m
that appear on both sides, square both sides, and set the reciprocals of the
two sides equal to each other to obtain
R
2
+
¶
ω
d
L
−
1
ω
d
C
·
2
=4
R
2
.
Thus
¶
ω
d
L
−
1
ω
d
C
·
2
=3
R
2
.
Now take the square root of both sides and multiply by
ω
d
C
to obtain
ω
2
d
(
LC
)
±
ω
d
¸
√
3
CR
¹
−
1=0
,
where the symbol
±
indicates the two possible signs for the square root. The last equation is a
quadratic equation for
ω
d
. Its solutions are
ω
d
=
±
√
3
CR
±
√
3
C
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 Spring '08
 Reich
 Capacitance, Current, Inductance, Resistance

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