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()
2
avg
30.0V
90.0W.
2 5.00
P
==
Ω
(d) At maximum power, the reactances are equal:
X
L
= X
C
. The phase angle
φ
in this case
may be found from
tan
,
=
−
=
XX
R
LC
0
which implies
= 0
°
.
(e) At maximum power, the power factor is cos
= cos 0° = 1.
(f) The minimum average power is
P
avg
= 0 (as it would be for an open switch).
(g) On the other hand, at minimum power
X
C
∝
1/
C
is infinite, which leads us to set
tan
=−∞
. In this case, we conclude that
= –90°.
(h) At minimum power, the power factor is cos
= cos(–90°) = 0.
59. We shall use
22
avg
2
2
2
.
2
21
/
mm
dd
RR
P
Z
RLC
εε
ωω
⎡
⎤
+−
⎣
⎦
where
ZR
L
C
=+
−
2
2
1
/
b
g
is the impedance.
(a) Considered as a function of
C, P
avg
has its largest value when the factor
2
2
1/
R
has the smallest possible value. This occurs for
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 Spring '08
 Any
 Physics

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