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35. (a) Now
X
L
= 0, while
R
= 200
Ω
and
X
C
= 1/2
π
f
d
C
= 177
Ω.
Therefore, the
impedance is
22
2
2
(200 )
(177
)
267 .
C
ZR
X
=+
=
Ω
+Ω
=
Ω
(b) The phase angle is
11
0 177
=tan
tan
41.5
200
LC
XX
R
φ
−−
⎛⎞
−
−Ω
=
=−
°
⎜⎟
Ω
⎝⎠
(c) The current amplitude is
36.0 V
0.135 A .
267
m
I
Z
ε
==
=
Ω
(d) We first find the voltage amplitudes across the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Current

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