This preview shows page 1. Sign up to view the full content.
37. (a) Now
X
C
= 0, while
R
= 200
Ω
and
X
L
=
ω
L
= 2
π
f
d
L
= 86.7
Ω
remain unchanged.
Therefore, the impedance is
22
2
2
(200
)
(86.7
)
218
.
L
ZR
X
=+
=
Ω
+
Ω
=Ω
(b) The phase angle is, from Eq. 3165,
11
86.7
0
tan
tan
23.4 .
200
LC
XX
R
φ
−−
⎛⎞
−
Ω−
==
=
°
⎜⎟
Ω
⎝⎠
(c) The current amplitude is now found to be
36.0 V
0.165 A .
218
m
I
Z
ε
=
Ω
(d) We first find the voltage amplitudes across the circuit elements:
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details