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38. (a) Now
X
L
= 0, while
R
= 160 Ω and
X
C
= 177 Ω remain as shown in the Sample Problem. Therefore,
the impedance, current amplitude and phase angle are
Z
=
q
R
2
+
X
2
C
=
p
(160 Ω)
2
+ (177 Ω)
2
= 239 Ω
,
I
=
E
m
Z
=
36
.
0V
239 Ω
=0
.
151 A
,
φ
=t
a
n
−
1
µ
X
L
−
X
C
R
¶
=tan
−
1
µ
0
−
177 Ω
160 Ω
¶
=
−
47
.
9
◦
.
(b) We Frst Fnd the voltage amplitudes across the circuit elements:
V
R
=
IR
=(0
.
151 A)(160 Ω)
≈
24 V
V
C
=
IX
C
=(0
.
151 A)(177 Ω)
≈
27 V
The circuit is capacitive, so
I
leads
E
m
. The phasor diagram is drawn to scale below.
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 Fall '08
 SPRUNGER
 Physics, Current

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