37. (a) Now
X
C
= 0, while
R
= 160 Ω and
X
L
=86
.
7 Ω remain unchanged. Therefore, the impedance is
Z
=
q
R
2
+
X
2
L
=
p
(160 Ω)
2
+(86
.
7Ω)
2
= 182 Ω
.
The current amplitude is now found to be
I
=
E
m
Z
=
36
.
0V
182 Ω
=0
.
198 A
.
The phase angle is, from Eq. 3365,
φ
=tan
−
1
µ
X
L
−
X
C
R
¶
=tan
−
1
µ
86
.
7Ω
−
0
160 Ω
¶
=28
.
5
◦
.
(b) We Frst Fnd the voltage amplitudes across the circuit elements:
V
R
=
IR
=(0
.
198 A)(160 Ω)
≈
32 V
V
L
=
IX
L
=(0
.
216 A)(86
.
7Ω)
≈
17 V
This is an inductive circuit, so
E
m
leads
I
. The phasor diagram is drawn to scale below.
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 Fall '08
 SPRUNGER
 Physics, Current, The Circuit, Phase angle, Phasor Diagram, current amplitude, voltage amplitudes

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