314
Chapter 31
31.12.
IDENTIFY:
Compare
C
v
that is given in the problem to the general form
sin
C
I
vt
C
ω
=
and determine
.
SET UP:
1
C
X
C
=
.
R
vi
R
=
and
cos .
iI
=
EXECUTE:
(a)
6
11
1736
(120 rad s)(4.80 10
F)
C
X
C
−
==
=
Ω
×
(b)
3
7.60 V
4.378 10 A
1736
C
C
V
I
X
−
=
×
Ω
and
3
cos
(4.378 10 A)cos[(120 rad/s) ].
t
t
−
×
Then
3
(4.38 10
A)(250
)cos((120 rad s) )
(1.10 V)cos((120 rad s) ).
R
R
t
t
−
×
Ω
=
EVALUATE:
The voltage across the resistor has a different phase than the voltage across the capacitor.
31.13.
IDENTIFY
and
SET UP:
The voltage and current for a resistor are related by
.
R
R
=
Deduce the frequency of the
voltage and use this in Eq.(31.12) to calculate the inductive reactance. Eq.(31.10) gives the voltage across the
inductor.
EXECUTE:
(a)
(3.80 V)cos[(720 rad/s) ]
R
=
3.80 V
, so
cos[(720 rad/s) ]
(0.0253 A)cos[(720 rad/s) ]
150
R
R
v
R i
t
t
R
⎛⎞
=
=
⎜⎟
Ω
⎝⎠
(b)
L
XL
=
720 rad/s,
0.250 H, so
(720 rad/s)(0.250 H) 180
L
LX
L
=
=
=
Ω
(c)
If
cos
t
=
then
cos(
90 )
LL
vV
t
=+
°
(from Eq.31.10).
(0.02533 A)(180
)
4.56 V
VI
L
I
X
=
Ω
=
(4.56 V)cos[(720 rad/s)
90 ]
L
°
But cos(
90 )
sin
aa
+°
=
−
(Appendix B), so
(4.56 V)sin[(720 rad/s) ].
L
= −
EVALUATE:
The current is the same in the resistor and inductor and the voltages are 90
°
out of phase, with the
voltage across the inductor leading.
31.14.
IDENTIFY:
Calculate the reactance of the inductor and of the capacitor. Calculate the impedance and use that
result to calculate the current amplitude.
SET UP:
With no capacitor,
22
L
Z
RX
and
tan
.
L
X
R
φ
=
.
L
=
.
V
I
Z
=
L
L
X
=
and
.
R
R
=
For an
inductor, the voltage leads the current.
EXECUTE:
(a)
(250 rad/s)(0.400 H) 100
.
L
=
Ω
(200 )
(100 )
224 .
Z
=
Ω+
Ω=
Ω
(b)
30.0 V
0.134 A
224
V
I
Z
=
Ω
(c)
(0.134 A)(200
)
26.8 V.
R
R
Ω
=
(0.134 A)(100 ) 13.4 V.
X
=
=Ω
=
(d)
100
tan
200
L
X
R
Ω
Ω
and
26.6
°.
Since
is positive, the source voltage leads the current.
(e)
The phasor diagram is sketched in Figure 31.14.
EVALUATE:
Note that
R
L
VV
+
is greater than
V
. The loop rule is satisfied at each instance of time but the voltages
across
R
and
L
reach their maxima at different times.
Figure 31.14
31.15.
IDENTIFY:
()
R
is given by Eq.(31.8).
( )
L
is given by Eq.(31.10).
SET UP:
From Exercise 31.14,
30.0 V
V
=
,
26.8 V
R
V
=
,
13.4 V
L
V
=
and
26.6
=
°
.
EXECUTE:
(a)
The graph is given in Figure 31.15.
(b)
The different voltages are
(30.0 V)cos(250
26.6 ),
(26.8 V)cos(250 ),
R
v
t
°
=
(13.4 V)cos(250
90 ).
L
°
At
20 ms:
t
=
20.5 V,
7.60 V,
12.85 V.
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 Spring '06
 Buchler
 Physics, Alternating Current, Cos, Electrical resistance, Electrical impedance

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