Alternating Current
319
(b)
The average power dissipated by the resistor is
2
2
rms
120 V
(350
)
7.32 W.
830
R
PIR
⎛⎞
==
Ω
=
⎜⎟
Ω
⎝⎠
EVALUATE:
Conservation of energy requires that the answers to parts (a) and (b) are equal.
31.27.
IDENTIFY:
The power factor is cos ,
φ
where
is the phase angle in Fig.31.13. The average power is given by
Eq.(31.31). Use the result of part (a) to rewrite this expression.
(a) SET UP:
The phasor diagram is sketched in Figure 31.27.
EXECUTE:
From the diagram
cos
,
R
VI
R
R
ZZ
===
as was to be shown.
Figure 31.27
(b)
rms
av
rms rms
rms rms
rms
cos
.
Z
RV
PV
I
V
I
IR
Z
⎛⎞⎛ ⎞
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
But
2
rms
rms
av
rms
, so
.
Z
V
I
EVALUATE:
In an
L

R

C
circuit, electrical energy is stored and released in the inductor and capacitor but none is
dissipated in either of these circuit elements. The power delivered by the source equals the power dissipated in the
resistor.
31.28.
IDENTIFY
and
SET UP:
av
rms rms
cos .
I
=
rms
rms
.
V
I
Z
=
cos
.
R
Z
=
EXECUTE:
rms
80.0 V
0.762 A.
105
I
Ω
75.0
cos
0.714.
105
Ω
Ω
av
(80.0 V)(0.762 A)(0.714)
43.5 W.
P
=
=
EVALUATE:
Since the average power consumed by the inductor and by the capacitor is zero, we can also
calculate the average power as
22
av
rms
(0.762 A) (75.0
)
43.5 W.
Ω
=
31.29.
IDENTIFY
and
SET UP:
Use the equations of Section 31.3 to calculate
rms
,
and
.
Z
V
The average power
delivered by the source is given by Eq.(31.31) and the average power dissipated in the resistor is
2
rms
EXECUTE:
(a)
2
2 (400 Hz)(0.120 H)
301.6
L
XLf
L
ω
ππ
=
=Ω
6
11
1
54.51
2
2 (400 Hz)(7.3 10 Hz)
C
X
Cf
C
ωπ
π
−
=
=
Ω
×
301.6
54.41
tan
, so
45.8 .
240
LC
XX
R
φφ
−Ω
−
Ω
=
+
°
Ω
The power factor is cos
0.697.
= +
(b)
222
2
(
)
(240 )
(301.6
54.51 )
344
ZRX
X
=+−
=
Ω
+
Ω
−
Ω
(c)
rms
rms
(0.450 A)(344
) 155 V
Z
Ω
=
(d)
av
rms
rms
cos
(0.450 A)(155 V)(0.697)
48.6 W
PI
V
=
(e)
av
rms
(0.450 A) (240
)
48.6 W
Ω
=
EVALUATE:
The average electrical power delivered by the source equals the average electrical power consumed
in the resistor.
(f)
All the energy stored in the capacitor during one cycle of the current is released back to the circuit in another
part of the cycle. There is no net dissipation of energy in the capacitor.
(g)
The answer is the same as for the capacitor. Energy is repeatedly being stored and released in the inductor, but
no net energy is dissipated there.
31.30.
IDENTIFY:
The angular frequency and the capacitance can be used to calculate the reactance
C
X
of the
capacitor. The angular frequency and the inductance can be used to calculate the reactance
L
X
of the inductor.
Calculate the phase angle
and then the power factor is cos .
Calculate the impedance of the circuit and then the
rms current in the circuit. The average power is
av
rms rms
cos .
I
=
On the average no power is consumed in the
capacitor or the inductor, it is all consumed in the resistor.
SET UP:
The source has rms voltage
rms
45 V
31.8 V.
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 Spring '06
 Buchler
 Physics, Current, Conservation Of Energy, Energy, Power, Electrical impedance, Pav, resonance frequency, L. Pav

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