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3114
Chapter 31
(b) EXECUTE:
power factor cos
φ
equals 1 so
0
=
and
.
CL
XX
=
Calculate the present value of
−
to
see how much more
L
X
is needed:
cos
(60.0
)(0.720)
43.2
RZ
=
=Ω
=
Ω
tan
so
tan
LC
XXR
R
φφ
−
=−
=
cos
0.720 gives
43.95
==
−
°
(
is negative since the voltage lags the current)
Then
tan
(43.2 )tan( 43.95 )
41.64 .
−=
=
Ω
−
°
=
−
Ω
Therefore need to add 41.64
of
.
L
X
Ω
41.64
2
and
0.133 H,
2
2 (50.0 Hz)
L
L
X
XLf
L
L
f
ωπ
ππ
Ω
=
=
=
amount of inductance to add.
EVALUATE:
From the information given we can&t calculate the original value of
L
in the circuit, just how much
to add. When this
L
is added the current in the circuit will increase.
31.46.
IDENTIFY:
Use
rms
rms
VI
Z
=
to calculate
Z
and then find
R.
2
av
rms
PIR
=
SET UP:
50.0
C
X
EXECUTE:
()
2
22
2
rms
rms
240 V
80.0
50.0
.
3.00 A
C
V
ZR
X
R
I
=+
Ω
Thus,
80.0
50.0
62.4
.
R
−
Ω
=
Ω
The average power supplied to this circuit is equal to the power dissipated
by the resistor, which is
2
2
rms
3.00 A
62.4
562 W.
Ω
=
EVALUATE:
50.0
tan
62.4
R
−−
Ω
Ω
and
38.7
°.
av
rms rms
cos
(240 V)(3.00 A)cos( 38.7 )
562 W,
PV
I
−
=
°
which checks.
31.47.
IDENTIFY:
The voltage and current amplitudes are the maximum values of these quantities, not necessarily the
instantaneous values.
SET UP:
The voltage amplitudes are
V
R
= RI
,
V
L
= X
L
I
, and
V
C
= X
C
I
, where
I = V/Z
and
2
2
1
.
L
C
ω
⎛⎞
=+−
⎜⎟
⎝⎠
EXECUTE:
(a)
= 2
π
f
= 2
π
(1250 Hz) = 7854 rad/s. Carrying extra figures in the calculator gives
X
L
=
L
=
(7854 rad/s)(3.50 mH) = 27.5
Ω
;
X
C
=
1/
C
= 1/[(7854 rad/s)(10.0
&
F)] = 12.7
Ω
;
ZRX
X
=
(50.0
)
(27.5
12.7
)
Ω+
Ω
−
Ω
= 52.1
Ω
;
I = V/Z =
(60.0 V)/(52.1
Ω
) = 1.15 A;
V
R
= RI =
(50.0
Ω
)(1.15 A) = 57.5 V;
V
L
= X
L
I
= (27.5
Ω
)(1.15 A) = 31.6 V;
V
C
= X
C
I
= (12.7
Ω
)(1.15 A) = 14.7 V.
The voltage amplitudes can add to more than 60.0 V because these voltages do not all occur at the same instant of
time. At any instant, the instantaneous voltages all add to 60.0 V.
(b)
All of them will change because they all depend on
.
X
L
=
L
will double to 55.0
Ω
, and
X
C
=
1/
C
will
decrease by half to 6.35
Ω
. Therefore
(50.0
)
(55.0
6.35
)
Z
=
Ω
−
Ω
= 69.8
Ω
;
I = V/Z
= (60.0 V)/(69.8
Ω
) =
0.860 A;
V
R
= IR =
(0.860 A)(50.0
Ω
) = 43.0 V;
V
L
= IX
L
= (0.860 A)(55.0
Ω
) = 47.3 V;
V
C
= IX
C
= (0.860 A)(6.35
Ω
) = 5.47 V.
EVALUATE:
The new amplitudes in part (b) are not simple multiples of the values in part (a) because the
impedance and reactances are not all the same simple multiple of the angular frequency.
31.48.
IDENTIFY
and
SET UP:
1
.
C
X
C
=
.
L
XL
=
EXECUTE:
(a)
1
1
1
L
C
=
and
2
1
1
.
LC
=
At angular frequency
2
,
2
21
2
1
(2
)
4.
1/
L
C
LC
XC
ωω
=
=
.
L
C
>
(b)
At angular frequency
3
,
2
2
1
3
2
1
11
.
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 Spring '06
 Buchler
 Physics, Power

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