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Homework 7
chapter 33: 21, 25, 49, 65
Problem 33.21
An RLC circuit consists of a 150-Ω resistor, a 21-μF capacitor and a 460-mH
inductor, connected in series with a 120-V, 60-Hz power supply. (a) What is the
phase angle between the current and the applied voltage? (b) Which reaches its
maximum earlier, the current or the voltage?
V
R
V
L
V
C
VL
I
ϕ
VC
a)
is
I
VR
V
From the definition, the inductive reactance of the inductor
XL = ωL = 2 π ⋅ 60 Hz ⋅ 0. 46 H = 173Ω
From the definition, the capacitive reactance of the
capacitor is
1
XC = 1 =
= 126Ω
ωC 2 π ⋅ 60 Hz ⋅ 21 ⋅10− 6 F
For the series connection, the instantaneous voltage across
the system is equal to the sum of voltage across each element.
Using the phasor diagram, we can find that the phase angle
between the voltage (across the system) and the current
(through the system) is:
ϕ = arctan XL − XC = arctan 174Ω − 126Ω = 17. 4o
R
150Ω
2
b) From the definition of phase angle, we find that the initial
phase of the voltage is greater than the initial phase of the
current.
δV = δI + ϕ
The voltage leads the current.
3
Problem 33.25
Draw to scale a phasor diagram showing Z, XL, XC, and ϕ for an ac series circuit
for which R = 300 Ω, C = 11 μF, L = 0.2 H, and f = 500/π Hz.
Before drawing the diagram, I will find the complex
impedance of each element. The complex impedance of a
resistor is equal to its resistance.
ZR = R = 300Ω
It is therefore a real number which I marked in the complex
plane on the Re axis.
The complex impedance of an inductor is an imaginary
number dependent on the angular frequency of the current and
the inductance of the inductor
500 −1
Z L = iωL = 2π ⋅
s ⋅ 0.2H ⋅ i = 200iΩ
π
This number is on the Im axis. Finally, the complex impedance
of the capacitor is also an imaginary number dependent on the
angular frequency of the current and the capacitance of the
capacitor
−i
−i
ZC =
=
= −90.9iΩ
500
ωC 2 π ⋅
s −1 ⋅ 11 ⋅ 10− 6 F
π
This number is also on the Im axis.
Since the elements are connected in series the equivalent
complex impedance is equal to the sum of the complex
impedance of each element.
Z = Z R + Z L + Z C = 300Ω + 200iΩ − 90.9iΩ = (300 + 109.1i )Ω
The complex impedance of each element is indicated in the
diagram below. One tic corresponds to 100Ω.
4
Im
Z
L
Z
ϕ
Z
Re
Z
C
R
The diagram also illustrates how to find the equivalent
complex impedance. Note that the phase of the complex
impedance of the resistor is 0, the phase of the complex
impedance of the capacitor is -90° and the phase of the complex
impedance of the inductor is 90°.
Supplement. Although it is not asked in the problem, let's see
in scale the relationship between the current (common for all
elements) and the voltages in the circuit. Let's assume that the
current has value of 20 mA and at a certain instant t it is
represented by the phasor indicated in the following figure. In
the figure one tic corresponds to 10 mA.
The complex voltage
across the resistor can be
found by multiplying the
complex current (through the
resistor) by the complex
impedance of the resistor
V R (t ) = I (t ) ⋅ Z R
Im
V
VL
V
Z
R
L
I
ϕ
Z
ϕ
Re
Z R
Z C
VC
Consistent with the given
values, the absolute value of
this voltage is
5
VR,m = Im ⋅ R = 20mA ⋅ 300Ω = 6V
Since the phase of the complex impedance of the resistor is zero,
the phase of the voltage across the resistor agrees with the
phase of the current. In the figure, one tic corresponds to 1V
The complex voltage across the inductor can be found by
multiplying the complex current (through the inductor) by the
complex impedance of the inductor
V L (t ) = I (t ) ⋅ Z L
Consistent with the given values, the absolute value of this
voltage is
VL,m = Im ⋅ XL = 20mA ⋅ 200Ω = 4V
Since the phase of the complex impedance of the inductor is 90°,
the voltage across the inductor leads with the current by a 90°
phase angle.
Finally, the complex voltage across the capacitor can be
found by multiplying the complex current (through the
capacitor) by the complex impedance of the capacitor
VC ( t ) = I ( t ) ⋅ Z C
Consistent with the given values, the absolute value of this
voltage is
VC,m = Im ⋅ XC = 20mA ⋅ 91Ω = 1.8V
Since the phase of the complex impedance of the capacitor is
-90°, the voltage across the capacitor lags behind the current by
90°.
If we want to find the complex voltage across the entire
system, we can add the complex voltages across all three
elements (they are connected in series). I indicated this
operation on the phasor diagram. That voltage can also be
6
found by multiplying the current by the equivalent impedance of
the system. Notice that the phase difference between the voltage
across the system and the current through the system is equal to
the phase angle of the system (the phase of the equivalent
impedance). The absolute value of the voltage is
Vm = I m ⋅ Z = 20mA ⋅
(300Ω)2 + (109Ω)2
7
≈ 6.4V
Problem 33.49
The resistor in the Figure P33.49 represents the midrange speaker in a
three-speaker system. Assume its resistance to be constant at 8 Ω. The source
represent an audio amplifier producing signals of uniform amplitude
ΔVin = 10V al all frequencies. The inductor and capacitor are to function as a
bandpass filter with ΔVout /ΔVin = ½ at 200 Hz and 4,000 Hz. (a) Determine
the required values of L and C. (b) Find the maximum value of the ratio
ΔVout/ΔVin. (c) Find the frequency f0 at which the ratio has its maximum
value. (d) Find the phase shift between ΔVin and ΔVout at 200 Hz, at f0, and at
4000 Hz. (e) Find the average power transferred to the speaker at 200 Hz, at
f0, and at 4000 Hz. (f) Treating the filter as a resonant circuit, find its quality
factor.
L
C
I
ΔVin
R
ΔVout
Complex current in the series RLC circuit is determined by the
input complex voltage and the complex impedance of the circuit
I=
Vin
Vin
Vin
=
=
1 ⎞
Z R + iω L + 1
⎛
R + i⎜ ωL −
⎟
i ωC
ωC ⎠
⎝
Complex voltage across the speaker is related to the complex
current through the speaker.
8
1 ⎞
⎛
R 2 − i ⋅ ⎜ ωL −
⎟R
R
ωC ⎠
⎝
Vout = I ⋅ R =
⋅ Vin
⋅V =
2
1 ⎞ in
⎛
1 ⎞
⎛
R + i ⋅ ⎜ ωL −
⎟
R 2 + ⎜ ωL −
⎟
ωC ⎠
⎝
ωC ⎠
⎝
The number relating complex output voltage to complex input
voltage is called the complex gain of the circuit.
a) From the relation between complex voltages, the relation
between the amplitudes is
ΔVout = Vout =
R
1 ⎞
⎛
R 2 + ⎜ ωL −
⎟
ωC ⎠
⎝
2
⋅ ΔVin
The number relating the amplitudes of voltage is called the gain
of the circuit. Gain G is a function of frequency
G=
R
1 ⎞
⎛
R + ⎜ ωL −
⎟
ωC ⎠
⎝
2
2
It assumes a particular value at specific frequencies, determined
by the difference between capacitive reactance and inductive
reactance
ω1L −
1
1
=−
−1 ⋅ R
ω1C
G2
and
9
ω2 L −
1
1
=
−1 ⋅ R
ω2 C
G2
Solving simultaneously for the unknowns
1
1
−
1
R
−1
2
2
8Ω ⋅ 4 − 1
G
G
L=
=
=
= 580μH
ω2 − ω1
2π(f 2 − f1 ) 2π ⋅ (4000Hz − 200Hz )
R
f 2 − f1
ω1 − ω2
=
=
1
1
−1
R ⋅ ω1ω2
− 1 2πR ⋅ f1f 2
2
2
G
G
4000Hz − 200Hz
=
= 54.6μF
2π ⋅ 8Ω ⋅ 200Hz ⋅ 4000Hz ⋅ 4 − 1
C=−
(b) The maximum gain occurs when the sum of inductive
reactance and capacitive reactance is zero
ωL −
1
=0
ωC
from which
G=
R
2
R +0
2
=1
(c) The sum of inductive reactance and capacitive reactance is
zero at the resonant frequency of this circuit
f0 =
ω0
1
=
=
2π 2π LC 2π ⋅
(5.8 ⋅10
10
1
−4
)(
−5
H ⋅ 5.46 ⋅ 10 F
)
= 894Hz
(d) Complex gain also includes the phase relation. The phase
difference α, between the output voltage and the input voltages
is
1 ⎞
⎛
1
− ⎜ ωL −
⎟
2πfL −
ωC ⎠
2πfC
α = arctan ⎝
= − arctan
R
R
Therefore at 200 Hz its value is
(
)
1
2π ⋅ 200Hz ⋅ 5.46 ⋅10 − 5 H
= 60°
8Ω
2π ⋅ 200Hz ⋅ 5.8 ⋅ 10 − 4 H −
α 200 Hz = − arctan
(
)
(The output voltage leads the input voltage.)
At 894 Hz its value is
α 894 Hz = − arctan
0
= 0°
8Ω
(The output voltage is in phase with the input voltage.)
At 4000 Hz its value is
(
)
1
2π ⋅ 4000Hz ⋅ 5.46 ⋅10 − 5 H
= −60°
8Ω
2π ⋅ 4000Hz ⋅ 5.8 ⋅ 10− 4 H −
α 4000 Hz = − arctan
(
(The output voltage lags behind the input voltage.)
11
)
(e) The average power delivered to an element depends on the
rms values of the voltage across the element and the current
through the element. Since the speaker acts like a resistor,
Ohm’s law can be used to relate the rms current (through the
speaker) to the voltage applied to the speaker. Knowing the gain
of the circuit the power can be directly related to the amplitude
of the input voltage
ΔVout ΔVout
(G ⋅ ΔVin )
⋅
⋅ cos 0 =
P = I rms Vrms cos ϕ =
2R
R 2
2
2
Hence
P200
2
(
0.5 ⋅ 10V )
=
P894
2
(
1 ⋅ 10V )
=
2 ⋅ 8Ω
P4000
2 ⋅ 8Ω
= 6.25W
2
(
0.5 ⋅ 10V )
=
2 ⋅ 8Ω
= 1.56 W
= 1.56 W
(f) Parameters of the RLC circuit (resistance, capacitance and
inductance) affect the quality factor
(
)
ω0 L 2πf 0 L 2π ⋅ 894Hz ⋅ 5.8 ⋅10 − 4 H
Q=
=
=
= 0.41
R
R
8Ω
12
Problem 33.65
Figure 33.69a shows a parallel RLC circuit, and the corresponding phasor diagram
is given in Figure 33.29b.
V
I L
V
R
L
IR
C
IC
The instantaneous voltage (and rms voltage) across each of the three circuit
elements is the same, and each is in phase with the current through the resistor.
The currents in the capacitor and the inductor lead or lag behind the current in the
resistor, as shown in Figure 33.66b. (a) Show that the rms current delivered by the
source is given by
2⎤
⎡
Irms = Vrms ⋅ ⎢ 12 + ⎛⎜⎝ ωC − 1 ⎞⎟⎠ ⎥
ωL
⎣R
⎦
1
2
(b) Show that the phase angle between the voltage and the current is given by
tan ϕ = −R⎛⎜⎝ 1 − 1 ⎞⎟⎠
XC XL
In fact, we are asked to find the impedance and the phase
angle for this system of elements connected in parallel. It will
be easier to analyze the complex impedance. For elements
connected in parallel the equivalent complex impedance of the
system is equal to the sum of inverse impedance of each element
1
1
1
1
1
1
1 ⎛
1 ⎞
=
+
+
= +
+ iωC = + i ⎜ ω C −
⎟
R ⎝
ωL ⎠
Zeq Z R Z L Z C R iωL
or
13
1 ⎞
1 ⎛
− i⎜ ωC −
⎟
1
R ⎝
ωL ⎠
Zeq =
=
2
1 ⎞ 1 ⎛
1 ⎛
1
⎞
+ i⎜ ωC −
⎟
ωL ⎠ R + i⎜⎝ ωC − ωL ⎟⎠
R ⎝
(a) Therefore
2
I rms
2
V
1 ⎞
1
1
1 V
1
⎛1⎞ ⎛
=
⋅ Im =
⋅I =
⋅
= m⋅
= ⎜ ⎟ + ⎜ ωC −
⎟ ⋅ Vrms
ωL ⎠
2
2
2 Zeq
2 Zeq
⎝R⎠ ⎝
(b) and
1 ⎞
⎛
− ⎜ ωC −
⎟
ωL ⎠
⎝
2
1 ⎛
1 ⎞
1 ⎞
⎛
+ i ⎜ ωC −
− ⎜ ωC −
⎟
⎟
Im Zeq
⎛ 1
1 ⎞
R ⎝
ωL ⎠
ωL ⎠
⎝
⎟⎟
=
= − R ⎜⎜
−
tan ϕ =
=
1
1
Re Zeq
X
X
⎝ C
L⎠
R
R
2
1 ⎛
1 ⎞
+ i⎜ ωC −
⎟
R ⎝
ωL ⎠
14