Download CH 21 - AC Circuits and Electromagnetic Waves

AC Circuits and Electromagnetic Waves
The first part of this chapter deals with circuits in which the currents and voltages vary
sinusoidally with time. The second part deals with the propagation of sinusoidally
varying electric and magnetic fields through space.
Alternating Current (AC) Circuits
We consider circuits consisting of combinations of resistors, capacitors, and inductors in
which the currents and voltages are sinusoidal. Assume the voltage source is given by
v  Vmax sin( 2 f t ) ,
Vmax is the peak voltage and f is the frequency of oscillation.
Resistor circuit
If the AC voltage source is connected across a resistor, then the current also varies
sinusoidally and is in phase with the voltage.
i
V
I max  max .
R
v
 I max sin( 2 f t ),
R
v (t) =
Vmaxsin(2ft)
R
iR
Imax
Vmax
vR
t
1
The power dissipated in the resistor is given by
P  i2R
Since i varies sinusoidally with time, then the power also varies with time. The average
power dissipated is
Pavg  i 2
avg
R
The average of sin2 (2ft) is ½. So,
2
Pavg  1 I max
R
2
We can write the average power as
2
Pavg  I rms
R,
where
I
I rms  max  0.707 I max .
2
Irms is called the ‘root-mean-square’ current. It is obtained by first squaring the current,
then finding the average (mean) of the square, and then taking the square root of this
average. It is the effective heating value (or DC value) of the time-varying current. The
rms voltage is defined in a similar way –
V
Vrms  max  0.707Vmax
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The rms current and rms voltage are related by
Vrms  I rms R
Example:
The line voltage in a house is nominally 120 volts rms. What is the maximum (peak)
voltage?
Vmax  2 Vmax  2( 120V )  170V
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The peak-to-peak voltage is the difference between the maximum positive and maximum
negative voltages and is 340 V.
Capacitor circuit
If a sinusoidal voltage source is connected across a capacitor, then the charge on the
capacitor and the current to the capacitor also vary sinusoidally with time. The voltage
and charge are related by v = q/C, so the charge and voltage are in phase. However, the
charge lags the current by 90o, so it is found that the voltage lags the current by 90o.
v (t) =
Vmaxsin(2ft)
C
iC
Imax
Vmax
vC
t
The current-voltage relationship for a capacitor is given by
Vrms  I rms X C ,
where
XC 
1
.
2 f C
XC is called the capacitive reactance and has units of ohms. It is the quantity that limits
the current to the capacitor, similar to resistance. However, unlike resistance power
cannot be dissipated in a capacitor. This is because the current and voltage are 90o out of
phase. This is analogous to pushing on a moving object. No work will be done by the
force if it is applied perpendicular to the displacement.
Note that the capacitive reactance decreases with increasing frequency and capacitance.
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Example:
A 0.02 F capacitor is connected to a 50-V rms AC voltage source which oscillates at 10
kHz. What is the rms current to the capacitor?
1
1

 796 
4
2 f C 2 ( 1x10 Hz )( 0.02 x10  6 F )
V
50V
I rms  rms 
 0.063 A
XC
796 
XC 
Inductor circuit
The voltage across an inductor depends on the time rate of change of the current and is
given by
v  L
i
t
This means that in an AC circuit, the voltage is sinusoidal and leads the current by 90o.
v (t) =
Vmaxsin(2ft)
Imax
Vmax
L
iL
vL
t
The current-voltage relationship for an inductor is given by
Vrms  I rms X L ,
where
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X L  2 f L
is called the inductive reactance.
As with the capacitor, no power can be dissipated in an ideal inductor (one with no
resistance). Note that the inductive reactance increases with increasing f and L.
Series LCR circuit
Consider an AC circuit containing a resistor, capacitor, and inductor in series. All have
the same current (which is true for any series circuit). As previously described, the
voltage across the resistor is in phase with the current, the voltage across the capacitor
lags the current by 90o, and the voltage across the inductor leads the current by 90o. This
means that the voltages across the inductor and capacitor are 180o out of phase. That is,
they subtract. The resulting voltage across the inductor and capacitor combination either
leads or lags the voltage across the resistor, depending on whether VL is greater than or
less than VC.
R
v (t) =
Vmaxsin(2ft)
L
Consequently, the rms voltage across the inductor-capacitor combination is
VLC ,rms  I rms | X L  X C |
Since the voltage across the LC combination is 90o out of phase with the voltage across
the resistor, then the total rms voltage must be obtained using the Pythagorean theorem –
Vrms  I rms R 2  ( X L  X C )2
The impedance of the circuit is defined as
Z  R 2  ( X L  X C )2 ,
So we can write
Vrms  I rms Z .
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Z has units of ohms and is a measure of the resistance of the circuit to the flow of current.
The total current and the power dissipated in a series RLC circuit depend on the phase
shift between the total current and the total voltage. This phase shift depends on the ratio
of the out-of-phase to the in-phase voltage. Thus,
V
I ( X  XC )
,
tan  LC  rms L
VR
I rms R
Or,
tan 
X L  XC
R
If XL = XC, then the total phase shift is zero and we get maximum current and maximum
power dissipated in the resistor. (No power is dissipated in the inductor and the
capacitor.)
Resonance in a series LCR circuit
The total current in the LCR circuit is given by
I rms 
Vrms
R 2  ( X L  X C )2
Since XL and XC depend on frequency, then Irms depends on frequency. There is a
particular frequency for which XL = XC, at which Irms has its maximum value. At this
resonance frequency the voltages across the inductor and capacitor exactly cancel, and all
the voltage drop is across the resistor. A plot of the current as a function of frequency
would look something like the following.
Irms
I0, rms = Vrms/R
f
f0 = 1/LC
The resonance frequency, f0, is given by
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X L  XC
2 f 0 L 
1
2 f 0 C
Or,
f0 
1
2 LC
Example:
Consider an RLC circuit for which R = 10 , L = 0.2 mH, C = 5 F and the applied
voltage is Vrms = 25 V?
What is the resonance frequency?
f0 
1
2 ( 0.2 x10  3 H )( 5 x10  6 F )
 5,033 Hz  5.033 kHz
What would be the current in the circuit if f = 3 kHz?
X L  2fL  2 ( 3 x103 )( 0.2 x10 3 )  3.77 
XC 
1
1

 10.6 
2fC 2 ( 3 x103 )( 5 x10  6 )
Z  R 2  ( X L  X C )2  ( 10 )2  ( 3.77  10.6 )2  12.1
V
25V
I rms  rms 
 2.06 A
Z
12.1
What is the power dissipation in the circuit?
2
Pavg  I rms
R  ( 2.06 )2 ( 10 )  42.6W
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Transformers:
A transformer consists of two coils which are closely coupled so that the flux generated
by one coil (the primary) passes mostly through the other coil (the secondary). The flux
coupling can be made nearly complete if the coils are wound around an easily
magnetizable core such as iron.
N1
V1
N2
R
According to Faraday’s law, the primary voltage is given by
V1   N1
 B
t
and the secondary voltage is given by
V2   N 2
 B
t
Thus, we have
V2 
N2
V1
N1
The secondary voltage can thus be larger or smaller than the primary voltage, depending
on the turns ratio. If we assume that the power delivered into the primary is the same as
the power delivered to the load by the secondary,
P1  P2
I1V1  I 2V2
then we find that
I2 
N1
I1
N2
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That is, a transformer which steps up the voltage must step down the current, and vice
versa.
Example:
A transformer has 20 primary turns and 100 secondary turns. If the primary voltage is 12
V, what is the secondary voltage?
V2 
N2
 100 
V1  
( 12V )  60V
N1
 20 
If the load resistor on the secondary is 50 , then
V
60V
I2  2 
 1.2 A
R 50 
The current in the primary is then
I1 
N2
 100 
I2  
 ( 1.2 A )  6 A
N1
 20 
Electromagnetic Waves
The Maxwell’s laws of electromagnetism
I)
( E )  Q /  0
II )
( B)  0
III )
 E l
i
i
 IDR  D  
i
  0 I   0 0
i
IV )
 B l
i
i

( B)
t

( E )
t
Note that Maxwell’s equations are nearly symmetric in E and B. Since there are
apparently no magnetic monopoles analogous the electric charges, then we have zero on
the right-hand side of the second equation and no monopole current term on the right
hand side of the third equation. The term  0 0  ( E ) was added by Maxwell into the
t
equation IV to allow for a varying in time E field to be a source of a B field as well. This
term was not motivated by any observations but simply by the request of consistency in
the sources of E and B as shown in the following table. So in term of the sources for E
and B we now have the complete picture:
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Source of E
Source of B
q: electric charge
ΔΦ(B)/Δt
I: electric current
ΔΦ(E)/Δt
direct source
induced source
And, because of Ampere’s law, magnets are not considered as fundamental sources of B.
Maxwell’s equations predict the existence of electromagnetic waves, which consist of
oscillating electric and magnetic fields that propagate through space at the speed of light.
Faraday’s law and the Ampere-Maxwell law show that it is impossible to have only one
oscillating fields without the need of the direct sources q and I. An oscillating E produces
an oscillation B, and vice versa.
By combing eq. III with eq. IV and setting q = I= 0 (no sources) we obtained two waves
equations (one for E and one for B) with solutions:
where
is the wave number and
is the angular frequency
Electromagnetic waves consist of oscillating electric and magnetic fields that propagate
through space at the speed of light. These waves can be produced by applying an
oscillating potential to an antenna. The
antenna could consist of a rod connected to
each side of an AC voltage source. The
voltage source would generate a sinusoidally
varying current and a sinusoidally varying
charge distribution in each rod. As a
consequence, the rods would generate
magnetic and electric fields which would be
perpendicular to each other and would
radiate from the rods.
At distances far away from the antenna, the
configuration of the electromagnetic wave
would look something like that given in the
figure to the right.
For a fixed frequency, both E and B vary sinusoidally in time and are in phase. Both E
and B are perpendicular to the direction of travel of the wave (c), and they are
perpendicular to each other. The relative orientations of E, B, and c are given by a right
hand rule (as, for example, y, z, and x for a Cartesian coordinate system).
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It can be shown that the speed of the electromagnetic waves is given by
c
1
 0 0
.
Using the accepted values of 0 = 4 x 10-7 Tm/A and 0 = 8.85 x 10-12 C2/(Nm2), this
equation gives c = 3 x 108 m/s, which is the speed of light. This is to be expected since
visible light is a part of the electromagnetic spectrum.
It can also be shown that the ratio of the magnitudes of E and B are always fixed and
given by
E
c
B
Electromagnetic waves carry energy. Specifically, the intensity of the wave, which is
power transmitted per unit area, is given by
E B
I  max max
2 0
Using the relationship between E and B and the expression for c, the intensity can also be
written as
E2
c
2
I  max 
Bmax
2 0 c 2 0
Electromagnetic waves also carry momentum, even though they don’t have mass. If a
surface A completely absorbs an amount of electromagnetic energy U = IA t in a time
 t, the momentum absorbed by the surface is
p 
U
c
(complete absorption)
If a surface completely reflects the radiation, then the momentum transferred to the
surface is
p 
2U
c
(complete reflection)
Thus, when electromagnetic radiation strikes a surface it imparts a force to the surface
(since force is rate of change of momentum).
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Example:
The intensity of sunlight incident upon the earth is about 1,400 W/m2. What are the
maximum values of the electric and magnetic fields associated with the radiation?
Solution:
E2
I  max
2 0 c
E max  2 0 cI  2( 4x10  7 )( 3 x108 )( 1400 )  1,027 V / m
E
1027
Bmax  max 
 3.4 x10  6 T
8
c
3 x10
How much solar power is incident upon the earth?
Solution:
The effective area of the earth seen by the sun is that of a circle with the earth’s radius.
So,
Power  P  IA  IR 2  ( 1400W / m 2 ) ( 6.38x106 m )2  1.79 x1017 W
If all the sun’s radiation is absorbed by the earth, then what is the force imparted by this
radiation on the earth?
Solution:
The power absorbed is the energy absorbed per second. This can be used to find the
momentum absorbed per second, which is the force.
p 
F
U
c
p U / t P 1.79 x1017 W

 
 6.0 x108 N
8
t
c
c
3x10 m / s
Although this is a large force, it has an imperceptible effect due to the large mass of earth.
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