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Transcript
Chapter 29
Electromagnetic Induction
and Faraday’s Law
Copyright © 2009 Pearson Education, Inc.
29-6 Transformers and Transmission
of Power
Energy must be conserved; therefore, in the
absence of losses, the ratio of the currents
must be the inverse of the ratio of turns:
Ip
Np
Vs


Is N s Vp
Copyright © 2009 Pearson Education, Inc.
29-6 Transformers and Transmission
of Power
Example 29-12: Cell phone charger.
The charger for a cell phone contains a
transformer that reduces 120-V ac to 5.0-V ac
to charge the 3.7-V battery. (It also contains
diodes to change the 5.0-V ac to 5.0-V dc.)
Suppose the secondary coil contains 30 turns
and the charger supplies 700 mA. Calculate
(a) the number of turns in the primary coil, (b)
the current in the primary, and (c) the power
transformed.
Copyright © 2009 Pearson Education, Inc.
29-6 Transformers and Transmission
of Power
Transformers work only if the current is
changing; this is one reason why electricity
is transmitted as ac.
Copyright © 2009 Pearson Education, Inc.
29-6 Transformers and Transmission
of Power
Example: Transmission lines.
An average of 120 kW of electric power is sent
to a small town from a power plant 10 km away.
The transmission lines have a total resistance
of 0.40 Ω. Calculate the power loss if the power
is transmitted at (a) 120 V and (b) 24,000 V.
Copyright © 2009 Pearson Education, Inc.
ConcepTest 29.12a Transformers I
1) 30 V
What is the voltage
2) 60 V
across the lightbulb?
3) 120 V
4) 240 V
5) 480 V
120 V
ConcepTest 29.12a Transformers I
1) 30 V
What is the voltage
2) 60 V
across the lightbulb?
3) 120 V
4) 240 V
5) 480 V
The first transformer has a 2:1 ratio
of turns, so the voltage doubles.
But the second transformer has a
1:2 ratio, so the voltage is halved
again. Therefore, the end result is
the same as the original voltage.
120 V
240 V
120 V
Summary of Chapter 29
• Magnetic flux:
• Changing magnetic flux induces emf:
• Induced emf produces current that
opposes original flux change.
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 29
• Changing magnetic field produces an electric
field.
• General form of Faraday’s law:
.
• Electric generator changes mechanical
energy to electrical energy; electric motor
does the opposite.
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 29
• Transformer changes magnitude of
voltage in ac circuit; ratio of currents is
inverse of ratio of voltages:
and
Copyright © 2009 Pearson Education, Inc.
Chapter 30
Inductance, Electromagnetic
Oscillations, and AC Circuits
Copyright © 2009 Pearson Education, Inc.
Units of Chapter 30
• Mutual Inductance
• Self-Inductance
• Energy Stored in a Magnetic Field
• LR Circuits
• LC Circuits and Electromagnetic Oscillations
• LC Circuits with Resistance (LRC Circuits)
• AC Circuits with AC Source
Copyright © 2009 Pearson Education, Inc.
Units of Chapter 30
• LRC Series AC Circuit
• Resonance in AC Circuits
• Impedance Matching
• Three-Phase AC
Copyright © 2009 Pearson Education, Inc.
30-1 Mutual Inductance
Mutual inductance: a changing current in one
coil will induce a current in a second coil:
And vice versa; note that the constant M,
known as the mutual inductance, is the same:
Copyright © 2009 Pearson Education, Inc.
30-1 Mutual Inductance
Unit of inductance: the henry, H:
1 H = 1 V·s/A = 1 Ω·s.
A transformer is an
example of mutual
inductance.
Copyright © 2009 Pearson Education, Inc.
30-1 Mutual Inductance
Example 30-1: Solenoid and coil.
A long thin solenoid of length l and cross-sectional
area A contains N1 closely packed turns of wire.
Wrapped around it is an insulated coil of N2 turns.
Assume all the flux from coil 1 (the solenoid)
passes through coil 2, and calculate the mutual
inductance.
Copyright © 2009 Pearson Education, Inc.
30-2 Self-Inductance
A changing current in a coil will also induce
an emf in itself:
Here, L is called the self-inductance:
Copyright © 2009 Pearson Education, Inc.
30-2 Self-Inductance
Example 30-3: Solenoid inductance.
(a) Determine a formula for the self-inductance
L of a tightly wrapped and long solenoid
containing N turns of wire in its length l and
whose cross-sectional area is A.
(b) Calculate the value of L if N = 100, l = 5.0 cm,
A = 0.30 cm2, and the solenoid is air filled.
Copyright © 2009 Pearson Education, Inc.
30-2 Self-Inductance
Conceptual Example 30-4: Direction of emf in
inductor.
Current passes through a coil from left to right
as shown. (a) If the current is increasing with
time, in which direction is the induced emf?
(b) If the current is decreasing in time, what
then is the direction of the induced emf?
Copyright © 2009 Pearson Education, Inc.
30-2 Self-Inductance
Example 30-5: Coaxial cable
inductance.
Determine the inductance per unit
length of a coaxial cable whose
inner conductor has a radius r1 and
the outer conductor has a radius r2.
Assume the conductors are thin
hollow tubes so there is no
magnetic field within the inner
conductor, and the magnetic field
inside both thin conductors can be
ignored. The conductors carry equal
currents I in opposite directions.
Copyright © 2009 Pearson Education, Inc.
30-3 Energy Stored in a Magnetic Field
Just as we saw that energy can be stored in
an electric field, energy can be stored in a
magnetic field as well, in an inductor, for
example.
Analysis shows that the energy density of the
field is given by
Copyright © 2009 Pearson Education, Inc.
30-4 LR Circuits
A circuit consisting of
an inductor and a
resistor will begin with
most of the voltage drop
across the inductor, as
the current is changing
rapidly. With time, the
current will increase less
and less, until all the
voltage is across the
resistor.
Copyright © 2009 Pearson Education, Inc.
30-4 LR Circuits
The sum of potential differences around the
loop gives
Integrating gives the current as a function
of time:
.
The time constant of an LR circuit is
.
Copyright © 2009 Pearson Education, Inc.
.
30-4 LR Circuits
If the circuit is then shorted across the
battery, the current will gradually decay away:
.
Copyright © 2009 Pearson Education, Inc.
30-4 LR Circuits
Example 30-6: An LR circuit.
At t = 0, a 12.0-V battery is
connected in series with a
220-mH inductor and a total of
30-Ω resistance, as shown. (a)
What is the current at t = 0? (b)
What is the time constant? (c)
What is the maximum current?
(d) How long will it take the
current to reach half its
maximum possible value? (e)
At this instant, at what rate is
energy being delivered by the
battery, and (f) at what rate is
energy being stored in the
inductor’s magnetic field?
Copyright © 2009 Pearson Education, Inc.
30-5 LC Circuits and
Electromagnetic Oscillations
An LC circuit is a charged capacitor
shorted through an inductor.
Copyright © 2009 Pearson Education, Inc.
30-5 LC Circuits and
Electromagnetic Oscillations
Summing the potential drops around the
circuit gives a differential equation for Q:
This is the equation for simple
harmonic motion, and has solutions
..
Copyright © 2009 Pearson Education, Inc.
30-5 LC Circuits and
Electromagnetic Oscillations
Substituting shows that the equation
can only be true for all times if the
frequency is given by
The current is sinusoidal as well:
Copyright © 2009 Pearson Education, Inc.
30-5 LC Circuits and
Electromagnetic Oscillations
The charge and current are both
sinusoidal, but with different phases.
Copyright © 2009 Pearson Education, Inc.
30-5 LC Circuits and
Electromagnetic Oscillations
The total energy in the circuit is
constant; it oscillates between the
capacitor and the inductor:
Copyright © 2009 Pearson Education, Inc.
30-5 LC Circuits and
Electromagnetic Oscillations
Example 30-7: LC circuit.
A 1200-pF capacitor is fully charged by a
500-V dc power supply. It is
disconnected from the power supply
and is connected, at t = 0, to a 75-mH
inductor. Determine: (a) the initial
charge on the capacitor; (b) the
maximum current; (c) the frequency f
and period T of oscillation; and (d) the
total energy oscillating in the system.
Copyright © 2009 Pearson Education, Inc.
30-6 LC Oscillations with Resistance
(LRC Circuit)
Any real (nonsuperconducting) circuit will
have resistance.
Copyright © 2009 Pearson Education, Inc.
30-6 LC Oscillations with Resistance
(LRC Circuit)
Now the voltage drops around the circuit give
The solutions to this equation are damped
harmonic oscillations. The system will be
underdamped for R2 < 4L/C, and overdamped
for R2 > 4L/C. Critical damping will occur
when R2 = 4L/C.
Copyright © 2009 Pearson Education, Inc.
30-6 LC Oscillations with Resistance
(LRC Circuit)
This figure shows the three cases of
underdamping, overdamping, and
critical damping.
Copyright © 2009 Pearson Education, Inc.
30-6 LC Oscillations with Resistance
(LRC Circuit)
The angular frequency for
underdamped oscillations is given by
.
The charge in the circuit as a
function of time is
.
Copyright © 2009 Pearson Education, Inc.
30-6 LC Oscillations with Resistance
(LRC Circuit)
Example 30-8: Damped oscillations.
At t = 0, a 40-mH inductor is placed in
series with a resistance R = 3.0 Ω and a
charged capacitor C = 4.8 μF. (a) Show
that this circuit will oscillate. (b)
Determine the frequency. (c) What is the
time required for the charge amplitude to
drop to half its starting value? (d) What
value of R will make the circuit
nonoscillating?
Copyright © 2009 Pearson Education, Inc.
30-7 AC Circuits with AC Source
Resistors, capacitors,
and inductors have
different phase
relationships between
current and voltage
when placed in an ac
circuit.
The current through
a resistor is in phase
with the voltage.
Copyright © 2009 Pearson Education, Inc.
30-7 AC Circuits with AC Source
The voltage across the
inductor is given by
or
.
Therefore, the current
through an inductor
lags the voltage by 90°.
Copyright © 2009 Pearson Education, Inc.
30-7 AC Circuits with AC Source
The voltage across the inductor is related
to the current through it:
.
The quantity XL is called the inductive
reactance, and has units of ohms:
Copyright © 2009 Pearson Education, Inc.
30-7 AC Circuits with AC Source
Example 30-9: Reactance of a coil.
A coil has a resistance R = 1.00 Ω and
an inductance of 0.300 H. Determine
the current in the coil if (a) 120-V dc is
applied to it, and (b) 120-V ac (rms) at
60.0 Hz is applied.
Copyright © 2009 Pearson Education, Inc.
30-7 AC Circuits with AC Source
The voltage across the
capacitor is given by
.
Therefore, in a capacitor,
the current leads the
voltage by 90°.
Copyright © 2009 Pearson Education, Inc.
30-7 AC Circuits with AC Source
The voltage across the capacitor is related
to the current through it:
.
The quantity XC is called the capacitive
reactance, and (just like the inductive
reactance) has units of ohms:
Copyright © 2009 Pearson Education, Inc.
30-8 LRC Series AC Circuit
Analyzing the LRC series AC circuit is
complicated, as the voltages are not in phase
– this means we cannot simply add them.
Furthermore, the reactances depend on the
frequency.
Copyright © 2009 Pearson Education, Inc.
30-8 LRC Series AC Circuit
We calculate the voltage (and current) using
what are called phasors – these are vectors
representing the individual voltages.
Here, at t = 0, the
current and
voltage are both at
a maximum. As
time goes on, the
phasors will rotate
counterclockwise.
Copyright © 2009 Pearson Education, Inc.
30-8 LRC Series AC Circuit
Some time t later,
the phasors have
rotated.
Copyright © 2009 Pearson Education, Inc.
30-8 LRC Series AC Circuit
The voltages across
each device are given
by the x-component of
each, and the current
by its x-component.
The current is the
same throughout the
circuit.
Copyright © 2009 Pearson Education, Inc.
30-8 LRC Series AC Circuit
We find from the ratio of voltage to
current that the effective resistance,
called the impedance, of the circuit
is given by
Copyright © 2009 Pearson Education, Inc.
Homework # 11
Chapter 29 – 54
Chapter 30 -0 2, 12, 20, 34, 43
Copyright © 2009 Pearson Education, Inc.