General Physics (PHY 2140)
6/12/2007
Reminder: Exam 2 this Wednesday 6/13
Lecture 10
1212-14 questions.
¾ Electricity and Magnetism
9Induced voltages and induction
9Self-Inductance
9RL Circuits
9Energy in magnetic fields
9AC circuits and EM waves
9Resistors, capacitors and
inductors in AC circuits
9The RLC circuit
9Power in AC circuits
Show your work for full credit.
Chapter 20-21
http://www.physics.wayne.edu/~alan/2140Website/Main.htm
Closed book.
You may bring a page of notes.
Bring a calculator.
Bring a pen or pencil.
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S
S
Lightning Review
Last lecture:
1. Induced voltages and induction
9 Generators and motors
9 SelfSelf-induction
mv
r=
qB
Φ = BA cos θ
ΔI
E = −L
Δt
L=
Review Problem: Charged particles passing
Review example
v
Determine the direction of current in the loop
for bar magnet moving down.
N
NΦ
I
Initial flux
Final flux
through a bubble chamber leave tracks consisting of
small hydrogen gas bubbles. These bubbles make
visible the particles’ trajectories. In the following
figure, the magnetic field is directed into the page,
and the tracks are in the plane of the page, in the
directions indicated by the arrows. (a) Which of the
tracks correspond to positively charged particles? (b)
If all three particles have the same mass and charges
of equal magnitude, which is moving the fastest?
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2
By Lenz’s law,
the induced field
is this
change
3
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1
20.6 Self-inductance
The magnetic flux is proportional to the magnetic field,
which is proportional to the current in the circuit
Thus, the selfself-induced EMF must be proportional to the
time rate of change of the current
When a current flows through a loop, the magnetic field
created by that current has a magnetic flux through the
area of the loop.
If the current changes, the magnetic field changes, and
so the flux changes giving rise to an induced emf. This
phenomenon is called self-induction because it is the
loop's own current, and not an external one, that gives
rise to the induced emf.
Faraday’s law states
E = −N
E = −L
where L is called the inductance of the device
Units: SI: henry (H)
ΔΦ
Δt
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Example: solenoid
L=
NΦB
N2A
= μ0
I
l
E = −L
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E
ΔI
ΔI
→
=−
Δt
Δt
L
L=N
A
ΔΦ N Φ
=
ΔI
I
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Inductor in a Circuit
Inductance can be interpreted as a measure of opposition to the rate
of change in the current
A solenoid of radius 2.5cm has 400 turns and a length of 20 cm. Find
(a) its inductance and (b) the rate at which current must change
through it to produce an emf of 75mV.
Φ B = BA = μ 0
1 H = 1V ⋅ s
If flux is initially zero,
5
N
B = μ 0 nI = μ 0 I
l
ΔI
Δt
„
Remember resistance R is a measure of opposition to the current
As a circuit is completed, the current begins to increase, but the
the
inductor produces an emf that opposes the increasing current
N
IA
l
„
„
= (4π x 10-7)(160000)(2.0 x 10-3)/(0.2) = 2 mH
Therefore, the current doesn’
doesn’t change from 0 to its maximum
instantaneously
Maximum current:
I max =
= (75 x 10-3)/ (2.0 x 10-3) = 37.5 A/s
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E
R
8
2
20.7 RL Circuits
RL Circuit (continued)
Recall Ohm’
Ohm’s Law to find the voltage drop on R
ΔV = IR
(voltage across a resistor)
We have something similar with inductors
E
L
= − L
Δ I
Δ t
(voltage across an inductor)
Similar to the case of the capacitor, we get an equation
for the current as a function of time (series circuit).
I=
E
R
(1 − e
− Rt / L
)
τ=
I=
L
R
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20.8 Energy stored in a magnetic field
)
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Example: stored energy
The battery in any circuit that contains a coil has to do
work to produce a current
Similar to the capacitor, any coil (or inductor) would store
potential energy
PEL =
(
V
1 − e − Rt / L
R
1 2
LI
2
A 24V battery is connected in series with a resistor and an inductor,
inductor,
where R = 8.0Ω
8.0Ω and L = 4.0H. Find the energy stored in the inductor
when the current reaches its maximum value.
Summary of the properties of circuit elements.
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Resistor
Capacitor
Inductor
units
ohm, Ω = V / A
farad, F = C / V
henry, H = V s / A
symbol
R
C
L
relation
V=IR
Q=CV
emf = -L (ΔI / Δt)
power dissipated
P = I V = I² R
= V² / R
0
0
energy stored
0
PEC = C V² / 2
PEL = L I² / 2
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3
A 24V battery is connected in series with a resistor and an inductor,
inductor, where R =
8.0Ω
8.0Ω and L = 4.0H. Find the energy stored in the inductor when the current
current
reaches its maximum value.
Chapter 21
Recall that the energy stored in the
inductor is
Given:
V = 24 V
R = 8.0 Ω
L = 4.0 H
PEL =
1 2
LI
2
The only thing that is unknown in
the equation above is current. The
maximum value for the current is
Find:
PEL =?
I max =
Alternating Current Circuits
and Electromagnetic Waves
V 24V
=
= 3.0 A
R 8.0Ω
Inserting this into the above expression for the energy gives
PEL =
1
2
( 4.0H )( 3.0 A) = 18J
2
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AC Circuit
„
„
Resistor in an AC Circuit
An AC circuit consists of a combination of
circuit elements and an AC generator or
source
The output of an AC generator is sinusoidal
and varies with time according to the
following equation
„
Δv = ΔVmax sin 2π
2πƒt
„
„
„
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Consider a circuit
consisting of an AC source
and a resistor
„ The graph shows the
current through and the
voltage across the resistor
„ The current and the
voltage reach their
maximum values at the
same time
„ The current and the
voltage are said to be in
„
Δv is the instantaneous voltage
ΔVmax is the maximum voltage of the generator
ƒ is the frequency at which the voltage changes, in Hz
phase
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4
More About Resistors in an AC
Circuit
„
„
rms Current and Voltage
„ The
The direction of the current has no effect on
the behavior of the resistor
The rate at which electrical energy is
dissipated in the circuit is given by
„ P = i2 R= (Imax sin 2π
2πƒt)2 R
„
„
„
that would dissipate the same amount
of energy in a resistor as is actually
dissipated by the AC current
where i is the instantaneous current
the heating effect produced by an AC current with a
maximum value of Imax is not the same as that of a DC
current of the same value
The maximum current occurs for a small amount of time
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Ohm’s Law in an AC Circuit
„ rms
values will be used when discussing
AC currents and voltages
AC ammeters and voltmeters are designed
to read rms values
„ Many of the equations will be in the same
form as in DC circuits
„
„ Ohm’s
circuit
„
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Law for a resistor, R, in an AC
ΔVrms = Irms R
„
rms current is the direct current
Also applies to the maximum values of v and i 19
Irms =
Imax
= 0.707 Imax
2
„ Alternating
voltages can also be
discussed in terms of rms values
ΔVrms =
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ΔVmax
= 0.707 ΔVmax
2
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Example: an AC circuit
An ac voltage source has an output of ΔV = 150 sin (377 t). Find
(a) the rms voltage output,
(b) the frequency of the source, and
(c) the voltage at t = (1/120)s.
(d) Find the rms current in the circuit when the generator is
connected to a 50.0Ω
50.0Ω resistor.
ΔVmax
= 0.707 ΔVmax = 0.707 x 150V = 106 V
2
ω = 377 rad/sec, ω = 2π f , f = ω / 2π = 377/ 2π = 60 Hz
ΔVrms =
ΔV = 150 sin (377 x 1/120) = 0 V
ΔVrms = Irms R thus, Irms = ΔVrms/R = 2.12 A
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5
More About Capacitors in an
AC Circuit
Capacitors in an AC Circuit
„
„
Consider a circuit containing a capacitor and
an AC source
The current starts out at a large value and
charges the plates of the capacitor
„
„
There is initially no resistance to hinder the flow of
the current while the plates are not charged
As the charge on the plates increases, the
voltage across the plates increases and the
current flowing in the circuit decreases
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Capacitive Reactance and
Ohm’s Law
„
The impeding effect of a capacitor on the
current in an AC circuit is called the capacitive
reactance and is given by
„
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„
„
When ƒ is in Hz and C is in F, XC will be in ohms
„
Ohm’
Ohm’s Law for a capacitor in an AC circuit
„
22
Inductors in an AC Circuit
1
XC =
2π ƒ C
„
The current reverses
direction
„ The voltage across
the plates decreases
as the plates lose
the charge they had
accumulated
„ The voltage across
the capacitor lags
behind the current
by 90°
90°
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„
ΔVrms = Irms XC
23
Consider an AC circuit
with a source and an
inductor
The current in the
circuit is impeded by
the back emf of the
inductor
The voltage across the
inductor always leads
the current by 90°
90°
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6
Inductive Reactance and
Ohm’s Law
Example: AC circuit with
capacitors and inductors
„ The
effective resistance of a coil in an
AC circuit is called its inductive
reactance and is given by
„
XL = 2π
2πƒL
„
ΔVrms = Irms XC , first we find XC:
When ƒ is in Hz and L is in H, XL will be in
ohms
„ Ohm’s
„
A 2.40mF capacitor is connected across an alternating voltage with
with an
rms value of 9.00V. The rms current in the capacitor is 25.0mA. (a) What
is the source frequency? (b) If the capacitor is replaced by an ideal coil
with an inductance of 0.160H, what is the rms current in the coil?
= 9.00V/25.0 x 10-3 A = 360 ohms
Law for the inductor
Now, solve for ƒ: ƒ = 1/ 2π XC C = 0.184 Hz
ΔVrms = Irms XL
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„
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Current and Voltage
Relationships in an RLC Circuit
The resistor,
inductor, and
capacitor can be
combined in a circuit
The current in the
circuit is the same at
any time and varies
sinusoidally with
time
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⎛
1 ⎞
⎜ XC =
⎟
2π ƒC ⎠
⎝
For and inductor XL = 2π
2πƒL, try solving for Irms = ΔVrms/ XL
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The RLC Series Circuit
„
ΔVrms/ Irms
„
„
„
27
The instantaneous
voltage across the
resistor is in phase with
the current
The instantaneous
voltage across the
inductor leads the
current by 90°
90°
The instantaneous
voltage across the
capacitor lags the
current by 90°
90°
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7
Phasor Diagram for RLC
Series Circuit
Phasor Diagrams
„
„
„
To account for the
different phases of the
voltage drops, vector
techniques are used
Represent the voltage
across each element as
a rotating vector, called
a phasor
The diagram is called a
„
„
„
phasor diagram
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„
The phasors are
added as vectors to
account for the
phase differences in
the voltages
ΔVL and ΔVC are on
the same line and so
the net y component
is ΔVL - ΔVC
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ΔVmax From the Phasor
Diagram
Phasor Diagram, cont
„
The voltage across the
resistor is on the +x
axis since it is in phase
with the current
The voltage across the
inductor is on the +y
since it leads the
current by 90°
90°
The voltage across the
capacitor is on the –y
axis since it lags behind
the current by 90°
90°
„
The voltages are not in phase, so they cannot
simply be added to get the voltage across the
combination of the elements or the voltage
source
2
ΔVmax = ΔVR + ( ΔVL − ΔVC )2
tan φ =
„
31
ΔVL − ΔVC
ΔVR
φ is the phase angle between the current and
the maximum voltage
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8
Impedance of a Circuit
„
Impedance and Ohm’s Law
„ Ohm’s
Law can be applied to the
impedance
The impedance, Z,
can also be
represented in a
phasor diagram
„
ΔVmax = Imax Z
Z = R 2 + ( XL − X C )2
tan φ =
XL − X C
R
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Summary of Circuit Elements,
Impedance and Phase Angles
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Problem Solving for AC
Circuits
„ Calculate
as many unknown quantities
as possible
For example, find XL and XC
„ Be careful of units -- use F, H, Ω
„
„ Apply
Ohm’s Law to the portion of the
circuit that is of interest
„ Determine all the unknowns asked for
in the problem
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9
Power in an AC Circuit
„
Power in an AC Circuit, cont
No power losses are associated with
capacitors and pure inductors in an AC circuit
„
„
In a capacitor, during oneone-half of a cycle energy is
stored and during the other half the energy is
returned to the circuit
In an inductor, the source does work against the
back emf of the inductor and energy is stored in
the inductor, but when the current begins to
decrease in the circuit, the energy is returned to
the circuit
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„ The
average power delivered by the
generator is converted to internal
energy in the resistor
Pav = IrmsΔVR = IrmsΔVrms cos φ
„ cos φ is called the power factor of the
circuit
„
„ Phase
shifts can be used to maximize
power outputs
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Resonance in an AC Circuit
„
Resonance occurs at
the frequency, ƒo,
where the current has
its maximum value
„
„
To achieve maximum
current, the impedance
must have a minimum
value
This occurs when XL = XC
ƒo =
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1
2π LC
39
10