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Induction - Spring 2006 1
What is going on?
There are only 2 more weeks after spring break. Spring break is next week (3/31-4/4), have a
good one and be safe. Projects are due the last class before the final
week Final exam.
Induction - Spring 2006 2
Chapter 31
Electromagnetic Oscillations and Alternating Current
In this chapter we will cover the following topics:
-Electromagnetic oscillations in an LC circuit -Alternating current (AC) circuits with capacitors -Resonance in RCL circuits -Power in AC-circuits -Transformers, AC power transmission
(31 - 1)
Induction - Spring 2006 3
LR Circuit
i
then0,ELet
0
equationcapacitor
theas form same
0
:0 drops voltageof sum
dt
dqR
C
qE
dt
diLiRE
Steady Source
Induction - Spring 2006 4
Max Current Rate ofincrease = max emf
VR=iR~current
Induction - Spring 2006 5
constant) (time
)1( /
R
L
eR
Ei LRt
Solve the lo
op equation.
Induction - Spring 2006 6
Time Dependent Result:
R
L
eR
Ei LRt
constant time
)1( /
Induction - Spring 2006 7
We also showed that
2
0
2
0
2
1
2
1
E
B
capacitor
inductor
u
u
Induction - Spring 2006 8
At t=0, the charged capacitor is now connected to the inductor. What would you expect to happen??
Induction - Spring 2006 9
LC
We will show that the charge on the capacitor plates as well as the current
1 in the inductor oscillate with constant amplitude at an angular frequency
The total energy in the circuit is t
q i
LCU
2 2
he sum of the energy stored in the electric field
of the capacitor and the magnetic field of the inductor. . 2 2
The total energy of the circuit does not change with time. Thus
E B
q LiU U U
CdU
2
2
2
2
0
0. 1
0
dt
dU q dq di dq di d qLi i
dt C d
d qL
t dt dtq
dtdt dt C
(31 - 2)
Induction - Spring 2006 10
LC
2
2
2
2
1 0 ( )
This is a homogeneous, second order, linear differential equation
which we have encountered previously. We used it to d
10
escribe
the simple harmonic oscillat
o
d q
L qdt C
d qq
dt LC
eqs.1
22
2 0
with sol
r (SHO)
( )
ution: ( ) cos( )
d xx
dtx t X t
eqs.2
If we compare eqs.1 with eqs.2 we find that the solution to the differential
equation that describes the LC-circuit (eqs.1) is:
1( ) cos where , and is the phase angle.
The current
q t Q tLC
sindq
i Q tdt
( ) co sq t Q t
1
LC
(31 - 3)
Induction - Spring 2006 11
LC
2 22
2 2 2 22 2
2
The energy stored in the electric field of the capacitor
cos2 2
The energy stored in the magnetic field of the inductor
sin sin2 2 2
The total energy
2
E
B
E B
q QU t
C C
Li L Q QU t t
CU U U
QU
2
2 2cos sin2
The total energy is constant;
Qt t
C C
energy is conserved
2
2
3The energy of the has a value of at 0, , , ,...
2 2 2
3 5The energy of the has a value of at , , ,...
2 4 4 4 When is maximum is zeE B
Q T Tt T
C
Q T T Tt
CU U
electric field maximum
magnetic field maximum
Note : ro, and vice versa (31 - 4)
Induction - Spring 2006 12
The Graph of that LC (no emf) circuit ..
Induction - Spring 2006 13
Induction - Spring 2006 14
Mass on a Spring Result
Energy will swap back and forth. Add friction
Oscillation will slow down Not a perfect analogy
Induction - Spring 2006 15
Induction - Spring 2006 16
LC Circuit
High
Q/CLow
Low
High
Induction - Spring 2006 17
The Math Solution (R=0):
LC
Induction - Spring 2006 18
New Feature of Circuits with L and C
These circuits produce oscillations in the currents and voltages
Without a resistance, the oscillations would continue in an un-driven circuit.
With resistance, the current would eventually die out.
Induction - Spring 2006 19
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8 9 10
Time
Vo
lts
Variable Emf Applied
emf
Sinusoidal
DC
Induction - Spring 2006 20
Sinusoidal Stuff
)sin( tAemf
“Angle”
Phase Angle
Induction - Spring 2006 21
Same Frequencywith
PHASE SHIFT
Induction - Spring 2006 22
Different Frequencies
Induction - Spring 2006 23
2
2
If we add a resistor in an RL cicuit (see figure) we must
modify the energy equation because now energy is
being dissipated on the resistor.
2E B
dUi R
dt
qU U U
C
Damped oscillations in an RCL circuit
22
2
Li dU q dq diLi i R
dt C dt dt
2
2
2
/ 2
2
2 2
10 This is the same equation as that
of the damped harmonics o 0 which hscillator:
The a
as the solution
( ) co ngul r fs a
:
bt mm
dq di d q d q dqi L R q
dt dt dt dt d
d x dxm b kx
dt dt
x t x e t
t C
2
2
2
2
/ 2 1 ( )
requency
For the damped RCL circuit the solut
cos
ion is:
The angular fre que4
ncy
4
Rt L Rq
k b
m m
t Qe tLC L
(31 - 6)
Induction - Spring 2006 24
/ 2Rt LQe
/ 2Rt LQe
( )q tQ
Q
( )q t / 2( ) cosRt Lq t Qe t
2
2
1
4
R
LC L
/ 2
2
2
The equations above describe a harmonic oscillator with an exponetially decaying
amplitude . The angular frequency of the damped oscillator
1 is always smaller than the angular f
4
Rt LQe
R
LC L
2
2
1requency of the
1undamped oscillator. If the term we can use the approximation
4
LC
R
L LC
(31 - 7)
Induction - Spring 2006 25
Note – Power is delivered to our homes as an oscillating source (AC)
Induction - Spring 2006 26
Producing AC Generator
x x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x x
Induction - Spring 2006 27
The Real World
Induction - Spring 2006 28
A
Induction - Spring 2006 29
Induction - Spring 2006 30
The Flux:
tAR
emfi
tBAemf
t
BA
bulb
sin
sin
cos
AB
Induction - Spring 2006 31
Source Voltage:
)sin(0 tVVemf
Induction - Spring 2006 32
In fig.a we show an ac generator connected to a resistor
From KLR we have: 0 sin
The current amplitude
The voltage across is equal to sin
The voltage
mR R
mR
R m
R
i R i tR R
IR
v R t
A resistive load
EEE
E
E
amplitude is equal to
The relation between the voltage and
current amplitudes is:
In fig.b we plot the resistor current and the
resistor voltage as function of time t.
Both quanti
m
R
R
R RV I R
i
v
E
ties reach their maximum values
at the same time. We say that voltage and
current are .in phase(31 - 10)
R RV I R
Induction - Spring 2006 33
2 2
We define the "root mean square" (rms) value of as follows:
The equation looks the same
as in the DC case. This power appears a
s2
heat on
rmsmrms
V
VP
R
R
V E
2
0
22
0
22
0
2
1( )
sin
1
1
si
1sin
n
22
Tm
Tm
mT
P P t dt P tT R
td
P tdtR
tT
T
PR
Average Power for R
E
E
E
2
2mPR
E 2
2
mrmsV
E
(31 - 12)
Induction - Spring 2006 34
In fig.a we show an ac generator connected to a
capacitor
From KLR we have: 0
sin
cos sin 90
The voltage amplitude equal to
The current am
C
C m
CC m
C m
C
q
Cq C C t
dqi C tdt t
dtV
A capacitive load
E
E E
E
E
plitude 1/
The quantity is known as the
In fig.b we plot the capacitor current and the capacitor
voltage as function of time t
1
. The current
the
v
/
CC
C
C
C
C
VI CV
C
i
v
X C
leads
capacitive reactance
oltage by a quarter of a period. The voltage and
current are .out of phase by 90
O
CX
(31 - 13)
1
CXC
Induction - Spring 2006 35
2
2
2
0 0
sin cos
sin 22
1 1( ) = sin 2 0
2
A capacitor does not dissipate any power
on the average. In some parts o
2sin cos sin 2
h
f t
mC C
C
m
C
T Tm
C
P V I t tX
P tX
P P t dt tdtT X T
Average Power for C
Note :
E
E
E
e cycle it absorbes
energy from the ac generator but at the rest of the cycle
it gives the energy back so that on the average no
power is used!
0CP
(31 - 14)
Induction - Spring 2006 36
In fig.a we show an ac generator connected to an inductor
From KLR we have: 0 sin
sin cos sin 90
The voltage amplitude equa
mL L
m m mL L
L
L
di diL t
dt dt L L
i di tdt tdt tL L L
V
An inductive load
EEE
E E E
l to
The current amplitude
The quantity is known as the
In fig.b we plot the inductor current and the
inductor voltage as function of time t.
The current
m
LL
L
L
L
VI
L
i
v
X L
E
inductive reactance
the voltage by a
quarter of a period. The voltage and
current are .
lags behind
out of phase by 90
O
LX
(31 - 15)
LX L
Induction - Spring 2006 37
2
2
2
0 0
Power sin cos
sin 22
1 1( ) = sin 2 0
2
A inductor does not dissipate any power
on
2sin cos sin 2
the average. In some p rt
a s
mL L
L
m
L
T Tm
L
P V I t tX
P tX
P P t dt tdtT X T
Note :
E
E
E
Average Power for L
of the cycle it absorbes
energy from the ac generator but at the rest of the cycle
it gives the energy back so that on the average no
power is used!
0LP
(31 - 16)
Induction - Spring 2006 38
Circuit element
Average Power
Reactance
Phase of current Voltage amplitude
Resistor
R
Current is in phase with the voltage
CapacitorC
Current leads voltage by a quarter of a period
Inductor
L
Current lags behind voltage by a quarter of a period
1CX
C
LX L
R RV I R
CC C C
IV I X
C
L L L LV I X I L
R
SUMMARY
2
2m
RP R
E
0CP
0LP
(31 - 17)
Induction - Spring 2006 39
An ac generator with emf is connected to
an in series combination of a resistor , a capacitor
and an inductor , as shown in the figure. The phasor
for the ac genera
sinm
R C
L
tThe series RCL circuit
E E
tor is given in fig.c. The current in
this circuit is described by the equation: sini I t sini I t
The current is for the resistor, the capacitor and the inductor
The phasor for the current is shown in fig.a. In fig.c we show the phasors for the
voltage across , the voltage across R C
i
v R v C
common
, and the voltage across .
The voltage is in phase with the current . The voltage lags behind
the current by 90 . The voltage leads ahead of the current by 90 .
L
R C
L
v L
v i v
i v i
(31 - 18)
Induction - Spring 2006 40
O
A B
Kirchhoff's loop rule (KLR) for the RCL circuit: . This equation
is represented in phasor form in fig.d. Because and have opposite directions
we combine the two in a single phasor
R C L
L C
L
v v v
V V
V
E
2 2 2 22 2 2 2
22
22
. From triangle OAB we have:
The denominator is known as the " "
of the circuit. The current amplitude
C
m R L C L C L C
m
L C
mL C
V
V V V IR IX IX I R X X
I ZR X X
Z R X X I
E
E
E
impedance
22
1
m
Z
I
R LC
E
22 L CZ R X X
mI
Z
E
sini I t
(31 - 19)
Induction - Spring 2006 41
O
A B
From triangle OAB we have: tan
We distinguish the following three cases depending on the relative values
of and .
0 The current phasor lags behind the generat
L C L C L C
R
L L
L C
V V IX IX X X
V IR R
X X
X X
1. or phasor.
The circuit is more inductive than capacitive
0 The current phasor leads ahead of the generator phasor
The circuit is more capacitive than inductive
0 The current phaso
C L
C L
X X
X X
2.
3. r and the generator phasor are in phase
sini I t
22 L CZ R X X
LX L
tan L CX X
R
1 CX
C
(31 - 20)
Induction - Spring 2006 42
Fig.a and b: 0
The current phasor lags behind
the generator phasor. The circuit is more
inductive than capacitive
L CX X 1.
Fig.c and d: 0 The current phasor leads ahead of the generator
phasor. The circuit is more capacitive than inductive
Fig.e and f: 0 The current phasor and the generator ph
C L
C L
X X
X X
2.
3. asor are
in phase (31 - 21)
Induction - Spring 2006 43
In the RCL circuit shown in the figure assume that
the angular frequency of the ac generator can
be varied continuously. The current amplitude
in the circuit is given by the equation:
mI
Resonance
E2
2
The current amplitude1
1has a maximum when the term 0
1This occurs when
R LC
LC
LC
The equation above is the condition for resonance. When its is satisfied
A plot of the current amplitude as function of is shown in the lower figure.
This plot is known as a "
mresI
RI
E
resonance c "urve
mresI
R
E1
LC
(31 - 22)
Induction - Spring 2006 44
2
We already have seen that the average power used by
a capacitor and an inductor is equal to zero. The
power on the average is consumed by the resistor.
The instantaneous power P i
Power in an RCL ciruit
2
0
222 2
0
sin
1The average power
1sin
2
cos
The term cos in the equation above is known as
the "
T
avg
T
avg rms
rmsavg rms rms rms rms rms rms rms
R I t R
P PdtT
I RP I R t dt I R
T
RP I RI I R I I
Z Z
E
E E
power fac " of the circuit. The average
power consumed by the circuit is maximum
when 0
tor
2avg rmsP I R cosavg rms rmsP I E (31 - 23)
Induction - Spring 2006 45
The transformer is a device that can change
the voltage amplitude of any ac signal. It
consists of two coils with different number
of turns wound around a common iron core.
The transformer
The coil on which we apply the voltage to be changed is called the " " and
it has turns. The transformer output appears on the second coils which is known
as the "secondary" and has turnsP
S
N
N
primary
. The role of the iron core is to insure that the
magnetic field lines from one coil also pass through the second. We assume that
if voltage equal to is applied across the primary then a voltagPV e appears
on the secondary coil. We also assume that the magnetic field through both coils
is equal to and that the iron core has cross sectional area A. The magnetic flux
through the primary
S
P
V
B
( )
The flux through the secondary ( )
PP P P
SS S S S
d dBN BA V N A
dt dtd dB
N BA V N Adt dt
eqs.1
eqs.2
(31 - 25)
Induction - Spring 2006 46
( )
( )
If we divide equation 2 by equation 1 we get:
P
S P
P P P P
SS S S S
SS S
P P S PP
d dBN BA V N A
dt dtd dB
N BA V N Adt dt
dBN AV Ndt
dBV NN A
V
Ndt
V
N
eqs.1
eqs.2
The voltage on the secondary
If 1 We have what is known a " " transformer
If 1 We have what is known a " " transformer
Both types of tran
SS P
P
SS P S P
P
SS P S P
P
NV V
N
NN N V V
N
NN N V V
N
step up
step down
sformers are used in the transport of electric power over large
distances.
S P
S P
V V
N N
(31 - 26)
Induction - Spring 2006 47
PI SI
If we close switch S in the figure we have in addition to the primary current
a current in the secondary coil. We assume that the transformer is " "
i.e. it suffers no losses due to heating
P
S
I
I ideal
then we have: (eqs.2)
If we divide eqs.2 with eqs.1 we get:
In a step-up transformer ( ) we have that
In a step-down transformer ( )
P P S
P P S S
S SP P
P S S P
PS P
S
S P S P
S P
S
V I V I
V IV I
V N V N
NI I
N
N
I N I N
N I I
N N
we have that S PI I
We have that:
(eqs.1)
S P
S P
S P P S
V V
N N
V N V N
S P
S P
V V
N N
S S P PI N I N
(31 - 27)