Download 31 - 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
1
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
2
LR Circuit
i
Steady Source
sum of voltage drops  0 :
di
 E  iR  L  0
dt
same form as the
capacitor equation
q
dq
E R
0
C
dt
Let E  0, then
Induction - Spring 2006
3
VR=iR
~current
Max Current Rate of
increase = max emf
Induction - Spring 2006
4
E
(1  e Rt / L )
R
L
  (time constant)
R
i
Induction - Spring 2006
5
Time Dependent Result:
E
 Rt / L
i  (1  e
)
R
time constant

L

R
Induction - Spring 2006
6
We also showed that
1
2
uinductor 
B
20
1
2
ucapacitor   0 E
2
Induction - Spring 2006
7
At t=0, the charged capacitor is now
connected to the inductor. What would
you expect to happen??
Induction - Spring 2006
8
L
C
We will show that the charge q on the capacitor plates as well as the current i
1
in the inductor oscillate with constant amplitude at an angular frequency  
LC
The total energy U in the circuit is the sum of the energy stored in the electric field
q 2 Li 2
of the capacitor and the magnetic field of the inductor. U  U E  U B   .
2C 2
dU
The total energy of the circuit does not change with time. Thus
0
dt
dU q dq
di
dq di d 2q
d 2q 1

 Li  0. i    2  L 2  q  0
dt C dt
dt
dt dt dt
dt C
(31 - 2)
Induction - Spring 2006
9
d 2q 1
d 2q  1 
L 2  q  0  2 
 q  0 (eqs.1)
dt
C
dt
 LC 
C
This is a homogeneous, second order, linear differential equation
which we have encountered previously. We used it to describe
L
the simple harmonic oscillator (SHO)
q(t )  Q cos t   

d 2x
2


x  0 (eqs.2)
2
dt
with solution: x(t )  X cos(t   )
1
LC
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
q (t )  Q cos t    where  
, and  is the phase angle.
LC
dq
The current i 
 Q sin t   
dt
(31 - 3)
Induction - Spring 2006
10
L
The energy stored in the electric field of the capacitor
C
q2 Q2
UE 

cos 2 t   
2C 2C
The energy stored in the magnetic field of the inductor
Li 2 L 2Q 2
Q2
2
UB 

sin t    
sin 2 t   
2
2
2C
The total energy U  U E  U B
 Q2 
Q2
2
2
U 
 cos t     sin t     
2C
 2C 
The total energy is constant; energy is conserved
Q2
T
3T
The energy of the electric field has a maximum value of
at t  0, , T , ,...
2C
2
2
Q2
T 3T 5T
The energy of the magnetic field has a maximum value of
at t  , , ,...
2C
4 4 4
Note : When U E is maximum U B is zero, and vice versa
(31 - 4)
Induction - Spring 2006
11
The Graph of that LC (no emf) circuit ..
Induction - Spring 2006
12
Induction - Spring 2006
13
Mass on a Spring Result


Energy will swap back and forth.
Add friction


Oscillation will slow down
Not a perfect analogy
Induction - Spring 2006
14
Induction - Spring 2006
15
LC Circuit
Low
High
Q/C
High
Low
Induction - Spring 2006
16
The Math Solution (R=0):
  LC
Induction - Spring 2006
17
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
18
Variable Emf Applied
1.5
1
Volts
emf
0.5
DC
0
0
1
2
3
4
5
6
7
8
9
10
-0.5
-1
Sinusoidal
-1.5
Tim e
Induction - Spring 2006
19
Sinusoidal Stuff
emf  A sin( t   )
“Angle”
Phase Angle
Induction - Spring 2006
20
Same Frequency
with
PHASE SHIFT

Induction - Spring 2006
21
Different Frequencies
Induction - Spring 2006
22
Damped oscillations in an RCL circuit
If we add a resistor in an RL cicuit (see figure) we must
modify the energy equation because now energy is
dU
 i 2 R
dt
q 2 Li 2
dU q dq
di
U  UE UB 



 Li  i 2 R
2C
2
dt C dt
dt
being dissipated on the resistor.
dq di d 2 q
d 2q
dq 1
i    2  L 2  R  q  0 This is the same equation as that
dt
dt dt
dt
dt C
d 2x
dx
of the damped harmonics oscillator: m 2  b  kx  0 which has the solut
dt
dt
2
k
b
x(t )  xm e bt / 2 m cos  t    The angular frequency   
 2
m 4m
For the damped RCL circuit the solution is:
q(t )  Qe
 Rt / 2 L
cos  t   
1 R2
The
angular frequency   
 (312- 6)23
Induction - Spring 2006
LC 4 L
q(t )
Qe
q(t )  Qe Rt / 2 L cos t   
 Rt / 2 L
q(t )
Q
 
Q
1
R2
 2
LC 4 L
Qe Rt / 2 L
The equations above describe a harmonic oscillator with an exponetially decaying
amplitude Qe  Rt / 2 L . The angular frequency of the damped oscillator
1
R2
1
 
 2 is always smaller than the angular frequency  
of the
LC 4 L
LC
R2
1
undamped oscillator. If the term
we can use the approximation    
2
4L
LC
Induction - Spring 2006
(31 - 7) 24
Note – Power is delivered to our homes
as an oscillating source (AC)
Induction - Spring 2006
25
Producing AC Generator
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Induction - Spring 2006
26
The Real World
Induction - Spring 2006
27
A
Induction - Spring 2006
28
Induction - Spring 2006
29
The Flux:
  B  A  BA cos 
  t
emf  BA sin t
emf
i
A sin t
Rbulb
Induction - Spring 2006
30
Source Voltage:
emf  V  V0 sin( t )
Induction - Spring 2006
31
VR  I R R
A resistive load
In fig.a we show an ac generator connected to a resistor R
E E
From KLR we have: E  iR R  0  iR   m sin t
R R
E
The current amplitude I R  m
R
The voltage vR across R is equal to Em sin t
The voltage amplitude is equal to Em
The relation between the voltage and
current amplitudes is: VR  I R R
In fig.b we plot the resistor current iR and the
resistor voltage vR as function of time t.
Both quantities reach their maximum values
at the same time. We say that voltage and
current are in phase.
(31 - 10)
Induction - Spring 2006
32
2
2
Em
Em
 P 
Vrms 
2R
2
Average Power for R
T
1
 P   P (t )dt
T 0
2
2
E
P  m sin 2 t 
R
T
E 1
2
 P  m
si
n
tdt

R T 0
T
1
1
2
sin

td
t

T 0
2
2
E
  P  m
2R
We define the "root mean square" (rms) value of V as follows:
2
2
Em
Vrms
Vrms 
  P 
The equation looks the same
R
2
as in the DC case. This power appears as heat on R
Induction - Spring 2006
(31 - 12)
33
A capacitive load
In fig.a we show an ac generator connected to a
capacitor C
XC
q
From KLR we have: E  C  0 
C
qC  EC  EmC sin t
O

dqC
iC 
 EmC cos tdt  sin t  90 
1
dt
XC 
The voltage amplitude VC equal to Em
C
VC
1/ C
The quantity X C  1 / C is known as the
The current amplitude I C  CVC 
capacitive reactance
In fig.b we plot the capacitor current iC and the capacitor
voltage vC as function of time t. The current leads the
voltage by a quarter of a period. The voltage and
(31 - 13)
current Induction
are out- Spring
of 2006
phase by 90 .
34
Average Power for C
PC  0
2
E
P  VC I C  m sin t cos t
XC
2
E
 P  m sin 2t
2XC
2sin  cos   sin 2
T
2
T
E 1
1
 P   P(t )dt = m
sin 2tdt  0

T 0
2XC T 0
Note : A capacitor does not dissipate any power
on the average. In some parts 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!
Induction - Spring 2006
(31 - 14)
35
An inductive load
In fig.a we show an ac generator connected to an inductor L
diL
diL E Em
From KLR we have: E  L
0
  sin t
dt
dt L L
E
E
E
iL   diL   m sin tdt   m cos tdt  m sin t  90 
L
L
L
The voltage amplitude VL equal to Em
XL
VL
The current amplitude I L 
L
The quantity X L   L is known as the

O
inductive reactance
In fig.b we plot the inductor current iL and the
X L  L
inductor voltage vL as function of time t.
The current lags behind the voltage by a
quarter of a period. The voltage and
current areInduction
out of
phase
- Spring
2006 by 90 .
(31 - 15)
36
PL  0
Average Power for L
2
Em
Power P  VL I L  
sin t cos t
XL
2
E
2sin  cos   sin 2  P   m sin 2t
2X L
T
2
T
Em 1
1
 P   P(t )dt = 
sin 2tdt  0

T0
2X L T 0
Note : A inductor does not dissipate any power
on the average. In some parts 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!
Induction - Spring 2006
(31 - 16)
37
SUMMARY
Circuit
element
Average
Power
Resistor
R
E
 PR   m
2R
 PC   0
Capacitor
C
Inductor
L
Reactanc
e
Voltage amplitude
R
Current is in phase
with the voltage
VR  I R R
1
XC 
C
Current leads
voltage by a quarter
of a period
IC
VC  I C X C 
C
X L  L
Current lags behind
voltage by a quarter VL  I L X L  I LL
of a period
2
 PL   0
Phase of current
Induction - Spring 2006
(31 - 17)
38
(31 - 18)
The series RCL circuit
An ac generator with emf E  Em sin t is connected to
an in series combination of a resistor R, a capacitor C
and an inductor L, as shown in the figure. The phasor
for the ac generator is given in fig.c. The current in
i  I sin t   
this circuit is described by the equation: i  I sin t   
The current i is common 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 vR across R, the voltage vC across C , and the voltage vL across L.
The voltage vR is in phase with the current i. The voltage vC lags behind
the current i by 90. The voltage vInduction
ahead of the current i by 90.
L leads
- Spring 2006
39
A
B
i  I sin t   
Z  R2   X L  X C 
O
I
2
Em
Z
Kirchhoff's loop rule (KLR) for the RCL circuit: E  vR  vC  vL . This equation
is represented in phasor form in fig.d. Because VL and VC have opposite directions
we combine the two in a single phasor VL  VC . From triangle OAB we have:
2
2
2
2
Em2  VR2  VL  VC    IR    IX L  IX C   I 2  R 2   X L  X C   


Em
I
The denominator is known as the "impedance" Z
2
2
R   X L  XC 
of the circuit. Z  R 2   X L  X C   The current amplitude I 
2
I
Em
Z
Em
1 

2
R  L 
C 

2
Induction - Spring 2006
(31 - 19)
40
i  I sin t   
(31 - 20)
1
XC 
C
A
B
Z  R2   X L  X C 
tan  
O
X L  XC
R
X L  L
From triangle OAB we have: tan  
VL  VC IX L  IX C X L  X C


VR
IR
R
We distinguish the following three cases depending on the relative values
of X L and X L .
1. X L  X C    0 The current phasor lags behind the generator phasor.
The circuit is more inductive than capacitive
2. X C  X L    0 The current phasor leads ahead of the generator phasor
The circuit is more capacitive than inductive
3. X C  X L    0 The current phasor and the generator phasor are in phase
Induction - Spring 2006
2
41
1. Fig.a and b: X L  X C    0
The current phasor lags behind
the generator phasor. The circuit is more
inductive than capacitive
2. Fig.c and d: X C  X L    0 The current phasor leads ahead of the generator
phasor. The circuit is more capacitive than inductive
3. Fig.e and f: X C  X L    0 The current phasor and the generator phasor are
(31 - 21)
in phase
Induction - Spring 2006
42

1
LC
I res
Em

R
Resonance
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:
I
Em
1 

R2    L 


C


2
The current amplitude
has a maximum when the term  L 
This occurs when  
1
0
C
1
LC
Em
The equation above is the condition for resonance. When its is satisfied I res 
R
A plot of the current amplitude I as function of  is shown in the lower figure.
This plot is known as a "resonance cInduction
urve"- Spring 2006
(31 - 22)
43
2
Pavg  I rms
R
Pavg  I rms Erms cos 
(31 - 23)
Power in an RCL ciruit
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 2 R   I sin t     R
2
T
The average power Pavg
1
  Pdt
T 0
1 T
 I 2R
2
2
Pavg  I R   sin t    dt  
 I rms
R
2
T 0

E
R
Pavg  I rms RI rms  I rms R rms  I rms Erms  I rms Erms cos 
Z
Z
The term cos in the equation above is known as
the "power factor" of the circuit. The average
power consumed by the circuit is maximum
whenInduction
  0- Spring 2006
44
2
(31 - 25)
The transformer
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 coil on which we apply the voltage to be changed is called the "primary" and
it has N P turns. The transformer output appears on the second coils which is known
as the "secondary" and has N S turns. 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 VP is applied across the primary then a voltage VS appears
on the secondary coil. We also assume that the magnetic field through both coils
is equal to B and that the iron core has cross sectional area A. The magnetic flux
dP
dB
through the primary  P  N P BA  VP  
 NP A
(eqs.1)
dt
dt
dS
dB
The flux through the secondary  S  N S BA  VS  
 NS A
(eqs.2)
dt
dt
45
Induction - Spring 2006
VS
V
 P
NS NP
dP
dB
 NP A
(eqs.1)
dt
dt
dS
dB
 S  N S BA  VS  
  NS A
(eqs.2)
dt
dt
If we divide equation 2 by equation 1 we get:
 P  N P BA  VP  
dB
VS
dt  N S  VS  VP

VP  N A dB
NP
NS NP
P
dt
NS A
The voltage on the secondary VS  VP
NS
NP
If N S  N P 
NS
 1  VS  VP We have what is known a "step up" transformer
NP
If N S  N P 
NS
 1  VS  VP We have what is known a "step down" transformer
NP
Both types of transformers are used in the transport of electric power over large
distances.
Induction - Spring 2006
(31 - 26)
46
VS
V
 P
NS NP
IS
IP
IS NS  I P NP
VS
VP
We have that:

NS NP
 VS N P  VP N S
(eqs.1)
If we close switch S in the figure we have in addition to the primary current I P
a current I S in the secondary coil. We assume that the transformer is "ideal"
i.e. it suffers no losses due to heating then we have: VP I P  VS I S
If we divide eqs.2 with eqs.1 we get:
IS 
(eqs.2)
VI
VP I P
 S S  IP NP  IS NS
VP N S
VS N P
NP
IP
NS
In a step-up transformer (N S  N P ) we have that I S  I P
In a step-down transformer (N S  N P ) we have that I S  I P
Induction - Spring 2006
(31 - 27)
47