Transcript PPT - LSU Physics & Astronomy

Physics 2102
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Jonathan Dowling
Lecture 21: THU 01 APR 2010
Ch. 31.4–7: Electrical Oscillations, LC
Circuits, Alternating Current
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LC Circuit: At t=0 1/3 Of Energy Utotal is on Capacitor C and Two
Thirds On Inductor L. Find Everything! (Phase =?)
1
2
U B (t)  L q0 sin( t   0 )
2
1
2
U E t  
q0 cos( t   0 )

2C
2
1


L
q

sin(

)
0 
U /3
U B (0) 2  0

 total
2
1
2U total / 3
UE 0
 q0 cos( 0 ) 
2C

tan( 0 ) 
1
2
 
 0  arctan 1 / 2  35.3
1
2
U B (0)  L q0 sin( 0 )  U total / 3
2
1
2
U E 0  
q
cos(

)

0
0   2U total / 3
2C
2
LC  q0 sin( 0 ) 
1   1 / LC

2
2 q0  VC
 q0 cos( 0 ) 
q  q0 cos( t   0 )
i(t)  q0 sin( t   0 )
i (t)   2 q0 cos( t   0 )
q
VL (t)   0 cos( t   0 )
C
q
VC (t)  0 cos( t   0 )
C
Damped LCR Oscillator
Ideal LC circuit without resistance: oscillations go
on forever;  =
C
(LC)–1/2
L
R
Real circuit has resistance, dissipates energy:
oscillations die out, or are “damped”
Math is complicated! Important points:
– Frequency of oscillator shifts away from
1.0
 = (LC)-1/2
0.8
 QLCR=2L/R
– For small damping, peak ENERGY decays with
time constant
 ULCR= L/R
UE
– Peak CHARGE decays with time constant =
U max
Q 2  RtL

e
2C
0.6
0.4
0.2
0.0
0
4
8
12
time (s)
16
20
Damped Oscillations in an RCL Circuit
If we add a resistor in an RL circuit (see figure) we must
modify the energy equation, because now energy is
dU
being dissipated on the resistor:
 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
dq 1
d 2q
di d 2 q
dq
  2  L 2  R  q  0. This is the same equation as that
i
dt C
dt
dt dt
dt
dx
d 2x
of the damped harmonics oscillator: m 2  b  kx  0, which has the solution:
dt
dt
b2
k
.

cos  t    . The angular frequency   
x(t )  xm e
2
m 4m
For the damped RCL circuit the solution is:
 bt /2 m
q(t )  Qe
 Rt /2 L
cos  t    . The angular frequency   
R2
1
 2.
LC 4 L
(31-6)
q(t )
Qe
q(t )  Qe Rt / 2 L cos t   
 Rt / 2 L
Q
q(t )
 
Q
1
R2
 2
LC 4 L
Qe Rt / 2 L
The equations above describe a harmonic oscillator with an exponentially 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
 RC  RC
 RL  L / R
 RCL  2L / R
Summary
• Capacitor and inductor combination produces
an electrical oscillator, natural frequency of
oscillator is =1/√LC
• Total energy in circuit is conserved: switches
between capacitor (electric field) and inductor
(magnetic field).
• If a resistor is included in the circuit, the total
energy decays (is dissipated by R).
Alternating Current:
To keep oscillations going we need to
drive the circuit with an external emf
that produces a current that goes back
and forth.
Notice that there are two frequencies
involved: one at which the circuit would
oscillate “naturally”. The other is the frequency at which we drive
the
oscillation.
However, the “natural” oscillation usually dies off quickly
(exponentially) with time. Therefore in the long run,
circuits actually oscillate with the frequency at which
they are driven. (All this is true for the gentleman trying
to make the lady swing back and forth in the picture too).
Alternating Current:
We have studied that a loop of wire,
spinning in a constant magnetic field
will have an induced emf that
oscillates with time,
E  Em sin( d t)
That is, it is an AC generator.
AC’s are very easy to generate, they are also easy to amplify and
decrease in voltage. This in turn
 makes them easy to send in distribution
grids like the ones that power our homes.
Because the interplay of AC and oscillating circuits can be quite
complex, we will start by steps, studying how currents and voltages
respond in various simple circuits to AC’s.
AC Driven Circuits:
1) A Resistor:
emf  vR  0
vR  emf  Em sin( d t)
v R Em
iR 

sin(  d t)
R
R
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Charles
Steinmetz

Resistors behave in AC very much as in DC,
current and voltage are proportional
(as functions of 
time in the case of AC),
that is, they are “in phase”.
For time dependent periodic situations it is useful to
represent magnitudes using Steinmetz “phasors”. These
are vectors that rotate at a frequency d , their
magnitude is equal to the amplitude of the quantity in
question and their projection on the vertical axis
represents the instantaneous value of the quantity.
AC Driven Circuits:
2) Capacitors:


vC  emf  Em sin( d t)
qC  Cemf  CEm sin( d t)
dqC
iC 
  d CEm cos( d t)
dt
iC   d CE m sin(  d t  90 0 )
Em
iC 
sin(  d t 
900 )
X
1 
where X 
" reactance"
d C
Em
im 
X
V
looks like i =
R
Capacitors “oppose a resistance” to AC
(reactance) of frequency-dependent magnitude 1/d C
(this idea is true only for maximum amplitudes,
the instantaneous
story is more complex).

AC Driven Circuits:
vL  emf  Em sin( d t)
3) Inductors:
d iL
 v L dt
vL  L
 iL 
dt
L


Em

im 
where
X
Em
Em
iL  
cos( d t) 
sin(  d t  900 )
L d
L d
Em
iL 
sin(  d t  900 )
X

X  L d
Inductors “oppose a resistance” to AC
(reactance) of frequency-dependent magnitude d L
(this idea is true only for maximum amplitudes,
the instantaneous story is more complex).
Transmission lines
Erms =735 kV , I rms = 500 A
Home
Step-down
transformer 110 V
Step-up
transformer
T2
T1
R = 220Ω
 1000 km
Power Station
The resistance of the power line R 
Heating of power lines Pheat
= “big”
Pheat=i2R
= “small”
Solution:
Big V!
Energy Transmission Requirements

Ptrans=iV
.
R is fixed (220  in our example).
A
2
 I rms
R. This parameter is also fixed
(55 MW in our example).
Power transmitted Ptrans  Erms I rms
(368 MW in our example).
In our example Pheat is almost 15 % of Ptrans and is acceptable.
To keep Pheat we must keep I rms as low as possible. The only way to accomplish this
is by increasing Erms . In our example Erms  735 kV. To do that we need a device
that can change the amplitude of any ac voltage (either increase or decrease).
(31-24)
The DC vs. AC Current Wars
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Thomas Edison pushed
for the development of a
DC power network.
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George Westinghouse
backed Tesla’s development
of an AC power network.
Nikola Tesla was
instrumental in developing
AC networks.
Edison was a brute-force experimenter, but was no mathematician. AC cannot be
properly understood or exploited without a substantial understanding of mathematics and
mathematical physics, which Tesla possessed.
The Tesla Three-Phase AC Transmission System
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The most common example is the Tesla three-phase power system used for
industrial applications and for power transmission. The most obvious advantage of
three phase power transmission using three wires, as compared to single phase
power transmission over two wires, is that the power transmitted in the three phase
system is the voltage multiplied by the current in each wire times the square root of
three (approximately 1.73). The power transmitted by the single phase system is
simply the voltage multiplied by the current. Thus the three phase system transmits
73% more power but uses only 50% more wire.
Niagara Falls and Steinmetz’s Turning of the Screw
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Against General Electric and Edison's proposal, Westinghouse, using Tesla's
AC system, won the international Niagara Falls Commission contract. Tesla’s
three-phase AC transmission became the World’s power-grid standard.
Transforming DC power from one voltage to another was difficult and
expensive due to the need for a large spinning rotary converter or motorgenerator set, whereas with AC the voltage changes can be done with simple
and efficient transformer coils that have no moving parts and require no
maintenance. This was the key to the success of the AC system. Modern
transmission grids regularly use AC voltages up to 765,000 volts.
The Transformer
The transformer is a device that can change
the voltage amplitude of any ac signal. It
consists of two coils with a 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 coil, which is known
as the "secondary" and has N S turns. The role of the iron core is to ensure 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
(eq. 1).
dt
dt
dS
dB
The flux through the secondary  S  N S BA  VS  
 NS A
dt
dt
(eq. 2).
(31-25)
VS
V
 P
NS NP
dP
dB
 NP A
(eq. 1)
dt
dt
dS
dB
 S  N S BA  VS  
 NS A
(eq. 2)
dt
dt
If we divide equation 2 by equation 1 we get:
dB
NS A
VS
dt  N S  VS  VP .

VP  N A dB
NP
NS NP
P
dt
 P  N P BA  VP  
The voltage on the secondary VS  VP
NS
.
NP
If N S  N P 
NS
 1  VS  VP , we have what is known as a "step up" transformer.
NP
If N S  N P 
NS
 1  VS  VP , we have what is known as a "step down" transformer.
NP
Both types of transformers are used in the transport of electric power over large
distances.
(31-26)
IS
IP
VS
V
 P
NS NP
IS NS  I P NP
VS
VP
We have that:

NS NP
 VS N P  VP N S
(eq. 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 eq. 2 with eq. 1 we get:
IS 
(eq. 2).
VI
VP I P
 S S  I P 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 .
(31-27)