lecture 30: fri 27 mar 09 ch. 31.4–7: electrical oscillations, lc circuits, alternating current
DESCRIPTION
Physics 2102 Jonathan Dowling. Lecture 30: FRI 27 MAR 09 Ch. 31.4–7: Electrical Oscillations, LC Circuits, Alternating Current. EXAM 03: 6PM THU 02 APR LOCKETT 5 EXAM 03 REVIEW: 6PM WED 01 APR NICHOLSON 130. - PowerPoint PPT PresentationTRANSCRIPT
Physics 2102
Jonathan Dowling
Lecture 30: FRI 27 MAR 09Lecture 30: FRI 27 MAR 09
Ch. 31.4–7: Electrical Ch. 31.4–7: Electrical Oscillations, LC Circuits, Oscillations, LC Circuits,
Alternating CurrentAlternating CurrentQuickTime™ and a
decompressorare needed to see this picture.
QuickTime™ and a decompressor
are needed to see this picture.
EXAM 03: 6PM THU 02 APR LOCKETT 5 EXAM 03 REVIEW: 6PM WED 01 APR NICHOLSON 130
The exam will cover: Ch.28 (second half) through Ch.32.1-3 (displacement current, and Maxwell's equations). The exam will be based on: HW08 – HW11.
Final Day to Drop Course: TODAY FRI 27 MAR
The formula sheet for the exam can be found here:http://www.phys.lsu.edu/classes/spring2009/phys2102/formulasheet.pdf
You can see examples of old exam IIIs here:http://www.phys.lsu.edu/classes/spring2009/phys2102/Test3.oldtests.pdf
Damped LCR OscillatorDamped LCR Oscillator
Ideal LC circuit without resistance: oscillations go
on for ever; = (LC)–1/2
Real circuit has resistance, dissipates energy:
oscillations die out, or are “damped”
Math is complicated! Important points:
– Frequency of oscillator shifts away from
= (LC)-1/2
– Peak CHARGE decays with time constant =
QLCR=2L/R
– For small damping, peak ENERGY decays with
time constant
ULCR= L/R 0 4 8 12 16 200.0
0.2
0.4
0.6
0.8
1.0
UE
time (s)
Umax =Q2
2Ce−
RtL
C
RL
2
2
If we add a resistor in an circuit (see figure) we must
modify the energy equation, because now energy is
being dissipated on the resistor: .
E B
RL
dUi R
dt
qU U U
=−
= + =
Damped Oscillations in an CircuitRCL
22
2 2Li dU q dq di
Li i RC dt C dt dt+ → = + =−
( )
2
2
2
/2
2
2 2
10. This is the same equation as that
of the damped harmonics osc 0, which has theillator:
The angular f
solution
re( ) c que
:
os
.bt mm
d x dxm b kxdt dt
x
dq di d q d q dqi L R qdt dt dt dt dt
t x e t
C
ω φ−
= → = → + + =
+ + =
′= +
( )2
/22
2
2ncy
For the damped circuit the solution is:
The angular freque1
( ) cos . .4
ncy
.
4
Rt L Rq
k b
RC
t Qe t
m
L
L
C L
m
ω φ ω
ω
− ′ ′ −
−
+
=
= =
′
(31-6)
/ 2Rt LQe−
/ 2Rt LQe−
( )q tQ
Q−
( )q t ( )/ 2( ) cosRt Lq t Qe tω φ− ′= +
2
2
1
4
R
LC L′= −
/2
2
2
The equations above describe a harmonic oscillator with an exponentially decaying
amplitude . The angular frequency of the damped oscillator
1 is always smaller than the angular
4
Rt LQe
R
LC L
−
′= −
2
2
1frequency of the
1undamped oscillator. If the term we can use the approximation .
4
LCRL LC
=
′≈=
RC = RC τ RL = L / R τ RCL = 2L / R
SummarySummary• 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 wouldoscillate “naturally”. The other is the frequency at which we drive theoscillation.
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).
We have studied that a loop of wire,spinning in a constant magnetic fieldwill have an induced emf that oscillates with time,
€
E =Em sin(ωd t)
That is, it is an AC generator.
Alternating Current:
AC’s are very easy to generate, they are also easy to amplify anddecrease in voltage. This in turn makes them easy to send in distributiongrids like the ones that power our homes.
Because the interplay of AC and oscillating circuits can be quitecomplex, we will start by steps, studying how currents and voltagesrespond in various simple circuits to AC’s.
AC Driven Circuits:
1) A Resistor:
0=− Rvemf
€
vR = emf = Em sin(ωd t)
€
iR =vRR
=Em
Rsin(ωd t)
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 torepresent magnitudes using “phasors”. These are vectorsthat rotate at a frequency d , their magnitude is equalto the amplitude of the quantity in question and theirprojection on the vertical axis represents the instantaneous value of the quantity under study.
AC Driven Circuits:
2) Capacitors:
€
vC = emf = Em sin(ωd t)
€
qC =Cemf =CEm sin(ωd t)
€
iC =dqCdt
=ωdCEm cos(ωd t)
€
iC =ωdCEm sin(ωd t + 900)
€
iC =Em
Xsin(ωd t + 900)
reactance"" 1
whereC
Xd
=
€
im =Em
X
€
looks like i =V
RCapacitors “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:
3) Inductors:
€
vL = emf = Em sin(ωd t)
Ldtv
idtid
Lv LL
LL
∫=⇒=
€
iL = −Em
Lωdcos(ωd t)
€
=Em
Lωdsin(ωd t − 900)
€
iL =Em
Xsin(ωd t − 900)
€
im =Em
X dLX = where
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).
Power Station
Transmission lines Erms =735 kV , I rms = 500 A Home
110 V
T1T2
Step-up transformer
Step-down transformer
R = 220Ω
1000 km=l
2heat rms
The resistance of the power line . is fixed (220 in our example).
Heating of power lines . This parameter is also fixed
(55 MW in our ex
R RA
P I R
ρ= Ω
=
Energy Transmission Requirements
l
trans rms rms
heat trans
heat rms
ample).
Power transmitted (368 MW in our example).
In our example is almost 15 % of and is acceptable.
To keep we must keep as low as possible. The onl
P I
P P
P I
=E
rms rms
y way to accomplish this
is by . In our example 735 kV. To do that we need a device
that can change the amplitude of any ac voltage (either increase or decrease).
=increasing E E
(31-24)
Ptrans=iV
= “big”
Pheat=i2R
= “small”
Solution:
Big V!
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 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 coil, which is known
as the "secondary" and has turnsP
S
N
N
primary
. 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 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 . The magnetic flux
through the primary
S
P
V
B A
Φ ( ).
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
Φ= → =− =−
ΦΦ = → =− =−
eq. 1
eq. 2(31-25)
( )
( )
If we divide equation 2 by equati
.
on 1 we get:
S P
S P
PP P P P
SS S S S
SS S
P PP
d dBN BA V N A
dt dtd dB
N BA V N Adt dt
dBN AV Ndt
dBV NN Adt
V V
N N
ΦΦ = → =− =−
ΦΦ = → =− =−
−= =−
=→
eq. 1
eq. 2
The voltage on the secondary .
If 1 , we have what is known as a " " transformer.
If 1 , we have what is known as a " " transformer.
Both type
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
s of transformers are used in the transport of electric power over large distances.
S P
S P
V V
N N=
(31-26)
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 heatin
P
S
I
I ideal,
g. Then we have: (eq. 2).
If we divide eq. 2 with eq. 1 we get:
In a step-up transformer ( ) we have that .
In a step-down transformer
.
(
P
P P S S
S SP P
P S S P
PS
P S S
PS
S P S P
I N
V I V I
V IV I
V N V N
N
I N
I IN
N N I I
N
=
= →
<
=
=
>) we have that .S P S PN I I< >
We have that:
(eq. 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)