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Chapter 31 Electromagnetic Oscillations and Alternating Current In this chapter we will cover the following topics: - PowerPoint PPT Presentation
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Chapter 31
Electromagnetic Oscillations and Alternating CurrentIn 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)
A horizontal power line carries a current of 5000 A from south to north. Earth's magnetic eld (60 μT) is directed towards north and inclined down-ward 50 degrees to the horizontal. Find the magnitude and direction of the magnetic force on 100 m of the line due to the Earth's field.(1)23 N, west (2) 23 N, east (3) 30N, west(4) 30N, east (5) none of these
50
BN
I
Suppose this page is perpendicular to a uniform magnetic field and the magnetic flux through it is 5Wb. If the page is turned by 30◦ around an edge the flux through it will be:
A. 2.5WbB. 4.3WbC. 5WbD. 5.8WbE. 10Wb
A car travels northward at 75 km/h along a straight road in a region where Earth’s magnetic field has a vertical component of 0.50 × 10−4 T. The emf induced between the left and rightside, separated by 1.7m, is:
A. 0B. 1.8mVC. 3.6mVD. 6.4mVE. 13mV
LC The circuit shown in the figure consists of a capacitor
and an inductor . We give the capacitor an initialchanrge and then abserve what happens. The capacitorwill discharge th
CL
Q
LC Oscillations
rough the inductor resulting in a timedependent current . i
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 Bq LiU U UC
dU
2
2
2
2
0
0. 1 0
dtdU q dq di dq di d qLi idt C d
d qLt dt dt
qdtdt dt C
(31 - 2)
LC
2
2
2
2
1 0 ( )
This is a homogeneous, second order, linear differential equationwhich we have encountered previously. We used it to d
1 0
escribethe simple harmonic oscillat
o
d qL qdt C
d q qdt LC
eqs.1
22
2 0
with sol
r (SHO)
( )
ution: ( ) cos( )
d x xdt
x t X t
eqs.2
If we compare eqs.1 with eqs.2 we find that the solution to the differentialequation that describes the LC-circuit (eqs.1) is:
1( ) cos where , and is the phase angle.
The current
q t Q tLC
sindqi Q tdt
( ) co sq t Q t
1 LC
(31 - 3)
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 tC C
Li L Q QU t tC
U U U
QU
2
2 2cos sin2
The total energy is constant;
Qt tC 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 TCQ T T TtC
U U
electric field maximum
magnetic field maximum
Note : ro, and vice versa(31 - 4)
0t
1
2
/ 8t T
3
/ 4t T
4
3 / 8t T
5
5/ 2t T
432
1
6
6
5 / 8t T
3 / 4t T
7 / 8t T
7
8
7
8
(31 - 5)
2
2
If we add a resistor in an RL cicuit (see figure) we mustmodify the energy equation because now energy is
being dissipated on the resistor.
2E B
dU i Rdt
qU U UC
Damped oscillations in an RCL circuit
22
2Li dU q dq diLi i R
dt C dt dt
2
2
2
/ 2
2
2 2
1 0 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 qdt dt dt dt d
d x dxm b kxdt 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 bm m
t Qe tLC L
(31 - 6)
/ 2Rt LQe
/ 2Rt LQe
( )q tQ
Q
( )q t / 2( ) cosRt Lq t Qe t
2
2
14R
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 f4
Rt LQe
RLC L
2
2
1requency of the
1undamped oscillator. If the term we can use the approximation 4
LCRL LC
(31 - 7)
A battery for which the emf is constant generates a current that has a constant direction. This typeof current is known as " " or " "In chapter 30 we encountered a d
Alternating Current
dcdirect current ifferent type
of sourse (see figure) whose emf is:
sin sin where , is the area of the generatorwindings, is the number of the windings, is the angular frequency of therotation of the windings, and is the magnetic field.
m mNAB t t NAB AN
B
E E E
This type of generator is known as " " or " " because the emf as well as the currentchange direction with a frequency 2 . In the US 60 Hz. Almost all commercial electrical
f f acalternating current
power used today is ac even though the analysis of ac circuits is more complicated than that of dc circuits.The reasons why ac power was adapted will be discussed at the end of thischapter.
(31 - 8)
sin m tE E
LC
Our objective is to analyze the circuit shown in thefigure ( circuit). The discussion will be greatly simplified if we examine what happens if we connecteach of the three elem
RCL
Three Simple Circuits
ents ( , , and ) separatelyto an ac generator.
R C L
From now on we will use the standard notation for ac circuitanalysis. Lower case letters will be used to indicate the
values of ac quantities. Upper case letterswill be used
A convention
instantaneous
to indicate the constant amplitudes of ac quantities.Example: The capacitor charge in an LC circuit was written as:
cos
The symbol is used for the instantaneous value of the charge The symbol
q Q t
q
is used for the constant amplitude of Q q
(31 - 9)
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 sinThe voltage
mR R
mR
R m
R
i R i tR R
IR
v R t
A resistive load
EEEE
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 Ri
v
E
ties reach their maximum values at the same time. We say that voltage andcurrent are .in phase
(31 - 10)
R RV I R
A convenient method for the representation of acquantities is that of phasors
The resistor voltage and the resistor current are representedby rotating vectors known as phasors using the following conventions:
Phasors rotate in the counterclockwise direction with angula
R Rv i
1. r speed The length of each phasor is proportional to tha ac quantity amplitude The projection of the phasor on the vertical axis gives the instantaneous
value of the ac quantity. The rotation
2.3.
4. angle for each phasor is equal to the phase of theac quantity ( in this example)t (31 - 11)
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
VPR
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)
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
CqC
q C C tdqi C tdt tdt
V
A capacitive load
EE E
EE
plitude 1/
The quantity is known as the
In fig.b we plot the capacitor current and the capacitorvoltage as function of time t
1
. The current
thev
/
CC
C
C
C
CVI CV
C
iv
X C
leads
capacitive reactance
oltage by a quarter of a period. The voltage andcurrent are .out of phase by 90
O
CX
(31 - 13)
1CX
C
2
2
2
0 0
sin cos
sin 22
1 1( ) = sin 2 02
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 absorbesenergy from the ac generator but at the rest of the cycleit gives the energy back so that on the average nopower is used!
0CP
(31 - 14)
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
Ldi diL tdt dt L L
i di tdt tdt tL L L
V
An inductive load
EEEE 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
VIL
iv
X L
E
inductive reactance
the voltage by a quarter of a period. The voltage andcurrent are .
lags behind
out of phase by 90
O
LX
(31 - 15)
LX L
2
2
2
0 0
Power sin cos
sin 22
1 1( ) = sin 2 02
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 absorbesenergy from the ac generator but at the rest of the cycleit gives the energy back so that on the average nopower is used!
0LP
(31 - 16)
Circuit element
Average Power
Reactance Phase of current Voltage amplitude
ResistorR
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 XC
L L L LV I X I L
R
SUMMARY
2
2m
RPR
E
0CP
0LP
(31 - 17)
An ac generator with emf is connected toan in series combination of a resistor , a capacitor and an inductor , as shown in the figure. The phasorfor the ac genera
sinm
R CL
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 inductorThe phasor for the current is shown in fig.a. In fig.c we show the phasors for thevoltage 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 behindthe current by 90 . The voltage leads ahead of the current by 90 .
L
R C
L
v Lv i v
i v i
(31 - 18)
O
A B
Kirchhoff's loop rule (KLR) for the RCL circuit: . This equationis represented in phasor form in fig.d. Because and have opposite directionswe combine the two in a single phasor
R C L
L C
L
v v vV 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
EE
E
impedance
22
1m
Z
I
R LC
E
22 L CZ R X X
mIZ
E
sini I t
(31 - 19)
O
A B
From triangle OAB we have: tan
We distinguish the following three cases depending on the relative valuesof 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 XV IR R
X XX X
1. or phasor.The circuit is more inductive than capacitive
0 The current phasor leads ahead of the generator phasorThe 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 XR
1 CXC
(31 - 20)
Fig.a and b: 0 The current phasor lags behind the generator phasor. The circuit is moreinductive than capacitive
L CX X 1.
Fig.c and d: 0 The current phasor leads ahead of the generatorphasor. 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)
In the RCL circuit shown in the figure assume thatthe 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
RE1
LC
(31 - 22)
2
We already have seen that the average power used bya 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
1 sin2
cos
The term cos in the equation above is known asthe "
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 RT
RP I RI I R I IZ Z
E E E
power fac " of the circuit. The averagepower consumed by the circuit is maximum when 0
tor
2avg rmsP I R cosavg rms rmsP I E (31 - 23)
Power Station
Transmission lines Erms =735 kV , I rms = 500 A home
110 V
T1T2
Step-up transformer
Step-down transformer
R = 220Ω1000 km
2
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 exa
heat rms
R RA
P I R
Energy Transmission Requirements
mple)Power transmitted (368 MW in our example)In our example is almost 15 % of and is acceptableTo keep we must keep as low as possible. The only wa
trans rms rms
heat trans
heat rms
P IP P
P I
E
y to accomplish thisis 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)
rms rms increasing E E
(31 - 24)
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 " " andit has turns. The transformer output appears on the second coils which is knownas the "secondary" and has turns
P
S
NN
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 coilsis 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 Adt dt
d dBN BA V N Adt dt
eqs.1
eqs.2
(31 - 25)
( )
( )
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 Adt dt
d dBN BA V N Adt dt
dBN AV NdtdBV NN A
VN
dt
VN
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 VN
NN N V VNNN N V VN
step up
step down
sformers are used in the transport of electric power over large distances.
S P
S P
V VN 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 heating
P
S
II 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 IV IV I
V N V NNI IN
N
I N I N
N I IN N
we have that S PI I
We have that:
(eqs.1)
S P
S P
S P P S
V VN N
V N V N
S P
S P
V VN N
S S P PI N I N
(31 - 27)
Hitt
A generator supplies 100 V to the primary coil of a transformer. The primary has 50 turns and the secondary has 500 turns. The secondary voltage is:A. 1000 VB. 500 VC. 250 VD. 100 VE. 10V
hitt
The main reason that alternating current replaced direct current for general use is:A. ac generators do not need slip ringsB. ac voltages may be conveniently transformedC. electric clocks do not work on dcD. a given ac current does not heat a power line as much as the same dc currentE. ac minimizes magnetic effects