Download Lecture 1410

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)
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. 2.5Wb
B. 4.3Wb
C. 5Wb
D. 5.8Wb
E. 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 right
side, separated by 1.7m, is:
A. 0
B. 1.8mV
C. 3.6mV
D. 6.4mV
E. 13mV
L
C
LC Oscillations
The circuit shown in the figure consists of a capacitor C
and an inductor L. We give the capacitor an initial
chanrge Q and then abserve what happens. The capacitor
will discharge through the inductor resulting in a time
dependent current i.
We will show that the charge q on the capacitor plates as well as the current i
1
LC
The total energy U in the circuit is the sum of the energy stored in the electric field
in the inductor oscillate with constant amplitude at an angular frequency  
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 2 q
d 2q 1

 Li  0. i 
  2  L 2  q  0 (31 - 2)
dt C dt
dt
dt
dt dt
dt
C
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   
(31 - 3)
dt
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)
2
t T /8
3
4
t T /4
t  3T / 8
5
1
t T /2
t 0
8
6
t  7T / 8
1
t  5T / 8
3
7
5
7
2
4
6
8
t  3T / 4
(31 - 5)
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 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
m 4m 2
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(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
(31 - 7)
Alternating Current
A battery for which the emf is constant generates
a current that has a constant direction. This type
of current is known as "direct current " or "dc"
E  Em sin t
In chapter 30 we encountered a different type
of sourse (see figure) whose emf is:
E   NAB sin t  Em sin t where Em   NAB, A is the area of the generator
windings, N is the number of the windings,  is the angular frequency of the
rotation of the windings, and B is the magnetic field. This type of generator
is known as "alternating current" or "ac" because the emf as well as the current
change direction with a frequency f  2. In the US f  60 Hz.
Almost all commercial electrical 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 this
chapter.
(31 - 8)
Three Simple Circuits
Our objective is to analyze the circuit shown in the
figure (RCL circuit). The discussion will be greatly
simplified if we examine what happens if we connect
each of the three elements (R, C , and L) separately
to an ac generator.
L
C
A convention
From now on we will use the standard notation for ac circuit
analysis. Lower case letters will be used to indicate the
instantaneous values of ac quantities. Upper case letters
will be used to indicate the constant amplitudes of ac quantities.
Example: The capacitor charge in an LC circuit was written as:
q  Q cos t   
The symbol q is used for the instantaneous value of the charge
The symbol Q is used for the constant amplitude of q
(31 - 9)
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)
A convenient method for the representation of ac
quantities is that of phasors
The resistor voltage vR and the resistor current iR are represented
by rotating vectors known as phasors using the following conventions:
1. Phasors rotate in the counterclockwise direction with angular speed 
2. The length of each phasor is proportional to tha ac quantity amplitude
3. The projection of the phasor on the vertical axis gives the instantaneous
value of the ac quantity.
4. The rotation angle for each phasor is equal to the phase of the
ac quantity (t in this example)
(31 - 11)
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
(31 - 12)
A capacitive load
In fig.a we show an ac generator connected to a
capacitor C
XC
qC
From KLR we have: E 
0
C
qC  EC  EmC sin t
O

dqC
iC 
 EmC cos tdt  sin t  90 
1
dt
XC 
C
The voltage amplitude VC equal to Em
VC
The current amplitude I C  CVC 
1/ C
The quantity X C  1 / C is known as the
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 are out of phase by 90 .
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!
(31 - 14)
An inductive load
In fig.a we show an ac generator connected to an inductor L
di
di
E E
From KLR we have: E  L L  0  L   m 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
V
The current amplitude I L  L
L

The quantity X L   L is known as the
O
inductive reactance
X L  L
In fig.b we plot the inductor current iL and the
inductor voltage vL as function of time t.
The current lags behind the voltage by a
quarter of a period. The voltage and
current are out of phase by 90 .
(31 - 15)
PL  0
Average Power for L
2
E
Power P  VL I L   m sin t cos t
XL
2sin  cos   sin 2
T
2
E
 P   m sin 2t
2X L
2
T
E 1
1
 P   P(t )dt =  m
sin 2tdt  0

T 0
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!
(31 - 16)
SUMMARY
Circuit
element
Average
Power
Resistor
R
E
 PR   m
2R
R
 PC   0
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
of a period
VL  I L X L  I LL
Capacitor
C
Inductor
L
Reactance
2
 PL   0
Phase of current
Current is in phase
with the voltage
Voltage amplitude
VR  I R R
(31 - 17)
(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 vL leads ahead of the current i by 90.
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
(31 - 19)
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  
2
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
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

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
The equation above is the condition for resonance. When its is satisfied I res 
A plot of the current amplitude I as function of  is shown in the lower figure.
This plot is known as a "resonance curve"
Em
R
(31 - 22)
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
when   0
2
Transmission lines
Erms =735 kV , I rms = 500 A
(31 - 24)
home
Step-down
transformer 110 V
Step-up
transformer
T2
T1
R = 220Ω
 1000 km
Power Station
Energy Transmission Requirements
The resistance of the power line R 
Heating of power lines Pheat

.
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 - 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
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.
(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
(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
(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 V
B. 500 V
C. 250 V
D. 100 V
E. 10V
hitt
The main reason that alternating current replaced
direct current for general use is:
A. ac generators do not need slip rings
B. ac voltages may be conveniently transformed
C. electric clocks do not work on dc
D. a given ac current does not heat a power line as
much as the same dc current
E. ac minimizes magnetic effects