Transcript Document

Physics 2102
Jonathan Dowling
Lecture 23
AC Circuits: Power
Electromagnetic waves
Driven RLC Circuits:
Summary
E = Em sin(dt); I = Imsin(dt - )
Em
Im 
Z
Z  R 2  ( X L  X C )2
X L  XC
tan  
R
XC=1/(dC), XL=dL
At resonance: XL=XC; d2=1/LC; =0;
Z =R (minimum), Em=Em/R (maximum)
The average of sin2 over
one cycle is ½:
Power in AC Circuits
•
E = Em sin(dt)
•
•
•
•
•
I = Imsin(dt - )
Power dissipated by R:
P = I2R
P = Im2R sin2(dt - )
Paverage = (1/2) Im2R 
= Irms2R
Erms
Irms 
Z
E rms 
Paverage  
Irms R
 Z 
R 
 E rms Irms   E rms Irms cos
Z 

cos  = “Power Factor”
Maximum power dissipated when cos  = 1.

•
Power in AC Circuits: Example
Series RLC Circuit:
 d  (2 )(550Hz)
R = 15W, C = 4.7 mF, L = 25mH.
The circuit is driven by a source of
Erms = 75 V and f = 550 Hz.
• At what average rate is energy
dissipated by the resistor?
 3457rad / s
1 2
Z  R  ( d L 
)
d C
2
 29W
Paverage  Irms R
2
E rms 
75V 
 
 R  
 (15W)  100.3W
 Z 
29W 
2
2
Power in AC circuits
What’s the average power dissipated in the following circuits?
A Very Real Example
River Bend
power plant
Transmission lines 735KV
Distance~30 miles~50km
Resistance 0.22W/km
Power capacity 936 MW
LSU
Current: P=IV, I=P/V=936MW/735kV=1300 A (!)
Power dissipated in wires:
Plost=I2R=(1300A)2x0.22W/km x 50km=19 MW (~2% of 936)
If the power delivered is constant, we want the highest voltage and
the lowest current to make the delivery efficient!
At home, however, we don’t want high voltages! We use transformers.
Transformers
Two coils (“primary and
secondary”) sharing the same
magnetic flux.
Faraday’s law:
d
 emf per turn
dt
VP VS


NP NS
VS 
NS
VP
NP
You can get any voltage you wish just playing with the number
of turns. For instance, the coil in the ignition system of a car goes
from 12V to thousands of volts. Or the transformers in most
consumer electronics go from 110V to 6 or 12 V.
NP
Energy is conserved : iS 
iP
NS
What you gain (lose) in voltage
you lose (gain) in current.
From the Power Plant
to Your Home
http://www.howstuffworks.com/power.htm: The Distribution Grid
155kV-765kV
few kV
7kV
120V
Electromagnetic Waves
A solution to Maxwell’s equations in free space:
E  Em sin( k x   t )

B  Bm sin( k x   t )
k
 c, speed of propagatio n.
Em
1
c

Bm
m00
m
 299,462,954  187,163m ps
s

Visible light, infrared, ultraviolet,
radio waves, X rays, Gamma
rays are all electromagnetic waves.
The Poynting Vector
Electromagnetic waves are able to transport energy from transmitter
to receiver (example: from the Sun to our skin).
The power transported by the wave and its
direction is quantified by the Poynting vector.
 1  
S
EB
m
John Henry Poynting (1852-1914)
For a wave, since
1
1 2
| S |
EB 
E
E is perpendicular to B:
m
cm
Units: Watt/m2
E
B
In a wave, the fields
change with time.
Therefore the Poynting
vector changes too!!The
direction is constant, but
the magnitude changes
S
from 0 to a maximum
value.
EM Wave Intensity, Energy Density
A better measure of the amount of energy in an EM wave is
obtained by averaging the Poynting vector over one wave cycle.
The resulting quantity is called intensity. Units are also Watts/m2.
I S 
1
cm 
___
2
E 
1
cm 
____________
2
2
Em sin (kx  t )
1
I
Em 2 or,
2cm
Both fields have the
same energy density.
I
1
cm
The average of sin2 over
one cycle is ½:
Erms 2
1
1
1
B2
1 B2
2
2
uE    E    (cB)   0

 uB
2
2
2   m 2 m
The total EM energy density is then
u   0 E  B / m0
2
2
Solar Energy
The light from the sun has an intensity of about 1kW/m2. What would
be the total power incident on a roof of dimensions 8x20m?
I=1kW/m2 is power per unit area.
P=IA=(103 W/m2) x 8m x 20m=0.16 MW!!
The solar panel shown (BP-275) has dimensions
47in x 29in. The incident power is then 880 W.
The actual solar panel delivers 75W (4.45A at
17V): less than 10% efficiency….
EM Spherical Waves
The intensity of a wave is power per unit area. If one has a source that
emits isotropically (equally in all directions) the power emitted by the
source pierces a larger and larger sphere as the wave travels outwards.
I
Ps
4r
2
So the power per unit area
decreases as the inverse of
distance squared.
Example
A radio station transmits a 10 kW signal at a frequency of 100
MHz. (We will assume it radiates as a point source). At a distance
of 1km from the antenna, find the amplitude of the electric and
magnetic field strengths, and the energy incident normally on a
square plate of side 10cm in 5min.
Ps
10kW
2
I


0
.
8
m
W
/
m
4r 2 4 (1km) 2
1
2
I
Em  Em  2cm I  0.775V / m
2cm
Bm  Em / c  2.58nT
Received
energy:
P U / t
S 
 U  SAt  2.4 mJ
A
A
Radiation Pressure
Waves not only carry energy but also momentum. The effect is
very small (we don’t ordinarily feel pressure from light). If light
is completely absorbed during an interval t, the momentum
transferred is given by p  u and twice as much if reflected.
c
A

p
Newton’s law: F 
t
I
Supposing one has a wave that hits a surface
of area A (perpendicularly), the amount of energy
transferred to that surface in time t will be
IA
IAt
F
U  IAt therefore p  c
c
Radiation
pressure:
I
2I
pr  (total absorption), pr 
(total reflection)
c
c
Radiation pressure: examples
Solar mills?
Not radiation pressure!!
Solar sails?
From the Planetary Society
Comet tails
Sun radiation: I= 1 KW/m2
Area 1km2 => F=IA/c=3.3 mN
Mass m=10 kg => a=F/m=3.3 10-4 m/s2
When does it reach 10mph=4.4 m/s?
V=at => t=V/a=1.3 104 s=3.7 hrs
EM Waves: Polarization
Radio transmitter:
If the dipole antenna
is vertical, so will be
the electric fields. The
magnetic field will be
horizontal.
The radio wave generated is said to be “polarized”.
In general light sources produce “unpolarized
waves”emitted by atomic motions in random directions.
Completely unpolarized light will have
equal components in horizontal and vertical
directions. Therefore running the light through
a polarizer will cut the intensity in half: I=I0/2
When polarized light hits a polarizing sheet,
only the component of the field aligned with the
sheet will get through.
E y  E cos( 
And therefore:
I  I 0 cos2 
Polarized sunglasses operate on this formula.
They cut the horizontally polarized light
from glare (reflections on roads, cars, etc).
Example
Initially unpolarized light of
intensity I0 is sent into a system
of three polarizers as shown.
Wghat fraction of the initial
intensity emerges from the
system? What is the
polarization of the exiting
light?
• Through the first polarizer: unpolarized to polarized, so I1=½I0.
• Into the second polarizer, the light is now vertically polarized. Then,
I2=I1cos26o= 1/4 I1 =1/8 I0.
• Now the light is again polarized, but at 60o. The last polarizer is
horizontal, so I3=I2cos23o3/4 I23/32 I0=0.094 I0.
• The exiting light is horizontally polarized, and has 9% of the
original amplitude.