Lecture 23Lecture 23
Physics 2102
Jonathan Dowling
AC Circuits: PowerAC Circuits: PowerElectromagnetic wavesElectromagnetic waves
Driven RLC Circuits: Driven RLC Circuits: SummarySummary
€
Im =Em
Z2 2( )L CZ R X X= + −
tan L CX X
Rφ −=
E = Em sin(dt); I = Imsin(dt - )
XC=1/(dC), XL=dL
At resonance: XL=XC; d2=1/LC; =0;
Z =R (minimum), Em=Em/R (maximum)
Power in AC CircuitsPower 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
€
Irms =Erms
Z
€
Paverage =Erms
Z
⎛
⎝ ⎜
⎞
⎠ ⎟IrmsR
€
=ErmsIrms
R
Z
⎛
⎝ ⎜
⎞
⎠ ⎟= ErmsIrms cosφ
cos = “Power Factor”Maximum power dissipated when cos = 1.
The average of sin2 overone cycle is ½:
Power in AC Circuits: ExamplePower in AC Circuits: Example• Series RLC Circuit:
R = 15, C = 4.7 F, 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?
=
−+=
29
)1
( 22
CLRZ
dd ω
ω
srad
Hzd
/3457
)550)(2(
== π
€
Paverage = Irms2R
=Erms
Z
⎛
⎝ ⎜
⎞
⎠ ⎟2
R =75V
29Ω
⎛
⎝ ⎜
⎞
⎠ ⎟2
(15Ω) =100.3W
Power in AC circuitsPower in AC circuitsWhat’s the average power dissipated in the following circuits?
River Bendpower plant
LSUTransmission lines 735KVDistance~30 miles~50km Resistance 0.22/kmPower capacity 936 MW
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.
A Very Real ExampleA Very Real Example
Current: P=IV, I=P/V=936MW/735kV=1300 A (!)Power dissipated in wires: Plost=I2R=(1300A)2x0.22/km x 50km=19 MW (~2% of 936)
TransformersTransformersTwo coils (“primary andsecondary”) sharing the samemagnetic flux.
Faraday’s law:
per turn emf=Φdtd
S
S
P
P
NV
NV ==
P
P
SS V
NN
V =
You can get any voltage you wish just playing with the numberof turns. For instance, the coil in the ignition system of a car goesfrom 12V to thousands of volts. Or the transformers in most consumer electronics go from 110V to 6 or 12 V.
P
S
PS i
NN
i = :conserved is Energy What you gain (lose) in voltageyou lose (gain) in current.
From the Power Plant From the Power Plant to Your Hometo Your Home
few kV
155kV-765kV
7kV
120V
http://www.howstuffworks.com/power.htm: The Distribution Grid
A solution to Maxwell’s equations in free space:
)sin( txkEE m −=
)sin( txkBB m −=n.propagatio of speed ,c
k=
€
c =Em
Bm
=1
μ0ε0
= 299,462,954m
s=187,163mps
Visible light, infrared, ultraviolet,radio waves, X rays, Gammarays are all electromagnetic waves.
Electromagnetic WavesElectromagnetic Waves
Electromagnetic waves are able to transport energy from transmitterto receiver (example: from the Sun to our skin).
The power transported by the wave and itsdirection is quantified by the Poynting vector. John Henry Poynting (1852-1914)
211|| E
cEBS
00==
The Poynting VectorThe Poynting Vector
E
BS
Units: Watt/m2
For a wave, sinceE is perpendicular to B: BES
rrr×=
01
In a wave, the fields change with time. Therefore the Poynting vector changes too!!The direction is constant, but the magnitude changes from 0 to a maximum value.
____________222
___)(sin
11tkxE
cEcSI m
−===
00
The average of sin2 overone cycle is ½:
2
21
mEc
I0
=
21rmsE
cI
0=
Both fields have the same energy density.
2 22 2
0
1 1 1 1( )
2 2 2 2E B
B Bu E cB uε ε ε
ε 0 00 0 0
= = = = =
or,
EM Wave Intensity, Energy DensityEM Wave Intensity, Energy DensityA 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.
The total EM energy density is then 022
0 / ε BEu ==
Solar EnergySolar EnergyThe 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 dimensions47in x 29in. The incident power is then 880 W. The actual solar panel delivers 75W (4.45A at 17V): less than 10% efficiency….
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.
24 r
PI s
π=
So the power per unit area decreases as the inverse of distance squared.
EM Spherical WavesEM Spherical Waves
ExampleExampleA 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.
222
/8.0)1(4
10
4mmW
km
kW
r
PI s ===
ππ
mVIcEEc
I mm /775.022
1 2 ==⇒= 00
nTcEB mm 58.2/ ==
mJSAtUA
tU
A
PS 4.2
/==Δ⇒
Δ==Received
energy:
Radiation PressureRadiation PressureWaves not only carry energy but also momentum. The effect is very small (we don’t ordinarily feel pressure from light). If lightis completely absorbed during an interval Δt, the momentum transferred is given by
cu
pΔ=Δ
tp
FΔΔ=Newton’s law:
Supposing one has a wave that hits a surfaceof area A (perpendicularly), the amount of energy transferred to that surface in time Δt will be
tIAU Δ=Δ therefore ctIA
pΔ=Δ
I
A
cIA
F =
)reflection (total 2
),absorption (total cI
pcI
p rr ==Radiation pressure:
and twice as much if reflected.
Radiation pressure: examplesRadiation pressure: examples
Not radiation pressure!!
Solar mills?
Solar sails?
From the Planetary Society
Sun radiation: I= 1 KW/m2
Area 1km2 => F=IA/c=3.3 mNMass 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
Comet tails
Radio transmitter:
If the dipole antennais vertical, so will bethe electric fields. Themagnetic field will behorizontal.
The radio wave generated is said to be “polarized”.
In general light sources produce “unpolarized waves”emitted by atomic motions in random directions.
EM Waves: PolarizationEM Waves: Polarization
Completely unpolarized light will have equal components in horizontal and verticaldirections. Therefore running the light througha 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 thesheet will get through.
)= θcos(EEy
And therefore: θ20 cosII =
Polarized sunglasses operate on this formula.They cut the horizontally polarized lightfrom glare (reflections on roads, cars, etc).
ExampleExampleInitially 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=I1cos20 = 1/4 I1 =1/8 I0.
• Now the light is again polarized, but at 60o. The last polarizer is horizontal, so I3=I2cos20 3/4 I0=0.094 I0. • The exiting light is horizontally polarized, and has 9% of the original amplitude.