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Lecture 23 Lecture 23 hysics 2102 onathan Dowling AC Circuits: Power AC Circuits: Power Electromagnetic waves Electromagnetic waves
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Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

Dec 22, 2015

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Page 1: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

Lecture 23Lecture 23

Physics 2102

Jonathan Dowling

AC Circuits: PowerAC Circuits: PowerElectromagnetic wavesElectromagnetic waves

Page 2: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic 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)

Page 3: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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:

Page 4: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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

Page 5: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

Power in AC circuitsPower in AC circuitsWhat’s the average power dissipated in the following circuits?

Page 6: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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)

Page 7: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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.

Page 8: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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

Page 9: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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

Page 10: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.
Page 11: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic 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.

Page 12: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

____________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 ==

Page 13: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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….

Page 14: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic 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.

24 r

PI s

π=

So the power per unit area decreases as the inverse of distance squared.

EM Spherical WavesEM Spherical Waves

Page 15: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic 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:

Page 16: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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.

Page 17: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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

Page 18: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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

Page 19: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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).

Page 20: Lecture 23 Physics 2102 Jonathan Dowling AC Circuits: Power Electromagnetic waves.

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.