Lecture 23Lecture 23

Physics 2102

Jonathan Dowling

AC Circuits: PowerAC Circuits: PowerElectromagnetic wavesElectromagnetic waves

Driven RLC Circuits:Driven RLC Circuits:SummarySummary

!

Im

=Em

Z

2 2( )L C

Z R X X= + !

tanL CX X

R!

"=

E = Em sin(ωdt); I = Imsin(ωdt - φ)

XC=1/(ωdC), XL=ωdL

At resonance: XL=XC; ωd2=1/LC; φ=0;

Z =R (minimum), Em=Em/R (maximum)

Power in AC CircuitsPower in AC Circuits• E = Em sin(ωdt)• I = Imsin(ωdt - φ)• Power dissipated by R:• P = I2R• P = Im

2R sin2(ωdt - φ)• Paverage = (1/2) Im

2R = Irms

2R

!

Irms

=Erms

Z

!

Paverage =Erms

Z

"

# $

%

& ' IrmsR

!

= ErmsIrms

R

Z

"

# $

%

& ' = E

rmsIrmscos(

cos φ = “Power Factor”Maximum power dissipated when cos φ = 1.

The average of sin2 overone cycle is ½:

Power in AC Circuits: ExamplePower in AC Circuits: Example• Series RLC Circuit:

R = 15Ω, C = 4.7 µF, L = 25mH.The circuit is driven by a source ofErms = 75 V and f = 550 Hz.

• At what average rate is energydissipated by the resistor?

!=

"+=

29

)1

( 22

CLRZ

d

d

##

srad

Hzd

/3457

)550)(2(

=

= !"

!

Paverage = Irms2R

=Erms

Z

"

# $

%

& '

2

R =75V

29(

"

# $

%

& '

2

(15() =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 0.22Ω/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)2x0.22Ω/km x 50km=19 MW (~2% of 936)

TransformersTransformersTwo coils (“primary andsecondary”) sharing the samemagnetic flux.

Faraday’s law:

per turn emf=!

dt

d

S

S

P

P

N

V

N

V==

P

P

S

SV

N

NV =

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

P

Si

N

Ni = :conserved isEnergy What you gain (lose) in voltage

you lose (gain) in current.

From the Power PlantFrom the Power Plantto 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( txkEEm

!"=

)sin( txkBBm

!"=

n.propagatio of speed ,ck=

!

!

c =Em

Bm

=1

µ0"0

= 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

==µµ

The The Poynting Poynting VectorVector

E

BS

Units: Watt/m2

For a wave, sinceE is perpendicular to B: BES

rrr!=

0µ

1

In a wave, the fieldschange with time.Therefore the Poyntingvector changes too!!Thedirection is constant, butthe magnitude changesfrom 0 to a maximumvalue.

____________222

___

)(sin11

tkxEc

Ec

SI m !µµ

"===00

The average of sin2 overone cycle is ½:

2

2

1

mEc

I

0

=µ

21

rmsEc

I

0

=µ

Both fields have the same energy density.

2 22 2

0

1 1 1 1( )

2 2 2 2E B

B Bu E cB u! ! !

! µ µ0 0

0 0 0

= = = = =

or,

EM Wave Intensity, Energy DensityEM Wave Intensity, Energy DensityA better measure of the amount of energy in an EM wave isobtained 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 then0

22

0/ µ! BEu ==

Solar EnergySolar EnergyThe light from the sun has an intensity of about 1kW/m2. What wouldbe 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 at17V): less than 10% efficiency….

The intensity of a wave is power per unit area. If one has a source thatemits isotropically (equally in all directions) the power emitted by thesource pierces a larger and larger sphere as the wave travels outwards.

24 r

PI

s

!=

So the power per unit areadecreases as the inverse ofdistance squared.

EM Spherical WavesEM Spherical Waves

ExampleExampleA radio station transmits a 10 kW signal at a frequency of 100MHz. (We will assume it radiates as a point source). At a distanceof 1km from the antenna, find the amplitude of the electric andmagnetic field strengths, and the energy incident normally on asquare plate of side 10cm in 5min.

2

22/8.0

)1(4

10

4mmW

km

kW

r

PI

s ===!!

mVIcEEc

Imm

/775.022

1 2

==!= 0

0

µµ

nTcEBmm

58.2/ ==

mJSAtUA

tU

A

PS 4.2

/==!"

!==

Receivedenergy:

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

c

up

!=!

t

pF

!

!=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 !=! thereforec

tIAp

!=!

I

A

c

IAF =

)reflection (total 2

),absorption (total c

Ip

c

Ip 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 “unpolarizedwaves”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.

)= !cos(EEy

And therefore: !20 cosII =

Polarized sunglasses operate on this formula.They cut the horizontally polarized lightfrom glare (reflections on roads, cars, etc).

ExampleExampleInitially unpolarized light ofintensity I0 is sent into a systemof three polarizers as shown.Wghat fraction of the initialintensity emerges from thesystem? What is thepolarization of the exitinglight?

• Through the first polarizer: unpolarized to polarized, so I1=½I0.• Into the second polarizer, the light is now vertically polarized.Then, I2=I1cos260ο = 1/4 I1 =1/8 I0.• Now the light is again polarized, but at 60o. The last polarizer ishorizontal, so I3=I2cos230ο =3/4 Ι2=3/32 I0=0.094 I0.• The exiting light is horizontally polarized, and has 9% of theoriginal amplitude.