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

2. Driven RLC Circuits: Summary E = E m sin(  d t); I = I m sin(  d t -  ) X C =1/(  d C), X L =  d L At resonance: X L =X C ;  d 2 =1/LC;  =0; Z =R (minimum), E m =E m /R (maximum)

3. Power in AC Circuits E = E m sin(  d t) I = I m sin(  d t -  ) Power dissipated by R: P = I 2 R P = I m 2 R sin 2 (  d t -  ) P average = (1/2) I m 2 R = I rms 2 R cos  = “Power Factor” Maximum power dissipated when cos  = 1. The average of sin 2 over one cycle is ½:

4. Power in AC Circuits: Example Series RLC Circuit: R = 15 , C = 4.7  F, L = 25mH. The circuit is driven by a source of E rms = 75 V and f = 550 Hz. At what average rate is energy dissipated by the resistor?

5. Power in AC circuits What’s the average power dissipated in the following circuits?

6. River Bend power plant LSU Transmission lines 735KV Distance~30 miles~50km Resistance 0.22  /km Power 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 Example Current: P=IV, I=P/V=936MW/735kV=1300 A (!) Power dissipated in wires: P lost =I 2 R=(1300A) 2 x0.22  /km x 50km=19 MW (~2% of 936)

7. Transformers Two coils (“primary and secondary”) sharing the same magnetic flux. Faraday’s law: 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. What you gain (lose) in voltage you lose (gain) in current.

8. From the Power Plant to Your Home few kV 155kV-765kV 7kV 120V http://www.howstuffworks.com/power.htmhttp://www.howstuffworks.com/power.htm: The Distribution Grid

9. A solution to Maxwell’s equations in free space: Visible light, infrared, ultraviolet, radio waves, X rays, Gamma rays are all electromagnetic waves. Electromagnetic Waves

11. 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. John Henry Poynting (1852-1914) The Poynting Vector E B S Units: Watt/m 2 For a wave, since E is perpendicular to B: 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.

12. The average of sin 2 over one cycle is ½: Both fields have the same energy density. or, 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/m 2. The total EM energy density is then

13. Solar Energy The light from the sun has an intensity of about 1kW/m 2. What would be the total power incident on a roof of dimensions 8x20m? I=1kW/m 2 is power per unit area. P=IA=(10 3 W/m 2 )  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….

14. 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. So the power per unit area decreases as the inverse of distance squared. EM Spherical Waves

15. 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. Received energy:

16. 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 Newton’s law: 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 therefore I A Radiation pressure: and twice as much if reflected.

17. Radiation pressure: examples Not radiation pressure!! Solar mills? Solar sails? From the Planetary Society Sun radiation: I= 1 KW/m 2 Area 1km 2 => F=IA/c=3.3 mN Mass m=10 kg => a=F/m=3.3 10 -4 m/s 2 When does it reach 10mph=4.4 m/s? V=at => t=V/a=1.3 10 4 s=3.7 hrs Comet tails

18. 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. EM Waves: Polarization 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=I 0 /2

19. When polarized light hits a polarizing sheet, only the component of the field aligned with the sheet will get through. And therefore: Polarized sunglasses operate on this formula. They cut the horizontally polarized light from glare (reflections on roads, cars, etc).

20. Example Initially unpolarized light of intensity I 0 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 I 1 =½I 0. Into the second polarizer, the light is now vertically polarized. Then, I 2 =I 1 cos 2   = 1/4 I 1 = 1/8 I 0. Now the light is again polarized, but at 60 o. The last polarizer is horizontal, so I 3 =I 2 cos 2    3/4      I 0 =0.094 I 0. The exiting light is horizontally polarized, and has 9% of the original amplitude.