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Lecture 23

Physics 2102 Jonathan Dowling. Lecture 23. AC Circuits: Power Electromagnetic waves. Driven RLC Circuits: Summary. E = E m sin(  d t); I = I m sin(  d t -  ). X C =1/( w d C), X L = w d L . At resonance: X L =X C ; w d 2 =1/LC; f =0; Z =R (minimum), E m =E m /R (maximum) .

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Lecture 23

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

  2. Driven RLC Circuits: Summary E= Em sin(dt); I = Imsin(dt - ) XC=1/(wdC), XL=wdL At resonance: XL=XC; wd2=1/LC; f=0; Z =R (minimum), Em=Em/R (maximum)

  3. 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 = Im2Rsin2(dt - ) • Paverage = (1/2) Im2R = Irms2R cos  = “Power Factor” Maximum power dissipated when cos  = 1.

  4. Power in AC Circuits: Example • Series RLC Circuit: 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?

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

  6. A Very Real Example Transmission lines 735KV Distance~30 miles~50km Resistance 0.22W/km Power capacity 936 MW River Bend power plant 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.

  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 http://www.howstuffworks.com/power.htm: The Distribution Grid 155kV-765kV few kV 7kV 120V

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

  10. E S B 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. John Henry Poynting (1852-1914) For a wave, since E is perpendicular to B: Units: Watt/m2 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.

  11. 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. The average of sin2 over one cycle is ½: or, Both fields have the same energy density. The total EM energy density is then

  12. 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….

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

  14. Received energy: 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.

  15. 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 Dt, the momentum transferred is given by and twice as much if reflected. A Newton’s law: I Supposing one has a wave that hits a surface of area A (perpendicularly), the amount of energy transferred to that surface in time Dtwill be therefore Radiation pressure:

  16. Radiation pressure: examples Solar mills? Not radiation pressure!! Comet tails Solar sails? 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 From the Planetary Society

  17. 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

  18. 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).

  19. 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=I1cos260o = 1/4 I1=1/8 I0. • Now the light is again polarized, but at 60o. The last polarizer is horizontal, so I3=I2cos230o =3/4 I2=3/32I0=0.094 I0. • The exiting light is horizontally polarized, and has 9% of the original amplitude.

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