The amazing world of self induction !

1. The amazing world of self induction !

2. 130 volts / 1.4 ohms = 93 amperes! VI = 12,000 watts ! What if this coil were connected to 130 v ac??? It would melt down!!!! The Resistance is 1.4 ohms Ohm meter

3. Now follows proof that the power converted into heat is LESS than expected.

4. A Square Wave input into a coil.

7. P T P T

8. The actual current is 3.45 amperes AC. Input voltage is 130 V. 450 watts ???????? Why is the current so small? And why doesn’t the coil even get warm? In fact, there are almost zero watts of heat !!! 130 v

9. Conservation of Energy around a loop.....in all circuits!

10. A little information to remember: 6 v 6 volts V R battery V Zero volts 0 volts Vbatt + Vresistor = 0 Gain 6 v + lose 6 v =0

11. V R AC Power Source Volt- meter Vapplied+Vinduced=0 Resistor Power Voltage Current Current and Voltage in in PHASE; therefore, the Power curve is in phase with them both.

12. The voltage supplied always equals the voltage consumed.

13. A sin wave input into a coil

14. Current in coil and resistor Induced voltage in coil O I Prop Flux in the coil follows this curve also to we expect maximum induced voltage to be where the flux is changing at the highest rate. VL = - t

15. Signal generator voltage Va Current in coil V Induced voltage L The voltages around a loop must add up to zero so the sum of Vsignal gen + VL = O

16. A coil in series with an AC power source: V AC coil POWER SOURCE Vapplied Voltmeter Vapplied + Vinduced=0 Vapplied current Vinduced

17. http://ostc.physics.uiowa.edu/~wkchan/ELECTENG/AC_POWER/AC.htmlhttp://ostc.physics.uiowa.edu/~wkchan/ELECTENG/AC_POWER/AC.html

18. Look at the POWER CURVE! Power = V battery x I battery

19. Output voltage of signal generator Current out of signal generator Power output taken from the signal generator would be: P = V I If these are multiplied the puzzle is solved.

20. = Power (Vmax Cos 0 ) (I max Sin 0) - - + + Flux building up Energy being stored and given back.

21. Why is there no heat being generated in the coil? …what energy is taken from the battery is given back so there is NO heating of the coil. Power = (Va )( I) cos O = zero! Theta is the angle between the two curves…in this case 90 degrees. Cos 90 degrees = zero

22. Now, why is the current so much lower than expected? If we use I=V/R it must be because somehow the resistance of the coil is larger to AC than to DC.

23. The resistance of a coil to AC current is given by the equation: Rcoil = 2 F L where F is the frequency and L the inductance of coil. L for the coil is 0.10 henrys. F = 60 hz R then = 38 ohms. I= V/R = 130 v/38 ohms = 3.4 amps. The resistance of the coil if proportional to both the frequency and L. As f increases the resistance gets larger…. Choking off the current.

24. Vbattery VIcos0= Power (Asin0)(Acos0)= Power Icurrent + _

25. Power = VI (+) (+) I (-) V The power taken from the signal generator (+) balances the power given back (-) and therefore no heat is produced in the coil even though the current is 3.4 amps!

26. Power Vapplied Current Time (s) Vinduced Vapplied

27. An arrangement for viewing the phase relationship between current in and voltage across a coil. A signal generator is connected in series with a coil and a resistor. An oscilloscope (for viewing how voltage changes in time) is connected (via computer) to both the coil and resistor. VL VR gnd R coil Signal Gen

28. VL VR Gnd

29. Current is max voltage is min. Current is min voltage is max. The current in coil and resistor are the same. The voltage across the resistor is in phase with the current so we can see the phase relationship between current and voltage of the coil.

30. A capacitor and AC voltage: ac

31. Power Capacitor

32. Power Vapplied Current Time (s) Vinduced Vapplied Inductor