1. Electric Power Physics Mr. Berman

2. Part I Electrical Power Joule’s Law
Thermal Energy Expended by a Resistor

3. Wind Generator

4. Generator

5. Generator

6. Niagara Falls Hydroelectric Plant

7. Hydroelectric Power Plant

8. Remember: P= W / t (Power=Work/time) Remember: W= Vq and I = q/t
So: P= I V

9. Electric Power, P= I V Known as Joule’s Law
P: is the power consumed by a resistor, R.
Unit: Joule/s= Watt

10. Derive the following using P=IV and Ohm’s Law:
P=V2/R P=I2R

11. Problem 1
The current through a car motor is 150A. The battery used is 12V. How much energy is supplied by the battery in 5s? Answer: 9,000J

12. What does the kWh measure, a) Energy or b) Power ?
kiloWatt hour
What does the kWh measure, a) Energy or b) Power ?

13. Problem 2
A small desktop radio has a resistance of 8,000 Ω. The voltage is 120V.
How much current does it draw?
How much power does it use?
How much does it cost to run the radio for 12 hours, if 1kWh costs $0.15?
Answers:
a)0.015A, b)1.8W, c) $

14. Resistors Expend Thermal Energy
Wasted heat energy is called “Joule Heating”
or “I2 R” loss.

15. Why is long distance power transmitted at high voltages?
Hint: P = I V
Answer:
For a given P,
keep the current, I, low to minimize “I2 R” loss in the transmitting wires, so increase V.

16. Electric heaters(Coil Heaters)
P= V2/R
The lower the R the greater the heat given off by the resistor for a given voltage.

17. Problem 3
An electric heating element has a resistance of 9 Ω. The voltage is 120V.
Calculate the current through the element.
Calculate the thermal energy supplied by the element in 10min.
Answer:
a)13.3A, b) 9.6x105 J

18. Part II
Brightness of a Lightbulb
Some Safety Issues

19. Light Bulb

20. Brightness of a Light bulb and Power
The greater the power actually used by a light bulb, the greater the brightness.
Note: the power rating of a light bulb is indicated for a given voltage and the bulb may be in a circuit that does not have that voltage.

21. Wattage and Thickness of Filament
For a given V, (P = IV) the higher the wattage of a light bulb, the larger the current and therefore the smaller the resistance of the filament (V=I R).
Thus, the higher wattage bulb will have a filament of lower resistance and therefore a larger cross-sectional area (R=ρ L / A). 

22. Question Which light bulb has a thicker filament, 60W or 75W?
Answer: 75W

23. Problem 4
A 60W light bulb is 22% efficient. How much energy is lost to heat in 5min? Answer: 1.4x104 J

24. Safety Issues
Why is it not safe to handle electric appliances with wet hands?
Hint:
A current of 15mA can cause serious injury.
The resistance of a dry human body is in the range of 100,000Ω. The resistance of a wet person can drop to the range of 5,000Ω.

25. Problem 5
A current of 1mA can be felt by a person. Will a person feel the current if the person has dry hands (a resistance of 100,000 Ω) and touches a 9V battery’s terminals?
Answer: I=0.09mA, No.

26. Problem 6
The same 9V battery is touched by a person with wet hands. Will this person feel the current if the resistance is 5,000Ω and the person touches the battery’s terminals?
Answer: I=1.8mA, Yes