1. Physics Section 17.4 Apply sources of potential difference and electric power Note: When a ball falls it loses potential energy, to restore its potential energy, work must be done to lift it. A charge loses energy as it moves from a high electric potential to a low potential energy. Work must be done to restore its potential energy. Sources of Electric Potential Difference 1.Piezoelectric Cell 2.Photovoltaic Cell 3.Thermal Electric Cell 4.Electro-chemical Cell 5.Electromagnetic Induction

2. Sources of Electric Potential Difference 1.Piezoelectric cell- pressure is applied to opposite sides of a crystal. One side becomes positively charged, the other side becomes negatively charged.

3. 2. Photovoltaic Cell- light energy is converted into a potential difference.

4. 3. Thermoelectric Cell- thermal energy is converted into a potential difference.

5. http://phet.colorado.edu/en/simulation/battery-voltage 4. Electro-chemical cell- chemical energy is converted into a potential difference. a. A battery consists of two electrodes which react with an electrolyte. One electrode becomes positive, the other becomes negative.

6. b. Fuel Cell- two reactants are continually supplied and create a positively and negatively charged electrode.

7. 5. Electromagnetic Induction- a conducting loop is rotated in a magnetic field, causing the electrons in the loop to move in one direction, then in the other direction.

8. Two types of electric current 1.Direct current – charges travel in one direction 2.Alternating current- charges travel in one direction then reverse and travel in the opposite direction P = I ∙V Electric Power is the rate at which electric carriers do work. P = power (W) I = current (A) V = potential difference (V)

9. Alternate Formula for Power P = V 2 R P = power (W) V = potential difference (V) R = resistance (Ω)

10. Examples A heater operates at a potential difference of 120 V with a resistance of 24 Ω. Find the power output and the current. A 150 W light bulb draws a current of.75 A. Find the resistance of the light bulb and the required potential difference.

11. A kilowatt-hour is the energy delivered in one hour at a rate of one kW per hour. 1 kW∙h x 10 3 W/ 1 kW x 60 min/1 h x 60 s/1 min = 3.6 x 10 6 W∙s = 3.6 x 10 6 J

12. Assignment page 623 Problems 1 - 4