1. -Electric Power AP Physics C Mrs. Coyle

2. Remember: P= W / t P= dW /d t Power=Work/time W= ΔV q and I = q/t P= I V

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

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

5. From P=I Δ V and Ohm’s Law: P=V 2 /R P=I 2 R

6. As a charge moves from a to b, the electric potential energy of the system increases by Q  V The chemical energy in the battery must decrease by this same amount The battery “pumps” energy to the +charges

7. As the current flows through the resistor (c to d), the system loses electric potential energy Energy is transformed into heat energy in the resistor

8. The power is the rate at which the energy is delivered to the resistor

9. Resistors Expend Thermal Energy Wasted heat energy is called “Joule Heating” or “I 2 R” loss.

10. 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 “I 2 R” loss in the transmitting wires, so increase V.

11. Electric heaters(Coil Heaters) P= V 2 /R The lower the R the greater the heat given off by the resistor for a given voltage.

12. 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.

13. 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).