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# -Electric Power

-Electric Power. AP Physics C Mrs. Coyle. Remember:. P= W / t P= dW /d t Power=Work/time W= Δ V q and I = q/t P= I V. Electric Power, P= I Δ V. Known as Joule’s Law P: is the power consumed by a resistor, R . Unit: Joule/s= Watt. kWh. kiloWatt hour

## -Electric Power

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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 ΔV Known as Joule’sLaw 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 ΔV and Ohm’s Law: P=V2/R P=I2R

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

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

9. 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.

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

11. 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.

12. 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).

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