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-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

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## -Electric Power

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**-Electric Power**AP Physics C Mrs. Coyle**Remember:**P= W / t P= dW /d t Power=Work/time W= ΔV q and I = q/t P= I V**Electric Power, P= I ΔV**Known as Joule’sLaw P: is the power consumed by a resistor, R. Unit: Joule/s= Watt**kWh**kiloWatt hour What does the kWh measure, a) Energy or b) Power ?**From P=I ΔV and Ohm’s Law:**P=V2/R P=I2R**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**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**The power is the rate at which the energy is delivered to**the resistor**Resistors Expend Thermal Energy**• Wasted heat energy is called “Joule Heating” or “I2 R” loss.**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.**Electric heaters(Coil Heaters)**• P= V2/R • The lower the R the greater the heat given off by the resistor for a given voltage.**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.**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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