Electric Power AP Physics C Mrs Coyle Remember
-Electric Power AP Physics C Mrs. Coyle
Remember: P= W / t P= d. W /d t Power=Work/time W= ΔV q and I = q/t P= I V
Electric Power, P= I ΔV Known as Joule’s Law P: is the power consumed by a resistor, R. Unit: Joule/s= Watt
k. Wh kilo. Watt hour What does the k. Wh measure, a) Energy or b) Power ?
From P=I ΔV and Ohm’s Law: P=V 2/R P=I 2 R
The battery “pumps” energy to the +charges n As a charge moves from a to b, the electric potential energy of the system increases by QDV n The chemical energy in the battery must decrease by this same amount
n As the current flows through the resistor (c to d), the system loses electric potential energy n Energy is transformed into heat energy in the resistor
n The power is the rate at which the energy is delivered to the resistor
Resistors Expend Thermal Energy n Wasted heat energy is called “Joule Heating” or “I 2 R” loss.
Why is long distance power transmitted at high voltages? n 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. n
Electric heaters(Coil Heaters) n P= V 2/R n The lower the R the greater the heat given off by the resistor for a given voltage.
Brightness of a Light bulb and Power n The greater the power actually used by a light bulb, the greater the brightness. n 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 n 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). n Thus, the higher wattage bulb will have a filament of lower resistance and therefore a larger cross-sectional area (R=ρ L / A).
- Slides: 13