-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’s Law

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

As a charge moves from a to b, the electric potential energy of the system increases by QV

The chemical energy in the battery must decrease by this same amount

The battery “pumps” energy to the +charges

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).