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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’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
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).