Equipotential Lines - Tenafly Public Schools
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Transcript Equipotential Lines - Tenafly Public Schools
-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).