# Electric Power

Document Sample

```					                       Electric Power

Physics
Mrs. Coyle

http://www.cnas.missouristate.edu/assets/cnas/Wind_Power.jpg
Part I
• Electrical Power
• Joule’s Law
• Thermal Energy Expended by a Resistor
Wind Generator
Generator
Generator
Niagara Falls Hydroelectric Plant

http://www.niagarafallslive.com/images/NY_Hydro_Electric_Plant.JPG
Hydroelectric Power Plant
Remember:
P= W / t
(Power=Work/time)

• Remember: W= Vq and I = q/t

So: P= I V
Electric Power, P= I V
Known as Joule’s Law

P: is the power consumed by a resistor, R.

Unit: Joule/s= Watt
Derive the following using P=IV
and Ohm’s Law:
P=V2/R

P=I2R
Problem 1
• The current through a car motor is
150A. The battery used is 12V. How
much energy is supplied by the
battery in 5s?

kWh
kiloWatt hour

What does the kWh measure,
a) Energy or b) Power ?
Problem 2
A small desktop radio has a resistance
of 8,000 Ω. The voltage is 120V.
a)How much current does it draw?
b)How much power does it use?
c)How much does it cost to run the radio
for 12 hours, if 1kWh costs \$0.15?

a)0.015A, b)1.8W, c) \$0.00324
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

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.
Problem 3
• An electric heating element has a
resistance of 9 Ω. The voltage is 120V.
a)Calculate the current through the element.
b)Calculate the thermal energy supplied by
the element in 10min.
a)13.3A, b) 9.6x105 J
Part II
• Brightness of a Lightbulb
• Some Safety Issues
Light Bulb
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).
Question
• Which light bulb has a thicker filament,
60W or 75W?

Problem 4
A 60W light bulb is 22% efficient. How
much energy is lost to heat in 5min?

Safety Issues
• Why is it not safe to handle electric
appliances with wet hands?
• Hint:
• A current of 15mA can cause serious
injury.
• The resistance of a dry human body is
in the range of 100,000Ω. The
resistance of a wet person can drop to
the range of 5,000Ω.
Problem 5
• A current of 1mA can be felt by a person.
Will a person feel the current if the person
has dry hands (a resistance of 100,000 Ω)
and touches a 9V battery’s terminals?

Problem 6
• The same 9V battery is touched by a
person with wet hands. Will this person feel
the current if the resistance is 5,000Ω and
the person touches the battery’s terminals?