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					                       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?



 Answer: 9,000J
          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?

 Answers:
 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

• 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.
               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.
Answer:
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?




• Answer: 75W
              Problem 4
 A 60W light bulb is 22% efficient. How
 much energy is lost to heat in 5min?




Answer: 1.4x104 J
             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?



• Answer: I=0.09mA, No.
               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?



• Answer: I=1.8mA, Yes

				
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posted:9/1/2011
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