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 A lot of devices are used to convert electrical energy into some other useable form ◦ Light bulb, hair dryer, stove  A lot of energy lost in heat ◦

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Presentation on theme: " A lot of devices are used to convert electrical energy into some other useable form ◦ Light bulb, hair dryer, stove  A lot of energy lost in heat ◦"— Presentation transcript:

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2  A lot of devices are used to convert electrical energy into some other useable form ◦ Light bulb, hair dryer, stove  A lot of energy lost in heat ◦ Light bulb & engines get warm

3  What are some appliances used to convert almost all energy into thermal energy ◦ Hair dryer, space heater, hot plate  To determine how much total energy is converted to thermal energy…. ◦ P = E/t  E = Pt ◦ P = IV  V = IR ◦ So… ◦ E = I 2 Rt

4  What is the current through the resistance  KnownUnknown ◦ R = 10.0 ΩI = ??? ◦ V = V  I = V / R  = V / 10.0 Ω  = 12.0 A

5  What thermal Energy is supplied by the heater in 10.0 s  KnownUnknown ◦ I = 12.0 AE = ??? ◦ R = 10.0 Ω ◦ t = 10.0 s  E = I 2 Rt  (12.0 A) 2 (10.0 Ω)(10.0 s)  =14.4 x 10 3 J  =14.4 kJ

6  Niagara Falls & Hoover dam produce a lot of electricity w/o much pollution ◦ Has to transmitted long distances ◦ Thermal energy lost is P=I 2 R ◦ Engineers want to reduce I or R ◦ **1km of wire = 0.20 Ω

7  Supposed a farm house is 3.5 km away and has a 41 A electric stove.  The power dissipated (loss) in the wires =  2(3.5 km)(0.20 Ω/km) ◦ = 1.4 Ω  P = I 2 R  = (41 A) 2 x 1.4Ω = 2400 W

8  Electric Companies charge you for energy  Energy used by any device is its rate of energy consumption ◦ Joules/second (J/s)(sec) ◦ Too small for commercial use so companies use the Kilowatt-Hour or kWh  = 1000 watts delivered continuously for 3600 sec  = 3.6 x 10 6 J

9  Other than heating appliances, most don’t need more than 1000-W of power ◦ Evident in your lab yesterday

10  A television set draws 2.0 A when operated on 120 V.  A) How much power does the TV use?  B) If the set operated for an average of 7.0 h/day, what energy in kWh does it consume/month(30 days)  C) At 11 cents per kWh, what is the cost of operating the set per month?  KnownUnkown  I = 2.0 AE = ???  V = VTotal Cost = ???  t = (7.0 h/day)(30 days)  Cost = 11 cents / kWh

11  P = IV  = (2.0 A)(120.0 V)  = 240 W

12  E = Pt  (240 W)(7.0 h/day)(30 d)  = 5.0 x 10 4 Wh  = 5.0 x 10 1 kWh

13  Cost = (5.0 x 10 1 kWh)($0.11/kWh)  = $5.50


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