Download presentation

Presentation is loading. Please wait.

1
**22-2 Using Electric Energy**

2
Energy Transfer 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
Heating a Resistor 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 = I2Rt

4
**A heater has a resistance of 10.0 Ω. It operates on 120.0 V**

What is the current through the resistance Known Unknown R = 10.0 Ω I = ??? V = V I = V / R = V / 10.0 Ω = 12.0 A

5
**A heater has a resistance of 10.0 Ω. It operates on 120.0 V**

What thermal Energy is supplied by the heater in 10.0 s Known Unknown I = 12.0 A E = ??? R = 10.0 Ω t = 10.0 s E = I2Rt (12.0 A)2(10.0 Ω)(10.0 s) =14.4 x 103 J =14.4 kJ

6
**Transmission of Electric Energy**

Niagara Falls & Hoover dam produce a lot of electricity w/o much pollution Has to transmitted long distances Thermal energy lost is P=I2R Engineers want to reduce I or R **1km of wire = 0.20 Ω

7
**Transmission of Electric Energy**

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 = I2R = (41 A)2 x 1.4Ω = 2400 W

8
**Kilowatt-Hour 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 106 J

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

10
**Cost of operating 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? Known Unkown I = 2.0 A E = ??? V = V Total Cost = ??? t = (7.0 h/day)(30 days) Cost = 11 cents / kWh

11
**A) How much power does the TV use?**

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

12
**E = Pt (240 W)(7.0 h/day)(30 d) = 5.0 x 104 Wh = 5.0 x 101 kWh**

B) If the set operated for an average of 7.0 h/day, what energy in kWh does it consume/month(30 days) E = Pt (240 W)(7.0 h/day)(30 d) = 5.0 x 104 Wh = 5.0 x 101 kWh

13
**Cost = (5.0 x 101 kWh)($0.11/kWh) = $5.50**

C) At 11 cents per kWh, what is the cost of operating the set per month? Cost = (5.0 x 101 kWh)($0.11/kWh) = $5.50

Similar presentations

Presentation is loading. Please wait....

OK

Circuits & Electronics

Circuits & Electronics

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Download ppt on rectification of errors Inner ear anatomy and physiology ppt on cells Ppt on introduction to object-oriented programming languages Ppt on real numbers for class 9th computer Ppt on principles of peace building definition Ppt on unity in diversity youtube Ppt on current account deficit in india Ppt on area of trapezium example Ppt on introduction to object-oriented programming tutorial Ppt on case study of drought in india