2. 9.2 The Power of Electricity

3.  the change in energy; the rate at which work is done or energy transformed  Energy is measured in joules (J)  One joule of energy transformed in one second is called one Watt (W) of power  Electrical power is the rate of change in electrical energy  A 60 W light bulb uses 60 J of energy every second and produces 60 J of heat and light energy  Instead of J/s, for electrical power it is calculated at P = VI

4. 1. A flashlight bulb operates on 3.0 V and draws a current of 4.0 A. What is the power of the bulbs? 2. A 60 W light bulb is connected to 120 V. What current passes through the light bulb? 3. A voltmeter measures 15 V across a 45 W resistor. What current is passing through the resistor?

5. 1. A flashlight bulb operates on 3.0 V and draws a current of 4.0 A. What is the power of the bulbs? P = VIP = 3.0 V x 4.0 A = 12 W 2. A 60 W light bulb is connected to 120 V. What current passes through the light bulb? I = P/VI = 60 W / 120 V = 0.5 A 3. A voltmeter measures 15 V across a 45 W resistor. What current is passing through the resistor? I = P/VI = 45 W/15 V = 3 A

6.  A power rating is a measurement of how much electrical energy an electrical device consumes for every second it is in use  P = E/t and E = Pt Questions 1. How much electrical energy is consumed by a 60 W light bulb if it is left on for 25 min? 2. A 1600 W kettle is turned on for 3.0 min. How much electrical energy does the kettle use in this time? 3. How much electrical energy is consumed by a 100 W light bulb left on for 4.0 h?

7. 1. How much electrical energy is consumed by a 60 W light bulb if it is left on for 25 min? E = PtE = 60 W x 25 min x 60 s/min = 9.0 x 10 4 J 2. A 1600 W kettle is turned on for 3.0 min. How much electrical energy does the kettle use in this time? E = PtE = 1600 W x 3.0 min x 60 s/min = 2.9 x 10 5 J 3. How much electrical energy is consumed by a 100 W light bulb left on for 4.0 h? E = PtE = 100 W x 4 h x 3600 s/h = 1.4 x 10 6 J

8.  Since the numbers can get pretty huge, kilowatt hour (kW-h) are used...that’s 1000 W in 3600 seconds  Power companies charge per kW-h...currently they charge around 7 or 8 cents per kW-h in BC  Electrical surges are brief increases in voltage that can fry electrical equipment; surge protectors absorb some of the surge and divert the rest to the ground