The document discusses electrical energy and power ratings. It explains that electrical energy is commonly used today but often inefficiently. Only about 2-10% of input energy is converted to light by incandescent and fluorescent bulbs, and fossil fuel power plants convert around 32% to electricity. Power ratings in Watts measure how much energy devices use over time. Common appliances' power ratings and calculating energy use in kilowatt-hours is also covered. The relationship between power (P), voltage (V) and current (I) is defined through different equations. Practice questions are provided to test understanding.

1. Electrical Power, Power
Rating, Cost of Electricity
 UsingUsing kWh to Measure Electricity Useto Measure Electricity Use
Nelson Reference Pages 504-507Nelson Reference Pages 504-507
(McGraw-Hill Reference Pages 652- 664)(McGraw-Hill Reference Pages 652- 664)

2. Electrical energy is the most common form ofElectrical energy is the most common form of
energy in use today. However, we usuallyenergy in use today. However, we usually
do not use this energy efficiently. Thedo not use this energy efficiently. The
common, but increasing less used,common, but increasing less used,
incandescent bulb converts only aboutincandescent bulb converts only about 2%2%
of the electrical energy to light energy. Aof the electrical energy to light energy. A
fluorescent bulb can convert aboutfluorescent bulb can convert about 10%10% ofof
the input energy to light. This isthe input energy to light. This is 55 timestimes
better but still not great. Even the way webetter but still not great. Even the way we
generate electrical energy is inefficient. Agenerate electrical energy is inefficient. A
fossil fueled power plant can only convertfossil fueled power plant can only convert
aboutabout 32%32% of its input energy to electricalof its input energy to electrical
energy.energy.

3. Power Rating (in Watts)
All electrical appliances will have a power ratingAll electrical appliances will have a power rating
which can be used to measure the amount ofwhich can be used to measure the amount of
electrical energy used for a specific amount ofelectrical energy used for a specific amount of
time. Devices thattime. Devices that generate heat (thermal(thermal
energy) typically have aenergy) typically have a high power rating, but air, but air
conditioners and microwave ovens also have highconditioners and microwave ovens also have high
ratings. (High rating means high energyratings. (High rating means high energy
consumption)consumption)
Typical Power Ratings (other values pg 505)
Hair Dryer: 1 200 W , Alarm Clock: 5 W
Electric Oven: 12 000 W
Clothes Dryer: 5 000 W

4. Calculation of Electrical Energy Consumption
For some electrical devices, such as light
bulbs, this is an easy calculation:
Energy Used (J) = Power Rating (W) x Time Used (s)
This equation will not directly work for ovens, airThis equation will not directly work for ovens, air
conditions, etc. which continuous turn on and off.conditions, etc. which continuous turn on and off.
Measuring Energy Used in kWh (kilowatt-hour)
This is similar to equation above, except power rating
must be in kilowatts (divide watts by ______ ) and
time used must be in ________ .

5. Example:
A 1200 W hair dryer is used for 20 minutes.A 1200 W hair dryer is used for 20 minutes.
Determine the energy used in kWh.Determine the energy used in kWh.
Ans. E = 1.200 kW x (20/60) h = 0.40 kWhE = 1.200 kW x (20/60) h = 0.40 kWh
If electricity costs $ 0.060/kWh, determineIf electricity costs $ 0.060/kWh, determine
the cost to run the dryer.the cost to run the dryer.
Ans.Ans. 0.40 kWh x 0.060 ($/kWh) = $ 0.0240.40 kWh x 0.060 ($/kWh) = $ 0.024

6. Relationship Between P, V and I
P =P = ΔΔEEQQ // ΔΔtt
But,But, V =V = ΔΔEEQQ / Q and/ Q and ΔΔEEQQ ==VQVQ
So,So, P = VP = VQ /Q / ΔΔtt
But,But, I = Q /I = Q / ΔΔtt
So,So, P = V I, this is the first power equation.
The other two equations are found by using
Ohm’s Law ( V = IR ) to substitute for V
and I in the first power equation to get:
P = I²R and P = V² /R

7. Practice Questions
Nelson Textbook:
Page 507 # 2- 4Page 507 # 2- 4
Workbook
Page 52 Q# 12 – 16Page 52 Q# 12 – 16 {Supply answers to these}{Supply answers to these}
Questions from McGraw-Hill Textbook
1.1. A light bulb designed for use with a 120 V powerA light bulb designed for use with a 120 V power
supply has a filament with a resistance of 240 Ω.supply has a filament with a resistance of 240 Ω.
a.a. What is the power output?What is the power output? (60 W)
b.b. If the bulb was accidentally connected to a 80.0 VIf the bulb was accidentally connected to a 80.0 V
power supply what would be the power output ofpower supply what would be the power output of
the bulb?the bulb? (27 W)

8. 2.2. A heater has a resistance of 15 Ω.A heater has a resistance of 15 Ω.
a.a. If the heater is drawing a current of 7.5 A what isIf the heater is drawing a current of 7.5 A what is
its power output?its power output? (840W)
b.b. If the current to the heater was cut in half, whatIf the current to the heater was cut in half, what
would happen to the power output?would happen to the power output? (Drops)
 An electric kettle operates on 120 V supply, and
has a heating element with a resistance of 10.0
Ω. If it takes 3.2 min to boil a litre of water andΩ. If it takes 3.2 min to boil a litre of water and
the cost of electricity is 6.5 cents per kWh,the cost of electricity is 6.5 cents per kWh,
determine:determine:
a.a. The power ratingThe power rating (1440 W)
b.b. The cost to boil the waterThe cost to boil the water (0.50 cents)

9. Additional Questions
Teacher May Give
McGraw-Hill questionsMcGraw-Hill questions
Page 655 #40 – 42Page 655 #40 – 42
Page 662 #46 – 50Page 662 #46 – 50
Test Review Questions McGraw-Hill:Test Review Questions McGraw-Hill:
Page 667 #24 to 28Page 667 #24 to 28
Workbook page 52 #8 (ρWorkbook page 52 #8 (ρFeFe = 10 x 10= 10 x 10 -8-8
Ωm),Ωm),
9,9,

10. Additional Questions
Teacher May Give
McGraw-Hill questionsMcGraw-Hill questions
Page 655 #40 – 42Page 655 #40 – 42
Page 662 #46 – 50Page 662 #46 – 50
Test Review Questions McGraw-Hill:Test Review Questions McGraw-Hill:
Page 667 #24 to 28Page 667 #24 to 28
Workbook page 52 #8 (ρWorkbook page 52 #8 (ρFeFe = 10 x 10= 10 x 10 -8-8
Ωm),Ωm),
9,9,