1. Equipment
This lesson has a small focus on a new equation, other than that it is a small recap and equation practice.
Calculators
A r

2. Recap: True or False
The earth wire in a three-core cable is usually brown
Appliances with metal casings are supplied with three-core cables
Fuses stop current flowing through a circuit by melting when the current flowing through them is above a certain value
A circuit breaker works by monitoring the difference in current between the live and earth wire
Mains electricity uses direct current at 100Hz.
True or false.. For the false answers students must add the correction! F T

3. 4 6 8 Lesson intentions: Watt is the unit of Power!
Date and title: Electrical Power
2 May, 2019
S t e p s
t o
S u c c e s s
Learning outcomes: 4
Calculate the power of an appliance from the current and the potential difference
Find the fuse required for an appliance based on its power rating
Perform calculations involving rearrangement of the power equation 6 12 8
Key words: Current, Potential Difference, Power, Watt, Joules

4. Bigger picture Topic Fuses and Circuit breakers ACDC Plugs
Electrical Power
Topic
Electricity in the home
Energy Transfer
Appliances and efficiency

5. Current, Charge and Power
Match up the electrical quantities, with their abbreviations and units. I V E
power
ampere Ω
Energy R
Voltage
volts
resistance
watts
joule P
current
For this section it is important that the students have everything matched up in terms of use of units
Power P Watts
Voltage V Volts
Resistance Ω Ohms
Current I Amps
Energy E Joule
Can you provide any definitions for these units?
Extension: can you describe how these factors are interrelated by giving equations or describing the links in sentences?

6. POWER (P) Power is the rate at which something transfers energy
Power (W) =
Energy Transferred (J)
Time taken for energy to be transferred (s) E t P=
REITTERATE TO STUDENTS THAT THEY HAVE SEEN THIS SLIDE BEFORE IN P1!
Discuss power, ask students whey they have heard the term powerful used and what it means in that context.. Powerful car or athlete…. Able to transfer a lot of energy in not much time!
Have students go back to their experiment data and now use the information to calculate their power output.
W = Watts
1W = 1J per second

7. Energy and Power
A portable fan transfers 1800 joules of energy in 60 seconds. What is its power rating? P=E/t = 1800J/60s = 30W A mains-powered hairdryer is rated at 500W. How much energy will it transfer in 1 minute? E=Pxt = 500 x60 = 30,000J

8. Power (W) = Current (A) x Voltage (V)
POWER (P)
It can also be calculated for the components within a circuited or appliances from the current and voltage!
Power (W) = Current (A) x Voltage (V)
P= I x V
EMPHASISE:
Using the 2 equations POWER IS INTERCHANGABLE
W = Watts
1W = 1J per second

9. For the 40W, 60W and 100W bulbs calculate the current flowing.
0.17A, 0.26A and 0.43A
What fuse should each of the lamps have?
They must show all of their working.
They will more than likely ask the question.. What is the voltage.. Ask them what they think it is (what are bulbs plugged into.. The mains.. 230v)

10. Rearrange it!
If you have a filament bulb and it operates at a power of 60W and it has a potential difference of 240V across it, what is the current running through the bulb?
P = IV
I = P/V
I = 60W / 240V
I = 0.25A

11. kV, kJ, kW
1 kV = 1000 V
1 kJ = 1000 J
1 kW = 1000 W
How many Volts in 6kV? _________ V
6 000
A reminder of uing Kilo and the traps that can be put into questions with different place values
12 300
How many Joules in 12.3kJ? _________ J
600
How many Watts in 0.6kW? _________ W

12. kV, kJ, kW
1 kV = 1000 V
1 kJ = 1000 J
1 kW = 1000 W
9.0
How many kiloVolts in 9 000V? _________ kV
23.5
How many kiloJoules in J? _________ kJ
0.325
How many kiloWatts in 325W? _________ kW

13. Power and Fuses
The table shows the power ratings of different electrical appliances.
A Calculate the current used by each appliance. (Hint: don’t forget to change kW to W before calculating your answer.)
B Work out what size fuse would be needed by the plug for each appliance. Choose from: 3 A, 5 A, 13 A.
Appliance
Power rating
Current (A)
Fuse
a Computer
67 W
b Fridge
63 W
c Fridge-freezer
100 W
d Lawnmower
900 W
e Toaster
850 W
f TV
90 W
g Hair dryer
1 kW
h Electric fire (2 bars)
2 kW
i Tumble dryer
1.5 kW
j Kettle
2.5 kW

14. Power and Fuses
The table shows the power ratings of different electrical appliances.
A Calculate the current used by each appliance. (Hint: don’t forget to change kW to W before calculating your answer.)
B Work out what size fuse would be needed by the plug for each appliance. Choose from: 3 A, 5 A, 13 A.
Appliance
Power rating
Current (A)
Fuse
a Computer
67 W
0.29 3A
b Fridge
63 W
0.27
c Fridge-freezer
100 W
0.43
d Lawnmower
900 W
3.91 5A
e Toaster
850 W
3.70
f TV
90 W
0.40
g Hair dryer
1 kW
4.34
h Electric fire (2 bars)
2 kW
8.70
13A
i Tumble dryer
1.5 kW
6.52
j Kettle
2.5 kW
10.87

15. Power Loss
The resistance of a component and the current flowing through it determine how much energy is dissipated from it as heat
Remember how we talked about the variables being the same across equations?
P= I x V
V = I x R
How many variables are there in these 2 equations?
What do they have in common?
Start off talking about the conservation of energy.. Work is done across a component therefore energy transfers happen
EMPHASISE:
Variables are interchangeable if they represent the same thing!
4 Variables, P, I, V and R
V and I in common

16. Power Loss
This means you can substitute variables with other equations!
What if you don’t have V and you want to calculate power??
P= I x V
The same! V
I x R =
EMPHASISE:
Variables are interchangeable if they represent the same thing!
4 Variables, P, I, V and R
V and I in common

17. P= I x V V I x R = P= I2xR Power Loss
This means you can substitute variables with other equations!
What if you don’t have V and you want to calculate power??
P= I x V
The same! V
I x R =
EMPHASISE:
Variables are interchangeable if they represent the same thing!
4 Variables, P, I, V and R
V and I in common
Substitute the IxR for the V in the other equation
P= I2xR
This takes a little explaining and can confuse some weaker groups
P= I2xR

18. Calculate the power wasted in the cable through heating
POWER (P)
A 6kW (6000w) oven is connected with a cable of resistance 0.25Ω. When it is switched on a current of 26A passes through it.
Calculate the power wasted in the cable through heating
P = ?
I = 26 A
R = 0.25 Ω
P= I2xR
P= I x I x R
P= 26 x 26 x 0.25
P= 169W

19. Questions
A light bulb is connected to a 2V supply and experiences a current of 6.4A. What is the power rating of the bulb?
A kettle has a power rating of 1500w. What is the potential difference that it must be supplied with to have a current flowing through it of 30A?
A student attaches a 10V supply to a bulb with a power rating of 100w. What is the current running through the bulb?
The student now connect a 25w bulb to the same supply. What is the difference between the current going through this bulb compared to the 100w bulb?
Bulb A transfers 1000J in 10seconds. Bulb B transfers 1500J in 3 seconds. Which bulb will have a higher current running through it when connected to a 12V supply?
Additional questions if needed