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

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

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

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

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?

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

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

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

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)

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

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

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 23 500J? _________ kJ 0.325 How many kiloWatts in 325W? _________ kW

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

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

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

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

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

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

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