AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER
Electrical power
Electricity · Lesson 6 of 10
Teacher copy - includes the notes for whoever is teaching from it.
Warm-up
Answer each one, then check.
1. What is the unit of power?
Watt (W)
2. Write the equation linking energy and time for power.
P = E ÷ t
3. What is V = IR used for?
Linking pd, current and resistance
4. Convert 2.3 kW to watts.
2300 W
5. Work out 3².
9
Learning Objectives
1. Explain how the power transfer in a device is related to the pd across it and the current through it.
2. Recall and apply P = VI.
3. Recall and apply P = I²R.
4. Rearrange to find current, pd or resistance.
Electrical Power
power = potential difference × current P = VI
Also P = I²R (power = current squared × resistance). Power is in watts (W), pd in volts (V), current in amperes (A) and resistance in ohms (Ω).
Power in a Circuit
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A device with a bigger pd or current has a greater power. |
A lamp with a voltmeter and ammeter and the power calculated as pd times current.
Using P = VI
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A kettle has a current of 10 A when connected to the 230 V mains. Calculate its power. |
1. Write the equation
P = VI
2. Substitute
P = 230 × 10
3. Answer
P = 2300 W
Answer: 2300 W (2.3 kW)
Using P = I²R
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A current of 6.0 A flows through a 5.0 Ω resistor. Calculate the power. |
1. Write the equation
P = I²R
2. Square the current first
6.0² = 36
3. Substitute
P = 36 × 5.0
4. Answer
P = 180 W
Answer: 180 W
Finding the Current
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A 60 W lamp is connected to the 230 V mains. Calculate the current. |
1. Rearrange
I = P ÷ V
2. Substitute
I = 60 ÷ 230
3. Answer
I = 0.26 A
Answer: 0.26 A
Which Equation?
Look at what you are given.
|
You have |
Use |
|---|---|
|
V and I |
P = VI |
|
I and R |
P = I²R |
|
V and R |
Find I first with I = V ÷ R, then P = VI |
|
E and t |
P = E ÷ t |
Common Mistakes
Take care with squares.
▸ Squaring. Square only the current, then multiply by the resistance.
▸ Units. Convert mA to A and kW to W first.
▸ Rating labels. The power rating of an appliance is at its normal pd.
Key Terms
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Power The rate at which energy is transferred. |
Watt One joule per second. |
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Power rating The power an appliance transfers in normal use. |
Kilowatt 1000 watts. |
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Potential difference Energy transferred per coulomb of charge. |
Current The rate of flow of charge. |
Your Task: Rating Plate
10 minutes
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A hair dryer is rated 1500 W at 230 V. Calculate the current. A toaster takes 4.0 A at 230 V; find its power. 1. Rearrange P = VI for current. 2. Then calculate the toaster power. |
A good answer shows: Hair dryer: 1500 ÷ 230 = 6.5 A. Toaster: 230 × 4.0 = 920 W.
Note: Ask which draws more current and what this means for the fuse.
Can I...?
☐ Recall P = VI.
☐ Recall P = I²R.
☐ Choose the right equation.
☐ Rearrange for current.
☐ Square the current correctly.
☐ Convert units to W and A.
☐ Explain what power means.
☐ Give the unit.
Summary
✓ P = VI.
✓ P = I²R.
✓ Power is in watts.
✓ Higher power means faster energy transfer.
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EXAM FOCUS A kettle draws 10 A from the 230 V mains. Calculate its power. (2 marks) Choose P = VI, substitute and give W. |
Exam Practice: Electrical power
Answer all questions. Use the mark allocation as a guide to how much to write. · 15 minutes
▸ Question 1 · 2 marks · Calculate. A kettle is connected to the 230 V mains and the current is 10 A. Calculate the power of the kettle. Use the equation: power = potential…
▸ Question 2 · 3 marks · Calculate. A current of 6.0 A flows through a resistor of resistance 5.0 Ω. Calculate the power. Use the equation: power = (current)² × resistance
▸ Question 3 · 3 marks · Calculate. A lamp has a power of 60 W and works at 230 V. Calculate the current in the lamp.
▸ Question 4 · 2 marks · Explain. A student says that the power of a device only depends on the current. Explain why the student is not correct.
▸ Question 5 · 4 marks · Calculate. An electric heater has a resistance of 46 Ω and is connected to a 230 V supply. Calculate the power of the heater.
Question 1 · 2 marks · Calculate
|
“A kettle is connected to the 230 V mains and the current is 10 A. Calculate the power of the kettle. Use the equation: power = potential difference × current” |
HOW TO ANSWER IT Command word: Calculate. Worth 2 marks, so plan before writing.
Question 1 · mark scheme
2 marks available. Award a mark for each point made.
▸ Correct substitution. 1 mark
▸ 2300 W. 1 mark
▸ Model answer. P = 230 × 10 = 2300 W
Question 2 · 3 marks · Calculate
|
“A current of 6.0 A flows through a resistor of resistance 5.0 Ω. Calculate the power. Use the equation: power = (current)² × resistance” |
HOW TO ANSWER IT Command word: Calculate. Worth 3 marks, so plan before writing.
Question 2 · mark scheme
3 marks available. Award a mark for each point made.
▸ Squares the current. 1 mark
▸ Correct substitution. 1 mark
▸ 180 W. 1 mark
▸ Model answer. P = 6.0² × 5.0 = 180 W
Question 3 · 3 marks · Calculate
|
“A lamp has a power of 60 W and works at 230 V. Calculate the current in the lamp.” |
HOW TO ANSWER IT Command word: Calculate. Worth 3 marks, so plan before writing.
Question 3 · mark scheme
3 marks available. Award a mark for each point made.
▸ Rearranges to I = P ÷ V. 1 mark
▸ Correct substitution. 1 mark
▸ 0.26 A. 1 mark
▸ Model answer. I = P ÷ V = 60 ÷ 230 = 0.26 A
Question 4 · 2 marks · Explain
|
“A student says that the power of a device only depends on the current. Explain why the student is not correct.” |
HOW TO ANSWER IT Command word: Explain. Worth 2 marks, so plan before writing.
Question 4 · mark scheme
2 marks available. Award a mark for each point made.
▸ Power = V × I. 1 mark
▸ Depends on pd (or resistance) as well as current. 1 mark
▸ Model answer. Power depends on both the current and the potential difference (P = VI), or on the current and the resistance (P = I²R).
Question 5 · 4 marks · Calculate
|
“An electric heater has a resistance of 46 Ω and is connected to a 230 V supply. Calculate the power of the heater.” |
HOW TO ANSWER IT Command word: Calculate. Worth 4 marks, so plan before writing.
Question 5 · mark scheme
4 marks available. Award a mark for each point made.
▸ Rearranges V = IR. 1 mark
▸ 5.0 A. 1 mark
▸ Correct substitution into P = VI. 1 mark
▸ 1150 W. 1 mark
▸ Model answer. I = 230 ÷ 46 = 5.0 A; P = 230 × 5.0 = 1150 W