AQA GCSE Physics Foundation (8463), Foundation tier · Electricity › Energy transfers
Practise Power. 8 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
How the power of a device depends on the potential difference across it and the current through it (P = V I), or on the current and its resistance (P = I² R). Expect 2 to 4 mark calculations, often with rearranging, a square root or a unit conversion.
Key facts
Power = rate of energy transfer; unit: watt (W) = J/s
P = V I: power (W) = pd (V) × current (A)
P = I2 R: power (W) = current2 (A2) × resistance (Ω)
I = P ÷ V and V = P ÷ I
1 kW = 1000 W
Notes
What power means
Power is the rate at which energy is transferred: the energy transferred each second.
Power is measured in watts (W). 1 W is 1 joule of energy transferred per second. 1 kW = 1000 W.
The power of a device depends on the pd across it and the current through it. A bigger pd or a bigger current means more energy is transferred each second.
P = V I
diagram
power = potential difference × current: P = V I (P in W, V in V, I in A).
Rearranged: I = P ÷ V and V = P ÷ I.
Example: a heater on the 230 V mains with a current of 8.0 A has a power of 230 × 8.0 = 1840 W.
Power from the meter readings: P = V × I.
P = I2 R
power = (current)2 × resistance: P = I2 R (P in W, I in A, R in Ω).
Use it when you know the current and the resistance but not the pd, e.g. for the energy transferred by heating in a cable.
Square the current first: for I = 3.0 A and R = 5.0 Ω, P = 3.02 × 5.0 = 9.0 × 5.0 = 45 W.
How to answer each type of question
Calculate power using P = V I
2 marksGrade 4
Write P = V I.
Substitute the pd in volts and the current in amps.
Give the answer in watts.
Example. A hairdryer is connected to the 230 V mains supply. The current through it is 5.0 A. Calculate the power of the hairdryer.
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P = 230 × 5.0 (1) P = 1150 W (1)
Calculate the current from a power rating
3 marksGrade 5
Convert kW to W.
Rearrange to I = P ÷ V.
Substitute and give the answer in amps.
Example. A kettle has a power rating of 2.3 kW. It is connected to the 230 V mains supply. Calculate the current through the kettle.
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P = 2.3 × 1000 = 2300 W (1) I = P ÷ V = 2300 ÷ 230 (1) I = 10 A (1)
Don’t lose marks
Forgetting to square the current in P = I2 R, or squaring I × R instead of just I.
Leaving the power in kW when using P = V I: convert to W first.
Forgetting to take the square root at the end when finding I from P = I2 R.
Choosing the wrong equation. Pick the one that uses the two quantities you are given.
More tips
Memory tricks
Formula triangle for P = V I: P on top, V and I underneath.
A rhyme for the second equation: 'Twinkle, twinkle, little star, power equals I squared R.'
Sense check: kettles and heaters are a few kilowatts, so on the 230 V mains their current is about 10 A.
Exam technique
P = V I and P = I2 R must both be learned: neither is on the equation sheet.
Write down the equation before substituting. This often earns the first mark.
Show intermediate values, such as I2 = 25, so the method marks are clear.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade 3
State the unit of powerPower is measured in watts (W); 1 W is 1 joule per second.
Grade 4
Explain what power meansPower is the rate of energy transfer: the energy transferred each second.
Grade 4
Calculate power using P = V IMultiply the pd in volts by the current in amps to get the power in watts.
Grade 5
Rearrange P = V IUse I = P ÷ V, e.g. to find the current an appliance takes from the mains.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Write down the equation that links current (I), power (P) and resistance (R).
P = I2 R
Write down the equation that links power (P), potential difference (V) and current (I).
P = V I
Which equation should the student use to calculate the power of the heater, and why?
P = I²R, because it uses only the current and the resistance, which are known.
A device transfers 40 J of energy every second. What is its power?
40 W
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
An electric heater is connected to the 230 V mains supply.
(a) What is the unit of power? Tick (✓) one box.[1]
joule
newton
volt
watt
(b) The current in the heater is 4.0 A. Calculate the power of the heater. Use the equation: power = potential difference × current[2]
(c) A lamp connected to the mains has a power of 60 W. Which statement is correct? Tick (✓) one box.[1]
The heater transfers more energy each second than the lamp.
The lamp transfers more energy each second than the heater.
The heater and the lamp transfer the same energy each second.
Show the answer and mark scheme
(a)Answer: watt
(b)Answer: 920 W
P = 230 × 4.0
P = 920 (W)
(c)Answer: The heater transfers more energy each second than the lamp.
Question 2Medium6 marks
Electrical appliances transfer energy at different rates.
(a) Write down the equation that links current (I), power (P) and resistance (R).[1]
(b) A 2.0 kW heater is connected to the 230 V mains supply. Calculate the current in the heater.[3]
(c) The current in a 40 Ω resistor is 0.25 A. Calculate the power transferred by the resistor.[2]