Practise Power. 14 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.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
State the unit of powerPower is measured in watts (W); 1 W is 1 joule per second.
4
Explain what power meansPower is the rate of energy transfer: the energy transferred each second.
4
Calculate power using P = V IMultiply the pd in volts by the current in amps to get the power in watts.
5
Rearrange P = V IUse I = P ÷ V, e.g. to find the current an appliance takes from the mains.
6
Calculate power using P = I² RSquare the current first, then multiply by the resistance.
7
Find current or resistance from P = I² RUse I = √(P ÷ R) or R = P ÷ I².
8
Combine power equations with V = I RLink P = V I, P = I² R and V = I R in multi-step problems.
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
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.
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.
Doubling the current makes the power four times bigger (for the same resistance). grade 7+
To find the current: I = √(P ÷ R). To find the resistance: R = P ÷ I2. grade 7+
How the equations link grade 8+
P = I2 R comes from P = V I with V = I R put in its place: P = (I R) × I = I2 R.
So if you know any two of V, I and R, you can find the third with V = I R and then the power with either equation.
Cheatsheet
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
I = √(P ÷ R) and R = P ÷ I2grade 7+
1 kW = 1000 W
How to answer each type of question
Calculate power using P = V I
2 marks4
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.
Show the model answer
P = 230 × 5.0 (1) P = 1150 W (1)
Calculate the current from a power rating
3 marks5
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.
Show the model answer
P = 2.3 × 1000 = 2300 W (1) I = P ÷ V = 2300 ÷ 230 (1) I = 10 A (1)
Calculate power using P = I² R
2 marks6
Write P = I2 R.
Square the current, then multiply by the resistance.
Give the answer in watts.
Example. The current through a resistor of resistance 20 Ω is 0.50 A. Calculate the power transferred by the resistor.
Show the model answer
P = 0.502 × 20 (1) P = 0.25 × 20 = 5.0 W (1)
Find the current using P = I² R
3 marks8
Rearrange to I2 = P ÷ R.
Calculate I2 and write it down.
Take the square root to find I, and give the unit.
Example. A heating element has a resistance of 48 Ω. The power of the heating element is 1200 W. Calculate the current through the heating element.
Show the model answer
I2 = P ÷ R = 1200 ÷ 48 (1) I2 = 25 (1) I = √25 = 5.0 A (1)
Shortcuts and 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.
Where marks are lost
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.
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.
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]
Show the answer and mark scheme
(a)Answer: P = I2 R
P = I2 × R / power = (current)2 × resistance
(b)Answer: 8.7 A
2000 = 230 × I
I = 2000 ÷ 230
I = 8.7 (A)
(c)Answer: 2.5 W
P = 0.252 × 40
P = 2.5 (W)
Question 3Hard8 marks
An extension lead contains two copper wires that carry the current to and from an appliance. The total resistance of the two wires is 0.15 Ω.
(a) A heater draws a current of 13 A through the extension lead. Calculate the power dissipated in the wires of the lead.[2]
(b) Explain why the wires in the extension lead get warm.[2]
(c) The heater is replaced by an appliance that draws a current of 6.5 A. Calculate the power dissipated in the wires now.[2]
(d) Halving the current reduces the power dissipated in the wires to a quarter. Explain why.[1]
(e) Suggest why an extension lead that is left tightly coiled can overheat.[1]
Show the answer and mark scheme
(a)Answer: 25 W
P = 132 × 0.15
P = 25 (W)
(b)Answer: As charge flows, work is done against the resistance of the wires (electrons collide with the ions in the metal), so energy is transferred to the thermal store of the wires.
work is done (by the current) against the resistance of the wires / electrons collide with the ions (in the copper)
so energy is transferred to the thermal energy store of the wires
(c)Answer: 6.3 W
P = 6.52 × 0.15
P = 6.3 (W)
(d)Answer: The power dissipated is proportional to the current squared (P = I²R with R constant), and (½)² = ¼.
power ∝ current2 / P = I2R with R constant, so (½)2 = ¼
(e)Answer: The energy dissipated in the wires cannot be transferred to the surroundings as easily, so the temperature rises.
energy is not transferred / dissipated to the surroundings as quickly (so the temperature of the wires rises)