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.
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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)