The Physics Question Bank

10 GCSE Physics questions about the electric blanket, from Foundation to Higher, each with a full worked solution.

An electric blanket has long, thin insulated heating wires sewn through it. When a current flows, the wires’ resistance makes them heat up, warming the bed directly. Because it only heats you and your bed – not the air in the whole room – it is one of the cheapest ways to keep warm at night.

Course: Combined Science + Separate Physics  |  Tier: Foundation + Higher  |  Topics: Electricity · Energy

Equations you’ll use

Quantity	Symbol	Unit
energy transferred	E	joules (J)
power	P	watts (W)
time	t	seconds (s)
current	I	amps (A)
potential difference	V	volts (V)
resistance	R	ohms (Ω)
charge flow	Q	coulombs (C)

Equations: V = I R  |  P = V I  |  P = I² R  |  E = P t  |  Q = I t  |  E = Q V

Use FIFA for every calculation

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (convert to standard units first).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and always give the unit.

Foundation: aim for Q1–8.   Higher: aim for Q4–10.

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Questions

Q1 (F) A single electric blanket has a power of 60 W and runs on the 230 V mains. Calculate the current in the blanket.

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F: P = V I
I: 60 = 230 × I
F: I = 60 ÷ 230
A: 0.26 A

Q2 (F) The current in a blanket’s heating wire is 0.25 A when the potential difference is 230 V. Calculate the resistance of the wire.

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F: V = I R
I: 230 = 0.25 × R
F: R = 230 ÷ 0.25
A: 920 Ω

Q3 (F) The 60 W blanket is switched on for 30 minutes to warm the bed. Calculate the energy transferred.

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Convert first: 30 minutes = 1800 s
F: E = P t
I: E = 60 × 1800
F: E is already the subject
A: 108 000 J

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Q4 (F/H) Each night the blanket runs at 60 W for 30 minutes, then on a low setting of 20 W for 7.5 hours. Electricity costs 25p per kWh. Calculate (a) the cost per night and (b) the cost for 120 nights over winter.

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Convert first: 60 W = 0.060 kW; 20 W = 0.020 kW; 30 minutes = 0.5 h
F: E = P t
I: E = (0.060 × 0.5) + (0.020 × 7.5) = 0.03 + 0.15
F: E = 0.18 kWh per night
A: (a) 0.18 × 25p = 4.5p per night; (b) 4.5p × 120 = 540p = £5.40

Q5 (F/H) A double electric blanket has a heating wire of resistance 460 Ω. The current is 0.50 A. Calculate the power of the blanket.

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F: P = I² R
I: P = 0.50² × 460
F: P is already the subject
A: 0.25 × 460 = 115 W

Q6 (F/H) On its low setting, a blanket has a power of 30 W and a resistance of 1760 Ω. Calculate the current, to 2 significant figures.

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F: P = I² R
I: 30 = I² × 1760
F: I² = 30 ÷ 1760 = 0.0170, so I = √0.0170
A: 0.13 A (2 s.f.)

Q7 (F/H) Explain why the heating wire in an electric blanket gets hot when there is a current in it. [3 marks]

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• The current is a flow of electrons through the metal wire.
• The electrons collide with the ions in the metal lattice (this is what causes resistance) …
• … which makes the ions vibrate more, so energy is transferred to the thermal energy store of the wire and its temperature rises.

Q8 (H) The control switch sets the blanket to “low” by increasing the resistance of the circuit. The potential difference stays at 230 V. Explain why this reduces the power. [3 marks]

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• V = I R: with V fixed, a larger resistance gives a smaller current.
• P = V I: with V fixed, a smaller current gives a smaller power.
• So energy is transferred to the blanket at a lower rate and it is less warm.

Q9 (H) On the low setting the current is 0.13 A. The blanket is left on for 8 hours. Calculate (a) the charge that flows and (b) the energy transferred. Give both to 2 significant figures.

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Convert first: 8 hours = 8 × 3600 = 28 800 s
(a) F: Q = I t; I: Q = 0.13 × 28 800; A: 3744 C = 3700 C (2 s.f.)
(b) F: E = Q V; I: E = 3744 × 230; A: 861 120 J = 8.6 × 10⁵ J (2 s.f.)

Q10 (H) Ali warms his bed with a 60 W electric blanket for 30 minutes. His sister heats her bedroom with a 2.0 kW fan heater for 1 hour. Electricity costs 25p per kWh. Use calculations to compare the costs, and explain why the electric blanket is a much cheaper way to keep warm. [6 marks]

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Model answer – any 6 of:
• Blanket: E = P t = 0.060 kW × 0.5 h = 0.03 kWh, cost = 0.03 × 25p = 0.75p.
• Heater: E = 2.0 kW × 1 h = 2.0 kWh, cost = 2.0 × 25p = 50p.
• The heater costs about 67 times as much.
• The blanket transfers energy directly to the bed and the person by conduction.
• The heater has to raise the temperature of all the air in the room (a much larger mass).
• The warm room loses energy to the surroundings through the walls, windows and gaps, so much of the energy is dissipated.
• The duvet acts as insulation above the blanket, reducing the rate of energy transfer away from the bed.

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⚠️ Where students lose marks

• Forgetting to square the current in P = I² R (0.50² = 0.25, not 1.0).
• Forgetting to take the square root when finding I from P = I² R.
• Mixing W and kW, or minutes and hours, in cost calculations – use kW and hours for kWh.
• Saying a bigger resistance gives a bigger power – at a fixed mains p.d., more resistance means less current and less power.

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