The Physics Question Bank

10 GCSE Physics questions about the wireless phone charger, from Foundation to Higher, each with a full worked answer. A wireless charger has a flat coil of wire inside its pad. An alternating current in this coil makes a magnetic field that keeps changing, and this changing field induces a potential difference in a second coil inside the back of your phone – so energy crosses the gap with no wires touching.

Course: Combined Science + Separate Physics  |  Tier: Foundation + Higher  |  Topics: Energy, Electricity, Magnetism and electromagnetism

Equations you’ll use

efficiency = useful power output ÷ total power input  |  E = P t  |  P = V I  |  Q = I t  |  E = Q V  |  V_p ÷ V_s = n_p ÷ n_s (S only · HT)  |  V_s × I_s = V_p × I_p (S only · HT)

Quantity	Symbol	Unit
energy transferred	E	joules (J)
power	P	watts (W)
time	t	seconds (s)
potential difference	V	volts (V)
current	I	amperes (A)
charge flow	Q	coulombs (C)
number of turns on a coil	n	no unit
efficiency	–	no unit (or %)

Use FIFA for every answer

• 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.

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Questions

Q1 (F) A wireless charging pad takes in 20 W of electrical power. It transfers 15 W usefully to the phone. Calculate its efficiency.

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F: efficiency = useful power output ÷ total power input
I: efficiency = 15 ÷ 20
F: efficiency is already the subject
A: 0.75 (75%)

Q2 (F) The charging pad is switched on for 2 hours with an input power of 20 W. Calculate the energy it takes in, in kJ.

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Convert first: 2 hours = 2 × 3600 = 7200 s
F: E = P t
I: E = 20 × 7200
F: E is already the subject
A: 144 000 J = 144 kJ

Q3 (F) The pad takes in 20 W and transfers 15 W usefully. Calculate the energy wasted in 30 minutes.

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Step 1 – wasted power: 20 − 15 = 5 W
Convert: 30 minutes = 30 × 60 = 1800 s
F: E = P t
I: E = 5 × 1800
F: E is already the subject
A: 9000 J (9 kJ)

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Q4 (F) After charging, the pad and the back of the phone feel warm. Explain what has happened to the wasted energy. [2 marks]

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• The current in the coils (and the changing magnetic field) heats the coils, the phone and the pad, so energy is transferred to their thermal stores. (1)
• This energy is then dissipated to the thermal store of the surroundings, where it spreads out and is no longer useful. (1)

Q5 (F/H) A fully charged phone battery can supply a current of 4.5 A for 1.0 hour. (a) Calculate the charge it stores in coulombs. (b) Calculate how long it takes to charge fully with a charging current of 1.5 A, in hours.

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Convert first: 1.0 hour = 1.0 × 3600 = 3600 s
(a) F: Q = I t
I: Q = 4.5 × 3600
A: 16 200 C
(b) F: Q = I t
I: 16 200 = 1.5 × t
F: t = 16 200 ÷ 1.5 = 10 800 s
A: 10 800 ÷ 3600 = 3.0 hours

Q6 (F/H) The battery stores a charge of 16 200 C at a potential difference of 3.85 V. Calculate the energy stored, in kJ, to 3 significant figures.

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F: E = Q V
I: E = 16 200 × 3.85
F: E is already the subject
A: 62 370 J = 62.4 kJ (3 s.f.)

Q7 (F/H) The battery needs 62 370 J to charge fully. The wireless charger is 75% efficient. Calculate the total energy the charger must take in, in kJ.

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Convert first: 75% = 0.75
F: efficiency = useful output energy ÷ total input energy
I: 0.75 = 62 370 ÷ total input
F: total input = 62 370 ÷ 0.75 = 83 160 J
A: 83 kJ (2 s.f.)

Q8 (H · S only · HT) Treat the pad and phone as a transformer. The coil in the pad has 12 turns and an alternating p.d. of 10 V across it. The coil in the phone has 9 turns. Calculate the p.d. induced across the phone’s coil. (Assume all the magnetic field passes through both coils.)

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F: V_p ÷ V_s = n_p ÷ n_s
I: 10 ÷ V_s = 12 ÷ 9
F: V_s = 10 × 9 ÷ 12
A: 7.5 V

Q9 (H · S only · HT) The current in the pad’s coil is 1.5 A at 10 V. The p.d. across the phone’s coil is 7.5 V. Assuming the transfer is 100% efficient, calculate the current in the phone’s coil.

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F: V_s × I_s = V_p × I_p
I: 7.5 × I_s = 10 × 1.5
F: I_s = 15 ÷ 7.5
A: 2.0 A

Q10 (H · S only · HT) Explain how a wireless charger causes a current in the coil inside the phone, and why it only works with an alternating current. Suggest why charging is slower and wastes more energy if the phone is not placed in the centre of the pad. [6 marks]

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Model answer – marking points:
• There is an alternating current in the coil in the pad.
• This produces a magnetic field around the coil that is constantly changing (in size and direction).
• The changing magnetic field passes through the coil in the phone.
• This induces a potential difference across the phone’s coil (electromagnetic induction); because the coil is part of a complete circuit, a current flows and charges the battery.
• A direct current would make a steady field – the field must change for a p.d. to be induced, so a steady current induces nothing (except briefly when switched on or off).
• If the phone is off-centre, less of the changing magnetic field passes through the phone’s coil, so a smaller p.d. is induced and charging is slower; more energy is wasted heating the pad, so efficiency is lower.

⚠️ Where students lose marks

• Writing that the wireless charger sends electricity through the air – it is a changing magnetic field that induces a p.d. in the phone’s coil.
• Forgetting that induction needs a changing field. A steady (d.c.) current in the pad would not charge the phone.
• Using milliamps or hours in Q = I t without converting to amps and seconds first.
• Giving an efficiency bigger than 1 (dividing input by output), or saying wasted energy is “lost” or “used up” instead of dissipated to the surroundings.

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