10 GCSE Physics questions about the power bank, from Foundation to Higher, each with a full worked solution. A power bank is a portable battery. You charge it from the mains, it stores the energy in its chemical energy store, and later it transfers that energy to your phone. Some energy is always dissipated as heat, so you never get back all the energy you put in.
Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Topics: Electricity · Energy
📌 KEY DEFINITIONS – learn them word for word
Charge flow: charge flow = current × time.
Potential difference: the energy transferred per unit charge passed between two points.
Efficiency: useful output energy transfer ÷ total input energy transfer.
Dissipated energy: energy transferred to the surroundings in a way that is not useful, usually heating them.
Equations you’ll use
Q = I t | E = Q V | P = V I | P = I² R | E = P t | efficiency = useful output ÷ total input
| Quantity | Symbol | Unit |
|---|---|---|
| charge | Q | coulombs (C) |
| current | I | amps (A) |
| time | t | seconds (s) |
| energy | E | joules (J) |
| potential difference | V | volts (V) |
| power | P | watts (W) |
| resistance | R | ohms (Ω) |
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.
Questions
Q1 (F) When fully charged, a power bank can supply a current of 10 A for 1 hour. Calculate the charge it stores in coulombs.
Show solution
Convert first: 1 hour = 3600 s
F: Q = I t
I: Q = 10 × 3600
F: Q is already the subject
A: 36 000 C
Q2 (F) The battery inside has a potential difference of 3.7 V. Calculate the energy stored.
Show solution
F: E = Q V
I: E = 36 000 × 3.7
F: E is already the subject
A: 133 200 J (about 133 kJ)
Q3 (F) The USB output supplies 5.0 V and 2.4 A. Calculate the output power.
Show solution
F: P = V I
I: P = 5.0 × 2.4
F: P is already the subject
A: 12 W
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Q4 (F) The power bank charges a phone using its 12 W USB output (from Q3) for 30 minutes. Calculate the energy transferred.
Show solution
Convert first: 30 minutes = 30 × 60 = 1800 s
F: E = P t
I: E = 12 × 1800
F: E is already the subject
A: 21 600 J
Q5 (F/H) The power bank stores 133 200 J but only delivers 100 000 J to phones. Calculate its efficiency as a percentage.
Show solution
F: efficiency = useful output ÷ total input
I: efficiency = 100 000 ÷ 133 200
F: already the subject
A: 0.751 × 100 = 75%
Q6 (F/H) A phone battery needs 55 000 J to charge fully. How many full charges can the power bank give, using the 100 000 J it delivers?
Show solution
100 000 ÷ 55 000 = 1.8
A: 1 full charge (and most of a second one – about 1.8 charges)
Q7 (F/H) The power bank is recharged at 10 W. Charging is 90% efficient. Calculate how long it takes to store 133 200 J. Give your answer in hours.
Show solution
Step 1 – energy supplied: 133 200 ÷ 0.90 = 148 000 J
Step 2 – time: t = E ÷ P = 148 000 ÷ 10 = 14 800 s
A: 14 800 ÷ 3600 = 4.1 hours
Q8 (H) The power bank charges a phone at 10 W through a 5.0 V USB cable for 2 hours. Calculate the charge that flows through the cable.
Show solution
Step 1 – current: I = P ÷ V = 10 ÷ 5.0 = 2.0 A
Step 2 – charge: Q = I t = 2.0 × 7200
A: 14 400 C
Q9 (H) The cable has a resistance of 0.20 Ω and carries 2.0 A. Calculate the power wasted heating the cable, and the energy wasted in 2 hours.
Show solution
F: P = I² R
I: P = 2.0² × 0.20
A: 0.80 W
Energy: E = P t = 0.80 × 7200 = 5760 J
Q10 (H) “Fast charge” power banks deliver the same power using a higher potential difference and a lower current. Explain why this wastes less energy in the cable. [4 marks]
Show answer
Marking points:
• power = p.d. × current, so a higher p.d. gives the same power with a lower current
• the power wasted heating the cable is P = I² R
• the wasted power depends on the square of the current, so halving the current makes the waste four times smaller
• less energy is dissipated, so charging is more efficient (the same idea is used in the National Grid)
⚠️ Where students lose marks
- Forgetting to change hours into seconds before using Q = I t or E = P t (1 hour = 3600 s).
- Dividing the wrong way round in efficiency – the answer can never be more than 1 (100%).
- Forgetting to square the current in P = I² R.
- Rounding too early in multi-step questions. Keep the full number until the end.
More devices: all 40 everyday devices · Electricity · Energy
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