The Physics Question Bank

Course: Combined Science + Separate Physics  |  Tier: Foundation + Higher  |  Equation sheet: given in the exam

The equation

energy transferred = charge flow × potential difference    E = Q V

Quantity	Symbol	Unit
energy transferred	E	joules (J)
charge flow	Q	coulombs (C)
potential difference	V	volts (V)

Rearranged: Q = E ÷ V  |  V = E ÷ Q  |  1 volt = 1 joule per coulomb

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.

Foundation: aim for Q1–8.   Higher: aim for Q4–10. Most questions need rearranging.

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Questions

Q1 (F) 20 C of charge flows through a 12 V car bulb. Calculate the energy transferred.

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F: E = Q V
I: E = 20 × 12
F: E is already the subject
A: 240 J

Q2 (F) 60 C of charge transfers 540 J in a radio. Calculate the p.d.

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F: E = Q V
I: 540 = 60 × V
F: V = 540 ÷ 60
A: 9 V

Q3 (F) A 230 V lamp transfers 4600 J. Calculate the charge that flows.

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F: E = Q V
I: 4600 = Q × 230
F: Q = 4600 ÷ 230
A: 20 C

Q4 (F) A 12 V car battery transfers 36 kJ to the starter motor. Calculate the charge that flows.

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Convert first: 36 kJ = 36 000 J
F: E = Q V
I: 36 000 = Q × 12
F: Q = 36 000 ÷ 12
A: 3000 C

Q5 (F) A 1.5 V AA cell transfers 2.7 kJ over its lifetime. Calculate the total charge that flows.

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Convert first: 2.7 kJ = 2700 J
F: E = Q V
I: 2700 = Q × 1.5
F: Q = 2700 ÷ 1.5
A: 1800 C

Q6 (F/H) A 3.7 V phone battery stores 37 kJ when fully charged. How much charge can it supply?

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Convert first: 37 kJ = 37 000 J
F: E = Q V
I: 37 000 = Q × 3.7
F: Q = 37 000 ÷ 3.7
A: 10 000 C

Q7 (F/H) 5000 C of charge flows through a kettle and transfers 1.15 MJ. Calculate the p.d.

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Convert first: 1.15 MJ = 1 150 000 J
F: E = Q V
I: 1 150 000 = 5000 × V
F: V = 1 150 000 ÷ 5000
A: 230 V

Q8 (F/H) A lightning strike transfers 25 C and 2.5 × 10⁹ J. Calculate the p.d. between cloud and ground.

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F: E = Q V
I: 2.5 × 10⁹ = 25 × V
F: V = 2.5 × 10⁹ ÷ 25
A: 1.0 × 10⁸ V

Q9 (H) A 12 V battery supplies a current of 2.5 A to a motor for 4 minutes. Calculate the energy transferred.

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Step 1 – charge (4 minutes = 240 s)
Q = I t = 2.5 × 240 = 600 C
Step 2 – energy
F: E = Q V
I: E = 600 × 12
A: 7200 J

Q10 (H) A 9.0 V battery in a smoke alarm tester supplies 0.045 A for 3 hours. Calculate the energy transferred, to 2 significant figures.

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Step 1 – charge (3 hours = 10 800 s)
Q = I t = 0.045 × 10 800 = 486 C
Step 2 – energy
F: E = Q V
I: E = 486 × 9.0 = 4374 J
A: 4400 J (2 s.f.)