The Physics Question Bank

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

The equation

power = (current)² × resistance    P = I² R

Quantity	Symbol	Unit
power	P	watts
current	I	amperes (A)
resistance	R	ohms (Ω)

Rearranged: R = P ÷ I²  |  I = √(P ÷ R)  |  Square the current before multiplying.

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) A current of 2 A flows through a 10 Ω heating wire in a hand warmer. Calculate the power.

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F: P = I² R
I: P = 2² × 10
F: P is already the subject
A: 4 × 10 = 40 W

Q2 (F) A current of 3 A flows through an electric blanket with a power of 180 W. Calculate its resistance.

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F: P = I² R
I: 180 = 3² × R, so 180 = 9 × R
F: R = 180 ÷ 9
A: 20 Ω

Q3 (F) A 500 W heater has a resistance of 20 Ω. Calculate the current.

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F: P = I² R
I: 500 = I² × 20
F: I² = 500 ÷ 20 = 25, so I = √25
A: 5 A

Q4 (F) A 50 Ω resistor in a speaker circuit dissipates 2 W. Calculate the current in mA.

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F: P = I² R
I: 2 = I² × 50
F: I² = 2 ÷ 50 = 0.04, so I = 0.2 A
A: 200 mA

Q5 (F) A 2.4 kW oven element carries a current of 10 A. Calculate its resistance.

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Convert first: 2.4 kW = 2400 W
F: P = I² R
I: 2400 = 10² × R, so 2400 = 100 × R
F: R = 2400 ÷ 100
A: 24 Ω

Q6 (F/H) An extension lead carrying 12 A wastes 72 W as heat. Calculate the resistance of the lead.

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F: P = I² R
I: 72 = 12² × R, so 72 = 144 × R
F: R = 72 ÷ 144
A: 0.5 Ω

Q7 (F/H) The current in a 5 Ω resistor doubles from 2 A to 4 A. By what factor does the power increase?

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F: P = I² R
I: before: P = 2² × 5 = 20 W; after: P = 4² × 5 = 80 W
F: factor = 80 ÷ 20
A: 4 times (doubling the current quadruples the power)

Q8 (F/H) A 20 Ω transmission line wastes 5.0 × 10⁶ W as heat. Calculate the current in the line.

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F: P = I² R
I: 5.0 × 10⁶ = I² × 20
F: I² = 2.5 × 10⁵, so I = √(2.5 × 10⁵)
A: 5.0 × 10² A

Q9 (H) A 40 Ω resistor has a p.d. of 20 V across it. Calculate the power it dissipates.

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Step 1 – current
F: V = I R
I: 20 = I × 40
F: I = 20 ÷ 40 = 0.5 A
Step 2 – power
F: P = I² R
I: P = 0.5² × 40
A: 10 W

Q10 (H) A 3.2 kW kettle element carries a current of 12.5 A. Calculate its resistance, to 2 significant figures.

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Convert first: 3.2 kW = 3200 W
F: P = I² R
I: 3200 = 12.5² × R, so 3200 = 156.25 × R
F: R = 3200 ÷ 156.25 = 20.48 Ω
A: 20 Ω (2 s.f.)