AQA GCSE Physics Physics Paper 1 (Foundation), June 2022: Question 11

8 marks · Standard Demand difficulty · Short Answer

Calculate current and useful power output using equations relating power, potential difference, current, and efficiency in the context of piezoelectric pavement tiles.

Practise this question

Question

Figure 15 shows a person walking on engineered pavement tiles that generate electricity. Questions 11.1 to 11.4 test understanding of electrical power equations and efficiency formulas, requiring students to select correct equations and perform multi-step calculations to find current and power output.
Question text

11 An engineering company has invented pavement tiles that generate electricity as

people walk on them.

Figure 15 shows someone walking on the pavement tiles.

Figure 15

Use the Physics Equations Sheet to answer questions 11.1 and 11.2.

11.1 What equation links current (I), potential difference (V) and power (P)?

[1 mark]

Tick ( ) one box.

V

P =

I

P = V × I

I = P × V

V = I × P 37

11.2 When a person walks on a tile, a potential difference of 40 V is induced across the tile.

The power output of the tile is 4.4 W.

Calculate the current in the tile.

[3 marks]

Current = A

Use the Physics Equations Sheet to answer questions 11.3 and 11.4.

11.3 What equation links efficiency, total power input and useful power output?

[1 mark]

Tick ( ) one box.

useful power output

Efficiency =

total power input

total power input

Efficiency =

useful power output

Efficiency = useful power output × total power input

11.4 The tiles are used to power LED lights in the pavement.

An LED light has a total power input of 4.0 W.

The efficiency of the LED light is 0.85

Calculate the useful power output of the LED light.

[3 marks]

Useful power output = W

Mark scheme

Show the mark scheme The mark scheme provides correct answers for identifying the power equation (P = V x I), calculating the current as 0.11 A, identifying the efficiency equation, and calculating the useful power output as 3.4 W across four sub-questions totaling 8 marks.

Question 11

AO /

Question Answers Extra information Mark

Spec. Ref.

11.1 P = V × I 1 AO1

4.2.4.1

AO /

Spec. Ref.

11.2 4.4 = 40 × I 1 AO2

4.2.4.1

4.4 1

I =

I = 0.11 (A) 1

AO /

Spec. Ref.

11.3 useful power output 1 AO1

efficiency = 4.1.2.2

total power input

AO /

Spec. Ref.

11.4 P AO2

0.85 = 1 4.1.2.2

4.0

P = 0.85 × 4.0 1

P = 3.4 (W) 1

Total Question 11 8

How to answer it

Study Guide: Pavement Tiles and Electrical Power

What this question tests

This question assesses your recall and application of key circuit equations linking power, potential difference, and current, alongside calculations involving device efficiency and energy transfers.

Question Part 11.1

Identifying the Power Equation

What equation links current (I), potential difference (V) and power (P)?

✅ Correct Answer

Tick the box for: P = V × I

Awarded: 1 Mark (AO1 - Specification 4.2.4.1)

💡 Key Knowledge

  • Power ( P ) is measured in watts (W).
  • Potential difference ( V ) is measured in volts (V).
  • Current ( I ) is measured in amperes or amps (A).
Question Part 11.2

Calculating Current from Power and Potential Difference

Calculate the current in the tile given a potential difference of 40 V and power of 4.4 W.

📐 Step-by-Step Calculation

  1. State the equation: P = V × I
  2. Substitute the numbers: 4.4 = 40 × I
  3. Rearrange and solve: I = 4.4 / 40
  4. Final Answer: I = 0.11 A
Awarded: 3 Marks total (1 for substitution, 1 for rearrangement, 1 for correct answer with unit) - AO2

❌ Common Errors & Exam Technique

  • Wrong way round: Dividing 40 by 4.4 instead of 4.4 by 40 will result in 9.09, which loses marks. Always check what the formula triangle looks like: I = P / V .
  • Missing units: Remember to write A or amperes on the answer line if it is not pre-printed.
Question Part 11.3

Identifying the Efficiency Equation

What equation links efficiency, total power input and useful power output?

✅ Correct Answer

Tick the box for:
Efficiency = useful power output / total power input

Awarded: 1 Mark (AO1 - Specification 4.1.2.2)

🧠 Exam Technique

Efficiency is a ratio of useful output to total input. Because efficiency cannot be greater than 1 (or 100%), the smaller number (useful) must always be on top of the fraction numerator!

Question Part 11.4

Calculating Useful Power Output

Calculate the useful power output of an LED light with a total power input of 4.0 W and an efficiency of 0.85.

📐 Step-by-Step Calculation

  1. State the equation: Efficiency = useful power output / total power input
  2. Substitute the numbers: 0.85 = P / 4.0
  3. Rearrange: P = 0.85 × 4.0
  4. Final Answer: P = 3.4 W
Awarded: 3 Marks total (1 for correct substitution, 1 for rearrangement, 1 for correct answer) - AO2

❌ Common Errors & Examiner Insights

  • Inverted division: Students who confused the equation layout sometimes tried to divide 4.0 by 0.85 instead of multiplying.
  • Significant figures: Keep your answer to 2 significant figures here to match the data provided in the question stem.

Topics

Physics · P1: Energy · P2: Electricity

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Foundation), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.