AQA GCSE Physics Physics Paper 1 (Foundation), November 2021: Question 4

8 marks · Standard Demand difficulty · Short Answer

Analyze a circuit investigation on parallel lamps, completing sentences on resistance and current, identifying errors, completing ammeter table readings, and calculating power output.

Practise this question

Question

The question presents an investigation into lamps connected in parallel, including circuit diagrams (Figures 4 and 6), a scatter graph showing current against the number of lamps (Figure 5), and four sub-questions involving sentence completion, identifying types of experimental error from fluctuating ammeter readings, reading values from a table for parallel circuit ammeters, and calculating electrical power using a provided formula.
Question text

04 A student investigated how the current in a circuit varied with the number of lamps

connected in parallel in the circuit.

Figure 4 shows the circuit with three identical lamps connected in parallel.

Figure 4

Figure 5 shows the results.

Figure 5

04.1 Complete the sentences.

Choose answers from the box.

Each answer can be used once, more than once or not at all.

*14* decreased stayed the same increased

[3 marks]

As the number of lamps increased, the current .

As the number of lamps increased, the total resistance of the

circuit .

As the number of lamps increased, the potential difference across the

battery .

04.2 When there were three lamps in the circuit the ammeter reading kept changing

between 0.35 A and 0.36 A.

What type of error would this lead to?

[1 mark]

Tick ( ) one box.

Random error

Systematic error

Zero error 16

Figure 6 shows a circuit with five ammeters and three identical lamps.

Figure 6

04.3 Complete Table 2 to show the readings on ammeters A2 and A5.

[2 marks]

Table 2

Ammeter A1 A2 A3 A4 A5

Current in amps 0.3617 0.12 0.12

04.4 The resistance of one lamp is 15 Ω.

The current in the lamp is 0.12 A.

Calculate the power output of the lamp.

Use the equation:

power = (current)2 × resistance

[2 marks]

Power = W

Mark scheme

Show the mark scheme The mark scheme for Question 4 lists correct answers for four parts: 04.1 requires 'increased', 'decreased', and 'stayed the same' (3 marks); 04.2 requires 'random error' (1 mark); 04.3 requires A2 = 0.12 A and A5 = 0.36 A (2 marks); and 04.4 requires substitution into P = 0.12^2 * 15 resulting in 0.216 W (2 marks), totaling 8 marks.

Question 4

AO /

Question Answers Extra information Mark

Spec. Ref.

04.1 increased 1 AO3

4.2.1.3

decreased 1 4.2.2

stayed the same 1

04.2 random error 1 AO3

4.2.1.3

4.2.2

04.3 A2 = 0.12 (A) 1 AO1

4.2.2

A5 = 0.36 (A) 1

04.4 P = 0.122 × 15 1 AO2

4.2.4.1

P = 0.216 (W) 1

Total 8

How to answer it

Current and Resistance in Parallel Circuits

What this question tests

This question assesses your understanding of parallel circuits, how adding more branches (lamps) affects total resistance, current, and potential difference, identifying types of measurement errors, applying Kirchhoff's First Law to currents at junctions, and calculating electrical power using the formula power = current² × resistance .

Question 04.1 [3 marks]

Parallel Circuit Relationships

✅ Correct Answers

  • As the number of lamps increased, the current increased.
  • As the number of lamps increased, the total resistance of the circuit decreased.
  • As the number of lamps increased, the potential difference across the battery stayed the same.

💡 Key Knowledge

  • Parallel circuits: Adding more loops/branches provides more paths for charge to flow, which decreases total resistance.
  • Ohm's Law / Current: Since resistance decreases while potential difference remains constant, total current drawn from the power supply increases.
Mark breakdown: 1 mark for each correct sentence selection. [Total: 3 marks]
Question 04.2 [1 mark]

Identifying Experimental Errors

✅ Correct Answer

  • Random error (Tick the top box)

🧠 Exam Technique & Examiner Insight

Fluctuating readings (values varying unpredictably between high and low limits, e.g., changing between 0.35 A and 0.36 A) are a classic textbook definition of a random error. Systematic errors cause readings to be skewed away from the true value by a consistent amount every time (like a zero error).

Mark breakdown: 1 mark for ticking 'Random error'. [Total: 1 mark]
Question 04.3 [2 marks]

Applying Current Rules in Parallel Circuits

✅ Correct Answers

  • A₂ = 0.12 A
  • A₅ = 0.36 A

💡 Key Knowledge

  • Identical lamps: Each branch contains an identical lamp, meaning the current shared equally among identical parallel branches is identical. Looking at A₃ and A₄ , we can see each branch carries 0.12 A . Therefore, A₂ must also read 0.12 A .
  • Total current: A₅ measures the total current leaving/entering the battery, which splits into the three branches ( 0.12 A + 0.12 A + 0.12 A = 0.36 A ). This matches A₁ .
Mark breakdown: 1 mark for A₂ = 0.12 (A), 1 mark for A₅ = 0.36 (A). [Total: 2 marks]
Question 04.4 [2 marks]

Calculating Power Output

📐 Step-by-Step Calculation

Equation: power = (current)² × resistance

  1. Substitute values into the formula:
    P = 0.12² × 15
  2. Square the current first (BIDMAS/BODMAS):
    0.12² = 0.0144
  3. Multiply by resistance:
    P = 0.0144 × 15 = 0.216 W

❌ Common Calculation Traps

  • Forgetting to square the current: Many students mistakenly calculate 0.12 × 15 = 1.8 instead of squaring 0.12 first.
  • Unit checking: Make sure your final answer includes the correct unit (W) if not already pre-printed on the answer line.
Mark breakdown: 1 mark for correct substitution and squaring (0.12² × 15), 1 mark for correct final answer evaluating to 0.216 (W). [Total: 2 marks]

Topics

Physics · Required Practicals · P2: Electricity · Required Practicals

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Foundation), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.