AQA GCSE Physics Physics Paper 2 (Higher), June 2023: Question 8

7 marks · Standard Demand difficulty · Short Answer

Identify the core material of a transformer and calculate the secondary coil current using the transformer equations and power formula.

Practise this question

Question

Figure 21 shows a transformer with a core, a primary coil with 2000 turns connected to a 230 V a.c. mains electricity supply, and a secondary coil with 40 turns connected to a lamp. Sub-question 08.1 asks for the core material and its reason (2 marks). Sub-question 08.2 asks to determine the current in the secondary coil for a power output of 6.9 W given the transformer is 100% efficient (5 marks).
Question text

08 Figure 21 shows a transformer used to power a lamp using the

mains electricity supply.

Figure 21

08.1 What material is used to make the core of the transformer?

Give the reason for using this material.

[2 marks]

Material

Reason

08.2 Determine the current in the secondary coil when the power output of the transformer

is 6.9 W.

The transformer is 100% efficient.

Use the Physics Equations Sheet.

[5 marks]

Current in the secondary coil = A

Mark scheme

Show the mark scheme The mark scheme for question 08 gives answers for 08.1 as iron and that it is easily magnetised and demagnetised (2 marks). For 08.2, it outlines calculation steps using the transformer turns equation and power equation to arrive at a secondary current of 1.5 A (5 marks).

Question 8

AO /

Question Answers Extra information Mark

Spec. Ref.

08.1 iron allow nickel / cobalt 1 AO1

do not allow steel 4.7.3.4

it is easily magnetised (and allow it is a magnetic material 1

demagnetised)

MP 2 is dependent on MP–HYSICS1 – –

AO /

Spec. Ref.

08.2 230 2000 1 AO2

=

Vs 40 4.7.3.4

40 subsequent marks can only be 1

Vs = × 230

2000 awarded if the first equation is

correct and has been used

Vs = 4.6 (V)

4.6 × Is = 6.9 this mark may be awarded if the 1

pd is incorrectly calculated

Is = 1.5 A allow a correctly calculated Is 1

using an incorrectly calculated

pd

OR

6.9 = Ip × 230 (1)

6.9 subsequent marks can only be

Ip = (1)

230 awarded if the first equation is

correct and has been used

Ip = 0.03 (A) (1)

2000 this mark may be awarded if Ip is

Is = 0.03 × (1)

40 incorrectly calculated

allow a correctly calculated Is

Is = 1.5 (A) (1) 23

using an incorrectly calculated Ip

Total Question 8 7

How to answer it

AQA GCSE Physics: Transformers Study Guide

Topic: Electromagnetism & Transformers

What this question tests

This question tests your understanding of transformer structure, core material properties, and multi-step calculations linking potential difference equations (using turns ratios) with power equations (assuming 100% efficiency).

Part 08.1: Transformer Core Material

Question: What material is used to make the core of the transformer? Give the reason for using this material. [2 marks]

✅ Correct Answers

  • Material: Iron (allow nickel / cobalt; do not allow steel)
  • Reason: It is easily magnetised and demagnetised (or is a magnetic material).

💡 Key Knowledge

  • The alternating current (a.c.) in the primary coil constantly changes direction, producing a changing magnetic field.
  • The soft iron core channels this magnetic field through the secondary coil.
  • Crucial rule: The second mark depends entirely on getting the first mark correct!

❌ Common Errors

  • Naming steel instead of iron. Steel retains its magnetism (hard magnetic material) and is used for permanent magnets, not transformers.
  • Vague reasons like "it conducts electricity" (the core must be insulated to prevent eddy currents, though that's higher tier). Keep it focused on magnetism.

🧠 Exam Technique

  • Keep answers concise. Examiners look for the exact phrasing: "easily magnetised and demagnetised".
Mark breakdown: 1 mark for naming iron + 1 mark for the magnetic property reason.

Part 08.2: Calculating Secondary Current

Question: Determine the current in the secondary coil when the power output is 6.9 W. The transformer is 100% efficient. [5 marks]

📐 Step-by-Step Calculation (Method 1: Find Voltage First)

  1. Recall the transformer equation:
    Vp / Vs = Np / Ns
  2. Substitute known values:
    230 / Vs = 2000 / 0
  3. Rearrange and solve for secondary voltage (Vs):
    Vs = (40 / 2000) × 230 = 4.6 V
  4. Recall the electrical power equation:
    P = V × I (or Power = Vs × Is )
  5. Substitute and solve for secondary current (Is):
    6.9 = 4.6 × Is
    Is = 6.9 / 4.6 = 1.5 A

🧠 Exam Technique & Alternative Method

Alternative Method (Power first):

  • Step 1: Primary power equals secondary power because efficiency is 100% ( Pp = 6.9 W ).
  • Step 2: Find primary current: Ip = P / V = 6.9 / 230 = 0.03 A .
  • Step 3: Use current ratio: Is = Ip × (Np / Ns) = 0.03 × (2000 / 40) = 1.5 A .

Error carried forward (ECF): Subsequent marks can still be awarded if your initial calculation had a math error, provided you used the correct physics formulas.

❌ Common Calculation Traps

  • Inverting the turns ratio ( Ns / Np instead of Np / Ns ). Remember: step-down transformers have fewer turns on the secondary coil, so the voltage must drop! Check that 4.6 V makes sense compared to 230 V.
  • Forgetting to state units ( A ) in the final answer line, though the unit is usually pre-printed on AQA papers.

✅ Final Answer

Current in the secondary coil = 1.5 A

Mark breakdown: 5 marks total distributed across correct formula substitution, voltage/current determination, and final correct answer.

Topics

Physics · P7: Magnetism and Electromagnetism

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.