AQA A-Level Physics Paper 2, June 2023: Question 27

1 mark ยท Medium difficulty ยท Multiple Choice

Calculate the current in the primary coil of a 90% efficient transformer connected to a 230 V supply given the secondary current and voltage.

Practise this question

Question

Multiple-choice question 27: A transformer for use in a 230 V ac supply is 90% efficient. The transformer provides a current of 3.00 A at 12.0 V. What is the current in the primary coil? Options are A: 0.141 A, B: 0.156 A, C: 0.174 A, D: 5.75 A.
Question text

27 A transformer for use in a 230 V ac supply is 90% efficient.

The transformer provides a current of 3.00 A at 12.0 V.

What is the current in the primary coil?

[1 mark]

A 0.141 A

B 0.156 A

C 0.174 A

D 5.75 A

Mark scheme

Show the mark scheme Mark scheme for question 27 showing key 'C' and value '0.174 A'.

27 C 0.174 A

How to answer it

Transformer Efficiency & Primary Current

๐Ÿ“Œ What this question tests

This question assesses your understanding of non-ideal transformers and the relationship between electrical power, voltage, current, and efficiency. You need to correctly link input power (primary coil) and output power (secondary coil) using the efficiency formula: Efficiency = (P_out / P_in) .

Question 27 • Multiple Choice

Calculation of Primary Coil Current

Question Analysis & Walkthrough

โœ… Correct Answer

Option C: 0.174 A

Award [1 mark] for selecting option C.

๐Ÿ’ก Key Knowledge

  • Electrical Power Formula: P = V ร— I
  • Transformer Efficiency (ฮท):
    ฮท = P_out / P_in = (V_s ร— I_s) / (V_p ร— I_p)
  • Input vs Output: Because efficiency < 100%, the input power P_in must always be greater than the useful output power P_out .

๐Ÿ“ Step-by-Step Calculation

  1. Identify the given quantities:
    • Primary voltage, V_p = 230 V
    • Secondary voltage, V_s = 12.0 V
    • Secondary current, I_s = 3.00 A
    • Efficiency, ฮท = 90% = 0.90
  2. Calculate the useful output power (secondary coil):
    P_s = V_s ร— I_s = 12.0 V ร— 3.00 A = 36.0 W
  3. Calculate the required input power (primary coil):
    Efficiency = P_s / P_p &implies; P_p = P_s / 0.90
    P_p = 36.0 W / 0.90 = 40.0 W
  4. Calculate the primary current (I_p):
    P_p = V_p ร— I_p &implies; I_p = P_p / V_p
    I_p = 40.0 W / 230 V = 0.17391... A
    Rounded to 3 significant figures: 0.174 A (matches Option C).

โŒ Common Errors & Distractor Analysis

  • Option A (0.141 A): Multiplying by efficiency instead of dividing.
    I_p = (36.0 ร— 0.90) / 230 = 0.141 A
    Examiner note: This implies the input power is less than the output power, which violates conservation of energy!
  • Option B (0.157 A / 0.156 A): Assuming 100% efficiency.
    I_p = 36.0 / 230 โ‰ˆ 0.1565 A
    Examiner note: Forgetting to apply the 90% efficiency altogether.
  • Option D (5.75 A): Inverting the voltage ratio.
    I_p = (230 / 12.0) ร— 3.00 / 0.90 โ‰ˆ 63.9 A or similar inverted step-down errors.

๐Ÿง  Exam Technique & Sanity Check

  • Sanity Check: A transformer that steps down voltage ( 230 V → 12 V ) will step up current. Therefore, I_p must be much smaller than I_s = 3.00 A . This immediately rules out D.
  • Efficiency Check: Since the transformer is only 90% efficient, it draws more power and current from the mains than an ideal transformer would.
    Ideal current = 36 / 230 โ‰ˆ 0.157 A .
    Since it's inefficient, I_p must be greater than 0.157 A → eliminates A and B, leaving only C!

Topics

Physics ยท 3.7 Fields and their consequences (A-level only)

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