AQA A-Level Physics Paper 2, June 2025: Question 23

1 mark ยท Medium difficulty ยท Multiple Choice

Calculate the efficiency of a transformer supplying a 100 W lamp from an ac supply with given peak potential difference and peak current.

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Question

Multiple choice question 23: A transformer connects an ac power supply to a 100 W filament lamp. When the lamp operates at its rated power, the power supply has a peak potential difference of 50 V and supplies a peak current of 5.0 A. Four percentage efficiency options are given: A 80%, B 57%, C 40%, and D 28%.

Mark scheme

Show the mark scheme Mark scheme for question 23 showing key 'A' with the answer 80% and assessment objective AO2.

How to answer it

Transformer Efficiency with AC Peak Values

๐Ÿ“Œ What this question tests

This question evaluates your understanding of alternating current (AC) power and the application of the efficiency formula in real transformers:

  • Distinguishing between peak values (Vโ‚€, Iโ‚€) and root-mean-square values (Vrms, Irms).
  • Calculating average AC power delivered from a sinusoidal supply: P = Vrms ร— Irms = ยฝ Vโ‚€Iโ‚€ .
  • Understanding device power ratings: a rated wattage (100 W) represents mean (useful) power output.
  • Determining efficiency as a percentage: ฮท = (useful power output / total power input) ร— 100% .

Question 23 Walkthrough

Multiple Choice [1 Mark] โ€” AO2

โœ… Correct Answer

Option A (80%)

Awarded 1 mark for correctly determining the average input power as 125 W and obtaining 80% efficiency.

๐Ÿ’ก Key Knowledge

  • Power rating: A 100 W bulb produces an average useful output power Pout = 100 W .
  • AC Root-Mean-Square (rms):
    Vrms = Vโ‚€ / โˆš2
    Irms = Iโ‚€ / โˆš2
  • Average AC Power:
    Pin = Vrms ร— Irms = (Vโ‚€ / โˆš2) ร— (Iโ‚€ / โˆš2) = ยฝ ร— Vโ‚€ ร— Iโ‚€

๐Ÿ“ Step-by-Step Calculation

  1. Identify useful power output:
    The lamp operates at its rated power:
    Pout = 100 W
  2. Determine input peak values:
    Peak voltage: Vโ‚€ = 50 V
    Peak current: Iโ‚€ = 5.0 A
  3. Calculate the average input power supplied to the transformer:
    Pin = Vrms ร— Irms = (Vโ‚€ ร— Iโ‚€) / 2
    Pin = (50 V ร— 5.0 A) / 2 = 250 / 2 = 125 W
  4. Calculate transformer efficiency:
    Efficiency = (Pout / Pin) ร— 100%
    Efficiency = (100 W / 125 W) ร— 100% = 0.80 ร— 100% = 80%

โŒ Common Traps & Distractor Analysis

  • Selecting C (40%): Multiplying peak values directly without converting to rms:
    P = 50 ร— 5.0 = 250 W โ†’ 100 / 250 = 40% . Remember: power is not peak voltage ร— peak current!
  • Selecting B (57%): Dividing by โˆš2 only once:
    P = 250 / โˆš2 โ‰ˆ 176.8 W โ†’ 100 / 176.8 โ‰ˆ 56.6% โ‰ˆ 57% . Both V and I are peak values, so โˆš2 ร— โˆš2 = 2 in the denominator.
  • Selecting D (28%): Misapplying the factor of โˆš2 to the 40% result ( 40% / โˆš2 โ‰ˆ 28% ).

๐Ÿง  Exam Technique & Examiner Tips

  • Spot the word "peak": Whenever an AC question mentions "peak", immediately write down / โˆš2 next to each value.
  • Rated power is always mean power: Domestic and lab appliances rated in watts (e.g., a 100 W bulb, 2 kW kettle) always refer to average/rms electrical power.
  • Sanity check: Modern practical transformers typically operate with relatively high efficiencies (frequently 70%โ€“95%). Seeing an efficiency of 80% is physically plausible.

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 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.