OCR GCSE Physics A: Transformers, microphones and loudspeakers (J249) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover statements P4.2g to P4.2j (transformers, microphones, loudspeakers and headphones) from the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026). These statements are in bold in the specification, so every question here is Higher Tier only. Read the study guide and revision notes first if you need to. Course pages: OCR GCSE Physics hub and the checklist.

Questions

1. (Higher Tier only) A student examines a simple transformer with two coils, labelled primary and secondary.

(a) State which coil is connected to the input supply. [1] (b) State the type of current the input supply must provide. [1] (c) State what links the two coils, as they are not connected by a wire. [1]

2. (Higher Tier only) A student connects the primary coil of a transformer to a 12 V battery. A voltmeter on the secondary coil shows no reading once the battery has been connected for a few seconds. Explain why. [3]

3. (Higher Tier only) A transformer has 250 turns on the primary coil and 1500 turns on the secondary coil. The primary potential difference is 18 V. Calculate the potential difference across the secondary coil and state whether the transformer is step-up or step-down. [3]

4. (Higher Tier only) A transformer changes a 90 V a.c. supply to 15 V. The primary coil has 1200 turns. Calculate the number of turns on the secondary coil. [3]

5. (Higher Tier only) A student builds a transformer. She keeps the primary coil at 40 turns and an input of 4.0 V a.c., and changes the number of turns on the secondary coil. Her results are in the table.

Secondary turns	Secondary p.d. (V)
20	1.9
40	3.8
80	7.6
120	11.3
160	14.8

(a) Calculate the secondary p.d. that you would expect for 120 turns if no energy were wasted. [2] (b) Describe the pattern in her results, using figures from the table. [2] (c) Calculate the percentage by which the measured p.d. is below the expected value for 160 turns. [2] (d) Suggest one reason why the measured values are lower than expected. [1]

6. (Higher Tier only) An ideal transformer has a primary p.d. of 240 V and a primary current of 0.50 A. The secondary p.d. is 15 V.

(a) Calculate the power supplied to the primary coil. [2] (b) Calculate the current in the secondary coil. [2]

7. (Higher Tier only) Explain how an alternating current in the primary coil of a transformer produces an alternating current in the secondary coil. [4]

8. (Higher Tier only) A transformer has 200 turns on the primary coil and 600 turns on the secondary coil. The primary p.d. is 10 V.

(a) Calculate the secondary p.d. [1] (b) The number of primary turns is doubled to 400 and nothing else changes. Calculate the new secondary p.d. [1] (c) Explain the change in terms of the numbers of turns. [1]

9. (Higher Tier only) Explain how a dynamic microphone converts the pressure variations in a sound wave into variations in current. [4]

10. (Higher Tier only) Explain how a loudspeaker produces a louder sound when the size of the current in its coil is increased. [4]

11. (Higher Tier only) A student plugs a pair of headphones into a sensitive meter instead of an audio player. When she claps near the headphones, the meter needle flickers.

(a) Explain why the meter shows a reading. [2] (b) State the name of the device that the headphones are behaving as. [1]

12. (Higher Tier only) A technician uses a transformer to run a lamp rated at 5.0 V and 3.0 A from a 400 V a.c. supply. The primary coil has 2000 turns. Assume the transformer is ideal.

(a) Calculate the number of turns needed on the secondary coil. [2] (b) Calculate the power of the lamp when it works as rated. [1] (c) Calculate the current in the primary coil. [2] (d) The technician replaces the secondary coil with one that has 50 turns. Calculate the new secondary p.d. and state what is likely to happen to the lamp. [3]

Answers

1. (a) The primary coil [1]. [1] (b) Alternating current [1]. [1] (c) A changing magnetic field (in the core) [1]. [1] Examiner insight: “Current” or “electricity” for part (c) names the wrong link. The two coils are joined by the magnetic field, and answers that say so gain credit.

2. The battery gives a direct current, so the current in the primary is steady [1]. A steady current produces a steady magnetic field, not a changing one [1]. A potential difference is induced only by a changing field, so none is induced in the secondary coil [1]. [3] Examiner insight: Say “steady” or “not changing” about the field. Writing only that “d.c. cannot flow” does not explain why there is no induced potential difference.

3. Vs = Vp × Ns / Np [1]. Vs = 18 × 1500 / 250 = 108 V [1]. The secondary has more turns and a larger p.d., so it is step-up [1]. [3] Examiner insight: Write the equation in symbols first and show the substitution. Showing the substitution lets a reader follow your method and lets you spot an inverted fraction.

4. Ns = Np × Vs / Vp [1]. Ns = 1200 × 15 / 90 [1]. Ns = 200 turns [1]. [3] Examiner insight: Rearranging before substituting reduces the chance of inverting the fraction. Check that the smaller p.d. goes with the smaller number of turns.

5. (a) Expected Vs = 4.0 × 120 / 40 [1] = 12.0 V [1]. [2] (b) The secondary p.d. increases as the number of turns increases [1]. It is roughly proportional: doubling the turns from 20 to 40 doubles the p.d. from 1.9 V to 3.8 V (or 40 to 80 turns, 3.8 V to 7.6 V) [1]. [2] (c) Expected value for 160 turns is 4.0 × 160 / 40 = 16.0 V. Difference = 16.0 - 14.8 = 1.2 V [1]. Percentage = 1.2 / 16.0 × 100 = 7.5% [1]. [2] (d) Any one: some energy is wasted as thermal energy in the wires or core; not all the magnetic field from the primary reaches the secondary [1]. [1] Examiner insight: For “describe the pattern”, quote at least two paired values from the table. A percentage difference is taken from the expected value, not the measured one.

6. (a) Power = Vp × Ip [1] = 240 × 0.50 = 120 W [1]. [2] (b) Vp × Ip = Vs × Is, so Is = 120 / 15 [1] = 8.0 A [1]. [2] Examiner insight: Always give the unit with each answer: here 120 W and 8.0 A. The current is larger on the lower p.d. side, which is a useful check.

7. The alternating current in the primary coil produces a magnetic field around it, which magnetises the core [1]. The current keeps changing size and direction, so the magnetic field is continually changing [1]. The secondary coil is in this changing field, so a potential difference is induced across it [1]. This drives an alternating current in the secondary circuit if the circuit is complete [1]. [4] Examiner insight: Each step of the chain is a separate point: current to field, changing field, induction in the secondary, current. Skipping the “changing” step is the most common reason an otherwise good answer falls short.

8. (a) Vs = 10 × 600 / 200 = 30 V [1]. [1] (b) Vs = 10 × 600 / 400 = 15 V [1]. [1] (c) The ratio Ns / Np has halved, so the secondary p.d. halves [1]. [1] Examiner insight: Link the change to the ratio of turns. “More turns gives more voltage” is not enough, because the primary turns went up and the output went down.

9. Sound waves are pressure variations that make the diaphragm vibrate [1]. The coil is attached to the diaphragm, so it moves back and forth in the magnetic field of a permanent magnet [1]. A moving coil in a magnetic field has a potential difference induced across it [1]. The coil changes direction with the vibration, so the induced p.d. changes direction and an alternating current flows, with the same frequency as the sound [1]. [4] Examiner insight: Name every part: diaphragm, coil, permanent magnet. Remember that induction is the effect. A force on the coil is the loudspeaker’s effect.

10. The coil is in the magnetic field of a permanent magnet, so a current in it experiences a force [1]. A larger current gives a larger force on the coil [1]. The coil and the attached cone move through a larger distance [1]. The cone makes larger pressure variations in the air, so the sound is louder [1]. [4] Examiner insight: “Louder” needs the chain from larger current to larger force to larger movement of the cone. Pitch depends on the frequency of the current, so do not confuse the two.

11. (a) The clap makes pressure variations that move the headphone’s cone and the coil attached to it [1]. The coil moves in the field of the permanent magnet, so a potential difference is induced and gives a small current in the meter [1]. [2] (b) A microphone [1]. [1] Examiner insight: Say that the coil moves in a magnetic field. The same device can run in either direction, which is why the two effects are written as a pair in the specification.

12. (a) Ns = Np × Vs / Vp [1] = 2000 × 5.0 / 400 = 25 turns [1]. [2] (b) Power = 5.0 × 3.0 = 15 W [1]. [1] (c) Vp × Ip = 15 [1]. Ip = 15 / 400 = 0.0375 A [1]. [2] (d) Vs = 400 × 50 / 2000 = 10 V [1]. This is double the rated p.d. [1], so the lamp is likely to overheat or burn out [1]. [3] Examiner insight: In part (c), state that the transformer is ideal, so the output power equals the input power, before you substitute. Work from the lamp’s rated figures rather than from an earlier rounded value.

Where marks are usually lost

• Writing “current” where “changing magnetic field” is needed to link the coils.
• Not saying the field is changing, or the supply is alternating.
• Inverting the turns ratio and placing the larger p.d. with the smaller number of turns.
• Dividing a difference by the measured value instead of the expected value in a percentage question.
• Forgetting units, or giving volts when the answer is in amperes.
• Saying a step-up transformer gives more power or more current.
• Describing a microphone with a force on the coil, or a loudspeaker with induction.
• Giving “more voltage” as a cause without naming the number of turns.

Next steps

• Revision notes
• Study guide
• OCR GCSE Physics hub
• Checklist
• All free 10-minute diagnostics
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Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249), specification Version 5.0 (July 2026), first assessment 2018, published by OCR. This page covers statements P4.2g to P4.2j.