OxfordAQA International GCSE Physics: Transformers and the National Grid (9203) – 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 follow section 3.6.2 (electricity transmission and distribution) of the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards. Treat every transformer as 100% efficient unless a question says otherwise.

Links: study guide, revision notes, course hub, printable checklist and the 10-minute diagnostics.

Questions

1. State where a step-up transformer is placed in the National Grid and what it does to the potential difference. [2]

2. A simple diagram of the National Grid reads: power station, then A, then transmission cables, then B, then homes.

(a) Name the devices A and B. [2] (b) State what each one does to the potential difference. [2]

3. Explain why electricity is transmitted over long distances at a high potential difference. [4]

4. Describe the structure of a basic transformer and explain how it produces an output when an alternating current flows in the primary coil. [4]

5. A technician powers a transformer from a cell. After a few seconds a voltmeter across the output coil reads 0 V. Explain why. [3]

6. A transformer has 1800 turns on its primary coil and 60 turns on its secondary coil. The primary potential difference is 3600 V. Calculate the secondary potential difference. Say which type of transformer this is. [4]

7. A transformer is to change 6.0 V into 90 V. The primary coil has 40 turns. Calculate the number of turns needed on the secondary coil. [3]

8. The secondary coil of a transformer supplies 48 V at 3.5 A. The primary is connected to a 240 V supply. Calculate the primary current. [4]

9. A solar farm sends 2.4 MW of power along a cable of total resistance 1.6 Ω.

(a) Calculate the current in the cable when the power is sent at 8 kV and at 40 kV. [2] (b) Calculate the power lost as heat in the cable at each potential difference. [2] (c) Compare the two losses, and explain the result in terms of the currents. [2]

10. Switch mode transformers are used in mobile phone chargers.

(a) State three features of a switch mode transformer. [3] (b) Explain why these features suit a mobile phone charger. [2]

11. A power station generates 72 MW at 16 kV. A step-up transformer with 600 turns on its primary coil raises the potential difference to 320 kV for transmission. The cable has a resistance of 4.0 Ω.

(a) Calculate how many turns the secondary coil needs. [2] (b) Calculate the current in the transmission cable. [2] (c) Calculate the power lost as heat in the cable. [2] (d) Calculate the power lost as a percentage of the 72 MW. [1] (e) The 72 MW could be sent at 16 kV with no transformer. Show that this is not a sensible choice. [2]

Answers

1. It is placed between the power station and the transmission cables, before the distribution lines [1]. It increases the potential difference [1]. [2] Examiner insight: Answers that name the position (after the generator) and the effect (potential difference increases) both get credit. “At the start” alone is vague.

2. (a) A is a step-up transformer [1] and B is a step-down transformer [1]. (b) A increases the potential difference [1]. B decreases it to a safer value for consumers [1]. [4] Examiner insight: “Transformer” on its own does not name the type. Say step-up or step-down, and match the effect to the position.

3. For a given power, P = V × I, so a higher potential difference means a lower current [1]. The cables have resistance, so a current heats them [1]. A lower current means less heating in the cables [1]. Less energy is wasted, so the system is more efficient [1]. [4] Examiner insight: Four linked steps are needed for four marks. “High voltage reduces energy loss” with no mention of current is a restatement of the question.

4. Two insulated coils, the primary and the secondary, share a soft iron core [1]. The alternating current makes the magnetic field in the core keep changing [1]. The core links that field to the secondary coil [1]. This induces a changing potential difference across the secondary coil, and an alternating current flows if the circuit is complete [1]. [4] Examiner insight: The word “changing” does most of the work in this answer. “Magnetic field” without “changing” loses the explanation.

5. A battery gives a steady direct current [1]. The field in the core then stays constant [1]. A potential difference is induced in the secondary coil only when the field changes, so the voltmeter reads zero [1]. [3] Examiner insight: Mention what the field does. Writing only “transformers need ac” repeats the fact but does not explain it.

6. Vp / Vs = np / ns, so 3600 / Vs = 1800 / 60 [1]. Vs = 3600 × 60 / 1800 [1]. Vs = 120 V [1]. It is step-down because the secondary potential difference is less than the primary [1]. [4] Examiner insight: Write the equation and the substitution so your method is visible. Give the unit.

7. Vp / Vs = np / ns, so ns = np × Vs / Vp [1]. ns = 40 × 90 / 6.0 [1]. ns = 600 turns [1]. [3] Examiner insight: Rearranging before substituting reduces slips. A secondary with fewer turns than the primary signals an inverted ratio.

8. Output power = 48 × 3.5 [1] = 168 W [1]. Input power equals output power for a 100% efficient transformer [1]. Ip = 168 / 240 = 0.70 A [1]. [4] Examiner insight: The primary current is smaller than the secondary current here because the primary voltage is higher. If your answer is bigger than 3.5 A, check the working.

9. (a) At 8 kV: I = 2 400 000 / 8000 = 300 A [1]. At 40 kV: I = 2 400 000 / 40 000 = 60 A [1]. (b) At 8 kV: loss = 300² × 1.6 = 144 000 W (144 kW) [1]. At 40 kV: loss = 60² × 1.6 = 5760 W (5.8 kW to 2 significant figures) [1]. (c) The loss at 8 kV is 144 000 / 5760 = 25 times as large [1]. The potential difference is 5 times higher, so the current is 5 times smaller, and the loss depends on current squared, giving 5² = 25 [1]. [6] Examiner insight: Convert kV and MW to V and W before dividing. Part (c) needs a reason, not only the ratio.

10. (a) Any three: operate at a high frequency, often 50 kHz to 200 kHz [1]; much lighter and smaller than traditional 50 Hz mains transformers [1]; use very little power when switched on with no load applied [1]. (b) A charger is carried around, so being light and small is useful [1]. It is often left plugged in with no phone attached, so low no-load power means little energy is wasted [1]. [5] Examiner insight: Each feature in part (b) needs to be tied to the charger. A bare list of features repeated from part (a) earns nothing new.

11. (a) ns = np × Vs / Vp = 600 × 320 000 / 16 000 [1] = 12 000 turns [1]. (b) I = P / V = 72 000 000 / 320 000 [1] = 225 A [1]. (c) Loss = I² × R = 225² × 4.0 [1] = 202 500 W (about 200 kW) [1]. (d) 202 500 / 72 000 000 × 100 = 0.28% [1]. (e) At 16 kV the current would be 72 000 000 / 16 000 = 4500 A [1]. The loss would be 4500² × 4.0 = 81 000 000 W = 81 MW, which is more than the 72 MW being sent, so it cannot work [1]. [9] Examiner insight: A “show that” answer needs the calculated values written out, not only the conclusion. Check that the power in part (e) is compared like for like, both in MW.

Where marks are usually lost

• Not saying that higher voltage means lower current in an explanation of grid losses.
• Calling a transformer just “a transformer” on a grid diagram, with no step-up or step-down.
• Leaving out the word “changing” when describing the magnetic field.
• Reading the turns ratio the wrong way up, so a step-down answer comes out larger than the input.
• Forgetting to convert kV, MW and kW to V and W before substituting.
• Using P = V × I for the power loss in a cable, when the loss is I² × R.
• Saying a step-up transformer increases power or energy.
• Giving no unit, or a unit that does not match the quantity.
• Listing switch mode features without linking them to the application.

Next steps

• Re-read the study guide for any question you missed.
• Condense it with the revision notes.
• Tick off the topic on the printable checklist and browse the course hub.
• Take one of the all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Physics (9203), specification for exams May/June 2018 onwards, published by OxfordAQA, section 3.6.2.