OCR GCSE Physics A: The national grid and transformers (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 P8.2c to P8.2f of topic P8.2, Powering Earth, in the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026). P8.2f is in bold in the specification, so it is Higher Tier only, and questions on it are labelled (Higher Tier only). Unlabelled questions can be met in both tiers. Use the study guide and the revision notes. Course pages: the hub and the checklist. Try the free 10-minute diagnostics for a quick check.

Questions

1. Put these in the order that electrical power passes through them, from first to last: homes; step-down transformer; power station; step-up transformer; high-voltage transmission lines. [3]

2. The national grid uses two types of transformer.

(a) State what a step-up transformer does and where it is used in the grid. [2] (b) State what a step-down transformer does and where it is used in the grid. [2]

3. A student says, “The grid sends power at 400 kV, so 400 kV is lost along the cable.” Explain what is wrong with this statement. [3]

4. A power station sends 18 MW along a cable of resistance 15 Ω at a potential difference of 300 kV.

(a) Calculate the current in the cable. [2] (b) Work out the heating power wasted in the cable. [2] (c) The same power is sent at 30 kV instead. State the new current, calculate the new wasted power and say what fraction of the 18 MW it is. [3]

5. Explain why transmitting electrical power at high potential difference makes the national grid efficient. [4]

6. A student sets up a model power line. An 18 W supply delivers power to a lamp through long thin wires of total resistance 0.80 Ω. She changes the potential difference put on to the line and keeps the power entering the line at 18 W.

Supply p.d. to line (V)	Current in line (A)	Power wasted in line (W)
9	2.0	3.2
18	1.0	0.80
36

(a) Calculate the current in the line at 36 V. [1] (b) Calculate the power wasted in the line at 36 V. [2] (c) Calculate the percentage of the 18 W that reaches the lamp at 36 V. [2] (d) Describe how the power wasted changes each time the supply potential difference is doubled. [1]

7. (Higher Tier only) A step-up transformer at a power station has 800 turns on the primary coil. The primary potential difference is 25 kV and the required output is 275 kV. The power is 50 MW. Assume no energy is wasted.

(a) Calculate the number of turns on the secondary coil. [2] (b) Calculate the primary current. [1] (c) Calculate the secondary current. [2]

8. (Higher Tier only) A step-down transformer has a primary coil of 6600 turns connected to 33 kV and a secondary potential difference of 11 kV. The secondary current is 90 A. Assume no energy is wasted.

(a) Find how many turns the secondary coil needs. [2] (b) Find the primary current. [2]

9. (Higher Tier only) Explain, using the idea of power, why the current in the secondary coil of a step-up transformer is smaller than the current in the primary coil. [3]

10. (Higher Tier only) A real substation transformer takes 42 A at 11 kV. It delivers 1950 A at 230 V.

(a) Calculate the input power. [1] (b) Calculate the output power. [1] (c) Calculate the efficiency of the transformer as a percentage. [2] (d) State how much power is wasted and what form the wasted energy takes. [1]

11. (Higher Tier only) A grid link must deliver 90 MW through cables of total resistance 6.0 Ω. Engineers compare transmitting at 90 kV and at 360 kV. The generator output is at 25 kV and the primary coil of the step-up transformer has 1000 turns.

(a) Calculate the current in the cables at each transmission potential difference. [2] (b) Calculate the power wasted in the cables at each potential difference. [2] (c) Calculate the wasted power as a percentage of 90 MW at each potential difference. [2] (d) Find the number of secondary turns needed to give 360 kV. [2] (e) Explain which transmission potential difference the engineers should choose. [2]

12. Explain why the potential difference is lowered by step-down transformers before power reaches homes instead of delivering the transmission potential difference. [3]

Answers

1. Power station [1], then step-up transformer followed by high-voltage transmission lines [1], then step-down transformer, then homes [1]. [3] Examiner insight: Check that step-up comes before the long lines and step-down after them.

2. (a) It increases the potential difference [1]. It is used at the power station, before the long transmission lines [1]. [2] (b) It decreases the potential difference [1]. It is used near towns and streets, before the power reaches homes [1]. [2] Examiner insight: Give both the function and the position. A bare “step-up increases voltage” answers only half of the question.

3. The 400 kV is the potential difference applied across the line, between the line and the ground [1]. The potential difference along the cable is only the current multiplied by the cable resistance [1]. This is much smaller than 400 kV, so only a small part of the energy is wasted as heat [1]. [3] Examiner insight: Use the words “across” and “along”. Writing that the potential difference is “used up” repeats the misconception rather than correcting it.

4. (a) I = P / V [1] = 18 000 000 / 300 000 = 60 A [1]. [2] (b) P = I²R = 60² × 15 [1] = 54 000 W (54 kW) [1]. [2] (c) Current = 18 000 000 / 30 000 = 600 A [1]. Wasted power = 600² × 15 = 5 400 000 W (5.4 MW) [1]. Fraction = 5.4 / 18 = 0.30 (30%) [1]. [3] Examiner insight: Convert megawatts and kilovolts to watts and volts before substituting, and set out each step. Noting the factor of 100 in wasted power shows you understand the squared term.

5. Power = potential difference × current, so for the same power a higher potential difference gives a smaller current [1]. Wasted power in a cable is I²R [1]. Reducing the current by a factor reduces the wasted power by the square of that factor [1]. So much less energy is transferred as thermal energy in the cables, and a larger share of the energy reaches the users [1]. [4] Examiner insight: Name the link in each step: higher potential difference, lower current, less heating. Answers that jump from “high voltage” to “less energy lost” leave out the reasoning that earns credit.

6. (a) I = 18 / 36 = 0.50 A [1]. [1] (b) P = I²R = 0.50² × 0.80 [1] = 0.20 W [1]. [2] (c) Power reaching the lamp = 18 − 0.20 = 17.8 W [1]. Percentage = 17.8 / 18 × 100 = 98.9% [1]. [2] (d) The wasted power falls to a quarter of its previous value [1]. [1] Examiner insight: In (d), quote figures from the table, such as 3.2 W to 0.80 W to 0.20 W, to support the pattern.

7. (a) Ns = Np × Vs / Vp [1] = 800 × 275 000 / 25 000 = 8800 turns [1]. [2] (b) Ip = P / Vp = 50 000 000 / 25 000 = 2000 A [1]. [1] (c) Is = P / Vs = 50 000 000 / 275 000 [1] = 182 A (3 s.f.) [1]. [2] Examiner insight: Check that the answer to (c) is smaller than the answer to (b). A step-up transformer must give a smaller secondary current when no energy is wasted, and this check catches inverted ratios.

8. (a) Ns = Np × Vs / Vp [1] = 6600 × 11 / 33 = 2200 turns [1]. [2] (b) Vp Ip = Vs Is, so Ip = 11 000 × 90 / 33 000 [1] = 30 A [1]. [2] Examiner insight: A step-down transformer has fewer secondary turns, so check that your answer to (a) is below 6600. Show the equation before substituting.

9. In a transformer in which no energy is wasted, the power in equals the power out, so Vp Ip = Vs Is [1]. In a step-up transformer, Vs is larger than Vp [1]. For the product to stay the same, Is must be smaller than Ip [1]. [3] Examiner insight: Start from the power statement and then compare the potential differences. A reason such as “because there are more turns” does not by itself explain why the current falls.

10. (a) P = 11 000 × 42 = 462 000 W (462 kW) [1]. [1] (b) P = 230 × 1950 = 448 500 W (448.5 kW) [1]. [1] (c) Efficiency = 448 500 / 462 000 [1] × 100 = 97.1% (3 s.f.) [1]. [2] (d) 462 000 − 448 500 = 13 500 W (13.5 kW), wasted as thermal energy [1]. [1] Examiner insight: Put the output power on the top of the efficiency fraction. If you get more than 100%, you have inverted it.

11. (a) At 90 kV: I = 90 000 000 / 90 000 = 1000 A [1]. At 360 kV: I = 90 000 000 / 360 000 = 250 A [1]. [2] (b) At 90 kV: 1000² × 6.0 = 6 000 000 W (6.0 MW) [1]. At 360 kV: 250² × 6.0 = 375 000 W (0.375 MW) [1]. [2] (c) At 90 kV: 6.0 / 90 × 100 = 6.7% [1]. At 360 kV: 0.375 / 90 × 100 = 0.42% [1]. [2] (d) Ns = Np × Vs / Vp [1] = 1000 × 360 000 / 25 000 = 14 400 turns [1]. [2] (e) Choose 360 kV [1]. The potential difference is four times larger, so the current is four times smaller and the wasted power is sixteen times smaller, so much less energy is wasted as heat in the cables [1]. [2] Examiner insight: Make the decision explicit and back it with a calculated comparison, such as 6.7% against 0.42%. Back the decision with figures and the reasoning chain.

12. A potential difference of hundreds of kilovolts would be dangerous in homes [1]. Household appliances are designed to work at a much lower potential difference, about 230 V [1]. A step-down transformer lowers the potential difference in stages to a safe, usable value [1]. [3] Examiner insight: Give two separate reasons (safety and suitability for appliances) rather than repeating “it is too high” twice.

Where marks are usually lost

• Writing “the voltage is used up along the cable” instead of separating the potential difference across the line from the small potential difference along it.
• Explaining the benefit of high voltage without mentioning that the current falls.
• Forgetting that the wasted power uses the current squared, so a tenfold change in current is a hundredfold change in wasted power.
• Leaving megawatts and kilovolts unconverted when using the equations.
• Placing the larger number of turns on the primary coil of a step-up transformer.
• Giving the secondary current as larger than the primary current for a step-up transformer.
• Saying a transformer increases the power, rather than trading potential difference for current.
• Answering a Higher Tier only question by recall, without linking turns, potential difference and power.

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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 P8.2c to P8.2f.