AQA GCSE Physics Physics Paper 1 (Foundation), June 2023: Question 8

10 marks · Standard Demand difficulty · Short Answer

Explain the role of a step-up transformer in the National Grid, calculate the resistance of a transmission cable using the power equation, and determine the useful energy output from the efficiency.

Practise this question

Question

Question 8 features Figure 11, a diagram showing the National Grid from a power station, through a step-up transformer labeled X, along transmission cables supported by pylons, through a step-down transformer, to consumers. Sub-question 08.1 asks to complete sentences about the effect of Transformer X on potential difference and current. Sub-question 08.2 asks to select the equation linking current, power, and resistance from four options. Sub-question 08.3 provides a cable power loss of 1.60 × 10⁹ W and a current of 2000 A, asking to calculate resistance. Sub-question 08.4 asks for the equation linking efficiency, total energy input, and useful energy output. Sub-question 08.5 gives an energy input of 34.2 GJ and efficiency of 0.992, asking to calculate the useful energy output in GJ.
Question text

08 Figure 11 shows how the National Grid connects a power station to consumers.

Figure 11

08.1 Complete the sentences.

[2 marks]

Transformer X causes the potential difference to .

Transformer X causes the current to .

Use the Physics Equations Sheet to answer questions 08.2 and 08.3.

08.2 Which equation links current (I), power (P) and resistance (R)?

[1 mark]

Tick ( ) one box.

I

P =

R

I

P = 2

R

P = I 2 R

P = IR 33

08.3 A transmission cable has a power loss of 1.60 × 109 W.

The current in the cable is 2000 A.

Calculate the resistance of the cable.

[3 marks]

Resistance = Ω

Use the Physics Equations Sheet to answer questions 08.4 and 08.5.

08.4 Write down the equation which links efficiency, total energy input and

useful energy output.

[1 mark]

08.5 The total energy input to the National Grid from one power station is 34.2 GJ.

The National Grid has an efficiency of 0.992

Calculate the useful energy output from this power station to consumers in GJ.

[3 marks]

Useful energy output = GJ

Mark scheme

Show the mark scheme Mark scheme for Question 8. Part 08.1 awards 1 mark for 'increase' and 1 mark for 'decrease' in order (AO1). Part 08.2 awards 1 mark for 'P = I²R' (AO1). Part 08.3 gives 3 marks: 1 for substitution (1.60 × 10⁹ = 2000² × R), 1 for rearrangement (R = 1.60 × 10⁹ / 2000²), and 1 for answer (R = 400 Ω) (AO2). Part 08.4 gives 1 mark for efficiency = useful energy output / total energy input (AO1). Part 08.5 gives 3 marks: 1 for substitution (0.992 = useful energy output / 34.2), 1 for rearrangement (useful energy output = 0.992 × 34.2), and 1 for 33.9 GJ (AO2). Total marks: 10.

Question 8

AO /

Question Answers Extra information Mark

Spec. Ref.

08.1 increase must be in this order 1 AO1

4.2.4.3

decrease 1

AO /

Spec. Ref.

08.2 P = I2R 1 AO1

4.2.4.1

AO /

Spec. Ref.

08.3 1.60 × 109 = 20002 × R 1 AO2

4.2.4.1

1.60×109

R = 2 1

2000

R = 400 (Ω)

AO /

Spec. Ref.

08.4 useful energy output 1 AO1

efficiency = 4.1.2.2

total energy input

or

efficiency =

useful output energy transfer – HYSICS – –

total input energy transfer

AO /

Spec. Ref.

08.5 useful energy output AO2

0.992 = 1 4.1.2.2

34.2

useful energy output

= 0.992 × 34.2 1

useful energy output allow a correct answer given to 1

= 33.9 (GJ) more than 3 s.f.

Total Question 8 10

How to answer it

The National Grid: Transformers, Power Loss & Efficiency

📌 What this question tests

This question assesses core understanding of how electricity is transmitted across the UK National Grid and how to perform related electrical power and efficiency calculations:

  • Step-up transformers: Recalling that they increase potential difference and decrease current to minimise thermal energy losses in transmission cables.
  • Power transmission equations: Identifying and applying the formula P = I²R to find electrical resistance.
  • Efficiency calculations: Recalling the word equation for efficiency and calculating useful energy output using decimal efficiency values and gigajoule ( GJ ) units.

Part 08.1: Function of Transformer X [2 Marks]

Sentence Completion • Role of a Step-Up Transformer

✅ Correct Answers

Transformer X causes the potential difference to increase. [1 mark]

Transformer X causes the current to decrease. [1 mark]

Mark Scheme Note: Answers must be given in this exact order to secure both marks.

💡 Key Knowledge

  • Transformer X is placed immediately after the power station; it is a step-up transformer.
  • Since electrical power is conserved ( P = V × I ), increasing voltage ( V ) causes the current ( I ) to decrease by the same proportion.
  • A lower current dramatically reduces heating losses in the cables because power lost is proportional to current squared ( I² ).

❌ Common Errors

  • Mixing up "step-up" and "step-down" transformers. The transformer near homes/consumers is step-down (decreases voltage to a safe 230 V).
  • Saying "increase" for both lines, wrongly assuming higher voltage means higher current in transmission lines.

🧠 Exam Technique

Always inspect where the transformer sits in the diagram. Next to the power station = step-up. Next to consumers = step-down.

Part 08.2: Formula Linking Current, Power, and Resistance [1 Mark]

Multiple Choice • Equation Selection

✅ Correct Answer

Tick the 3rd box down:

P = I² R [1 mark]

🧠 Exam Technique

  • The rubric explicitly states: "Use the Physics Equations Sheet".
  • Look directly under the "Electricity" section of your sheet to confirm whether the current is squared, avoiding silly memory slips.

Part 08.3: Calculating Cable Resistance [3 Marks]

Quantitative Problem • Standard Form & Rearrangement

📐 Step-by-Step Calculation

  1. Substitute values into the formula:
    1.60 × 10⁹ = 2000² × R [1 mark]
  2. Rearrange to make resistance (R) the subject:
    R = (1.60 × 10⁹) / 2000² [1 mark]
    Note: 2000² = 4,000,000 (or 4.0 × 10⁶)
  3. Solve for R:
    R = 1,600,000,000 / 4,000,000 = 400
    Resistance = 400 Ω [1 mark]

❌ Common Calculation Traps

  • Forgetting to square the current: Dividing 1.60 × 10⁹ by 2000 instead of 2000² gives 800,000 Ω , losing 2 marks!
  • Calculator bracket errors: If typing 1.60 × 10⁹ / 2000² , make sure your calculator squares only the denominator.
  • Rearrangement slip: Multiplying P × I² instead of dividing.

Part 08.4: Efficiency Equation [1 Mark]

Recall • Word Equation

✅ Correct Equation

efficiency = useful energy output / total energy input [1 mark]

Alternative acceptable wording:
efficiency = useful output energy transfer / total input energy transfer

❌ Common Errors

  • Upside down: Writing total input / useful output . Efficiency must always be ≤ 1 (or ≤ 100%).
  • Omitting "useful": Writing just "energy output / total input" is not precise enough and may not be awarded.

Part 08.5: Useful Energy Output Calculation [3 Marks]

Calculation • Rearranging Efficiency with Units

📐 Step-by-Step Calculation

  1. Substitute given values into the formula:
    0.992 = useful energy output / 34.2 [1 mark]
  2. Rearrange to isolate useful energy output:
    useful energy output = 0.992 × 34.2 [1 mark]
  3. Calculate and round:
    = 33.9264 GJ
    Useful energy output = 33.9 GJ [1 mark]
    (Answers given to more than 3 s.f., e.g. 33.93, are also accepted)

🧠 Exam Technique & Units

  • Unit Check: Notice that the input is given in GJ (gigajoules) and the answer line specifies GJ. There is no need to convert to joules!
  • Sanity Check: The useful output ( 33.9 GJ ) must be slightly less than the total input ( 34.2 GJ ) because efficiency is 0.992 (99.2%). If your answer is bigger than 34.2, you multiplied/divided the wrong way around!

Topics

Physics · P1: Energy · P2: Electricity

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Foundation), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.