AQA GCSE Physics Physics Paper 1 (Higher), June 2023: Question 1
10 marks · Low Demand difficulty · Short Answer
Answer questions about the National Grid, including the effect of step-up transformers, calculating transmission cable resistance using power loss and current, and calculating useful energy output using efficiency.
Practise this questionQuestion
Question text
01 Figure 1 shows how the National Grid connects a power station to consumers.
Figure 1
01.1 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 01.2 and 01.3.
01.2 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 3
01.3 A transmission cable has a power loss of 1.60 × 109 W.
*02* 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 01.4 and 01.5.
01.4 Write down the equation which links efficiency, total energy input and
useful energy output.
[1 mark]
01.5 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
Question 1
AO /
Question Answers Extra information Mark
Spec. Ref.
01.1 increase must be in this order 1 AO1
4.2.4.3
decrease 1
AO /
Spec. Ref.
01.2 P = I2R 1 AO1
4.2.4.1
AO /
Spec. Ref.
01.3 1.60 × 109 = 20002 × R 1 AO2
4.2.4.1
1.60×109
R = 2 1
2000
R = 400 (Ω) 1
AO /
Spec. Ref.
01.4 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.
01.5 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 1 10
How to answer it
National Grid Transformers, Power Loss & Efficiency
This question assesses core recall, equation selection, and calculation skills from Paper 1: Energy & Electricity:
- Step-up transformers: Recall how potential difference and current change before transmission cables.
- Power transmission equations: Identifying and applying P = I²R to calculate resistance.
- Efficiency: Recalling the formula and rearranging it to find useful output energy (handling gigajoule units).
Question 01.1: Step-Up Transformer Function
2 Marks • Recall (AO1) • Spec Reference: 4.2.4.3
✅ Correct Answer
- Transformer X causes the potential difference to increase. [1 mark]
- Transformer X causes the current to decrease. [1 mark]
Note: Answers must be in this exact order.
💡 Key Knowledge
Transformer X is situated between the power station and transmission cables, meaning it is a step-up transformer:
- Increases potential difference (voltage) to very high levels (e.g. 400 kV).
- Decreases the current in the transmission cables.
- Lower current means significantly less thermal energy is wasted as heat in the cables ( P = I²R ).
🧠 Exam Technique
Always identify the position of the transformer first:
- Near power station: Step-up (Voltage goes UP, current goes DOWN).
- Near consumers/homes: Step-down (Voltage goes DOWN, current goes UP to safe, usable levels).
❌ Common Errors
- Swapping the words around (writing "decrease" then "increase").
- Writing "step up" or "higher" instead of complete verbs like "increase" or "rise".
Question 01.2: Equation for Power, Current & Resistance
1 Mark • Equation Recall (AO1) • Spec Reference: 4.2.4.1
✅ Correct Box to Tick
P = I² R
Ticking the 3rd box scores [1 mark]. All other options are incorrect.
💡 Physics Equations Sheet Check
The question reminds you to use the Physics Equations Sheet. Even if you have memorised it, double-check:
- P = I² R relates power ( P ), current ( I ), and resistance ( R ).
- P = VI is power, potential difference, and current.
- V = IR is Ohm's Law (potential difference, current, resistance).
Question 01.3: Calculating Cable Resistance
3 Marks • Application & Calculation (AO2) • Spec Reference: 4.2.4.1
📐 Step-by-Step Calculation
Given: Power loss, P = 1.60 × 10⁹ W | Current, I = 2000 A
- Substitute values into the equation:
1.60 × 10⁹ = 2000² × R [1 mark] - Rearrange to make R the subject:
2000² = 4,000,000 = 4.0 × 10⁶
R = (1.60 × 10⁹) / (2000²)
R = (1.60 × 10⁹) / (4.0 × 10⁶) [1 mark] - Calculate the final value:
R = 400 Ω [1 mark]
🧠 Exam Technique
Always write out your substitution before rearranging. If you make a slip on your calculator, the examiner can still award 1 mark for substitution and 1 mark for rearrangement.
❌ Common Errors
- Forgetting to square the current: Dividing 1.60 × 10⁹ by 2000 instead of 2000² (gives 800,000 Ω — completely incorrect!).
- Bracket mistakes on calculators: Typing 1.60 × 10⁹ / 2000² without brackets can sometimes lead to order-of-operation errors depending on the calculator model.
Question 01.4: Efficiency Equation
1 Mark • Recall (AO1) • Spec Reference: 4.1.2.2
✅ Correct Answer
efficiency = useful energy output / total energy input
Also accepted: useful output energy transfer / total input energy transfer
[1 mark]
🧠 Top Tip
Remember that efficiency is always a fraction or percentage where output is compared to input:
Useful OUT / Total IN
Since efficiency cannot exceed 1 (or 100%), the smaller number (useful output) is always on top!
Question 01.5: Calculating Useful Energy Output
3 Marks • Application & Calculation (AO2) • Spec Reference: 4.1.2.2
📐 Step-by-Step Calculation
Given: Total energy input = 34.2 GJ | Efficiency = 0.992
- Substitute given values into the efficiency equation:
0.992 = useful energy output / 34.2 [1 mark] - Rearrange to solve for useful energy output:
useful energy output = 0.992 × 34.2 [1 mark] - Calculate the final answer:
useful energy output = 33.9 GJ (or 33.9264 GJ) [1 mark]
💡 Unit Watch: GJ (Gigajoules)
Notice the question already has GJ on the answer line and gives the input in GJ. You do not need to convert to joules ( × 10⁹ ). Keeping it in GJ saves time and prevents standard form errors!
❌ Common Errors
- Dividing instead of multiplying: Calculating 34.2 / 0.992 = 34.48 GJ . This gives an output larger than the input, which is physically impossible (efficiency cannot exceed 100%).
- Multiplying by 100: The efficiency is already given as a decimal ( 0.992 ), not a percentage ( 99.2% ). Multiplying by 100 here leads to an answer 100 times too big.
Topics
Physics · P1: Energy · P2: Electricity
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Higher), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.