AQA GCSE Physics Physics Paper 1 (Higher), November 2021: Question 5

📐 Step-by-Step Calculation

1. Convert time to seconds:
   t = 24 × 60 × 60 = 86 400 s
2. State formula & substitute:
   Q = I × t
   27 000 = I × 86 400
3. Rearrange to make I the subject:
   I = 27 000 / 86 400
4. Calculate final value:
   I = 0.3125 A (or 0.31 A / 0.313 A )

✅ Mark Scheme Breakdown

• Mark 1: Converting 24 hours to seconds: 86 400 (s)
• Mark 2: Correct substitution: 27 000 = I × 86 400
• Mark 3: Correct rearrangement: I = 27 000 / 86 400
• Mark 4: Correct final value: 0.3125 (A)

🧠 Exam Technique: Unit Traps

Physics equations require standard SI units. Time must always be converted to seconds ( s ) before substituting into electrical formulas unless specifically told otherwise.

Method Mark Safety Net: Even if you forget to convert time, substituting t = 24 allows you to still gain marks 2 and 3 through error carried forward (ecf).

❌ Common Errors

• Multiplying 24 by only 60 (converting to minutes instead of seconds: 1440 s ).
• Dividing time by charge ( 86 400 / 27 000 ), resulting in 3.2 A .
• Leaving time as 24 hours, giving 1125 A (a massive current for a small solar setup!).

📐 Step-by-Step Calculation

1. Convert total power input into Watts (W):
   7.8 kW = 7.8 × 1000 = 7800 W
2. Substitute into efficiency equation:
   Efficiency = Useful power output / Total power input
   0.15 = Useful power output / 7800
3. Rearrange and solve:
   Useful power output = 0.15 × 7800 = 1170 W

✅ Mark Scheme Breakdown

• Mark 1: Correct substitution using converted power: 0.15 = output / 7800
• Mark 2: Correct rearrangement: output = 0.15 × 7800
• Mark 3: Correct answer: 1170 (W) (also accepts 1200 W rounded to 2 s.f.)

💡 Key Knowledge: Efficiency

• Efficiency is a ratio and has no units.
• It can be written as a decimal ( 0.15 ) or a percentage ( 15% ).
• Since efficiency is less than 1 (or 100%), the useful output must always be smaller than the total input!

❌ Common Errors

• Multiplying by 100 mistakenly because of efficiency: 0.15 × 100 .
• Dividing input power by efficiency: 7800 / 0.15 = 52 000 W (impossible output).
• Forgetting to convert kW to W, ending up with 1.17 on the answer line labelled W.

✅ Marking Points (Choose ONE complete pair)

Route 1 (Land vs Output):

• A really large area of land would need to be covered with solar cells [1]
• Due to the low useful power output / low efficiency of solar cells [1]

Route 2 (Environmental / Geographical):

• Low number of daylight hours (in the UK / during winter) [1]
• Low solar intensity / solar radiation is too low in the UK [1]

Alternative Point:

• Materials for manufacturing solar panels or lithium backup batteries are in limited supply [1]

🧠 Exam Technique: "Explain Why"

A 2-mark "Explain" question requires a cause and effect or a fact and consequence:

• Merely saying "the UK is cloudy" is too vague to score the scientific explanation mark.
• Use specific scientific terms: mention low solar intensity / radiation, seasonal daylight variation, or low efficiency requiring large land area.