AQA GCSE Combined Science: Trilogy Physics Paper 1 (Higher), 2023: Question 3

✅ Correct answer

• Transformer X increases the potential difference.
• The current decreases.
• This reduces power loss in the transmission cables, so less energy is wasted as heat.

💡 Key knowledge

• Transformer X is a step-up transformer.
• For the same power, a larger voltage means a smaller current.
• Losses in cables happen because they have resistance, so current causes heating.

🧠 Exam technique

• Use the words potential difference, current, and losses.
• Don’t just write “it is more efficient” — say why.
• A top-level answer links the transformer to reduced heating in the cables.

❌ Common errors

• Writing only “it increases efficiency” with no explanation.
• Confusing voltage and current.
• Not mentioning that cable losses are reduced.

📐 Calculation steps

1. Use the transformer relationship: Vₚ × Iₚ = Vₛ × Iₛ
2. Substitute the values:
   400 000 × 660 = 11 000 × Iₛ
3. Rearrange:
   Iₛ = (400 000 × 660) / 11 000
4. Calculate:
   Iₛ = 24 000 A

✅ Correct answer

24 000 A

🧠 Exam technique

• Always use the correct units: volts and amps.
• Show substitution clearly to pick up method marks.
• Write the final answer with the unit A.

❌ Common errors

• Using the ratio the wrong way round.
• Forgetting to rearrange before calculating.
• Leaving the answer in amps but without a unit.

✅ Correct answer

The street lamps are connected in parallel.

💡 Key knowledge

• In a parallel circuit, the current splits between branches.
• So each lamp takes only part of the total current.

🧠 Exam technique

• This is a one-mark tick-box question: choose the exact statement from the options.
• “Connected in parallel” is the only correct option here.

❌ Common errors

• Choosing “current is used up” — current is not used up.
• Thinking resistance makes current disappear.
• Choosing the wrong statement because it sounds realistic.

✅ Correct answer

• The resistance of the LDR increases.
• The current in the circuit decreases.
• The potential difference across resistor R decreases.

💡 Key knowledge

• An LDR has higher resistance in dim light.
• As resistance increases, current decreases.
• In a series circuit, the supply p.d. is shared between components.

🧠 Exam technique

• Use a clear chain: light decreases → LDR resistance increases → current decreases → p.d. across R decreases.
• Top responses gained marks by linking each stage logically.
• You can also say the LDR takes a greater share of the potential difference.

❌ Common errors

• Saying the p.d. across R increases.
• Missing the connection between resistance and current.
• Forgetting this is a series circuit, so the p.d. is shared.

📐 Calculation steps

1. Convert the current to amps:
   20 mA = 0.020 A
2. Use the power equation:
   P = I²R
3. Substitute values:
   6.0 = 0.020² × R
4. Rearrange:
   R = 6.0 / 0.020²
5. Calculate:
   R = 15 000 Ω

✅ Correct answer

15 000 Ω

🧠 Exam technique

• This is a 4-mark calculation, so show full working.
• Convert mA to A before using the formula.
• Use the equation sheet carefully: P = I²R is the key relationship.

❌ Common errors

• Not converting 20 mA into 0.020 A.
• Using P = IV but not rearranging correctly.
• Forgetting the squared current in I²R .
• Losing marks for no unit or a wrong final unit.

💡 What top answers did well

• Explained the physics chain, not just the final idea.
• Used correct electrical language: p.d., current, resistance, losses.
• Showed clear substitutions and correct units in calculations.

❌ Biggest traps

• Mixing up current and voltage in transformers.
• Forgetting that losses in cables are mainly heating losses due to resistance.
• Missing the LDR resistance change when light level falls.
• Not converting milliamps to amps.