AQA GCSE Combined Science: Trilogy Physics Paper 1 (Higher), 2018: Question 6

✅ Correct answer

• A curve in the first and third quadrants only.
• The graph must pass through the origin.
• The line should have a decreasing gradient as potential difference increases.

💡 Key knowledge

• A filament lamp is non-ohmic, so current is not directly proportional to potential difference.
• As current increases, the filament gets hotter.
• Higher temperature means higher resistance, so the current rises less quickly.

🧠 Exam technique

• For 2 marks, one mark is usually for the correct shape / quadrant and one for the decreasing gradient.
• Make sure your curve is smooth, not a straight line.
• Use the axes correctly: current on the vertical axis, potential difference on the horizontal axis.

❌ Common errors

• Drawing a straight line like an ohmic conductor.
• Putting the curve in the wrong quadrants.
• Forgetting that it must go through the origin.
• Making the gradient increase instead of decrease.

✅ Correct answer

• I₂ = I₃
• I₁ = I₂ + I₃
• So I₁ = 2I₂ and I₁ = 2I₃

💡 Key knowledge

• The lamps are identical.
• They are connected in parallel, so each branch gets the same potential difference.
• Identical components in parallel carry the same current.
• The current from the cell splits between the branches.

🧠 Exam technique

• Use a clear comparison sentence or equations.
• Marks are for the relationship, not just saying “same” or “double” with no explanation.
• If writing in words, say that the total current is the sum of the branch currents.

❌ Common errors

• Saying all three currents are equal.
• Forgetting that I₁ is the total current in the main loop.
• Mixing up series and parallel rules.

📐 Calculations: step-by-step

1. Use power = current² × resistance
   P = I²R
2. Substitute values for one lamp:
   3 = I² × 12
3. Rearrange:
   I² = 3 ÷ 12 = 0.25
4. Take the square root:
   I = √0.25 = 0.5 A
5. There are two identical lamps in parallel, so total current is:
   I_total = 0.5 + 0.5 = 1.0 A
6. Use charge = current × time
   Q = I × t
7. Convert time: 1 minute = 60 s
   Q = 1.0 × 60 = 60 C

✅ Correct answer

• Charge = 60 C
• Unit = coulombs or C

💡 Key knowledge

• P = I²R is useful when power and resistance are given.
• Q = I × t tells you charge from current and time.
• In a parallel circuit, the total current is the sum of the branch currents.
• Always use seconds for time in charge calculations.

🧠 Exam technique

• Write the equations first to secure method marks.
• Show the substitution clearly.
• Include the unit in your final answer.
• Round sensibly: 0.5 A and 1.0 A are both acceptable, but keep consistency.

❌ Common errors

• Using P = IV without finding the current or voltage correctly.
• Forgetting that there are two lamps, so the cell current is doubled.
• Using 1 minute as 1 second.
• Giving the answer in amps instead of coulombs.
• Missing the final unit mark.

✅ Correct answer

When the environmental conditions change so that light intensity increases, the resistance of the LDR decreases. This makes the total resistance of the circuit decrease, so the current in the ammeter increases.

The potential difference is shared between components in series. Because the LDR now has a lower resistance, it takes a smaller share of the potential difference, so the voltmeter reading decreases.

💡 Key knowledge

• An LDR has lower resistance when light intensity increases.
• In a series circuit, increasing one component’s resistance affects the whole circuit.
• Current increases when total resistance decreases, assuming the battery voltage stays the same.
• Potential difference is shared between series components.

🧠 Exam technique

• For 5–6 marks, write a linked explanation: change in light → change in LDR resistance → change in total resistance → change in current → change in voltmeter reading.
• Use the meter names correctly: ammeter increases, voltmeter decreases.
• Top-level answers explained why the voltmeter changes, not just what happens.

❌ Common errors

• Saying only “the readings change” with no explanation.
• Mixing up the meters: ammeter measures current, voltmeter measures potential difference.
• Saying the voltmeter increases when the LDR resistance falls.
• Forgetting to mention that the supply potential difference stays the same.

✅ 06.1

Curve in first and third quadrants, through the origin, decreasing gradient.

✅ 06.2

I₂ = I₃ and I₁ = I₂ + I₃

✅ 06.3

60 C and unit C

✅ 06.4

Ammeter reading increases; voltmeter reading decreases when light intensity increases.