AQA A-Level Physics Paper 1, November 2020: Question 5

✅ Correct Answer

R = 96 Ω (using P = V² / R or combining P = IV and V = IR)

📐 Step-by-Step Calculation

1. Identify given values: V = 12 V, P = 1.5 W.
2. Select the power equation linking V, P, and R: P = V² / R .
3. Rearrange for resistance: R = V² / P .
4. Substitute values: R = 12² / 1.5 = 144 / 1.5 = 96 Ω .

❌ Common Errors

• Using reverse arguments (starting from the target value instead of deriving it).
• Failing to show working (the mark scheme explicitly states "Must see some working").

✅ Correct Answer

Current = 0.38 A (accept 0.36 A to 0.4 A depending on intermediate rounding or 100 Ω use)

📐 Step-by-Step Calculation

1. Method 1: Find current through one lamp using I = P / V (1.5 / 12 = 0.125 A). Multiply by 3 for three lamps in parallel: 0.125 × 3 = 0.375 A (rounds to 0.38 A).
2. Method 2: Find combined resistance of three identical 96 Ω lamps in parallel: 96 / 3 = 32 Ω . Then use I = V / R = 12 / 32 = 0.375 A .

🧠 Exam Technique

Always keep extra significant figures in your intermediate calculator steps and round only at the final stage to avoid rounding errors.

✅ Correct Answer

Resistance = 49 Ω (accept 48 Ω)

💡 Key Knowledge

The resistivity equation is R = ρL / A , where cross-sectional area of a round wire is A = πd² / 4 or A = πr² .

📐 Step-by-Step Calculation

1. Convert wire diameter to metres: d = 0.19 mm = 0.19 × 10⁻³ m .
2. Calculate cross-sectional area: A = π × (0.19 × 10⁻³)² / 4 = 2.835 × 10⁻⁸ m² .
3. Substitute into resistivity formula: R = (4.9 × 10⁻⁷ × 2.8) / (2.835 × 10⁻⁸) .
4. Calculate final resistance: R = 48.37 Ω (rounds to about 49 Ω).

❌ Common Errors

Power-of-ten (POT) errors when converting millimeters to meters ( 10⁻³ ) or dealing with area calculations. Don't forget to square the radius/diameter!

✅ Correct Answer

Yes, the demonstration works. Substantial voltage is dropped across the long constantan transmission wires, leaving insufficient potential difference across the parallel lamps.

📐 Step-by-Step Calculation & Proof

1. Total wire resistance (two long wires): 49 + 49 = 98 Ω (or using 46 Ω if using 48 Ω).
2. Parallel lamp combination resistance: 96 / 3 = 32 Ω .
3. Total circuit resistance: 98 + 32 = 130 Ω .
4. New circuit current: I = 12 / 130 = 0.092 A (or ~0.096 A).
5. Conclusion: Current is far below the 0.38 A required for full brightness, proving lamps become significantly dimmer or virtually off.

🧠 Exam Technique

For "Discuss whether..." questions, always provide a balanced numerical proof first, then state a clear qualitative conclusion linking back to the question statement ("significantly dimmer").

✅ Correct Answer

Advantage: Zero electrical resistance / resistivity, eliminating power/thermal losses in transmission cables.
Difficulty: Maintaining extremely low temperatures (critical temperature) over long distances is technologically challenging and expensive.

💡 Key Knowledge

A superconductor is a material that has zero electrical resistivity below a critical temperature ( T_c ). Examiners strictly accept references to zero resistance/resistivity and cooling difficulties; ignore mentions of critical field strength.

❌ Common Errors

Vague answers such as "it is too cold" without explaining that the difficulty lies in *maintaining* that low temperature continuously across long distribution networks.