AQA A-Level Physics Paper 1, June 2023: Question 3

💡 Key Knowledge

• Emf is defined as the work done (or energy transferred) per unit charge flowing through the source.
• Energy supplied by the cell must equal the total energy dissipated across all components in the circuit (including internal resistance).
• Equation format: ε = V₁ + V₂ + Ir

✅ Expected Answer Points

• 1 C of charge gains ε joules of energy when passing through the cell.
• Energy transferred (or work done) by 1 C in resistors R₁ and R₂ are V₁ and V₂ respectively.
• Energy transferred by 1 C in the internal resistance r is Ir (lost volts).
• Total energy out equals total energy in: ε = V₁ + V₂ + Ir

❌ Common Errors

• Referring to "voltage across r " instead of correctly identifying it as "lost volts" or Ir .
• Treating conservation of energy purely as a formula without explaining the physical movement of one coulomb of charge around the loop as requested.

🧠 Exam Technique

Always structure your explanation following the path of charge: start at the source (gaining energy), move through each external component (dissipating potential differences), and finish with the internal resistance.

📐 Step-by-Step Calculation

1. Identify circuit state: When contact C is moved to end B, the entire resistance of the wire ( R = 125 Ω ) is connected across the terminals, and the voltmeter measures the terminal p.d. ( V = 2.89 V ).
2. State the relevant equation: ε = V + Ir or 3.00 = 2.89 + Ir .
3. Calculate circuit current ( I ):
   I = V / R = 2.89 / 125 = 0.02312 A .
4. Rearrange for r :
   r = (ε - V) / I = (3.00 - 2.89) / 0.02312 = 4.76 Ω (approximately 4.8 Ω ).

❌ Common Calculation Traps

• Forgetting to use the stated emf of 3.00 V instead of just working with terminal p.d.
• Rounding intermediate values too early, which leads to rounding errors on the final answer check. Examiner note: you must see at least 3 significant figures leading to 4.76 Ω .

📐 Calculation Steps

1. Find segment resistance: Contact C is at 1/5 of the distance, so R_AC = (1/5) × 125 Ω = 25.0 Ω .
2. Apply potential divider formula:
   V_out = ε × [R_AC / (R_total + r)]
   Where R_total = 125 Ω and
   r ≈ 4.76 Ω (or use I × R_AC if current was successfully found).
3. Evaluate:
   V_out = 3.00 × [25.0 / (125 + 4.76)] = 3.00 × (25.0 / 129.76) = 0.58 V .

🧠 Exam Technique & Alternative Methods

You can also solve this by finding current I first from the total circuit resistance ( I = 3.00 / 129.76 = 0.0231 A ), then using V = I × R = 0.0231 × 25.0 = 0.58 V .

Top tip: If you missed part 03.2, examiners allow error carried forward (ECF) if you substitute your previous values consistently.

💡 Understanding the Faults

• From A to P: Wire is thinner, meaning higher resistance per unit length. The potential increases more steeply than normal (steeper gradient).
• At Q (scratch): A complete break or severe constriction across the full width causes an abrupt step-up/discontinuity in potential as current crosses the gap.
• From R to B (connector laid over): The metal connector has near-zero resistance, so no potential difference drops across this region (horizontal line at 3.0 V ).

✅ Expected Graph Features (Any 4 required)

• Straight line starting from origin (0 V, A) up to point P .
• A change in gradient between P and Q (steeper slope due to thinner wire AP making PB behave differently).
• A short, sharp vertical/steep increase in potential at point Q due to the scratch.
• A return to a non-zero gradient between Q and R .
• A flat horizontal line from R to B remaining constant at 3.0 V .