AQA A-Level Physics Paper 2, June 2023: Question 5

📐 Step-by-Step Derivation

1. State the charging equation:
   V = V₀(1 − e^{−t / R₁C}) , where V₀ = 6.0 V .
2. Find t₁ (time to reach 2.0 V):
   2.0 = 6.0(1 − e^{−t₁ / R₁C})
   1/3 = 1 − e^{−t₁ / R₁C} ⇒ e^{−t₁ / R₁C} = 2/3
   t₁ = −R₁C ln(2/3) ≈ 0.405 R₁C
3. Find t₂ (time to reach 4.0 V):
   4.0 = 6.0(1 − e^{−t₂ / R₁C})
   2/3 = 1 − e^{−t₂ / R₁C} ⇒ e^{−t₂ / R₁C} = 1/3
   t₂ = −R₁C ln(1/3) ≈ 1.099 R₁C
4. Calculate the time difference (Δt):
   Δt = t₂ − t₁ = −R₁C ln(1/3) − (−R₁C ln(2/3))
   Δt = R₁C [ln(3) − ln(1.5)] = R₁C ln(2)
   Δt = 0.693 R₁C ≈ 0.69 R₁C ≈ 0.7 R₁C

❌ Common Traps & Pitfalls

• Using the discharge equation: Using V = V₀ e^−t/RC with V₀ = 4.0 V and V = 2.0 V gives ln(2)RC ≈ 0.69RC accidentally! The mark scheme awards 0 marks for this because the capacitor is charging from a 6.0 V supply, not discharging.
• Assuming a 4.0 V maximum: Setting V₀ = 4.0 V in the charging equation only scores at most 1 mark. The battery EMF is 6.0 V.
• Premature rounding: Writing just 0.7R₁C without showing 0.69R₁C or ln(2)R₁C will lose the final mark on a "show that" question.

📐 Method 1: Area Under I–t Graph (Charge Q)

1. Scale factor per 1 cm × 1 cm large grid block:
   Time width = 1 s, Current height = 0.5 × 10⁻⁵ A
   Area of 1 large box = 1 s × 0.5 × 10⁻⁵ A = 0.5 × 10⁻⁵ C
2. Count total grid blocks under curve (0 to 11 s):
   Total blocks ≈ 80 (acceptable range: 78 to 82).
3. Calculate total charge Q:
   Q = 80 × 0.5 × 10⁻⁵ C = 4.0 × 10⁻⁴ C
   (acceptable: 3.9 × 10⁻⁴ C to 4.1 × 10⁻⁴ C)
4. Find capacitance:
   At t = 11 s, pd across capacitor is 4.0 V.
   C = Q / V = (4.0 × 10⁻⁴ C) / 4.0 V = 1.0 × 10⁻⁴ F

📐 Method 2: Resistor & Time Constant

1. Find resistance R₁:
   At t = 0 , capacitor pd = 0 V, so full 6.0 V is across R₁:
   R₁ = V / I₀ = 6.0 V / (6.0 × 10⁻⁵ A) = 1.0 × 10⁵ Ω
2. Use current decay equation at t = 11 s:
   At t = 11 s , I = 2.0 × 10⁻⁵ A .
   I = I₀ e^{−t / R₁C} ⇒ 2.0 × 10⁻⁵ = 6.0 × 10⁻⁵ e^{−11 / R₁C}
   ln(3) = 11 / (R₁C) ⇒ R₁C = 11 / 1.0986 = 10.0 s
3. Calculate C:
   C = (R₁C) / R₁ = 10.0 s / (1.0 × 10⁵ Ω) = 1.0 × 10⁻⁴ F

❌ Major Misconception

Dividing by 6.0 V instead of 4.0 V: Charging stops at t = 11 s when the potential difference across the capacitor reaches 4.0 V (not the battery EMF of 6.0 V). Dividing by 6.0 V gives C = 0.67 × 10⁻⁴ F , losing the final 2 marks.

💡 Examiner Insight

Students who simply estimated average current by sight (e.g. taking midway current I = 4.0 × 10⁻⁵ A and doing Q = I·t ) were capped at a maximum of 1 mark. You must show rigorous counting or use exponential equations.

💡 Circuit Analysis during Discharge

Look carefully at Figure 8 when the two-way switch is flicked to the right terminal:

• The battery is completely disconnected.
• The discharge current leaves the top plate of capacitor C, travels through the switch into R₂, then flows through the ammeter, through R₁, and back to the bottom plate of C.
• Therefore, R₁ and R₂ are in series!
• R_Total = R₁ + R₂

📐 Step-by-Step Calculation

1. Extract discharge time from Figure 10:
   Discharge starts at: t₁ = 11 s (at pd = 4.0 V)
   Discharge ends at: t₂ = 32 s (at pd = 2.0 V)
   Δt = 32 − 11 = 21 s
2. Determine R_Total using half-life:
   The voltage halves from 4.0 V to 2.0 V in 21 s:
   T_½ = ln(2) · R_Total · C ≈ 0.693 R_Total C
   R_Total = 21 / (0.693 × 1.0 × 10⁻⁴) = 3.03 × 10⁵ Ω ≈ 3.0 × 10⁵ Ω
3. Subtract R₁ to find R₂:
   From Part 05.2, R₁ = 1.0 × 10⁵ Ω .
   R₂ = R_Total − R₁ = 3.0 × 10⁵ − 1.0 × 10⁵ = 2.0 × 10⁵ Ω

❌ The Deadly Trap in Part 05.3

Many candidates correctly found the total resistance 3.0 × 10⁵ Ω but wrote this directly as their final answer for R₂ , forgetting that R₁ remains in the discharge loop. Always trace the complete closed circuit loop!

✅ Final Answer

R₂ = 2.0 × 10⁵ Ω  (or 200 kΩ)