AQA A-Level Physics Paper 2, June 2025: Question 18

1 mark · Medium difficulty · Multiple Choice

Calculate the resistance of a variable resistor in a constant-current capacitor charging circuit after 30 seconds.

Practise this question

Question

A circuit diagram showing a 2.0 V cell in series with an ammeter reading 6.0 mA, a variable resistor R, and a 300 mF capacitor. The text explains that the capacitor is charged with a constant current by decreasing the resistance of R. It asks for the resistance of R at time t = 30 s, with four options: A (58 Ω), B (100 Ω), C (233 Ω), and D (333 Ω).

Mark scheme

Show the mark scheme Mark scheme table indicating question number 18 has answer C (233 Ω) and tests assessment objective AO2.

How to answer it

Constant Current Capacitor Charging

📌 What this question tests

This question assesses your ability to analyse a non-standard capacitor circuit where current is kept constant rather than decaying exponentially. It requires applying Q = I × t for constant current, the definition of capacitance C = Q / V, Kirchhoff's second law (conservation of energy in a series loop), and Ohm's law to determine the dynamic resistance of a variable resistor at a specific time.

Question 18 • Multiple Choice [1 mark]

Resistance Calculation at t = 30 s

AQA A-Level Physics • Capacitors • AO2 (Application)

✅ Correct Answer

C — 233 Ω

At t = 30 s , the potential difference across the capacitor rises to 0.60 V , leaving 1.40 V across the variable resistor. With a constant current of 6.0 mA , R = 1.40 / 0.006 = 233 Ω .

💡 Key Knowledge

  • Constant Current Charging: Because current is held constant by adjusting R, the exponential decay equations do not apply. Instead, Q = I × t .
  • Capacitor Potential Difference: VC = Q / C .
  • Kirchhoff's Second Law: ε = VC + VR (since internal resistance is negligible).
  • Ohm's Law for Resistor: R = VR / I .

📐 Step-by-Step Calculation

Step 1: Convert given values to standard SI units
  • Emf (ε) = 2.0 V
  • Capacitance (C) = 300 mF = 300 × 10⁻³ F = 0.300 F
  • Current (I) = 6.0 mA = 6.0 × 10⁻³ A
  • Target time (t) = 30 s
Step 2: Calculate charge stored on the capacitor at t = 30 s

Because the current is constant:
Q = I × t = (6.0 × 10⁻³ A) × 30 s = 0.18 C

Step 3: Calculate the p.d. across the capacitor (VC)

Using V = Q / C :
VC = 0.18 C / 0.300 F = 0.60 V

Step 4: Calculate the p.d. across the variable resistor (VR)

Using Kirchhoff's voltage law ( ε = VC + VR ):
VR = 2.0 V − 0.60 V = 1.40 V

Step 5: Calculate the resistance of R

Using Ohm's law ( R = VR / I ):
R = 1.40 V / (6.0 × 10⁻³ A) = 233.3 Ω ≈ 233 Ω

🧠 Exam Technique & Distractor Breakdown

  • Option D (333 Ω): This is the initial resistance at t = 0 s ( 2.0 V / 6.0 mA ). Spotting this lets you immediately rule it out, as the resistance must decrease over time.
  • Option B (100 Ω): Obtained if you calculate VC / I = 0.60 V / 6.0 mA . This is an error of finding an "effective resistance" of the capacitor rather than the resistor.
  • Option A (58 Ω): Represents the resistance much later in the charging cycle (at around t = 82.5 s ).

❌ Common Errors to Avoid

  • Reaching for the exponential formula: Students see a capacitor and automatically try to use Q = Q0(1 − e−t/RC) . That formula only applies when resistance is fixed and current falls exponentially!
  • Unit conversion traps: Mixing up millifarads ( mF = ×10⁻³ F ) with microfarads ( μF = ×10⁻⁶ F ), or forgetting to convert mA to A .
  • Neglecting VR: Forgetting that VR = ε − VC , resulting in dividing the cell emf or capacitor voltage directly by the current.

Topics

Physics · 3.7 Fields and their consequences (A-level only) · 3.5 Electricity

Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.