AQA A-Level Physics: Capacitor Charging and Discharging Through Resistors — mark scheme explained

The question

Mark scheme, decoded

How the examiner actually awards the marks on this topic.

Gradora's own decode of the marking approach — not the exam board's published mark scheme.

Model answer

A full-mark response to the question above, worked through step by step.

⏱Timing: Allocate approximately 5-7 minutes per mark to ensure you have enough time to complete each question accurately.

1. Identify the values of R and C. R = 2 kΩ = 2000 Ω C = 10 μF = 10 × 10 -6 F0 marks
2. Calculate the time constant (τ). τ = RC τ = 2000 Ω × 10 × 10 -6 F τ = 0.02 s or 20 ms3 marks
3. The time it takes for the charge to reach 63.2% of its maximum value is equal to one time constant. Time = τ = 0.02 s1 mark

Final answer: The time constant (τ) is 0.02 s, and the time it takes for the charge to reach 63.2% of its maximum value is 0.02 s.

✓Work through every step correctly and you earn all 4 marks.

Common mistakes

• Confusing the time constant (τ) with the time to halve the charge, voltage, or current.

  Why it happens: Students often mix up the time constant (τ = RC) with the time it takes for the charge, voltage, or current to halve (T = 0.69RC).

  Fix: Remember that the time constant (τ) is the time it takes for the charge, voltage, or current to reach approximately 63.2% of its final value during charging, or to fall to 36.8% of its initial value during discharging. The time to halve these values is given by T = 0.69RC.

• Incorrectly interpreting the gradient of a Q-t graph as voltage instead of current.

  Why it happens: Students sometimes confuse the physical meaning of the gradient in different graphs, leading to incorrect interpretations.

  Fix: The gradient of a Q-t graph represents the current I at any time t. Since I = dQ/dt, a steeper slope indicates a higher current.

• Forgetting to use exponential functions when calculating charge, voltage, or current during charging and discharging.

  Why it happens: Students may rely on linear relationships instead of the correct exponential formulas for RC circuits.

  Fix: Use the appropriate exponential formulas for charging and discharging: - Charging: Q = Q 0 (1 - e -t/RC ), V = V 0 (1 - e -t/RC ), I = I 0 e -t/RC - Discharging: Q = Q 0 e -t/RC , V = V 0 e -t/RC , I = I 0 e -t/RC

• Incorrectly calculating the time constant from a log-linear plot.

  Why it happens: Students may misinterpret the slope of the ln(Q) vs. t graph or use incorrect units.

  Fix: To determine the time constant (RC) from a log-linear plot, calculate the slope of the ln(Q) vs. t graph and use the formula RC = -1 / (slope). Ensure that all units are consistent.

• Using the wrong initial conditions in exponential equations.

  Why it happens: Students may forget to use the correct initial values for charge, voltage, or current when applying exponential formulas.

  Fix: Always check the initial conditions (Q 0 , V 0 , I 0 ) and ensure they are correctly substituted into the exponential equations.

• Wrongly assuming the area under a V-t graph represents energy stored in the capacitor.

  Why it happens: Students may mix up the physical meaning of areas under different graphs, leading to incorrect interpretations.

  Fix: The area under an I-t graph represents the total charge that has flowed through the circuit. The area under a V-t graph does NOT represent energy stored; the energy stored in a capacitor is ½QV = ½CV² = ½Q²/C, equal to the area under a charge–voltage (Q–V) graph.

Where the marks go

The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.

Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.

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