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

The question

A capacitor with a capacitance of 10 μF is charged through a resistor of 2 kΩ. Calculate the time constant (τ) and the time it takes for the charge to reach 63.2% of its maximum value. [Paraphrased for study — not reproduced from any exam paper.]

Mark scheme, decoded

What each mark is really for — in plain English — and the wording trap that loses it.

• S1

  Step 1: Identify the values of R and C. R = 2 kΩ = 2000 Ω C = 10 μF = 10 × 10 -6 F

• S2

  Step 2: Calculate the time constant (τ). τ = RC τ = 2000 Ω × 10 × 10 -6 F τ = 0.02 s or 20 ms

• S3

  Step 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 s

Model answer

Worked through, with each step tagged to the mark it earns.

1. S1

   Step 1: Identify the values of R and C. R = 2 kΩ = 2000 Ω C = 10 μF = 10 × 10 -6 F

2. S2

   Step 2: Calculate the time constant (τ). τ = RC τ = 2000 Ω × 10 × 10 -6 F τ = 0.02 s or 20 ms

3. S3

   Step 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 s

4. 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.

Common mistakes

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

• Full worked solution (all marking points)4 marks

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