OxfordAQA International AS and A-level Physics: Capacitor charge and discharge through resistors (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.9.1, Capacitor charge and discharge, of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, including Required practical 6. The specification marks section 3.9 as International A-level only. Learn the method in the study guide; recall it with the revision notes.

Questions

1. (a) Explain what the time constant tells you about a capacitor discharging through a resistor. [1] (b) Calculate the time constant for a 3300 µF capacitor and a 1.5 kΩ resistor. [2]

2. An uncharged capacitor is connected in series with a resistor to a supply of constant e.m.f. V₀. Describe how the current varies with time. [2]

3. A 100 µF capacitor is charged to 20 V and then discharged through a 330 kΩ resistor. (a) Calculate the initial charge and the initial current. [2] (b) Calculate the p.d. across the capacitor 50 s after discharge starts. [2]

4. A 1200 µF capacitor discharges through a 39 kΩ resistor. (a) Calculate the time for the charge to halve. [2] (b) Calculate the time for the charge to fall to one eighth of its initial value. [2]

5. An uncharged 1800 µF capacitor is connected through an 8.2 kΩ resistor to a 7.5 V supply of negligible internal resistance. (a) Calculate the time constant and the final charge on the capacitor. [2] (b) Calculate the charge on the capacitor after 10 s. [2] (c) Calculate the current in the resistor after 10 s. [2]

6. A 1500 µF capacitor discharges through a resistor. The initial current is 480 µA. A tangent drawn to the current–time graph at t = 0 meets the time axis at t = 25 s. (a) State the time constant. [1] (b) Calculate the initial charge on the capacitor, explaining the method you use. [2] (c) Find the p.d. across the capacitor at t = 0. [1] (d) Calculate the resistance of the resistor. [1]

7. A capacitor charged to 9.0 V discharges through a 91 kΩ resistor. The p.d. V is recorded at 30 s intervals.

t / s	0	30	60	90	120	150
V / V	9.0	5.0	2.8	1.6	0.88	0.49
ln of V value	2.20	1.61	1.03	0.47

(a) Show that a graph of ln V against t should be a straight line, and state what its gradient represents. [2] (b) Complete the last two entries in the table. [1] (c) Use the end points of the table to find the gradient. [2] (d) Find the time constant. [1] (e) Find the capacitance. [1]

8. Describe an investigation of the discharge of a capacitor through a resistor that finds RC from a log-linear graph. Include one precaution. [6]

9. A 100 µF capacitor discharges through a 1.0 MΩ resistor. A voltmeter of resistance 10 MΩ is connected across the capacitor to measure its p.d. (a) Calculate the total resistance the capacitor discharges through. [2] (b) Calculate the new time constant and the percentage by which it is smaller than 100 s. [2]

10. A capacitor charges through a resistor from a supply of constant e.m.f. V₀. The p.d. V across the capacitor is measured. A graph of ln(V₀ − V) against t is a straight line with gradient −0.0250 s⁻¹ and intercept 2.303 on the ln axis. (a) Show that ln(V₀ − V) = ln V₀ − t/RC. [2] (b) Find V₀. [1] (c) Find the time constant. [1] (d) The resistance is 80 kΩ. Find the capacitance. [1]

11. A 1000 µF capacitor, initially uncharged, is charged for 15 s through a 10 kΩ resistor from a 12.0 V supply of negligible internal resistance. The supply is then removed and the capacitor is discharged through a 22 kΩ resistor. (a) Calculate the p.d. across the capacitor at the end of the charging. [2] (b) Calculate the charge stored at this time. [1] (c) Calculate the p.d. 20 s after discharge starts. [2] (d) Calculate the time after discharge starts for the p.d. to fall to 1.0 V. [2] (e) Describe the graph of capacitor p.d. against time for the whole process. [2]

12. A 500 µF capacitor charged to 12 V discharges through a 40 kΩ resistor. The experiment is repeated with the resistor replaced by one of 80 kΩ. State and justify the change, with new values, in (i) the initial current, (ii) the initial charge, (iii) the time constant, (iv) the area under the current–time graph and (v) the time for the p.d. to halve. [5]

Answers

1. (a) The time constant is RC. It is the time in which the charge, p.d. or current drops to 1/e of where it started (roughly 0.37 of it) [1]. (b) RC = 3300 × 10⁻⁶ × 1.5 × 10³ [1] = 4.95 s [1]. [3] Examiner insight: Stating what happens at t = RC shows what the quantity means; a bare “RC” does not.

2. The current starts at its maximum value V₀/R [1] and falls exponentially towards zero, steepest at the start [1]. [2] Examiner insight: Give the starting value, shape and final value, not shape alone.

3. (a) Q₀ = CV₀ = 100 × 10⁻⁶ × 20 = 2.0 × 10⁻³ C [1]; I₀ = V₀/R = 20 / 330 000 = 6.1 × 10⁻⁵ A [1]. (b) RC = 330 000 × 100 × 10⁻⁶ = 33 s [1]. V = 20e^(−50/33) = 4.4 V [1]. [4] Examiner insight: Convert µF and kΩ first and give the unit with each answer.

4. (a) RC = 1200 × 10⁻⁶ × 39 × 10³ = 46.8 s; T½ = 0.693 × 46.8 [1] = 32 s [1]. (b) One eighth is three halvings [1]: 3 × 32.4 = 97 s [1]. [4] Examiner insight: One eighth is three half-times, a quick route; solving the exponential also works.

5. (a) RC = 8200 × 1800 × 10⁻⁶ = 14.8 s [1]. Final charge = CV₀ = 1800 × 10⁻⁶ × 7.5 = 1.35 × 10⁻² C [1]. (b) Q = 1.35 × 10⁻²(1 − e^(−10/14.76)) [1] = 6.6 × 10⁻³ C [1]. (c) I₀ = 7.5 / 8200 = 9.15 × 10⁻⁴ A [1]. I = 9.15 × 10⁻⁴ × e^(−10/14.76) = 4.6 × 10⁻⁴ A [1]. [6] Examiner insight: Check that the capacitor p.d. (3.7 V) and resistor p.d. (3.8 V) add to 7.5 V; this catches a swapped equation.

6. (a) The tangent at t = 0 meets the time axis at t = RC, so RC = 25 s [1]. (b) The area under the I–t graph is the charge, and for an exponential this is I₀ × RC [1] = 480 × 10⁻⁶ × 25 = 1.2 × 10⁻² C [1]. (c) V₀ = Q₀/C = 1.2 × 10⁻² / 1500 × 10⁻⁶ = 8.0 V [1]. (d) R = RC / C = 25 / 1.5 × 10⁻³ = 1.7 × 10⁴ Ω (17 kΩ) [1]. [5] Examiner insight: Say why I₀ × RC gives the charge; the reasoning about the area matters, not just the product.

7. (a) V = V₀e^(−t/RC), so ln V = ln V₀ − t/RC [1], which has the form y = mx + c, so plotting ln V on the vertical axis against t gives a line of gradient −1/RC [1]. (b) ln 0.88 = −0.13, ln 0.49 = −0.71 [1]. (c) Gradient = (−0.71 − 2.20) / (150 − 0) [1] = −0.0194 s⁻¹ [1]. (d) RC = 1 / 0.0194 = 52 s [1]. (e) C = RC / R = 51.5 / 91 000 = 5.7 × 10⁻⁴ F (570 µF) [1]. [7] Examiner insight: Keep the minus sign and the unit s⁻¹ on the gradient; then take the reciprocal and drop the sign when stating RC.

8. Connect a capacitor, resistor, supply, two-way switch and voltmeter (or voltage sensor with data logger) [1]. Charge the capacitor fully, then switch to discharge and start timing at the switch [1]. Record V at regular intervals over several time constants [1]. Calculate ln V and plot it against t [1]; a straight line confirms an exponential decay and the gradient is −1/RC, so RC = −1/gradient [1]. Precaution: use a high-resistance voltmeter or sensor so it does not drain the capacitor [1]. [6] Examiner insight: Say what is plotted and what the gradient means; “plot a graph” alone is too vague.

9. (a) The voltmeter and resistor are in parallel across the capacitor: 1/R = 1/1.0 + 1/10 [1], so R = 0.91 MΩ [1]. (b) RC = 0.909 × 10⁶ × 100 × 10⁻⁶ = 91 s [1]. It is 9.1 s shorter, so 9.1% smaller than 100 s [1]. [4] Examiner insight: A common slip is adding the two resistances as if in series; the current has two routes, so they are in parallel.

10. (a) V = V₀(1 − e^(−t/RC)), so V₀ − V = V₀e^(−t/RC) [1]. Taking natural logs: ln(V₀ − V) = ln V₀ − t/RC [1]. (b) ln V₀ = 2.303, so V₀ = e^2.303 = 10 V [1]. (c) RC = 1 / 0.0250 = 40 s [1]. (d) C = 40 / 80 000 = 5.0 × 10⁻⁴ F (500 µF) [1]. [5]

Examiner insight: In a “show that”, write every step: start from the charging equation and rearrange before taking logs.

11. (a) RC = 10 s [1]. V = 12.0(1 − e^(−15/10)) = 9.3 V [1]. (b) Q = CV = 1000 × 10⁻⁶ × 9.32 = 9.3 × 10⁻³ C [1]. (c) RC = 22 s [1]. V = 9.32 × e^(−20/22) = 3.8 V [1]. (d) 1.0 = 9.32e^(−t/22) [1]; t = 22 ln 9.32 = 49 s [1]. (e) The p.d. rises from 0 as a curve that flattens towards 9.3 V at 15 s [1]; then it falls from 9.3 V to 3.8 V at 35 s, more gently than it rose because the time constant is larger (22 s against 10 s) [1]. [9] Examiner insight: In a two-stage question, the end value of stage one is the start value of stage two, and the time constant changes with the resistor.

12. (i) I₀ = V₀/R halves, from 300 µA to 150 µA [1]. (ii) Q₀ = CV₀ is unchanged at 6.0 mC [1]. (iii) RC doubles, from 20 s to 40 s [1]. (iv) The area under the I–t graph is the charge, which is unchanged at 6.0 mC, so the lower current lasts longer [1]. (v) T½ = 0.693 RC doubles from 13.9 s to 27.7 s [1]. [5] Examiner insight: Justify each change with a formula and give new values; “it takes longer” alone is too vague.

Where marks are usually lost

• Not converting µF and kΩ to F and Ω, which gives an answer wrong by a power of ten.
• Mixing up the charging and discharging equations.
• Giving the gradient of ln V against t as +1/RC.
• Plotting ln V instead of ln(V₀ − V) for a charging capacitor.
• Saying the capacitor is discharged after one time constant; 37% is left.
• Describing a curve without its start value, or without saying it never reaches the end value.

Next steps

• Recall the formulas in the revision notes or re-read the study guide.
• Go to the OxfordAQA A-level Physics hub or tick topics off on the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by OxfordAQA. These questions follow section 3.9.1 and Required practical 6.