OCR A Level Physics A: Charging, discharging and exponential decay in RC circuits (H556) – 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 6.1.3, Charging and discharging capacitors, of the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. All data needed are given in the questions. Revise first with the study guide or the revision notes. Course pages: OCR A Level Physics hub and checklist.

Questions

1. (a) Show that the unit of CR is the second. [2] (b) Calculate the time constant for a 4.7 µF capacitor and a 6.8 MΩ resistor. [2]

2. A 330 µF capacitor charged to 15 V discharges through a 27 kΩ resistor. (a) Calculate the time constant and the initial charge. [2] (b) Calculate the initial current. [1] (c) Calculate the p.d. across the capacitor 12 s after discharge begins. [2]

3. An uncharged capacitor is charged from a cell through a resistor. Describe how the current and the p.d. across the capacitor change from the moment the switch is closed. [4]

4. An uncharged 47 µF capacitor is connected through a 100 kΩ resistor to a 6.0 V supply. (a) Find the time constant. [1] (b) Calculate the time taken for the p.d. across the capacitor to reach 4.5 V. [3] (c) Calculate the current at that time. [1]

5. A capacitor discharges through a 47 kΩ resistor. The table shows the p.d. V at time t.

t / s	0	20	40	60	80	100
V / V	8.0	4.3	2.3	1.2	0.66	0.35
ln(V / V)	2.08	1.46	0.83		−0.42

(a) Calculate ln(V / V) for t = 60 s and t = 100 s. [2] (b) Use the points at t = 0 and t = 100 s to find the gradient of the ln V against t graph. [2] (c) Determine the capacitance. [2]

6. The p.d. across a discharging capacitor is 8.0 V at t = 0 and 5.0 V at t = 20 s. (a) Predict the p.d. at t = 40 s. [1] (b) State the property of the decay that justifies your prediction. [1] (c) Calculate the time constant. [2] (d) Calculate the time for the p.d. to halve. [1]

7. A student investigates the discharge of a capacitor. (a) Describe how a data-logger can be used to obtain the data, and how they find the time constant from it. [3] (b) State two advantages of a data-logger over a voltmeter and a stopwatch. [2] (c) Explain why the voltmeter used must have a very large resistance. [1]

8. A capacitor C discharges through a resistor R. (a) Show that ΔQ/Δt = −Q/CR. [3] A spreadsheet models the discharge with C = 400 µF, R = 15 kΩ, an initial p.d. of 6.0 V and a time step Δt = 1.5 s. (b) Calculate Q at t = 1.5 s, 3.0 s and 4.5 s. [3] (c) The exact charge at 4.5 s is 1.13 mC. Compare it with your result and state how to improve the model. [2]

9. A discharge current I (in µA) is plotted as lg I against t. The line has gradient −0.0145 s⁻¹ and intercept 2.00 on the lg I axis. The resistor is 47 kΩ. (a) Calculate CR. [2] (b) Calculate the initial current. [1] (c) Calculate the initial p.d. across the capacitor. [1] (d) Calculate the capacitance. [2]

10. A 2200 µF capacitor, initially uncharged, charges from a 9.0 V supply through a 4.7 kΩ resistor. After 15 s the supply is replaced by a 12 kΩ resistor only, which discharges the capacitor. (a) Calculate the charging time constant. [1] (b) Calculate the p.d. across the capacitor after 15 s. [2] (c) Calculate the charge on the capacitor at 15 s. [1] (d) Calculate the charging current at 15 s. [2] (e) Calculate the time for the p.d. to fall to 2.0 V once discharge through the 12 kΩ resistor begins. [3]

11. (a) Show that the time for a discharging capacitor to lose half its charge is CR ln 2. [2] (b) A 120 µF capacitor loses half its charge in 3.6 s. Calculate R. [2] (c) Calculate the percentage of the charge left after 10.8 s. [1]

Answers

1. (a) C = Q/V and R = V/I, so CR = (Q/V)(V/I) = Q/I [1]. Charge divided by current is time (Q = It), so the unit is the second [1] [2] (b) CR = 4.7 × 10⁻⁶ × 6.8 × 10⁶ [1] = 32 s [1] [2] Examiner insight: A “show that” answer should start from the definitions of C and R and reach the unit; stating “CR is a time” scores nothing by itself.

2. (a) τ = 330 × 10⁻⁶ × 27 × 10³ = 8.9 s [1]; Q₀ = CV₀ = 330 × 10⁻⁶ × 15 = 4.95 × 10⁻³ C [1] [2] (b) I₀ = V₀/R = 15 / 27 000 = 5.6 × 10⁻⁴ A [1] [1] (c) V = 15 e^(−12/8.91) [1] = 3.9 V [1] [2] Examiner insight: Convert µF and kΩ before multiplying, and write the unit on each answer. Using the unrounded τ in part (c) avoids a rounding drift.

3. The current is at its maximum (cell p.d. ÷ R) at the start [1], and falls with time, quickly at first and then more slowly [1]. The p.d. across the capacitor starts at zero and rises towards the cell p.d., more slowly as time goes on [1]. Cell p.d. = p.d. across R + p.d. across C, so as the p.d. across C rises, that across R and the current fall [1] [4] Examiner insight: Mention both quantities and say “approaches” a limit; a statement that the current “stops” at a definite time misdescribes an exponential.

4. (a) τ = 47 × 10⁻⁶ × 100 × 10³ = 4.7 s [1] [1] (b) 4.5 = 6.0(1 − e^(−t/τ)), so e^(−t/τ) = 0.25 [1]; t = τ ln 4 [1] = 6.5 s [1] [3] (c) V_R = 6.0 − 4.5 = 1.5 V, so I = 1.5 / 100 000 = 1.5 × 10⁻⁵ A [1] [1] Examiner insight: In part (b), isolate the exponential before taking logs. Part (c) is quickest through the p.d. across R, not the exponential.

5. (a) ln 1.2 = 0.18; ln 0.35 = −1.05 [1] [1] [2] (b) Gradient = (−1.05 − 2.08) / (100 − 0) [1] = −0.0313 s⁻¹ [1] [2] (c) CR = 1 / 0.0313 = 32 s [1]; C = 32 / 47 000 = 6.8 × 10⁻⁴ F [1] [2] Examiner insight: Keep the minus sign in the gradient and invert it for CR. Reading CR as 0.0313 s, or as a negative time, loses the answer.

6. (a) Ratio 5.0/8.0 = 0.625, so V = 5.0 × 0.625 = 3.1 V [1] [1] (b) An exponential decay has a constant ratio over equal time intervals [1] [1] (c) 0.625 = e^(−20/τ) [1]; τ = 20 / ln(1.6) = 43 s [1] [2] (d) t½ = 0.693 × 42.6 = 29 s [1] [1] Examiner insight: Name the property in (b), not just the arithmetic. A linear fall would give 2.0 V at 40 s, so state why that is wrong.

7. (a) Connect a voltage sensor across the capacitor, connected to the data-logger [1]. Set a sampling interval short compared with CR and record V as the capacitor discharges [1]. Plot ln V against t; the gradient is −1/CR [1] [3] (b) Any two of: no reaction-time error in timing; many closely spaced readings; data plotted and processed immediately [1] [1] [2] (c) A low-resistance voltmeter would take current from the capacitor, discharging it faster, so the measured CR would be too small [1] [1] Examiner insight: “More accurate” is too vague for part (b). State the specific source of error removed.

8. (a) I = −ΔQ/Δt (charge leaves the capacitor) [1]. V = IR and V = Q/C [1]. So Q/C = −R ΔQ/Δt, which gives ΔQ/Δt = −Q/CR [1] [3] (b) CR = 400 × 10⁻⁶ × 15 × 10³ = 6.0 s; Q₀ = 400 × 10⁻⁶ × 6.0 = 2.4 mC [1]. ΔQ = −(2.4/6.0) × 1.5 = −0.60 mC, so Q = 1.8 mC at 1.5 s [1]. Then 1.35 mC at 3.0 s and 1.0125 mC (1.01 mC) at 4.5 s [1] [3] (c) The model gives 1.01 mC, about 10% below the exact 1.13 mC [1]. Use a smaller time step, which makes the assumed rate closer to the true rate during each step [1] [2] Examiner insight: Compare numbers with a percentage or a difference, and state the direction of the error.

9. (a) lg I = lg I₀ − (lg e) t/CR, so gradient = −0.434/CR [1]. CR = 0.434 / 0.0145 = 30 s [1] [2] (b) I₀ = 10^2.00 = 100 µA [1] [1] (c) V₀ = I₀R = 100 × 10⁻⁶ × 47 000 = 4.7 V [1] [1] (d) C = CR/R = 30 / 47 000 [1] = 6.4 × 10⁻⁴ F [1] [2] Examiner insight: Check the log base before inverting the gradient. A base-10 plot needs the factor 0.434.

10. (a) τ = 2200 × 10⁻⁶ × 4.7 × 10³ = 10.3 s [1] [1] (b) V_C = 9.0(1 − e^(−15/10.34)) [1] = 9.0 × 0.7656 = 6.9 V [1] [2] (c) Q = CV = 2200 × 10⁻⁶ × 6.89 = 1.5 × 10⁻² C [1] [1] (d) V_R = 9.0 − 6.89 = 2.11 V [1]; I = 2.11 / 4700 = 4.5 × 10⁻⁴ A [1] [2] (e) New τ = 2200 × 10⁻⁶ × 12 × 10³ = 26.4 s [1]. 2.0 = 6.89 e^(−t/26.4) [1]; t = 26.4 ln(6.89/2.0) = 33 s [1] [3] Examiner insight: In a multi-stage problem, the final value of one stage becomes x₀ for the next, and the time constant changes with the resistor.

11. (a) Q₀/2 = Q₀ e^(−t/CR) gives e^(−t/CR) = 1/2 [1]. Taking logs, t/CR = ln 2, so t = CR ln 2 [1] [2] (b) R = t½ / (C ln 2) = 3.6 / (120 × 10⁻⁶ × 0.693) [1] = 4.3 × 10⁴ Ω [1] [2] (c) 10.8 s is three half-times, so (1/2)³ = 0.125, which is 12.5% [1] [1] Examiner insight: Part (c) is quickest by the constant-ratio property. It needs no value of R or C.

Where marks are usually lost

• Leaving C in µF or R in kΩ inside CR.
• Using x₀ as the start value in a charging equation when it is the final value.
• Applying the charging equation to the current.
• Reading CR as the gradient of a ln graph, or dropping its negative sign.
• Using a base-10 log graph without the 0.434 factor.
• Writing “more accurate” for a data-logger without the specific reason.
• Not stating the direction of error in a spreadsheet model.
• Treating one time constant as complete discharge.

Next steps

• Revise with the revision notes and the study guide.
• Check the OCR A Level Physics hub and the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

This page is aligned to the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), first assessment 2017, published by OCR, section 6.1.3.