OCR A Level Physics A: Capacitance, capacitor combinations and stored energy (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 follow the OCR A Level Physics A (H556) specification, Version 3.0 (March 2026), sections 6.1.1(a)-(e) and 6.1.2(a)-(c): capacitance, electron flow, series and parallel capacitors, circuits with resistors, practical investigation and stored energy. Time-dependent charging through a resistor is taught separately.

Links: study guide, revision notes, OCR A Level Physics hub, printable checklist and the free diagnostics.

Questions

1. (a) State what is meant by one farad. [1] (b) A 47 µF capacitor holds 0.28 mC on its positive plate. Calculate the p.d. across it. [2]

2. An uncharged capacitor is connected through a resistor to a 9.0 V battery. Explain, in terms of electrons, how the capacitor becomes charged and why the current eventually stops. [4]

3. The battery in question 2 is removed and the two plates are joined through the resistor. Describe the movement of electrons and explain what happens to the current. [3]

4. A 12 µF capacitor and a 24 µF capacitor are connected in series across a 27 V supply. (a) Calculate the total capacitance. [2] (b) Calculate the charge on the 12 µF capacitor. [1] (c) Calculate the p.d. across each capacitor. [2]

5. An 8.0 µF capacitor is connected in parallel with a combination of a 6.0 µF capacitor and a 12 µF capacitor in series. The network is connected across a 10 V supply. (a) Calculate the total capacitance of the network. [3] (b) Calculate the total charge supplied by the source. [1] (c) Calculate the charge on the 12 µF capacitor. [2]

6. A 9.0 V cell of negligible internal resistance is connected in series with a 1.5 kΩ resistor and a 3.3 kΩ resistor. A 220 µF capacitor is connected across the 3.3 kΩ resistor and has been connected for a long time. (a) Calculate the p.d. across the capacitor. [3] (b) Calculate the charge on the capacitor. [2] (c) State the p.d. across the 1.5 kΩ resistor. [1]

7. Readings of p.d. V and charge Q for a capacitor are shown.

Q / µC	0	20	40	60	80
V / V	0	1.0	2.0	3.0	4.0

(a) Use a gradient to find the capacitance. [2] (b) Calculate the energy stored at 4.0 V. [2] (c) Calculate the extra energy needed to raise the p.d. from 2.0 V to 4.0 V. [2]

8. A 4.0 µF capacitor is charged to 9.0 V. It is then disconnected from the supply and connected in parallel with an uncharged 2.0 µF capacitor. (a) Calculate the initial charge on the 4.0 µF capacitor. [1] (b) Calculate the final p.d. across the pair. [2] (c) Calculate the energy stored before and after the connection. [2] (d) Calculate the energy difference and suggest where it goes. [2]

9. A student investigates capacitors using a reed switch of frequency 50 Hz that connects the capacitor alternately to a 5.0 V supply and to a microammeter. (a) Explain how the capacitance of one capacitor can be found from the meter reading. [3] (b) The student has a 2.2 µF and a 4.7 µF capacitor. Calculate the readings expected when the two are connected in series and then in parallel, to 2 significant figures. [3] (c) State two precautions when using capacitors in this investigation. [2]

10. A 6.0 µF capacitor and a 3.0 µF capacitor are connected in series across a 12 V supply. (a) Calculate the charge on each capacitor. [2] (b) Calculate the p.d. across each capacitor. [2] (c) Show that the energy stored in the 3.0 µF capacitor is twice that in the 6.0 µF capacitor, and that the total equals ½CV² for the combination. [2]

11. A camera flash unit uses a 0.90 mF capacitor to store 15 J. (a) Calculate the p.d. to which it must be charged, to 2 significant figures. [3] (b) Explain why a capacitor is suitable for this use. [2] (c) The energy is released in 2.0 ms. Calculate the mean power. [2]

Answers

1. (a) One farad is the capacitance when a charge of one coulomb is stored per volt of p.d. (1 F = 1 C V⁻¹) [1]. (b) V = Q/C = 0.28 × 10⁻³ / 47 × 10⁻⁶ [1] = 6.0 V (5.96 V) [1]. [3] Examiner insight: Convert mC and µF to base units before dividing, and write the substituted line so your working is visible.

2. Electrons flow from the negative terminal of the battery onto the plate connected to it [1]. Electrons leave the other plate and flow towards the positive terminal, leaving that plate positive [1]. No electrons cross the insulating gap between the plates [1]. The current stops when the p.d. across the capacitor equals the battery e.m.f., so there is no net p.d. driving charge round the circuit [1]. [4] Examiner insight: Name the direction of the electrons and what happens on each plate. “Charge flows through the capacitor” is the typical error.

3. Electrons flow from the negative plate, through the resistor, to the positive plate [1]. This is the opposite direction to the charging current [1]. The current decreases to zero as the plates become neutral and the p.d. falls to zero [1]. [3] Examiner insight: State both that the direction reverses and that the current decreases, since they are separate points.

4. (a) 1/C = 1/12 + 1/24 = 3/24 [1], so C = 8.0 µF [1]. (b) Q = CV = 8.0 × 10⁻⁶ × 27 = 2.2 × 10⁻⁴ C (216 µC) [1]. (c) The charge is the same on each: V = Q/C. For 12 µF: 216 / 12 = 18 V [1]. For 24 µF: 216 / 24 = 9.0 V [1]. [5] Examiner insight: Check that the p.d.s sum to the supply (18 + 9.0 = 27 V). The smaller capacitor takes the larger p.d.

5. (a) Series pair: 1/C = 1/6.0 + 1/12 = 3/12, so C = 4.0 µF [1]. Parallel with 8.0 µF: C = 8.0 + 4.0 [1] = 12 µF [1]. (b) Q = CV = 12 × 10⁻⁶ × 10 = 1.2 × 10⁻⁴ C [1]. (c) The series branch has C = 4.0 µF, so its charge is 4.0 × 10⁻⁶ × 10 = 4.0 × 10⁻⁵ C [1]. The same charge is on each capacitor in series, so the 12 µF capacitor holds 4.0 × 10⁻⁵ C (40 µC) [1]. [6] Examiner insight: In part (c) the 12 µF capacitor does not have 10 V across it. Reduce the branch first, then use “same charge in series”.

6. (a) Total resistance = 1500 + 3300 = 4800 Ω [1]. Steady current I = 9.0 / 4800 = 1.875 × 10⁻³ A [1]. V = IR = 1.875 × 10⁻³ × 3300 = 6.2 V (6.19 V) [1]. (b) Q = CV = 220 × 10⁻⁶ × 6.19 [1] = 1.4 × 10⁻³ C [1]. (c) 9.0 − 6.19 = 2.8 V [1]. [6] Examiner insight: The capacitor charges to the p.d. across the resistor it is connected to, not the cell e.m.f. Say that no steady current enters its branch, so the resistors behave as a simple series pair.

7. (a) Gradient of V against Q = 4.0 / 80 µC = 0.050 V µC⁻¹ [1]. C = 1/gradient = 20 µF [1]. (b) W = ½QV = 0.5 × 80 × 10⁻⁶ × 4.0 [1] = 1.6 × 10⁻⁴ J (160 µJ) [1]. (c) The area between V = 2.0 V and 4.0 V is a trapezium: ½(2.0 + 4.0) × (80 − 40) × 10⁻⁶ [1] = 1.2 × 10⁻⁴ J (120 µJ) [1]. [6] Examiner insight: The gradient of a V-Q graph is 1/C, not C.

8. (a) Q = CV = 4.0 × 9.0 = 36 µC [1]. (b) Total capacitance = 4.0 + 2.0 = 6.0 µF [1]. Charge is conserved, so V = 36 / 6.0 = 6.0 V [1]. (c) Before: ½ × 4.0 × 10⁻⁶ × 9.0² = 1.6 × 10⁻⁴ J (162 µJ) [1]. After: ½ × 6.0 × 10⁻⁶ × 6.0² = 1.1 × 10⁻⁴ J (108 µJ) [1]. (d) Difference = 162 − 108 = 54 µJ [1]. It is transferred as thermal energy in the connecting wires (and some is radiated) as charge flows between the capacitors [1]. [7] Examiner insight: The stored energy falls, so say where the 54 µJ goes. Saying only that energy is conserved is incomplete.

9. (a) Each cycle the capacitor is charged to V and then discharged through the meter [1]. The meter reads the mean current I = fQ = fCV [1]. So C = I / (fV), using the known frequency and supply p.d. [1]. (b) Series: C = (2.2 × 4.7) / (2.2 + 4.7) = 1.5 µF [1]; I = fCV = 50 × 1.4986 × 10⁻⁶ × 5.0 = 3.7 × 10⁻⁴ A [1]. Parallel: C = 6.9 µF, so I = 50 × 6.9 × 10⁻⁶ × 5.0 = 1.7 × 10⁻³ A [1]. (c) Any two from: check the polarity of electrolytic capacitors; do not exceed the rated voltage; discharge capacitors before changing the circuit; use a high-resistance voltmeter across the capacitor [1][1]. [8] Examiner insight: Part (a) asks you to explain, so link the meter reading to charge per second rather than only describing the apparatus.

10. (a) 1/C = 1/6.0 + 1/3.0 = 3/6.0, so C = 2.0 µF [1]. Q = CV = 2.0 × 10⁻⁶ × 12 = 2.4 × 10⁻⁵ C, the same on each capacitor [1]. (b) 6.0 µF: V = Q/C = 24 / 6.0 = 4.0 V [1]. 3.0 µF: 24 / 3.0 = 8.0 V [1]. (c) W = ½QV: 6.0 µF: ½ × 24 × 4.0 = 48 µJ; 3.0 µF: ½ × 24 × 8.0 = 96 µJ, which is twice 48 µJ [1]. Total = 144 µJ, and ½CV² = 0.5 × 2.0 × 10⁻⁶ × 12² = 144 µJ [1]. [6] Examiner insight: On a “show that” question, give both energies and the final comparison. Saying “the answers match” demonstrates nothing.

11. (a) W = ½V²C, so V = √(2W/C) [1] = √(2 × 15 / 0.90 × 10⁻³) [1] = 1.8 × 10² V (183 V) [1]. (b) It can release its stored energy in a very short time [1], giving a high-power pulse that a small battery could not supply directly [1]. (c) P = W/t = 15 / 2.0 × 10⁻³ [1] = 7.5 × 10³ W [1]. [7] Examiner insight: Use the energy equation containing W and C, and rearrange before substituting.

Where marks are usually lost

• Leaving µF or mF values unconverted.
• Adding capacitances for a series pair, or using 1/C = C₁ + C₂.
• Quoting 1/C as the final answer.
• Using the supply p.d. across a capacitor that sits across only part of the circuit.
• Saying that charge crosses the dielectric.
• Using W = QV or forgetting the square in ½CV².
• Using the gradient of a V-Q graph as C instead of 1/C.
• Describing a practical method without saying how C is calculated.

Next steps

• Review the revision notes and re-read the study guide.
• Tick topics off on the printable checklist and see the OCR A Level Physics hub.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, Advanced GCE in Physics A (H556) specification, Version 3.0, March 2026, first assessment 2017, sections 6.1.1 and 6.1.2.