OCR A Level Physics A: Capacitance, capacitor combinations and stored energy (H556) – Revision Notes

Condensed recall for OCR A Level Physics A (H556), specification Version 3.0 (March 2026), sections 6.1.1(a)-(e) and 6.1.2(a)-(c). Fuller explanations and longer examples are in the study guide. Test yourself with the practice questions, tick topics off on the printable checklist, find the course on the OCR A Level Physics hub, and try the free diagnostics.

Key equations

Quantity	Equation	Notes
Capacitance	C = Q/V	Q is the charge on one plate; unit farad, F = C V⁻¹
Series	1/C = 1/C₁ + 1/C₂ + …	Same Q on each; p.d.s add; total below the smallest
Parallel	C = C₁ + C₂ + …	Same V across each; charges add; total above the largest
Energy	W = ½QV	Area under the V-Q graph
Energy	W = ½Q²/C	Use when Q and C are known
Energy	W = ½V²C	Use when V and C are known

Prefixes: µ = 10⁻⁶, n = 10⁻⁹, p = 10⁻¹². Convert to farads before substituting.

6.1.1(a) Capacitance

• C = Q/V: charge stored per unit p.d.
• 1 F = 1 C V⁻¹.
• Q is proportional to V for a given capacitor. The Q-V graph is a straight line from the origin whose gradient is C.
• The capacitor as a whole is neutral. +Q on one plate, −Q on the other.

Reminder. 220 µF holding 0.68 mC:

V = Q/C = 0.68 × 10⁻³ / 220 × 10⁻⁶ = 3.1 V

6.1.1(b) Electron flow

Charging from a supply

• Electrons move from the supply’s negative terminal onto the connected plate.
• Electrons leave the other plate towards the positive terminal.
• No charge crosses the gap.
• Current falls as the capacitor’s p.d. rises. It stops when that p.d. equals the supply e.m.f.

Discharging through a resistor

• Electrons flow from the negative plate round the external circuit to the positive plate.
• This is the reverse direction to the charging current.
• Current and p.d. fall to zero as the plates become neutral.

6.1.1(c) and (d) Combinations

Series

• Charge: the same on every capacitor.
• P.d.: V = V₁ + V₂ + …
• Smaller capacitance takes the larger share of the p.d.
• Two capacitors: C = C₁C₂/(C₁ + C₂).

Parallel

• P.d.: the same across every capacitor.
• Charge: Q = Q₁ + Q₂ + …
• Add the capacitances.

Method for a mixed network

1. Find the innermost series or parallel group.
2. Replace it with one equivalent capacitor.
3. Repeat until one capacitor remains.
4. Use Q = CV with the supply p.d. to find the total charge.
5. Work back outwards, using “same charge in series” and “same p.d. in parallel”.

Reminders.

Series 2.2, 6.8, 1.0 µF:   1/C = 0.4545 + 0.1471 + 1.0 = 1.6016    C = 0.62 µF
Parallel 2.2, 6.8, 1.0 µF: C = 10 µF
3.0 µF and 6.0 µF in series across 12 V: C = 2.0 µF, Q = 24 µC,
   V(3.0 µF) = 8.0 V, V(6.0 µF) = 4.0 V

6.1.1(e)(i) Circuits with resistors

For a d.c. circuit in the steady state:

• A fully charged capacitor carries no current, so its branch behaves like a break.
• A resistor in series with a charged capacitor has no p.d. across it.
• A capacitor in parallel with a resistor has the same p.d. as that resistor.
• Find that p.d. with V = IR or a potential divider, then use Q = CV.

Joining capacitors. Charged capacitor joined in parallel to an uncharged one: total Q is conserved, final V = total Q / total C. Some energy is transferred to the surroundings as thermal energy and radiation, so the total stored falls.

6.1.1(e)(ii) Practical investigation

Reed-switch method

1. A reed switch of frequency f alternately connects the capacitor to a supply of p.d. V and to a microammeter.
2. The mean current is I = fQ = fCV.
3. So C = I/(fV). A graph of I against V has gradient fC.
4. Repeat with two capacitors in series, then in parallel. Compare with the combination rules.

Points to state

• A voltmeter across the capacitor must have a very high resistance.
• Check the polarity of electrolytic capacitors and stay below the rated voltage.
• Discharge capacitors before rewiring.
• Compare measured and predicted C within the uncertainties.

6.1.2(a) The V-Q graph

• Plot p.d. (vertical) against charge (horizontal). The line starts at the origin and its gradient is 1/C.
• Work done to add charge ΔQ at p.d. V is V ΔQ, a thin strip of area.
• Energy stored = area under the graph.
• From zero: triangle, ½QV.
• Between Q₁ and Q₂: trapezium, ½(V₁ + V₂)(Q₂ − Q₁).

6.1.2(b) Energy stored

Three forms: W = ½QV, W = ½Q²/C, W = ½V²C.

Q = 1.5 mC, C = 250 µF:  W = Q²/2C = (1.5 × 10⁻³)² / (2 × 250 × 10⁻⁶) = 4.5 × 10⁻³ J
W = 0.45 J at 30 V:      C = 2W/V² = 2 × 0.45 / 30² = 1.0 × 10⁻³ F

• At fixed C, doubling V gives four times the energy.
• At fixed Q, W = ½Q²/C, so halving C doubles the energy.
• A battery of e.m.f. V that charges a capacitor to V supplies QV. Half is stored and half is transferred as thermal energy in the circuit.

6.1.2(c) Uses as energy stores

• Defibrillator: slow charge, short high-power pulse through the patient.
• Camera flash: charged over seconds, discharged in a fraction of a second.
• Back-up supply for memory circuits during a short interruption.
• Supercapacitors: store energy from braking and release it quickly.
• Compared with a battery of the same mass a capacitor stores far less energy, but it can release it much faster.

Must-know distinctions

• Series versus parallel. Series: same charge, p.d.s add, reciprocals add. Parallel: same p.d., charges add, capacitances add.
• Charge stored versus charge flowing. The capacitor holds +Q and −Q, so its net charge is zero. The charge that flows is Q.
• Energy stored versus energy supplied. Stored: ½QV. Supplied by a battery: QV.
• Which equation for W. Choose the one that uses your two known quantities.

Quick self-test

1. State the definition of capacitance.
2. A 1.2 mF capacitor is charged to 5.0 V. What charge is on each plate?
3. Find the total capacitance of 3.0 µF and 6.0 µF in parallel, then in series.
4. Two capacitors are in series. Which takes the larger p.d., the larger or the smaller?
5. In terms of electrons, describe what happens at the plates when a capacitor is charged by a cell.
6. A capacitor stores 4.5 mJ when carrying 1.5 mC. What is its capacitance?
7. What does the area under a V-Q graph represent?
8. A fully charged capacitor is in series with a 10 kΩ resistor and a 6.0 V cell. What is the p.d. across the resistor?
9. A battery charges a capacitor to 9.0 V and transfers 0.12 J in total. How much energy is stored?
10. Why is a capacitor better than a battery for a camera flash?

Answers

1. Charge stored per unit p.d. across the capacitor, C = Q/V.
2. Q = CV = 1.2 × 10⁻³ × 5.0 = 6.0 × 10⁻³ C (6.0 mC).
3. Parallel: 9.0 µF. Series: 1/C = 1/3.0 + 1/6.0 = 1/2.0, so 2.0 µF.
4. The smaller capacitor, because V = Q/C with the same Q.
5. Electrons flow onto the plate joined to the negative terminal and off the other plate towards the positive terminal. No electrons cross the gap.
6. C = Q²/2W = (1.5 × 10⁻³)² / (2 × 4.5 × 10⁻³) = 2.5 × 10⁻⁴ F (250 µF).
7. The energy stored in the capacitor.
8. Zero. No current flows in the series branch when the capacitor is fully charged, so V = IR gives 0 V.
9. Half of 0.12 J: 0.060 J.
10. It can release its stored energy in a very short time, giving a high power pulse.

Where marks are usually lost

• Leaving capacitance in µF or mF when substituting.
• Adding capacitances in series, or adding reciprocals in parallel.
• Stopping at 1/C and giving the reciprocal as the answer.
• Saying that electrons flow through the capacitor or across the dielectric.
• Using the cell e.m.f. as the p.d. across a capacitor that sits across only one resistor.
• Using W = QV for the energy stored.
• Dropping the square when using W = ½V²C or W = ½Q²/C.
• Giving the series total as larger than the smallest capacitor and not noticing.
• Describing a practical method without saying what is measured and how C is calculated.

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.