A 1000\ μF capacitor discharges through a 2.0\ k resistor. Find the time constant. [2 marks]

OCR OCR 2018-style practice: A 2200\ μF capacitor is charged to 12\ V. Calculate the charge stored and the energy stored.

Charge: Q = CV = (2200 × 10^-6)(12) = 2.64 × 10^-2\ C, about 0.026\ C. Energy: E = (1)/(2)CV^2 = (1)/(2)(2200 × 10^-6)(12)^2 = (1)/(2)(2200 × 10^-6)(144) = 0.16\ J. Markers reward Q = CV giving about 0.026\ C, and E = (1)/(2)CV^2 giving about 0.16\ J. The half is essential, since energy is the area under the charge-voltage graph.

OCR OCR 2021-style practice: A 470\ μF capacitor charged to 9.0\ V is discharged through a 10\ k resistor. Calculate the time constant and the potential difference across the capacitor 3.0\ s after discharge begins.

Time constant: = RC = (10 × 10^3)(470 × 10^-6) = 4.7\ s. Discharge: V = V_0 e^-t/RC = 9.0\, e^-3.0/4.7. Exponent: (-3.0)/(4.7) = -0.638, so e^-0.638 = 0.528. V = 9.0 × 0.528 = 4.8\ V. Markers reward = RC = 4.7\ s, the exponential form V = V_0 e^-t/RC, and the value about 4.8\ V.

EnglandPhysicsSyllabus dot point

How does a capacitor store charge and energy, and how does it discharge through a resistor?

Capacitors: capacitance and the farad, capacitors in series and parallel, the energy stored in a capacitor, and the exponential charge and discharge through a resistor with the time constant.

A focused answer to the OCR H556 capacitors content, covering capacitance and the farad, combining capacitors in series and parallel, the energy stored in a capacitor, and the exponential charging and discharging of a capacitor through a resistor with the time constant.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Capacitance is the charge stored per unit potential difference, $C = \frac{Q}{V}$ , measured in farads. Capacitors in parallel add directly, $C_{\text{total}} = C_1 + C_2$ , while in series the reciprocals add, $\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}$ (the opposite of resistors). The energy stored is the area under a charge-voltage graph, $E = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$ . When a capacitor discharges through a resistor, the charge, current and voltage all decay exponentially, for example $V = V_0 e^{-t/RC}$ , with time constant $\tau = RC$ , the time to fall to $\frac{1}{e}$ (about 37 per cent) of the initial value.

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

OCR wants you to define capacitance and the farad, use $Q = CV$ , combine capacitors in series and parallel, find the energy stored in a capacitor, and analyse the exponential charge and discharge of a capacitor through a resistor using the time constant.

The answer

Capacitance and the farad

Capacitors in series and parallel

Energy stored

Charging and discharging through a resistor

Examples in context

Capacitors smooth the output of rectified power supplies, provide the energy burst in a camera flash and a defibrillator, and store charge in the memory cells of computers. The time constant sets the timing in flashing-indicator and timer circuits, and the slow discharge of a capacitor through a resistor is used in touchscreens and in audio filters that block or pass particular frequencies. Supercapacitors store large amounts of charge for rapid energy delivery in vehicles.

Try this

Q1. Define capacitance and state its unit. [2 marks]

Cue. The charge stored per unit potential difference, $C = \frac{Q}{V}$ , in farads.

Q2. Two $4.0\ \mu\text{F}$ capacitors are connected in parallel. Find the total capacitance. [1 mark]

Cue. In parallel, $C = 4.0 + 4.0 = 8.0\ \mu\text{F}$ .

Q3. A $1000\ \mu\text{F}$ capacitor discharges through a $2.0\ \text{k}\Omega$ resistor. Find the time constant. [2 marks]

Cue. $\tau = RC = (2.0 \times 10^{3})(1000 \times 10^{-6}) = 2.0\ \text{s}$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20184 marksA

2200\ \mu\text{F}

capacitor is charged to

12\ \text{V}

. Calculate the charge stored and the energy stored.

Show worked answer →

Charge: $Q = CV = (2200 \times 10^{-6})(12) = 2.64 \times 10^{-2}\ \text{C}$ , about $0.026\ \text{C}$ .

Energy: $E = \frac{1}{2}CV^2 = \frac{1}{2}(2200 \times 10^{-6})(12)^2 = \frac{1}{2}(2200 \times 10^{-6})(144) = 0.16\ \text{J}$ .

Markers reward $Q = CV$ giving about $0.026\ \text{C}$ , and $E = \frac{1}{2}CV^2$ giving about $0.16\ \text{J}$ . The half is essential, since energy is the area under the charge-voltage graph.

OCR 20215 marksA

470\ \mu\text{F}

capacitor charged to

9.0\ \text{V}

is discharged through a

10\ \text{k}\Omega

resistor. Calculate the time constant and the potential difference across the capacitor

3.0\ \text{s}

after discharge begins.

Show worked answer →

Time constant: $\tau = RC = (10 \times 10^{3})(470 \times 10^{-6}) = 4.7\ \text{s}$ .

Discharge: $V = V_0 e^{-t/RC} = 9.0\, e^{-3.0/4.7}$ .

Exponent: $\frac{-3.0}{4.7} = -0.638$ , so $e^{-0.638} = 0.528$ .

$V = 9.0 \times 0.528 = 4.8\ \text{V}$ .

Markers reward $\tau = RC = 4.7\ \text{s}$ , the exponential form $V = V_0 e^{-t/RC}$ , and the value about $4.8\ \text{V}$ .

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)