A 100\,μF capacitor is charged to 12\,V. Find the energy stored. [2 marks]

A capacitor discharges through a 2.0\,k resistor with capacitance 50\,μF. Find the time constant. [2 marks]

WJEC WJEC 2020-style practice: A 470\,μF capacitor is charged to 9.0\,V and then discharged through a 15\,k resistor. Calculate the initial charge stored, the time constant, and the voltage across the capacitor 10\,s after discharge begins.

Initial charge: Q_0 = CV = 470 × 10^-6 × 9.0 = 4.23 × 10^-3\,C. Time constant: = RC = 15 × 10^3 × 470 × 10^-6 = 7.05\,s. Voltage after 10\,s using V = V_0 e^-t/RC: V = 9.0 × e^-10/7.05 = 9.0 × e^-1.42 = 9.0 × 0.242 = 2.2\,V. Markers reward Q = CV, the time constant as RC, and the exponential decay giving about 2.2\,V.

WalesPhysicsSyllabus dot point

How does a capacitor store charge and energy, and why are charge and discharge exponential?

Capacitance and the charge stored, the energy stored in a capacitor, and exponential charge and discharge through a resistor with time constant RC.

A focused answer to WJEC A-Level Physics Unit 4 capacitance, covering capacitance and the charge stored, the energy stored in a capacitor, and the exponential charge and discharge of a capacitor through a resistor with time constant RC.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

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

What this dot point is asking

WJEC wants you to define capacitance, calculate the charge and energy stored, and analyse the exponential charge and discharge of a capacitor through a resistor using the time constant. The exponential decay is the same mathematics that appears in radioactive decay, so mastering it here pays off twice, and the examiners reward fluent use of the time constant.

The answer

Capacitance and charge

A capacitor stores charge by separating it onto two plates: positive on one, an equal negative on the other, with an electric field in the gap. The farad is a very large unit, so practical capacitors are usually quoted in microfarads or picofarads.

Energy stored

The energy stored is the area under the charge-voltage graph, which is a triangle, giving the factor of one half.

Exponential charge and discharge

A larger resistance or capacitance gives a longer time constant and a slower process. After about five time constants the capacitor is essentially fully discharged.

Time to fall to a given voltage

A $220\,\mu\text{F}$ capacitor charged to $12\,\text{V}$ discharges through a $33\,\text{k}\Omega$ resistor. How long until the voltage falls to $3.0\,\text{V}$ ?

step Find the time constant

$\tau = RC = 33 \times 10^{3} \times 220 \times 10^{-6} = 7.26\,\text{s}$ .

step Set up the exponential

$V = V_0 e^{-t/\tau}$ , so $\dfrac{3.0}{12} = e^{-t/7.26}$ , giving $0.25 = e^{-t/7.26}$ .

step Take logarithms and solve

$\ln(0.25) = -t/7.26$ , so $-1.386 = -t/7.26$ and $t = 1.386 \times 7.26 = 10.1\,\text{s}$ .

Examples in context

Example 1. A camera flash: A flashgun charges a large capacitor slowly from a small battery, then discharges it through the flash tube almost instantly. Because the stored energy is $\tfrac{1}{2}CV^2$ , a high voltage and large capacitance store enough energy for a bright burst, while the brief discharge through the low-resistance tube has a tiny time constant, giving the sharp flash.
Example 2. Smoothing a power supply: A capacitor placed across the output of a rectifier charges up on each voltage peak and discharges slowly through the load between peaks. Choosing $RC$ to be much larger than the time between peaks keeps the output voltage nearly steady, turning bumpy rectified AC into smooth DC. This is the same exponential discharge described above, exploited to hold a voltage up.
Example 3. A defibrillator: A medical defibrillator charges a capacitor to several thousand volts, storing energy $E = \tfrac{1}{2}CV^2$ of around $200\,\text{J}$ . On command it discharges that energy through the patient's chest in a few milliseconds, a very short time constant set by the low resistance of the body and electrodes. The brief, intense current pulse can restart a fibrillating heart. The whole device hinges on the capacitor's ability to store energy slowly and release it almost instantly.

Try this

Q1. A $100\,\mu\text{F}$ capacitor is charged to $12\,\text{V}$ . Find the energy stored. [2 marks]

Cue. $E = \frac{1}{2}CV^2 = \frac{1}{2}\times100\times10^{-6}\times12^2 = 7.2\times10^{-3}\,\text{J}$ .

Q2. A capacitor discharges through a $2.0\,\text{k}\Omega$ resistor with capacitance $50\,\mu\text{F}$ . Find the time constant. [2 marks]

Cue. $\tau = RC = 2000\times50\times10^{-6} = 0.10\,\text{s}$ .

Exam-style practice questions

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

WJEC 20206 marksA

470\,\mu\text{F}

capacitor is charged to

9.0\,\text{V}

and then discharged through a

15\,\text{k}\Omega

resistor. Calculate the initial charge stored, the time constant, and the voltage across the capacitor

10\,\text{s}

after discharge begins.

Show worked answer →

Initial charge: $Q_0 = CV = 470 \times 10^{-6} \times 9.0 = 4.23 \times 10^{-3}\,\text{C}$ .

Time constant: $\tau = RC = 15 \times 10^{3} \times 470 \times 10^{-6} = 7.05\,\text{s}$ .

Voltage after $10\,\text{s}$ using $V = V_0 e^{-t/RC}$ :

$V = 9.0 \times e^{-10/7.05} = 9.0 \times e^{-1.42} = 9.0 \times 0.242 = 2.2\,\text{V}$ .

Markers reward $Q = CV$ , the time constant as $RC$ , and the exponential decay giving about $2.2\,\text{V}$ .

WJEC 20184 marksShow that the energy stored in a

2200\,\mu\text{F}

capacitor charged to

24\,\text{V}

is about

0.63\,\text{J}

, and explain why this is half of the product

QV

Show worked answer →

Energy stored: $E = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 2200 \times 10^{-6} \times (24)^2$ .

$(24)^2 = 576$ , so $E = \tfrac{1}{2} \times 2200 \times 10^{-6} \times 576 = 0.63\,\text{J}$ , as required.

The factor of one half arises because the voltage across the capacitor rises from $0$ to $V$ as it charges. The energy is the area under the charge-voltage graph, which is a triangle of base $Q$ and height $V$ , giving $\tfrac{1}{2}QV$ . If the voltage stayed at $V$ throughout the energy would be $QV$ , but the average voltage during charging is only $V/2$ . Markers reward the calculation and the area-under-graph explanation of the half.

Related dot points

Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)