A 220\ μF capacitor is charged to 5.0\ V. Find the charge stored. [2 marks]

A 10\ μF capacitor discharges through a 100\ k resistor. Find the time constant. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A 470\ μF capacitor is charged to 9.0\ V and then discharged through a 22\ k resistor. Calculate the time constant, the energy initially stored, and the voltage across the capacitor after 20\ s.

Time constant (up to 2 marks): = RC = 22000 × 470 × 10^-6 = 10.3\ s. Energy stored (up to 2 marks): E = (1)/(2)CV^2 = (1)/(2) × 470 × 10^-6 × 9.0^2 = (1)/(2) × 470 × 10^-6 × 81 = 1.9 × 10^-2\ J. Voltage after 20\ s (up to 2 marks): V = V_0 e^-t/RC = 9.0\, e^-20/10.3 = 9.0\, e^-1.94 = 9.0 × 0.144 = 1.3\ V. Markers reward = 10.3\ s, E = 1.9 × 10^-2\ J, and the discharge value 1.3\ V using V = V_0 e^-t/RC.

EnglandElectronicsSyllabus dot point

How do capacitors and inductors store energy, and how do they charge and discharge through a resistor?

Capacitors and inductors: capacitance and stored energy, the RC time constant and exponential charge and discharge, inductance and stored energy, and combining capacitors in series and parallel.

An Eduqas A-Level Electronics answer on capacitors and inductors: capacitance and the energy stored in a capacitor, the RC time constant and exponential charge and discharge, inductance and the energy stored in an inductor, and how capacitors combine in series and parallel (the reverse of resistors).

Generated by Claude Opus 4.814 min answerUpdated 2026-06-02

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

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Quick answer

Capacitance $C = \frac{Q}{V}$ (farads) measures the charge stored per volt; the energy stored is $E = \frac{1}{2}CV^2 = \frac{1}{2}QV$ . A capacitor charges and discharges through a resistor following an exponential law with time constant $\tau = RC$ : on discharge $V = V_0 e^{-t/RC}$ , and one time constant is the time to fall to $\frac{1}{e}$ (about 37 per cent) of the start. After about 5 time constants it is effectively fully charged or discharged. Inductance $L$ (henries) relates the voltage to the rate of change of current, $V = L\frac{\Delta I}{\Delta t}$ , and an inductor stores energy $E = \frac{1}{2}LI^2$ in its magnetic field. Capacitors combine the reverse of resistors: they add in parallel and combine by reciprocals in series.

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What this dot point is asking

Eduqas wants you to define capacitance and the energy stored in a capacitor, use the RC time constant with the exponential charge and discharge equations, define inductance and the energy stored in an inductor, and combine capacitors in series and parallel. These energy-storing components set timing, smooth supplies and shape signals throughout the course.

The answer

Capacitance and stored energy

The RC time constant and exponential charge and discharge

Inductance and stored energy

Combining capacitors

Finding a discharge time from a target voltage

A $100\ \mu\text{F}$ capacitor charged to $12\ \text{V}$ discharges through a $47\ \text{k}\Omega$ resistor. How long until the voltage falls to $3.0\ \text{V}$ ?

step 1: Find the time constant

$\tau = RC = 47000 \times 100 \times 10^{-6} = 4.7\ \text{s}$ .

step 2: Write the discharge equation for the target

$3.0 = 12\, e^{-t/4.7}$ , so $e^{-t/4.7} = \frac{3.0}{12} = 0.25$ .

step 3: Take logarithms and solve

$-\frac{t}{4.7} = \ln(0.25) = -1.386$ , so $t = 1.386 \times 4.7 = 6.5\ \text{s}$ .

Examples in context

Capacitors set the timing of 555 astables and monostables, smooth the rectified output of a mains power supply, couple audio between amplifier stages, and decouple the supply pins of logic chips. The 5-time-constant rule decides how long a timing circuit takes to settle and how big a smoothing capacitor a power supply needs. Inductors and capacitors together tune radio receivers and form the energy-transfer element of switch-mode supplies.

Try this

Q1. A $220\ \mu\text{F}$ capacitor is charged to $5.0\ \text{V}$ . Find the charge stored. [2 marks]

Cue. $Q = CV = 220 \times 10^{-6} \times 5.0 = 1.1 \times 10^{-3}\ \text{C}$ .

Q2. A $10\ \mu\text{F}$ capacitor discharges through a $100\ \text{k}\Omega$ resistor. Find the time constant. [2 marks]

Cue. $\tau = RC = 100000 \times 10 \times 10^{-6} = 1.0\ \text{s}$ .

Q3. Two $4.7\ \mu\text{F}$ capacitors are connected in parallel. Find the total capacitance. [1 mark]

Cue. $C = 4.7 + 4.7 = 9.4\ \mu\text{F}$ (parallel adds).

Exam-style practice questions

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

Eduqas 20196 marksA

470\ \mu\text{F}

capacitor is charged to

9.0\ \text{V}

and then discharged through a

22\ \text{k}\Omega

resistor. Calculate the time constant, the energy initially stored, and the voltage across the capacitor after

20\ \text{s}

Show worked answer →

Time constant (up to 2 marks): $\tau = RC = 22000 \times 470 \times 10^{-6} = 10.3\ \text{s}$ .

Energy stored (up to 2 marks): $E = \frac{1}{2}CV^2 = \frac{1}{2} \times 470 \times 10^{-6} \times 9.0^2 = \frac{1}{2} \times 470 \times 10^{-6} \times 81 = 1.9 \times 10^{-2}\ \text{J}$ .

Voltage after $20\ \text{s}$ (up to 2 marks): $V = V_0 e^{-t/RC} = 9.0\, e^{-20/10.3} = 9.0\, e^{-1.94} = 9.0 \times 0.144 = 1.3\ \text{V}$ .

Markers reward $\tau = 10.3\ \text{s}$ , $E = 1.9 \times 10^{-2}\ \text{J}$ , and the discharge value $1.3\ \text{V}$ using $V = V_0 e^{-t/RC}$ .

Eduqas 20224 marksExplain what the time constant of an RC circuit represents, and state how many time constants are needed for a capacitor to be considered fully discharged.

Show worked answer →

Meaning (up to 3 marks): the time constant $\tau = RC$ is the time taken for the charge (or voltage, or current) on a discharging capacitor to fall to $\frac{1}{e}$ , about 37 per cent, of its initial value. Equivalently it is the time to rise to about 63 per cent of the final value when charging. A larger resistance or capacitance gives a slower response.

Fully discharged (up to 1 mark): after about 5 time constants the capacitor has fallen to under 1 per cent of its initial value and is treated as fully discharged.

Markers reward the $\frac{1}{e}$ (37 per cent) definition, the link to $RC$ , and the "5 time constants" rule of thumb.

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Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)