A sinusoid has a peak value of 6.0\ V. Find its RMS value. [2 marks]

A signal has a frequency of 250\ Hz. Find its period. [1 mark]

WJEC Eduqas Eduqas 2020-style practice: A sinusoidal signal has a peak voltage of 8.0\ V and a period of 2.0\ ms. Calculate its frequency, its peak-to-peak voltage, and its root-mean-square voltage.

Frequency (up to 2 marks): f = 1T = 12.0 × 10^-3 = 500\ Hz. Peak-to-peak (up to 1 mark): V_pp = 2 V_0 = 2 × 8.0 = 16\ V. Root-mean-square (up to 2 marks): V_rms = V_0sqrt(2) = 8.0sqrt(2) = 5.7\ V. Markers reward f = 500\ Hz, V_pp = 16\ V and V_rms = 5.7\ V using V_rms = (V_0)/( 2).

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How do we describe an alternating signal, and why does the opposition of a capacitor or inductor depend on frequency?

AC signals and reactance: amplitude, peak-to-peak, period and frequency of a sinusoid, root-mean-square values, and the frequency-dependent reactance of capacitors and inductors.

An Eduqas A-Level Electronics answer on alternating signals and reactance: amplitude, peak-to-peak, period and frequency of a sinusoid, the root-mean-square value and its relation to the peak, and the frequency-dependent reactance of capacitors and inductors that underlies all filtering.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

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Quick answer

A sinusoid is described by its amplitude (peak value $V_0$ ), peak-to-peak value ( $V_\text{pp} = 2V_0$ ), period $T$ (time for one cycle) and frequency $f = \frac{1}{T}$ (cycles per second, hertz). The root-mean-square (RMS) value is the equivalent steady value that delivers the same average power: for a sinusoid $V_\text{rms} = \frac{V_0}{\sqrt{2}}$ . A capacitor and an inductor oppose AC with a frequency-dependent reactance (in ohms). Capacitive reactance $X_C = \frac{1}{2\pi f C}$ falls as frequency rises (a capacitor passes high frequencies). Inductive reactance $X_L = 2\pi f L$ rises as frequency rises (an inductor passes low frequencies). Reactance stores and returns energy rather than dissipating it.

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What this dot point is asking

Eduqas wants you to describe a sinusoidal signal with amplitude, peak-to-peak, period and frequency, relate the root-mean-square value to the peak, and calculate the frequency-dependent reactance of a capacitor and an inductor. Reactance is what makes filters, coupling capacitors and tuned circuits work.

The answer

Describing a sinusoidal signal

Root-mean-square values

Capacitive and inductive reactance

Reactance versus resistance

Finding the current drawn by a capacitor from an AC source

A $220\ \text{nF}$ capacitor is connected across a $5.0\ \text{V}$ peak, $2.0\ \text{kHz}$ sinusoidal supply. Find the peak current it draws.

step 1: Find the reactance

$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 2000 \times 220 \times 10^{-9}} = \frac{1}{2.76 \times 10^{-3}} = 362\ \Omega$ .

step 2: Apply Ohm's law for reactance

$I_0 = \frac{V_0}{X_C} = \frac{5.0}{362} = 1.38 \times 10^{-2}\ \text{A} \approx 14\ \text{mA}$ .

step 3: Sense-check the frequency dependence

Doubling the frequency would halve $X_C$ and double the current, confirming the capacitor passes higher frequencies more readily.

Examples in context

Reactance explains why a coupling capacitor passes the audio signal but blocks the DC bias between amplifier stages, why a decoupling capacitor shorts high-frequency noise to ground, and why a radio's tuned circuit selects one station. RMS values let you size a power supply and compare an AC source with a battery. The frequency dependence of $X_C$ and $X_L$ is the starting point for the passive-filter topic.

Try this

Q1. A sinusoid has a peak value of $6.0\ \text{V}$ . Find its RMS value. [2 marks]

Cue. $V_\text{rms} = \frac{6.0}{\sqrt{2}} = 4.2\ \text{V}$ .

Q2. A signal has a frequency of $250\ \text{Hz}$ . Find its period. [1 mark]

Cue. $T = \frac{1}{f} = \frac{1}{250} = 4.0\ \text{ms}$ .

Q3. State how the reactance of an inductor changes as the frequency increases. [1 mark]

Cue. It increases ( $X_L = 2\pi f L$ is proportional to frequency).

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 20205 marksA sinusoidal signal has a peak voltage of

8.0\ \text{V}

and a period of

2.0\ \text{ms}

. Calculate its frequency, its peak-to-peak voltage, and its root-mean-square voltage.

Show worked answer →

Frequency (up to 2 marks): $f = \dfrac{1}{T} = \dfrac{1}{2.0 \times 10^{-3}} = 500\ \text{Hz}$ .

Peak-to-peak (up to 1 mark): $V_\text{pp} = 2 V_0 = 2 \times 8.0 = 16\ \text{V}$ .

Root-mean-square (up to 2 marks): $V_\text{rms} = \dfrac{V_0}{\sqrt{2}} = \dfrac{8.0}{\sqrt{2}} = 5.7\ \text{V}$ .

Markers reward $f = 500\ \text{Hz}$ , $V_\text{pp} = 16\ \text{V}$ and $V_\text{rms} = 5.7\ \text{V}$ using $V_\text{rms} = \frac{V_0}{\sqrt 2}$ .

Eduqas 20225 marksA

100\ \text{nF}

capacitor is used at a frequency of

1.0\ \text{kHz}

. Calculate its reactance, and explain how the reactance changes if the frequency is increased.

Show worked answer →

Reactance (up to 3 marks): $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi \times 1000 \times 100 \times 10^{-9}} = \dfrac{1}{6.28 \times 10^{-4}} = 1.6 \times 10^{3}\ \Omega = 1.6\ \text{k}\Omega$ .

Effect of frequency (up to 2 marks): capacitive reactance is inversely proportional to frequency, $X_C \propto \frac{1}{f}$ , so increasing the frequency decreases the reactance. At high frequency the capacitor offers little opposition (it "passes" high frequencies); at low frequency it offers high opposition.

Markers reward $X_C = 1.6\ \text{k}\Omega$ , the inverse-frequency relationship, and the conclusion that reactance falls as frequency rises.

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Sources & how we know this

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