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Sinusoidal Alternating Current & Voltage

Key idea: Use x = x0 sin(ωt) and ω = 2πf to find phase, period, frequency, peak values and instantaneous values in A Level sinusoidal a.c. questions.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Represent sinusoidal a.c. and use peak and r.m.s. values.
  • Analyse mean power in resistive a.c. loads and half-wave rectification.

1. Definitions (Must Know)

A. Alternating current (a.c.)

Alternating current (a.c.) is a current that reverses direction periodically.

B. Period and frequency

  • Period, T: time for one complete cycle (s).
  • Frequency, f: number of cycles per second (Hz).

f = 1/T

C. Angular frequency, ω

Angular frequency, ω (rad s⁻¹):

ω = 2π f

D. Peak value (amplitude), x₀

Peak value (amplitude), x₀, is the maximum magnitude of a sinusoid (e.g. I₀ for current, V₀ for voltage).

E. Root-mean-square (r.m.s.) value

The r.m.s. value is the equivalent d.c. value that gives the same mean power in a resistor.

2. Key Ideas (What Earns Marks)

  • A sinusoidal a.c. quantity can be represented by:
    • x = x₀ sin(ω t)
    • where x can be I or V.
  • Converting between T, f and ω:
    • T = 1/f, ω = 2π f, f = ω/2π
  • For a sinusoidal wave:
    • Iᵣₘₛ = I₀/(square root of 2), Vᵣₘₛ = V₀/(square root of 2)
  • Quick checks:
    • at t = 0: x = 0 (for the pure sin form)
    • one full cycle: t = T = 2π/ω
When to use r.m.s.

Mains voltages quoted on appliances (e.g. 230 V) are almost always r.m.s. values, unless the question explicitly says “peak”.

3. Detailed Explanations

A. The sinusoidal model

Many a.c. supplies can be modelled as a sinusoid: x = x₀ sin(ω t)

Meaning:

  • x₀ sets the maximum value
  • ω sets how fast the oscillation happens

One cycle of sinusoidal a.c. (50 Hz example)

A single 50 Hz cycle of a sine wave showing peak values and the period.

Scroll across the graph to read all labels.

A single 50 Hz cycle of a sine wave showing peak values and the period.A single 50 Hz cycle of a sine wave showing peak values and the period.
For 50 Hz, the period is T = 20 ms. The wave reaches +x0 at T/4 and −x0 at 3T/4.
Open full-size graph
View figure data
Values for One cycle of sinusoidal a.c. (50 Hz example)
Time, t (ms)x = x0 sin(ωt)
00
2.50.707
51
7.50.707
100
12.5-0.707
15-1
17.5-0.707
200

B. Connecting angular frequency and frequency

One full cycle corresponds to a phase change of 2π radians.

If the angular frequency is ω rad s⁻¹, then the period is: T = 2π/ω

So: f = 1/T = ω/2π ⇒ ω = 2π f

C. Peak vs r.m.s.

For a sinusoidal a.c.: Iᵣₘₛ = I₀/(square root of 2), Vᵣₘₛ = V₀/(square root of 2)

These are the values used when calculating mean power in a purely resistive load: ⟨P⟩ = Iᵣₘₛ²R = Vᵣₘₛ²/R = IᵣₘₛVᵣₘₛ

The factors I₀/square root of 2 and V₀/square root of 2 apply to a sinusoidal waveform. For the general r.m.s. definition and non-sinusoidal boundary, continue to Root-Mean-Square Values.

AC Waveform & Power Explorer

Connect sinusoidal timing, r.m.s. values, resistive power and ideal half-wave output on synchronized graphs.

BetaA LevelElectricityBest for: A Level current electricity
  • Waveform Reading
  • r.m.s. Values
  • Resistive Power
  • Rectification

Open the full interactive simulation on its own page

Use the standalone simulation page for the live controls, SVG scene, run modes, and scoring flow.

Open Interactive Simulation

The lesson stays lightweight and links out to the dedicated simulation page.

4. Common Mistakes

  • Using ω = 2f instead of ω = 2π f.
  • Treating ω as “frequency” (it is rad s⁻¹, not Hz).
  • Mixing degrees and radians when using calculator trig functions.
  • Using peak values inside mean-power formulas instead of r.m.s. values.

5. Exam Tips

  • State whether you are using peak or r.m.s. before substituting numbers.
  • If you are asked for an equation, include the unit of ω (rad s⁻¹) in your working.

6. Worked Examples

Modelled example 1

Write the a.c. equation from peak value and frequency

Core

Problem

A current has peak I₀ = 5.0 A and frequency 50 Hz. Write I(t) for zero phase.
Study the worked solution

  1. Convert frequency to angular frequency

    Method

    Use ω = 2π f.

    Reason

    Each cycle advances phase by 2π radians.

    Working

    ω = 2π(50) = 100π rad s⁻¹

  2. Build the equation

    Method

    Insert peak and angular frequency into I = I₀ sin ω t.

    Reason

    Zero phase starts at zero and increases positively.

    Working

    I(t) = 5.0 sin(100π t) A

Guided practice 2

Find frequency from an a.c. equation

About 5 min

Problem

For V = 12 sin(400π t) in SI units, find frequency and period.

Try this before viewing the solution

Hints

Hint 1: read the coefficient of time
Compare with V = V₀ sin(ω t), then use f = ω/(2π).
View solution step by step

  1. Extract frequency

    Method

    Read ω = 400π and divide by 2π.

    Reason

    Angular frequency counts phase radians per second.

    Working

    f = 400π/2π = 200 Hz

  2. Find period

    Method

    Take the reciprocal of frequency.

    Reason

    Period is seconds per cycle.

    Working

    T = 1/200 = 5.0 × 10⁻³ s

Common misconception 3

Convert between r.m.s. and peak voltage

Find and correct the mistake

Learner working

Given Vᵣₘₛ = 230 V, a learner divides by square root of 2 to find peak voltage. Locate the first error and calculate V₀.

Try this before viewing the solution

Correct relation

View solution step by step

  1. Reverse the r.m.s. relation

    Method

    Use V₀ = square root of 2 Vᵣₘₛ.

    Reason

    Vᵣₘₛ = V₀/square root of 2 for a sinusoid.

    Working

    V₀ = square root of 2 (230).

  2. Evaluate

    Method

    Obtain approximately 325 V.

    Reason

    Peak magnitude must exceed the r.m.s. value.

    Working

    V₀ = 3.25 × 10² V.

Examiner practice 4

Instantaneous current at a given time

3 marks

Examination question

For I = 4.0 sin(200π t) A, find I at t = 2.5 ms. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Convert time

    1 mark

    Method

    Use t = 2.5 × 10⁻³ s.

    Reason

    The equation uses SI seconds.

    Working

    t = 0.0025 s.

  2. Find phase

    1 mark

    Method

    Calculate 200π t = π/2.

    Reason

    The sine argument is the phase in radians.

    Working

    200π(0.0025) = π/2.

  3. Evaluate current

    1 mark

    Method

    Use sin(π/2) = 1.

    Reason

    This instant is the positive peak.

    Working

    I = 4.0 A.

Challenge 5

Time to reach half the peak value

Minimal support

Inverse-time transfer

For x = x₀ sin(ω t), find the first t > 0 when x = (1/2)x₀.

Try this before viewing the solution

Hints

Hint 1: solve for phase first
Divide by x₀ and identify the first positive angle whose sine is 1/2.
View solution step by step

  1. Find the first phase

    Method

    Set sin(ω t) = 1/2 and use ω t = π/6.

    Reason

    π/6 is the first positive sine solution.

    Working

    ω t = π/6.

  2. Solve for time

    Method

    Divide by angular frequency.

    Reason

    Phase grows at ω radians per second.

    Working

    t = π/6ω

7. Mind Stretchers

Mind stretcher 1: Time for the first peakExtension

For x = x₀ sin(ω t), find the first time t > 0 when x = x₀.

Show Answer

Need sin(ω t) = 1, which first occurs at ω t = π/2. t = (π/2)/ω = π/2ω

Mind stretcher 2: Sine vs cosine formExtension

Show that x = x₀ cos(ω t) can be written as a sine function with a phase shift.

Show Answer

Use the identity cos(θ) = sin(θ + π/2): x = x₀ cos(ω t) = x₀ sin(ω t + π/2)

Mind stretcher 3: Optional (Enrichment)Extension

A. Phase constant (beyond the syllabus form)

Sometimes the sinusoid is written as:

x = x₀ sin(ω t + φ)

where φ is a phase constant that shifts the graph left/right. In this syllabus, you are usually given or asked to use the simpler form x = x₀ sin(ω t).

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027