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AC generators and alternators - TCE Physics (Tasmania)

Syllabus dot point

“Explain the operation of an AC generator and the sinusoidal EMF it produces.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min read

Quick answer

An AC generator rotates a coil in a magnetic field so the flux through it changes continuously, inducing a sinusoidal EMF by Faraday's law. Slip rings keep the connection to the same coil ends, so the output alternates direction. It is essentially a motor run in reverse, turning rotation into electricity.

What this dot point is asking

This dot point applies Faraday's law to a rotating coil, the device that produces almost all the world's electricity.

How the EMF is generated

A generator has a coil that is spun inside a magnetic field. As the coil turns, the angle between the field and the coil face changes, so the flux Φ=BAcos⁡θ\Phi = BA\cos\theta through the coil rises and falls. Faraday's law says the induced EMF is proportional to the rate of change of flux, so a steadily rotating coil produces a continuously varying EMF.

The EMF is greatest when the coil is moving parallel to the field lines, that is when the coil plane is parallel to the field, because then the flux is changing fastest. The EMF is zero when the coil is perpendicular to the field, where the flux is at a maximum but momentarily not changing. The result is a smoothly varying, sinusoidal output.

The sinusoidal output

For a coil of NN turns and area AA rotating at angular frequency ω\omega in a field BB, the induced EMF varies as:

ε=NBAωsin⁡(ωt)\varepsilon = N B A \omega \sin(\omega t)

The peak EMF is ε0=NBAω\varepsilon_0 = NBA\omega, so a faster rotation, a stronger field, more turns or a larger area all raise the peak voltage. The output reverses direction every half turn, which is what makes it alternating current. The angular frequency relates to the rotation rate by ω=2πf\omega = 2\pi f, so a generator turning at 50 Hz50\ \text{Hz} has ω≈314 rad s−1\omega \approx 314\ \text{rad s}^{-1}.

EMF peaks when flux changes fastest

The induced EMF is largest at the instant the flux through the coil is changing most rapidly, which is when the coil plane is parallel to the field. It is zero when the flux is momentarily at its maximum and not changing, when the coil plane is perpendicular to the field.

Slip rings and the link to motors

A generator connects the coil to the external circuit through slip rings, two continuous rings that each stay connected to one end of the coil. Because the connection never swaps, the output naturally alternates as the coil rotates, giving AC. This contrasts with a DC motor's split-ring commutator, which deliberately swaps the connection every half turn.

A generator is structurally the same as a motor: a coil, a magnetic field and a means of connection. A motor uses an input current to produce rotation; a generator uses input rotation to produce a current. This is why the same machine can often run either way, and why a spinning motor produces a back EMF that opposes the supply.

Alternator

An alternator is an AC generator: a device that converts mechanical rotational energy into alternating electrical energy by rotating a coil (or magnet) so the magnetic flux through a winding changes continuously, inducing a sinusoidal EMF.

Peak and RMS values

Because AC voltage varies, we describe it by two numbers. The peak value ε0\varepsilon_0 is the maximum, while the root-mean-square (RMS) value is the steady value that would deliver the same average power. For a sinusoidal output they are related by:

εrms=ε02.\varepsilon_\text{rms} = \frac{\varepsilon_0}{\sqrt{2}}.

Mains voltage quoted as 230 V230\ \text{V} is the RMS value, with a higher peak of about 325 V325\ \text{V}. The RMS value matters because the average power delivered to a resistor is set by εrms\varepsilon_\text{rms}, not the peak.

Worked example: peak EMF of a generator

A generator coil has 5050 turns, area 0.020 m20.020\ \text{m}^2, and spins at ω=314 rad s−1\omega = 314\ \text{rad s}^{-1} in a field of 0.30 T0.30\ \text{T}. Find the peak EMF and the RMS output voltage.

Step 1: Identify the formula for peak EMF

When a coil rotates at constant angular frequency in a uniform field, Faraday's law gives a sinusoidal EMF whose peak value occurs when the coil plane is parallel to the field and flux is changing most rapidly. The peak EMF depends on how many turns the coil has, how strong the field is, the area swept by the coil, and how fast it spins:

ε0=NBAω\varepsilon_0 = NBA\omega

Each factor multiplies the effect: more turns, a larger area, a stronger field, or a faster rotation all increase the maximum output voltage.

Step 2: Substitute the given values and evaluate

Reading off N=50N = 50, B=0.30 TB = 0.30\ \text{T}, A=0.020 m2A = 0.020\ \text{m}^2, and ω=314 rad s−1\omega = 314\ \text{rad s}^{-1}, and substituting:

ε0=(50)(0.30)(0.020)(314)=94 V\varepsilon_0 = (50)(0.30)(0.020)(314) = 94\ \text{V}

Step 3: Convert peak to RMS

The RMS value is the equivalent steady voltage that would deliver the same average power to a resistive load. For a sinusoidal waveform the RMS value is always smaller than the peak by a factor of 2\sqrt{2}:

εrms=ε02=942=66 V\varepsilon_\text{rms} = \frac{\varepsilon_0}{\sqrt{2}} = \frac{94}{\sqrt{2}} = 66\ \text{V}

Final answer: Peak EMF =94 V= 94\ \text{V}; RMS output =66 V= 66\ \text{V}.

Confusing slip rings with a commutator

A common error is to mix up the AC generator's slip rings with the DC motor's split-ring commutator. Slip rings keep the same connection so the output alternates (AC). A commutator reverses the connection every half turn to keep current flowing one way (motor) or to produce DC output. Naming the wrong component loses marks in explanation questions.

In the exam, link the size of the EMF to how fast the flux is changing, and identify slip rings as the feature that gives alternating output. If asked, state that the EMF is zero when the coil is perpendicular to the field and maximum when the coil plane is parallel to it, and convert between peak and RMS values with the 2\sqrt{2} factor.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice2 marks
A 3030-turn coil is disconnected from a power source and rotated at 50 Hz50\ \text{Hz} in a uniform 0.2 T0.2\ \text{T} magnetic field. Sketch the potential difference at the outputs against time, starting from the position where the coil plane is parallel to the field, and explain its shape.
Show worked answer →

As the coil rotates the flux varies as Φ=BAcos⁡(ωt)\Phi = BA\cos(\omega t). By Faraday's law the induced EMF is the rate of change of flux, ε=−NdΦdt=NBAωsin⁡(ωt)\varepsilon = -N\dfrac{d\Phi}{dt} = NBA\omega\sin(\omega t), which is sinusoidal.

The graph is a sine curve. Starting with the coil plane parallel to the field, the flux is momentarily minimum but its rate of change is maximum, so the EMF starts at its peak value and traces a smooth sinusoid, crossing zero a quarter-turn later (when the coil plane is perpendicular to the field and flux is maximum but unchanging).

Period: T=1f=150=0.02 sT = \dfrac{1}{f} = \dfrac{1}{50} = 0.02\ \text{s}, so one full cycle occupies 0.02 s0.02\ \text{s}, with equal positive and negative peaks each half cycle (the current reverses every half turn). Markers want a labelled sinusoid of period 0.02 s0.02\ \text{s}, EMF maximum when the coil plane is parallel to the field, and the reasoning that EMF is proportional to the rate of change of flux.

Practice3 marks
A laboratory alternator has a coil of 8080 turns and area 1.5×10−2 m21.5 \times 10^{-2}\ \text{m}^2 spinning in a 0.25 T0.25\ \text{T} field. It produces a peak EMF of 94 V94\ \text{V}. Calculate the angular frequency and rotation rate of the coil, then state the RMS value of the output.
Show worked answer →

The peak EMF is ε0=NBAω\varepsilon_0 = NBA\omega, so rearrange for ω\omega:

ω=ε0NBA=94(80)(0.25)(1.5×10−2)=940.30=313 rad s−1.\omega = \frac{\varepsilon_0}{NBA} = \frac{94}{(80)(0.25)(1.5\times10^{-2})} = \frac{94}{0.30} = 313\ \text{rad s}^{-1}.

Rotation rate: f=ω2π=3132π=49.8 Hzf = \dfrac{\omega}{2\pi} = \dfrac{313}{2\pi} = 49.8\ \text{Hz}, about 50 Hz50\ \text{Hz}.

RMS value: εrms=ε02=941.414=66.5 V\varepsilon_\text{rms} = \dfrac{\varepsilon_0}{\sqrt{2}} = \dfrac{94}{1.414} = 66.5\ \text{V}.

Markers want ω=ε0NBA\omega = \dfrac{\varepsilon_0}{NBA}, the conversion f=ω2πf = \dfrac{\omega}{2\pi}, and εrms=ε02\varepsilon_\text{rms} = \dfrac{\varepsilon_0}{\sqrt{2}}.

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