How to Label the Rotor, Stator and Brushes in a Generator

Look at a simple generator diagram: a rectangular coil sits between north and south magnetic poles, a shaft turns the coil, and two carbon blocks press against metal rings attached to the shaft. Which part is the rotor, which is the stator, and where do the brushes belong?

In the common rotating-coil generator, the coil, shaft, and rings form the rotor. The fixed magnetic poles and supporting frame form the stator. The brushes are stationary carbon contacts that touch the rotating rings and connect the generator to the external circuit.

The important detail is that the brushes do not rotate with the rings. They maintain electrical contact while the rotor turns.

01Start with what moves

The easiest way to label a generator is to ignore what each component does electrically for a moment.

Instead, ask one question:

Which components rotate, and which components remain stationary?

This gives you the correct labels without needing to memorise the appearance of every generator.

The rotor

The rotor is the rotating part of the generator.

In a typical HSC rotating-coil generator, it includes:

• The rectangular conducting coil.
• The shaft attached to the coil.
• The slip rings or split-ring commutator attached to the shaft.

These components rotate together.

If the diagram shows the coil turning between two fixed magnets, the coil is part of the rotor.

The stator

The stator is the stationary part of the generator.

In the same arrangement, it includes:

• The north and south magnetic pole pieces.
• The supporting frame or housing.

The stator holds the magnetic field assembly in position while the rotor turns.

The brushes

Brushes are stationary electrical contacts, usually made from carbon or graphite.

Each brush presses against a rotating metal ring. This allows electrical current to pass between the rotating generator components and the stationary external circuit.

The brushes are often mounted on a holder attached to the generator frame.

Although they are physically stationary, label them separately rather than simply calling them part of the stator.

Quick identification table

The most useful rule is:

Rotor = rotating assembly. Stator = stationary assembly. Brushes = stationary electrical contacts.

One qualification matters: not every generator has a rotating coil. Some generators rotate the magnetic field assembly instead. We will examine that variation later.

02Don’t confuse brushes, slip rings, and split rings

This is where generator diagrams become confusing.

The brushes and rings touch each other, but they are different components with different jobs.

Both slip rings and split-ring commutators rotate with the rotor. The brushes remain stationary.

The difference is the construction of the rings.

Slip rings: two complete rings

An AC generator commonly uses two separate, continuous slip rings.

Each end of the rotating coil connects to a different ring.

As the rotor turns:

• The first ring remains connected to the same end of the coil.
• The second ring remains connected to the other end.
• Each stationary brush maintains contact with its corresponding ring.

The external circuit therefore receives an alternating voltage.

The rings do not swap the coil connections.

Split-ring commutator: two half-rings

A simple DC generator uses a split-ring commutator.

Instead of two complete rings, there are two conducting semicircular segments separated by insulating gaps.

Each segment connects to one end of the rotating coil.

As the rotor completes each half-turn, the segments exchange contact with the stationary brushes. This reverses the coil’s connections to the external circuit.

The result is a unidirectional, pulsating output rather than an alternating output.

Notice that the split-ring commutator is still part of the rotor. The brushes are still stationary.

A common examination mistake is to label the split-ring commutator as the brushes.

Remember that the metal segments rotate. The carbon contacts touching them do not.

For a more detailed explanation of the electrical consequences, see How a Split-Ring Commutator Changes a Generator’s Waveform.

03Connect the coil’s geometry to magnetic flux

Now that the components are identified, consider why the coil’s orientation matters.

In a rotating-coil generator, the stator provides a magnetic field, while the rotor carries the coil through different orientations.

As the coil rotates, the magnetic flux through it changes. This changing flux produces an induced electromotive force (emf).

To understand the connection, you need to identify one important geometric feature: the coil’s area normal.

The area normal is an imaginary line perpendicular to the flat surface of the coil.

It is not a line running along the coil’s surface.

Which angle should you use?

Let \(\theta\) be the angle between the magnetic field \(B\) and the coil’s area normal.

For a flat coil in a uniform magnetic field:

\[ \Phi=BA\cos\theta \]

where:

• \(\Phi\) is the magnetic flux through one turn, measured in webers (Wb).
• \(B\) is the magnetic flux density, measured in teslas (T).
• \(A\) is the area enclosed by one turn, measured in square metres (\(\mathrm{m^2}\)).
• \(\theta\) is the angle between the magnetic field and the area normal.

Consider three important positions.

Position 1: The normal is parallel to the magnetic field.

Here, \(\theta=0^\circ\).

The magnetic field passes directly through the coil’s surface, so the flux is at its maximum positive value.

\[ \Phi=BA \]

Position 2: The normal is perpendicular to the magnetic field.

Here, \(\theta=90^\circ\).

The magnetic field runs parallel to the plane of the coil, so the flux is zero.

\[ \Phi=0 \]

Position 3: The normal points opposite to the magnetic field.

Here, \(\theta=180^\circ\).

The flux reaches its maximum negative value.

\[ \Phi=-BA \]

The negative sign indicates that the magnetic field passes through the coil in the opposite direction to our chosen positive normal.

Why maximum flux does not mean maximum emf

Here’s a prediction worth making before looking at the equation.

If the magnetic flux is greatest when the coil’s normal points along the magnetic field, should the induced emf also be greatest at that instant?

No. The induced emf is actually zero.

The reason is that Faraday’s law depends on the rate of change of magnetic flux, not simply the amount of flux.

\[ \mathcal{E}=-N\frac{d\Phi}{dt} \]

Here:

• \(\mathcal{E}\) is the induced emf in volts (V).
• \(N\) is the number of turns in the coil.
• \(\Phi\) is the magnetic flux through one turn, in webers.
• \(d\Phi/dt\) is the rate of change of magnetic flux.

Imagine a ball rolling over the top of a hill. Its height is greatest at the top, but its instantaneous vertical velocity is zero.

Magnetic flux behaves differently from height, but the useful distinction is the same: the largest value of a quantity does not necessarily correspond to its greatest rate of change.

At maximum positive flux, the flux is momentarily changing at zero rate. Therefore, the induced emf is zero.

A quarter-turn later, the flux is zero but changing most rapidly. The induced emf is then at its maximum magnitude.

For a coil rotating at constant angular velocity in a uniform magnetic field:

\[ \theta=\omega t \]

where \(\omega\) is angular velocity in radians per second and \(t\) is time in seconds.

Substituting the rotating coil’s flux into Faraday’s law gives:

\[ \begin{aligned} \Phi&=BA\cos(\omega t)\\ \mathcal{E}&=-N\frac{d\Phi}{dt}\\ \mathcal{E}&=NBA\omega\sin(\omega t) \end{aligned} \]

The maximum emf is therefore:

\[ \mathcal{E}_{\max}=NBA\omega \]

For the sign convention used here, the angle increases from \(0^\circ\) in the direction of rotation, and the positive emf direction is chosen consistently with Faraday’s law. Reversing the voltmeter leads reverses the measured signs, but not the magnitudes.

Use the coil’s position to predict the emf

This table is particularly useful when interpreting generator diagrams and matching coil positions to voltage graphs.

It also explains why the rotating coil and stationary magnetic poles have different jobs: the stator provides the magnetic field, while the rotor changes the coil’s orientation relative to that field.

04How an experiment tests the arrangement

Suppose you want to investigate how the orientation of a rotating coil affects its induced emf.

You need more than a coil and a magnet. You need a way to rotate the coil, maintain electrical contact, and measure its position and voltage.

Set up the apparatus

Consider a rectangular coil mounted on a horizontal axle between two fixed magnetic pole pieces.

The north pole is on one side, the south pole is on the other, and the magnetic field points across the gap from north to south.

An electric motor turns the axle at a constant speed.

Now follow the electrical connections:

1. One end of the rotating coil connects to the first slip ring.
2. The other end connects to the second slip ring.
3. Two stationary carbon brushes press against the separate rings.
4. Wires connect the brushes to a voltmeter or oscilloscope.

The signal travels from the rotating coil, through the slip rings and brushes, and into the measuring instrument.

An angle sensor or marked scale on the axle records the coil’s orientation.

What should the experiment show?

Before collecting data, predict what the oscilloscope should display.

If the coil rotates at constant speed in a uniform magnetic field, the magnetic flux varies as a cosine function, while the induced emf varies as a sine function.

The ideal predictions are:

• At \(0^\circ\), the flux is maximum and the emf is zero.
• At \(90^\circ\), the flux is zero and the emf is maximum.
• At \(180^\circ\), the flux is minimum and the emf is zero.
• At \(270^\circ\), the flux is zero and the emf is maximum in the opposite direction.

With slip rings, the oscilloscope should show alternating positive and negative voltage.

These are theoretical predictions for an ideal generator, not claimed measurements from a particular experiment.

A real apparatus may produce a less perfect trace because of contact resistance, friction, variations in rotational speed, and a non-uniform magnetic field.

What does the experiment actually establish?

The experiment demonstrates that the induced emf depends on the rate of change of magnetic flux.

It also shows why the slip rings and brushes are necessary in a rotating-coil arrangement: the coil rotates, but the measuring instrument and external wires remain stationary.

However, this arrangement has one important limitation.

It only demonstrates a generator in which the coil rotates and the magnetic poles remain fixed. It does not establish that every generator must be built this way.

Revise the arrangement: rotate the magnetic field instead

Now imagine a different generator.

Keep the coil stationary and connect its ends directly to the measuring instrument. Instead of rotating the coil, rotate a permanent-magnet assembly so that the magnetic field direction at the coil changes with time.

The magnetic flux through the stationary coil can still change.

Faraday’s law still applies, so an induced emf is produced.

The labels are now different:

• The rotating magnetic assembly is the rotor.
• The stationary coil and supporting structure form the stator.
• The coil’s output wires connect directly to the external circuit.
• Brushes are not needed to collect the output from the stationary coil.

This is the crucial distinction: the rotor is defined by what rotates, not by whether it contains the coil or the magnets.

05Worked examples

06Questions and solutions

Question 1

A simple generator contains:

• Fixed north and south magnetic pole pieces.
• A rectangular coil attached to a rotating shaft.
• Two complete metal rings attached to the same shaft.
• Two stationary carbon blocks touching the metal rings.

The carbon blocks connect to an external circuit.

Identify the following:

A: The fixed magnetic pole assembly and frame.

B: The rotating coil, shaft, and attached metal rings.

C: The two carbon blocks.

State which assembly rotates and which remains stationary.

Solution 1

A is the stator, B is the rotor, and C represents the brushes. The rotor turns with the shaft, while the stator and brushes remain stationary.

The magnetic pole pieces and supporting frame do not move, so they form the stator.

The coil is attached to the shaft, and the rings are attached to the same shaft. All three rotate together, making them part of the rotor.

The carbon blocks maintain contact with the rings without rotating. They are the brushes.

The important distinction is between the ring and the brush touching it. The ring rotates, while the brush stays in position.

Question 2

An 80-turn rectangular coil has an area of \(0.012\,\mathrm{m^2}\). It rotates at \(5.0\) revolutions per second in a uniform magnetic field of \(0.25\,\mathrm{T}\).

At a particular instant, the angle between the magnetic field and the coil’s area normal is \(120^\circ\).

The angle is increasing in the direction of rotation.

Calculate:

(a) The magnetic flux through one turn.

(b) The instantaneous induced emf, using the positive sign convention defined in this article.

Solution 2

(a) The magnetic flux is \(-1.50\times10^{-3}\,\mathrm{Wb}\) per turn. (b) The induced emf is approximately \(+6.53\,\mathrm{V}\).

The negative flux indicates that the coil’s area normal points partly opposite to the magnetic field.

Part (a): Calculate the magnetic flux.

Use:

\[ \Phi=BA\cos\theta \]

Substitute:

\[ \begin{aligned} \Phi&=(0.25)(0.012)\cos120^\circ\\ &=0.0030(-0.5)\\ &=-0.00150\,\mathrm{Wb} \end{aligned} \]

Therefore:

\[ \boxed{\Phi=-1.50\times10^{-3}\,\mathrm{Wb}} \]

Part (b): Calculate the induced emf.

First convert the rotational frequency into angular velocity.

\[ \omega=2\pi f \]

\[ \omega=2\pi(5.0) \]

\[ \omega=31.42\,\mathrm{rad\,s^{-1}} \]

Differentiate the flux equation with respect to time:

\[ \frac{d\Phi}{dt}=-BA\omega\sin\theta \]

Substitute:

\[ \begin{aligned} \frac{d\Phi}{dt} &=-(0.25)(0.012)(31.42)\sin120^\circ\\ &=-0.0816\,\mathrm{Wb\,s^{-1}} \end{aligned} \]

Now apply Faraday’s law:

\[ \mathcal{E}=-N\frac{d\Phi}{dt} \]

\[ \mathcal{E}=-80(-0.0816) \]

\[ \boxed{\mathcal{E}=+6.53\,\mathrm{V}} \]

The positive emf occurs because the flux is becoming more negative as the coil continues rotating. Its rate of change is negative, so the induced emf is positive under our chosen convention.

A common mistake is to see negative flux and immediately assume negative emf. These are different quantities. The flux describes the magnetic field passing through the coil, while the emf depends on how quickly that flux changes.

Question 3

An experimental generator uses a 40-turn rectangular coil with an area of \(0.010\,\mathrm{m^2}\).

The coil rotates at \(2.0\) revolutions per second between fixed magnetic pole pieces. The magnetic field strength is \(0.15\,\mathrm{T}\).

The coil is connected to two complete slip rings, which contact stationary brushes connected to an oscilloscope.

The angle \(\theta\) is measured between the magnetic field and the coil’s area normal. At \(\theta=0^\circ\), the normal points along the magnetic field.

Answer the following:

(a) Identify the rotor, stator, and brushes.

(b) Calculate the maximum induced emf.

(c) Predict the induced emf at \(0^\circ\), \(90^\circ\), and \(180^\circ\).

(d) Explain why the emf is greatest at \(90^\circ\), even though the magnetic flux is zero.

Solution 3

(a) The coil, shaft, and slip rings form the rotor. The fixed magnetic poles and frame form the stator. The carbon contacts are the stationary brushes.

(b) The maximum induced emf is approximately \(0.754\,\mathrm{V}\).

(c) The emf is zero at \(0^\circ\), \(+0.754\,\mathrm{V}\) at \(90^\circ\), and zero again at \(180^\circ\).

(d) At \(90^\circ\), the magnetic flux is zero but changing at its maximum rate.

Part (a): Identify the components.

The coil and slip rings are attached to the rotating shaft, so they belong to the rotor.

The magnetic pole pieces remain fixed, so they form the stator.

The brushes remain stationary while maintaining electrical contact with the slip rings.

Part (b): Calculate the maximum emf.

Use:

\[ \mathcal{E}_{\max}=NBA\omega \]

First calculate angular velocity:

\[ \omega=2\pi f \]

\[ \omega=2\pi(2.0) \]

\[ \omega=12.57\,\mathrm{rad\,s^{-1}} \]

Substitute:

\[ \begin{aligned} \mathcal{E}_{\max} &=40(0.15)(0.010)(12.57)\\ &=0.754\,\mathrm{V} \end{aligned} \]

Therefore:

\[ \boxed{\mathcal{E}_{\max}=0.754\,\mathrm{V}} \]

Part (c): Predict the instantaneous emf.

For the chosen sign convention:

\[ \mathcal{E}=NBA\omega\sin\theta \]

At \(0^\circ\):

\[ \mathcal{E}=0.754\sin0^\circ=0\,\mathrm{V} \]

At \(90^\circ\):

\[ \mathcal{E}=0.754\sin90^\circ \]

\[ \mathcal{E}=+0.754\,\mathrm{V} \]

At \(180^\circ\):

\[ \mathcal{E}=0.754\sin180^\circ=0\,\mathrm{V} \]

The ideal oscilloscope trace therefore passes through zero, reaches a positive maximum, and returns to zero over the first half-turn.

Part (d): Explain the maximum emf.

At \(0^\circ\), the coil’s normal is parallel to the magnetic field. The flux is at its maximum positive value, but its instantaneous rate of change is zero.

At \(90^\circ\), the flux is zero because the magnetic field runs parallel to the plane of the coil. However, the coil is passing through this orientation at the point where the flux changes most rapidly.

Faraday’s law therefore predicts maximum emf at \(90^\circ\).

The tempting prediction that zero flux must mean zero emf confuses the value of flux with its rate of change.

Question 4

An ideal generator has a peak induced emf of \(12\,\mathrm{V}\).

Two versions of the generator are tested:

• Arrangement I has two separate, complete metal rings attached to the rotating shaft.
• Arrangement II has two semicircular metal segments separated by insulating gaps, also attached to the rotating shaft.

Both arrangements use stationary carbon brushes.

At a particular instant, the coil has rotated to \(\theta=210^\circ\).

Assume the external positive brush polarity is chosen so that the output is positive during the first half-turn.

Answer the following:

(a) Identify the ring arrangement that uses slip rings and the arrangement that uses a split-ring commutator.

(b) Calculate the induced emf in the rotating coil at \(210^\circ\).

(c) Determine the ideal external voltage for each arrangement at this instant.

(d) Explain why the two external voltages have different signs.

Solution 4

(a) Arrangement I uses slip rings, while Arrangement II uses a split-ring commutator.

(b) The induced emf in the coil is \(-6.0\,\mathrm{V}\).

(c) Arrangement I produces an ideal external voltage of \(-6.0\,\mathrm{V}\), while Arrangement II produces \(+6.0\,\mathrm{V}\).

(d) The split-ring commutator reverses the coil’s connections to the external circuit every half-turn, while the slip rings preserve the original connections.

Part (a): Identify the rings.

Two complete, separate rings indicate slip rings.

Two semicircular conducting segments separated by insulating gaps indicate a split-ring commutator.

Both are attached to the rotating shaft, and both make contact with stationary brushes.

Part (b): Calculate the induced emf.

Use the sinusoidal emf equation:

\[ \mathcal{E}=\mathcal{E}_{\max}\sin\theta \]

Substitute:

\[ \begin{aligned} \mathcal{E}&=12\sin210^\circ\\ &=12(-0.5)\\ &=-6.0\,\mathrm{V} \end{aligned} \]

The negative result means the induced emf has reversed direction relative to the first half-turn.

Part (c): Calculate the external voltage.

For Arrangement I, the slip rings maintain the same connections between the coil ends and the external brushes.

Therefore, the external voltage retains the sign of the coil emf:

\[ \boxed{V_{\mathrm{I}}=-6.0\,\mathrm{V}} \]

For Arrangement II, the split-ring commutator reverses the coil connections to the external circuit after half a revolution.

This reverses the external polarity relative to the coil’s instantaneous polarity:

\[ \boxed{V_{\mathrm{II}}=+6.0\,\mathrm{V}} \]

Part (d): Explain the difference.

The coil itself produces an alternating emf in both arrangements.

The difference is what happens when that emf reaches the external circuit.

Slip rings preserve the connection between each coil end and its corresponding brush. The external voltage therefore alternates.

The split-ring commutator swaps the connections every half-turn. This makes the external voltage maintain the same polarity, although its magnitude still falls to zero twice per revolution.

The commutator does not eliminate the alternating emf inside the coil. It changes how that emf is connected to the external circuit.

Question 5

Two ideal generators, P and Q, use identical rectangular coils.

Each coil has:

• \(100\) turns.
• An area of \(0.010\,\mathrm{m^2}\).
• A magnetic field strength of \(0.30\,\mathrm{T}\).

In both generators, the angle between the magnetic field and the coil’s area normal changes at a constant angular velocity of \(40\pi\,\mathrm{rad\,s^{-1}}\).

The two arrangements differ mechanically.

Generator P:

The coil rotates between stationary magnetic pole pieces. The coil ends connect to two slip rings on the shaft, which contact stationary brushes connected to an external circuit.

Generator Q:

The coil remains stationary. A permanent-magnet assembly rotates around it, producing a changing magnetic field direction at the coil. The field magnitude remains \(0.30\,\mathrm{T}\), and the angle between the field and the coil’s normal follows the same angular relationship as in Generator P.

The stationary coil connects directly to the external circuit.

Assume both generators have uniform magnetic fields across their coils, constant rotational speed, and no commutator.

A student makes three claims:

1. The coil must be the rotor in both generators because it is the component in which emf is induced.
2. Generator Q cannot produce an emf because its coil does not physically rotate.
3. Both generators need brushes to transfer their output to the external circuit.

Evaluate all three claims.

Then calculate the induced emf in each generator when \(\theta=90^\circ\).

Solution 5

All three student claims are incorrect. Both generators produce the same instantaneous emf, approximately \(+37.7\,\mathrm{V}\), at \(90^\circ\). However, their rotors, stators, and electrical connections are different.

This problem tests an important distinction between mechanical motion and electromagnetic induction.

Step 1: Identify the rotor and stator in each generator.

In Generator P, the coil rotates while the magnetic poles remain fixed.

Therefore:

• Rotor: coil, shaft, and slip rings.
• Stator: fixed magnetic pole pieces and frame.
• Brushes: stationary contacts touching the slip rings.

In Generator Q, the magnetic assembly rotates while the coil remains stationary.

Therefore:

• Rotor: permanent-magnet assembly and its rotating support.
• Stator: stationary coil and supporting structure.
• Brushes: not required for collecting the output from the stationary coil.

The first claim is incorrect because the coil is not necessarily the rotor.

The rotor is whichever assembly physically rotates.

Step 2: Determine whether Generator Q produces an emf.

The second claim is also incorrect.

Faraday’s law does not require the coil itself to rotate. It requires the magnetic flux through the coil to change.

In Generator P, the coil rotates relative to the magnetic field.

In Generator Q, the magnetic field rotates relative to the stationary coil.

Both arrangements produce the same changing angle between the magnetic field and the coil’s area normal.

The problem explicitly states that both generators have the same field magnitude and the same angle as a function of time.

Therefore, their magnetic flux functions are identical:

\[ \Phi=BA\cos\theta \]

Their rates of change of flux are also identical.

Consequently, their induced emfs are identical under the stated sign convention.

Step 3: Determine whether both generators require brushes.

The third claim is incorrect.

Generator P has a rotating coil, so its electrical connections must cross from the rotating assembly to the stationary external circuit.

It uses slip rings and stationary brushes to maintain that connection.

Generator Q has a stationary coil. Its output wires do not rotate, so they can connect directly to the external circuit.

No brushes are needed to collect its output.

This is why the mechanical design matters when labelling a generator diagram. A brush is needed when an electrical connection must cross a rotating-stationary boundary, not simply because a generator produces electricity.

Step 4: Calculate the emf at \(90^\circ\).

For both generators:

\[ \mathcal{E}=NBA\omega\sin\theta \]

At \(90^\circ\):

\[ \sin90^\circ=1 \]

Therefore, the induced emf is at its maximum positive value.

Substitute the supplied values:

\[ \begin{aligned} \mathcal{E} &=100(0.30)(0.010)(40\pi)\\ &=12\pi\\ &=37.7\,\mathrm{V} \end{aligned} \]

Thus:

\[ \boxed{\mathcal{E}_{P}=\mathcal{E}_{Q}=+37.7\,\mathrm{V}} \]

Both generators have zero magnetic flux at this instant because the magnetic field is parallel to the plane of the coil.

However, the flux is changing at its greatest rate, so the induced emf is maximum.

Step 5: Compare the external outputs.

Neither generator uses a commutator, so both produce alternating voltage.

Generator P transfers its output through slip rings and brushes.

Generator Q transfers its output through stationary wires connected directly to the coil.

The induced emf is the same because the magnetic flux changes in the same way, but the electrical connections are different because the rotating components are different.

This is the key result: the changing magnetic flux determines the induced emf, while the mechanical arrangement determines which components are the rotor, stator, and brushes.

07What to remember when labelling a generator

When you encounter an unfamiliar generator diagram in an HSC Physics question, work through these decisions:

1. Find the shaft or other rotating assembly.
2. Identify every component attached to it. These form the rotor.
3. Identify the components that remain fixed. These form the stator.
4. Find the stationary contacts touching any rotating metal rings. These are the brushes.
5. Check whether the rings are complete slip rings or a split-ring commutator.
6. Identify the magnetic field direction and the coil’s area normal before using the flux equation.

The biggest trap is assuming that the coil must always be the rotor. It is the rotating component in the common HSC rotating-coil model, but generators can also use a rotating magnetic field and stationary coils.

Once you can identify the moving components and the angle between the magnetic field and the coil’s normal, you can predict the magnetic flux and induced emf without guessing.

For the broader treatment of rotating coils, sinusoidal emf, and slip rings, continue with AC Generators: Sinusoidal EMF and Slip Rings Explained.