Brushed DC vs AC Induction Motors: How They Compare

Two motors can turn at almost exactly the same speed and still produce torque in completely different ways. One might have brushes touching a rotating commutator, while the other produces current inside its rotor without any electrical connection to it.

So, how do you compare a brushed DC motor with an AC induction motor?

The key is to follow the current. A brushed DC motor receives current directly in its rotating armature through brushes and a split-ring commutator. An AC induction motor produces current in its rotor through electromagnetic induction: AC in the stator creates a rotating magnetic field, which induces current in the rotor.

Both motors use the motor effect to produce torque, but they create and control their rotor currents differently.

01The essential differences at a glance

Before examining either motor, establish the central comparison.

The most important distinction is not simply that one uses DC and the other uses AC.

It is that the brushed DC motor receives its rotor current electrically, whereas the induction motor generates its rotor current through electromagnetic induction.

Let’s follow that difference through the actual machines.

02First, establish a fair comparison

Imagine investigating both motors using a supervised laboratory test rig.

The aim is to compare their operating mechanisms and examine how each responds to a mechanical load.

Each motor is mounted on a separate test stand. Its output shaft connects to a torque sensor and an adjustable dynamometer, which applies a controlled resisting torque.

A tachometer measures the rotational speed of the shaft.

The electrical connections are different:

• The brushed DC motor receives power from a DC supply, with voltage and current measured at its terminals.
• The induction motor receives three-phase AC through its stator windings, with its real electrical input power measured using an appropriate power meter.

The rotor of the induction motor has no direct electrical connection to the supply.

What should the investigation show?

Before collecting data, make two predictions.

For the brushed DC motor, increasing the mechanical load should cause the motor to slow slightly. Its back emf will decrease, allowing more current to flow through the armature. The increased current produces additional torque.

For the induction motor, increasing the mechanical load should also cause the rotor to slow slightly. However, the stator’s rotating magnetic field continues at its synchronous speed. The increased difference between field speed and rotor speed produces greater slip, allowing the motor to develop additional torque.

Both motors respond to the increased load, but their electrical mechanisms are different.

Why the first comparison might be misleading

Suppose you simply connect both motors to a 24 V supply and compare their current readings.

That is not a fair comparison.

The motors have different electrical designs and operating requirements. Their rated voltages may differ, and AC current readings alone do not tell you the real electrical power entering an induction motor. Its power factor also matters.

A better investigation operates each motor within its rated conditions, applies the same shaft torque, and measures:

• rotational speed,
• shaft torque,
• real electrical input power, and
• mechanical output power.

This allows you to compare their performance without confusing electrical supply characteristics with motor mechanisms.

The following values are illustrative test data, not measurements from a published experiment.

Notice that both motors produce the same measured shaft torque, but their rotational speeds differ.

This means their mechanical output powers will also differ slightly.

We will calculate those values shortly. First, let’s understand why the motors behave differently.

03How a brushed DC motor produces torque

Start with a rectangular coil placed between the north and south poles of a magnet.

The magnetic field is directed from north to south. When current passes through the coil, its two sides experience magnetic forces in opposite directions.

These forces form a couple, producing a turning effect called torque.

This is the motor effect.

For a straight current-carrying conductor, the magnetic force is

\[ F=BIL\sin\theta \]

where:

• \(F\) is magnetic force in newtons,
• \(B\) is magnetic flux density in tesla,
• \(I\) is current in amperes,
• \(L\) is the conductor length in metres, and
• \(\theta\) is the angle between the conductor and magnetic field.

The forces on opposite sides of the coil produce rotation.

However, there is a problem.

Why does the commutator need to reverse the current?

Imagine the coil completes half a revolution.

Its two sides have now exchanged positions in the magnetic field. If the current remained in the same direction through the coil, the forces would also change their turning effect.

The coil would experience torque in the opposite direction.

The motor would not maintain continuous rotation.

This is where the split-ring commutator becomes essential.

The commutator consists of two conducting segments connected to opposite ends of the armature coil. Stationary carbon brushes maintain electrical contact with these rotating segments.

Every half-turn, the commutator swaps the connections between the armature coil and the DC supply.

This reverses the current through the coil at the appropriate moment.

Because the coil has also changed orientation, the resulting magnetic forces continue producing torque in the same rotational direction.

The important point is that the external supply remains DC. The commutator mechanically switches the current direction within the rotating armature coil.

Without this switching, the motor could not maintain continuous rotation in the simple arrangement.

Where does the armature current come from?

The armature current is supplied directly by the DC source.

It travels through:

1. The supply connection.
2. The stationary carbon brushes.
3. The rotating commutator.
4. The armature windings.
5. The return brush and supply connection.

This is a direct electrical connection to the rotating part.

For a fixed magnetic field, the motor’s torque is proportional to its armature current:

\[ \tau=k_t I_a \]

where:

• \(\tau\) is electromagnetic torque in newton metres,
• \(k_t\) is the motor’s torque constant, and
• \(I_a\) is armature current in amperes.

Increasing armature current therefore increases the torque available from the motor, assuming the magnetic field remains constant.

But the DC motor also uses electromagnetic induction

There is an important detail that students sometimes miss.

As the armature rotates through the magnetic field, its conductors also experience electromagnetic induction.

This produces an induced voltage that opposes the applied supply voltage. It is called back emf.

The back emf is not what initially supplies the armature current. It is an induced voltage that affects how much current the DC supply can drive through the armature.

For a fixed magnetic field, we can model it as

\[ \mathcal{E}_b=k_e\omega \]

where:

• \(\mathcal{E}_b\) is back emf in volts,
• \(k_e\) is a motor constant, and
• \(\omega\) is angular speed in radians per second.

This gives an important relationship:

\[ \text{Higher speed} \rightarrow \text{Greater back emf} \]

If the motor slows under an increased load, its back emf decreases. More voltage is then available to drive current through the armature resistance, increasing the current and producing additional torque.

This is the DC motor’s response to a changing load.

For a more detailed treatment of this mechanism, see Back emf in a DC motor: HSC Physics explained.

04How an AC induction motor produces torque

An induction motor works differently.

Instead of supplying current directly to the rotor, it uses a rotating magnetic field to induce current inside the rotor.

For this comparison, we will use the common three-phase squirrel-cage induction motor.

How does the stator create a rotating magnetic field?

The stator is the stationary outer section of the motor. It contains three sets of windings arranged around the rotor.

These windings are separated by \(120^\circ\) electrically in their three-phase arrangement.

When connected to a balanced three-phase AC supply, the currents in the three windings also differ in phase by \(120^\circ\).

Each winding produces its own magnetic field.

Because the currents change continuously, the individual magnetic fields also change. Their combined effect is a resultant magnetic field that rotates around the stator.

This is called a rotating magnetic field.

An important distinction is that the stator windings themselves do not physically rotate. The combined magnetic field rotates because the three currents change in a coordinated way.

The rotating field is the key to the induction motor’s operation.

How does current appear in the rotor?

The rotor in a common squirrel-cage induction motor contains conducting bars running along its length.

These bars are joined together by conducting rings at both ends, forming closed conducting paths.

The rotating magnetic field passes the rotor bars, producing a changing magnetic flux through the rotor circuits.

According to Faraday’s law, this changing magnetic flux induces an emf.

\[ \mathcal{E}=-N\frac{\Delta\Phi}{\Delta t} \]

where:

• \(\mathcal{E}\) is induced emf in volts,
• \(N\) is the number of turns in a coil,
• \(\Delta\Phi\) is the change in magnetic flux in webers, and
• \(\Delta t\) is the time interval in seconds.

The negative sign represents Lenz’s law.

Because the rotor bars and end rings form closed circuits, the induced emf drives current through the rotor.

There is no need for brushes to deliver current to the rotor.

Now the rotor contains current-carrying conductors inside a magnetic field.

The motor effect produces magnetic forces on these conductors.

The combined forces produce torque, causing the rotor to rotate in the same direction as the rotating magnetic field.

The complete process is:

\[ \begin{aligned} &\text{AC supply to stator}\\ &\downarrow\\ &\text{Rotating magnetic field}\\ &\downarrow\\ &\text{Induced rotor emf}\\ &\downarrow\\ &\text{Induced rotor current}\\ &\downarrow\\ &\text{Magnetic force on rotor conductors}\\ &\downarrow\\ &\text{Torque and rotation} \end{aligned} \]

The crucial difference is that the rotor current is induced rather than supplied through brushes.

05Why does an induction motor need slip?

The induction motor has one more important feature that separates it from a brushed DC motor.

Its rotor cannot normally reach the speed of the rotating magnetic field while producing motor torque.

The speed of the rotating magnetic field is called synchronous speed.

It depends on the AC supply frequency and the number of magnetic poles.

\[ n_s=\frac{120f}{P} \]

where:

• \(n_s\) is synchronous speed in revolutions per minute (rpm),
• \(f\) is supply frequency in hertz (Hz), and
• \(P\) is the total number of magnetic poles.

For example, a four-pole induction motor connected to a 50 Hz supply has a synchronous speed of

\[ n_s=\frac{120(50)}{4} =1500\text{ rpm} \]

The rotor, however, must rotate slightly slower than this speed when operating as a motor under load.

Why?

If the rotor reached exactly \(1500\text{ rpm}\), the rotating magnetic field would no longer move relative to the rotor.

The magnetic flux through the rotor circuits would no longer change because of relative motion.

Consequently, the induced rotor emf would fall to zero in the idealised model. Without induced rotor current, there would be no electromagnetic torque.

The motor therefore needs a difference between the rotating field speed and rotor speed.

This difference is measured by slip.

\[ s=\frac{n_s-n_r}{n_s} \]

where:

• \(s\) is slip as a decimal,
• \(n_s\) is synchronous speed, and
• \(n_r\) is actual rotor speed.

To express slip as a percentage:

\[ s(\%)=\frac{n_s-n_r}{n_s}\times100 \]

For normal motoring operation, slip is positive.

A larger mechanical load generally causes the rotor to slow slightly, increasing slip. This increases the relative motion between the rotating field and rotor, allowing the induction motor to develop additional torque.

The frequency of the induced rotor current is also related to slip:

\[ f_r=sf \]

where \(f_r\) is the rotor-current frequency in hertz.

This is another useful difference from the brushed DC motor: the rotor-current frequency in an induction motor depends on the relative motion between the rotor and rotating field.

For a fuller explanation of rotating fields, rotor currents, and slip, see AC Induction Motors: Rotating Fields, Rotor Currents and Slip.

06Compare how each motor responds to an increased load

Both motors slow slightly when a larger resisting load is applied, but the reason they produce additional torque is different.

Brushed DC motor

The sequence is:

1. The mechanical load increases.
2. The motor slows slightly.
3. Its back emf decreases.
4. More voltage is available across the armature resistance.
5. Armature current increases.
6. The motor produces greater torque.

AC induction motor

The sequence is:

1. The mechanical load increases.
2. The rotor slows slightly.
3. The rotating stator field continues at synchronous speed.
4. Slip increases.
5. The changing magnetic flux in the rotor produces a greater induction effect.
6. The motor develops additional torque.

The two mechanisms are similar in that a small speed decrease helps the motor respond to a larger load.

However, do not confuse back emf with slip.

Back emf is an induced voltage in the rotating DC armature that opposes the supply. Slip measures the relative speed difference between the induction motor’s rotating magnetic field and its rotor.

They describe different physical effects.

07A direct comparison of the two mechanisms

A common misconception: both motors have slip

A student might see that both motors rotate slightly below some expected speed and conclude that both have slip.

That is incorrect.

Slip is specifically defined using the synchronous speed of a rotating magnetic field.

A conventional brushed DC motor does not have a rotating stator field whose speed is set by an AC supply frequency and pole number.

Its speed is instead influenced by the applied voltage, magnetic field strength, and mechanical load.

The DC motor does experience back emf, but back emf is not slip.

This is an important distinction when interpreting experimental data or answering a comparison question.

08Worked examples

09A reliable method for answering HSC comparison questions

When asked to compare a brushed DC motor with an AC induction motor, work through these five decisions.

1. Identify the source of rotor current.

• Brushed DC: current is delivered through brushes and a commutator.
• Induction: current is induced in the rotor by the rotating magnetic field.

2. Identify how the magnetic field behaves.

• Brushed DC: the stator produces a stationary magnetic field.
• Induction: the three-phase stator currents produce a rotating magnetic field.

3. Explain how torque is produced.

Both motors use the motor effect.

The difference is whether the rotor current is supplied directly or induced.

4. Use the correct speed relationship.

For an induction motor, use

\[ n_s=\frac{120f}{P} \]

and

\[ s=\frac{n_s-n_r}{n_s} \]

For a brushed DC motor, consider its back emf, armature current, and applied voltage instead.

5. Explain the response to an increased load.

• Brushed DC: reduced back emf allows greater armature current.
• Induction: increased slip allows the rotor to develop greater electromagnetic torque.

If you can explain these five points, you have the central comparison covered.

10Questions and solutions

Question 1

A student examines two electric motors.

Motor A contains a rotating armature connected to carbon brushes and a split-ring commutator. Motor B contains a squirrel-cage rotor surrounded by three-phase stator windings.

Explain how current reaches the rotor in each motor and identify the purpose of the commutator in Motor A.

Solution 1

Motor A receives its armature current directly from the DC supply through its brushes and commutator. Motor B develops induced rotor currents through electromagnetic induction, without a direct electrical connection to its squirrel-cage rotor.

In Motor A, the brushes maintain contact with the rotating commutator, allowing current to enter and leave the armature.

The split-ring commutator reverses the current through the armature coil every half-turn. This ensures that the magnetic forces continue producing torque in the same rotational direction.

Without the commutator, the torque on a simple rotating coil would reverse after each half-turn, preventing continuous rotation in the required direction.

Motor B works differently. Its three-phase stator windings produce a rotating magnetic field. This field induces an emf in the closed conducting paths of the squirrel-cage rotor, producing current.

The stator magnetic field then exerts forces on the current-carrying rotor conductors, producing torque.

The key distinction is that Motor A receives rotor current through electrical contacts, while Motor B produces its rotor current through induction.

Question 2

A brushed DC motor operates from a \(24\text{ V}\) supply. Its armature resistance is \(0.40\,\Omega\).

Initially, it draws \(3.0\text{ A}\). A larger mechanical load is applied, causing the current to increase to \(6.0\text{ A}\).

Assume the magnetic field remains constant and back emf is directly proportional to angular speed.

A separate four-pole induction motor operates from a 50 Hz supply. Its rotor speed decreases from \(1470\text{ rpm}\) to \(1440\text{ rpm}\) under an increased mechanical load.

For both motors:

a. Calculate the initial and final back emf of the DC motor.

b. Determine the final DC motor speed as a percentage of its original speed.

c. Calculate the initial and final slip of the induction motor.

d. Calculate the corresponding rotor-current frequencies.

e. Explain how both motors respond to the increased mechanical load, identifying the different electromagnetic mechanisms.

Solution 2

The DC motor’s back emf decreases from \(22.8\text{ V}\) to \(21.6\text{ V}\), and its final speed is approximately \(94.7\%\) of its original speed. The induction motor’s slip increases from \(2.0\%\) to \(4.0\%\), while its rotor-current frequency increases from \(1.0\text{ Hz}\) to \(2.0\text{ Hz}\).

The calculations show that both motors respond to the increased load by slowing down, but they use different electromagnetic feedback mechanisms.

Part a: Calculate the DC motor’s back emf.

Use the DC motor voltage relationship:

\[ V_s=\mathcal{E}_b+I_aR \]

Rearranging:

\[ \mathcal{E}_b=V_s-I_aR \]

Initially:

\[ \mathcal{E}_{b1}=24-(3.0)(0.40) \]

\[ \mathcal{E}_{b1}=22.8\text{ V} \]

After the load increases:

\[ \mathcal{E}_{b2}=24-(6.0)(0.40) \]

\[ \mathcal{E}_{b2}=21.6\text{ V} \]

The back emf decreases because the motor slows down.

Part b: Calculate the change in DC motor speed.

For a constant magnetic field,

\[ \mathcal{E}_b\propto\omega \]

Therefore,

\[ \frac{\omega_2}{\omega_1} = \frac{\mathcal{E}_{b2}}{\mathcal{E}_{b1}} \]

Substituting:

\[ \frac{\omega_2}{\omega_1} = \frac{21.6}{22.8} \]

\[ \frac{\omega_2}{\omega_1}=0.947 \]

Therefore, the final angular speed is

\[ 94.7\% \]

of the original angular speed.

The motor has slowed by approximately \(5.3\%\).

The increased current also means the electromagnetic torque increases. For a fixed magnetic field, torque is proportional to armature current:

\[ \frac{\tau_2}{\tau_1} = \frac{6.0}{3.0} =2.0 \]

The electromagnetic torque has doubled under the stated assumptions.

Part c: Calculate the induction motor’s slip.

First calculate synchronous speed:

\[ n_s=\frac{120f}{P} \]

\[ n_s=\frac{120(50)}{4} \]

\[ n_s=1500\text{ rpm} \]

Initially, the rotor speed is \(1470\text{ rpm}\).

\[ s_1=\frac{1500-1470}{1500} \]

\[ s_1=0.020 \]

Therefore,

\[ s_1=2.0\% \]

After the load increases, the rotor speed is \(1440\text{ rpm}\).

\[ s_2=\frac{1500-1440}{1500} \]

\[ s_2=0.040 \]

Therefore,

\[ s_2=4.0\% \]

The induction motor’s slip has doubled.

Part d: Calculate the rotor-current frequencies.

Use

\[ f_r=sf \]

Initially:

\[ f_{r1}=(0.020)(50) \]

\[ f_{r1}=1.0\text{ Hz} \]

After the load increases:

\[ f_{r2}=(0.040)(50) \]

\[ f_{r2}=2.0\text{ Hz} \]

The rotor-current frequency doubles.

Part e: Compare the two responses.

The tempting conclusion is that both motors slow down because their electrical supplies provide less energy.

That is not what the calculations show.

The DC motor slows, reducing its back emf. This allows more current to flow from the DC supply through the armature, increasing the torque.

The induction motor slows relative to its rotating stator field. This increases slip and changes the induction effect in the rotor, allowing the motor to develop additional torque.

Both motors respond to the increased mechanical load, but one uses back-emf feedback while the other relies on slip and induced rotor currents.

Question 3

A student builds a simple brushed DC motor using a rectangular coil between two magnetic poles.

The original motor has a split-ring commutator. The student replaces it with two continuous slip rings but leaves the DC supply unchanged.

a. Explain what happens to the current through the armature coil after half a revolution in each arrangement.

b. Predict the effect on the direction of torque in Version B.

c. Explain why the original split-ring commutator is required for continuous rotation in the simple DC motor.

Solution 3

Version A reverses the armature current every half-turn, while Version B maintains the same current direction through the coil. Consequently, the torque in Version B reverses after half a revolution, preventing the motor from maintaining continuous rotation in the same direction.

In Version A, the split-ring commutator swaps the connections between the armature coil and the DC supply every half-turn.

The coil current therefore reverses just as the two sides of the coil exchange positions in the magnetic field.

These two changes work together to maintain torque in the same rotational direction.

In Version B, continuous slip rings maintain the same electrical connection between each coil terminal and its corresponding supply terminal.

The current through the rotating coil therefore does not reverse when the coil completes half a revolution.

However, the coil sides have exchanged positions in the magnetic field. The magnetic forces now produce torque in the opposite rotational direction.

The motor cannot maintain continuous rotation in the original direction.

The tempting idea is that both arrangements provide a continuous electrical connection, so they should work equally well.

The missing detail is that the DC motor needs the current direction within the rotating coil to change at precisely the correct point in its rotation.

Slip rings do not perform this switching. The split-ring commutator does.

Question 4

A three-phase squirrel-cage induction motor contains 24 conducting rotor bars joined by two conducting end rings.

One rotor bar fractures near one end, making that particular bar electrically open. All other bars and both end rings remain electrically continuous.

The motor remains connected to a 50 Hz, four-pole supply and continues rotating.

a. Will all induced rotor currents disappear because one bar is broken? Explain.

b. Will the stator’s synchronous speed change?

c. Explain how the motor may be affected when operating under the same mechanical load.

d. Can the new rotor speed or slip be calculated from the information provided? Justify your answer.

Solution 4

No. The motor can continue producing torque because the remaining intact rotor bars and end rings still provide closed conducting paths. Its synchronous speed remains \(1500\text{ rpm}\), although its rotor behaviour may be affected by the broken bar.

Part a: Determine whether rotor current disappears.

The tempting prediction is that breaking one conducting bar interrupts the rotor circuit and therefore stops all induced current.

However, a squirrel-cage rotor contains many conducting bars connected through the end rings.

The broken bar can no longer participate in its normal closed conducting paths, but the remaining intact bars can still form closed circuits.

The rotating magnetic field therefore continues inducing emf and current in the intact rotor circuits.

The motor can still produce torque.

Part b: Determine the synchronous speed.

Synchronous speed depends on supply frequency and the number of stator poles.

It does not depend on whether an individual rotor bar is damaged.

\[ n_s=\frac{120f}{P} \]

\[ n_s=\frac{120(50)}{4} \]

\[ n_s=1500\text{ rpm} \]

The synchronous speed remains \(1500\text{ rpm}\).

Part c: Explain the effect of the broken bar.

The broken bar changes the distribution of induced currents throughout the rotor.

The remaining intact paths can still produce torque, but the rotor’s electromagnetic behaviour is no longer as symmetrical as it was before the damage.

The motor may experience reduced torque capability, increased torque fluctuations, or additional vibration.

For the same mechanical load, the motor may need to operate at a different slip to establish a new steady operating condition.

Part d: Determine whether the new speed can be calculated.

No. The number of broken bars alone is insufficient to calculate the new rotor speed or slip.

We would also need information about the motor’s electrical characteristics, rotor resistance, load torque, and how the broken bar changes the distribution of rotor currents.

The important distinction is between the stator field and rotor behaviour.

The synchronous speed remains fixed by the AC supply and pole number, while the actual rotor speed can change because the rotor’s ability to develop torque has been affected.

A broken rotor bar does not automatically stop an induction motor. The remaining closed conducting paths still matter.

Question 5

A brushed DC motor is mechanically coupled to a three-phase squirrel-cage induction machine through a common shaft.

The DC motor is supplied by a controlled DC drive and remains in motoring operation.

The induction machine is connected to a three-phase electrical grid that can provide the magnetising current required by the machine and accept real electrical power.

The induction machine has four poles and operates on a 50 Hz supply.

An adjustable brake applies a constant resisting torque to the common shaft.

Initially, the common shaft rotates clockwise at \(1470\text{ rpm}\). A controlled drive gradually increases the shaft speed, passing through \(1500\text{ rpm}\) and reaching \(1530\text{ rpm}\).

Assume:

• The stator phase sequence remains unchanged throughout.
• The direction of rotation remains clockwise.
• The shaft speed is controlled at each stated operating point.
• The induction machine can operate as a generator when driven above synchronous speed.
• Neglect mechanical and electrical losses when analysing the idealised energy transfers, but identify where losses would occur in a real system.

a. Calculate the synchronous speed of the induction machine.

b. At \(1470\text{ rpm}\), determine its slip and explain the direction of its electromagnetic torque. Describe the main energy transfers occurring in the coupled system.

c. At exactly \(1500\text{ rpm}\), determine the induced rotor emf, rotor current, and electromagnetic torque in the idealised model. Explain why the common shaft does not stop rotating.

d. At \(1530\text{ rpm}\), calculate the slip and explain the direction of electromagnetic torque. Describe the energy transfers through both machines.

e. Explain why the induction machine can change from a motor to a generator without reversing its stator magnetic field. Why does this not create energy from nothing?

Solution 5

The induction machine has a synchronous speed of \(1500\text{ rpm}\). At \(1470\text{ rpm}\), it operates as a motor and contributes torque to the common shaft. At exactly \(1500\text{ rpm}\), its idealised rotor-induced torque is zero. At \(1530\text{ rpm}\), it operates as a generator, opposing the shaft’s rotation and transferring mechanical energy back to the AC grid.

The key is that the relative motion between the rotor and rotating magnetic field changes sign as the shaft passes through synchronous speed.

Part a: Calculate synchronous speed.

Use

\[ n_s=\frac{120f}{P} \]

Substitute \(f=50\text{ Hz}\) and \(P=4\):

\[ n_s=\frac{120(50)}{4} \]

\[ \boxed{n_s=1500\text{ rpm}} \]

The stator magnetic field continues rotating clockwise at \(1500\text{ rpm}\) throughout the entire process.

Part b: Analyse the machine at \(1470\text{ rpm}\).

The rotor speed is below synchronous speed.

Using

\[ s=\frac{n_s-n_r}{n_s} \]

gives

\[ s=\frac{1500-1470}{1500} \]

\[ s=0.020 \]

Therefore,

\[ \boxed{s=+2.0\%} \]

The positive slip means the stator magnetic field is moving ahead of the rotor.

The relative motion induces emf and current in the rotor. The resulting magnetic forces produce torque in the clockwise direction.

The induction machine is therefore operating as a motor.

The main energy transfers are:

• Electrical energy from the AC grid is converted into mechanical energy by the induction machine.
• Electrical energy from the DC supply is converted into mechanical energy by the brushed DC motor.
• Both machines can contribute torque to the common shaft.
• The brake converts mechanical energy into thermal energy.

The DC drive can adjust its torque to maintain the required shaft speed while the induction motor assists with the mechanical load.

The important point is that the induction machine is consuming electrical energy and contributing mechanical torque.

Part c: Analyse the machine at exactly \(1500\text{ rpm}\).

At this speed,

\[ n_r=n_s \]

Therefore,

\[ s=\frac{1500-1500}{1500}=0 \]

The rotor and rotating magnetic field move together.

In the idealised model, there is no relative motion between the rotor and the magnetic-field pattern.

Consequently,

\[ \mathcal{E}_{\text{rotor}}=0 \]

The induced rotor current is therefore zero:

\[ I_{\text{rotor}}=0 \]

And the electromagnetic torque is

\[ \boxed{\tau_{\text{induction}}=0} \]

A tempting prediction is that the entire coupled system must stop because the induction machine is no longer producing torque.

That is incorrect.

The brushed DC motor is still connected to its supply and continues producing torque. It can maintain the shaft’s rotation and provide the torque needed to overcome the brake.

The induction machine has simply reached the point where its idealised rotor induction mechanism produces no torque.

In a real induction machine, the stator may still draw magnetising current, and real losses still occur.

Part d: Analyse the machine at \(1530\text{ rpm}\).

The rotor is now moving faster than the rotating magnetic field.

Calculate slip:

\[ s=\frac{1500-1530}{1500} \]

\[ s=-0.020 \]

Therefore,

\[ \boxed{s=-2.0\%} \]

The negative slip indicates that the rotor is overtaking the rotating magnetic field.

This reverses the relative motion responsible for the induction process.

According to Lenz’s law, the induced electromagnetic effect opposes this relative motion.

The resulting electromagnetic torque therefore acts anticlockwise, opposing the clockwise rotation of the common shaft.

The induction machine is now operating as a generator.

The energy transfers have changed:

• The DC supply provides electrical energy to the brushed DC motor.
• The DC motor converts that energy into mechanical energy at the common shaft.
• The induction machine absorbs mechanical energy from the shaft.
• The induction machine converts that mechanical energy into electrical energy and returns real power to the AC grid.
• The brake continues converting some mechanical energy into thermal energy.

The DC motor must now provide enough torque to overcome both the brake and the opposing electromagnetic torque of the induction generator.

The induction machine is no longer helping drive the shaft. It is resisting the shaft while generating electrical power.

Part e: Explain the change in operating mode.

The stator magnetic field never changes direction.

It continues rotating clockwise at \(1500\text{ rpm}\), because the AC frequency and number of poles remain unchanged.

What changes is the rotor’s speed relative to that field.

Below synchronous speed, the field overtakes the rotor. The induction machine produces torque in the direction of rotation and operates as a motor.

Above synchronous speed, the rotor overtakes the field. The induced electromagnetic torque opposes the rotor’s rotation, and the machine can operate as a generator.

The important result is that a change in relative motion reverses the direction of electromagnetic torque without requiring the stator field to reverse.

This also explains why there is no energy created from nothing.

When the induction machine generates electrical power, the energy comes from mechanical work supplied through the common shaft. That work ultimately comes from the DC supply driving the brushed DC motor, with additional energy dissipated through losses in a real system.

The induction machine can return electrical energy to the grid because the DC motor is supplying mechanical energy to it.

The complete comparison therefore comes down to two different ways of producing rotor current:

• A brushed DC motor uses brushes and a commutator to supply and switch current in its rotating armature.
• An AC induction motor uses a rotating magnetic field to induce current in its rotor, with slip providing the relative motion needed for torque.

Once you understand that difference, the next useful step is studying motor torque-speed relationships. Back emf explains how a brushed DC motor responds to load, while slip explains how an induction motor adjusts its torque. Those two ideas provide the foundation for understanding why different motors behave differently under changing operating conditions.