Can a Motor Return Electrical Energy to Its Supply?

An electric bike is travelling downhill. Its wheels are turning the motor, rather than the motor turning the wheels. Could the motor send electrical energy back into the battery?

Yes. A motor can return electrical energy to its supply when it is driven mechanically fast enough to operate as a generator. This is the principle behind regenerative braking.

But there is a catch. A spinning motor does not automatically return energy. Its back emf must overcome the supply voltage and relevant circuit resistance, allowing the current to reverse. The supply must also be capable of accepting that electrical energy.

To understand why, we need to look at something that makes a motor quite different from an ordinary resistor: back emf.

01Why a motor does not behave like a fixed resistor

Imagine connecting a DC motor to a 12 V power supply.

Initially, the motor is stationary. Current flows through its windings, producing a magnetic field that interacts with the external magnetic field. This produces a torque, causing the motor to rotate.

As the motor speeds up, something else happens.

The rotating windings move through the magnetic field, inducing an electromotive force (emf). According to Lenz’s law, the induced emf acts in a direction that opposes the electrical effect producing it.

This opposing emf is called back emf.

It is not a separate voltage supplied by another battery. It is generated by the motor’s own rotation.

What happens as the motor speeds up?

Predict what happens to the current if the supply voltage stays at 12 V but the motor begins rotating faster.

You might initially expect the current to remain constant because the supply voltage has not changed.

That prediction would be correct for an ideal fixed resistor. It is not correct for a motor.

As rotational speed increases, back emf increases. This reduces the voltage available to drive current through the windings.

For a simple permanent-magnet DC motor:

\[ V_s=E_b+IR_w \]

where:

• \(V_s\) is the supply voltage across the motor, in volts (V).
• \(E_b\) is the motor’s back emf, in volts (V).
• \(I\) is the current through the windings, in amperes (A).
• \(R_w\) is the winding resistance, in ohms (\(\Omega\)).

Rearranging:

\[ \boxed{I=\frac{V_s-E_b}{R_w}} \]

This equation describes the motor under steady-state conditions, assuming a constant winding resistance and negligible resistance in the external circuit.

We define positive current as current flowing from the supply’s positive terminal into the motor’s positive terminal.

The equation reveals something important: the current depends on both winding resistance and rotational speed.

For a motor with constant magnetic field strength, back emf is approximately proportional to angular speed:

\[ E_b=k_e\omega \]

Here, \(k_e\) is the motor’s back-emf constant, and \(\omega\) is its angular speed in radians per second.

As the motor spins faster, \(E_b\) increases.

Three possible operating conditions

The relationship between supply voltage and back emf determines the direction of current.

The third condition is the interesting one.

When back emf exceeds supply voltage, the induced emf can drive current in the opposite direction. The motor is now acting as a generator.

Notice that back emf exists during ordinary motor operation too. Its presence alone does not mean the motor is generating electricity for the supply.

The important change is that the current reverses.

02Why winding resistance still matters

Back emf explains why a motor can reverse its current, but winding resistance determines how much current flows.

A motor is not an ideal generator. Its copper windings have resistance, so some of the electrical energy is always converted into thermal energy.

The power dissipated as heat in the windings is:

\[ \boxed{P_{\text{heat}}=I^2R_w} \]

Notice that the current is squared.

This means winding resistance produces heating whether current flows forwards or backwards.

The direction of current changes the direction of energy transfer at the motor terminals, but it does not make resistive heating negative.

03How can we experimentally show that a motor returns energy?

We can investigate this by mechanically driving a motor while monitoring the direction of current through its electrical terminals.

The important distinction is between measuring an induced voltage and demonstrating an actual transfer of electrical energy.

The first experiment: Does rotation generate an emf?

Start with a small permanent-magnet DC motor.

Connect a voltmeter across its two terminals, leaving the motor otherwise disconnected.

Mechanically rotate its shaft using a separate drive motor.

The voltmeter should register a potential difference across the motor terminals. Increasing the rotational speed should increase the magnitude of the generated emf.

This supports the prediction that rotation through a magnetic field produces an induced emf.

However, there is a limitation.

Because the motor’s terminals are effectively open-circuit, almost no current flows.

Electrical power transfer is:

\[ P=VI \]

If current is approximately zero, the electrical power delivered to an external circuit is also approximately zero.

The first experiment demonstrates voltage generation, but it does not demonstrate energy being returned to a supply.

Revising the experiment: Measure current direction

To test energy transfer, the circuit needs to be completed.

Use the following arrangement:

• A test motor is mounted securely on a bench.
• Its shaft is mechanically coupled to a separate drive motor.
• Its two electrical terminals are connected to a bidirectional DC supply through a signed current sensor.
• A voltmeter is connected across the test motor’s terminals.
• A speed sensor measures the rotational speed of the test motor.

The bidirectional supply is important. It must be capable of both supplying electrical power and absorbing returned electrical power.

An ordinary bench power supply cannot necessarily do both.

What should happen?

Suppose the supply voltage is held at 12 V and the test motor has a winding resistance of \(2.0\ \Omega\).

As the drive motor increases the test motor’s rotational speed, its back emf increases.

We can predict the current using:

\[ I=\frac{12-E_b}{2.0} \]

The following table contains illustrative calculated values, not measurements from an actual experiment.

The negative current values indicate that current is flowing out of the motor’s positive terminal and towards the supply’s positive terminal.

This is the decisive result.

As the externally driven motor spins faster, the back emf eventually exceeds the supply voltage. The current reverses, allowing electrical power to flow back into the supply.

The revised experiment addresses the original limitation: it measures not just whether voltage is generated, but whether electrical energy actually flows back into the supply.

05Regenerative braking: Where does this happen in real life?

One of the most useful applications is regenerative braking in electric vehicles.

During ordinary driving, the battery supplies electrical energy to the motor. The motor produces torque, which turns the wheels.

During regenerative braking, the process reverses.

The vehicle is still moving, so its wheels continue rotating. Instead of the motor driving the wheels, the wheels mechanically drive the motor.

The motor generates an emf, and if the electrical conditions allow reverse current, energy flows towards the battery.

Why does the vehicle slow down?

This is where Lenz’s law becomes important.

The induced current creates a magnetic effect that opposes the mechanical motion responsible for generating it.

Consequently, the motor produces an electromagnetic torque that opposes the wheel’s rotation.

The vehicle slows down.

The kinetic energy of the moving vehicle is converted into electrical energy, with some energy also being dissipated through resistive heating and other losses.

Three situations that students often confuse

A motor that generates an emf is not necessarily returning energy to its supply.

Consider what happens in three different circuit arrangements.

The second situation is called dynamic braking.

It can slow a motor, but it does not recharge the battery. The electrical energy is dissipated as heat, mainly in the connected resistance and the motor windings.

Regenerative braking is different because the electrical energy is transferred to a supply that can accept it.

Does all the braking energy return to the battery?

No.

Even when regenerative braking works, some energy is lost through:

• Heating in the motor windings.
• Internal resistance of the battery.
• Magnetic losses within the motor.
• Friction in the mechanical system.
• Energy conversion losses in the controller.

There are also practical limits on how much current a battery can accept.

If a battery is fully charged or its charging conditions do not permit additional current, regenerative braking may be restricted.

A conventional DC power supply may also be unable to absorb reverse current. A suitable regenerative drive or bidirectional power converter is needed.

The key distinction is that generating an emf, transferring electrical power, and storing energy in a battery are three related but different things.

For the underlying relationship between power and total electrical energy, see Electrical Energy in HSC Physics: Using E = Pt.

A related but different energy-conversion process occurs in How Solar Cells Convert Light into Electrical Energy. A motor in generating mode converts mechanical energy into electrical energy, whereas a solar cell converts energy from light.

06Questions and solutions

Question 1

A permanent-magnet DC motor is connected to an ideal 9.0 V supply.

At its current rotational speed, the motor has a back emf of 6.0 V and a winding resistance of \(1.0\ \Omega\).

Calculate the current through the motor and determine whether it is consuming or returning electrical energy.

Assume positive current flows from the supply into the motor.

Solution 1

The motor draws a current of \(+3.0\text{ A}\), so it consumes electrical energy from the supply.

Use the motor equation:

\[ I=\frac{V_s-E_b}{R_w} \]

Substitute:

\[ I=\frac{9.0-6.0}{1.0} \]

\[ \boxed{I=+3.0\text{ A}} \]

The positive sign means current flows into the motor from the supply.

We can also calculate the electrical power entering the motor:

\[ P=V_sI \]

\[ P=9.0\times3.0 \]

\[ P=27\text{ W} \]

Therefore, the supply transfers electrical energy to the motor at 27 J per second.

The back emf is smaller than the supply voltage, so the motor is operating as a motor rather than returning energy to the supply.

Question 2

A DC motor has a winding resistance of \(0.60\ \Omega\) and is connected to a constant 12 V supply.

When the motor is held stationary, its back emf is zero.

When it reaches its operating speed, its back emf increases to 9.6 V.

Calculate the current in both situations, assuming steady-state conditions and ignoring inductance and external circuit resistance.

Explain why the current decreases as the motor speeds up, even though the supply voltage remains constant.

Solution 2

The current is 20 A when stationary and 4.0 A at operating speed.

The difference occurs because the rotating motor generates an increasing back emf.

Stationary motor

When the motor is stationary:

\[ E_b=0 \]

Therefore:

\[ I=\frac{12-0}{0.60} \]

\[ \boxed{I=20\text{ A}} \]

This is the idealised steady stall current. In an actual motor, the current may be limited by the supply and other circuit effects.

Motor at operating speed

When the back emf reaches 9.6 V:

\[ I=\frac{12-9.6}{0.60} \]

\[ I=\frac{2.4}{0.60} \]

\[ \boxed{I=4.0\text{ A}} \]

The current has decreased significantly.

As the motor speeds up, its rotation generates a larger opposing emf. This leaves a smaller net voltage across the winding resistance, reducing current.

A common mistake is to calculate current using \(I=V/R\) for both situations.

That treats the entire motor as a fixed resistor. The winding resistance is the same in both cases, but the back emf is different.

This is precisely why a motor needs a model that includes both resistance and induced emf.

Question 3

A DC motor is mechanically driven by the wheels of an electric vehicle during regenerative braking.

At a particular instant:

• The motor’s back emf is 15 V.
• The battery has an emf of 12 V.
• The motor winding resistance is \(1.0\ \Omega\).
• The battery’s internal resistance is \(0.50\ \Omega\).

Assume the battery can accept charge, and ignore all other circuit resistance and energy losses.

Calculate:

(a) The charging current entering the battery.

(b) The battery’s terminal voltage while charging.

(c) The electrical power being stored chemically in the battery.

(d) The power dissipated as heat in the motor windings and the battery’s internal resistance.

(e) The total energy stored chemically in the battery over 10 seconds, assuming all quantities remain constant.

Solution 3

The charging current is 2.0 A, the battery terminal voltage is 13 V, and the battery stores 240 J of chemical energy in 10 seconds.

(a) Charging current

The motor’s generated emf must overcome the battery emf and both series resistances.

The total resistance is:

\[ R_{\text{total}}=1.0+0.50 \]

\[ R_{\text{total}}=1.50\ \Omega \]

Therefore:

\[ I=\frac{E_b-\mathcal{E}_{\text{bat}}} {R_{\text{total}}} \]

where \(\mathcal{E}_{\text{bat}}\) is the battery’s emf.

Substitute:

\[ I=\frac{15-12}{1.50} \]

\[ \boxed{I=2.0\text{ A}} \]

The positive charging current flows into the battery’s positive terminal.

(b) Battery terminal voltage

During charging, the battery’s terminal voltage is greater than its emf because current passes through its internal resistance.

Use:

\[ V_{\text{terminal}} = \mathcal{E}_{\text{bat}}+Ir_{\text{bat}} \]

Substitute:

\[ V_{\text{terminal}} = 12+(2.0)(0.50) \]

\[ \boxed{V_{\text{terminal}}=13\text{ V}} \]

The motor terminal voltage must also be 13 V.

We can verify this using the motor:

\[ V_{\text{motor}}=E_b-IR_w \]

\[ V_{\text{motor}}=15-(2.0)(1.0) \]

\[ \boxed{V_{\text{motor}}=13\text{ V}} \]

Both sides of the circuit agree.

(c) Power stored chemically in the battery

The battery’s chemical energy increases at a rate determined by its emf multiplied by the charging current.

\[ P_{\text{chemical}}=\mathcal{E}_{\text{bat}}I \]

\[ P_{\text{chemical}}=12\times2.0 \]

\[ \boxed{P_{\text{chemical}}=24\text{ W}} \]

Notice that we use the battery’s emf here, not its terminal voltage. The difference accounts for energy dissipated inside the battery.

(d) Heating losses

For the motor windings:

\[ P_{\text{motor heat}}=I^2R_w \]

\[ P_{\text{motor heat}}=(2.0)^2(1.0) \]

\[ \boxed{P_{\text{motor heat}}=4.0\text{ W}} \]

For the battery’s internal resistance:

\[ P_{\text{battery heat}}=I^2r_{\text{bat}} \]

\[ P_{\text{battery heat}}=(2.0)^2(0.50) \]

\[ \boxed{P_{\text{battery heat}}=2.0\text{ W}} \]

The energy generated electromagnetically inside the motor is:

\[ P_{\text{generated}}=E_bI \]

\[ P_{\text{generated}}=15\times2.0 \]

\[ P_{\text{generated}}=30\text{ W} \]

The energy balance is:

\[ 30=24+4+2 \]

The 30 W of generated electrical power is divided into 24 W stored chemically, 4 W of winding heating, and 2 W of battery internal heating.

(e) Energy stored over 10 seconds

Using:

\[ E=Pt \]

\[ E=24\times10 \]

\[ \boxed{E=240\text{ J}} \]

The battery stores 240 J of chemical energy over the 10-second interval.

This example demonstrates why regenerative braking cannot return all the mechanical energy converted by the motor. Even in this simplified model, some energy is dissipated as heat.

Question 4

A student spins a DC motor mechanically using a rotating flywheel.

The motor is tested under three conditions:

• Condition A: The motor terminals are disconnected.
• Condition B: The motor is connected across a resistor.
• Condition C: The motor is connected to a rechargeable battery whose emf is lower than the motor’s generated emf.

For each condition, explain:

(a) Whether an induced emf exists.

(b) Whether current flows through the motor.

(c) Where the mechanical energy goes.

(d) Whether the motor produces an electromagnetic braking torque.

Explain why measuring a voltage across the disconnected motor does not prove that electrical energy is being returned to a supply.

Solution 4

The motor generates an emf in all three conditions, but electrical current and energy transfer depend on whether a complete circuit exists and what is connected to it.

Condition A: Disconnected terminals

An induced emf exists because the motor is rotating through a magnetic field.

However, the terminals are open-circuit, so almost no current flows.

Therefore:

\[ P=VI\approx0 \]

The motor does not return significant electrical energy to an external supply.

In the simplified model, there is also no significant electromagnetic braking torque from armature current.

The flywheel can continue rotating, although mechanical friction and other losses will gradually reduce its speed.

Condition B: Connected to a resistor

An induced emf still exists.

However, this time there is a complete electrical circuit, so current flows through the motor and resistor.

The generated electrical energy is converted into thermal energy through resistance.

The motor also develops an electromagnetic torque opposing the flywheel’s rotation, causing it to slow down.

This is dynamic braking.

The energy is dissipated as heat rather than stored in a battery.

Condition C: Connected to a rechargeable battery

The induced emf is greater than the battery’s emf, so current can flow into the battery’s positive terminal, provided the battery can accept the charging current.

The battery stores some of the transferred energy chemically.

The motor also produces an opposing electromagnetic torque, slowing the flywheel.

This is regenerative operation.

Why the voltmeter alone is insufficient

A voltmeter measures potential difference, not power.

A motor can generate a measurable voltage while its terminals are open-circuit. But without significant current, there is no significant electrical power transfer.

To demonstrate energy being returned, we need evidence of both a potential difference and current flowing in the correct direction.

Question 5

An electric vehicle uses a permanent-magnet DC motor connected directly to a rechargeable 12.0 V battery.

Assume the following:

• The battery can accept charging current.
• The motor winding resistance is \(1.0\ \Omega\).
• The battery’s internal resistance is \(0.50\ \Omega\).
• The total series resistance is therefore \(1.50\ \Omega\).
• The motor’s back emf is 18 V when it rotates at 3000 rpm.
• The motor’s back emf is directly proportional to its rotational speed.
• There is no voltage-converting controller between the motor and battery.
• The motor speed is proportional to the vehicle’s forward speed.
• The vehicle’s total kinetic energy is proportional to the square of its speed.

During regenerative braking, the wheels drive the motor.

(a) Calculate the charging current at 3000 rpm.

(b) Determine the rotational speed at which regenerative charging stops.

(c) Calculate the fraction of the vehicle’s original kinetic energy that remains when regenerative charging first stops.

(d) A student claims:

“The motor is generating electricity at 3000 rpm, so it will keep returning energy to the battery until the vehicle stops. If there are no conventional friction brakes, all the vehicle’s original kinetic energy must end up stored in the battery.”

Explain why this reasoning is incorrect.

In your explanation, determine what would happen to the current if the motor remained directly connected to the battery at 1000 rpm.

Solution 5

Regenerative charging initially occurs at 4.0 A, but stops at 2000 rpm. At that point, the vehicle still has \(4/9\) of its original kinetic energy.

The student’s conclusion is incorrect because back emf decreases with rotational speed, and the motor cannot continue driving charging current into a fixed-voltage battery once its generated emf falls below the battery’s emf.

(a) Charging current at 3000 rpm

The motor’s back emf is 18 V, while the battery’s emf is 12 V.

The charging current is:

\[ I=\frac{E_b-\mathcal{E}_{\text{bat}}} {R_{\text{total}}} \]

Substitute:

\[ I=\frac{18-12}{1.50} \]

\[ \boxed{I=4.0\text{ A}} \]

Current flows from the motor into the battery, so regenerative charging occurs.

(b) Speed at which regenerative charging stops

Regenerative charging stops when the motor’s back emf equals the battery’s emf.

At this point:

\[ E_b=12\text{ V} \]

We know that back emf is proportional to rotational speed.

Therefore:

\[ \frac{E_{b,2}}{E_{b,1}} = \frac{\omega_2}{\omega_1} \]

Substitute:

\[ \frac{12}{18} = \frac{\omega_2}{3000} \]

Rearrange:

\[ \omega_2 = 3000\times\frac{12}{18} \]

\[ \boxed{\omega_2=2000\text{ rpm}} \]

At 2000 rpm, the induced emf exactly balances the battery emf.

There is no net voltage to drive charging current through the circuit.

(c) Fraction of kinetic energy remaining

This is where the final condition in the question becomes important.

Kinetic energy is:

\[ E_k=\frac12 mv^2 \]

Since vehicle speed is proportional to motor rotational speed, kinetic energy is proportional to the square of rotational speed:

\[ E_k\propto\omega^2 \]

Therefore:

\[ \frac{E_{k,2}}{E_{k,1}} = \left(\frac{\omega_2}{\omega_1}\right)^2 \]

Substitute:

\[ \frac{E_{k,2}}{E_{k,1}} = \left(\frac{2000}{3000}\right)^2 \]

\[ = \left(\frac23\right)^2 \]

\[ \boxed{\frac{E_{k,2}}{E_{k,1}}=\frac49} \]

The vehicle still has \(4/9\), or approximately 44.4%, of its original kinetic energy when regenerative charging stops.

The fraction of its original kinetic energy that has been removed is:

\[ 1-\frac49=\frac59 \]

So \(5/9\), or approximately 55.6%, of its initial kinetic energy has been removed by the time the charging current reaches zero.

Importantly, this does not mean that \(5/9\) of the original kinetic energy has been stored in the battery. Some energy has already been dissipated through resistance and other losses.

(d) What happens below 2000 rpm?

The student’s assumption that regenerative charging continues until the vehicle stops is incorrect.

Consider the motor at 1000 rpm.

Because back emf is proportional to rotational speed:

\[ E_b=18\times\frac{1000}{3000} \]

\[ \boxed{E_b=6.0\text{ V}} \]

The battery still has an emf of 12 V.

Therefore, if the motor remains directly connected, the battery now drives current through the motor:

\[ I=\frac{12-6.0}{1.50} \]

\[ \boxed{I=4.0\text{ A}} \]

This current flows into the motor rather than into the battery.

The motor has returned to motoring operation. Its electromagnetic torque now tends to drive the wheel in its original direction rather than oppose its rotation.

This is the crucial limitation of the direct-connection model.

As the vehicle slows down, the motor’s generated emf falls. Eventually it is no longer large enough to overcome the battery’s emf, so regenerative charging stops.

A real vehicle’s control system must prevent the battery from simply driving the motor below this operating threshold. The remaining kinetic energy must be handled through other mechanisms, such as conventional friction braking, unless a suitable power converter allows regenerative operation under different voltage conditions.

There is another important limitation: even before the cutoff speed is reached, winding resistance, battery resistance, and mechanical losses prevent all the vehicle’s lost kinetic energy from being stored chemically.

The tempting prediction was that a motor acting as a generator could keep returning energy until the vehicle stopped. The overlooked constraint was that the generated emf depends on rotational speed, and rotational speed decreases during braking.

That is why regenerative braking can recover some kinetic energy, but cannot automatically recover all of it.

07The key relationship to remember

A motor can return electrical energy to its supply, but only when its operating conditions allow current to flow in the generating direction.

For a simple permanent-magnet DC motor:

\[ \boxed{I=\frac{V_s-E_b}{R_w}} \]

The sign of the current tells you which way electrical energy is flowing.

• If \(V_s>E_b\), the motor consumes electrical energy.
• If \(V_s=E_b\), the idealised armature current is zero.
• If \(E_b>V_s\), current reverses and the motor can return electrical energy to a compatible supply.

Remember that winding resistance still dissipates energy as heat, even during regenerative operation.

The next useful step is to connect this model to electromagnetic induction and Lenz’s law. Those ideas explain not only why a motor can generate electricity, but also why the resulting electromagnetic torque opposes the motion that produces it.