Why a Stalled DC Motor Overheats: Back EMF and Current

A small robot drives into a wall. Its wheels stop turning, but the motor is still connected to the battery. You might expect it to use less electricity because it is no longer moving anything. Instead, the motor can become dangerously hot.

A stalled DC motor can overheat quickly because stopping the rotor removes its back electromotive force (back EMF). Without that opposing voltage, a much larger current can flow through the motor’s low-resistance copper windings. Since electrical heating increases with the square of current, even a short stall can produce substantial heat.

Here’s the strange part: the motor may still be producing considerable torque, but because its shaft isn’t rotating, it delivers no mechanical power to the load. Much of the incoming electrical energy is instead converted into heat.

01Why stopping a motor increases its current

Imagine a small brushed DC motor turning a wheel.

Inside the motor, copper windings rotate between magnetic poles. Current passes through the windings, and the magnetic field exerts forces on them. These forces produce the torque that turns the shaft.

But a second process happens while the motor rotates.

The moving conductors also experience electromagnetic induction. They generate an electromotive force that opposes the applied voltage during normal motor operation.

This is called back EMF, or counter-EMF.

Predict what happens when the wheel becomes stuck. Does the back EMF stay the same, or does it disappear?

It disappears once the rotor stops. The conductors are no longer moving through the magnetic field, so the rotationally induced EMF falls to zero.

The copper windings still have resistance, but that resistance is usually small. Without back EMF opposing the supply, the current can become much larger.

The equation that explains the change

For a DC motor under steady electrical conditions, the terminal voltage is shared between the winding’s resistive voltage drop and its back EMF:

\[ V_t=E_b+IR_w \]

Here:

• \(V_t\) is the voltage across the motor terminals, measured in volts (V).
• \(E_b\) is the back EMF, measured in volts (V).
• \(I\) is the armature current, measured in amperes (A).
• \(R_w\) is the effective armature winding resistance, measured in ohms (\(\Omega\)).

Take current entering the motor’s positive terminal as positive. During ordinary forward motoring, \(E_b\) is positive in this equation because the induced EMF opposes the supply.

Rearranging:

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

This is why you cannot treat a running motor as a fixed resistor. Even if its winding resistance stays constant, its current changes with rotational speed because its back EMF changes.

For a motor with a fixed magnetic field, the average back EMF is approximately proportional to its angular speed:

\[ E_b=k_e\omega \]

Here, \(k_e\) is the motor’s back-EMF constant, measured in volt seconds per radian (\(\mathrm{V\,s\,rad^{-1}}\)), and \(\omega\) is the angular speed in radians per second (\(\mathrm{rad\,s^{-1}}\)).

When the rotor stops:

\[ \omega=0,\qquad E_b=0 \]

The current then approaches:

\[ I_{\text{stall}}=\frac{V_t}{R_w} \]

This is the steady-state stall-current estimate at a specified winding resistance and terminal voltage.

It doesn’t mean current jumps instantly to that value. Motor windings also have inductance, which resists rapid changes in current. And a real power supply may reduce its output voltage or activate current limiting.

For the underlying electromagnetic induction, see Back emf in a DC motor: HSC Physics explained.

Why the extra current becomes heat

Copper conducts electricity well, but it isn’t a perfect conductor. Current passing through the windings transfers electrical energy into thermal energy.

The rate of this heating is:

\[ P_w=I^2R_w \]

where \(P_w\) is the electrical power converted into heat by the winding resistance, in watts (W).

Notice the square.

If the current doubles while resistance stays constant, the winding heating increases by a factor of four.

If the current triples, the heating increases by a factor of nine.

A motor doesn’t need to draw an enormous current for the heating increase to be serious. It only needs to draw substantially more current than its normal operating value.

02How could we investigate heating during a stall?

Suppose we want to find out why a motor heats up after its shaft is blocked.

A first attempt might use a small DC motor driving a wheel, with a temperature probe attached to the outside of its metal casing.

Run the wheel, obstruct it briefly, and watch the temperature reading.

What should happen if the stopped motor has genuinely stopped converting electrical energy into heat?

Its temperature should stop rising, apart from heat already moving through the motor.

But there is a problem with that test. The thermometer measures the casing, not the copper windings. Heat takes time to travel from the windings through the rest of the motor.

The casing might show almost no immediate temperature change even while substantial heating is occurring inside.

We need a better way to investigate the electrical energy entering the windings.

A revised arrangement: measure current, voltage, and speed

Consider a proposed investigation using a small permanent-magnet brushed DC motor and a protected, current-limited DC supply.

Connect the positive supply terminal through a switch and a current sensor to the motor’s positive brush terminal. Inside the motor, current passes through the commutator and the armature windings. It returns through the other brush to the negative supply terminal.

Connect a voltmeter directly across the motor terminals.

An optical speed sensor observes the rotating shaft, while a temperature probe remains on the casing. The shaft drives a small wheel whose mechanical load can be increased.

The investigation would compare the motor while freely rotating, heavily loaded, and briefly stopped.

If the motor behaved like a fixed resistor, keeping its terminal voltage and resistance unchanged would keep its current unchanged.

The DC motor model predicts something different: as speed falls, back EMF falls, current rises, and winding heating increases.

Illustrative results: what the model predicts

The following values are calculated examples, not measurements from an actual experiment.

Assume:

• Motor terminal voltage remains \(9.0\,\mathrm{V}\).
• Effective winding resistance stays at \(1.50\,\Omega\).
• The magnetic field remains constant.
• The current has settled electrically at each operating condition.
• Brush voltage drops and other electrical losses are negligible.

The supply voltage hasn’t changed, and neither has the assumed winding resistance.

Yet the predicted winding heating has increased from \(1.5\,\mathrm{W}\) to \(54\,\mathrm{W}\).

That’s 36 times the original heating rate.

The experimental observation to look for would be a sharp rise in current as the shaft slows, followed by delayed warming of the casing. Whether the predicted stall current is actually reached depends on the supply, protection circuitry, and changing winding resistance.

This is why waiting for the outside of a motor to feel hot is a poor way to detect excessive winding heating. A protected investigation would use short, controlled measurements and electrical power estimates rather than holding a motor stalled at full operating voltage.

03Why a stalled motor can produce torque without doing mechanical work

There’s another piece of the puzzle.

Current flowing through the motor windings interacts with the magnetic field and produces torque. A larger current can produce a larger electromagnetic torque, provided the magnetic field is approximately constant.

So a stalled motor can push hard against whatever is blocking it.

But mechanical power delivered through a rotating shaft is:

\[ P_{\text{shaft}}=\tau\omega \]

where:

• \(P_{\text{shaft}}\) is mechanical shaft power in watts,
• \(\tau\) is torque in newton metres (\(\mathrm{N\,m}\)), and
• \(\omega\) is angular speed in radians per second.

At stall:

\[ \omega=0 \]

Therefore:

\[ P_{\text{shaft}}=0 \]

The motor can exert torque, but the point where that torque acts has no angular displacement. No mechanical work is being delivered to the load through rotation.

Now examine the electrical input.

Multiplying the motor voltage equation by current gives:

\[ V_tI=E_bI+I^2R_w \]

The terms represent:

• \(V_tI\): electrical power entering the armature.
• \(E_bI\): electromagnetic conversion power.
• \(I^2R_w\): resistive heating in the windings.

In the worked example, the running motor receives \(9.0\,\mathrm{W}\). Of that, \(1.5\,\mathrm{W}\) becomes winding heat, and \(7.5\,\mathrm{W}\) is converted electromagnetically towards mechanical power.

At stall, \(E_b=0\).

Under the simplified assumptions, the full \(54\,\mathrm{W}\) of electrical input to the armature becomes resistive heating.

That explains the apparent contradiction. The stopped motor isn’t harmless because it has stopped doing useful work. Its electrical energy has lost its normal route into rotational mechanical power.

In real motors, brush losses, electronic components, and other effects can change where energy is dissipated. But the main conclusion remains: a continuously powered stall can produce substantial heating without useful shaft rotation.

04Why the temperature may keep rising even when current falls

A real motor’s winding resistance isn’t perfectly constant.

Copper resistance generally increases as temperature increases.

That creates a useful distinction between two questions:

1. How much heat is the motor generating now?
2. Is the motor’s temperature rising or falling?

Those are not the same question.

The first depends strongly on electrical current and resistance. The second also depends on how quickly the motor loses heat to its surroundings.

A motor producing \(40\,\mathrm{W}\) of internal heat may still be getting hotter if it is losing heat at less than \(40\,\mathrm{W}\).

And a motor with a fan attached to its shaft may lose some of its normal forced-air cooling when the shaft stops. This can make sustained stalls more damaging than the electrical calculation alone suggests.

05A starting motor is not the same as a motor that stays stalled

When a motor first starts, its rotor is also stationary. Its rotational back EMF is initially zero.

So why don’t motors always overheat when switched on?

The difference is how long they remain at low speed.

During normal starting, current produces torque, the rotor accelerates, and back EMF builds. As it builds, the current generally decreases.

A jam prevents that normal acceleration.

The motor can remain at low or zero speed, keeping the back EMF small and the current high. Thermal energy continues accumulating.

The distinction is between a brief starting surge and a sustained stall. A motor designed to tolerate the first may not tolerate the second.

What to check when answering a stalled-motor question

The most important distinction is between a motor’s electrical heating rate and its actual temperature.

Current and resistance tell you how quickly electrical energy is being converted into winding heat. Heat capacity, thermal conduction, cooling, and time determine the resulting temperature.

06Questions and solutions

For Questions 1-5, use the simplified brushed DC motor model unless otherwise stated. Take current entering the positive motor terminal as positive, and assume steady electrical conditions at each specified operating point.

Question 1

A small DC motor operates from a constant \(6.0\,\mathrm{V}\) supply. Its effective winding resistance is \(2.0\,\Omega\).

While running normally, it produces a back EMF of \(4.0\,\mathrm{V}\). The wheel then becomes completely jammed.

Calculate:

a. The current while running normally.

b. The current when stalled.

c. The winding heating power in both conditions.

d. How many times greater the winding heating is during stall.

Solution 1

The current increases from \(1.0\,\mathrm{A}\) to \(3.0\,\mathrm{A}\), making the stalled winding heating nine times greater.

a. Running current

\[ \begin{aligned} I_r&=\frac{V_t-E_b}{R_w}\\ &=\frac{6.0-4.0}{2.0}\\ &=1.0\,\mathrm{A} \end{aligned} \]

b. Stalled current

At stall, the rotor no longer generates rotational back EMF.

\[ \begin{aligned} I_s&=\frac{6.0}{2.0}\\ &=3.0\,\mathrm{A} \end{aligned} \]

c. Winding heating

\[ \begin{aligned} P_r&=(1.0)^2(2.0)\\ &=2.0\,\mathrm{W}\\[4pt] P_s&=(3.0)^2(2.0)\\ &=18\,\mathrm{W} \end{aligned} \]

d. Heating ratio

\[ \frac{P_s}{P_r}=\frac{18}{2.0}=9 \]

The tempting mistake is to calculate \(I=V_t/R_w\) for both conditions. That ignores the back EMF present while the motor is rotating.

The motor’s winding resistance has not changed. Its current has increased because the opposing induced EMF has disappeared.

Question 2

A motor powers a small conveyor belt from a regulated \(8.0\,\mathrm{V}\) supply.

Its winding resistance is \(1.0\,\Omega\). When the belt moves normally, the motor’s back EMF is \(6.0\,\mathrm{V}\).

Over a nine-second interval, the conveyor:

• runs normally for \(6.0\,\mathrm{s}\),
• becomes completely jammed for \(3.0\,\mathrm{s}\), and
• is then switched off.

Assume constant resistance and terminal voltage while energised, negligible heating losses, and instantaneous changes between the two steady electrical operating conditions for this calculation.

a. Calculate the total thermal energy generated in the windings during the nine seconds.

b. Find the average current over the nine seconds while the motor is energised.

c. A student uses this average current in \(P=I^2R_w\), then multiplies by nine seconds. Explain why this method gives the wrong thermal energy, and calculate the error.

d. What fraction of the total winding heating occurs during the three-second jam?

Solution 2

The windings generate \(216\,\mathrm{J}\) of heat. Using the average current gives only \(144\,\mathrm{J}\), underestimating the actual heating by \(72\,\mathrm{J}\). The error occurs because heating depends on current squared, not current itself.

a. Calculate the energy separately for each interval.

During normal operation:

\[ \begin{aligned} I_r&=\frac{8.0-6.0}{1.0}\\ &=2.0\,\mathrm{A}\\ P_r&=(2.0)^2(1.0)\\ &=4.0\,\mathrm{W} \end{aligned} \]

For the first six seconds:

\[ Q_r=4.0(6.0)=24\,\mathrm{J} \]

During stall:

\[ \begin{aligned} I_s&=\frac{8.0}{1.0}\\ &=8.0\,\mathrm{A}\\ P_s&=(8.0)^2(1.0)\\ &=64\,\mathrm{W} \end{aligned} \]

For the next three seconds:

\[ Q_s=64(3.0)=192\,\mathrm{J} \]

The total is:

\[ Q_{\text{total}}=24+192=216\,\mathrm{J} \]

b. Calculate average current.

The time-weighted average current is:

\[ \begin{aligned} I_{\text{avg}} &=\frac{2.0(6.0)+8.0(3.0)}{9.0}\\ &=4.0\,\mathrm{A} \end{aligned} \]

c. Test the student’s method.

Using the average current as though it flowed continuously:

\[ \begin{aligned} Q_{\text{wrong}} &=(4.0)^2(1.0)(9.0)\\ &=144\,\mathrm{J} \end{aligned} \]

The difference is:

\[ 216-144=72\,\mathrm{J} \]

The method underestimates heating by \(72\,\mathrm{J}\), or one-third of the true amount.

The issue is mathematical and physical.

Because current is squared in \(P=I^2R_w\), a short period of high current contributes disproportionately to heating.

In general, the square of the average current is not equal to the average of the squared current.

d. Find the fraction produced during the jam.

\[ \begin{aligned} \frac{Q_s}{Q_{\text{total}}} &=\frac{192}{216}\\ &=0.889 \end{aligned} \]

Approximately \(88.9\%\) of the winding heat is generated during the jam, even though the motor is stalled for only one-third of the time.

This is why a current reading averaged over several seconds can conceal damaging short-duration heating.

Question 3

A motor is connected to an ideal \(15\,\mathrm{V}\) battery through long leads with a combined resistance of \(1.0\,\Omega\).

The motor itself has an effective winding resistance of \(2.0\,\Omega\).

While running normally, it produces \(9.0\,\mathrm{V}\) of back EMF. Its shaft then becomes completely blocked.

Assume all resistances remain constant, and the battery’s internal resistance is negligible.

a. Determine the running and stalled currents.

b. Calculate the motor terminal voltage in each condition.

c. Calculate the winding heating power before and after the jam, and the power dissipated in the leads during stall.

d. A student argues that the motor must become cooler because its terminal voltage decreases when it jams. Evaluate this claim.

Solution 3

The motor terminal voltage falls from \(13\,\mathrm{V}\) to \(10\,\mathrm{V}\), but its winding heating increases from \(8.0\,\mathrm{W}\) to \(50\,\mathrm{W}\). A lower terminal voltage does not necessarily mean less heating.

a. Calculate currents using the total series resistance.

Let \(V_s=15\,\mathrm{V}\) be the ideal battery voltage and \(R_s=1.0\,\Omega\) be the lead resistance.

The circuit equation is:

\[ V_s=E_b+I(R_s+R_w) \]

While running:

\[ \begin{aligned} I_r&=\frac{15-9.0}{1.0+2.0}\\ &=2.0\,\mathrm{A} \end{aligned} \]

At stall, back EMF vanishes:

\[ \begin{aligned} I_s&=\frac{15}{1.0+2.0}\\ &=5.0\,\mathrm{A} \end{aligned} \]

b. Find the motor terminal voltages.

The battery voltage is divided between the leads and the motor.

While running:

\[ \begin{aligned} V_{t,r}&=V_s-I_rR_s\\ &=15-2.0(1.0)\\ &=13\,\mathrm{V} \end{aligned} \]

At stall:

\[ \begin{aligned} V_{t,s}&=15-5.0(1.0)\\ &=10\,\mathrm{V} \end{aligned} \]

The greater current produces a larger voltage drop along the connecting leads.

c. Calculate the heating powers.

For the motor windings:

\[ \begin{aligned} P_r&=(2.0)^2(2.0)\\ &=8.0\,\mathrm{W}\\[4pt] P_s&=(5.0)^2(2.0)\\ &=50\,\mathrm{W} \end{aligned} \]

For the leads during stall:

\[ \begin{aligned} P_{\text{leads}} &=(5.0)^2(1.0)\\ &=25\,\mathrm{W} \end{aligned} \]

The battery supplies:

\[ P_{\text{battery}}=15(5.0)=75\,\mathrm{W} \]

All \(75\,\mathrm{W}\) is dissipated resistively in this stalled-circuit model: \(50\,\mathrm{W}\) in the windings and \(25\,\mathrm{W}\) in the leads.

d. Evaluate the student’s claim.

The claim is incorrect.

Before the stall, \(9.0\,\mathrm{V}\) of the motor’s terminal voltage balances the back EMF, while \(4.0\,\mathrm{V}\) appears across the winding resistance.

At stall, the back EMF disappears. Although the motor terminal voltage falls to \(10\,\mathrm{V}\), all of that voltage now appears across the winding resistance.

The higher current produces much greater winding heating.

The tempting mistake is to look only at terminal voltage. To predict resistive heating, you must also know the current and the voltage contribution from back EMF.

Question 4

A motor remains completely stalled while its effective copper winding resistance rises from \(2.0\,\Omega\) to \(3.0\,\Omega\) as it warms.

Compare two possible power supplies.

Supply A maintains a constant \(12\,\mathrm{V}\) directly across the motor terminals.

Supply B maintains a constant motor current of \(4.0\,\mathrm{A}\), adjusting its output voltage as needed.

Assume both supplies can maintain their specified conditions and that other electrical losses are negligible.

a. Calculate the current and winding heating power at both resistance values with Supply A.

b. Calculate the terminal voltage and winding heating power at both resistance values with Supply B.

c. Explain why increasing resistance has opposite effects on heating in these two cases.

d. Can you conclude that the motor’s temperature falls whenever its winding heating power decreases? Explain.

Solution 4

Increasing the winding resistance reduces heating from \(72\,\mathrm{W}\) to \(48\,\mathrm{W}\) with the constant-voltage supply, but increases heating from \(32\,\mathrm{W}\) to \(48\,\mathrm{W}\) with the constant-current supply.

The crucial difference is what the power supply holds constant.

a. Supply A: constant terminal voltage

At stall:

\[ I=\frac{V_t}{R_w} \]

For \(R_w=2.0\,\Omega\):

\[ \begin{aligned} I_1&=\frac{12}{2.0}=6.0\,\mathrm{A}\\ P_1&=(6.0)^2(2.0)\\ &=72\,\mathrm{W} \end{aligned} \]

For \(R_w=3.0\,\Omega\):

\[ \begin{aligned} I_2&=\frac{12}{3.0}=4.0\,\mathrm{A}\\ P_2&=(4.0)^2(3.0)\\ &=48\,\mathrm{W} \end{aligned} \]

b. Supply B: constant current

Here, \(I=4.0\,\mathrm{A}\) in both conditions.

For \(R_w=2.0\,\Omega\):

\[ \begin{aligned} V_1&=4.0(2.0)=8.0\,\mathrm{V}\\ P_1&=(4.0)^2(2.0)\\ &=32\,\mathrm{W} \end{aligned} \]

For \(R_w=3.0\,\Omega\):

\[ \begin{aligned} V_2&=4.0(3.0)=12\,\mathrm{V}\\ P_2&=(4.0)^2(3.0)\\ &=48\,\mathrm{W} \end{aligned} \]

c. Explain the opposite trends.

For a stalled motor with a constant terminal voltage:

\[ P_w=\frac{V_t^2}{R_w} \]

Increasing resistance reduces power because the resulting fall in current more than offsets the increased resistance.

With constant current:

\[ P_w=I^2R_w \]

Increasing resistance directly increases heating power.

The misconception is that greater resistance must always produce greater heating. That is only necessarily true when current is held constant.

d. Explain the temperature behaviour.

A decrease in heating power does not prove that the motor is cooling.

Its temperature depends on the balance between internal heating and heat loss to the surroundings.

Even after the heating drops to \(48\,\mathrm{W}\), the motor can continue getting hotter if it is losing heat at less than \(48\,\mathrm{W}\).

The equations predict heat generation, not temperature by themselves.

Question 5

Extension challenge: Can a moving motor overheat faster than a stalled motor?

A permanent-magnet brushed DC motor is connected to an ideal \(12\,\mathrm{V}\) battery through an external \(1.0\,\Omega\) series resistor.

The motor’s effective winding resistance is \(2.0\,\Omega\).

The battery polarity remains fixed throughout the experiment.

Consider three separate, electrically steady conditions:

1. The motor rotates forward under its own power, producing a back EMF of \(+8.0\,\mathrm{V}\).
2. The motor’s shaft is locked, so its back EMF is zero.
3. An external powered winch forces the motor shaft to rotate backwards at constant speed, producing a signed back EMF of \(-6.0\,\mathrm{V}\).

A positive back EMF opposes the applied supply during normal forward motoring. The negative value in Condition 3 means the induced voltage has reversed polarity.

Assume the magnetic field remains constant, the external winch maintains the specified backward speed, and all electrical resistance values remain unchanged. Ignore mechanical friction, brush losses, and other motor losses.

a. Calculate the current and motor winding heating power in each condition.

b. Identify which condition produces the greatest winding heating.

c. For Condition 3, calculate the motor terminal voltage and the electrical power entering the motor through its terminals.

d. In Condition 3, the motor’s winding heating exceeds the electrical power entering through its terminals. Explain how this is possible without violating conservation of energy. Account for the external resistor and the battery as well.

Solution 5

The backwards-driven motor produces the greatest winding heating: \(72\,\mathrm{W}\), compared with \(32\,\mathrm{W}\) at stall. The result is possible because the external winch supplies mechanical energy that is converted into additional heat.

This is a case where the usual statement that stall produces the greatest current needs an important qualification: it applies to ordinary forward motoring at fixed voltage and resistance, not every possible direction of rotation.

a. Calculate current in each condition.

For all three conditions:

\[ 12=E_b+I(1.0+2.0) \]

Rearrange:

\[ I=\frac{12-E_b}{3.0} \]

For forward rotation:

\[ \begin{aligned} I_f&=\frac{12-8.0}{3.0}\\ &=1.33\,\mathrm{A} \end{aligned} \]

For stall:

\[ \begin{aligned} I_s&=\frac{12-0}{3.0}\\ &=4.0\,\mathrm{A} \end{aligned} \]

For forced backward rotation:

\[ \begin{aligned} I_b&=\frac{12-(-6.0)}{3.0}\\ &=6.0\,\mathrm{A} \end{aligned} \]

The negative back EMF doesn’t reduce the current. Under these conditions, its reversed polarity makes the induced EMF assist the battery’s drive of current through the resistances.

The winding heating powers are:

Forward rotation, using the exact current of \(4/3\,\mathrm{A}\):

\[ \begin{aligned} P_f&=2.0\left(\frac{4}{3}\right)^2\\ &=3.56\,\mathrm{W} \end{aligned} \]

Stall:

\[ P_s=(4.0)^2(2.0)=32\,\mathrm{W} \]

Forced backward rotation:

\[ P_b=(6.0)^2(2.0)=72\,\mathrm{W} \]

b. Identify the greatest heating.

The forced-backward condition produces the greatest heating.

A student might confidently choose stall because back EMF becomes zero there, allowing a large current.

That prediction makes sense for forward rotation. But when the shaft is forced backwards, the induced EMF changes polarity.

The current can then exceed the ordinary stall current because the battery and the induced EMF drive current in the same circuit direction.

c. Calculate terminal voltage and electrical input during backward rotation.

The current is \(6.0\,\mathrm{A}\), so the external resistor has a voltage drop of:

\[ V_R=6.0(1.0)=6.0\,\mathrm{V} \]

The motor terminal voltage is:

\[ \begin{aligned} V_t&=12-6.0\\ &=6.0\,\mathrm{V} \end{aligned} \]

The electrical power entering the motor through its terminals is:

\[ \begin{aligned} P_{\text{in}}&=V_tI\\ &=6.0(6.0)\\ &=36\,\mathrm{W} \end{aligned} \]

But the winding heating is \(72\,\mathrm{W}\).

Where does the additional \(36\,\mathrm{W}\) come from?

d. Account for the energy.

The key is that the motor is being mechanically forced to rotate against the direction of its electromagnetic torque.

The current still produces forward electromagnetic torque, but the winch forces the shaft backwards.

The winch therefore does work on the motor.

Recall the armature power relationship:

\[ V_tI=E_bI+I^2R_w \]

During reverse rotation:

\[ \begin{aligned} E_bI&=(-6.0)(6.0)\\ &=-36\,\mathrm{W} \end{aligned} \]

The negative electromagnetic conversion power means that energy is being transferred from the mechanical side into the electrical system.

The external winch supplies \(36\,\mathrm{W}\) of mechanical power to the motor.

The motor also receives \(36\,\mathrm{W}\) electrically through its terminals.

Together, these provide the \(72\,\mathrm{W}\) dissipated in the windings.

Now check the complete circuit.

The battery supplies:

\[ P_{\text{battery}}=12(6.0)=72\,\mathrm{W} \]

The external resistor dissipates:

\[ P_R=(6.0)^2(1.0)=36\,\mathrm{W} \]

So the full energy balance is:

Energy is conserved.

The deeper lesson is that back EMF is a signed induced voltage, not a permanent current-reducing property of a motor.

During ordinary forward motoring, it opposes the battery and limits current. At stall, it disappears. During forced backward rotation, its polarity reverses, and the battery and external mechanical drive can both contribute energy that ultimately becomes heat.

07What determines whether a stalled motor actually overheats?

A stalled motor doesn’t have a single universal heating rate. Its behaviour depends on the electrical conditions and the motor’s ability to lose heat.

The most useful chain of reasoning is:

• Stalling stops the rotor and removes its rotational back EMF.
• With fixed terminal voltage and winding resistance, current rises.
• Resistive heating increases according to \(I^2R_w\).
• With no rotation, the shaft delivers no mechanical power, despite the motor potentially producing torque.
• If heating exceeds heat loss, the motor’s temperature rises.
• Sustained high temperatures can damage winding insulation and other motor components.

This explains why a motor can tolerate starting from rest but struggle with a prolonged jam: startup normally allows back EMF to build, while a jam prevents it.

When estimating overheating, the next useful decision is whether the motor really receives its full supply voltage during stall. Battery resistance, connecting leads, and current-limiting circuits can all change the result.

For the electrical side of that calculation, see How to Calculate DC Motor Stall Current: HSC Physics.

The practical distinction to carry forward is simple: stall current predicts how severely the windings may be heated electrically, but current alone cannot tell you how hot they will become. You must also account for the duration of the stall and the motor’s thermal behaviour.