How to Calculate DC Motor Stall Current: HSC Physics

A DC motor draws \(2.0\,\mathrm{A}\) while running from a \(12.0\,\mathrm{V}\) supply. Suddenly, its shaft becomes jammed and stops rotating. Will the motor keep drawing \(2.0\,\mathrm{A}\), or will the current increase?

If its armature winding resistance is \(0.80\,\Omega\), and the supply maintains \(12.0\,\mathrm{V}\) across the motor, the predicted stall current is \(15\,\mathrm{A}\). That’s more than seven times its running current.

To calculate a DC motor’s stall current, divide the voltage across its terminals by its effective armature resistance:

\[
I_{\mathrm{stall}}=\frac{V_{\mathrm{t}}}{R_{\mathrm{w}}}
\]

This works because a stationary motor produces no rotational back emf. The full terminal voltage is then available to drive current through the winding resistance.

The result is a steady-state estimate at a specified winding temperature. It is not necessarily the current measured at the exact instant the motor stops, and it does not mean the motor can safely carry that current continuously.

01Why does a stopped motor draw more current?

A DC motor contains copper windings. Like any conducting wire, these windings have electrical resistance.

If the motor were simply a resistor, you could use Ohm’s law to calculate its current from the applied voltage and resistance. But a running motor does something else: its conductors move through a magnetic field and generate an induced electromotive force (emf).

During normal forward motoring, this induced voltage acts against the supply. It’s called back emf, or counter-emf.

Picture the motor as having two electrical effects operating together:

• The copper winding creates a resistive voltage drop when current flows.
• The rotating armature generates a voltage that opposes the applied supply.

The first effect remains when the motor stops. The second disappears.

The voltage equation

For a simple DC motor operating under steady electrical conditions:

\[
V_{\mathrm{t}}=E_{\mathrm{b}}+IR_{\mathrm{w}}
\]

Here:

• \(V_{\mathrm{t}}\) is the voltage across the motor terminals, in volts (V).
• \(E_{\mathrm{b}}\) is the back emf, in volts (V).
• \(I\) is the armature current, in amperes (A).
• \(R_{\mathrm{w}}\) is the effective armature winding resistance, in ohms (\(\Omega\)).

For now, take current as positive when it flows from the supply into the motor. Back emf is positive when it opposes that current during normal forward rotation.

Rearranging:

\[
I=\frac{V_{\mathrm{t}}-E_{\mathrm{b}}}{R_{\mathrm{w}}}
\]

Notice that the current depends on two things besides resistance: terminal voltage and back emf.

As the motor spins faster, the induced back emf generally increases. Less voltage remains across the winding resistance, so less current flows.

If the shaft stops:

\[
E_{\mathrm{b}}=0
\]

Therefore:

\[
\begin{aligned}
I_{\mathrm{stall}}
&=\frac{V_{\mathrm{t}}-0}{R_{\mathrm{w}}}\\
&=\frac{V_{\mathrm{t}}}{R_{\mathrm{w}}}
\end{aligned}
\]

The motor hasn’t lost its resistance. It has lost the opposing voltage produced by rotation.

Why back emf disappears at stall

Back emf comes from electromagnetic induction. As the armature conductors move through the magnetic field, they experience an induced emf.

For a fixed-field DC motor, the average back emf is approximately proportional to angular speed:

\[
E_{\mathrm{b}}=k_{\mathrm{e}}\omega
\]

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

At zero speed, \(\omega=0\), so the speed-generated emf is zero.

This is a consequence of electromagnetic induction, not an arbitrary rule for motors. For a fuller explanation of where the induced voltage comes from, see Back emf in a DC motor: HSC Physics explained.

02How would you measure stall current?

Suppose you’re given a motor without a published stall-current specification.

You connect it to a \(12.0\,\mathrm{V}\) supply, measure a running current of \(2.0\,\mathrm{A}\), and calculate:

\[
R=\frac{12.0}{2.0}=6.0\,\Omega
\]

It looks like you’ve found the motor’s resistance.

But have you?

The first test: voltage and current while running

Imagine the following apparatus.

A DC supply’s positive terminal connects through a switch and an ammeter to the motor’s positive brush. The motor’s other brush returns to the negative supply terminal.

A voltmeter is connected directly across the two motor terminals. A speed sensor measures the shaft’s rotation.

The motor can run freely or have its speed changed by a mechanical load.

If the motor really behaved like a fixed \(6.0\,\Omega\) resistor, the same \(12.0\,\mathrm{V}\) would produce \(2.0\,\mathrm{A}\) whether the shaft rotated or not.

Now consider these invented, illustrative readings. Assume the supply maintains \(12.0\,\mathrm{V}\) at the motor terminals, the effective winding resistance remains \(0.80\,\Omega\), and the electrical current has settled at each condition.

The current changes dramatically even though the terminal voltage is unchanged.

That contradicts the fixed-resistor interpretation.

What the first test couldn’t tell us

When the motor runs, the terminal voltage is shared between the resistive voltage drop and back emf.

Dividing terminal voltage by current therefore gives an apparent resistance, not the actual winding resistance.

For the running measurement:

\[
\frac{V_{\mathrm{t}}}{I}
=\frac{12.0}{2.0}
=6.0\,\Omega
\]

But the effective winding resistance is only \(0.80\,\Omega\).

The remaining voltage is explained by back emf:

\[
\begin{aligned}
E_{\mathrm{b}}&=V_{\mathrm{t}}-IR_{\mathrm{w}}\\
&=12.0-(2.0)(0.80)\\
&=10.4\,\mathrm{V}
\end{aligned}
\]

While running, the motor generates \(10.4\,\mathrm{V}\) opposing the applied voltage.

When stationary, that opposing voltage disappears.

A better measurement

To determine stall current accurately, you need to isolate the winding resistance from the speed-generated voltage.

A better approach is to disconnect the motor from its operating supply, keep the rotor stationary, and measure its effective armature resistance using an appropriately low-current resistance test.

For a brushed motor, resistance readings can depend on commutator position and brush contact. Small resistances can also be difficult to measure accurately using ordinary test leads.

Once you have an estimate of the effective armature resistance at a known temperature, combine it with the expected terminal voltage to predict stall current.

A brief, controlled locked-rotor measurement could then test that prediction, but switching transients, source voltage changes, and winding heating must be considered. Holding a motor stalled at full operating voltage can cause rapid overheating and damage; it’s not a suitable casual classroom test.

The revised measurement separates what the first test mixed together: copper resistance and rotational back emf.

04What if the supply voltage drops when the motor stalls?

The simple formula contains a condition students often miss:

\[
I_{\mathrm{stall}}=\frac{V_{\mathrm{t}}}{R_{\mathrm{w}}}
\]

It uses the voltage across the motor terminals, not automatically the voltage printed on the battery or power supply.

Suppose a battery provides \(9.0\,\mathrm{V}\), but the motor is connected through leads, switches, or other components with significant resistance.

When current increases, those components develop their own voltage drops.

That leaves less voltage across the motor.

For an ideal source of voltage \(V_{\mathrm{s}}\) with an additional series resistance \(R_{\mathrm{s}}\), the locked-rotor circuit gives:

\[
V_{\mathrm{s}}
=I_{\mathrm{stall}}(R_{\mathrm{w}}+R_{\mathrm{s}})
\]

Therefore:

\[
I_{\mathrm{stall}}
=\frac{V_{\mathrm{s}}}{R_{\mathrm{w}}+R_{\mathrm{s}}}
\]

This is still Ohm’s law, but now you’re applying it to the complete resistive circuit at zero rotational speed.

05A reliable decision process for stall-current questions

Before substituting numbers, work out what the question means by its stated voltage.

In each case, the simple calculation assumes the rotor is stationary, resistance is known, and the electrical current has settled.

Four details that can change the result

1. Starting current is not instantly equal to stall current.

A real armature has inductance, which opposes sudden changes in current. At switch-on, current takes time to rise. If the shaft starts accelerating, back emf grows and may reduce the current before it reaches the theoretical locked-rotor value.

2. Winding temperature matters.

Copper resistance increases as temperature rises. If the winding gets hot during a stall, its resistance increases, reducing the current predicted for a constant terminal voltage. That doesn’t make prolonged stalling safe; substantial heating may continue.

3. A controller may limit the current.

A motor controller can reduce the effective voltage applied to the motor to prevent the current exceeding a set value. The actual stalled current may therefore be much smaller than the unrestricted \(V/R\) estimate.

4. Mechanical stall and electrical disconnection are not the same.

A stalled motor has a stationary shaft but remains electrically connected. A disconnected motor may also be stationary, but without an applied voltage there is no sustained supply current.

One useful boundary remains: under the same constant positive supply voltage and resistance, ordinary forward motoring current cannot exceed the zero-speed stall current in this model. That conclusion assumes the rotor is turning in its usual motoring direction and the induced emf opposes the supply.

A motor mechanically driven in the opposite direction can behave differently, as the final question explores.

06Questions and solutions

Question 1

A brushed DC motor is connected directly to an ideal \(6.0\,\mathrm{V}\) supply. Its effective armature resistance is \(0.75\,\Omega\).

The shaft is prevented from rotating.

a. Calculate its predicted stall current.

b. Calculate the power converted into heat in the winding at stall.

Assume constant resistance and negligible brush voltage drop.

Solution 1

The predicted stall current is \(8.0\,\mathrm{A}\), and the winding produces \(48\,\mathrm{W}\) of heat.

Part a

At stall, the shaft is stationary, so the back emf is zero.

\[
I_{\mathrm{stall}}=\frac{V_{\mathrm{t}}}{R_{\mathrm{w}}}
\]

Substitute:

\[
\begin{aligned}
I_{\mathrm{stall}}
&=\frac{6.0}{0.75}\\
&=8.0\,\mathrm{A}
\end{aligned}
\]

Part b

Use resistive heating power:

\[
P_{\mathrm{heat}}=I^2R_{\mathrm{w}}
\]

\[
\begin{aligned}
P_{\mathrm{heat}}
&=(8.0)^2(0.75)\\
&=48\,\mathrm{W}
\end{aligned}
\]

The winding converts electrical energy into heat at \(48\,\mathrm{J}\) per second.

A tempting mistake is to assume that because the shaft isn’t moving, no energy is being transferred. The motor produces no rotational shaft power at stall, but electrical current still flows through its resistance.

Question 2

A \(12.0\,\mathrm{V}\) battery is connected to a brushed DC motor. The battery’s stated voltage is its open-circuit emf, and its internal resistance is unknown.

The motor’s effective armature resistance is \(0.50\,\Omega\).

During a brief, steady locked-rotor test, the current is \(8.0\,\mathrm{A}\), and a voltmeter connected directly across the motor reads \(4.0\,\mathrm{V}\).

The shaft is then allowed to rotate. In the new steady state, the current decreases to \(2.0\,\mathrm{A}\).

Assume the battery’s internal resistance and the motor’s winding resistance remain constant. Neglect other circuit resistance.

a. Calculate the battery’s internal resistance.

b. Calculate the motor’s terminal voltage while running.

c. Determine the running back emf.

d. Explain why predicting a stall current of \(24\,\mathrm{A}\) would be incorrect.

Solution 2

The battery’s internal resistance is \(1.0\,\Omega\), the running motor terminal voltage is \(10.0\,\mathrm{V}\), and its running back emf is \(9.0\,\mathrm{V}\). The predicted \(24\,\mathrm{A}\) is incorrect because it ignores the voltage drop inside the battery.

Part a: Battery internal resistance

At stall, the back emf is zero.

The battery supplies an emf of \(12.0\,\mathrm{V}\), but the motor receives only \(4.0\,\mathrm{V}\).

The remaining voltage is dropped across the battery’s internal resistance \(r\):

\[
V_r=12.0-4.0=8.0\,\mathrm{V}
\]

Since the current is \(8.0\,\mathrm{A}\):

\[
\begin{aligned}
r&=\frac{V_r}{I}\\
&=\frac{8.0}{8.0}\\
&=1.0\,\Omega
\end{aligned}
\]

Part b: Running terminal voltage

When the motor is running, the current falls to \(2.0\,\mathrm{A}\).

The battery’s internal voltage drop is now:

\[
V_r=(2.0)(1.0)=2.0\,\mathrm{V}
\]

Therefore:

\[
V_{\mathrm{t}}=12.0-2.0=10.0\,\mathrm{V}
\]

The motor receives more voltage while running because the smaller current produces a smaller internal voltage drop.

Part c: Running back emf

Use:

\[
E_{\mathrm{b}}=V_{\mathrm{t}}-IR_{\mathrm{w}}
\]

\[
\begin{aligned}
E_{\mathrm{b}}
&=10.0-(2.0)(0.50)\\
&=9.0\,\mathrm{V}
\end{aligned}
\]

Part d: The incorrect stall prediction

The incorrect method gives:

\[
I=\frac{12.0}{0.50}=24\,\mathrm{A}
\]

But the complete circuit contains both the motor resistance and the battery’s internal resistance.

At stall:

\[
\begin{aligned}
I_{\mathrm{stall}}
&=\frac{12.0}{0.50+1.0}\\
&=8.0\,\mathrm{A}
\end{aligned}
\]

This matches the locked-rotor observation.

The crucial distinction is between the battery’s emf and the actual terminal voltage delivered to the motor. A motor cannot use voltage that’s being dropped elsewhere in the circuit.

Question 3

A permanent-magnet DC motor runs from an ideal \(12.0\,\mathrm{V}\) supply.

Initially, its winding resistance is \(0.60\,\Omega\), and its running current is \(5.0\,\mathrm{A}\).

After operating under a heavier load, the windings become hotter. Their effective resistance rises to \(0.90\,\Omega\). At this new steady operating point, the current is still \(5.0\,\mathrm{A}\).

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

a. Calculate the back emf in both operating conditions.

b. Determine the ratio of the second angular speed to the first.

c. Calculate the theoretical stall current at each winding temperature.

d. A student argues that because the current hasn’t changed, the motor’s speed must also be unchanged. Evaluate this claim.

Solution 3

The back emf decreases from \(9.0\,\mathrm{V}\) to \(7.5\,\mathrm{V}\), so the motor slows to about \(83.3\%\) of its original speed despite drawing the same current. The theoretical stall current also falls, from \(20.0\,\mathrm{A}\) when cooler to \(13.3\,\mathrm{A}\) when hotter.

Part a: Initial and final back emf

Initially:

\[
\begin{aligned}
E_{\mathrm{b1}}
&=V_{\mathrm{t}}-I_1R_1\\
&=12.0-(5.0)(0.60)\\
&=9.0\,\mathrm{V}
\end{aligned}
\]

After heating:

\[
\begin{aligned}
E_{\mathrm{b2}}
&=12.0-(5.0)(0.90)\\
&=7.5\,\mathrm{V}
\end{aligned}
\]

The higher winding resistance produces a larger resistive voltage drop, even though the current is unchanged.

Part b: Speed ratio

For a permanent-magnet motor with a constant magnetic field:

\[
E_{\mathrm{b}}=k_{\mathrm{e}}\omega
\]

The constant \(k_{\mathrm{e}}\) is unchanged, so:

\[
\frac{\omega_2}{\omega_1}
=\frac{E_{\mathrm{b2}}}{E_{\mathrm{b1}}}
\]

Substitute:

\[
\begin{aligned}
\frac{\omega_2}{\omega_1}
&=\frac{7.5}{9.0}\\
&\approx0.833
\end{aligned}
\]

The second speed is approximately \(83.3\%\) of the original speed.

Part c: Stall currents

For the cooler motor:

\[
\begin{aligned}
I_{\mathrm{stall,cold}}
&=\frac{12.0}{0.60}\\
&=20.0\,\mathrm{A}
\end{aligned}
\]

For the hotter motor:

\[
\begin{aligned}
I_{\mathrm{stall,hot}}
&=\frac{12.0}{0.90}\\
&\approx13.3\,\mathrm{A}
\end{aligned}
\]

These are predictions at the two specified resistances. They do not imply the motor would remain at either temperature during a prolonged stall.

Part d: Evaluate the student’s conclusion

The claim is incorrect.

An unchanged current does not guarantee unchanged speed when another electrical property has changed.

Here, increasing resistance raises the winding voltage drop from \(3.0\,\mathrm{V}\) to \(4.5\,\mathrm{V}\).

Because the supply stays at \(12.0\,\mathrm{V}\), the back emf must decrease from \(9.0\,\mathrm{V}\) to \(7.5\,\mathrm{V}\).

Since back emf is proportional to angular speed, the motor is rotating more slowly.

The tempting argument considers current alone. The correct reasoning considers current, resistance, and back emf together.

Question 4

A permanent-magnet DC motor has a winding resistance of \(0.40\,\Omega\).

It is powered through a motor controller connected to an ideal \(12.0\,\mathrm{V}\) supply.

The controller operates as follows:

• While the motor current is below \(10.0\,\mathrm{A}\), it maintains \(12.0\,\mathrm{V}\) across the motor.
• When the current would exceed \(10.0\,\mathrm{A}\), it reduces the motor’s effective terminal voltage to maintain a current of \(10.0\,\mathrm{A}\).

Assume ideal current regulation, negligible switching ripple, and constant winding resistance.

The motor initially has a back emf of \(10.0\,\mathrm{V}\). Its mechanical load is then gradually increased until the shaft stops.

a. Calculate its initial current.

b. Determine the back emf at which the controller first reaches its current limit.

c. Find the effective motor terminal voltage and winding heating power when the shaft is stalled.

d. Explain why the motor does not draw \(30\,\mathrm{A}\) at stall, even though \(12.0/0.40=30\).

e. Explain why increasing the mechanical load further cannot make the controller produce more sustained motor torque under these assumptions.

Solution 4

The initial current is \(5.0\,\mathrm{A}\). The controller reaches its limit when back emf falls to \(8.0\,\mathrm{V}\), and at stall it maintains \(10.0\,\mathrm{A}\) by reducing the effective motor terminal voltage to \(4.0\,\mathrm{V}\). The winding then dissipates \(40\,\mathrm{W}\).

Part a: Initial current

The controller initially applies the full \(12.0\,\mathrm{V}\).

\[
I=\frac{V_{\mathrm{t}}-E_{\mathrm{b}}}{R_{\mathrm{w}}}
\]

\[
\begin{aligned}
I
&=\frac{12.0-10.0}{0.40}\\
&=5.0\,\mathrm{A}
\end{aligned}
\]

This is below the \(10.0\,\mathrm{A}\) limit, so the controller doesn’t need to reduce the voltage.

Part b: Back emf when limiting begins

At the current threshold:

\[
I=10.0\,\mathrm{A}
\]

The controller is still applying \(12.0\,\mathrm{V}\) at the point where limiting begins.

\[
E_{\mathrm{b}}=V_{\mathrm{t}}-IR_{\mathrm{w}}
\]

\[
\begin{aligned}
E_{\mathrm{b}}
&=12.0-(10.0)(0.40)\\
&=8.0\,\mathrm{V}
\end{aligned}
\]

As the motor slows further, the back emf decreases below \(8.0\,\mathrm{V}\).

Without regulation, that would increase the current beyond \(10.0\,\mathrm{A}\). The controller instead reduces the effective applied voltage.

Part c: Stalled terminal voltage and heating

At stall:

\[
E_{\mathrm{b}}=0
\]

The controller maintains:

\[
I=10.0\,\mathrm{A}
\]

Therefore:

\[
\begin{aligned}
V_{\mathrm{t}}
&=IR_{\mathrm{w}}\\
&=(10.0)(0.40)\\
&=4.0\,\mathrm{V}
\end{aligned}
\]

The winding heating is:

\[
\begin{aligned}
P_{\mathrm{heat}}
&=(10.0)^2(0.40)\\
&=40\,\mathrm{W}
\end{aligned}
\]

Part d: Why the current isn’t \(30\,\mathrm{A}\)

A direct, unrestricted \(12.0\,\mathrm{V}\) supply would produce:

\[
I_{\mathrm{stall}}
=\frac{12.0}{0.40}
=30\,\mathrm{A}
\]

But the controller does not maintain \(12.0\,\mathrm{V}\) across the stalled winding.

It reduces the effective terminal voltage to \(4.0\,\mathrm{V}\), keeping the current at \(10.0\,\mathrm{A}\).

The \(30\,\mathrm{A}\) result describes the hypothetical unrestricted circuit, not this controlled circuit.

Part e: Why further loading cannot increase sustained motor torque

For a permanent-magnet motor with a fixed magnetic field, electromagnetic torque is proportional to armature current.

Once the controller fixes current at \(10.0\,\mathrm{A}\), it also fixes the maximum sustained electromagnetic torque available under this model.

If the applied mechanical load requires greater torque than the motor can provide, the shaft can slow to a stop.

The controller cannot generate the missing torque simply by keeping the current at its limit.

The difficult part of this question is recognising that the controller changes an assumption used in the ordinary stall-current calculation. The motor no longer receives a fixed terminal voltage.

Question 5

A permanent-magnet brushed DC machine is connected to an ideal \(12.0\,\mathrm{V}\) DC source through a lead resistance of \(0.25\,\Omega\).

The machine’s winding resistance is \(0.50\,\Omega\).

The DC source can both deliver and absorb electrical energy. A separate mechanical drive can force the machine’s shaft to rotate in either direction.

Define:

• Positive angular velocity as the direction the motor would normally rotate when powered by the source.
• Positive current as current flowing from the DC source into the machine.
• Positive induced emf as the polarity associated with positive angular velocity.

The induced emf obeys:

\[
E_{\mathrm{b}}=k_{\mathrm{e}}\omega
\]

where \(k_{\mathrm{e}}=0.060\,\mathrm{V\,s\,rad^{-1}}\).

The shaft is mechanically driven at a steady speed in two separate trials. Its direction and speed are not initially known.

The signed ammeter readings are:

Assume constant winding resistance, constant magnetic field, steady electrical conditions, negligible mechanical losses, and negligible brush voltage drop. The mechanical drive maintains the necessary shaft speed in each trial.

a. Calculate the predicted stall current and motor terminal voltage if the shaft is held stationary.

b. Determine the shaft’s angular velocity, including its direction, in trials A and B.

c. For each trial, calculate the electrical power supplied by or returned to the DC source, the total resistive heating power, and the mechanical power that the external drive must supply.

d. A student argues that because the stall current is the largest possible current, neither trial is physically possible. Evaluate the claim and state precisely when stall current is an upper bound.

Solution 5

The stationary stall current is \(+16.0\,\mathrm{A}\), but both measured trials are physically possible.

In trial A, the shaft is being driven backwards at \(-100\,\mathrm{rad\,s^{-1}}\). Its induced emf assists the supply, so the machine draws more than its ordinary stall current.

In trial B, the shaft is driven forwards at \(+500\,\mathrm{rad\,s^{-1}}\). Its induced emf exceeds the applied supply voltage, reversing the current and returning electrical energy to the source.

The essential point is that stall current is an upper bound for ordinary forward motoring under fixed electrical conditions, not for every possible mechanically driven state.

Part a: Stationary stall current

First calculate the total circuit resistance:

\[
\begin{aligned}
R_{\mathrm{total}}
&=R_{\mathrm{w}}+R_{\mathrm{s}}\\
&=0.50+0.25\\
&=0.75\,\Omega
\end{aligned}
\]

At zero speed, the induced emf is zero.

\[
\begin{aligned}
I_{\mathrm{stall}}
&=\frac{12.0}{0.75}\\
&=+16.0\,\mathrm{A}
\end{aligned}
\]

The motor terminal voltage at stall is:

\[
\begin{aligned}
V_{\mathrm{t,stall}}
&=I_{\mathrm{stall}}R_{\mathrm{w}}\\
&=(16.0)(0.50)\\
&=8.0\,\mathrm{V}
\end{aligned}
\]

The remaining \(4.0\,\mathrm{V}\) is dropped across the lead resistance.

Part b: Infer the shaft motion

The signed voltage equation for the entire circuit is:

\[
12.0=E_{\mathrm{b}}+I(0.75)
\]

For trial A, \(I=+24.0\,\mathrm{A}\).

\[
\begin{aligned}
E_{\mathrm{bA}}
&=12.0-(24.0)(0.75)\\
&=-6.0\,\mathrm{V}
\end{aligned}
\]

Using \(E_{\mathrm{b}}=k_{\mathrm{e}}\omega\):

\[
\begin{aligned}
\omega_A
&=\frac{-6.0}{0.060}\\
&=-100\,\mathrm{rad\,s^{-1}}
\end{aligned}
\]

The negative angular velocity means the external drive is rotating the shaft opposite to the machine’s usual motoring direction.

That reverse rotation reverses the polarity of the induced emf.

Instead of opposing the applied voltage, the induced emf now assists it.

This explains how the positive current can exceed the normal stall current.

For trial B, \(I=-24.0\,\mathrm{A}\).

\[
\begin{aligned}
E_{\mathrm{bB}}
&=12.0-(-24.0)(0.75)\\
&=30.0\,\mathrm{V}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
\omega_B
&=\frac{30.0}{0.060}\\
&=+500\,\mathrm{rad\,s^{-1}}
\end{aligned}
\]

The shaft rotates in the normal forward direction, but so quickly that the induced emf exceeds the source voltage.

The current consequently reverses.

The machine is operating as a generator, returning electrical energy to the DC source.

Part c: Account for the energy transfers

In both trials, the current magnitude is \(24.0\,\mathrm{A}\).

The total resistive heating is:

\[
P_R=I^2R_{\mathrm{total}}
\]

\[
\begin{aligned}
P_R
&=(24.0)^2(0.75)\\
&=432\,\mathrm{W}
\end{aligned}
\]

This is the combined heating in the winding and lead resistance.

Now define source power \(P_{\mathrm{src}}=V_{\mathrm{s}}I\) as positive when the source delivers energy and negative when it absorbs energy.

For trial A:

\[
\begin{aligned}
P_{\mathrm{src,A}}
&=(12.0)(+24.0)\\
&=+288\,\mathrm{W}
\end{aligned}
\]

The source delivers \(288\,\mathrm{W}\).

But the total resistive heating is \(432\,\mathrm{W}\).

Where does the extra energy come from?

It comes from the external mechanical drive, which is forcing the shaft backwards against the electromagnetic torque.

The mechanical input must be:

\[
\begin{aligned}
P_{\mathrm{mech,A}}
&=432-288\\
&=144\,\mathrm{W}
\end{aligned}
\]

Thus:

\[
288+144=432\,\mathrm{W}
\]

The electrical source and the mechanical drive together supply the energy dissipated as heat.

For trial B:

\[
\begin{aligned}
P_{\mathrm{src,B}}
&=(12.0)(-24.0)\\
&=-288\,\mathrm{W}
\end{aligned}
\]

The negative sign means the DC source absorbs \(288\,\mathrm{W}\).

The external mechanical drive must now supply enough power for both the heating and the electrical energy returned to the source:

\[
\begin{aligned}
P_{\mathrm{mech,B}}
&=432+288\\
&=720\,\mathrm{W}
\end{aligned}
\]

The energy balance is:

\[
720=432+288\,\mathrm{W}
\]

The external drive supplies \(720\,\mathrm{W}\), of which \(432\,\mathrm{W}\) becomes resistive heat and \(288\,\mathrm{W}\) is delivered to the DC source.

Both mechanical power results are also consistent with the signed electromagnetic conversion relation \(P_{\mathrm{em}}=E_{\mathrm{b}}I\). In these two trials, \(E_{\mathrm{b}}I\) is negative, indicating energy conversion from mechanical to electrical form.

Part d: Is stall current always the maximum?

The student’s claim is incorrect because it assumes the induced emf must always oppose the supply with a positive magnitude smaller than the supply voltage.

That is true during ordinary forward motoring, but not for every possible shaft motion.

For this circuit:

\[
I=\frac{12.0-E_{\mathrm{b}}}{0.75}
\]

Consider three different regimes.

There is also a fourth regime: sufficiently fast forward rotation can produce an induced emf greater than the supply voltage. Then current becomes negative, as in trial B.

In the idealised model, even the magnitude of that reverse current can exceed the ordinary stall current.

The precise statement is:

For a fixed positive supply voltage and fixed circuit resistance, stall current is the maximum positive current drawn from the source during ordinary forward motoring, provided the back emf opposes the supply.

It is not an unrestricted maximum for a DC machine that can be mechanically driven backwards or into generator operation. Real equipment also has current, temperature, speed, and voltage limits that restrict these operating states.

The conceptual trap is treating \(E_{\mathrm{b}}\) as something that can only range from zero to the supply voltage. Once shaft direction and energy flow are allowed to change, the signs become part of the physics.

07What stall current tells you next

A stall-current calculation is useful for predicting the electrical demand when a motor stops turning, but it also reveals something about energy conversion.

At stall, the speed-generated back emf is zero and the winding heating can be very large. During normal rotation, part of the terminal voltage balances back emf, which is associated with electromagnetic energy conversion into mechanical form.

That leads naturally to How to Calculate Mechanical Conversion Power in a DC Motor.

The next question is no longer just how much current the motor draws. It’s how much of the supplied electrical energy is converted electromagnetically, and how much is lost as heat.