Why a DC Motor Draws More Current When It Starts

Connect a small brushed DC motor to a fixed DC supply. Its current can be much higher during start-up than during normal running, even though the resistance of its copper windings has not suddenly changed.

The reason is that a stationary motor produces no rotational back emf. Almost the entire voltage across the motor is therefore available to drive current through its relatively low-resistance windings. As the motor accelerates, back emf increases, opposing the supply voltage and reducing the current.

There is one important detail: the current does not jump instantly to its maximum at the exact moment the switch closes. The windings also have inductance, which affects how quickly current builds. Understanding both effects gives you a much more accurate picture of what happens when a DC motor starts.

01Why is the starting current so high?

Imagine a DC motor connected to a 12 V supply. Before switching it on, its shaft is stationary.

Predict what happens to the current when you close the switch. Will it be the same as the current when the motor is spinning normally?

A simple resistor model might suggest that it will be the same. The winding resistance has not changed, and the supply voltage is unchanged.

But a motor is not just a resistor.

Its rotating armature also behaves like a generator. As the armature turns through the magnetic field, electromagnetic induction produces a voltage that opposes the supply. This is called back electromotive force, or back emf.

The armature is the rotating assembly containing the motor’s windings.

Before the motor starts moving, there is no rotational back emf. Once it starts rotating, back emf develops and increases with rotational speed.

The sequence is:

1. The supply voltage drives current through the stationary motor windings.
2. The current produces a magnetic force on the armature conductors.
3. These forces produce torque, causing the rotor to accelerate.
4. As the armature rotates, it generates an opposing back emf.
5. The increasing back emf reduces the voltage available across the winding resistance.
6. The current falls towards its normal running value.

The important point is that the winding resistance stays approximately constant. The changing back emf is what changes the current.

For a closer explanation of how the rotating conductors generate this opposing voltage, see Back emf in a DC motor: HSC Physics explained.

02The voltage equation behind starting current

Now that the physical process makes sense, we can describe it mathematically.

For a DC motor operating under approximately steady electrical conditions, Kirchhoff’s voltage law gives:

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

where:

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

We take current entering the motor’s positive terminal as positive. Back emf is defined as positive when it opposes this current during normal motoring operation.

Rearranging gives:

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

This is the key relationship.

Notice that the current depends on more than just the supply voltage and winding resistance. It also depends on the back emf.

What happens at different stages?

Consider a motor connected to a constant 12.0 V supply. Assume its winding resistance remains constant.

At the instant before the motor moves, its angular speed is zero. Therefore, its rotational back emf is also zero.

In the simplified steady electrical model:

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

So:

\[ I_{\mathrm{start}}\approx\frac{V_{\mathrm t}}{R_{\mathrm w}} \]

This can produce a large current because motor windings are designed to have relatively low resistance.

Once the motor is rotating, the back emf takes up part of the supply voltage. Less voltage remains across the winding resistance to drive the running current, so the current decreases.

Why does back emf increase with speed?

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

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

where:

• \(k_{\mathrm e}\) is the motor’s back-emf constant, in \(\mathrm{V\,s\,rad^{-1}}\).
• \(\omega\) is angular speed, in \(\mathrm{rad\,s^{-1}}\).

This relationship explains why increasing rotational speed produces a larger opposing voltage.

The motor does not need its winding resistance to change. Its rotation changes the voltage balance inside the circuit.

03What actually happens in the first few milliseconds?

There is a subtlety worth understanding before using the starting-current equation.

The motor windings are coils of wire. Coils have inductance, which means they oppose rapid changes in current.

Consequently, the current cannot instantly jump from zero to \(V_{\mathrm t}/R_{\mathrm w}\) when the switch closes.

A more complete voltage equation is:

\[ V_{\mathrm t} = E_{\mathrm b} + IR_{\mathrm w} + L\frac{dI}{dt} \]

Here, \(L\) is the effective armature inductance, measured in henries (H), and \(dI/dt\) describes how quickly the current changes.

At the instant of switching:

• The rotor is stationary, so \(E_{\mathrm b}=0\).
• The initial current is zero, assuming it was zero before switching.
• The current begins increasing because of the voltage across the inductance.

For an idealised supply step:

\[ \left.\frac{dI}{dt}\right|_{t=0} = \frac{V_{\mathrm t}}{L} \]

As current builds, the motor develops torque and starts accelerating. Its back emf then increases, opposing the supply and reducing the current.

This means the actual current-time graph can initially rise, reach a peak, and then fall.

This distinction is important:

• The simplified HSC starting-current calculation ignores inductance and estimates the current with zero back emf.
• A real motor has inductance, so its current takes time to rise.
• As the motor accelerates, increasing back emf causes the current to fall.

Unless a question specifically provides inductance or asks about the switching transient, the simplified motor equation is generally the appropriate starting point for HSC-style calculations.

04How could we investigate the starting current experimentally?

Suppose we want to test whether the current decreases because the motor develops back emf, rather than because its winding resistance changes.

We need to measure three things together:

• The current entering the motor.
• The voltage across its terminals.
• The rotational speed of its shaft.

Setting up the experiment

Use a regulated, current-limited DC supply.

Connect its positive terminal through a switch and an ammeter to the motor’s positive brush. Connect the motor’s other brush back to the negative supply terminal.

Place a voltmeter directly across the two motor terminals. Aim an optical tachometer at a reflective marker on the motor shaft to measure rotational speed.

Connect the measuring instruments to a data logger so that current, voltage, and speed are recorded against the same time axis.

Before collecting data, we should also determine the motor’s effective winding resistance independently.

This can be estimated with the motor disconnected from its supply, using an appropriate low-current resistance measurement. The rotor should remain stationary, and the winding temperature should be recorded because copper resistance changes with temperature.

What should the experiment show?

There are two competing predictions.

Prediction 1: The motor behaves like a fixed resistor.

If this model were correct, a constant terminal voltage and constant resistance would produce a constant current, regardless of shaft speed.

Prediction 2: The motor generates back emf.

If this model is correct, current should decrease as the motor accelerates and back emf increases.

The second prediction is what we expect for a normally operating DC motor.

Consider the following illustrative data.

Illustrative readings only; these are not measurements from a particular commercial motor.

Assume:

• Motor terminal voltage is maintained at \(12.0\,\mathrm V\).
• Effective winding resistance is \(2.0\,\Omega\).
• The magnetic field remains approximately constant.
• The winding temperature changes negligibly during the short measurement.

Notice two different behaviours.

First, the current rises from zero to \(4.8\,\mathrm A\). This is consistent with the effect of winding inductance.

Second, as the motor accelerates further, the current falls to \(0.7\,\mathrm A\). This is consistent with the development of back emf.

We can use the final reading to estimate the back emf.

At approximately 1000 ms, assume the current is changing slowly enough that the inductive voltage is negligible.

Using:

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

Substitute the measured values:

\[ \begin{aligned} E_{\mathrm b} &=12.0-(0.70)(2.0)\\ &=12.0-1.4\\ &=10.6\,\mathrm V \end{aligned} \]

The motor is producing an opposing back emf of approximately \(10.6\,\mathrm V\).

Only \(1.4\,\mathrm V\) remains across the winding resistance to drive the running current.

This is strong evidence for the back-emf explanation.

What would a simpler experiment miss?

Suppose we only connected an ammeter and measured the current after the motor was already running.

We might calculate:

\[ R=\frac{V_{\mathrm t}}{I} \]

But this would not give the actual winding resistance. It would combine the effects of winding resistance and back emf.

A better experiment records speed, current, and terminal voltage simultaneously. It also measures winding resistance independently.

This allows us to distinguish the resistive voltage drop from the voltage generated by rotation.

One limitation remains: the winding resistance can increase as the motor heats up. Keeping the investigation short and monitoring temperature helps prevent this from being mistaken for the main starting-current mechanism.

05Starting current versus stall current

There is an important difference between a motor starting normally and a motor whose shaft is jammed.

Both situations involve a stationary or nearly stationary rotor. Both therefore have approximately zero rotational back emf.

However, their behaviour after that point is very different.

A motor starting normally

The motor initially has no rotational back emf, so its current increases.

The resulting magnetic torque accelerates the rotor. As speed increases, back emf increases and the current falls.

The high-current period is temporary, provided the motor accelerates successfully.

A motor that is stalled

Now imagine that the same motor is running normally, but its shaft suddenly becomes jammed.

Its rotational speed falls towards zero. Consequently, its back emf decreases towards zero.

The current therefore increases towards the stationary-rotor value.

If the shaft remains jammed, the motor cannot accelerate and restore its back emf. The high current can persist, producing substantial heating in the windings.

Electrical heating is given by:

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

This is why a stalled motor can overheat rapidly.

The important distinction is:

• Starting: current is initially high, but acceleration builds back emf and reduces it.
• Stalling: current becomes high, but the motor cannot accelerate, so back emf does not recover.

For a more detailed treatment of locked-rotor current calculations, see How to Calculate DC Motor Stall Current: HSC Physics.

06Worked examples

07Questions and solutions

Question 1

A DC motor is connected to an ideal \(10.0\,\mathrm V\) supply. Its effective winding resistance is \(2.5\,\Omega\), and its rotor is initially stationary.

Calculate the simplified starting-current estimate, assuming negligible inductance, source resistance, and brush voltage drop.

Explain why the current decreases once the motor reaches normal operating speed.

Solution 1

The simplified stationary-rotor current is \(4.0\,\mathrm A\).

Step 1: Identify the back emf.

The motor is stationary, so:

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

Step 2: Apply the motor voltage equation.

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

Substitute:

\[ \begin{aligned} I&=\frac{10.0-0}{2.5}\\ &=4.0\,\mathrm A \end{aligned} \]

Step 3: Explain what happens as the motor accelerates.

Once the rotor starts moving, its conductors rotate through the magnetic field and generate back emf.

This opposes the supply voltage, reducing the voltage across the winding resistance.

Consequently, the current falls.

The winding resistance has not disappeared or increased. The change comes from the additional opposing voltage generated by rotation.

Question 2

A DC motor operates from an \(18.0\,\mathrm V\) supply and has an effective winding resistance of \(1.2\,\Omega\).

During normal operation, it draws \(2.5\,\mathrm A\).

Calculate:

a. The back emf during normal operation.

b. The simplified stationary-rotor current.

c. How many times greater the stationary-rotor current is than the normal running current.

d. Explain why dividing the supply voltage by the running current does not give the actual winding resistance.

Assume a constant terminal voltage and negligible inductive and brush effects.

Solution 2

The running back emf is \(15.0\,\mathrm V\), and the stationary-rotor current estimate is \(15.0\,\mathrm A\), which is six times the running current.

Step 1: Calculate the running back emf.

Use:

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

Substitute:

\[ \begin{aligned} E_{\mathrm b} &=18.0-(2.5)(1.2)\\ &=18.0-3.0\\ &=15.0\,\mathrm V \end{aligned} \]

The motor generates a \(15.0\,\mathrm V\) opposing back emf during normal operation.

Step 2: Calculate the stationary-rotor current.

At zero rotational speed:

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

Therefore:

\[ \begin{aligned} I_{\mathrm{start}} &=\frac{18.0}{1.2}\\ &=15.0\,\mathrm A \end{aligned} \]

Step 3: Calculate the current ratio.

\[ \begin{aligned} \frac{I_{\mathrm{start}}}{I_{\mathrm{run}}} &=\frac{15.0}{2.5}\\ &=6.0 \end{aligned} \]

The stationary-rotor estimate is six times the normal running current.

Step 4: Explain why \(V/I\) does not give the winding resistance.

If we divide the supply voltage by the running current:

\[ \frac{18.0}{2.5}=7.2\,\Omega \]

This is not the actual winding resistance.

The motor is generating \(15.0\,\mathrm V\) of back emf while running. Only \(3.0\,\mathrm V\) is available across the winding resistance.

The actual resistance is:

\[ R_{\mathrm w}=\frac{3.0}{2.5} =1.2\,\Omega \]

The mistake is treating a running motor as an ordinary resistor and ignoring its generated back emf.

Question 3

A student investigates the start-up behaviour of a DC motor using a time-resolved current sensor and a rotational speed sensor.

The following illustrative readings are recorded:

The student argues:

“A faster motor draws less current because its back emf increases. Therefore, the current should decrease continuously from the instant the switch closes. These results must be incorrect because the current initially increases.”

Evaluate the student’s argument.

Solution 3

The measurements are consistent with the behaviour of a real DC motor. The student’s explanation ignores winding inductance.

Step 1: Explain the initial increase in current.

Immediately after switching, the motor is stationary and produces no rotational back emf.

However, its windings have inductance. Inductance opposes rapid changes in current, so the current begins at zero and rises over a finite time.

As the current increases, the motor develops electromagnetic torque.

Step 2: Explain why the current eventually decreases.

The increasing current produces torque, accelerating the rotor.

As the rotational speed increases, back emf develops and becomes larger.

The motor voltage equation is:

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

once the inductive voltage is negligible.

As \(E_{\mathrm b}\) increases, less voltage remains across the winding resistance. The current therefore decreases.

Step 3: Evaluate the student’s conclusion.

The student is correct that increasing rotational speed increases back emf and reduces current.

However, this does not mean the current must decrease from the exact instant of switching.

The initial current rise is explained by inductance, while the later decrease is explained by increasing back emf.

The two effects occur during different stages of the start-up process.

Question 4

A DC motor is powered by a battery with an electromotive force of \(12.0\,\mathrm V\) and an internal resistance of \(0.40\,\Omega\).

The motor has a winding resistance of \(0.80\,\Omega\).

Calculate:

a. The current drawn when the motor is stationary.

b. The voltage across the motor terminals when stationary.

c. The current drawn when the motor is running with a back emf of \(8.0\,\mathrm V\).

d. The voltage across the motor terminals when running.

Explain why the motor’s starting current is lower than the value predicted by dividing the battery emf by the winding resistance alone.

Assume constant winding resistance and negligible inductive and brush effects.

Solution 4

The stationary motor draws \(10.0\,\mathrm A\), with \(8.0\,\mathrm V\) across its terminals. During running, it draws approximately \(3.33\,\mathrm A\), with approximately \(10.67\,\mathrm V\) across its terminals.

Step 1: Calculate the stationary-rotor current.

When stationary:

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

The battery’s internal resistance and motor winding resistance are in series.

Therefore:

\[ R_{\mathrm{total}}=r+R_{\mathrm w} \]

where \(r\) is the battery’s internal resistance.

Substitute:

\[ \begin{aligned} I_{\mathrm{start}} &=\frac{12.0}{0.40+0.80}\\ &=\frac{12.0}{1.20}\\ &=10.0\,\mathrm A \end{aligned} \]

Step 2: Calculate the motor terminal voltage when stationary.

The internal voltage drop is:

\[ V_r=Ir \]

Therefore:

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

The motor receives \(8.0\,\mathrm V\), not the full battery emf of \(12.0\,\mathrm V\).

Step 3: Calculate the running current.

The back emf is now \(8.0\,\mathrm V\).

The combined circuit equation is:

\[ \mathcal E=E_{\mathrm b}+I(R_{\mathrm w}+r) \]

where \(\mathcal E\) is the battery emf.

Rearrange:

\[ I=\frac{\mathcal E-E_{\mathrm b}}{R_{\mathrm w}+r} \]

Substitute:

\[ \begin{aligned} I_{\mathrm{run}} &=\frac{12.0-8.0}{0.80+0.40}\\ &=\frac{4.0}{1.20}\\ &=3.33\,\mathrm A \end{aligned} \]

Step 4: Calculate the running terminal voltage.

\[ \begin{aligned} V_{\mathrm t} &=\mathcal E-Ir\\ &=12.0-(3.33)(0.40)\\ &\approx10.67\,\mathrm V \end{aligned} \]

We can check this against the motor voltage equation:

\[ \begin{aligned} V_{\mathrm t} &=E_{\mathrm b}+IR_{\mathrm w}\\ &=8.0+(3.33)(0.80)\\ &\approx10.67\,\mathrm V \end{aligned} \]

Both calculations agree.

Step 5: Explain the difference.

The battery has internal resistance, so some voltage is lost inside the battery when current flows.

At start-up, the current is large, producing a substantial internal voltage drop.

As the motor accelerates, its back emf increases and current decreases. The internal voltage drop then becomes smaller, allowing the motor terminal voltage to rise.

The important lesson is that the battery emf is not necessarily equal to the voltage across the motor terminals.

Question 5

Two identical permanent-magnet DC motors are connected separately to the same ideal \(12.0\,\mathrm V\) supply.

Both motors have:

• Winding resistance \(R_{\mathrm w}=1.0\,\Omega\).
• Armature inductance \(L=0.20\,\mathrm H\).
• Back-emf constant \(k_{\mathrm e}=0.020\,\mathrm{V\,s\,rad^{-1}}\).
• Torque constant \(k_{\mathrm t}=0.020\,\mathrm{N\,m\,A^{-1}}\).

Both begin at rest with zero initial current.

Motor A has a rotor moment of inertia:

\[ J_A=2.0\times10^{-3}\,\mathrm{kg\,m^2} \]

Motor B has an identical motor rotor but carries an additional frictionless flywheel. Its total moment of inertia is:

\[ J_B=1.0\times10^{-2}\,\mathrm{kg\,m^2} \]

Assume:

• There is no external mechanical load or friction.
• The only difference between the motors is their total moment of inertia.
• Motor torque is \(\tau=k_{\mathrm t}I\).
• Back emf is \(E_{\mathrm b}=k_{\mathrm e}\omega\).
• The motor constants remain unchanged.
• Both motors are allowed to accelerate until they reach \(100\,\mathrm{rad\,s^{-1}}\).

Answer the following:

a. What is the current in each motor at the exact instant the supply is connected?

b. Calculate the initial rate of increase of current in each motor.

c. After their rotational speeds begin to separate, which motor draws more current at the same elapsed time? Explain your reasoning using both the electrical and mechanical behaviour.

d. Calculate the total electrical charge that passes through each motor before it reaches \(100\,\mathrm{rad\,s^{-1}}\).

e. A student claims that Motor B must draw five times the initial current because its moment of inertia is five times larger. Explain precisely why this prediction is incorrect, even though Motor B does require more total charge to reach the specified speed.

Solution 5

Motor B does not have a larger current at the exact instant of switching. Both currents begin at zero and have the same initial rate of increase, \(60\,\mathrm{A\,s^{-1}}\).

However, Motor B accelerates more slowly, so its back emf builds more slowly. It therefore draws more current after their speeds begin to separate.

Motor A requires \(10\,\mathrm C\) of charge to reach the specified speed, while Motor B requires \(50\,\mathrm C\).

Step 1: Determine the current at the exact instant of switching.

Both motors begin with:

\[ I(0)=0 \]

A current cannot jump instantaneously in an ideal inductive winding.

Therefore:

\[ I_A(0)=I_B(0)=0\,\mathrm A \]

Step 2: Calculate the initial rate of increase of current.

The full transient equation is:

\[ V_{\mathrm t} = E_{\mathrm b} + IR_{\mathrm w} + L\frac{dI}{dt} \]

At the instant of switching:

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

and:

\[ I=0 \]

Therefore:

\[ \frac{dI}{dt}=\frac{V_{\mathrm t}}{L} \]

Substitute:

\[ \begin{aligned} \frac{dI}{dt} &=\frac{12.0}{0.20}\\ &=60\,\mathrm{A\,s^{-1}} \end{aligned} \]

Both motors have the same supply voltage and inductance.

Consequently, both initially increase their current at the same rate.

Their different moments of inertia do not affect this initial electrical response.

Step 3: Determine which motor draws more current after their speeds begin to separate.

Motor A has a smaller moment of inertia, so the same electromagnetic torque produces a greater angular acceleration.

Motor A therefore accelerates faster.

Its increasing rotational speed produces an increasing back emf:

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

Motor B accelerates more slowly because its larger moment of inertia resists changes in rotational speed.

Its back emf therefore increases more slowly.

Consider the transient equation again:

\[ L\frac{dI}{dt} = V_{\mathrm t}-IR_{\mathrm w}-E_{\mathrm b} \]

For the same instantaneous current, the motor with the smaller back emf has a larger rate of increase of current.

Therefore, once their speeds begin to separate, Motor B draws more current at the same elapsed time.

This is the crucial interaction between the electrical and mechanical parts of the motor.

The heavier flywheel does not directly increase the initial current. It changes how quickly the motor accelerates, which changes how quickly back emf develops.

Step 4: Calculate the total charge required to reach the target speed.

This is where the problem becomes particularly interesting.

The electromagnetic torque is:

\[ \tau=k_{\mathrm t}I \]

Newton’s second law for rotational motion gives:

\[ \tau=J\frac{d\omega}{dt} \]

Combining these equations:

\[ J\frac{d\omega}{dt}=k_{\mathrm t}I \]

Rearrange:

\[ I\,dt=\frac{J}{k_{\mathrm t}}d\omega \]

Total electrical charge is:

\[ Q=\int I\,dt \]

Therefore:

\[ Q=\frac{J}{k_{\mathrm t}}\Delta\omega \]

This relationship shows that the total charge depends on the change in angular momentum when the motor torque is proportional to current.

For Motor A:

\[ \begin{aligned} Q_A &=\frac{J_A\Delta\omega}{k_{\mathrm t}}\\ &=\frac{(2.0\times10^{-3})(100)} {0.020}\\ &=10\,\mathrm C \end{aligned} \]

For Motor B:

\[ \begin{aligned} Q_B &=\frac{J_B\Delta\omega}{k_{\mathrm t}}\\ &=\frac{(1.0\times10^{-2})(100)} {0.020}\\ &=50\,\mathrm C \end{aligned} \]

Motor B requires five times as much electrical charge to reach the same rotational speed.

Step 5: Evaluate the student’s prediction.

The student is correct that Motor B requires more electrical charge, but incorrect about the initial current.

The tempting prediction is that five times the inertia must mean five times the current.

However, current is initially governed by the electrical properties of the circuit, particularly the supply voltage and winding inductance.

The difference in inertia becomes important as the motors accelerate.

Motor B accelerates more slowly, develops back emf more slowly, and consequently draws a larger current for longer.

The total charge is greater because more angular momentum must be transferred to the heavier rotating system.

We can also check that the target speed is physically reachable. At \(100\,\mathrm{rad\,s^{-1}}\):

\[ E_{\mathrm b} = (0.020)(100) = 2.0\,\mathrm V \]

This is below the \(12.0\,\mathrm V\) supply, so the motor can still develop positive electromagnetic torque at the target speed.

The final distinction is important: inertia determines how much angular momentum must be gained, while back emf and inductance help determine how the current changes during that process.

This connects directly to the next useful motor concept: how an external mechanical load changes rotational speed, back emf, current, and the torque available at the shaft.