How to Calculate DC Motor Back EMF Using Voltage and Current

A DC motor draws \(2.0\,\text{A}\) from a \(12.0\,\text{V}\) supply. Does that mean its resistance is \(6.0\,\Omega\)?

Not necessarily. If the motor’s armature winding resistance is \(1.0\,\Omega\), only \(2.0\,\text{V}\) is lost across the copper winding. The remaining \(10.0\,\text{V}\) is the motor’s back emf.

The calculation is:

\[ \boxed{E_b=V_m-IR_w} \]

where \(E_b\) is the back emf, \(V_m\) is the voltage across the motor terminals, \(I\) is the armature current, and \(R_w\) is the armature winding resistance.

There is an important catch: you cannot calculate back emf from supply voltage and current alone. You also need the winding resistance. You must also check that the supply voltage is actually the voltage across the motor, rather than the voltage before cables, resistors, or a motor controller.

01The equation for calculating back emf

A DC motor has two electrical effects that you need to keep separate:

• The copper winding has electrical resistance, which causes a voltage drop when current flows.
• The rotating armature generates an induced emf that opposes the applied supply voltage. This is the back emf.

For a motor operating steadily, Kirchhoff’s voltage law gives:

\[ V_m=E_b+IR_w \]

Rearranging:

\[ \boxed{E_b=V_m-IR_w} \]

Here is what each symbol means.

We take current as positive when it enters the motor’s positive terminal. During normal motoring operation, the back emf has a polarity that opposes the applied supply.

The equation tells you something important. The supply voltage is shared between the back emf and the voltage drop across the winding.

For example, if the motor receives \(12\,\text{V}\) and the winding loses \(2\,\text{V}\), the back emf must be \(10\,\text{V}\).

The motor is not simply a resistor. Its rotation generates a voltage of its own.

02Why you cannot treat a motor as a fixed resistor

This is one of the easiest mistakes to make in a motor calculation.

Suppose a motor has a terminal voltage of \(12\,\text{V}\) and draws \(2.0\,\text{A}\). You might calculate:

\[ R=\frac{V}{I} =\frac{12}{2.0} =6.0\,\Omega \]

But this is not the winding resistance.

The ratio \(V/I\) includes the effect of the back emf as well as the winding resistance. From the motor equation:

\[ \frac{V_m}{I} = R_w+\frac{E_b}{I} \]

The winding resistance is only one part of the electrical behaviour.

If the actual winding resistance is \(1.0\,\Omega\), the back emf is:

\[ E_b=12-(2.0)(1.0) \]

\[ E_b=10\,\text{V} \]

The motor’s \(6.0\,\Omega\) voltage-to-current ratio does not mean its copper winding has a resistance of \(6.0\,\Omega\).

What changes when the motor starts rotating?

Consider a motor connected to a constant \(12\,\text{V}\) supply. Assume its winding resistance remains \(1.0\,\Omega\).

Compare three operating conditions.

These are illustrative values calculated using the steady-state motor model.

When the motor is stationary, there is no rotational back emf. The full supply voltage acts across the winding.

Once the motor rotates, it generates an opposing emf. Less voltage is available to drive current through the winding, so the current decreases.

Notice that the winding resistance has not changed in this simplified comparison. The changing current is explained by the changing back emf.

This is why the motor cannot be analysed as though it were an ordinary fixed resistor.

03How to calculate back emf: a reliable method

Use this process whenever an HSC Physics question asks you to calculate the back emf from motor voltage and current.

Step 1: Identify the motor terminal voltage.

Use the voltage measured across the motor’s terminals.

If the question states that the motor is connected directly to an ideal supply, the supply voltage and motor terminal voltage are equal.

If there are other components between the supply and motor, you may need to calculate the voltage drop across them first.

Step 2: Find the armature winding resistance.

Look for the resistance given in the question. It may be called:

• armature resistance;
• winding resistance;
• internal motor resistance.

Use the effective resistance of the motor’s armature winding, not the ratio of supply voltage to operating current.

Step 3: Calculate the winding voltage drop.

Use Ohm’s law:

\[ V_R=IR_w \]

Multiply the armature current by the winding resistance.

Step 4: Subtract the winding voltage drop from the motor terminal voltage.

\[ \boxed{E_b=V_m-IR_w} \]

Your answer should be in volts.

A useful final check is that, during normal motoring operation, the back emf should be positive and smaller than the motor terminal voltage.

How do you find the winding resistance?

In many HSC questions, the winding resistance will be provided. In a practical measurement, however, you may need to determine it.

For a disconnected DC motor, a low-resistance measurement can be made across its armature terminals using a suitable ohmmeter.

There are a few details to watch:

• Disconnect the motor from its power supply before measuring resistance.
• Account for the resistance of the meter leads, which can be significant when measuring small resistances.
• Brush and commutator contact can cause readings to vary with rotor position.
• The winding resistance increases as the copper heats up, so a cold measurement may not represent the resistance during normal operation.

For an accurate running calculation, use the winding resistance at the operating temperature, if it is available.

If you use a cold resistance that is smaller than the actual hot resistance, you will underestimate the winding voltage drop and overestimate the back emf.

04Worked examples

05When the simple equation needs extra care

The equation

\[ E_b=V_m-IR_w \]

is the steady-state model for a DC motor.

It assumes that:

• the current is steady;
• the effective winding resistance is known;
• the voltage is measured across the motor terminals;
• brush voltage drops and other small effects are ignored.

There are three situations where you should be particularly careful.

1. The motor winding has heated up

Copper resistance increases with temperature.

If a question gives both cold and operating-temperature resistance, use the operating-temperature value when calculating the back emf during normal running.

2. The current is changing rapidly

Immediately after switching on a motor, the current may still be changing.

The armature also has inductance, which affects the instantaneous voltage balance. The simple equation is not a complete description of that brief transient.

For ordinary steady-running HSC-style calculations, use the steady-state equation unless the question specifically introduces additional effects.

3. The motor is connected through an electronic controller

A motor controller may cause the voltage across the motor to differ from the supply voltage.

In that situation, do not automatically substitute the battery’s rated voltage for \(V_m\). Use the voltage across the motor terminals, taking the controller’s operation into account.

06Questions and solutions

Question 1

A DC motor operates from a \(12.0\,\text{V}\) supply. Its armature winding resistance is \(0.75\,\Omega\), and the motor draws a steady current of \(3.0\,\text{A}\).

Calculate:

a. the voltage drop across the winding;

b. the back emf of the motor.

Solution 1

The voltage drop across the winding is \(2.25\,\text{V}\), and the back emf is \(9.75\,\text{V}\).

Part a: Winding voltage drop

Using Ohm’s law:

\[ V_R=IR_w \]

Substituting:

\[ V_R=(3.0)(0.75) \]

\[ \boxed{V_R=2.25\,\text{V}} \]

Part b: Back emf

Since the motor is connected directly to the supply:

\[ V_m=12.0\,\text{V} \]

Using the motor voltage equation:

\[ E_b=V_m-IR_w \]

\[ E_b=12.0-2.25 \]

\[ \boxed{E_b=9.75\,\text{V}} \]

The back emf accounts for the remaining voltage after the voltage drop across the winding.

The key is to subtract the winding voltage drop, not the current itself or the motor’s calculated \(V/I\) ratio.

Question 2

A \(24.0\,\text{V}\) ideal supply is connected to two parallel branches.

The first branch contains a DC motor with an armature winding resistance of \(1.5\,\Omega\). An ammeter in this branch reads \(4.0\,\text{A}\).

The second branch contains a \(12\,\Omega\) resistor.

A third ammeter, positioned before the current splits between the branches, reads \(6.0\,\text{A}\).

Calculate:

a. the current through the resistor;

b. the back emf of the motor;

c. the incorrect back emf that would be calculated if the total supply current were used instead of the motor branch current.

Explain why the third ammeter reading cannot be used directly in the motor equation.

Solution 2

The resistor branch carries \(2.0\,\text{A}\). The motor back emf is \(18.0\,\text{V}\). Incorrectly using the total current would give \(15.0\,\text{V}\).

Part a: Current through the resistor

The two branches are connected in parallel, so the resistor has the full supply voltage across it.

Using Ohm’s law:

\[ I_R=\frac{V}{R} \]

\[ I_R=\frac{24.0}{12} \]

\[ \boxed{I_R=2.0\,\text{A}} \]

This also agrees with Kirchhoff’s current law:

\[ I_{\text{total}}=I_{\text{motor}}+I_R \]

\[ 6.0=4.0+2.0 \]

Part b: Motor back emf

The motor receives the full \(24.0\,\text{V}\) because it is connected directly across the ideal supply.

The motor branch current is \(4.0\,\text{A}\).

Therefore:

\[ E_b=V_m-IR_w \]

\[ E_b=24.0-(4.0)(1.5) \]

\[ E_b=24.0-6.0 \]

\[ \boxed{E_b=18.0\,\text{V}} \]

Part c: Incorrect calculation

If the total current of \(6.0\,\text{A}\) is substituted:

\[ E_b=24.0-(6.0)(1.5) \]

\[ E_b=24.0-9.0 \]

\[ \boxed{E_b=15.0\,\text{V}} \]

This is incorrect.

The motor only carries \(4.0\,\text{A}\). The remaining \(2.0\,\text{A}\) flows through the parallel resistor.

The tempting mistake is to assume that every current measurement in the circuit represents the motor current. It does not.

Always identify which ammeter measures the current through the motor’s armature winding before substituting into the back-emf equation.

Question 3

A DC motor operates from an \(18.0\,\text{V}\) supply.

Its winding resistance is initially measured as \(1.2\,\Omega\) at room temperature. After the motor has been running for some time, its winding resistance increases to \(1.5\,\Omega\) because of heating.

At its operating temperature, the motor draws a steady current of \(4.0\,\text{A}\).

Calculate:

a. the correct back emf using the operating-temperature resistance;

b. the back emf that would be calculated using the original cold resistance;

c. the percentage by which the cold-resistance calculation overestimates the back emf.

Explain why the two answers differ.

Solution 3

The correct back emf is \(12.0\,\text{V}\). Using the cold resistance gives \(13.2\,\text{V}\), an overestimate of \(10\%\).

Part a: Correct back emf

The motor is running at its operating temperature, so we must use:

\[ R_w=1.5\,\Omega \]

The motor terminal voltage is:

\[ V_m=18.0\,\text{V} \]

Using:

\[ E_b=V_m-IR_w \]

\[ E_b=18.0-(4.0)(1.5) \]

\[ E_b=18.0-6.0 \]

\[ \boxed{E_b=12.0\,\text{V}} \]

Part b: Calculation using cold resistance

If we incorrectly use \(1.2\,\Omega\):

\[ E_b=18.0-(4.0)(1.2) \]

\[ E_b=18.0-4.8 \]

\[ \boxed{E_b=13.2\,\text{V}} \]

Part c: Percentage error

The absolute error is:

\[ \Delta E_b=13.2-12.0 \]

\[ \Delta E_b=1.2\,\text{V} \]

Percentage error:

\[ \text{Percentage error} = \frac{\text{absolute error}}{\text{correct value}} \times100\% \]

\[ =\frac{1.2}{12.0}\times100\% \]

\[ \boxed{10\%} \]

The cold-resistance calculation overestimates the back emf.

Why?

The actual hot winding has a higher resistance, so it produces a larger voltage drop at the same current.

Using the smaller cold resistance underestimates this voltage drop. Since the winding voltage drop is subtracted from the terminal voltage, the calculated back emf becomes too large.

This is a useful reminder that the resistance value must match the operating condition described in the question.

Question 4

A DC motor is connected directly to an \(18.0\,\text{V}\) supply.

A laboratory test sheet records two measurements for the same motor:

• During a brief locked-rotor test, the motor draws \(12.0\,\text{A}\) after the initial electrical transient has settled.
• During normal running, the motor draws \(4.0\,\text{A}\).

Assume the winding temperature is the same for both measurements and ignore brush voltage drops.

Calculate:

a. the armature winding resistance;

b. the back emf during normal running;

c. the ratio of the locked-rotor current to the normal running current.

Explain why the locked-rotor measurement can be used to determine winding resistance, but the normal running measurement cannot be used in the same way.

Solution 4

The armature winding resistance is \(1.5\,\Omega\), the running back emf is \(12.0\,\text{V}\), and the locked-rotor current is three times the normal running current.

Part a: Winding resistance

During the locked-rotor test, the motor is stationary.

Since there is no rotation, there is no rotational back emf:

\[ E_b=0 \]

The steady-state voltage equation becomes:

\[ V_m=IR_w \]

Rearranging:

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

Substituting the locked-rotor measurements:

\[ R_w=\frac{18.0}{12.0} \]

\[ \boxed{R_w=1.5\,\Omega} \]

This works because the back emf is zero when the rotor is stationary.

Part b: Back emf during normal running

The motor now draws \(4.0\,\text{A}\).

The terminal voltage remains \(18.0\,\text{V}\), and the winding resistance is unchanged.

Using:

\[ E_b=V_m-IR_w \]

\[ E_b=18.0-(4.0)(1.5) \]

\[ E_b=18.0-6.0 \]

\[ \boxed{E_b=12.0\,\text{V}} \]

Part c: Current ratio

\[ \text{Current ratio} = \frac{I_{\text{locked}}}{I_{\text{running}}} \]

\[ =\frac{12.0}{4.0} \]

\[ \boxed{3:1} \]

The motor draws three times as much current when stationary as it does during normal running.

The important distinction is why the resistance calculation works in one case but not the other.

When stationary, the back emf is zero, so the entire terminal voltage is across the winding. Therefore, \(V/I\) gives the winding resistance.

During normal running, part of the terminal voltage is associated with back emf. Consequently, \(V/I\) is not the winding resistance.

A real motor should not be held in a stalled condition for an extended test, because the large current can cause substantial heating. The measurements here are supplied data for the calculation.

Question 5

A permanent-magnet DC machine is connected to a \(12.0\,\text{V}\) electrical source that can both supply and absorb energy.

Its armature winding resistance is \(0.40\,\Omega\).

An external mechanical drive forces the shaft to rotate in the motor’s normal direction at a speed that produces a back emf of \(13.2\,\text{V}\).

Take current entering the motor’s positive terminal as positive.

Assume the external drive maintains constant rotational speed throughout the electrical measurements.

Calculate:

a. the signed armature current;

b. the actual direction of current flow;

c. whether the electromagnetic torque assists or opposes the imposed rotation;

d. the electrical power returned to the source;

e. the power dissipated as heat in the winding.

Finally, explain what happens if the external mechanical drive is disconnected while the \(12.0\,\text{V}\) source remains connected. Can the back emf be assumed to remain at \(13.2\,\text{V}\)?

Solution 5

The signed armature current is \(-3.0\,\text{A}\). Therefore, \(3.0\,\text{A}\) flows out of the machine towards the source. The electromagnetic torque opposes the imposed rotation, \(36.0\,\text{W}\) is returned to the source, and \(3.6\,\text{W}\) is dissipated as heat.

Part a: Signed armature current

Start with the motor voltage equation:

\[ V_m=E_b+IR_w \]

Rearrange:

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

Substitute:

\[ I=\frac{12.0-13.2}{0.40} \]

\[ I=\frac{-1.2}{0.40} \]

\[ \boxed{I=-3.0\,\text{A}} \]

The negative sign is important.

It does not mean the calculation has failed. It means the actual current flows opposite to the direction we defined as positive.

Part b: Actual current direction

We defined positive current as entering the motor’s positive terminal.

Since the calculated current is negative, the actual current leaves the motor’s positive terminal and flows towards the electrical source.

The machine is now delivering electrical energy rather than consuming it.

Part c: Direction of electromagnetic torque

In a fixed-field DC machine, electromagnetic torque is proportional to armature current.

Reversing the current reverses the electromagnetic torque.

Therefore, the electromagnetic torque opposes the externally imposed rotation.

This is consistent with Lenz’s law: the electromagnetic effect opposes the mechanical process producing the induced emf.

Part d: Electrical power returned to the source

Electrical power entering the machine is:

\[ P_{\text{terminal}}=V_mI \]

Substituting:

\[ P_{\text{terminal}}=(12.0)(-3.0) \]

\[ P_{\text{terminal}}=-36.0\,\text{W} \]

The negative sign means power is leaving the machine.

Therefore:

\[ \boxed{P_{\text{returned}}=36.0\,\text{W}} \]

The machine returns \(36.0\,\text{W}\) to the electrical source.

Part e: Power dissipated as heat

Resistive heating is calculated using:

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

Substituting:

\[ P_{\text{heat}}=(-3.0)^2(0.40) \]

\[ P_{\text{heat}}=(9.0)(0.40) \]

\[ \boxed{P_{\text{heat}}=3.6\,\text{W}} \]

Notice that the current is squared, so the heating power remains positive even though the current direction has reversed.

The electromagnetic power conversion can also be checked:

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

\[ P_{\text{em}}=(13.2)(-3.0) \]

\[ P_{\text{em}}=-39.6\,\text{W} \]

The negative sign indicates that mechanical energy is being converted into electrical energy.

The energy balance is:

\[ 39.6=36.0+3.6 \]

The external drive supplies \(39.6\,\text{W}\) of mechanical power. Of this, \(36.0\,\text{W}\) is returned electrically, and \(3.6\,\text{W}\) becomes heat in the winding.

What happens if the external drive is disconnected?

The back emf cannot be assumed to remain at \(13.2\,\text{V}\).

The electromagnetic torque is initially opposing the rotation, so the shaft begins to slow down. As its rotational speed decreases, the induced back emf also decreases.

When the back emf falls to \(12.0\,\text{V}\), the simplified model predicts zero armature current.

If the back emf falls below \(12.0\,\text{V}\), the current reverses direction again. The electrical source now drives the machine as a motor.

The final operating speed cannot be calculated from the information provided. We would also need information about the mechanical load and other losses.

This is the important conceptual result: back emf is determined by the motor’s rotation, not simply capped at the supply voltage.

07What to learn next

The calculation \(E_b=V_m-IR_w\) gives you the back emf at a particular operating point. The next step is understanding why that value changes when the motor speeds up or slows down.

In particular, connect these three relationships:

• Rotational speed determines the back emf.
• Back emf and winding resistance determine the armature current.
• Armature current determines the electromagnetic torque.

For a fuller explanation of the physical origin of back emf, how it changes with speed, and why heavier mechanical loads increase motor current, see Back emf in a DC motor: HSC Physics explained.

Being able to calculate back emf is the first step. Being able to use it to predict how a motor responds to changing conditions is where the electrical and mechanical parts of the problem come together.