How to Calculate Mechanical Conversion Power in a DC Motor

A 24.0 V DC motor draws 3.0 A. Multiplying voltage by current gives \(72.0\,\text{W}\). Is that the mechanical power being produced?

Not quite. Some of that electrical power heats the copper winding. The rest is converted electromagnetically into mechanical power.

Suppose the winding resistance is \(0.50\,\Omega\). The voltage drop across the winding is \(IR=(3.0)(0.50)=1.5\,\text{V}\). This leaves a back emf of \(22.5\,\text{V}\).

The mechanical conversion power is therefore:

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

\[ P_{\text{conv}}=22.5\times3.0 =\boxed{67.5\,\text{W}} \]

The remaining \(4.5\,\text{W}\) becomes thermal energy in the winding.

This is the key method: multiply back emf by armature current, not supply voltage by current. If back emf is not provided, calculate it first using the winding resistance.

01Understand the motor’s voltage balance

Before calculating power, you need to distinguish two voltages inside a running DC motor.

• The supply voltage pushes current through the armature winding.
• The back emf is induced as the armature rotates. It acts in opposition to the supply voltage.

The winding also has electrical resistance, so some voltage is lost across the copper.

For a motor operating with steady current, the voltage balance is:

\[ \boxed{V_s=E_b+IR_w} \]

Where:

Rearranging gives:

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

This is the equation you need when a question gives you the supply voltage, current, and winding resistance instead of back emf directly.

Why the motor is not just a resistor

It is tempting to treat a DC motor like an ordinary resistor.

For a resistor, the relationship between voltage and current is determined by its resistance:

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

But a running motor has an additional voltage: its back emf.

Its current is determined by:

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

The winding resistance is approximately constant at a particular temperature, but the back emf changes with rotational speed.

As the motor spins faster, its back emf increases. This leaves less voltage across the winding resistance, so the current decreases.

As the motor slows under a heavier load, the back emf decreases, allowing more current to flow.

The motor therefore cannot be modelled as a fixed resistor across the supply.

If you need to revisit why rotation produces an opposing emf, see Back emf in a DC motor: HSC Physics explained.

02Derive the mechanical conversion power equation

Now we can connect the voltage balance to power.

Start with:

\[ V_s=E_b+IR_w \]

Multiply every term by the armature current \(I\):

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

Each term now represents a power transfer, measured in watts.

Rearranging gives the central result:

\[ \boxed{P_{\text{conv}}=E_bI} \]

This works because voltage represents energy transferred per coulomb, while current represents coulombs passing each second.

Multiplying them gives energy transferred per second, which is power.

The three power terms also satisfy:

\[ \boxed{P_{\text{in}}=P_{\text{conv}}+P_{\text{copper}}} \]

This is the energy balance for the simplified armature model.

What exactly does mechanical conversion power mean?

There is an important distinction between mechanical conversion power and useful shaft output.

\(E_bI\) represents the electromagnetic power converted from electrical to mechanical form inside the motor. It is also called the gross mechanical power developed.

However, a real motor experiences additional losses, including friction and air resistance.

Consequently:

\[ P_{\text{shaft}}<P_{\text{conv}} \]

unless those losses are negligible.

If a question asks for mechanical conversion power, use \(E_bI\). If it asks for useful mechanical power delivered by the shaft, you may need additional information about the motor’s mechanical losses.

04Measuring back emf when it is not given

In an experimental situation, you might not have a direct measurement of back emf.

Instead, you can determine it from the supply voltage and winding resistance.

Imagine a simple motor test bench.

A DC supply connects to the motor through an ammeter. The ammeter measures the current entering the armature. A voltmeter connects across the motor terminals to measure the applied voltage.

The motor shaft drives a small mechanical load.

With the supply disconnected, an ohmmeter can be used to measure the armature winding resistance.

The measurement sequence is important:

1. Measure the winding resistance with the motor isolated from the supply.
2. Connect the motor to the DC supply.
3. Measure the terminal voltage and armature current while the motor operates steadily.
4. Calculate the voltage drop across the winding.
5. Subtract this drop from the supply voltage to find back emf.
6. Multiply back emf by armature current.

Why measuring only voltage and current is insufficient

Suppose a student measures:

• Supply voltage: \(24.0\,\text{V}\)
• Armature current: \(3.0\,\text{A}\)

They calculate:

\[ P_{\text{in}}=VI=24.0\times3.0 =72.0\,\text{W} \]

This is correct electrical input power.

However, it does not tell us how much power is being converted electromagnetically.

We need one more piece of information: either the back emf or the winding resistance.

Adding a winding-resistance measurement allows us to separate the two power transfers.

05The important misconception: High current does not always mean high mechanical power

Consider what happens when a DC motor is prevented from rotating.

At the instant the motor is stationary, its rotational speed is zero. Therefore, it produces approximately zero back emf.

What happens to the current?

Using:

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

When \(E_b=0\):

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

The current can become very large if the winding resistance is small.

Consider an illustrative motor with:

• Supply voltage: \(24.0\,\text{V}\)
• Winding resistance: \(0.50\,\Omega\)

At stall:

\[ I=\frac{24.0-0}{0.50} =48.0\,\text{A} \]

That is a very large current.

However, its electromagnetic conversion power is:

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

\[ P_{\text{conv}}=0\times48.0 \]

\[ \boxed{P_{\text{conv}}=0\,\text{W}} \]

Despite drawing enormous current, the stationary motor is not converting power into mechanical motion.

Its electrical input is being dissipated as winding heat in this simplified model:

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

\[ P_{\text{copper}}=(48.0)^2(0.50) \]

\[ P_{\text{copper}}=1152\,\text{W} \]

This is why a stalled motor can overheat rapidly. The calculation is an idealised model, not a recommendation to stall a real motor.

The underlying misconception is that more current must mean more mechanical power.

That is not true.

Current contributes to electromagnetic torque, but mechanical power also depends on rotational speed:

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

where:

• \(\tau\) is torque in newton metres (\(\text{N m}\)).
• \(\omega\) is angular velocity in radians per second (\(\text{rad/s}\)).

At stall, the motor can produce substantial torque, but its angular velocity is zero. Therefore, its mechanical power is zero.

You need both the electromagnetic action and the rotational motion to produce mechanical power.

06Conversion power versus useful shaft power

One final distinction is worth understanding before tackling exam questions.

The equation \(P_{\text{conv}}=E_bI\) gives the electromagnetic power developed inside the motor.

If a question gives you shaft torque and rotational speed, it is asking about the mechanical power delivered at the shaft.

Use:

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

If rotational speed is given in revolutions per minute (rpm), convert it into radians per second:

\[ \omega=\frac{2\pi n}{60} \]

where \(n\) is rotational speed in rpm.

For example, suppose a motor converts \(120\,\text{W}\) electromagnetically, but friction and other mechanical losses account for \(10\,\text{W}\).

Its useful shaft output is:

\[ P_{\text{shaft}}=120-10 \]

\[ P_{\text{shaft}}=110\,\text{W} \]

The conversion power and shaft output are different quantities.

If the question asks for efficiency, check which two power values it wants you to compare. Overall motor efficiency uses useful output divided by total electrical input, while the ratio of shaft power to electromagnetic conversion power describes the losses between conversion and the shaft.

07A quick decision process for HSC questions

Use this table to decide which calculation to perform.

A few final checks can prevent common exam mistakes:

• Use armature current, not a separately supplied field-winding current.
• Use winding resistance, not \(V_s/I\) as though the entire motor were a resistor.
• Subtract the winding voltage drop when calculating back emf.
• Use back emf, not supply voltage, in the conversion power equation.
• Include units in every calculation.
• Do not automatically equate electromagnetic conversion power with useful shaft output.

08Questions and solutions

The following questions progress from a straightforward calculation to more demanding applications of the same principles.

Question 1

A DC motor operates at a steady speed with a back emf of \(16.0\,\text{V}\) and an armature current of \(3.0\,\text{A}\).

Calculate the electromagnetic conversion power of the motor.

Solution 1

The electromagnetic conversion power is \(48.0\,\text{W}\).

Since both back emf and armature current are provided, use:

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

Substitute:

\[ P_{\text{conv}}=16.0\times3.0 \]

\[ \boxed{P_{\text{conv}}=48.0\,\text{W}} \]

The motor converts electrical energy into mechanical energy at a rate of \(48.0\,\text{J/s}\).

There is no need to calculate winding resistance or supply voltage because the two required quantities are already given.

Question 2

A permanent-magnet DC motor is connected to a \(36.0\,\text{V}\) supply.

During steady operation, it draws an armature current of \(5.0\,\text{A}\). Its winding resistance is \(0.80\,\Omega\).

Calculate:

a) The back emf.

b) The electromagnetic conversion power.

c) The power dissipated in the winding.

Solution 2

The back emf is \(32.0\,\text{V}\), the conversion power is \(160\,\text{W}\), and the winding dissipates \(20.0\,\text{W}\) as heat.

Part a: Calculate back emf

Start with the voltage balance:

\[ E_b=V_s-IR_w \]

Substitute:

\[ E_b=36.0-(5.0)(0.80) \]

\[ \boxed{E_b=32.0\,\text{V}} \]

Part b: Calculate conversion power

Use:

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

\[ P_{\text{conv}}=32.0\times5.0 \]

\[ \boxed{P_{\text{conv}}=160\,\text{W}} \]

Part c: Calculate winding heating

Use:

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

\[ P_{\text{copper}}=(5.0)^2(0.80) \]

\[ \boxed{P_{\text{copper}}=20.0\,\text{W}} \]

The electrical input power is:

\[ P_{\text{in}}=36.0\times5.0 =180\,\text{W} \]

Check:

\[ 160+20=180\,\text{W} \]

The result is consistent with conservation of energy.

Question 3

A permanent-magnet DC motor operates from a constant \(24.0\,\text{V}\) supply. Its winding resistance is \(0.60\,\Omega\).

Two steady operating conditions are recorded:

• Condition A: The motor draws \(2.0\,\text{A}\).
• Condition B: A heavier mechanical load is applied, and the motor draws \(8.0\,\text{A}\).

Assume the winding resistance and magnetic field remain constant.

a) Calculate the back emf in each condition.

b) Calculate the electromagnetic conversion power in each condition.

c) Explain why the current increases fourfold without the conversion power increasing fourfold.

Solution 3

The conversion powers are \(45.6\,\text{W}\) in Condition A and \(153.6\,\text{W}\) in Condition B.

The second condition has approximately \(3.37\) times the conversion power, rather than four times.

Part a: Calculate the back emf in each condition

For Condition A:

\[ E_{b,A}=24.0-(2.0)(0.60) \]

\[ \boxed{E_{b,A}=22.8\,\text{V}} \]

For Condition B:

\[ E_{b,B}=24.0-(8.0)(0.60) \]

\[ \boxed{E_{b,B}=19.2\,\text{V}} \]

Notice that the back emf decreases when the heavier load is applied.

This happens because the motor slows down, reducing the induced back emf.

Part b: Calculate conversion power

For Condition A:

\[ P_A=22.8\times2.0 \]

\[ \boxed{P_A=45.6\,\text{W}} \]

For Condition B:

\[ P_B=19.2\times8.0 \]

\[ \boxed{P_B=153.6\,\text{W}} \]

Part c: Explain the change

The tempting prediction is that quadrupling the current should quadruple the conversion power.

That would only be true if back emf remained constant.

But it does not.

The heavier load slows the motor, causing back emf to decrease from \(22.8\,\text{V}\) to \(19.2\,\text{V}\).

The current increases, but the back emf decreases. Both changes must be included when calculating power.

The conversion power therefore increases by:

\[ \frac{153.6}{45.6}=3.37 \]

The current has quadrupled, but the conversion power has increased by a smaller factor.

This is a useful example of why you must calculate both back emf and current at each operating condition.

Question 4

A DC motor operates at a steady speed with:

• Back emf: \(20.0\,\text{V}\)
• Armature current: \(6.0\,\text{A}\)
• Shaft torque: \(0.70\,\text{N m}\)
• Shaft rotational speed: \(1500\,\text{rpm}\)

Calculate:

a) The electromagnetic conversion power.

b) The useful mechanical power delivered at the shaft.

c) The power lost between electromagnetic conversion and the shaft.

d) The efficiency of transferring converted power to the shaft.

Assume the measured torque is the useful shaft torque.

Solution 4

The electromagnetic conversion power is \(120\,\text{W}\), while the useful shaft power is approximately \(110\,\text{W}\). The difference is approximately \(10.0\,\text{W}\).

Part a: Calculate electromagnetic conversion power

Use:

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

\[ P_{\text{conv}}=20.0\times6.0 \]

\[ \boxed{P_{\text{conv}}=120\,\text{W}} \]

Part b: Calculate useful shaft power

First, convert rotational speed into radians per second:

\[ \omega=\frac{2\pi n}{60} \]

\[ \omega=\frac{2\pi(1500)}{60} \]

\[ \omega=157.1\,\text{rad/s} \]

Now use:

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

Substitute:

\[ P_{\text{shaft}}=0.70\times157.1 \]

\[ \boxed{P_{\text{shaft}}=110\,\text{W}} \]

Part c: Calculate mechanical losses

The difference between conversion power and shaft output is:

\[ P_{\text{loss}}=P_{\text{conv}}-P_{\text{shaft}} \]

\[ P_{\text{loss}}=120-110 \]

\[ \boxed{P_{\text{loss}}=10.0\,\text{W}} \]

Part d: Calculate conversion-to-shaft efficiency

The efficiency of transferring converted power to the shaft is:

\[ \eta=\frac{P_{\text{shaft}}}{P_{\text{conv}}}\times100 \]

\[ \eta=\frac{110}{120}\times100 \]

\[ \boxed{\eta=91.7\%} \]

The motor converts \(120\,\text{W}\) electromagnetically, but only approximately \(110\,\text{W}\) reaches the shaft.

This is not necessarily the motor’s overall electrical efficiency. That would require the total electrical input power as well.

Question 5

A DC motor is connected to a \(24.0\,\text{V}\) supply. Its armature winding resistance is \(0.50\,\Omega\).

An external mechanical drive now spins the motor faster than its normal operating speed, producing a back emf of \(27.0\,\text{V}\).

Use the following sign convention:

• Positive current flows from the supply into the motor.
• Positive conversion power represents electrical-to-mechanical energy conversion.
• Negative conversion power represents mechanical-to-electrical energy conversion.

Assume the supply can accept electrical energy returned by the motor, and neglect additional mechanical losses.

Determine:

a) The signed armature current.

b) The signed electromagnetic conversion power.

c) The winding copper loss.

d) The electrical power returned to the supply.

Explain why the motor is no longer operating as an ordinary motor.

Solution 5

The signed current is \(-6.0\,\text{A}\), and the signed conversion power is \(-162\,\text{W}\). The motor is operating as a generator, converting mechanical energy into electrical energy.

Part a: Calculate the signed armature current

Start with the motor voltage equation:

\[ V_s=E_b+IR_w \]

Rearrange:

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

Substitute:

\[ I=\frac{24.0-27.0}{0.50} \]

\[ \boxed{I=-6.0\,\text{A}} \]

The negative sign is important.

It means the actual current flows opposite to our chosen positive direction. Instead of entering the motor from the supply, current flows out of the motor towards the supply.

Part b: Calculate signed conversion power

Use:

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

Substitute:

\[ P_{\text{conv}}=27.0\times(-6.0) \]

\[ \boxed{P_{\text{conv}}=-162\,\text{W}} \]

The negative sign tells us that the energy conversion direction has reversed.

The motor is no longer converting electrical energy into mechanical energy. It is converting mechanical energy into electrical energy.

Part c: Calculate winding copper loss

Copper heating is always positive because current is squared:

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

\[ P_{\text{copper}}=(-6.0)^2(0.50) \]

\[ \boxed{P_{\text{copper}}=18.0\,\text{W}} \]

Even though the current has reversed, the winding still heats up.

Part d: Calculate electrical power returned to the supply

The signed electrical power entering the motor is:

\[ P_{\text{in}}=V_sI \]

\[ P_{\text{in}}=24.0\times(-6.0) \]

\[ P_{\text{in}}=-144\,\text{W} \]

A negative input power means the motor is delivering electrical power back to the supply.

Therefore:

\[ \boxed{P_{\text{returned}}=144\,\text{W}} \]

Check the energy balance

The signed power equation remains:

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

Substituting:

\[ -144=-162+18 \]

\[ -144=-144 \]

The balance is satisfied.

The external mechanical drive supplies energy to the motor. Of the \(162\,\text{W}\) crossing the electromagnetic conversion boundary, \(144\,\text{W}\) is returned electrically to the supply, while \(18\,\text{W}\) heats the winding.

The important conceptual point is that the motor’s operating mode depends on the relationship between back emf and supply voltage.

Normally, \(E_b<V_s\), so current enters the motor and electrical energy becomes mechanical energy.

Here, \(E_b>V_s\), so current reverses and the motor behaves as a generator.

This is the same voltage and power model, but with the direction of energy transfer reversed.

Once you can distinguish electrical input, electromagnetic conversion, and useful shaft output, the next useful calculation is overall motor efficiency. That requires comparing the mechanical power delivered by the shaft with the complete electrical power supplied to the motor.