Year 12 Physics Module 6 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 17 of 21

DC Motors in Depth

Consider a 600 V DC traction motor with 0.30 Ω winding resistance. Without rotational back emf, its zero-speed current would be 600 ÷ 0.30 = 2000 A, so a starting resistor or controller is essential. If its running current is 8.0 A, the circuit model gives a back emf of 597.6 V. The numbers illustrate the model; actual tram motors also include series/parallel controls and other losses.

Today's hook: Melbourne's W-class trams (Melbourne and Metropolitan Tramways Board, 1923 fleet) ran on 600 V DC with coil resistance 0.3 Ω. At startup, the driver had to engage a starting resistor. Why? Calculate the start-up current without the resistor, and explain what physics causes the running current to drop to just 8 A at full speed.
0/5TASKS
1
Retrieve motor current, predict back-emf behaviour and set up the finite startup-current model.
Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you read, predict

A DC motor is connected to a 12 V battery. When it starts from rest, it draws 6.0 A. When running at full speed, it draws 1.0 A.

  1. Why does the current drop as the motor speeds up?
  2. What would happen to the current if the motor were suddenly loaded so it slowed down?
  3. Calculate the back emf when the motor is running at full speed. The coil resistance is 2.0 ohms.

Warm-up, when a DC motor is first switched on (stationary), what is the back emf?

Learning Intentions
goals

Know, Back emf in Detail

  • Back emf is proportional to motor speed: $\varepsilon_{\text{back}}=k_e\omega$
  • Net voltage driving current: $V_{net} = V_{applied} - \varepsilon_{back}$
  • Stall current $= V_{applied}/R$ (maximum, dangerous)

Understand, Torque-Speed Relationship

  • Maximum torque at startup (back emf = 0)
  • Torque decreases as speed increases
  • At no load, speed rises until back emf nearly equals applied voltage

Can Do, Analyse Performance

  • Calculate back emf, current, and torque at different speeds
  • Explain why starting resistors are used in large DC motors
  • Analyse the energy and power relationships in a DC motor
Scan these before reading
vocab
Back emfThe voltage induced in a motor's rotating coil, opposing the applied voltage. Proportional to speed.
Stall currentThe current drawn when the motor is prevented from rotating: $I_{stall} = V/R$, because back emf is zero.
Starting resistorA resistor placed in series with a motor at startup to limit the initial surge current.
No-load speedThe maximum speed a motor reaches with no external load. Back emf nearly equals applied voltage.
Cross-lesson links: L16 introduced the AC induction motor. L17 examines the DC motor's hidden physics, back EMF (Faraday's law working in reverse) limits the current and prevents the motor burning out at start-up. Understanding back EMF is essential for explaining why DC motors draw high current at low speed.

Core Content

2
Use the equivalent circuit to relate speed, back emf and winding current.
1
Back emf and Motor Current
+5 XP

The self-regulating nature of DC motors

Connect a DC motor to a battery and clip an ammeter in series. At the instant you complete the circuit, before the shaft starts moving, the ammeter reads a large current. Within a fraction of a second, as the motor spins up to speed, the ammeter reading drops dramatically and settles at a much lower steady value. Nothing changed about the circuit resistance; the "missing" current is accounted for by back emf, the voltage induced in the spinning coil by Faraday's Law, which opposes the applied voltage and reduces the net voltage driving current through the coil.

Motor current and back emf

$I = \dfrac{V - \varepsilon_{back}}{R}$  , net voltage divided by resistance

$\varepsilon_{\text{back}}=k_e\omega$  , general speed relation for fixed flux

I = current (A) · V = applied voltage (V) · R = winding resistance (Ω) · $k_e$ = machine emf constant (V s rad$^{-1}$) · ω = angular speed (rad/s)

Equivalent circuit and speed graphs

The DC motor circuit is modelled as the supply voltage opposing a speed-dependent generator voltage in series with winding resistance: $V=\varepsilon_{\text{back}}+IR$.

supply V R εback opposes V stall no-load speed I and torque speed ω

For a fixed field, torque is proportional to current. In the simple model, $I=(V-k_e\omega)/R$, so current and torque have maximum intercepts at stall and fall toward the no-load speed where $\varepsilon_{\text{back}}$ nearly equals $V$.

Startup: When stationary, $\omega = 0$, so $\varepsilon_{back} = 0$. The current is maximum: $I = V/R$. This is why motors draw a large surge current when starting.

Running: As speed increases, $\varepsilon_{back}$ grows, reducing the net voltage and current. At full speed, $\varepsilon_{back}$ is nearly equal to $V$, and the current drops to just enough to overcome friction and drive the load.

Loading: If the motor is loaded and slows down, $\varepsilon_{back}$ decreases, so the current increases. The motor automatically draws more current when it needs more torque.

Back emf and Conservation of Energy

Back emf is not just a mathematical inconvenience, it is required by conservation of energy. Inside the motor:

  • The applied voltage V delivers electrical power $P_{in} = VI$.
  • Some power is lost as heat in the coil resistance: $P_{heat} = I^2R$.
  • The remainder crosses the electromagnetic conversion boundary: $P_{\text{em}}=\varepsilon_{\text{back}}I$.
  • Shaft output is $P_{\text{shaft}}=P_{\text{em}}-P_{\text{mechanical losses}}$.

If there were no back emf, the stationary circuit current would be limited to $V/R$ by the winding resistance. The deeper contradiction is that a model with no speed-dependent back emf would fail to account consistently for electromagnetic power conversion as speed changes. Back emf makes the power ledger explicit: electrical input = electromagnetic conversion + resistive heating, with mechanical losses then separating converted power from shaft output.

HSC Tip

In extended response questions, always link back emf to Lenz's Law and conservation of energy. Explain that the mechanical work done by the motor comes from the energy supplied minus heat losses, back emf is how the motor "self-reports" how much mechanical energy it is producing.

Stop & Check

A DC motor has coil resistance 4.0 Ω and is connected to 24 V. At full speed it draws 2.0 A. Calculate the back emf. If the motor is suddenly loaded so its speed halves, what happens to the back emf and the current?

Motor current: $I=(V-\varepsilon_\text{back})/R$; generally $\varepsilon_\text{back}=k_e\omega$. Startup: $\omega=0$, so $I=V/R$ is a finite maximum. Power ledger: $VI=I^2R+\varepsilon_\text{back}I$, where $\varepsilon_\text{back}I$ is electromagnetic conversion power; subtract mechanical losses to obtain shaft output.

Pause, copy the highlighted back emf formula and three operating states into your book before moving on.

A DC motor is running steadily. The motor is then loaded so it slows down. What happens to back emf and current?

3
Explain why starting resistors or controllers limit the finite but dangerous startup surge.
2
Starting Resistors and Motor Protection
+5 XP

Managing the startup surge

We just saw that $I = V/R$ at startup because back emf is zero. That raises a question: in a real large motor, this startup surge can destroy the coils, how do engineers protect against it? This card answers it → a starting resistor in series limits the current to $V/(R_\text{coil}+R_\text{start})$ until back emf builds up.

The startup current ($I = V/R$) can be 5–10 times the running current. In large motors, this surge can overheat the coils, damage the commutator and brushes, and cause voltage dips in the power supply.

A starting resistor is placed in series with the motor at startup. It limits the initial current to a safe value. As the motor speeds up and back emf increases, the starting resistor is gradually bypassed (using a switch or relay) until the motor runs at full voltage.

Current with a starting resistor

$I_{start} = \dfrac{V}{R_{coil} + R_{start}}$

Adding $R_{start}$ in series reduces the startup surge current to a safe level.

Modern electronic controllers (like those in electric vehicles) use pulse-width modulation (PWM) to vary the effective voltage smoothly, eliminating the need for mechanical starting resistors.

Real world, electric vehicle motors

Tesla's drive motors are AC induction motors controlled by high-frequency PWM inverters. The inverter ramps the effective voltage from near-zero at startup, preventing damaging surge currents, the electronic equivalent of a starting resistor, but far more efficient.

Starting resistor: $I_\text{start} = V/(R_\text{coil}+R_\text{start})$ (A), limits startup surge. As speed builds, back emf grows, current falls; resistor is bypassed when no longer needed. Modern alternative: PWM controller ramps effective voltage smoothly from zero.

Add the highlighted starting resistor equation and purpose to your notes before the check below.

A starting resistor is placed in series with the motor to limit the initial current surge.

The starting resistor increases the back emf at startup.

Modern EV controllers use PWM to achieve the same protection as a starting resistor.

4
Calculate startup/running current and distinguish conversion power from shaft output.
WE
Worked Example, DC Motor Analysis
+5 XP

Complete analysis of back emf, power and efficiency

Problem

A DC motor has coil resistance 3.0 Ω and is connected to an 18 V supply. When running at full speed the back emf is 15 V. Find: (a) startup and full-speed currents; (b) electrical power supplied at full speed; (c) heat dissipated in the coil; (d) electromagnetic conversion power and conversion efficiency. Then state what extra information is needed for shaft output.

Step 1, Currents

At startup, $\varepsilon_{back} = 0$:

$I_{start} = \dfrac{V}{R} = \dfrac{18}{3.0} = 6.0 \text{ A}$

At full speed:

$I_{run} = \dfrac{V - \varepsilon_{back}}{R} = \dfrac{18 - 15}{3.0} = 1.0 \text{ A}$

Step 2, Electrical power in

$P_{in} = VI = 18 \times 1.0 = 18 \text{ W}$

Step 3, Heat in coil

$P_{heat} = I^2R = (1.0)^2 \times 3.0 = 3.0 \text{ W}$

Step 4, Conversion power and efficiency

$P_{em} = P_{in} - P_{heat} = 18 - 3.0 = 15 \text{ W}$

Check: $P_{em} = \varepsilon_{back} I = 15 \times 1.0 = 15 \text{ W}$ ✓

$\eta_{conversion} = \dfrac{P_{em}}{P_{in}} \times 100 = \dfrac{15}{18} \times 100 = 83.3\%$

Shaft output is less than 15 W if bearing, windage or other mechanical losses are present; those losses must be supplied to calculate shaft efficiency.

A DC motor (coil resistance 2.5 Ω, supply 15 V) runs at full speed with back emf = 12 V. The running current in amperes is _____.

5
Apply the current-speed and torque-speed model to a loaded DC motor.
Activity 1, Analyse Motor Performance
ApplyBand 4

A DC motor has: V = 12 V, R = 2.0 Ω, and at full speed $\varepsilon_{back}$ = 10 V.

  1. Calculate the stall current and the running current at full speed.
  2. Calculate the power supplied by the battery at full speed.
  3. Calculate the power dissipated as heat in the coil at full speed.
  4. Calculate the electromagnetic conversion power at full speed.
  5. Calculate the conversion efficiency, then explain why shaft efficiency cannot be found without mechanical-loss data.

Activity check, for the motor above (V = 12 V, R = 2.0 Ω, $\varepsilon_{back}$ = 10 V at full speed), what is the electromagnetic conversion efficiency before mechanical losses?

6
Explain seizure/protection scenarios and consolidate the back-emf energy story.
Activity 2, Explain and Apply
UnderstandBand 5

Explain a classic motor scenario

A DC motor is running at full speed with $\varepsilon_{back}$ = 10 V on a 12 V supply with coil resistance 0.5 Ω. The motor suddenly seizes (stops completely).

  1. Calculate the new current when the motor is seized.
  2. Explain why this current could damage the motor.
  3. What design feature would protect against this scenario in a large industrial motor?
Synthesis, connect the ideas
  • Back emf increases with motor speed, reducing net current.
  • At startup, back emf is zero and current is maximum ($I = V/R$).
  • Loading the motor reduces speed → reduces back emf → increases current.
  • Shaft efficiency = shaft power output / electrical power input; conversion efficiency uses $\varepsilon_{\text{back}}I$ before mechanical losses.
  • In calculations that use $\varepsilon_{\text{back}}I$, label it as electromagnetic conversion power unless mechanical losses are explicitly negligible.
  • Starting resistors protect large motors from excessive startup current.
7
Check DC motor current, back emf, conversion power and protection with mixed practice.
Quick recall, DC motors and back emf
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 7 marks
+5 XP

ApplyBand 4(3 marks) 1. A DC motor has coil resistance 2.5 Ω and is connected to 15 V. At full speed, the back emf is 12 V. Calculate: (a) the current at startup, (b) the current at full speed, and (c) the electromagnetic conversion power at full speed.

1 mark: $I_{start} = V/R$ correctly · 1 mark: $I_{run} = (V-\varepsilon_{back})/R$ correctly · 1 mark: $P_{em} = \varepsilon_{back} I_{run}$ correctly

AnalyseBand 5(4 marks) 2. Explain why back emf is necessary for a consistent energy account in a DC motor. In your answer, describe the finite stall current, how back emf grows with speed, and why $\varepsilon_{\mathrm{back}}I$ is electromagnetic conversion power rather than automatically shaft output.

1 mark: with zero back emf, winding resistance limits current to $V/R$ · 1 mark: back emf rises with angular speed · 1 mark: electromagnetic conversion power is $\varepsilon_{\mathrm{back}}I$, not automatically shaft output · 1 mark: shaft output is smaller after mechanical losses

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (3 marks): (a) At startup $\varepsilon_{back} = 0$, so $I_{start} = V/R = 15/2.5 = 6.0$ A (1 mark). (b) At full speed: $I_{run} = (V - \varepsilon_{back})/R = (15-12)/2.5 = 3.0/2.5 = 1.2$ A (1 mark). (c) $P_{em} = \varepsilon_{back} I_{run} = 12 \times 1.2 = 14.4$ W (1 mark). This is conversion power; shaft output would be smaller if mechanical losses are present.

Q2 (4 marks): At zero speed, $\varepsilon_{\mathrm{back}}=0$, so resistance limits the current to the stall value $I=V/R$ (1 mark). As angular speed rises, the induced back emf rises and the current becomes $I=(V-\varepsilon_{\mathrm{back}})/R$ (1 mark). The product $\varepsilon_{\mathrm{back}}I$ is electromagnetic conversion power, not automatically shaft output (1 mark). Mechanical losses must be subtracted before identifying shaft output, so the complete ledger is electrical input = resistive heating + electromagnetic conversion, then conversion = shaft output + mechanical losses (1 mark).

8
Revisit the tram current puzzle and complete the retrieval check.
Check what actually stuck
Take the full module quiz
quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →
How did your thinking change?

At the start you were asked about Melbourne's W-class trams (Melbourne and Metropolitan Tramways Board, 1923 fleet) on 600 V DC with 0.3 Ω coil resistance. What is the start-up current without the starting resistor, and why does the running current drop to just 8 A?

The answer: without back emf, $I_{start} = V/R = 600/0.3 = 2000$ A, enough to destroy the motor immediately. The starting resistor limited this to a safe value. At full speed, back emf $\varepsilon_{back} = V - IR = 600 - 8 \times 0.3 = 597.6$ V, 99.6% of supply voltage. Only 2.4 V net drives the 8 A running current. The "missing" current was never needed once back emf took over: this is exactly Lenz's Law, the spinning coil generates a counter-emf that resists the applied voltage.

Extend your thinking: A motor often operates efficiently near its design speed because back emf is high and $I^2R$ losses are comparatively low. Why must an engineer still account for mechanical losses before claiming a shaft efficiency?

← Previous: AC Induction Motors and Generators Next: Module Consolidation →

Module overview · Physics · Checkpoint 4 · Module quiz