DC Motors in Depth
Melbourne's W-class trams (operated by the Melbourne and Metropolitan Tramways Board, 1923–2014) ran on 600 V DC. Maintenance records show back emf at full speed was approximately 550 V — 92% of supply voltage. Without back emf, the start-up current through the 0.3 Ω coil resistance would have been 600 ÷ 0.3 = 2,000 A. At full running speed the actual current was about 8 A — a factor of 250 reduction, entirely due to back emf.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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.
- Why does the current drop as the motor speeds up?
- What would happen to the current if the motor were suddenly loaded so it slowed down?
- 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?
Know — Back emf in Detail
- Back emf is proportional to motor speed: $\varepsilon_{back} = NBA\omega$
- 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
Core Content
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.
$I = \dfrac{V - \varepsilon_{back}}{R}$ — net voltage divided by resistance
$\varepsilon_{back} = NBA\omega$ — back emf proportional to angular speed
I = current (A) · V = applied voltage (V) · R = coil resistance (Ω) · ω = angular speed (rad/s)
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 rest does mechanical work: $P_{mech} = \varepsilon_{back} \cdot I$.
If there were no back emf, the motor would draw unlimited current and produce unlimited mechanical energy — violating conservation of energy. Back emf ensures that electrical energy in = mechanical energy out + heat losses.
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.
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: $I = (V - \varepsilon_\text{back})/R$; $\varepsilon_\text{back} = NBA\omega \propto \omega$. Startup: $\omega = 0$ → $\varepsilon_\text{back} = 0$ → $I = V/R$ (maximum surge). Full speed: $\varepsilon_\text{back} \approx V$ → minimum $I$. Power: $P_\text{in} = VI$; $P_\text{heat} = I^2R$; $P_\text{mech} = \varepsilon_\text{back}I$.
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?
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.
$I_{start} = \dfrac{V}{R_{coil} + R_{start}}$
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.
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.
Complete analysis of back emf, power and efficiency
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) mechanical power output and efficiency.
At startup, $\varepsilon_{back} = 0$:
$I_{start} = \dfrac{V}{R} = \dfrac{18}{3.0} = 6.0 \text{ A}$
At full speed:
$I_{run} = \dfrac{V - \varepsilon_{back}}{R} = \dfrac{18 - 15}{3.0} = 1.0 \text{ A}$
$P_{in} = VI = 18 \times 1.0 = 18 \text{ W}$
$P_{heat} = I^2R = (1.0)^2 \times 3.0 = 3.0 \text{ W}$
$P_{mech} = P_{in} - P_{heat} = 18 - 3.0 = 15 \text{ W}$
Check: $P_{mech} = \varepsilon_{back} \cdot I = 15 \times 1.0 = 15 \text{ W}$ ✓
$\eta = \dfrac{P_{mech}}{P_{in}} \times 100 = \dfrac{15}{18} \times 100 = 83.3\%$
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 _____.
A DC motor has: V = 12 V, R = 2.0 Ω, and at full speed $\varepsilon_{back}$ = 10 V.
- Calculate the stall current and the running current at full speed.
- Calculate the power supplied by the battery at full speed.
- Calculate the power dissipated as heat in the coil at full speed.
- Calculate the mechanical power output of the motor at full speed.
- Calculate the motor's efficiency at full speed.
Activity check — for the motor above (V = 12 V, R = 2.0 Ω, $\varepsilon_{back}$ = 10 V at full speed), what is the efficiency?
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).
- Calculate the new current when the motor is seized.
- Explain why this current could damage the motor.
- What design feature would protect against this scenario in a large industrial motor?
- 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.
- Motor efficiency = mechanical power output / electrical power input.
- Starting resistors protect large motors from excessive startup current.
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.
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 mechanical power output at full speed.
1 mark: $I_{start} = V/R$ correctly · 1 mark: $I_{run} = (V-\varepsilon_{back})/R$ correctly · 1 mark: $P_{mech} = \varepsilon_{back} \times I_{run}$ correctly
AnalyseBand 5(4 marks) 2. Explain why back emf is a necessary consequence of conservation of energy in a DC motor. In your answer, describe what would happen to the current and mechanical power if there were no back emf, and identify what physical principle would be violated.
1 mark: without back emf, all applied voltage drives current through $R$; current would be $V/R$ at all speeds · 1 mark: this would produce arbitrarily large currents and thus unlimited mechanical power · 1 mark: unlimited mechanical energy from a fixed supply violates conservation of energy · 1 mark: back emf (from Lenz's Law) accounts for the mechanical energy extracted, ensuring electrical energy in = mechanical out + heat losses
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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) $P_{mech} = \varepsilon_{back} \times I_{run} = 12 \times 1.2 = 14.4$ W (1 mark).
Q2 (4 marks): Without back emf, the current at every speed would be $I = V/R$ — the maximum stall current — regardless of how fast the motor is spinning (1 mark). Since torque and thus mechanical power depend on current, the motor would produce its maximum mechanical power at all speeds simultaneously, which is impossible for a fixed input (1 mark). This would mean the system generates unlimited mechanical energy from a fixed voltage source — a perpetual-motion scenario (1 mark). Back emf (induced by Lenz's Law as the coil rotates) reduces the net voltage available to drive current. As the motor speeds up, back emf rises, current falls, and the power balance is maintained: electrical energy in = mechanical energy out + resistive heat losses. Conservation of energy is satisfied (1 mark).
Five timed questions on DC motors and back emf. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaAt 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 operates most efficiently at high speed (high back emf, low current, low $I^2R$ losses). Why do engineers design motors to run near their no-load speed under normal operating conditions?