DC motors: practice problems with step-by-step solutions

Free DC motors practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: back-EMF, current and torque, speed–torque and efficiency, starting and speed control.

Free practice problemEasy
+−240 VRa​M+−E = 216 Vn = 1800 rpmT

A DC motor turns at 1800 rpm, and its back-EMF at that speed is 216 V. Find the motor constant k in V·s/rad.

E = 216 Vn = 1800 rpm
V·s/rad
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What you practise

  1. Easyback-EMF, current and torque
  2. Mediumspeed–torque and efficiency
  3. Hardstarting and speed control

How to solve DC motors problems

  1. Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
  2. Use V = E + Ia·Ra to link voltage, current and back-EMF.
  3. Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
  4. For power and efficiency, start from V·Ia, subtract the copper loss Ia²·Ra and then the rotational losses.

Common mistakes

  • Leaving the speed in rpm in E = k·ω or P = T·ω.
  • Forgetting that the back-EMF is zero at standstill, which makes the starting current very large.
  • Adding the armature drop to the supply instead of subtracting it.
  • Keeping the current constant after the field is weakened: the same torque needs more current.

Key formulas

Armature circuit
V = E + IaRa
Back-EMF and torque
E = k·ω, T = k·Ia
Speed
ω = 2π·n/60, ω = (V − IaRa)/k
No-load speed and stall torque
ω0 = V/k, Tstall = k·V/Ra
Powers
Pin = V·Ia, Pdev = E·Ia = T·ω, Pout = Pdev − Prot
Starting
Istart = V/(Ra + Rst)
Speed regulation
SR = (n0 − nFL)/nFL
Maximum developed power
Pmax = V²/(4Ra)
Chopper drive
Va = D·Vdc

Learn DC motors from scratch

Before this: Series & parallel resistors

1What a DC motor is

A DC motor turns electrical power into rotation. Current in the rotating winding, the armature, sits in a magnetic field and feels a force, which gives a torque on the shaft.

As the armature turns, it also acts as a generator. It produces a voltage that opposes the supply: the back-EMF E. The faster it turns, the larger E.

+−VRa​M+−EωT

Electrically the armature is a resistance Ra in series with the back-EMF. That simple model explains how speed, torque and current depend on each other.

Self-check What is the back-EMF of a DC motor at the moment it is switched on, before it turns?
  1. Equal to the supply voltage
  2. Half the supply voltage
  3. The same as at full speed
  4. Zero

Answer: Zero. The back-EMF is proportional to the speed, E = k·ω. At standstill ω = 0, so E = 0 and only Ra limits the current.

2The four motor equations

Walk round the armature circuit:

V = E + Ia·Ra

The back-EMF grows with speed and the torque with current, with the same motor constant k in SI units (V·s/rad = N·m/A):

E = k·ω T = k·Ia

Speeds are often given in revolutions per minute. Convert before you use the equations:

ω = 2π·n/60

The power converted to mechanical form is the developed power E·Ia = T·ω. The input is V·Ia, and the difference Ia²Ra is lost as heat in the armature.

Self-check A motor has k = 0.5 N·m/A and draws an armature current of 8 A. What torque does it develop?
  1. 0.0625 N·m
  2. 16 N·m
  3. 4 N·m
  4. 8 N·m

Answer: 4 N·m. T = k·Ia = 0.5·8 = 4 N·m. Dividing the current by k gives 16, and dividing k by the current gives 0.0625.

3The method step by step

  1. Draw the armature circuit and list what is given: V, Ra, k, and a load torque, a current or a speed.
  2. If a torque is given, find the current from Ia = T/k.
  3. Find the back-EMF from V = E + IaRa.
  4. Find the speed from ω = E/k and convert to rpm if asked: n = 60·ω/(2π).
  5. For power and efficiency, start from V·Ia and subtract the copper loss and then the rotational losses.

4A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A DC motor with Ra = 0.5 Ω and k = 0.8 V·s/rad runs from 120 V and drives a load torque of 8 N·m. Find its armature current, its speed in rpm and its developed power.

1

Look at the circuit

+−120 VRa​ = 0.5 ΩIa​ ?M+−En ?T = 8 N·m

The load torque fixes the current; the current fixes the voltage drop; the back-EMF then fixes the speed.

2

Armature current

Ia = T/k = 8/0.8 = 10 A
3

Back-EMF

E = V − IaRa = 120 − 10·0.5 = 115 V
4

Speed

ω = E/k = 115/0.8 = 143.75 rad/s
n = 60·143.75/(2π) = 1373 rpm
5

Developed power and a check

Pdev = E·Ia = 115·10 = 1150 W

Check with the shaft side: T·ω = 8·143.75 = 1150 W. The input is 120·10 = 1200 W, so 50 W is lost in Ra.

Answer

Ia = 10 A, n = 1373 rpm and Pdev = 1150 W.

5Speed–torque line, starting and speed control

Combining the equations gives a straight line: the speed falls a little as the load torque rises.

ω = V/k − Ra·T/k²
Tn14321928 N·m, 1373
  • No-load speed (T = 0): ω0 = V/k. In the example 150 rad/s, or 1432 rpm.
  • Stall torque (ω = 0): T = k·V/Ra = 0.8·120/0.5 = 192 N·m.
  • Starting current: V/Ra = 240 A, far above the running current. A series starting resistor keeps it down.
  • Speed control: lower the armature voltage (for example with a chopper, Va = D·Vdc), add series resistance, or weaken the field to run above the base speed.
Self-check The 120 V motor with Ra = 0.5 Ω must start with at most 30 A. What starting resistor is needed?
  1. 0.5 Ω
  2. 3.5 Ω
  3. 4 Ω
  4. 4.5 Ω

Answer: 3.5 Ω. The total resistance must be 120/30 = 4 Ω. The armature already gives 0.5 Ω, so the resistor is 4 − 0.5 = 3.5 Ω. Adding Ra instead of subtracting it gives 4.5 Ω.

6Common mistakes

  • Leaving the speed in rpm in E = k·ω or P = T·ω.
  • Adding the armature drop to the supply instead of subtracting it: in a motor E is smaller than V.
  • Forgetting that the back-EMF is zero at standstill.
  • Keeping the current unchanged after weakening the field: with a smaller k, the same torque needs more current.
Self-check A motor has a back-EMF of 100 V at 1000 rpm. What is its motor constant k?
  1. 0.0159 V·s/rad
  2. 0.1 V·s/rad
  3. 0.955 V·s/rad
  4. 6 V·s/rad

Answer: 0.955 V·s/rad. ω = 2π·1000/60 = 104.7 rad/s, so k = 100/104.7 = 0.955. Dividing by the rpm directly gives 0.1, and forgetting the 60 gives 0.0159.

7How exam questions look

  • Easy: back-EMF, current and torque. You find E, Ia, the torque from k, the motor constant from a speed, the speed, the starting current or the developed power.
  • Medium: speed–torque and efficiency. You find the no-load speed, the speed at a load, the stall torque, the efficiency with rotational losses, the speed regulation or the shaft torque from a power.
  • Hard: starting and speed control. You size a starting resistor, find the armature voltage or series resistance for a speed, the speed after field weakening, the duty cycle of a chopper drive, the maximum power, or the speed after a load change.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
+−120 VRa​Ia​ = 8 AM+−EωT = ?

A permanent-magnet DC motor has the torque constant k = 0.6 N·m/A. Find the torque T it develops at an armature current of 8 A.

k = 0.6 N·m/AIa = 8 A

Find T (N·m)

Show hints
  1. The torque constant links torque and armature current.
  2. T = k·Ia.
  3. T = 0.6·8.
Show the answer and the step-by-step solution

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

Torque

T = k·Ia = 0.6·8 = 4.8 N·m

Answer

T = 4.8 N·m

In SI units the same k is the back-EMF constant in V·s/rad.

Example 2Medium
T (N·m)n (rpm)20 N·mn0​n = ?

A DC motor (k = 0.8 N·m/A, Ra = 0.4 Ω) runs from 110 V and drives a load torque of 20 N·m. Find its speed n in rpm.

V = 110 Vk = 0.8 N·m/ARa = 0.4 ΩT = 20 N·m

Find n (rpm)

Show hints
  1. The torque sets the current: Ia = T/k.
  2. Ia = 20/0.8 = 25 A.
  3. ω = (V − IaRa)/k.
Show the answer and the step-by-step solution

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

Armature current

Ia = T/k = 20/0.8 = 25 A

Back-EMF and speed

E = 110 − 25·0.4 = 100 V
ω = E/k = 125 rad/s → n = 1194 rpm

Answer

n = 1194 rpm
Example 3Hard
+−120 VRa​ = 1 ΩM+−En = ?T = 6 N·m

A separately excited DC motor (k = 1.5 V·s/rad at full field, Ra = 1 Ω, V = 120 V) drives a constant load torque of 6 N·m. The field flux is reduced to 75 % of its full value. Find the new speed n.

V = 120 Vk = 1.5 V·s/radRa = 1 ΩT = 6 N·mΦ = 75 %

Find n (rpm)

Show hints
  1. Both constants scale with the flux: k₂ = 0.75·k = 1.125.
  2. Ia = T/k₂ = 5.333 A.
  3. ω = (V − IaRa)/k₂.
Show the answer and the step-by-step solution

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

Weakened motor constant

k₂ = 0.75·1.5 = 1.125 V·s/rad

New armature current

Ia = 6/1.125 = 5.333 A

New speed

ω = (120 − 5.333·1)/1.125 = 101.9 rad/s → n = 973.3 rpm

Answer

n = 973.3 rpm

Field weakening is how a DC drive runs above its base speed.

Questions and answers

What is DC motors?

A DC motor turns electrical power into rotation. Current in the rotating winding, the armature, sits in a magnetic field and feels a force, which gives a torque on the shaft.

As the armature turns, it also acts as a generator. It produces a voltage that opposes the supply: the back-EMF E. The faster it turns, the larger E.

How do I solve DC motors problems step by step?
  1. Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
  2. Use V = E + Ia·Ra to link voltage, current and back-EMF.
  3. Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
  4. For power and efficiency, start from V·Ia, subtract the copper loss Ia²·Ra and then the rotational losses.
What are common mistakes in DC motors?
  • Leaving the speed in rpm in E = k·ω or P = T·ω.
  • Forgetting that the back-EMF is zero at standstill, which makes the starting current very large.
  • Adding the armature drop to the supply instead of subtracting it.
  • Keeping the current constant after the field is weakened: the same torque needs more current.
What do the DC motors exercises cover?
  • Easy: back-EMF, current and torque
  • Medium: speed–torque and efficiency
  • Hard: starting and speed control

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Series & parallel resistors

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Printable worksheet

DC motors worksheet with answers
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