DC motors worksheet with answers

Free printable DC motors worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
+−220 VRa​ = 2 ΩIa​ = ?M+−E = 0n = 0T

A DC motor with an armature resistance of 2 Ω is switched directly onto a 220 V supply. Its normal full-load current is 4 A. Find the armature current Istart at the moment of starting.

V = 220 VRa = 2 Ω
Istart = A
2.Easy
+−240 VRa​ = 1 ΩIa​ = 20 AM+−EωT

A DC motor runs from 240 V with an armature current of 20 A. The armature resistance is 1 Ω. Find the developed (electromagnetic) power Pdev.

V = 240 VIa = 20 ARa = 1 Ω
Pdev = W
3.Medium
T (N·m)n (rpm)50 N·mn0​n = ?

A DC motor (k = 2 N·m/A, Ra = 0.4 Ω) runs from 240 V and drives a load torque of 50 N·m. Find its speed n in rpm.

V = 240 Vk = 2 N·m/ARa = 0.4 ΩT = 50 N·m
n = rpm
4.Medium
+−120 VRa​ = 0.25 ΩIa​ = 50 AM+−EωT

A DC motor draws 50 A from a 120 V supply. Its armature resistance is 0.25 Ω and its rotational (friction, windage and iron) losses are 269 W. Find its efficiency η.

V = 120 VIa = 50 ARa = 0.25 ΩProt = 269 W
η = %
5.Hard
T (N·m)n (rpm)2 N·m1000 rpm

A DC motor with k = 0.4 V·s/rad and Ra = 1.5 Ω must drive a load of 2 N·m at 1000 rpm. Find the armature voltage V it needs.

k = 0.4 V·s/radRa = 1.5 ΩT = 2 N·mn = 1000 rpm
V = V
6.Hard
+−120 VRa​ = 2 ΩM+−En = ?T = 4.8 N·m

A separately excited DC motor (k = 0.8 V·s/rad at full field, Ra = 2 Ω, V = 120 V) drives a constant load torque of 4.8 N·m. The field flux is reduced to 85 % of its full value. Find the new speed n.

V = 120 Vk = 0.8 V·s/radRa = 2 ΩT = 4.8 N·mΦ = 85 %
n = rpm

Answer key

  1. 1. Istart = 110 A
  2. 2. Pdev = 4400 W
  3. 3. n = 1098 rpm
  4. 4. η = 85.1 %
  5. 5. V = 49.39 V
  6. 6. n = 1487 rpm

Full step-by-step solutions: practise this topic on examproblems.com.