A voltage of 24 V is applied to a DC motor with an armature resistance of 0.5 Ω and a back-EMF constant of 0.1 V/(rad/s), and the armature current is 8 A. Which of the following is the angular speed of rotation at this point?

Answer: B. 200 rad/s

In a DC motor the applied voltage equals the voltage drop across the armature resistance plus the back EMF (V = R × I + E). The back EMF is E = 24 − 0.5 × 8 = 20 V, and from E = K_e × ω, ω = 20/0.1 = 200 rad/s (about 1910 min⁻¹). In SI units the torque constant has the same value as the back-EMF constant, so the torque is 0.1 × 8 = 0.8 N·m. When the load and therefore the current increase, the resistive drop increases and the back EMF decreases, so the motor slows slightly.

• A. This divides the resistive voltage drop (4 V) by the back-EMF constant.
• B. Correct. (24 − 4) ÷ 0.1 = 200 rad/s.
• C. This is the no-load angular speed, ignoring the resistive voltage drop.