Question

The input of a 98% efficiency, 2000 rpm separately excited dc motor is 5 kW. The torque output is

• A. 28.1 Nm
• B. 23.9 Nm
• C. 20.3 Nm
• D. 19.1 Nm

Hint:

Torque = Power ÷ Angular Speed.

Answer 1

The Correct Option is B

Output power = 0.98 × 5000 = 4900 W
Angular speed \( \omega = \frac{2\pi N}{60} = \frac{2\pi \times 2000}{60} \approx 209.44 \, \text{rad/s} \)
Torque \( T = \frac{P}{\omega} = \frac{4900}{209.44} \approx 23.4 \, \text{Nm} \)