Question

For a dc shunt motor of 5 kW, running at 1000 rpm, the induced torque will be ____________

(a) 47.76 N

(b) 57.76 N

(c) 35.76 N

(d) 37.76 N

I got this question during an interview.

I'd like to ask this question from Operating Characteristics of DC Motors topic in division DC Motors of Electrical Machines

Answer 1

Correct option is (a) 47.76 N

Easiest explanation: Torque = Power/(speed in rev/s)

                    = 5000/(2*pi*1000/60)

      = 47.76 N.