1.

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 NI 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» CORRECT OPTION is (a) 47.76 N

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

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

= 47.76 N.


Discussion

No Comment Found

Related InterviewSolutions