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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. |
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