a 3-phase induction motor is running at 2% slip. if the input to rotor is 1000 w, then mechanical power developed by the motor is
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The formula is given by :
Mechanical power = (1 - S) × Power input.
Power input = 1000 W
S = 2%
S = 2/100 = 0.02
Doing the substitution we have :
(1 - 0.02) = 0.98
= 0.98 × 1000 = 980W
Answer :
= 980 W
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