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A 230 V, 50 Hz, 4-pole, single-phase induction motor is rotating in the clockwise (forward) direction at a speed of 1425 rpm. If the rotor resistance at standstill is 7.8 Ω, then the effective rotor resistance in the backward branch of the equivalent circuit will be
- (A) 2 Ω
- (B) 4 Ω
- (C) 78 Ω
- (D) 156 Ω
Correct answer: (A) 2 Ω
Solution
Ns=1500 rpm, so slip s=(1500-1425)/1500=0.05 and backward slip 2-s=1.95. The effective rotor resistance in the backward branch is R2/[2(2-s)] = 7.8/(2 x 1.95) = 2 ohm.
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