A 60 pole, 50 Hz induction motor has an equivalent rotor resistance of 0.01 Ω/phase. If its stalling speed is 900 rpm, the resistance that must be included in rotor windings per phase to obtain maximum torque at starting is
A
0.1 Ω/phase
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B
0.33 Ω/phase
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C
0.09 Ω/phase
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D
0.03 Ω/phase
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Solution
The correct option is C 0.09 Ω/phase Synchronous speed Ns=120×506=1000 rpm
Stalling speed =900 rpm
Slip at stalling torque, s=1000−9001000=0.1
Slip at maximum torque, SmT=R2X2=0.01X2
∵0.1=0.01X2 X2=0.1Ω
To obtain maximum torque at starting,
Let rotor resistance =R2′
At starting, slip, s = 1