Question

# A 6 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 inserted in rotor windings per phase to obtain maximum torque at starting is

A
0.1 Ω/phase
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B
0.09 Ω/phase
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C
0.33 Ω/phase
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D
0.03 Ω/phase
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Solution

## The correct option is B 0.09 Ω/phaseSynchronous speed, Ns=120×506=1000 rpm Stalling speed = 900 rpm Slip at stalling speed, s=1000−9001000=0.1 Slip at maximum torque; SmT=R2X2=0.01X2 ∵ 0.1=0.01X2 X2=0.1 Ω To obtain maximum torque at starting, Let rotor resistance =R′2 at starting s = 1 SmT=R′2X2 ⇒ 1=R′20.1 ⇒ R′2=0.1 Ω/phase The external resistance to be added Rext=0.1−0.01 =0.09 Ω/phase

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