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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/phase
Synchronous speed, Ns=120×506=1000 rpm

Stalling speed = 900 rpm

Slip at stalling speed, s=10009001000=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 =R2
at starting s = 1

SmT=R2X2

1=R20.1

R2=0.1 Ω/phase

The external resistance to be added

Rext=0.10.01

=0.09 Ω/phase

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