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Question

# A 3-phase induction motor has 5% slip at full load torque and at normal voltage. The rotor resistance and rotor standstill leakage reactance per phase are 0.1 Ω and 0.6 Ω respectively. What should be the percentage reduction in supply voltage if the motor has to develop full-load torque at 34th of normal full-load speed. (Neglect stator impedance)

A
20.35
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B
36.57
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C
63.43
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D
79.64
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Solution

## The correct option is A 20.35Tef1∝V21(r2s)2+x22×r2s∝V21(0.100.05)2+(0.6)2×0.100.05∝0.4587 V21 When speed is 34th of normal full speed Full load speed = (1 - s) Ns Full load speed = 0.95 Ns 34th of full load speed = 0.7125 Ns New slip =Ns−NrNs=0.2875 So, Tef 2∝(xV1)2(0.100.2875)2+(0.6)2×0.010.2875∝0.7231 (xV21) Now as both torques are equal, 0.4587 V21=0.7231 x2V21 x2=0.6343 x=0.7964 % reduction in voltage =(1−0.7964)1×100=20.35%

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