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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 20.35
Tef1V21(r2s)2+x22×r2sV21(0.100.05)2+(0.6)2×0.100.050.4587 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 =NsNrNs=0.2875

So, Tef 2(xV1)2(0.100.2875)2+(0.6)2×0.010.28750.7231 (xV21)

Now as both torques are equal,
0.4587 V21=0.7231 x2V21
x2=0.6343

x=0.7964

% reduction in voltage =(10.7964)1×100=20.35%

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