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A 500 hp, 6-pole, 3-phase, 440 V, 50 Hz induction motor has a speed of 950 rpm on full-load. The full load slip and the number of cycles the rotor voltage makes per minute will be respectively
1. 10% and 150
2. 10% and 125
3. 5% and 150
4. 5% and 125

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Correct Answer - Option 3 : 5% and 150

Number of poles, P = 6

Frequency, f = 50 Hz

Rotor speed, Nr = 950 rpm at full load

Synchronous speed, \({N_s} = \frac{{120f}}{P} = \frac{{120\; \times \;50}}{6} = 1000\;rpm\)

Full load slip = \(\frac{{{N_s} - {N_{\rm{r}}}}}{{{N_s}}} = \;\frac{{1000 - 950}}{{1000}} = 0.05 = 5\% \) 

Rotor frequency, fr = sf = 0.05 × 50 = 2.5 Hz

Rotor voltage speed \(= \frac{{120f}}{P} = \frac{{120\; \times \;2.5}}{P} = \frac{{300}}{P}\)

Since, P should be an integer hence,

P = 2 then speed of voltage \(= \frac{{300}}{2} = 150\;{\bf{rpm}}\)

P = 4 then speed of voltage \(= \frac{{300}}{4} = 75\;rpm\)

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