Question

A 4-pole, 50 Hz, 3-phase induction motor with a rotor resistance of 0.25 Ω develops a maximum torque of 25 N.m at 1400 rpm. The rotor reactance x₂ and slip at maximum torque s_max,T respectively would be
1. \(2.0\;and\frac{1}{{15}}\)
2. \(3.75\;and\frac{1}{{12}}\)
3. \(2.0\;and\frac{1}{{12}}\)
4. \(3.75\;and\frac{1}{{15}}\)

Answer 1

Correct Answer - Option 4 : \(3.75\;and\frac{1}{{15}}\)

Synchronous speed, \({N_s} = \frac{{120f}}{P} = \frac{{120 \times 50}}{4} = 1500\;rpm\)

\({s_m} = \frac{{{N_s} - {N_r}}}{{{N_s}}}\)

\( = \frac{{1500 - 1400}}{{1500}} = \frac{1}{{15}}\)

Also, S_m = R₂ / X₂

Therefore, X₂  = 0.25 × 15 = 3.75