The figure is the per-phase equivalent circuit of a two-pole, three-phase 50 Hz induction motor. Air-gap voltage Vg is 210 V rms and slip is 0.05. The torque in N·m is ________.
The rotor branch has R2/s=0.05/0.05=1 Ω and reactance 0.22 Ω. Rotor current squared is 210²/(1²+0.22²).
Air-gap power is 3I2²(R2/s). Electromagnetic torque is this power divided by synchronous mechanical angular speed, ωs=2π50 for a two-pole motor.
Thus $T=3(210)^2/[1.0484(100\pi)]=$ **401.68 N·m**, in 400–403. The given air-gap voltage already accounts for the stator drop.
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