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The figure shows the per-phase equivalent circuit of a t wo-pol e t hr ee-phase i nduct i on mot or operating at 50 Hz. The “air-gap” voltage, Vg across the magnetizing inductance, is 210V rms, and the slip, is 0.05. The torque (in Nm) produced by the motor is _______.
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- 200 – 303
- 303
- 400 – 403
- 40
Correct Option: C
VOC across j6.28 Ω = Vg = 210 V
Now from the given circuit,
| Rth = | ||
| (0.04 + j0.22) + j6.28 |
= 0.216 ∠ 80.04
= 0.0375 + j0.2127
| Now , I2 = | = | |||
| z | (0.0375 + 1) + j(0.2127 + j0.22) |
∴ |I2| = 186.81 A
| τ = | × I22 . | |||
| ωs | s |
| = | × 186.812 × 1 = 333.27 N-m | |
| 314.15 |
