Discussion
A synchronous motor developing 2,000 N·m at 500 rpm delivers a shaft power of nearest
Answer: D
Correct answer: 105 kW
\[ \omega = \frac{2\pi \times 500}{60} = 52.4\ \mathrm{rad/s}, \qquad P = T\omega = 2000 \times 52.4 = 105\ \mathrm{kW}. \] Multiplying the torque by the speed in revolutions per minute without converting gives a figure ten times too large, which is the usual slip in this calculation.
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