Discussion
A synchronous motor with a terminal voltage of 1.0 per unit, an excitation emf of 1.5 per unit and a synchronous reactance of 1.2 per unit has a pull-out power of
Answer: D
Correct answer: 1.25 per unit
\[ P_{max} = \frac{EV}{X_s} = \frac{1.5 \times 1.0}{1.2} = 1.25\ \mathrm{per\ unit}, \] reached at a load angle of ninety degrees. In practice the machine is never loaded near that point, since the synchronising stiffness falls to zero there and any disturbance would pull it out of step.
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