Discussion
A generator with a synchronous reactance of 1.0 per unit, an excitation emf of 1.4 per unit and rated terminal voltage has a pull-out power of
Answer: C
Correct answer: 1.4 per unit
\[ P_{max} = \frac{EV}{X_s} = \frac{1.4 \times 1.0}{1.0} = 1.4\ \mathrm{per\ unit}, \] reached at a load angle of ninety degrees. The machine cannot be loaded to that point in practice: the synchronising coefficient falls to zero there, so any disturbance pulls it out of step.
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