Correct Answer - Option 2 : 921.9 kW

Synchronizing power of a synchronous machine is given by,

\({P_{syn}} = 3\frac{{{E_f}{V_t}}}{{{X_s}}}\sin \delta \)

Where E_{f} is the internally generated emf

V_{t} is the terminal voltage

X_{S} is the synchronous reactance

δ is the torque angle

**Calculation:**

Given that,

\({E_f} = {V_t} = \frac{{6600}}{{\sqrt 3 }}V\)

X_{S} = 1.65 Ω

Mechanical angle (θ_{m}) = 1°

Number of poles (P) = 12

Electrical angle, \({\theta _e} = \frac{{{\theta _m}}}{2} \times P = \frac{1}{2} \times 12 = 6^\circ = \frac{\pi }{{30}}rad\)

\(P = 3 \times \frac{{\frac{{6600}}{{\sqrt 3 }} \times \frac{{6600}}{{\sqrt 3 }}\; \times \;\cos 0^\circ }}{{1.65}}\; \times \;\frac{\pi }{{30}}\)

= 2764.6 kW

= 921.53 kW per phase