New current and pf when back emf rises 25%

Assumptions: star connection; the shaft load (input power 50 kW) stays the same when the excitation is raised; $R_a = 0$.

Original operating point

$$\begin{aligned} V &= \frac{660}{\sqrt3} = 381.05\ \text{V},\quad I_a = \frac{50000}{\sqrt3 \times 660 \times 0.8} = 54.67\ \text{A} \\ \vec I_a &= 54.67\angle -36.87^\circ = 43.74 - j32.80\ \text{A} \\ jX_s\vec I_a &= j3(43.74 - j32.80) = 98.41 + j131.22 \\ \vec E_f &= \vec V - jX_s\vec I_a = 282.64 - j131.22 = 311.61\angle -24.90^\circ\ \text{V} \end{aligned}$$

New back emf (+25 %)

$$E_f' = 1.25 \times 311.61 = 389.52\ \text{V per phase}$$

Power is unchanged, so $P = \dfrac{3VE_f'}{X_s}\sin\delta'$:

$$\begin{aligned} \sin\delta' &= \frac{P X_s}{3VE_f'} = \frac{50000 \times 3}{3 \times 381.05 \times 389.52} = 0.3369 \\ \delta' &= 19.69^\circ \end{aligned}$$

New current and power factor

$$\begin{aligned} \vec E_f' &= 389.52\angle -19.69^\circ = 366.75 - j131.22 \\ \vec E_r &= \vec V - \vec E_f' = 14.30 + j131.22 \\ \vec I_a' &= \frac{\vec E_r}{jX_s} = \frac{14.30 + j131.22}{j3} = 43.74 - j4.77 \\ &= 44.00\angle -6.22^\circ\ \text{A} \end{aligned}$$

Power factor $= \cos6.22^\circ = 0.9941$ lagging.

Check: $\sqrt3 \times 660 \times 44.00 \times 0.9941 = 50.0$ kW.

Raising the excitation reduced the reactive (lagging) part of the current, so the current fell and the pf moved close to unity.

Answer: New current $= 44.00$ A at a power factor of 0.994 lagging (originally 54.67 A at 0.8 lagging).