Generator on bus with 20% excitation increase

Given: star, 5 kVA, 208 V, 50 Hz, $X_s = 8\ \Omega$, $R_a \approx 0$, in parallel with a 208 V bus (infinite bus).

(i) Excitation voltage and power angle at rated kVA, 0.8 pf lagging

$$\begin{aligned} V &= \frac{208}{\sqrt3} = 120.09\ \text{V} \\ I &= \frac{5000}{\sqrt3\times208} = 13.88\ \text{A},\quad \bar I = 13.88\angle -36.87^\circ \\ \bar E &= \bar V + jX_s\bar I = 120.09 + j8\times13.88(0.8 - j0.6) \\ &= (120.09 + 66.62) + j88.82 = 186.71 + j88.82 \\ E &= 206.76\ \text{V per phase},\quad \delta = 25.44^\circ \end{aligned}$$

Check: $P = \dfrac{3EV}{X_s}\sin\delta = \dfrac{3\times206.76\times120.09}{8}\sin 25.44^\circ = 4000$ W $= 5\times0.8$ kW.

(ii) Excitation increased by 20%, same input power

$$\begin{aligned} E' &= 1.2\times206.76 = 248.11\ \text{V} \\ E'\sin\delta' &= E\sin\delta = 88.82 \quad (P \text{ constant}) \\ \sin\delta' &= \frac{88.82}{248.11} = 0.3580,\quad \delta' = 20.98^\circ \\ \bar E' &= 248.11\angle 20.98^\circ = 231.66 + j88.82 \\ \bar I' &= \frac{\bar E' - \bar V}{jX_s} = \frac{111.58 + j88.82}{j8} = 11.10 - j13.95 \\ I' &= 17.83\ \text{A},\quad \angle -51.48^\circ \\ \text{pf} &= \cos 51.48^\circ = 0.623\ \text{lagging} \end{aligned}$$

The active current component stays 11.10 A (power unchanged); extra excitation only increases the lagging reactive current, so the generator supplies more reactive power ($Q$ rises from 3 kVAR to 5.02 kVAR).

Answer: (i) E = 206.8 V/phase (358.1 V line), δ = 25.44°; (ii) I = 17.83 A, pf = 0.623 lagging.