Electrical – Testing AC generator on a Unity power factor (Resistive Load)

Yes. The ratings essentially say that the generator is rated 1043 amps maximum and the engine is rated 600 kW plus the generator losses. The engine and generator are matched for 600 kW at 0.8 pf, 1043 A. At any other operating point, either the generator or engine is working at more or less than its rated operating point.

There is a sign error in your calculation. For unloaded operation, you have : V = E. So: $$E_{nl} = 2000V$$

You want to calculate the variation of E you need to have in order to ensure that V stay constant when the alternator is loaded :

$$ E_l = V + R \times I + j \times X \times I $$ $$ E_l = 2000 + (0.8 + j \times 4) \times I = 1824 + 368j $$ $$ \left | E_l \right | = 1861 V $$

Note that E_l is lower than E_nl because of the leading power factor. Finally : $$ {{1861 - 2000} \over {2000}} = -6.96\% $$