• The saturation (magnetisation) curve is a graph of generated e.m.f. E against field current If, with the armature on open circuit and driven at constant rated speed.
• Starting from zero field current, a small residual e.m.f. is generated due to residual magnetism in the pole cores.
• As field current increases, the e.m.f. rises almost linearly at first (the iron is unsaturated, so flux is proportional to field current).
• As the field current is increased further, the iron begins to saturate and the e.m.f. rises more slowly, the curve bending over (knee of the curve).
• At high field currents the curve flattens off as the iron is fully saturated and further field current produces little increase in flux or e.m.f.
• The curve is used to determine the operating point, the field current required for a given voltage, and the effect of armature reaction and demagnetisation. The knee represents the most economical operating point (maximum e.m.f. per ampere of field current).

• Synchronous impedance Zs = open-circuit e.m.f. / short-circuit current (same field current) = 50 / 200 = 0.25 ohm.
• Synchronous reactance Xs = sqrt(Zs^2 - Ra^2) = sqrt(0.25^2 - 0.1^2) = sqrt(0.0625 - 0.01) = sqrt(0.0525) = 0.229 ohm.
• Induced voltage to deliver 100 A at 0.8 p.f. lagging with terminal voltage 200 V:

Assume a three-phase star-connected alternator. Phase voltage Vph = 200 / root 3 = 115.5 V. Load current per phase I = 100 A (line = phase for star). cos phi = 0.8, sin phi = 0.6.

• E = sqrt[(Vph cos phi + I Ra)^2 + (Vph sin phi + I Xs)^2]
• = sqrt[(115.5 x 0.8 + 100 x 0.1)^2 + (115.5 x 0.6 + 100 x 0.229)^2]
• = sqrt[(92.4 + 10)^2 + (69.3 + 22.9)^2]
• = sqrt[102.4^2 + 92.2^2] = sqrt[10486 + 8501] = sqrt[18987] = 137.8 V per phase.
• Line value of induced e.m.f. = 137.8 x root 3 = 238.6 V.

So the alternator must be excited to give an induced e.m.f. of about 137.8 V per phase (238.6 V line).