Why terminal voltage of the self-excited shunt generator lower than that of the separately excited shunt generato?

Answer 1

Some generators are self excited; this means their terminal voltage is fed back to the excitation system to supply power to the rotor of the generator (which makes it into an electromagnet); the more power that is fed back, the stronger the electromagnet becomes, which makes it harder to turn the generator, which causes the generator to push out more power (simplified, really quick version).

If there is a fault electrically near the terminal of a self excited generator, the terminal voltage will sage to near zero; this means the voltage supplied to the excitation system will drop by the same percentage (say the terminal voltage is 30% of what it should be, then the maximum supplied voltage to the excitation system drops to 30% of what it normally is, since P = V*I). Since the input power is less, the output of the generator will decrease (current will decrease). The terminal voltage is determined by the impedance between the generator and the fault such that V = I*Z; As I decreases, V will also continue to fall, causing the terminal voltage to sag even more.

A non-self excited generator gets its' excitation power from the grid, specifically from a location that is electrically separated from its' terminal voltage. If the terminal voltage sagged to 30% (same fault location as above example), the excitation system voltage may be impacted slightly (say 2%) so the excitation system power is near maximum (98% for this example). Since the excitation system is much farther removed from the terminal voltage, it is not dependent upon it, thus the terminal voltage will not continue to sag as with a self excited system.