What factors affect the voltage drop in a basic AC generator when an additional load is added in parallel

Question

As a preface, I am in school to become an engineer on a ship so my question relates to a self-contained circuit/grid and not a home.

We are going over generator theory and how an automatic voltage regulation circuit is used to control voltage. I know AC generator voltage is proportional to the magnetic field and relative speed between the magnetic field and the armature (output) winding. I get this and how the voltage regulation works. What I don't understand is how adding a parallel load to an otherwise steady state generator without a voltage regulator causes a noticeable voltage drop.

One of the laws in a parallel circuits is that the voltage drop across loads in parallel is constant. So if you add a load in parallel the resistance goes down, resulting in current going up. And therefore wattage goes up. Voltage isn't affected. This is an ideal circuit so let's take it a step further.

Say you have a engine turning a permanent magnet next to a coil of wire. This is the most basic generator I can think of. Now if I have that generator operating in steady state with an electrical load where everything is constant and if I add a load in parallel then my total resistance goes down again, and this would cause current to go up again. But winding/coil impedance is a thing and Lenz's law would would result in a counter force to slow down the engine and reduce the RPM and thus frequency. But, engine feedback will correct this returning to steady state RPM and frequency. Therefore, neither the speed nor the magnetic field of the generator has changed. Generator current has gone up and therefore power has too, but I still can't see how this would drop voltage.

Finally I know the generator coil windings will have some kind of resistance, but this is always called out as conductor resistance it is effectively zero.

Effectively zero is not zero though, and an increase in current would result in an increased voltage drop for internal resistance. The actual voltage drop seems too large for this alone to account for the change. Is it this internal resistance causing the noticeable voltage drop at output, or is there some other factor, law, or equation that I am missing that would account for a generators output voltage dropping?

Answer 1

Since you are asking about an AC generator, you need to consider the generator's inductive reactance. That is responsible for most of the voltage drop that remains when the speed is maintained to provide the rated frequency. With a wound-field synchronous generator, the voltage can be more closely maintained by adjusting the field excitation. That is not an option with a permanent-magnet generator.

Answer 2

If you are adding a load in parallel then you are effectively increasing the armature current which would increase armature reaction causing demagnetization or cross magnetization of the field flux.As field flux affects induced emf the deterioration of the field flux caused the induced em to decrease. But this is not the case if the power factor is purely leading or zero leading, where induced emf would increase.