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?

  • \$\begingroup\$ I am pretty sure the voltage drop at full load due to winding resistance would be something like 10 percent. In order to have this loss be negligible at full load, you would need to make the generator much bigger. I think that is why they have regulation in the first place. \$\endgroup\$
    – user57037
    Feb 18, 2021 at 1:36
  • \$\begingroup\$ Re, "Effectively zero is not zero." Um,... "effectively zero" does not mean "close to zero." It means, having the same effect as zero. It means that you can safely pretend that the resistance is zero, and you still will get a satisfactory answer. If, in reality, ignoring the resistance gives you an unsatisfactory answer, then it would be wrong to call the resistance "effectively zero." \$\endgroup\$ Apr 9, 2022 at 16:35

1 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.


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