On Jan 15, 12:43*pm, harry wrote:
On Jan 15, 2:02*pm, "
wrote:





On Jan 15, 2:21*am, harry wrote:

On Jan 14, 9:22*pm, "
wrote:

On Jan 14, 3:23*pm, harry wrote:

On Jan 14, 2:17*pm, "
wrote:

On Jan 14, 3:37*am, harry wrote:

On Jan 13, 10:18*pm, "MarkK" wrote:

Ill repeat my reply here for the sake of those that filter google groups

thats actually a *real interesting question...

with a pure reactive (inductive or capacitive) load, the current flow
in the windings will cause a voltage drop but if i'm not mistaken
since there is no (or very little) actual power flow, there will not
be a load on the engine so the engine speed will not be a factor.

with a pure resistive load, the current through the windings will
cause a voltage drop AND there will be a load on the engine that will
try to slow it down. *It is the job of the speed governor on the
engine to keep the speed constant. *If the speed drops the voltage and
frequency will drop due to the engine speed drop.

So for a given amps, you may get more of a drop with a resistive load
depending on how tight the governor speed control the engine.

Mark

You neglect the resistance of the generator windings.-

I don't see that he neglected the resistance of the
generator windings. *In a generator the resistance of
the windings is very small, negligible compared to
the resistance of any real load.

If there is no resistance in the load (ie purely inductive/
capacitive), it would be the only resistance in the circuit and hence
hugely important.

There is no such thing as a pure inductor BTW.

Uh, huh. *So, why did you just bring such a load
into the discussion?

Someone else did.

There are almost pure capacitors.- Hide quoted text -

- Show quoted text -

Uh, no. *The generator winding resistance is still small. * Small is
still small, whether you have a purely resistance
load or a load that has a substantial component of
inductance or capcitance, like the welder in question
or a motor.

Uh............Yes. The resistance however small will be the only thing
in the circuit consuming any energy.

One considers the Inductive element of a circuit separately when doing
any calculations.- Hide quoted text -

And there you have it folks, after telling others here
that they have s*** for brains, the village idiot has once
again demonstrated that he's the one that is clueless.

Resistance is most certainly *not* the only place
energy is consumed. *Take a simple electric motor,
for example., like the one powering a water pump.
If resistance is the only place that consumes energy,
how do you account for the work done by the motor?
The motor isn't a resistance heater and the energy
is being used pushing the water. *According to your
logic, if we had a 1hp motor, all the power would
have to be in the form of resistance and we'd
actually have a 750watt heater.

We were discussing the electrical losses in a generator.
And the heating losses in the generator *arise from the resistance of
the windings (copper losses) and the magnetic hysterisis losses of the
iron core.
If there was no resistance. there would be no copper losses.- Hide quoted text -

- Show quoted text -

Nice try at back peddling. We weren't discussing the
resistance of anything. Mark made a post about the
voltage regulating characteristics of the generator.
Then you claimed that the resistance of the generator
matters. The guy is hooking up a frigging welder.
THAT is significant in regard to voltage regulation.
The very small resistance of the generator is not.

Then you went even further off the rails:

"Uh............Yes. The resistance however small will be the only
thing in the circuit consuming any energy.
One considers the Inductive element of a circuit separately when doing
any calculations. "

Clearly a circuit, is, well a circuit. It's *not* just
the generator windings. It's the generator plus the
load. Your statement means that
there is no other energy transfer in a circuit with that
welder or a motor, other than through resistance.
THAT is just wrong.