complaints when using a hughes autoformer

ScottG wrote:

Lantley wrote:
P=I*E Power = current x volts

Lets use a 30 amp circuit operating properly
P= I x E 3600 =30 x 120
However under low voltage conditions
there is only 100 volts instead of 120
P= I x E or 30 x 100 =3000
the autoformer takes that same 3000 watts
and keeps the voltage consistent but the available current drops
3000 = 25 x 120 .(note current dropped from 30 amps to 25 amps).
Power is still 3000 watts
Now there are some heat losses and losses do to impedance. But over all that is the fundamental function of the autoformer.
It ensures adequate voltage, however in the process less current is available.
On a 50 amp circuit we don't necessarily need the full 6000 watts.
50 amps x 120 volts=6000 watts
So if the current drops a bit it doesn't matter as long as adequate voltage is maintained.

What??
The voltage stays consistent through the autoformer?
Maybe you meant wattage? You'd still be wrong.

Some heat losses due to impedance?
Impedance has nothing to do with it. The heat comes from flux generation.

Current doesn't drop in the autoformer, it goes UP as it raises the voltage.

There's much wrong with your understanding of electricity.

I think Lantley is talking about input voltage/current where you're thinking output and vice-versa. The autoformer is designed or intended to maintain--or at least enable one to maintain--a more or less constant output voltage, despite fluctuating input voltages. In the process, the output current you can use goes down if the input voltage goes down since the input current is limited by the circuit breaker etc.

I think you're all agreeing, more or less, on the basic concepts without realizing it. ;)