Yon knucklehead is assuming 100% efficiency or doesn't know the meaning of
the term.

So ask Yon knucklehead how one builds an amplifier that yields 1200 Watts
of DC Power into 1200 Watts of RF power ??
Lots of folks would like to know how to do that.

Power out divided by power in times 100 = % efficiency

"Richard" anom@anom wrote in message ...
Some knucklehead in misc.transport.trucking is showing off what a real ass
he is in the way of knowledge.



That should take you to the proper thread. If not, look for the post from
"mad dog" posted at 9:35 am.

I kid you not, these are his words:

quote
Keep the P=IxE formula in mind when choosing a mobile amplifier.
Power = Current x Voltage
or Current = Power / Voltage
P=Power or Wattage
I =Current draw
E=Energy or Voltage
Example: Tommy truck driver wants to install a 1200 watt spurious
emissions generator on his Kenworth with a 12VDC electrical
system. We just learned that Current draw is calculated by dividing
the Voltage into the Wattage so if the 1200 watt cataract producer has
a Input Voltage of 12 Volts then the Current draw will
be...............................
One Hundred Amps (100). Yes folks you read that right,
that is more current draw than most alternators can produce.So please
be realistic when adding accessories to your already marginal electrical
system or you may find yourself playing the "Prince of Darkness"
on a very wet and rainy night.

--
714 Sandpile,
The Mad Dog wavin' good bye

/quote

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is only
an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything other
than what he has written.
So if one of you could come in to the thread and explain how damn wrong he
is, I would appreciate it.

thanks for the time.