Excessive Current draw
 

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

Mad Dog,
You 'done good', but you're not quite finished, your
answer doesn't take the efficiency of the amplifier into
account. Since the efficiency of most solid state devices
is only about 50%, the amplifier would draw half again as
much more current as your almost correct answer...
'Doc

True that Doc..thanx 4 the heads up

--
714 Sandpile,
The Mad Dog wavin' good bye
"'Doc" wrote in message ...


Mad Dog,
You 'done good', but you're not quite finished, your
answer doesn't take the efficiency of the amplifier into
account. Since the efficiency of most solid state devices
is only about 50%, the amplifier would draw half again as
much more current as your almost correct answer...
'Doc

Slower traffic KEEP right, and broke down alternators keep even MORE right!

Hey Mad Dog---do you know that IN a Communist country, there is only 1
trucking company! Do you know what it's name is? JBSnydWerner !
Pass it on!
Dave!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

"Mad Dog" wrote in message
...
True that Doc..thanx 4 the heads up

--
714 Sandpile,
The Mad Dog wavin' good bye
"'Doc" wrote in message ...


Mad Dog,
You 'done good', but you're not quite finished, your
answer doesn't take the efficiency of the amplifier into
account. Since the efficiency of most solid state devices
is only about 50%, the amplifier would draw half again as
much more current as your almost correct answer...
'Doc






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