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#1
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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 |
#2
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![]() 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 |
#3
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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 |
#4
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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 -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
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