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#1 |
Banned
Oct 2009
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![]() Hi to all, This is all Information I myself have gathered from the MECP Installers Guide book. Alright Here's the Question! I am trying to figure out how much pull on my alternator I am getting from the amp. Here is the formula We use to figure this out I = (Po - n*Po + Po) / V Now I = amperes Po=RMS n=efficiency V=voltage So for me its: I = (3000 - .8*3000 + 3000) / 13.8 So I do this 3000*.8 and i get 2400 3000-2400=600 600+3000 3600/13.8 260.87 This is my amperage draw correct? and if I'm correct for everyone else heres an easier equation for this: I = (2-n)Po/V So for me this would go 1. 2-.8 =1.2 2. 1.2*3000=3600 3. 3600/13.8 And there we go. 260.87!!!!! Now, We must remember that Most Class D Amplifiers have a 80% Efficiency Rating and Most Class A/B Amplifiers have a 60-70% efficiency Rating! |
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#2 | |
Active Member
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![]() Quote:
But if you are pulling 3000 watts rms that looks like a fitting amount of amps. I dont know what amp you have to tell you how things are running. IE how many watts RMS you are pulling at what ohms it's set to. |
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