I have a 1991 Chrysler Newyorker Fifth Avenue Wondering how good my altenator is. Gonna be Running a Class D bp1200.1 Jbl or where i can find out how much it supplies
Incorrect, Mr. Bates. That's Ohm's Law at work. P (watts) = I (amperes) * E (volts) If the amplifier is accurately rated for it's RMS power, then the calculation is easy. In the case of underrated amplifiers, the law still holds precisely true, and RMS output can be found through the use of a dummy load and an oscilloscope. JL Audio amplifiers are a good example. The 500/1 actually puts out about 670wRMS. I was giving him a ballpark for current draw so he'd know how much of an increase he'll need in alternator current. What I can't calculate for him is how much of his stock alternator's current is used by the car itself. That changes from car to car.
Until you can find me an amp that is 100% efficient, you can't rely on these calculations to determine the power demands of an amp.
How about heat? Ever seen an amp that does not waste power producing heat? The heat comes from somewhere and that's consumed power....which Ohm's law does not take into account. The 80A + that my 500w amp draws is just one example of many.
You still can't determine the actual power draw of an amplifier based on output...not outside of the classroom anyway.
Ok, let's recap this... based on the claimed power output of an amp and some fog lights you determine the consumption of an electrical system.
Wolf: "your system and those lights are drawing 47A of current."
Moto: "You cannot determine power demand based on claimed power output. Sorry."
Wolf: "Incorrect, Mr. Bates. That's Ohm's Law at work. P (watts) = I (amperes) * E (volts)" 'blah blah'
Moto: "Until you can find me an amp that is 100% efficient, you can't rely on these calculations to determine the power demands of an amp." 'blah blah'
Wolf: "heat is still energy. you cannot create, or destroy energy. it merely changes forms. Now please stop being a troll."
Glasswolf, these formulas take neither heat nor the substancial energy consumed by heat into account so I say again, "You cannot determine power demand based on claimed power output. Sorry."
Saying my post is incorrect is innacurate and misleading information, that is my only issue here besides the personal and equipment attacks you have been passing off on others seeking help.
Call me a troll all you want, doesn't change anything. I've been helping people on here for over a year and if you can find me a bit of information I have passed off as fact which is untrue I'd love to hear about it. Good luck.
well, I can agree to disagree. I can estimate efficiency easily enough based on the amplifier class, which is simple to do. I did miscalculate though since I didn't account for two driving lights at 55W each. I was thinking 55W total when I added that up. anyway, that's that.
On a side note, another way to estimate max current draw for an amp is to look at the fuse rating that came with the amplifier. no matter how inefficient the amp may be, it cannot draw more than the fuse will allow without blowing the fuse. If the amp maker has a decent product, they will match the fuse fairly close to the max current demand of the amplifier. Then you just need to divide down for the load on the amp.
Blah
Unregistered guest
Posted on
I say, just hook everything up. When crap starts blowing up, then you know you need a bigger supply of power.