Assuming 100% efficiency of the amp (all power dissipated in speakers), total power input to amp is 340W peak
Watts = Volts x Amps
rearrange:
Amps = Watts/Volts = 340W/14V = 24 Amps peak
jsayre914
Mar 7 2008, 07:48 PM
QUOTE(Spoke @ Mar 7 2008, 08:34 PM)
Assuming you will power 6 channels @ 40W RMS,
Total power is 6 x 40W RMS = 240W RMS
Change to peak power: 240W x sqrt(2) = 340W peak
Assuming 100% efficiency of the amp (all power dissipated in speakers), total power input to amp is 340W peak
Watts = Volts x Amps
rearrange:
Amps = Watts/Volts = 340W/14V = 24 Amps peak
lol
Jeffs9146
Mar 7 2008, 08:12 PM
Thanks!! I had just finished a Google search on that exact question and came up with the same info. I did forget about the peak part though!
Thanks
Scott Carlberg
Mar 7 2008, 09:28 PM
Jeff, I asked my buddy who manages & does the install at a car stereo shop here in Modesto.
Here's what he said: The Maximum amount of Current that the Amp can draw from the Battery is EQUAL to the amount/Size of the Fuse in the Amp.
NO amp should be able to draw more current than the Fuse in the Amp.
sww914
Mar 7 2008, 10:23 PM
20 amp fuse X 12v = 240 watts 30 amp fuse X 12v = 360 watts The 5 channel US amps amplifier I have in my dually uses 3 30 amp fuses. It's plenty.
Jeffs9146
Mar 8 2008, 01:38 AM
QUOTE
The Maximum amount of Current that the Amp can draw from the Battery is EQUAL to the amount/Size of the Fuse in the Amp.
That part I knew! And most people dont know that if you put a smaller fuse in the ground wire it will also restrict the amp load as if it was a resistor.
There is still a MAX load and the fuse is 10 - 20% above the rated amp max at full volume.
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