determining wattage produced

Hi all! How would I be able to determine the wattage a given amplifier produces at a given voltage? For instance: If most car audio amps are rated and tested at 14.4volts. That given, if such amp produces 2500watts at 14volts into "X" ohms - HOW then would I be able to calculate what that amp would produce if it saw less than 14volts ("Z"<14v)????? I want to be able to figure out what an amp would produce at 11volts {into "X" ohms). Granted, for audio systems that require a lot of draw on the electrical systems measures must be taken to provide the juice needed: IE: alternator upgrades or multiple ones, 1/0 gauge wire, multiple batteries, etc. etc. I just want to know what an amp will produce if it isn't running at 14.4volts. Given that I know that amp was tested and rated at 14.4volts. Oh and I don't have an amp that is currently installed that is running at 11volts or a multimeter to try and go test to see what wattage is produced. I've just been saving money and doing research for a build and want to get an idea on paper of how things would look given a voltage drop does happen or it runs a little low - with an amp that has been rated and tested at 14.4v. Thanks for your input! Take care yo!

The formula is voltage squared divided by resistance. But the voltage is the A/C output voltage at the amp not the D/C input voltage. You would have to play a steady note in order to get the measurement.

Example: 40VAC x 40VAC / 4ohms = 400 Watts

That would only work if you were using a dummy load.  The impedance of a speaker varies tremendously depending on the frequency of the note being played. 

You would have to have an Oscilloscope and 2 meters to pull this off.  At clipping as determined by the display of the scope, Voltage x Current = Wattage.

Power (Watts) is VI = VV/R irrespective of ac or dc, but ac must be RMS. (For pure DC, average = peak = RMS; and DC with ripple is DC with AC superimposed, or metered as true-RMS etc.) Amplifiers greater than a ~ 200 Watts (into 1Ω; 100W@2Ω); 50W@Ω; peak!) require SMPS, hence their power output should not vary with voltage input. However it is obvious that many do. And for many that I have seen, their output is directly proportional to the input voltage-squared - ie, as if the amp itself is a perfect resistive load - eg at 11V compared to 14.4V is (11/14.4)**2 = 0.763888 x 0.763888 = 0.584 times the 14.4V power. PS - above; VV/R is VxV/R; ie, P = VV/R = IIR.

Without knowing the true efficiency of the amplifier, any calculations trying to determine how the output wattage changes based on input voltage are guesswork.  The only thing you can be certain of is the amplifier will produce less wattage than the manufacturer's rating and likely with higher distortion.Support the12volt.com