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 Power measurement
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 7th June 2002, 04:42 PM #1 diyAudio Member   Join Date: Dec 2001 Location: Earth Power measurement I was wondering on how to measure the power of an amp i was told that the amp should be connected to a dummy load then the voltage is measure across the 8 ohm dummy load. The voltage measured is AC then the voltage is divided by the 8 ohm to get current then P=IV for to calculate the power for the amp Is this possible or is there any other way?
 7th June 2002, 05:38 PM #3 diyAudio Member     Join Date: Jan 2002 Location: Sweden Ohm my god I like the law... Also for a lot of people the formula U = RI is a dear friend (all of them are variations of the basic Ohm's law). If then P=IU (current multipled by voltage) then why not use P = U(sqr)/R as I solved from U=RI is I=U/R giving P=UI moving on to P=UU/R. Since this is peak and the RMS it usually more interesting and the RMS value for sinusoidal is the peak divided by 2(sqrt) (which is why FETHead says 2.83 for peak-to-peak which is twice the peak and twice the root of 2 is 2.83) is is easy to calculate as the root of 2 squared is 2 again... Hence for RMS values using peak voltage: P=U(sqr)/2R This is the quick one I think and even works in my head since it usually (or sometimes) is things like 20 Volts at 8 Ohms which is 20*20/2*8 = 25 W, nice and easy numbers... Sorry for the lengthy reply/UrSv
 7th June 2002, 05:52 PM #4 diyAudio Member   Join Date: Dec 2001 Location: Pittsburgh, PA, USA None of the DMM I have are good much above 120Hz. Using a tone generator at 120Hz works fine of course. As FEThead alluded some amps may do nasty things (like power limit, or break) if operated at full power for any length of time. Commercial amps in particular often aren't designed for much above a 15% duty cycle....you'll note that the "5x100W" amp is rated for only 265W from the line.

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