Members Gravity Posted April 5, 2005 Members Share Posted April 5, 2005 Maybe I'm nuts, but I swore I read a long time ago that you used 1.414 when doing a calculation with the impedance to find your actuall wattage? I remember something like "Suppose you have an amp that puts out 100 watts at 4 ohms and you decide to hook up an 8 ohm load. To find the wattage dissipated with 8 ohms, you would divide 100 by 1.414." And you would divide by 1.414 for each time the impedance doubled. So if it were 16 ohms, you'd divide by (1.414)^2 Am I nuts, or is that a true statement? Link to comment Share on other sites More sharing options...
Members Scodiddly Posted April 5, 2005 Members Share Posted April 5, 2005 The 1.414 comes up in conversions between RMS and peak voltages. Link to comment Share on other sites More sharing options...
Members agedhorse Posted April 5, 2005 Members Share Posted April 5, 2005 Power is simply I x V, which with Ohm's law yields the other forms like V**2/R, I**2R etc. Since R is a linear term (not squared) then the power is related linearly (and inversely) to it, so double the resistance and you half the power. Link to comment Share on other sites More sharing options...
Members Gravity Posted April 5, 2005 Author Members Share Posted April 5, 2005 Originally posted by agedhorse so double the resistance and you half the power. ok ... that makes more sense, because I did look at the voltage, current, impedance equations recently, and that is what I deduced as well ... twice the resistance = 0.5 the power It was a long time ago in an article in a guitarworld magazine ... I must have confused it with something else then. But this brings up another question.My power amp is power rated like this per channel in stereo: @2 ohms = 700 watts@4 ohms = 450 watts@8 ohms = 280 watts Why is the power not reduced by 1/2 each time the resistance is doubled? Link to comment Share on other sites More sharing options...
Members agedhorse Posted April 5, 2005 Members Share Posted April 5, 2005 Originally posted by Gravity But this brings up another question.My power amp is power rated like this per channel in stereo:@2 ohms = 700 watts@4 ohms = 450 watts@8 ohms = 280 wattsWhy is the power not reduced by 1/2 each time the resistance is doubled? And it's a good question too! In theory, it would behave as you guessed, but non-linearities in the amplifier, plus limitations in the power supply (transformer impedance mostly) combine to cause the non-theoritical results you see in the specs. Link to comment Share on other sites More sharing options...
Guest Anonymous Posted April 5, 2005 Share Posted April 5, 2005 Originally posted by Gravity ok ... that makes more sense, because I did look at the voltage, current, impedance equations recently, and that is what I deduced as well ... twice the resistance = 0.5 the powerIt was a long time ago in an article in a guitarworld magazine ... I must have confused it with something else then.But this brings up another question.My power amp is power rated like this per channel in stereo:@2 ohms = 700 watts@4 ohms = 450 watts@8 ohms = 280 wattsWhy is the power not reduced by 1/2 each time the resistance is doubled? Layperson's understanding here: In addition to what Andy said, generally those maximum wattage specs are qualified by the THD limitation cut-off point for the tests. This is where folks get into trouble "the amp only puts out 1000 watts in bridge/mono at 4 ohms".... well that might be true if the test determined the maximum wattage output not to exceed 0.1% THD. At +6dB above clip the power output might be considerably higher (breifly). Link to comment Share on other sites More sharing options...
Members Gravity Posted April 5, 2005 Author Members Share Posted April 5, 2005 ok ... so we're not dealing with a "perfect" system here ... that does makes sense thx fellas Link to comment Share on other sites More sharing options...
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