On Sunday 17 February 2008 19:58, Aaron Kulkis wrote:
Randall R Schulz wrote:
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However, the point is that by squaring the primitive (and time-varying) quantity first (voltage or current, but not power which depends on both current and voltage)) then taking the average over a cycle and then taking the square root of that value, you'll get the average power.
This is clearly incorrect. The result of applying the RMS algorithm to voltage is a value with units of voltage. This value squared and divided by the resistance into which the voltage drives the current resulting in some power dissipation gives that power.
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The units of RMS voltage is the volt, not volt^2.
True. But another correction (to one of your earlier statements) is required. On Sunday 17 February 2008 19:56, Aaron Kulkis wrote:
... RMS is just as statistical method, which is useful for making sense of any time-variant function.
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While RMS is the name for a statistical technique, as it's used it circuit analysis, it's an analytical technique. E.g., if the waveform is a sine wave, then one can compute the definite integral analytically for the function sin(x)^2. Taking the square root of this value gives the RMS value. For hard-to-integrate functions or those with no a priori analytical characterization, a numeric technique can be used. Randall Schulz -- To unsubscribe, e-mail: opensuse+unsubscribe@opensuse.org For additional commands, e-mail: opensuse+help@opensuse.org