On Fri, Sep 14, 2001 at 06:27:01PM +0800, Johann Vincent Paul U. Tagle wrote:
> AFAICR (As far as I can remember) from my EE subjects, the 1.414 is
> needed because the computation (for 115V and 0.3amps) involves
> alternating current (AC). P=VI is okay for DC computations.
>
The way to get real power for an arbitrary AC waveform would be to
integrate voltage*current over the period of your AC waveform. If
both voltage and current are sinusoidal you get a factor of sqrt(2),
which is ~1.414:
P = \int_0^1 0.3 \sin(2\pi t) 115 \sin(2\pi t) dt
= 0.3(115) \int_0^1 \sin^2 (2\pi t) dt
= 0.3(115)\sqrt(2)
I think...
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