why does RMS = amount of power?

[Alan M. Horowitz]

I understand quite clearly how RMS voltage/current is computed.

I don't understand why it equals the amount of power delivered.

[Jussi Ohenoja]

In article <1994Dec14.0...@nosc.mil>, horo...@nosc.mil (Alan M. Horowitz) says:
>
>I understand quite clearly how RMS voltage/current is computed.
>
>I don't understand why it equals the amount of power delivered.

When you are calculating RMS, you are actually measuring the area
between voltage (or current) plotted against time and 0-axle.

This integral represents the mean voltage that would produce
the same amount of heat if driven into resistive load.
The values are squared and rooted to get the whole plot on
the positive side of the 0-axle, and then averaged.
(RMS == Root Mean Square ;)

- Juice -

Jussi Ohenoja
ohe...@ncsvax.ntc.nokia.com
ohe...@ncsbsr03ou.ntc.nokia.com

[Steve Bennett]

In article <1994Dec14.0...@nosc.mil>, horo...@nosc.mil (Alan M. Horowitz) writes:
|> I understand quite clearly how RMS voltage/current is computed.
|>
|> I don't understand why it equals the amount of power delivered.

If memory serves me ,the quick and dirty answer is :
RMS voltage is used for finding the *average* power
that a *sin* wave (or cos wave) delivers to a load
as oposed to *peak* *instantaneous* power you would get if
you used the peak value of the sin in your calculation.
By _average_ I mean the average over a full cycle of the sin wave.
If you want to know the average power delivered to the
load when the input voltage is something other than a sin
wave then you have to dig out your calculus book and derive
a new formula. Vrms= Vpeak*.707 is only valad for sin or
cos waves.

The basic idea is 1 Volt rms will delivers the same power
to a resistive load (on average over a complete cycle)
that 1 volt D.C. would deliver.

Hope this answers your question.

[George Jefferson]

you have a periodic voltage V(t) applied to a resistive load,
then the current is also periodic I(t)=V(t)/r
the time varying power is thus P(t)=V(t)^2/r

the average power delivered over the period is:
(where Int()->integral over one period divided by the period)
P = Int( P(t) ) = Int( V(t)^2 / r ) = (1/r) Int( V(t)^2 )

so now we ask what are the 'effective' values for V and I? Well
we want Veff*Ieff = Power and Veff=Ieff*r, so you arrive at
Veff^2/r = (1/r) Int( V(t)^2 )
thus:
Veff= Sqrt[ Int ( V(t)^2 ) ] ( likewise for the current )

This is the 'root' of the 'mean' of the 'square' or rms

The V,I need not be sinusoidal..

--
george
geo...@mech.seas.upenn.edu