Re: watts vs VA Re: Primary- vs. secondary-current feedback

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: <davep@xxxxxxxx>


> *apparent* power is RMS V * RMS I  (that is, if you hook up a
> voltmeter and ammeter, that's what you'll get)
>
> However, if the current isn't in phase with the voltage (that is, if
> you have a load that is partly inductive or capacitive), then the
> actual power (in terms of something that can turn into heat or
> mechanical work) will be less than the apparent power.  Active power  is
> what a wattmeter measures.
    Generally concur, however a few further thoughts may be of use:
    1) Wattmeters _try_ to measure Watts, real watts, and
    more or less succeed.  However they can fail, more or less,
    especially with 'interesting' (nonsinusoidal) waveforms.
    Which brings up:
    2) In addition to the VA vs W being affected by the 'phase'
    as usually discussed: VA vs W is affected by the wave shape.
    loads with 'interesting' waveforms (notably: current into
    spark gaps....) can do odd things to VA value, W Value and
    the indications on any meter attempting to measure them.
    A fully specified meter will try to discuss these influences,
    if only by stating: 'accurate for sine waves' ehich means:
    less accurate for NOT sine waves.

> The ratio of active to apparent power is the "power factor"... =1 for  a
> resistive load
>
> Apparent power is in VoltAmps (VA)
> Active power is in Watts
> Reactive power is in VAR (volt amp reactive)..
>
> Since transformer ratings depend on things like the voltage (flux in
> the core) and current (heating of the windings), fairly independent  of
> the power factor, they are rated in VA.
    best
     dwp