Re: 1600 watts magnetizing current?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Hydrogen18" <hydrogen18-at-bellsouth-dot-net> 

I know how power factor works, and I know that you techincally should use
VA, but seeing as how this is a tesla coil board involving lots of AC work I
figured we all coudl use watts and VA interchangeably. whenever I add some
PFC its idle current does drop considerably. You mention your equations
assumes sinuosidal current draw. How do you work out power factor when your
voltage and current are non sinuosidal(we have baad waveform distortion
here)?

---Eric
----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Tuesday, August 03, 2004 8:06 AM
Subject: 1600 watts magnetizing current?


 > Original poster: "Steve Conner" <steve.conner-at-optosci-dot-com>
 >
 >  >The 1600 watts isnt going to heat, it is magnetizing the core.
 >
 > The word "watts" should only be used when you're talking about real power.
 > For reactive power (like the power associated with magnetizing current)
use
 > VA.
 >
 > If you want to be really pedantic, use "VA" for apparent power and VAr for
 > reactive power.
 >
 > If we denote real power by "P", reactive power by "Q", apparent power by
 > "S", and power factor by cos(phi) (phi is the phase shift between voltage
 > and current) then the following relations hold:
 >
 > P=Vrms*Irms*cos(phi)
 > Q=Vrms*Irms*sin(phi)
 > S=Vrms*Irms
 > S^2=P^2+Q^2
 >
 > These assume sine waves. When we start to deal with seriously
non-sinusoidal
 > currents, as in SSTC rectifier/filter power supplies, there are two power
 > factors- the original "displacement" power factor plus a "harmonic" one.
 >
 > You probably couldn't care less about this stuff, but at least try to
 > remember that when you measure line current, and multiply it by 120, the
 > answer you get is the apparent power S (whereas in most cases you actually
 > want to know P)
 >
 > Steve C.
 >
 >