Re: Piezo HV Power Supply

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

Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>" <Mddeming-at-aol-dot-com>

In a message dated 4/11/02 12:08:56 AM Eastern Daylight Time, tesla-at-pupman-dot-com
writes:



>
> Original poster: "David Thomson by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> <dave-at-volantis-dot-org>
>
> Hi Pete,
>
> Dave T.
> >>Your formula for energy is wrong in that it should be:
> >>joule = farad * volt^2
>
> Pete K.
> >Your formula for energy is incorrect.  The energy in a cap is:
> >Joules = farads times volts squared divided by two.
>
> Well, let's take a look at this.
>
> >From NIST http://physics.nist.gov/cuu/Units/units.html
>
> Farad = m-2*kg-1*s4*A2
> Volt = m2*kg*s-3*A-1
>
> Farad * Volt^2 = m-2*kg-1*s4*A2*m2*kg*s-3*A-1*m2*kg*s-3*A-1
> Farad * Volt^2 = s-2*m2*kg
> Farad * Volt^2 = joule
>
> Where does the "1/2" come in?
>
> Can you direct me to the source that you referenced?  Perhaps your source is
> wrong?
>
> Dave
>



Energy= 0.5 CV^2
The work done in charging the capacitor, and therefore the energy available, is
proportional to the AVERAGE potential difference encountered during the
charging process V(avg)= (V+0)/2=0.5Q/C
Therefore Work (energy)= 0.5Q/C *Q = 0.5Q^2/C
Now since V=Q/C, this is equivalent to W=0.5CV^2
Understanding the physical process takes more than just playing with units out
of a reference table.

References: 
College Physics 3rd ed., Sears & Zemansky, Addison Wesley Publ. 1960 pg 522
Elements of Physics 3rd ed., Shortly & Williams,Printice-Hall 1961 pg 677
Any of a hundred advanced high school or elementry college physics texts or
almost any basic electronics text.
Hope this helps, 

Matt D.
G3-1085