Re: Who is right?

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

Original poster: "Malcolm Watts by way of Terry Fritz <twftesla-at-uswest-dot-net>" <m.j.watts-at-massey.ac.nz>

On 13 Mar 01, at 11:39, Tesla list wrote:

> Original poster: "ebyng by way of Terry Fritz <twftesla-at-uswest-dot-net>"
> <ebyng-at-netlimit-dot-com>
> 
> I hate to ask this, but...
> 
> I was browsing through some old books at a used book store recently
> (yesterday) and i came accross a book devoted to the study of
> capacitors. It states that when stringing capacitors, in serial or in
> parallel, only the capacitance rateing changes, not the voltage
> rateing.  I cross referenced with a newer book, and it states that
> Both the capacitance and voltage ratings change in proportion to the
> number of capacitors connected. So... Who is right?  I'm tempted to
> stick with the newer book, but i'd just like to double check...

The newer book is closest to the truth but leaves out important 
details. 
Paralleled caps: - voltage rating = lowest voltage rated cap 
                 - total capacitance = sum of all capacitances

Series'd caps: - voltage rating = tricky. How much voltage is 
impressed upon a particular cap in the string depends on the relative 
capacitances of the caps in the string in an AC circuit. In a DC 
situation, this is further complicated by dielectric leakage currents 
which may also be a problem at low frequencies. It can be calculated 
of course (and knowing the leakage current of each cap is a big help) 
but is not easy to put in a generalized form. A good rule is that if 
you are going to run a bunch of caps in series, make them all the 
same value and voltage rating. Then the working voltage is n*working 
voltage of a single cap where "n" = number of caps.
               - total capacitance = calculate as for paralelled 
resistors i.e. 1/[ (1/C1)+(1/C2)+....+(1/Cn) ]
     If all caps have the same value, the total capacitance = 
(capacitance of one cap)/n.

Malcolm