# Re: Multiple Caps: How and Why

```While on the subject of series and parallel caps I would like to hear more
series caps of differing cap values and the attentuation effects (voltage
dividing or multiplication effects) and associated current effects --as this
seems to be a subject of very little attention....
..Thanks.... JJ
Not that I think many coilers are using series caps of differing values- I just
like the subject.  If people think that it is irrelevant then I will beg to
differ.....

Tesla List wrote:
>
> Original Poster: "Reinhard Walter Buchner" <rw.buchner-at-verbund-dot-net>
>
> Hi All,
>
> I just wrote to someone privately about caps in
> series and parallel, because he didn´t understand
> the equations for seriesed C. It might well be that
> some beginners on the list are experiencing the
> same problems, so here goes:
>
> Resistors:
> --------------
> If you SERIES two resistors -at- 100 ohm you will get:
> 1.) a total value of 200 ohms Rt = R1+R2
> 2.) a total voltage capability of 2x the single R (Vr1+Vr2)
>
> If you PARALLEL two resistors -at- 100 ohm you will get:
> 1.) a total value of 50 ohms  1/Rt= 1/((1/R1)+(1/R2))
> 2.) a total voltage capability of 1x the single R
>
> Caps:
> --------
> Very easy to remember, is the fact, that caps are the
> exact opposite of resistors, so:
>
> If you PARALLEL two caps (1µF/1000Vdc each) you will get:
> 1.) a total value of 2µF at 1kV  Ct = C1+C2
> 2.) a total voltage capability of 1x the single C, because each C
>      sees the full voltage.
>
> If you SERIES two caps (1µF/1000Vdc each) you WILL get:
> 1.) a total value of 0,5µF 1/Ct = 1/(1/C1)+(1/C2))
> 2.) a total voltage capability of 2x the single C, because the cap
>      effectively acts as a voltage divider (Ohm´s Law again), so
>      each cap will only see 0.5x the total V.
>
> To make it fast and simple (I am too lazy to get into differential
> equations. This is a bi.ch via ASCII. Plus it might be "over the
> heads" of some beginners). A Farad is defined as 1 Ampere-
> second per 1 Volt. The nag with the capacitance is very simple.
> C is defined as q/V. This simply means a given capacitance is
> achieved, when two metal plates are spaced apart with "x" cm
> and while carrying a charge "q" (standard physics definition).
>
> Lets take 3 caps and place them in series. Now we attach a
> battery to one end of C1 and the other pole to C3. A charge of
> +q will be established on the outer pole of C1 and a charge of
> -q will be established on the outer pole of C3. Electrons will
> attracted to the inner plate of C1, leaving the outer plate of C2
> negative.  Similar goes for C3 in our string. Thus (this is the
> important part) each cap aquires the SAME magnitude of
> charge q (charge is equal across all plates). The potential V
> across C1-C3 is the combination of all the potentials across
> each cap: Vt=V1+V2+V3 or rearranging it a little bit, we get:
> V=(q/C1+q/C2+q/C3). Now let´s solve for q, which leads us
> to q = V / (1/C1+1/C2+1/C3). q=CV is a given fact. To explain
> this, we would need to get into Gauss´ Law and lots of
> differential equations, but from this fact (q=CV), we can see
> that 1/Ct = ((1/C1)+(1/C2)+(1/C3)) or that the math.txt  (the guy
> was referring to the RQ docs) is correct in stating the total
> capacitance in a seriesed circuit MUST BE smaller than any
> of the caps in the string.
>
> Charged greets from Germany,
> Reinhard

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