Re: Multiple Caps: How and Why

To: tesla-at-pupman-dot-com
Subject: Re: Multiple Caps: How and Why
From: Bob James <nrgdude-at-pond-dot-net> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)
Date: Wed, 03 Nov 1999 17:22:08 -0700
Approved: twftesla-at-uswest-dot-net
Delivered-To: fixup-tesla-at-pupman-dot-com-at-fixme
While on the subject of series and parallel caps I would like to hear more
about
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