Hi Declan,

Please choose a fixed font as Courier to view the following text.

> Imagine a number of capacitors with one plate in common
> 
>                           | To +DCV
> _______________|_________________________
> ____     ____     ____     ____     ____     ____
>   |            |           |            |           |            |
> 
> 

In trying to answer I have some handicaps:
- Your draft is got messy because the symbol pitch is not right, even 
when I use a fixed font.
- You did not draft the resistors
- It is not sure that the pulling down of the capacitors is through 
the footresister or directly from the downside of the condensor?

I take the following assumptions:
- 2 condensors are pulled down directly from the undersides of the 
condensors.
- 2 condensors are pulled up directly from the uppersides of the 
condensors to DC.

> Each one with a resistor in series (including a small resistor in 
> series with the common). If I pull 2 capacitors down and two up (i.e. 
> short them) at the same time,  then the two being charged seem to 
> charge a lot faster. Is ther any good reason why this is so, and can it 
> be calculated? The capacitors are large, and most are equally sized.

The charging can be calculated in detail. The clue is the theorem of 
Thevenin. By the connections made in the described situation you 
create the following circuit:

           Rtop
           ____
      ----|____|----------------------
     |             |                  |
     |            | | 2 R's //        | 2 C's // 
     = Your DC    | | downside from   = to be charged
     |            | | shortcut C's    |
     |             |  =Rfoot          |
     |_____________|__________________|


Thevenin states that this circuit is equivalent with the following 
circuit:

          Rfoot//Rtop
           ____
      ----|____|----------------------
     |                                |
     |    Rfoot                       |
     = ----------*DC                  = Same C as above
     | Rfoot+Rtop                     |
     |                                |
     |________________________________|
     
So what you can conclude is that
- the voltage with which the capacitor is charged is lower than the 
original voltage   
- the RC-time is C*(Rfoot//Rtop), so lower than that of a single C 
which sees a single Rfoot in series with Rtop.

Where you stated that the top resistor is small (I assume, with 
respect to Rfoot) the influence of Rfoot can be negligible. 

> 
> For the curious, the model above is based on a device being researched 
> locally and is stripped of any unnecessary complications.

Are you sure the stripping did not have any influence on the 
functioning of the circuit?

Regards, Harry 
-- 
Author: H.C. Croon
  INET: [EMAIL PROTECTED]

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