On 201611/14/ 1:58 PM, wrote:
On Monday, November 14, 2016 at 4:46:42 PM UTC-5, Ralph Mowery wrote:
In article ,
says...

On Monday, November 14, 2016 at 4:13:07 PM UTC-5, wrote:
"Two caps in series must be calculated, but keeping it very simple, if two 20uF @ 450V are connected in series, the actual capacitance will be 5uF @ 900V. "

That's not what I "heard". It should be 10 uF.

(C1 x C2) / (C1 + C2) = Working Capacitance is the equation I have used these many years.

Example 1:

10 x 10 = 100
10 + 10 = 20
100 / 20 = 5

Example 2 (from the link):

10 x 22 = 220
10 + 22 = 32
220 / 32 = 6.875

At least where I come from.

Peter Wieck
Melrose Park, PA


If equal value capacitors are in series (same for resistors in parallel)
all you have to do is devide the value of one of them by the number.
That would be 20 devided by 2 for 10.

You can also take the long hard way and do the product over the sum as
in your example. You just need to LOOK AT THE VALUE IN THE EXAMPLE AS
BEING 20 INSTEAD OF THE 10 FOR EACH ONE THAT YOU USED.

For equal values of capacitance the voltage will add, if the voltages
are not equal then there are problems. Resistors are needed across the
capacitors when equal values of capacitors are used. This is mainly due
to the wide variation of capacitance in most of them. Helps to bleed
off the stored charge also.

OK - I have enough stray caps at home to do a lash-up and I have enough very accurate meters to test the actual results of that lash-up. But, I just rebuilt the power-supply of my Dynaco ST-70 and my actual working results were as the equation implied, as I remember.

Nor would I ever run caps of differing voltage in series. Or parallel for that matter.

I will report my results.

Again, here is the link. I am NOT making this up.

https://www.kitronik.co.uk/blog/how-...-and-parallel/

Peter Wieck
Melrose Park, PA


The previous poster was simply pointing out that you made an error in
your math. You had a 10uf and a 20uf cap as your math components based
on the link, but were speaking of two 20uf caps in series in the
preceding text.

20*20=400
20+20=40
400/40 = 10uf

For two identical value caps in series divide be two for the resulting
capacitance. Voltage is added. If there are differences in the
capacitance between the two caps then the formula is needed.

Hey, we all make mistakes, even I've made one or two (a minute)...

John ;-#)#

--
(Please post followups or tech inquiries to the USENET newsgroup)
John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
(604)872-5757 or Fax 872-2010 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."