RE: Capacitor - series?

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Original poster: "Philip Chalk" <phil-at-apsecurity-dot-com.au> 


Subject: RE: Capacitor - series?

Original poster: "Randy & Lori" <rburney6-at-comcast-dot-net>

Luke,
I think you can put a number of different sized caps in series and
expect the same voltage to be felt on each.  I explained why in one of
the other replies.


Randy,

I suppose it's tempting to think so, but it's not right.

Firstly, at ac, just pick a frequency, calculate Xc, treat them as two
(or 5) resistors & see what you get.  Better still wire it up & get out
your meter - this you can measure.

The voltage will divide across capacitors in inverse proportion to their
capacitance - i.e. in direct proportion to their reactances, just as if
they were resistors.

It is the same for DC, can be explained by conservation of charge, but
is harder to measure.  Well not really, just use a DMM & uF range caps.
[Time constant = (10M Ohm x 10uF) = 100sec - long enough to measure]

Because the circuit does not see the full
capacitance of any single capacitor in series, you can't use that
individual cap value to figure reactance because that caps true value is
never "felt" by the circuit.  See my other reply.

It matters less in this discussion what the circuit sees than what the
individual capacitors see. In any case whichever way you do the maths
you get the same result.

The two capacitors -are- two discrete components. The electrons flow out
of one & into the other. It makes sense to discuss their parameters
individually, which was the point of the original question.

If the original question's basic intent is something like: "Will one or
both capacitors blow up when used in a TC tank circuit ?" then my answer
would be along the lines of 'Not necessarily with those two examples
(equal C.V products), but often yes.'

Sure the smaller cap governs the charging characteristics & limits the
maximum charge of the combination - when it's charged the current stops
flowing & they both stop charging. So it is more nearly fully charged,
and the larger cap less fully charged (To its capacity, I mean. The
absolute Charge contained in each cap is the same.)

In any given instant or cycle or whatever, the same current has flowed
through each capacitor. So they each hold an identical charge - no of
electrons. But a given amount of charge, if you like, concentrated in a
smaller volume (capacitance) results in a higher pressure (voltage) &
vice-versa.

Q = C.V  ->  V = Q/C    That pretty much says it.


Phil Chalk.


Randy
Savannah, GA



-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Friday, February 06, 2004 10:56 PM
To: tesla-at-pupman-dot-com
Subject: RE: Capacitor - series?


Original poster: "Luke" <Bluu-at-cox-dot-net>

The only way you would get 5KV across both caps at 10KV is if the two
cap values were equal and therefore their reactances equal at the
operating frequency.  If you have unequal capacitance values you cannot
simply divide the total voltage by the number of caps to see what
voltage they will see across them.

Luke Galyan
Bluu-at-cox-dot-net