Re: Capacitor - series?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: Mddeming-at-aol-dot-com 


Hi Phil,  (see interspersed comments)

    In a message dated 2/8/04 2:25:06 PM Eastern Standard Time, 
tesla-at-pupman-dot-com writes:

SNIPPPP

 >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.

Yeah! Scientific method!  Unfortunately, far too many people are willing to 
jump right to opinion without this step.

 >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.
Correct.
 >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.

Yes!!

 >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.)
Yes!!

 >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.

  Of course, you are right. You obviously have a sound understanding of 
first principles, but it is amazing how much mis-information one can find, 
even on this list. What frightens me is the number of Newbies and 
Not-so-newbie "self-ordained experts" who are actively working with high 
voltages and advising others, with only a minimal (or less)  understanding 
of basic principles of electricity. This capacitor question is only one 
small example of the larger problem.

There was a time when cost of publication and peer review acted as B.S. 
filters, but now anyone with an opinion and the price of a keyboard can 
"fertilize" the whole Internet in a matter of minutes. While I am in no way 
in favor of censorship of the list, it would be nice, (but practically 
impossible) if statements that are blatantly contrary to all known physical 
laws could be annotated as such before being rebroadcast. Oh well, I guess 
our after-the-fact reviews are the best that can be done. I just hope that 
newbies can figure out who is "blowing smoke" before they blow their coils, 
their instruments, or themselves up in smoke.

Matt D.