I
Ian Macmillan
Have a look at my blog - there might be something of interest there:
merlins-mutterings.blogspot.com
All the best
Ian Macmillan
merlins-mutterings.blogspot.com
All the best
Ian Macmillan
Have a look at my blog - there might be something of interest there:
merlins-mutterings.blogspot.com
Phil Allison said:"Ian Macmillan"
** Oh dear - your very first item has a serious math error.
The formula for charge Q on a capacitor is simply:
Q = CV
The formula for the energy E stored in a capacitor is:
E = 0.5CV*2
In the case where a charged cap is connected in parallel with an uncharged
one of the same rating - the resulting voltage of the combination is found
by considering the law of " conservation of charge " - that is the CV of
the combination is the SAME as the CV of the first cap.
So, the combination will have exactly half the V across it - seeing as C
has now doubled.
The mystery in the puzzle is explaining what happened to the missing ENERGY
!!
..... Phil
"Phil Allison"
ENERGY !!
Silly me, confusing symbols mainly. But I did point out that the mystery
energy loss was in the discharge path,
which was the point of the article.
"Ian Macmillan"
Well, it obviously needs revision.Phil Allison said:"Phil Allison"
** Dear me - I must have upset the man.
He's pulled the item ...
.. Phil
What really happens depends on the resistance of the path between thePhil Allison said:"Ian Macmillan" >
** Where is that pointed out ?
What sort of losses are you alleging go on ?
** Fraid it does not have much of one.
Be instructive to connect a scope across the uncharged cap and observe what
really happens when the charged one is first connected. Try a couple of 1uF
film caps.
.... Phil
"Phil Allison"
What really happens depends on the resistance of the path between the
charged and uncharged capacitors. Initially when connected the uncharged
capacitor is at 0 volts, so the current that initially flows is E/R,
asymtoting to zero when the capacitor voltages match. With your 1uF film
caps, there could be an oscillation, particularly if you just shorted them
together.
If you just short the
capacitors together, the resistance is in the connection or internal in
the
caps - its low and the current is high but it works out the same.
I am
unhappy with my piece which was badly done, and I'll re-do it in due
course.
As to the point of it, obviously none for you, but could be relevant for
others.
By the way, your response also has a math error.
** Oh dear Phil, your 2nd formula has a serious math error;
E = 0.5CV^2 NOT E = 0.5CV*2
Dennis said:Good on you Ian. Without guys like yourself willing to stick some interesting stuff up on the net, the net would not be what it is
today.