On Sat, 24 Jul 2010 20:18:53 -0500, flipper wrote:

On Sat, 24 Jul 2010 15:29:45 -0400, ehsjr
wrote:

flipper wrote:
On Fri, 23 Jul 2010 00:12:50 -0400, ehsjr
wrote:


flipper wrote:

On Thu, 22 Jul 2010 13:43:51 -0400, ehsjr
wrote:



Jim Thompson wrote:


On Thu, 22 Jul 2010 01:40:54 -0500, flipper wrote:




On Wed, 21 Jul 2010 22:45:33 -0400, ehsjr
wrote:


[snip]



Well, I don't see any inconsistancy in putting "stuff" in at
one rate, and getting that "stuff" out at another, whether it's
water into bottles, cookies into bags, electrons into conductors,
whatever. You get the same amount of "stuff" out as you put in
(assuming you empty the container), but the time taken to do it
is different.

That only applies if your method of getting and putting is lossless
but the general case in electronics is 100%.




Ed



"stuff" is the source of the error.

...Jim Thompson

Exactly. A coulomb is massless. It is not a fixed physical thing


An electrostatic field is massless. Are you trying to suggest an
electrostatic field is not 'real'?



that occupies some particular volume,


In reality it does.

Nope. I guess I didn't make my point clear.
You can charge 2 different size (physical volume)
caps with a coulomb. There is no particular volume
associated with "a coulomb".


I don't understand what you are trying to get at.

Are you of the opinion that there is one particular volume,
and one particular volume only, that a 1 coulomb charge
will occupy? If so, what is that volume?
If not, then you and I have no disagreement on the point.

Simply repeating the same thing doesn't explain it any more than the
first time did.

Changing the area/spacing ratios would, indeed, change the capacitance
but it's irrelevant as the physical parameters are subsumed into the
given capacitances and further irrelevant as none of the parameters
change anyway.

To go further. A coulomb is a numerical result of a mathematical
operation, not a physical object. There is no physical object
that is an "ampere second". A coulomb is massless, and volumeless.
The puzzle misdirection appeared to be treating the coulomb as an
object that could be divided by dividing that coulomb's container.

It can.

As you yourself pointed out, the story might as well begin with two 2s
in parallel and skip the fluff of 'splitting' the 4 in half as the
result is the same and nothing mysterious happens. You have two 2F
capacitors each with a 'charge' (Q) of 1 coulomb.

And as I pointed out, the entire parallel step is entirely superfluous
as you have the same supposed 'mystery' if you just look at two 2's in
series. Each have 1 coulomb and the total has one coulomb.

Of course, as I also explained, there is no 'mystery' because coulombs
do not add in series. They each and all have the same Q as the whole
because the exact same current goes through each and all as the whole.
I.E. instead of the 'mystery' posited by the diversion 1 + 1 = 1, "huh
where's the missing coulomb?" the correct answer is 1 = 1 = 1, just as
it should.

Area and spacing, which define a volume, the dielectric of the volume
and the applied volts determine the coulombs. It's not an 'arbitrary'
thing and, in that sense, yes it is 'fixed' (by the particulars).
Otherwise one would not be able to write an equation for it.

It's even more the case when simply alternating between parallel or
series connections as none of their parameters change, not even V
(hence Q), and that's about as 'fixed' as one can get.


which seems to be the
misconception at the heart of the so called puzzle.


If you are referring to the, so called, 'missing coulomb' there isn't
one.

Correct, there is no missing coulomb. I've already said that,
earlier in the discussion.


Ok. So I threaded back and, unless I misunderstood, you seem to be
suggesting there is some unexplained 'difference' between whatever you
deem a 'real' 1F capacitor and two 2s in series, despite their
identical behavior, so that two 2s in series are "not a 1F cap" and 1F
"doesn't matter."

Not what I was suggesting.

Ok. Then what did the two 2s are "not a 1F cap" and 1F
"doesn't matter" mean?

In the puzzle the 2 coulombs is a number that tells you how much
current got stuffed into a container and how long the "stuffing"
took. Magically dividing the container into two containers
(or N containers) does not change the amount of current got
stuffed in, nor the amount of time it took to do that stuffing.

Exactly. So what is the machination on "volume" about?


There's no need for mysterious 'differences' as, assuming all else is
equal, a single 1F, two 2s in series, four 4s in series, eight 8s in
series, and so on (although equal values are not required, just that
the resulting series comes to 1F), are black box indistinguishable. At
1V overall they'll each and every one have 1 coulomb because they all
have the exact same charging current go through.


The puzzle's magic division creates the "mysterious differences"
(it results in Vcap = V/N for each cap),

Not so. Capacitors in parallel have the same V and still do when then
'disconnected' from each other (since they're 'ideal' with no
leakage)..

hiding the fact that
total ampere seconds is the same.

There's nothing 'hidden' about 2 parallel coulombs divided into two
being 1 coulomb each.

Albeit superfluous the parallel step is entirely accurate. You begin
with .5V on 4F for 2 coulombs and you end up with two 2s each with .5V
and 1 coulomb.

'Volume', dielectric, and all other physical parameters are irrelevant
as they are subsumed into the 'given' capacitances. It doesn't matter
how they are made as 2F is 2F. In parallel it's 4F and in series it's
1F. That's it.

You may have missed the context of the whole starting from
the beginning, when a "puzzle" (which wasn't a puzzle at all) was
presented.

Paul Hovnanian started the thread with a proposed 'solution' to the
'cap charging a cap' puzzle and mine was the first reply. All I did
was point out that the 'solution' was equivalent to the oft posited
voltage source charging an 'ideal' cap 'puzzle' and that substituting
a cap for the voltage source simply makes the puzzle appear more
mysterious than it is. (A sub thread then ensued about R=0)

You introduced the 'parallel to series' puzzle and I think I am the
only one in this thread, so far, to explain it.

And I don't think it matters - the puzzle was just "mental
gymnastics" that caused confusion instead of making a point, in my
opinion.

The purpose of a 'puzzle' is to posit what appears to be a conundrum
for the reader to then figure out.

The 'mystery' of the original puzzle is in leaving series R out of the
model and the 'mystery' of the second is being diverted into thinking
coulombs should 'add' when, in series, they do not.

Your paragraph above shows you understand the principle,
so the puzzle is irrelevant. And the op clarified his meaning:
"My point all along is that, in actual circuit design, the generalism
"charge is conserved" is dangerous, given that "charge" is
ampere-seconds that you can actually measure and use." His "puzzle"
was misdirection to show that a coulomb was somehow lost. No
coulomb was lost.

That was John Larkin and he was neither the OP nor did he posit either
of the 'puzzles'.

His first entry was to me and I did not respond because he hacked my
words to imply I said things I did not and changed the model to boot
so it wasn't even relevant to the discussion.


Perhaps what I've posted isn't clear to you, but if you
think that no coulomb was lost in the puzzle, and that a
coulomb of charge does not require one and only one specific
volume, then we're in agreement.

You still seem to be positing some 'mysterious' thing about 'volume'
that has nothing to do with solving the puzzle. Maybe if you put it to
numbers I could gather what you mean.

Here's my math

Q(Coulombs) = CV, E(Joules) = .5QV (or, by substitution, .5CV^2)

Each: 2F, .5V, 1C, .25J
Parallel: 2F + 2F = 4F, .5V = .5V = .5V, 1C + 1C = 2C, .5J
Series: 1/(1/2F + 1/2F) = 1F, .5V + .5V = 1V, 1C = 1C = 1C, .5J

Note that 'Q per volume' (whatever it might be) of the 2F caps remains
'fixed' because neither Q nor 'volume' change and neither does V, or
anything else. Or at least we can chose to 'pick' that simpler case
because 'volume' is not mentioned anywhere since, being subsumed into
F, it's completely irrelevant.

Energy is conserved, 'charge' (Q) is conserved, and all is right with
the world.

There is no 'puzzle' unless one is fooled by the riddle into thinking
series coulombs should add.

Ed


Ed

snip

Unless your business is located in San Fransicko, Californica

...Jim Thompson
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