B
Bob Myers
- Jan 1, 1970
- 0
Don Bowey said:
In a sense, though, there's no real difference between
the two from the "dielectric barrier" angle. To examine
this further, let's for the moment ignore the "resistor"
question and just look at the difference between AC
conduction through a capacitor vs. a plain conductor.
Electrical energy passes through any "conductive" path by
virtue of the interaction of the fields of charged particles.
In a conductor, this occurs at the atomic/subatomic
level; in a capacitor, the interaction-through-fields
clearly happens on a physically gross level, through the
dielectric, and individual charge carriers cannot pass
through the dieletric. But that really doesn't matter -
to pass electrical energy or an electrical signal, it is
the motion of carriers "downstream," induced by a
similar motion "upstream," that matters - not that a
particular carrier physically passes through the entire
length of the conductive path.
In the case of AC, what goes on within a conductor
in terms of the charge carrier motion is very interesting.
Imagine a perfect conductor as being a frictionless pipe
filled with ping-pong balls which just fit inside the pipe.
If you push in a ball at one end, a ball pops out the other
end (with the time between these two events governed
by the physical attributes - elasticity and such - of the
balls). There has been a transfer of energy, even though
the ball you put in at the one end really didn't get very
far and is certainly not the same ball that popped out
at the far end. If we model AC this way, then you take
one ball at the near end and alternately push it in and
somehow suck it out AT THAT END. And at the far
end, we have a ball which is behaving in exactly the same
way, alternately popping out and being sucked back in.
We can again transfer energy (or information) through this
process, even though the ball we're pushing/pulling on
at the near end NEVER makes it beyond that point!
Going back to actual electricity, let's further note that if we
have a capacitor of sufficient size, there is no way at all to
distinguish the capacitor-in-series case from the "straight
conductor" case, if all we have to look at is the situation
"downstream" of the capacitor. The only way the two
cases could be distinguished in any event is through the
capacitor's effect on the phase relationship between
current and voltage, which, for a sufficiently large
capacitance, becomes negligible. (The only other means
you could use to distinguish these cases at all, given
access to any information you want, would be to somehow
"tag" individual electrons at the "upstream" side, and then
wait a sufficiently long time on the "downstream" side to
see if those particular carriers are coming through. But
you'd have to wait a very long time to be certain...)
Another way to say this is that an infinite capacitance is
indistinguishable from a "short circuit" (a "perfect"
conductive path), again unless you have the ability to
tag individual charge carriers. This makes sense because
a truly infinite capacitance would always have the ability
to make the corresponding change in charge on the
"downstream" plate as ANY amount of charge enters
or leaves the "upstream" plate. You can envision an
"infinite" capacitance as either possessing plates of
infinite area (if you can ignore concerns re the propagation
times across the plate itself) or (possibly better) as
having an infinitely thin dielectric (which equates to
saying we have a zero-thickness "magic barrier"
inserted between two conductors, such that individual
carriers cannot pass through but still have an effect
on the carriers on the other side, as if the barrier were
not there).
In the real world, of course, we can't have infinite capacitances
(or at the very least, you can't easily go down to Radio
Shack and buy one...
), so we have to rely on thefrequency of the AC, relative to the capacitance, to
make the effects of the capacitor effectively "drop out"
of the circuit. In more familiar terms, we would say that
for a sufficiently low capacitive reactance has no
significant effect when inserted in series into an AC
circuit; there is no way to discern its presence by looking
at the conditions "downstream" of the capacitor. In any
practical sense of the words, then, we would have to say
that yes, a capacitor DOES "pass AC."
Bringing resistance into the picture, as opposed to a perfect
conductor, is only relevant if we want to compare the effects
of resistance to a comparable capacitive reactance. And
in this case, we would fall back to the fundamental difference
between these two forms of impedance: resistances dissipate
energy, while reactances merely store it and return it to the
circuit later in time (which results in the voltage/current phase
effects in a reactive circuit). But there's still nothing going on
here that would cause us to say that the resistor is actually
"passing AC," while the capacitor (reactance) is not.
Bob M.
