[Prev][Next][Index][Thread]
RE: 50%
I've been watching this fascinating discussion of charge, energy,
capacitors, dielectrics, and philosophy for a while. After
this last exchange, I thought I would toss in my two cents,
and see if I can add to the confusion!!
> From: Malcolm Watts <MALCOLM-at-directorate.wnp.ac.nz>
> Subject: RE: 50%
>
> I think there might be some possible confusion between
> conservation of charge and conservation of energy in your reply.....
>
> > What are you talikng about? Some specific circuit I presume. Your
> > For instance,
< Classic two-capacitor problem snipped C>
> > This is close to 100% transmission of 50% of the charge. The
> > remaining charge stayed on the first capacitor. The losses to the
> > **circuit** resistive and magnetic are tiny. Most of the energy is
> > still stored *in* the two capacitors.
> >
> > Fred W. Bach , Operations Group | Internet: music-at-triumf.ca
>
> If the circuit settles down to a steady state (no oscillation),
> 50% of the energy is indeed lost, but charge is indeed conserved.
> Elaborating: E=0.5CV^2 You can see that a cap at half its original
> voltage has only 1/4 of its original energy. Two identical caps at
> half the original voltage of one has still only half the original
> energy.
> BUT, since Q=CV, you can see that if you double C and halve V,
> Q (charge) remains conserved.
>
> Malcolm
Perfectly correct, Malcolm, although I sympathize with those
who don't see this non-intuitive result! Here is a mechanical
analogy that MAY help explain it:
Fill a cylinderical tank, of say 10 gallons, with water. Place
another, identical, but EMPTY container next to it, on the same
plane. I think we can all agree that 1) there is a certain,
fixed amount of liquid in the "system", and that 2) there is
a certain AVAILABLE potential energy, derived from the water's
measurable elevation; it could be let out of the container
through a turbine, and could do work.
Now, open a valve connecting the FULL cointainer with the EMPTY
one, at the bottom. Water will flow from the first tank into
the second, sloshing back and forth a bit, but eventually
settling down to a steady state. What do you see now? Pretty
clearly, you will have two identical tanks, each containing
5 gallons of water, each one half as deep as the original,
full tank. OK so far?
Clearly, no water (=CHARGE) has been lost. Cetainly VERY little
energy was given up to resistive losses - the water will NOT
be measurably warmer! But, just how much potential energy
is available to do work? You will find that you have the same
mass, but at ONE HALF THE ELEVATION. Where did the energy GO?
It didn't go anywhere, except that it is no longer available
to you, because it's spread out at a lower potential (=VOLTAGE).
Should this discussion, and we ourselves, live long enough,
we would see the same thing happen to the universe as a
whole. All the heat energy will be spread throughout the
volume of space, with all the matter at a uniform few degrees
Kelvin. This is called the "heat death" of the universe.
All the mass/energy we started with will still be there,
but, with no DIFFERENCE in energy level from one point
to another, no WORK can be done.
Dave