On Apr 18, 10:48*am, Cecil Moore wrote:
Keith Dysart wrote:
So you are saying that ideal voltage sources work differently
in distributed networks than they do in lumped circuits.

"Work differently" is a loaded expression. Since both
lumped circuit and distributed network models exist,
it is safe to say that the lumped circuit model and
the distributed network model indeed "work differently".
If they didn't "work differently", there would be no
need for both of them to exist.

Your previous error is obvious. You were using the
lumped circuit model on the left side of Rs and
using the distributed network model on the right
side of Rs. If it is necessary to use the distributed
network model for part of the network, then it is
absolutely necessary to be consistent for all of
the network.

When you switched to the lumped circuit model on
the right side of Rs, the energies balanced. When
you switch to the distributed network model on
the left of Rs, the energies will also balance.

The lumped circuit model is a subset of the distributed
network model. See:http://www.ttr.com/corum/andhttp://w...1-MASTER-1.pdf

Here's some quotes: "Lumped circuit theory fails because
it's a *theory* whose presuppositions are inadequate.
Every EE in the world was warned of this in their first
sophomore circuits course. ... Lumped circuit theory
isn't absolute truth, it's only an analytical *theory*.
... Distributed theory encompasses lumped circuits and
always applies."

It is a good thing I checked the original references, otherwise
I would have had to assign Corum and Corum immediately to the
flake bucket where they could join some of the other contendors
on this group. But no, it turns out they have the appropriate
qualifications on all their statements about when it is
appropriate to use a lumped analysis and when it is not. And
when is lumped okay, when the physical dimensions of the
elements can be measured in small fractions of a wavelength.
Nothing new there, most of us know that.

Let me remind you of the circuit at hand:

50 ohms
+----------\/\/\/\/-----------+
+| +|
Vsl=100 VDC Vsr=50 VDC
| |
+-----------------------------+

Firstly, there are no inductors to cause any sort of difficulty.
Secondly, it is constructed of ideal components, which have the
luxury of being infinitely small.
And thirdly, it is a DC circuit so the two points above do not
matter any way.

So we are back to the question you keep dodging...

Where does the energy being absorbed by these ideal voltage
sources go?

0+j0 ohms cannot absorb energy.

Now that is a non-sequitor. The element absorbing energy is an
ideal voltage source, not a resistor.

As Eugene Hecht said:
"If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

It is a DC circuit. This means the instantaneous value is
the average value.

(But poor Hecht, here he is saying power is more useful than
cumulative energy, and you misinterpret him to be comparing
instantaneous to average. And is it the distribution over
the area that is being averaged, or the distribution over
time?)

You seem to have discovered that "limited utility".
You have an instantaneous power calculated over an
infinitesimally small amount of time being dissipated
or stored in an impedance of 0+j0. Compared to that
assertion, the Virgin Birth seems pretty tame. :-)

Again, it is a DC circuit since we have had to go back
to learning the fundamentals of ideal voltage sources.

But back to the simple question...

Where does the energy that is flowing into the ideal
DC voltage source on the right go?

If no energy is flowing into the ideal voltage source
on the right, where is the energy that is being
continuously provided by the ideal voltage source on
left going? Of the 100 W being provided by the ideal
source on the left, only 50 W is being dissipated in
the resistor. Where goes the remaining 50 W, if not
into the ideal voltage source on the right?

...Keith