On Jan 15, 2:24*am, Roy Lewallen wrote:
Keith Dysart wrote:

You don't need Poynting vectors to realize that when
the instantaneous power is always 0, no energy is flowing.
And when the instantaneous power is always 0, it is
unnecessary to integrate and average to compute the
net energy flow, because no energy is flowing at all.

And if by your response you really do mean that energy
can be flowing when the instantaneous power is always 0,
please be direct and say so.

But then you will have to come up with a new definition
of instantaneous power for it can not be that it is
the rate of energy transfer if energy is flowing when
the instantaneous power is zero.

The little program I wrote shows that, on the line being analyzed, the
energy is changing -- moving -- on both sides of a point of zero power.
Energy is flowing into that point from both directions at equal rates,
then flowing out at equal rates. This causes the energy at that point to
increase and decrease. What zero power at a given point means is that
there is no *net* energy moving in either direction past that point.

"*net* energy moving" seems to be a bit of a dangerous notion.

If "*net* energy moving" is the time averaged power, then
it is zero at *every* point on the line under consideration.
And I do not mind this definition.

But at the points where the current or voltage is always
zero, it seems to me unnecessary to use the qualifier "*net*"
since the power IS always zero [from p(t)=v(t)*i(t)]. That
is, unless you are introducing another interpretation of
"*net*".

...Keith