On Mar 14, 5:51*pm, Cecil Moore wrote:
Keith Dysart wrote:
On Mar 14, 7:59 am, Cecil Moore wrote:
P(t).reactance = [V(t).reactance][I(t).reactance]
Where is that term in your equations?

It is unnecessary. But if you believe me wrong, show me
where it goes, compute the values, and show how it
accounts for the energy that is not dissipated in the
source resistor.

It is unnecessary to account for all of the instantaneous
power???? Your problem is greater than just a simple
misunderstanding of the laws of physics by which we must
all abide.

The DC energy is stored in your vehicle's battery until
it is needed to start your vehicle. That delay between
stored energy and needed energy is related to the
(undefined) wavelength. Think about it.

In an AC circuit, the reactance has no say as to when
to store the energy and when it is delivered back to the
system. It is also related to wavelength which is defined.

When the source voltage is zero at its zero crossing
point/time, the instantaneous power dissipation in
the source resistor is NOT zero! Doesn't that give
you pause to wonder where the instantaneous power
is coming from when the instantaneous power delivered
by the source is zero????

Previously explained, as you may have forgotten, with
the energy cyclically returned from the line.

For your convenience, I copy from a previous post:
----
Recall that
Pg(t) = 32 + 68cos(2wt)

For some of the cycle, energy flow is from the line
towards the resistor and the voltage source.

But this is not the energy in the reflected wave
which has the function
Pr.g(t) = -18 + cos(2wt)
and only flows in one direction, towards the source.
And it is this supposed energy that can not be
accounted for in the dissipation of the source
resistor.
----

And, as expected
Ps(t) = Prs(t) + Pg(t)

Energy is conserved.

...Keith