On Mar 25, 10:28 am, Cecil Moore wrote:
Keith Dysart wrote:
Read it as Pr.correction(t) to emphasize that it is not
average power of which I am writing. Then it is not
interference.

That statement makes it obvious that you don't understand
interference. When instantaneous values are being used,
if [V1(t)^2 + V2(t)^2] NOT= [V1(t) + V2(t)]^2, then
interference is present. Did you miss Physics 201?

Let us build a slightly better example that complies with
your "NOT =" expression above.

+-----------------------------------+
| |
Vs1(t) = 141.4cos(wt) \
| = 100 Vrms Rload /
| 50 ohms \
Vs2(t) = 70.7cos(wt) /
| = 50 Vrms |
+-----------------------------------+

Using superposition the contribution from source 1 is
Vload.s1 = 100 Vrms
Iload.s1 = 2 Arms
and from source 2 is
Vload.s2 = 50 Vrms
Iload.s2 = 1 Arms
combining
Vload = 150 Vrms
Iload = 3 Arms

From
Pload = Vload * Iload
= 150 * 3
= 450 Waverage

As can be seen, this example satisfies your requirement
for interference:
[V1(t)^2 + V2(t)^2] NOT= [V1(t) + V2(t)]^2

Computing the imputed powers for the waves from each source
we have
Pload.s1 = 100 * 2
= 200 Waverage
Pload.s2 = 50 * 1
= 50 Waverage
To obtain the power in the load from these imputed powers
we need to use
Pload = Pload.s1 + Pload.s2 + Pload.correction
450 = 200 + 50 + Pload.correction
200 = Pload.correction

From previous analysis
Pcorrection = 2 * sqrt(P1 * P2)cos(theta)
(the cos(theta) term is appropriate here because these
are average powers being used)
Pcorrection = 2 * sqrt(10000) * 1
= 200
as required from above.

So according to your energy analsysis, the power in
the load comes from
the wave from source 1 = 200 W
the wave from source 2 = 50 W
"interference energy" = 200 W
for a total of
450 W as required.

Now if the 200 W from the wave from source 1 and the
50 W from the wave from source 2 represent actual
energy flows, then the "interference energy" must
also be an actual energy flow to satisfy conservation
of energy.

What element provides the energy for this "interference
energy" flow? Note that in this analysis, this energy
is an average energy flow, so it can not be saved during
part of the cycle and returned during another part.

In other posts, you have suggested that this would be
a constructive interference energy and that there would
be an equal destructive interference energy to provide
it. If you still claim this, where is this destructive
interference happening?

....Keith