Cecil Moore wrote:
The following is from an email to which I replied today.

Given two coherent EM waves superposed in a 50 ohm environment
at considerable distance from any source:

Wave#3 = Wave#1 superposed with Wave#2

Wave#1: V = 50v at 0 deg, I = 1.0a at 0 deg, P = 50 joules/sec

yes, lets say source S1 supplies a voltage V1 into a load L1, where L1
is a pure 50 Ohm resistance.


Wave#2: V = 50v at 45 deg, I = 1.0a at 45 deg, P = 50 joules/sec

Source S2 supplies a voltage V2 puts into a load L2, where L2 is a pure
50 Ohm resistance.

These two waves superpose to V = 92.38v and I = 1.85a
Note: P = 171 joules/sec

You have changed the circuit.

Source S1 is no longer connected to a load L1 consisting of a 50 Ohm
load. It is connected to a load L3, consisting of a pure resistance in
series with a voltage source.

Since you have changed the circuit source 1 is connected to, you should
not be surprised it supplies a different power.

Move the phase difference to 180 degrees, and source S1 would supply no
power at all.


*During each second*, Wave#1 supplies 50 joules of energy and
Wave#2 supplies 50 joules of energy for a total of 100 joules of
energy being supplied *every second* to the superposition process.
Yet the results of that superposition process yields 171 joules
of energy *during each second*, 71 joules more than is being
supplied to the process. Where are the extra 71 joules per second
coming from?

35.5 J/s (Watts) comes from the source S1 and 35.5 J/s (Watts) comes
from the source S2. It's different to the first case, as they are
connected to a different circuit.

You can do the same with DC - you don't need to use AC at all. Put a 50
V battery in series with the pure 50 Ohm load and it supplies 50 W. Put
it in series with another load, consisting of a 50 Ohm voltage source in
series with a 50 Ohm load, and it is no surprise it delivers a different
power. Depending on what way you connect the two batteries, the current
would be 0 A or 4 A, and so the power 0 or 200W.