Roy Lewallen wrote:
I think you've put your finger right on Cecil's conceptual problem. When
we solve the battery example by superposition, we get the right answer,
zero current. But now let's put a resistor between the two batteries and
repeat the solution. When we "turn off" the left hand battery, we have a
lot of power being dissipated in that resistor. Using Cecil's view, we
would assign this to be the power associated with the current flowing to
the left. Then we "turn on" the left hand battery and "turn off" the
right hand battery. Again we have a lot of power being dissipated. This
would be the power associated with the current flowing to the right. The
problem comes in having to somehow manipulate these powers to get zero,
which is what we actually see. The mistake, as you continually point
out, is attributing a power to each current -- or each wave -- in the
first place.

Roy, the intensity equation from the field of optics actually
handles that situation perfectly. The batteries are hooked
up in opposition to each other. Mathematically, their voltages
are 180 degrees out of phase.

Let P1 be the power dissipated in the resistor when battery
#1 is on. Let P2 be the power dissipated in the resistor
when battery #2 is on. P1 = P2

When both batteries are on, the total power in the resistor is:

Ptot = P1 + P2 + SQRT(P1*P2)cos(180) = 0

The intensity equation gets the same total power as you do. If
it's wrong then you are wrong. If it's right then you are wrong.
Seems that you are wrong either way. :-)
--
73, Cecil http://www.w5dxp.com