Roger Sparks wrote:
The equation should apply at all locations on the transmission line.

What you say is true but there is a caveat. Remember
that I have previously said that interference can exist
without wave cancellation. As long as Z0 remains constant
up and down a transmission line, there is no forward and
reflected wave interaction. They pass each other like
"ships in the night".

The resistor is a physical object. When the forward and
reflected waves superpose at the *resistor*, there is
an obvious interaction, i.e., a permanent interference.
One only has to observe the power dissipated in the resistor
to ascertain that it is not the same as the forward power
plus the reflected power unless the E-fields of the two
waves are 90 degrees apart, a condition for which zero
interference exists.

What is interesting is a procedure for determining the
current through Rs using only powers. We have to be
pretty accurate with the powers to do that. We know
the forward power is 25w. The power reflection coefficient
at the load is [(50-1)/(50+1)]^2 = 0.92311. That makes
the reflected power from the load equal to 23.077 watts.
When the forward wave and reflected wave superpose at
the 50 ohm source resistor, they are 180 degrees out of
phase which makes the interference term minus and
therefore destructive.

25w + 23.077w - 2*SQRT(25w*23.077w) = 0.03845w

That's the dissipation in the 50 ohm source resistor.
So the RMS current is SQRT(0.03845w/50ohm)= 0.02773a

Ig(t) = 1.414(0.02773)cos(wt) = 0.039216 cos(wt)

And that is, indeed, the current through the 50 ohm
source resistor, Rs, using only an energy analysis.

All energy, except 0.03845w, is "redistributed", i.e.
reflected back toward the load. This is all explained
in my energy analysis article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com