On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)
Keith Dysart wrote:

clip
But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
Esource.50[90..91] = 0.03046 J
Ers.50[90..91] = 0.01523 J
Eline.50[90..91] = 0.01523 J
Esource = Ers + Eline
as expected.
Efor = Eline
since Eref is 0.
I come up with 141.4v across 50 plus 50 ohms. The current should be 1.414a. Power to each 50 ohm resistor would be 50*1.4142^2 = 200w. For the 1 degree interval of 1 second, that would be 200 joules.

Right? Peak current flows at 90 degrees?
--
73, Roger, W7WKB