Jim Kelley wrote:
Or maybe it was just too hard for you.

It was impossible for anyone to solve without having
a math model for the sources.

How about a simpler problem then:

This is not simpler - it is contradictory. It provides a
constant power source with nowhere for the source power
to go during steady-state. If you change the source to
a Thevenin equivalent then it can be solved.

XMTR--X--one second long lossless feedline---infinite ohms
PS=100W

Please explain how the XMTR can supply 100W during steady-
state? Where does the 100 joules/sec go? As stated, this
problem, like your other one, is impossible to resolve.

If you change the source to a Thevenin equivalent 141.42V
and 50 ohm source impedance and specify Z=50 ohms feedline
impedance then the problem becomes solvable. The answer is
that 200 joules exist in the feedline and, contrary to what
W7EL states, will dissipate in the source resistance after
the source voltage goes to zero. Note the source resistance
is the only resistance in the entire circuit.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
If you change the source to a Thevenin equivalent 141.42V
and 50 ohm source impedance and specify Z=50 ohms feedline
impedance then the problem becomes solvable.

Going with this configuration which is similar to the 1/2
WL stub suggested by W7EL in his Food For Thought #1:

SW
Source nc c Pfor=100W-- --Pref=100W
141.4V---o---o---one second long lossless 50 ohm stub--open
50 ohm
o---/\/\/\/\/\--Gnd
no 50 ohms

For the first two seconds after power up, the Thevenin
source delivers 200 joules into the stub while dissipating
200 joules as heat in the source resistance. Those 200
joules of energy in the stub must be conserved.

During steady-state, assuming the stub is an exact integral
number of wavelengths, the source will see an open circuit
and be delivering zero power. A wattmeter calibrated for
50 ohms will read 100 watts forward power and 100 watts
reflected power during steady-state.

After steady-state is reached, we throw the switch and connect
the stub to a 50 ohm dummy load. 100 watts will be supplied
to the dummy load for two seconds for a total of 200 joules.

Who was it who said that reflected power cannot be
recovered? (Food For Thought #1 on www.eznec.com).
--
73, Cecil http://www.qsl.net/w5dxp