Keith Dysart wrote:

Cecil Moore wrote:
1/2WL 50 ohm
SGCL1---1---2--+--lossless line--+--2---1---SGCL2
\ / \ /
3 3
| |
R1 R2

Your zero current "branch" is now 1/2WL long and in
the center of that zero current "branch", the current
is at a maximum value of 0.4 amps for 50 ohm signal
generator voltages of 10 volts as in your original example.

Using the distributed network approach, you have added
an infinite number of branches, and now there are two
branches which are appropriate places to make the cut.

The point is that in the above example, there are absolutely
no reflections. When you cut the line you cause reflections
where none existed before. It is obviously invalid to completely
change the operation of the circuit in that manner. We could
be sending data from SG1 to R2 and from SG2 to R1. Those data
streams stop when you cut the line.

How can the current in the middle of the line be 0.4 amps
when the current at both points '+' is zero? Does that
0.4 amps survive a cut at point '+'?

Absolutely, if the line is lossless. Cut both "+" and the
current in middle of the line still remains.

We both know that is a physical impossibility. Sometimes you
are just forced to accept reality and get on with it.
--
73, Cecil http://www.w5dxp.com