On Fri, 01 May 2009 18:00:40 -0700, Roy Lewallen
wrote:

The resistance is transformed in the ratio of N^2:1, which is 400:1 for
a 20:1 turns ratio. So the insertion resistance is 100/400 = 0.25 ohm,

Hi Roy,

Yes, quite so.

not 5. I'm confident it works as Tom claims.

It works as it claims is obvious, but not for SWR (neither claimed,
nor something you show interest in) - not that it should, but could.
We aren't going there either (few would as such an application has no
obvious merit). Pursuing the topic of reported phase does not appear
to be productive. A curiousity all around.

If one were to claim to have made ANY current measurements, and then
wholly ignore the contribution of errors, then the discussion of phase
in a system such as a 5 Ohm radiator with a 5 Ohm instrument loss is
going to be absurd. Such are the fruits of poor reporting.

No, this is the fruit of poor reporting.

No, that you caught me in a detail with that detail mishandled by me.
Complements! The detail was quite accurate; so much so that you
accepted it (it was also sourced by Tom) in your reply. That is the
value of good reporting. The net result is a very short transaction
of point-counter point that doesn't demand 300 posts of careful
parsing. My fault was poor application, certainly. You probably
stole Art and Cecil's delight at catching me there. ;-0

73's
Richard Clark, KB7QHC