Jim, as I read the arguement, (and these guys are RIGHT), you are confusing
current being constant, in a media, with POWER being constant in a media
(minus insertion loss's, of corse). for example, visualize a 1/4 wave (or
a 1/2 wave dipole) , at the feedpoint, what is the current? even at a 1:1
swr, in coax, the current is constant, as you predict! But, WHAT, PREY, is
the current, at the far END (S) of that 1/4 wave (DIPOLE) ? answer is 0 !
but the voltage has increased to that necessary to equal the amount of power
impressed on it (I realize that any number times, 0 = 0, but, obviously,
there is a physical limit approaching this, and also, for the power to stay
the same, you would have an INFINANT voltage!! This is the stuff that R.F.
BURNS are made of! And, when SWR is measured in a cable, current (and
voltage) will vary, depending upon WHERE , in the cable that you measure
current, (or voltage, even tho, the POWER must stay the same, again
considering losses in the transmission line: As information, another Jim,
NN7K


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Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:
Yep, and for the same power level, a higher impedance usually means
a lower current and vice versa.

But the impedance *at* such points does not affect the current *at*
those points?

Cause and effect, Jim. Hint: The impedance equals v/i and is completely
virtual, i.e. clearly a result, not a cause.

Little v over little i, Cecil? Are we talking instantaneous impedances
now also? ;-)

73, Jim AC6XG