gareth wrote:
Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central
conductor protruding.

Now there's no difficulty in feeding all that power into that
1/4 inch because it is so short compared to a wavelength
that there is a uniformity of voltage and current along it,
and it will be essentially the same as that existing in the last
gnat's cock of the coax.

Attach a hi-impedance scope probe to the end of that
1/4 inch and all the power being delivered through the
coax will be detectable right at the tip of that 1/4 inch.

Now, will that 1/4 inch antenna radiate all the power that
is being successfully fed to it at 3.6MHz, or will the
configuration behave merely as an open-circuit with all
the power being reflected back down the coax?

A number of contributors to this NG claim that the 1/4 inch
stub antenna will radiate the full 100W at 3.6MHz, but
I fear that they are sadly mistaken and that their associated
infantile outbursts are because they are in denial either
about their error, or cannot face up to a challenge to
their seemingly-religious faith as to what is happening.

The resistive impedance of your 1/4 inch stub will be a small fraction
of an Ohm.

Let us assume 0.001 Ohms though it would likely be much smaller than
that and I do not want to spend the effort on getting the actual value
for your nonsense.

Since P=I^2R and power in your scenario is 100 W and the total resistance
is 50.001 Ohms, the current is 1.414 A.

Again since P=I^R and the antenna has a resistive impedance of 0.001 Ohms,
the power in the antenna is 0.002 W.

Do look up impedance matching and why it is used.




--
Jim Pennino