I just read one of Cebik's lessons on common mode currents and am
looking at a diagram he posts. The diagram shows I1 as the RF current
on the center conductor, I2 as RF current on the inside of the braid
and I3 as commond mode current on the outside of the braid.

Since I have not seen my query posted anywhere, I am a bit reluctant to
ask because the answer is probably very obvious. I hope I don't get
chided that I an an Extra, asking a simple, fundamental question. I
have studied hard most every day but without the benefit of an
electrical/electronics/engineering background, it is going to take the
rest of my lifetime to acquire all that I am curious about.

This particular bit of confusion comes from knowing how I attach a
PL-259 to the braid of coax. The OUTSIDE surface of the braid fits
flush to the PL-259 shell, not the INSIDE surface of the braid where I2
is portrayed to be. Subsequently, the shell of the PL-259 gets coupled
to one side of the antenna. I understand the potential for I3. And, I
reasoned that the RF current could be initially placed on the inside
surface of the braid at the transmitter end of the coax run. But then,
back at the antenna end of the coax and PL-259 connector, I only see a
dead end for the I2 current. I added a Balun component to my thinking
about the issue, but the confusion did not go away.

I just cannot understand how I2 gets coupled to the dipole antenna
element since the coax gets coupled at the PL-259 via the outside
surface of the braid.

k1drw wrote:
I just cannot understand how I2 gets coupled to the dipole antenna
element since the coax gets coupled at the PL-259 via the outside
surface of the braid.

It is the physics of the coaxial cable that causes I1
and I2 to be equal in magnitude and opposite in phase.
The internal coax fields force that condition. At the
antenna, I1 has only one choice of paths - into the
antenna (if matched). But I2 has two choices - into
the antenna (if matched) or down the outside of the
coax. It will divide at that point according to Ohm's
law most of it taking the path of least impedance.
A choke increases the impedance down the outside of
the coax making the antenna impedance more attractive
to the current.
--
73, Cecil http://www.w5dxp.com

Cecil:

Thank you for the quick reply and clear explanation. I now
understand the coupling of rf current from inside the braid to
potentially, depending on impedance, both paths.

Sincere thanks..!

You might get a little more insight from
http://eznec.com/Amateur/Articles/Baluns.pdf.

Roy Lewallen, W7EL