On 8/4/2015 11:19 AM, John S wrote:
On 8/4/2015 2:30 AM, Jeff wrote:

But there is a difference. With the resistor load the current flowing
from the inner lead of the coax *must* (by Kirchoff's 1st law) balance
the current flowing on the shield inner surface. So there is no current
remaining to flow on the shield outer surface. No diagram needed.

For the case of the dipole, the load is no longer balanced with the
shield outer surface connected to one side and there is no requirement
for the current in the two arms to be equal. So in this case current
flows on the shield outer surface in an amount inversely proportional to
the impedance seen by the current in the antenna element and the shield.


No, the dipole is still balanced, just as it would be if it were fed by
twin wire balanced feeder. The power will ALL be applied to the dipole
and, if perfectly matched, will ALL be radiated (or lost as heat).

The only current that will flow on the coax outer is due to the induced
currents from the radiated field, or that caused if the dipole is not
truly balanced.


If the dipole is truly balanced (and the coax is perpendicular to the
antenna), how does current get induced into the coax? My understanding
is that perpendicular wires have little current induction due to this
orientation.

Current does not need to be "induced" in the shield outside. There is
a voltage on the shield at the antenna feed point and a path for the
current on the shield outside. Current flows. Why would the current
prefer the antenna element over the shield outside?

Yes, the antenna has not changed and it is still "ballanced", but the
antenna is not the only load seen by the feed line. The current in the
two antenna elements are not equal because some of the current flows on
the shield having nothing to do with induction from the antenna.

--

Rick