It has to do with the coupling between the center conductor and the
shield of coax -- when current flows in the center, an equal current
flows in the shield, and we're stuck with the loss of both of those
currents.
The loss in cables are primarily I^2 x R losses in the conductors (at
least at HF) where we tend to use them.
If one creates a 25 Ohm feedline by paralleling two 50 Ohm lines, the
current in *each* of the two parallel lines has to increase by SQRT(2)
in order to maintain the same *total power* (P = I^2 / Z). That means
the current in each cable is now 1.4*I/2.
Now, (1.4*I/2)^2 is 2*I^2/4 or I^2/2. The resistance of each cable is
the same as the single cable, so the LOSS in each cable is (I^2/2)*R.
Sum the loss of each cable and you're back to I^2*R just as it was with
the original single cable!
It's really simply arithmetic, Jim!
73,
... Joe, W4TV
On 8/4/2016 2:07 PM, Jim Brown wrote:
On Thu,8/4/2016 10:50 AM, Hans Hammarquist via TowerTalk wrote:
That's something I don't really understand. Why do you get the same
loss in coax if you run them parallel?
That's something I also had trouble getting my head around too. EMC
expert Henry Ott, who retired many years ago from AT&T Bell Labs, was
they guy who explained it to me. It has to do with the coupling between
the center conductor and the shield of coax -- when current flows in the
center, an equal current flows in the shield, and we're stuck with the
loss of both of those currents.
73, Jim K9YC
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