[TowerTalk] Losses in parallel coax

[Joe Subich, W4TV]

> 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
>
> _______________________________________________
>
>
>
> _______________________________________________
> TowerTalk mailing list
> TowerTalk at contesting.com
> http://lists.contesting.com/mailman/listinfo/towertalk
>