Now I am really confused.. enough of the facile quick stuff.. time for the
rigorous analysis.. stay tuned, but more facile analysis below
At 12:55 PM 7/29/2003 -0400, Chuck Counselman wrote:
>At 9:34 AM -0700 7/29/03, Jim Lux wrote:
>>...Looking at loss as a dB/foot for the coax, you divide the power into
>>two pieces of coax, so the absolute loss (in watts) will be half in each
>>piece of coax, but you've got two coaxes, so the total loss is exactly
>>the same....
>
>
>No. To deliver the _same_power_ as a single coaxial line, two coaxial
>lines operating in "push-pull" deliver half the current at twice the voltage.
But half the power flows through each line, no? and, half the power isn't
half the current.
> At HF, virtually all of the loss is ohmic, and the power dissipated per
> unit length of conductor is equal to I^2*R', where I is the current in
> the conductor, I^2 is I squared, and R' is the resistance per unit length
> of conductor.
Yup..
> With half the current, the power dissipated per unit length of conductor
> is quartered;
Ok.. but, is the current halved, or 0.707'd... P/Z0 = I^2... current would
go as the square root of the power.
>the length of conductor is doubled; so the total power dissipated in the
>pair of push-pull coaxial lines is one-half that dissipated in a single
>coaxial line carrying the _same_power_.
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