Good explanation.

I would like to add that this applies to coaxial cables only because
the fields of the individual lines do not interact.

For parallel lines the situation is quite different due to the
overlapping fields and much more difficult to calculate.

The calculation described rlate to a broad band impedance transformer
that can easliy be made with coax. The coaxes are connected in
parallel on one end and in series on the other. This yields a
transformer from 25 ohms unbalanced to 100 ohms balance in the case of
50 ohm line.

It is necessary to make the impedance high for the currents that will
flow on the shields. There are two ways to accomplish this. One is to
wind the lines onto a ferrite core. The other is to coil the lines to
form an air core inductor. The latter is the best approach for high
power coax which is a bit big to wind on a core. Also it eliminates
any issues of core heating and saturation effects.

73 de K9GXC, Jim

On Wed, 01 Jun 2005 07:50:06 -0500, Cecil Moore
wrote:

Asimov wrote:
This is a hypothetical question, if one needs to send more power down
a line than its capacity can 2 or more lines be paralleled? Is the
equivalent characteristic impedance of the combined lines the same as
that of a single individual line?

Let's say we parallel two pieces of lossless 50 ohm coax such
that the voltages at the load (V1 and V2) are equal magnitude
and phase. The currents at the load (I1 and I2) will be equal
magnitude and phase.

For matched line operation, V1/I1 = V2/I2 = 50 ohms.
Pload = (V1*I1)+(V2*I2) = 2*(V1*I1)
Vload = V1 = V2
Iload = I1 + I2
Rload *for matched line operation* = V1/(I1+I2) = V1/(2*I1).
Rload = 1/2(V1/I1) = 50/2 = 25 ohms. So 25 ohms is the
equivalent Z0 of two pieces of 50 ohm coax in *parallel*.

Conversely, if you use one piece of 50 ohm coax going to the
load and one piece of 50 ohm coax coming back from the load
to achieve a shielded balanced feedline, the Z0 of that
feedline is 2*Z0 = 100 ohms and the two lines are in *series*.