Reg Edwards wrote:
Or even more simple, for Zo to be purely resistive, G = C*R/L

In "Transmission Lines" by Chipman, he gives an example where
R = 0.1 ohm/m and G = 0.9 micromhos/m. For Z0 to be a purely
resistive 50 ohms, G would have to be 40 micromhos/m making
the transmission line considerably more lossy just to achieve
a purely resistive Z0. Real world transmission lines rarely
have a purely resistive characteristic impedance.

The formula for the attenuation factor is R/2*Z0 + G*Z0/2
That's 0.001 + 0.0000225, so you can see that G has negligible
effect on losses, i.e. virtually all losses in the above
example are series I^2*R losses.

The attenuation factor is 0.0010225 for both the voltage and
current so it's obvious that the current attenuation is caused
by the series I^2*R losses, the same thing that causes the
voltage attenuation.
--
73, Cecil, W5DXP