Since IR losses dominate at these frequencies, reducing current reduces loss.
The higher Z means more voltage and less current for the same power. Loss
will be 1/36th, assuming all the conductor sizes are the same.

On another thread I was arguing that, if I am correct, the current ratio should
be the square root of 6, not 6. Accordingly the loss would be 1/6, instead of
1/36.

73

Tony I0JX