Since E/I is simply a ratio, R is also a ratio. And we know that a ratio
cannot dissipate power, or turn
electrical energy into heat, thus the output resistance R is non-dissipative.
I have made many measurements
that prove this.

Walt, W2DU

Hello Walter,

thanks for your explanation.

I remember having read your excellent articles on QST magazine many years ago,
in which you also explained, among many other things, why reflected power cannot
dissipate in the final stage of a transmitter.

I am not at all opposing your explanation of "non disspative resistance" (on
the other hand how may I contradict a person named Maxwell, hi), but I have some
difficulties to appreciate it.

In my understanding resistance just means that current and voltage are in phase.

There are two possibilities for this to occur:

1) dissipative resistance. Example is a pure resistor, in which power is
converted into heat.
2) non-dissipative resistance: Example is a DC motor, that converts electrical
power into mechanical power. An ideal motor would convert all absorbed power
into mechanical power, producing no heat

But I am unable to see how the second case could be fitted into the transmitter
model.

Thanks and 73

Tony I0JX - Rome, Italy