I agree equivalent circuits are invalid when it comes to calculating either
efficiency or internal power loss. The circuit in my example was the ACTUAL
circuit, i.e., a 10 volt source with a series 50 ohm resistor, not a Thevenin
nor a Norton equivalent. With this restriction, I don't see anything wrong with
my statement. Depending on what the ACTUAL circuit is in a power amplifier, I
believe the current drawn can either increase of decrease when the load is removed.

Ron

W5DXP wrote:

Ron wrote:

It helps me understand reflected power to think of a 50 ohm source of
10 volts connected to a half wave lossless line. In this situation the
line can be removed from the equation and the load can be considered
connected directly to the 50 ohm 10 volt source. If the load R is
either a short or open circuit, there will be zero power transferred
to the load, but there will be a big difference in the power
dissipated in the source, two watts with the short and zero watts with
the open.


Not with a Norton source. :-) Quoting _Fields_and_Waves_in_Communication_
Electronics_, by Ramo, Whinnery, & Van Duzer, page 721: "It must be
emphasized, as in any Thevenin equivalent circuit, that the equivalent
circuit was derived to tell what happens in the *LOAD* under different
load conditions, and significance cannot be automatically attached to
a calculation of power loss in the internal impedance of the equivalent
circuit." Seems your above assertion violates that sage admonition.