Richard Clark wrote:
"What is the source resistance of any power amplifier?"

Most power amplifiers are designed for a specified load but include
controls to adjust their outputs, both resistance and reactance, so that
a load which deviates can be made to match the 50-ohm or 600-ohm load
specified for the transmitter.

Am broadcast stations in the USA have a point at the transmitter output
just before power branches out to antenna system elements called the
"common point". Impedance here has been adjusted to a specified
resistance and a thermocouple ammeter is routinely installed which is
used to determine that the specified power output is being delivered by
the station by I squared R according to the FCC allocation. There is a
multiplier specified of 0.92 for stations of 5 KW or less. For stations
more powerful than 5 KW the multiplier is 0.947. This multiplier allows
for a slight increase in power to offset misc. errors and losses.

John E. Cunningham writes in "The Complete Broadcast Antenna Book" on
page 51:
"Suppose for example, we have a transmitter that is designed to work
into a 50-ohm load at an efficiency of 70% (Fig. 1-26). The efficiency
of this circuit expressed as a decimal, is given by

eff = RL / RS + RL

Rearranging

RS = RL [1/eff - 1]

Substituting the numbers inti this, we get

RS = 50 [1/0.7 - 1] = 50(1/0,7 - 1) = 50(0.43) = 21.4 ohms

You have an example in numbers. Hope you are happy.

Best regards, Richard Harrison, KB5WZI