All of this assumes transformer or inductive loading of an SE or a push-pull amplifier. Inductive loading means using an inductor (choke) to load the output device. Both tube and solid-state amplifiers, whether single-ended or push-pull, can be inductively loaded as long as they run in strict Class A. The inductor dissipates virtually no heat, yet provides a current source like functioning for audio frequencies. Furthermore the inductor does not subtract from the available B+ voltage as a current source would. You can readily see how brilliant a decision it was to use a inductive load so as to increase the efficiency of a Class A amplifier. But as very few amplifiers used today are Class A designs, the inductive load has almost been forgotten. (A big mistake.)

   Consequently, we know that the active device, at idle, so as to match the current source, must also see all the voltage and current that the load will see at peak output. Thus, the maximum theoretical efficiency of a optimally designed current source loaded Class A amplifier is half that of the inductive loaded amplifier: only 25%. For example, delivering 2 amps into a speaker requires that current source and tube draw 2 amps at idle. So that when the tube approaches cutoff the current source can provide 2 amps into the load. And the tube must draw peaks of 4 amps so that it can over come 2 amps from the current source and deliver 2 amps into the load impedance. If the B+ voltage is 32 volts, the total dissipation of the amplifier's output stage comes in at 64 watts, which when divided into the 16 watts RMS that is given into the speaker, equals 25%. This example assumes the active devices are themselves perfectly efficient; they aren't.
   With the resistor loaded output stage, the efficiency is even worse. Normally, we are not much bothered by the  low efficiency of a resistor loaded grounded cathode amplifier or cathode follower, as our aim is usually voltage amplification, not power delivery. But consider the woeful efficiency of an amplifier that dissipates 2 watts in the load resistor the tube (100 volts each at 10 mA), yet only delivers 1 mW into a load resistance (10 volts RMS into a 100k resistor). The recommended practice of specifying a plate resistor or cathode resistor one fifth the value of the load resistance means that maximum effective efficiency of that amplifier can only be on fifth of its theoretical maximum.

    Resistor or current source loading, on the other hand, necessarily decreases the efficiency of the output stage, as the resistor or current source parasitically drag on the output stage, whereas, the transformer or inductive load displaces no voltage and gives back what current it takes. Remember a perfect transformer or inductor, like a perfect capacitor, dissipates no energy, as it see no voltage potential across its leads at idle. An inductor stressed by a DC current flow of 4 amps, gives up that 4 amps when unstressed.  The resistor and the current source, on the other hand, do dissipate heat, as they do see both a voltage differential and a current flow at idle. In fact, the current source at idle must see all the voltage and current that the load will see at peak output, so that the current source can pull the load resistance to that peak when the active device nears cutoff.