Just as in the analysis of the White Cathode Follower, the first step to understanding a circuit is to redraw it so the output works into a short to ground, i.e. a load impedance of zero ohms. Ideally, the current should flow symmetrically in and out of the output into the ground. In other words, when the grid of the bottom tube is presented with a 1 volt pulse in grid-to-cathode voltage, the ensuing increase in conduction by the bottom tube should be met by the equal decrease conduction from the top tube. Now, a schematic for the bottom tube can be redrawn to show the equivalent circuit at zero grid voltage.

   Immediately we see that the maximum currents do not equal. To restore balance, an additional resistor that equals Rak must be placed in series with the top triode's plate and the B+ power supply. Now the maximum current for the top tube equals that  of the bottom tube. (This is not new, as the plate resistor was part of the 1943 circuit.)
   The next step is to determine the optimal value for Rak for a given load impedance and idle current.

Optimal Rak Value
   Too large a value for resistor Rak unbalances the push-pull aspect of the SRPP circuit, but too little a value not only reduces the  circuit's gain and increases the output impedance, but also unbalances the amplifier. For example, if we reduce Rak to zero ohms, then the top tube effectively becomes a plate resistor equal to rp in value and the circuit loses all its push-pull attributes. So the question is: What would be the optimal value this resistor for a specified load and a desired output voltage swing?
     Once again, for any push-pull tube amplifier to perform well, there must be a symmetrical current delivery into the load impedance. Normally in a push-pull power amplifier, matched tubes and equal drive voltages are necessary. Surprisingly, in the SRPP neither is necessary. In this circuit, if the bottom  triode sees a 1 volt increase in its grid-to-cathode voltage, then the top triode might see a 2 volt decrease in its grid-to-cathode voltage and still symmetrically deliver current into the load impedance.
      To see how this might be, imagine that we have replaced the triodes with devices that do  not have a plate resistance, such as a pentode or a MOSFET. Now the math becomes trivial. The transconductance (Gm) of the top device is the only key variable to defining the value of Rak:
    Rak = 1 / Gm.
Now if a bottom MOSFET is used with a Gm of 1 amp per volt, it little matters if the top

  We see that the maximum potential current flow for the bottom triode equals the one half the B+ voltage divided by the sum of the rp of the tube and the value of resistor Rak:
   V1 Max. Current = Vb / 2(Rak + rp).
The maximum current flow for the top triode under zero grid voltage is equal to the one half the B+ voltage divided by the rp of the tube:
   V2 Max. Current = Vb / 2rp.