Maybe I'm just dense, but I can't figure out why the plate and cathode resistors to the reverb driver shown (left side) are called out to be 5W and 2W respectively.
(heavy dashed line at bottom of the first file uploaded is ground bus, at 0 V)
I didn't make these values, ratings or voltages up -- straight from the original schematic and service manual. If I do the math, based on the resistor values and the voltage drops across them, I get
Plate - P = Vē/R = 40 x 40 / 2200 = 0.727 W (spec'd at 5, or nearly 7 x calculated wattage)
Cathode - P = Vē/R = 4.2 x 4.2 / 330 = 0.054 W (spec'd at 2, or 37 times calculated wattage)
Now, that's at quiescence, and I don't know how much swing, or change in voltages, there is in operation, but those factors still give me pause. I can see that the reverb triode on the left has about 2 x the plate voltage as its brother the recovery triode. And unlike the tremolo circuit, where voltages are given for both trem on and trem off (much higher when off than when on), there's only one voltage state given, and I don't even know which state (though the switch is drawn open -- reverb on -- on Ampeg's schematic too.
If anyone can explain why Ampeg chose those values, I'd appreciate it! Obviously I'm a reverb newbie.