I know that, if I have a voltage amplifier, the output resistance Rout should be as small as possible: indeed, if the circuit is replaced by its equivalent Thevenin's circuit (1st picture), the voltage on a (generic) load is maximized if Rout goes to zero.
1st question: if I replace the voltage amplifier with Norton's circuit (2nd picture), the output resistance Rout doesn't change, then if Rout goes to zero, I should be able to see the same behaviour of the voltage amplifier: I would expect, indeed, that the more Rout goes to zero, the more the current flows in the load resistance RL, thus increasing the output voltage (that is, a voltage amplifier). But I'm not able to show this: if Rout goes to zero, I get a current source which is shorted by itself, and all the current flows in Rout=0 (and the output voltage is of course 0).
2nd question Is it correct to say that, in general, if the output resistance Rout of a generic circuit is very great, a circuit behaves like a current amplifier (if I put it in Norton's circuit this is evident, if I put it in Thevenin's circuit I don't know how to show this) and if the output resistance is very small, a circuit behaves like a voltage amplifier (if I put it in the Thevenin's circuit this is evident, if I put it in the Norton's circuit I don't know how to to show this)
Thanks!