437V is accounting for the voltage drop [across the generator resistance]. 437V is what is required to push 190A through 2.3 Ohms, some of which is the load and some of which is the generator's resistance. 57V is the voltage you would need not accounting for the voltage drop since If the generator had no resistance you would only need the generator to produce 57V to get 190A through the 0.3 Ohm load.
By the way, you would not actually measure 437V if you took probes to the generator terminals while it is powering the load. You would measure 57V, because the remaining 380V has already been dropped across the gen's resistance is inside the generator by the time it reaches the output terminals.
In your scenario you're losing way more power and voltage inside the generator than is actually getting the load. This is pretty much what happens when you connect a generator that is too small to a load that is too large.
Normally you choose a generator with an internal resistance much less than your load resistance so that the power (and voltage) delivered to the load is the vast majority and that which is lost inside the generator is the minority. Otherwise, it's a bit self-defeating. The example you gave would probably qualify as shorting out the generator since the load resistance is so small relative to the generator's own resistance.