Generator burnout

It should have said safely provide.

But also, there's often a confusion between drawing power and drawing current.

For example, you can get the maximum power out of any voltage source if your load has exactly the resistance of the internal resistance of that source.

You can of course also plug in a much, much lower (practically, a "short") resistance load. The power flowing into that load will then be lower than the maximum power of the source, but the current will be much higher than what's flowing at the max power point (at the expense of the voltage dropping very significantly).

As an example, take an AA battery. (THOUGHT experiment. Don't do this at home, kids.) Let's say that has an internal resistance of 0.2 Ω. Now, when you take a 0.2 Ω resistor (which would probably be something like a piece of thin steel wire), that resistor would get very hot – the maximum amount of power the battery can supply gets converted to heat.

Next, really short the battery with a thick piece of copper and very good contacts (that's where you usually get into trouble - 0.2Ω isn't all that much, and making a contact way better than that is hard) and short-circuit your battery. What will now get hot?

Right, the battery, and not so much the copper wire. The power converted to heat in the copper wire isn't very high, and most of the power is "wasted" across the internal resistance of the battery.

I'd recommend not shorting batteries, by the way. It's dangerous. They can get extremely hot, depending on the type catch fire, or even explode.