A capacitor, as another example, can supply extremely high currents (compared to batteries), but they store charge, and are not a charge pump, as a battery is. As such, they're sort of like super-high-speed batteries with extremely limited capacity. Different battery compositions will have different amounts of real-world "impure" limitations. Internal resistance, temperature versus performance characteristics, "memory" and recovery effects, and so on. One of the difficult times I had learning about electronics was doing calculations and then wondering why the physical components on the breadboard were different. The figures on paper say I should measure 9 volts. I'm actually measuring 8.654 volts. What gives? Let's take, for example, a 9 V battery. Forgetting about internal resistance or any temperature restrictions, what is the maximum current I can draw from this? Some batteries are capable of some extremely high current. Consider automotive "wet cell" lead batteries. You'll find that they're capable of 1000 amperes or more, especially for turning over huge engines during start. In electronics and physics, many things are a trade off. If you want super high current, you may have to accept lower voltage, lower battery life, or extremely high cost.

The answer is right there in your question summary. You're overlooking the fact that while in theory, theory and practice are the same, in actual practice, they're not. In physics, similiar to the maximum speed of light, there is a maximum power through a surface of any size; it is c When we consider a battery to be a voltage source, that assumes that you are using it in the regime where that is a good approximation, because an approximation it is. Over a range of discharge currents it will be a good approximation, but at very low or very high currents it will fail. Similarly, when we draw a wire in a schematic we consider it to be zero resistance most of the time, assuming that the voltage drop will be small enough not to matter. A short length of wire might well be only 5 mΩ, but when you connect the battery using only the wire, it doesn't vaporize the wire with a massive surge of almost 2000 amperes. Why? Because the battery is limited by real-world physics. Inductance prevents the curent through the wires to change too fast (e.g. you will have difficulty to empty the theoretical battery in 1ns)Drawing this much current at 9 V would require around 5 milliohms according to my calculations. I know this isn't possible in the real world, but theoretically maybe? It was the biggest eye-opener for me as a kid in school to realize that applying Ohm's law to components was not exactly straightforward. You have to take the physics into consideration, and it's messy. A capacitor isn't just a capacitor: it has some resistance and inductance as well. The best way to think about components and batteries, I think, is that any component is a mixture of a bunch of other components, but imagine a control panel with sliders. A resistor might have its "resistance" slider at a large amount, but the "capacitance" and "inductance" sliders can't be at zero. A wirewound resistor, for example, will have more inductance than say a carbon composition resistor.

Even theoretically ignoring the temperature restrictions and internal resistance, this seems obviously impossible. Where am I going wrong? If you "forget about" internal resistance, then the maximum current is infinite. An "ideal" component, non-existent in the real world, can provide mathematically "pure" infinite or zero amounts of resistance, voltage, current, and all the rest. You can't ignore the internal resistance (actually, the real-world electrochemical processes that we model as resistance), because it directly limits the short-circuit current, and for high-performance batteries it can lead to battery damage (sometimes spectacular) due to overtemperature.