I'm having trouble understanding how Ohm's Law applies to a battery that is being short-circuited. If the terminals are both connected by a low-resistance wire, then they should be at the same potential, since there's no resistor to cause a drop in potential. But the terminals of the battery are certainly not at the same potential, by definition. So what will happen to the voltage and current in this circuit? Will it simply be that the battery maintains its potential difference (say, 9V) and the current goes to infinity (with ~0 resistance)? Would the power then go to infinity also? Or is Ohm's law not valid in this case, as we're no longer operating in the intended regime for the battery? Does the internal resistance of the battery play a role here? Thanks.