electric current - What happens in a circuit, when the wire and the battery are superconducting. And shorted

When a wire with no resistance is connected to the terminals of an ideal battery, will a current exist in the circuit? If a capacitor is added to the circuit, will it be charged by the battery or will it remain uncharged?

In all cases there is absolutely no resistance through the circuits.

Answer

To be clear on the setup, we have an ideal battery (DC voltage source) and an ideal wire (zero resistance.

In ideal circuit theory, asking what happens when one connects an ideal wire across an ideal voltage source is essentially asking what happens when $1 = 0$.

However, if we allow that the wire has non-zero radius and non-zero length, then, when the wire is connected to the battery, there is a non-zero associated inductance $L$.

Thus, there will be a current through the circuit formed by the battery and wire and that current will change at a constant rate given by

$$\frac{di}{dt} = \frac{V_{bat}}{L}$$

Now, said inductance is likely to very small and thus, the current will rapidly become enormous so this model is of little physical relevance.

For example, if the ideal battery produces 1V and the circuit inductance is $L = 1nH$, the current 1 second after the wire is connected would be $i = 1GA$ which is roughly 1,000,000 times larger than the electric current associated with lightning.

Any physical battery has an associated short circuit current and, assuming one can manage to keep the battery intact until the stored energy is depleted, the battery will produce the short circuit current through the connected wire.

For example, the short circuit current of a typical 9V battery is roughly $4A$.