# 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.

• First answer this: what happens when an irresistible force meets an immovable object? – garyp Apr 17 '14 at 15:50
• Related: physics.stackexchange.com/q/1060 – jinawee Apr 17 '14 at 18:54

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 $4A$.

When a wire with no resistance is connected to the terminals of an ideal battery, will a current exist in the circuit?

yes, it would be infinite

If a capacitor is added to the circuit, will it be charged by the battery or will it remain uncharged?

yes, the capacitor would be charged.

• I would argue that $\frac{1}{0} != \infty$. $\frac{1}{0^+}$ does. – evil999man Apr 17 '14 at 15:53
• @Awesome I agree, but I don't see why that invalidates the original statement. Since $R$ is positive for all reasonable circuits, we take the limit of $V/R$ as $R \rightarrow 0^+$, which is $\infty$ for positive voltage. – Shivam Sarodia Apr 17 '14 at 17:18

It depends on inductivity:

I suspect the circuit would explode, with or without capacitor.

With just the battery - which also has zero inner resistance - the current would get infinite during the first 0 seconds. Except - while no energy is lost in heating wires - that would make the whole capacity of the battery.going into creating a magnetic field around the wire.

I'm not sure whether that would reduce the current in the wire, or not; Note this is unrelated to resistance.

If it does, the battery's energy goes into creating a magnetic field, during some time greater than zero.

And that's it!
We now have a completely normal superconducting electromagnet.
Nice!

It will just store the energy of the battery forever (which is still part of the circuit),

In case the inductivity - causing the magnetic field to build up - does not reduce the current:

The infinite curent would create an infinite magnetic field, that creates infinite forces on the wire and resistor. These have a mass greater than zero. So we have the - similar to the resistor - a perfect explosion.
Some people would call... wait, would have called it Apocalypse.
Not nice!

Regarding inserting the capacitor added after that:

In the first case, we will open the circuit - which makes our magnetic field break down, during finite time. This causes a voltage to be induced into the wire, with or without capacitor, but also make the wire move and bend a little, because of the changing magnetic field. There will be some more conversions between magnetic field and voltage/current, until all of the energy was spend on bending the wire.

So: "It depends."

• You might want to put a disclaimer : Don't try this at home, school or anywhere else – evil999man Apr 17 '14 at 17:21
• @Awsome Yes, the "anywhere else" part is important. I wouldn't care whether Apocalypse happens at school - not only because I had no time. – Volker Siegel Apr 17 '14 at 17:24
• @Awsome Ok, if someone comes up with a law, like that "contains peanuts" thing, I'll add a disclaimer. – Volker Siegel Apr 17 '14 at 17:26
• I liked the phrase : first 0 seconds. – evil999man Apr 17 '14 at 17:26
• Thanks :) Hmm... maybe I try to make the question title a little more "catchy for physicists"... Hope we can find a 29th reader then :) – Volker Siegel Apr 17 '14 at 17:46

The battery would be shorted in first case.

In the second case:

Although, resistance is never zero. You see that while charge on a charging capacitor varies with time as $$q=CV(1-e^{-\frac{t}{RC}})$$

When you set $R=0 \Omega$, the result is undefined. Although you can talk of $R\mapsto 0^{+}\Omega$

In that case, the capacitor will be charged instantly.

• Your idea in the question comment right, se my answer :) - turns out that the issue may involve Apocalypse indeed! – Volker Siegel Apr 17 '14 at 17:16