# Safety switch “breaking capacity” vs. voltage

In the context of safety switch rating (fuses, circuit breakers, etc.), the concept of breaking capacity refers to the maximum electric current which can be safely interrupted by a tripping safety switch, i.e. interrupted without the formation of an electric arc. The article Breaking capacity on Wikipedia states that

The current breaking capacity corresponds to a certain voltage, so an electrical apparatus may have more than one breaking capacity current, according to the actual operating voltage.

I am curious as to what a breaking capacity vs. operating voltage plot would look like, but I haven't been able to find one. I suspect breaking capacity is negatively correlated with open circuit voltage, but that is just a guess.

For my question, consider the following case:

Let there be a circuit with a switch, where $$\A\$$ and $$\B\$$ denote the contactors of the switch. A voltage of $$\U_a\$$ is the maximum voltage between $$\A\$$ and $$\B\$$ such that no electric arc can form between them when the switch is open.

If the switch is originally open, a voltage of $$\U_a\$$ between $$\A\$$ and $$\B\$$ will not generate an arc between $$\A\$$ and $$\B\$$.

If the switch is originally closed and a current of $$\I_b\$$ flows, when suddenly the switch is opened, is the voltage $$\U_a\$$ now enough to generate an arc between $$\A\$$ and $$\B\$$?

If the flow of current reduces the voltage required for electric arc formation in the example above, why does it happen? Does it have to do with momentum of the charge carriers?

I hope my question is clear enough for being answered. If not, please point out where it is unclear.

If the switch is originally closed and a current of $$\I_b\$$ flows, when suddenly the switch is opened, is the voltage $$\U_a\$$ now enough to generate an arc between $$\A\$$ and $$\B\$$?