I don't think so. The energy generated by an arc is dependent only on the current and the resistance of the arc (which is extremely low). IIRC typical arc voltages (voltage across arc) are of the order of 10-20V/cm, so when the contacts have only opened by, say, 1mm, that will be 1-2V, regardless of of what the 'supply voltage' is. In fact, the contacts (and what's happening between them) haven't got a clue what the 'supply voltage' is whilst the arc is present (since the arc effectively keeps the contacts 'closed'). Only when the arc is broken (and the contacts become truly 'open') will the voltage across the (now open) contacts rise to the supply voltage - but then, of course, it no longer matters. AFAICS, a given current will therefore result in the same arc energy (when contacts attempt to open), regardless of what 'supply voltage' is responsible for that current.Not necessarily. When breaking the supply on a fault, there's a high power supply so when the contacts open, there's the voltage and energy to initiate a spark. Basically, you've got a supply capable of putting 240V and hundreds of amps into the spark. ...I'll bet that an AC rated MCB switching DC would do the contacts in.
Kind Regards, John