Sobering fact that with a 6 Amp MCB protecting the circuit the plasma discharge will be creating at least 1.5kW of heat until the MCB trips.
I'm not sure how you worked that out, or what you think the rating of the MCB has got (directly) to do with it.
As I need not tell you, an MCB does nothing to limit the amount of current that can flow which, in the case of a 'dead short', will be determined only by the supply voltage and (L-N) loop impedance.
If the circuit has been designed correctly, then the
L-E loop resistance (as measured from the place of interest) needs to be low enough for the MCB to trip magnetically in response to a 'dead short' at that location - so, to cater for the worst case, 5 x 6A = 30A for a 6A MCB. However, since CPCs are invariably smaller than the corresponding N conductors, the L-N loop impedance will be appreciably lower than the L-E loop impedance - so, if the circuit (protected by a 6A MCB) is correctly designed (in relation to L-E faults), the current resulting from a L-N 'dead short' will be appreciably greater than 30A, and that current will flow until the MCB trips.
Hence, in that situation, the power generated (primarily in the cables) will be well in excess of 6.9 kW, rather than the modest 'at least 1.5 kW' you suggest. Having said that, the MCB will trip very quickly under those conditions, and 6.9+ kW dissipated (primarily in the cables) for a handful of milliseconds will not amount to very much energy, so probably not even a perceptible temperature rise.
Kind Regards, John