Anyone in this forum have an idea what the surge current is to charge the Input capacitor on an Inverter?
Yes, my estimate is 20,000A
That comes from the internal resistance (and cell voltage) of a single string of 280 Ah cells for one LiFePO4 battery. Possibly its short-circuit current is actually less (I think someone did publish a bit of measured data), but that's what I have to go on.
Connect several banks in parallel, and multiply that figure.
Capacitors in an inverter present a dead short, so closing the switch is like a "bonded short" on output of battery. Wire resistance limits current a little, but cell IR dominates. Of course, capacitor charges up in a short time so no wires burn up. But BMS FETs may take a hit, also switch/relay contacts.
Capacitors in an inverter will be about 10,000 to 100,000 uF, 0.01 to 0.1 Farad.
A Farad is an ampere-second per volt. 0.1 F charged 1V is 0.1A for 1 second; charged to 50V it is 5.0A for 1 second. 5000A for a millisecond.
It might peak around 10,000A for an instant, taper down to zero in a fraction of a second.
That is enough to burn contacts. Relays tend to close, bounce open, close again. During the time open, it melts the contact point, then the molten points fuse when they touch again. Next time it opens (if you're lucky), and tears out a bit of metal.