Thsi topic is being discussed in two other threads but here is a different question.
Assume 10 120V circuits each with their own neutral. #12 all in one conduit to a juction box with one ground wire and a ground bus in the junction box. Separate MC from there to the devices, lighting etc.
50% of 25A is 12.5 amps. Does that mean that the branch circuit breaker needs to be less than 12.5 amp? Or just that the connected load can not exceed 12.5 amps?

The load on each circuit would need to be less than 12.5A, but you could use the next larger standard breaker, eg. a 15A breaker.

Note that for derating based upon number of conductors in a conduit, you get to work from the temperature rating of the wire, even if it is higher than that of the terminations. So you would probably use THHN 90C wire, with an ampacity of 30A, 15A after derating with a 15A breaker.

-Jon

Not if the circuit supplies a multioutlet branch circuit supplying receptacles for cord-and-plug-connected portable loads.
See 240.4(B)(1).
Don

I agree Don, if the circuits are feeding receptacles you looking at 10a circuit breakers.

OK, so for the receptacle circuits we can not go to the next higher size circuit breaker.
But, if the wire is 90C we can use 30 amp which means 15 amp and a 15 amp receptacle?
(regardless of the termination rating?)

If you have a 90° wire (THHN) #12 AWG using it in a ambient of 86° damp or dry location and derate to 15a due to raceway fill, you could feed 15 receptacles and have a 15a overcurrent device protecting it.