Derating

[e57]

Alright - I think I'm starting to put this together....

For the sake of easy numbers:

15aA, zeroN, 15aB total zero current lost or added....
vs
10aA, 10aN, 20aB - total zero current lost or added.....

Are you telling me that they are the same in terms of heat effect on the conductors for derating purposes?

I get this part.... I think?

A neutral conductor that carries only the unbalanced current from other conductors of the same circuit shall not be required to be counted when applying the provisions of 310.15(B)(2)(a).

Are you trying to say that the language of 210.4a is permissive????

210.4 Multiwire Branch Circuits.
(A) General. Branch circuits recognized by this article shall be permitted as multiwire circuits. A multiwire branch circuit shall be permitted to be considered as multiple circuits. All conductors shall originate from the same panelboard.

If so - considering it "permisive" - allows it to be considered as "multiple circuits" (IMO they way I'm reading it) - like I was saying before.... If and once considered as "Multiple circuits" in 210.4 - They are no longer the "Same" in 310.15B4a are they? At which point the neutral is counted - right? This seems to be the local AHJ interpetation of this - just trying to get a better grip on it - work with me please....:wink:

 

[Bob NH]

Alright - I think I'm starting to put this together....

For the sake of easy numbers:

15aA, zeroN, 15aB total zero current lost or added....
vs
10aA, 10aN, 20aB - total zero current lost or added.....

Are you telling me that they are the same in terms of heat effect on the conductors for derating purposes?

I get this part.... I think?

They are not the same in terms of the heating effect in the conductors.

The power delivered by the circuit is the same (not counting voltage drop) but the power lost due to voltage drop (and therefore the heating) is greater in Case 2.

Your Case 1: 15 Amp, zero Amp, 15 Amp:
Must use sum of Amps squared for each conductor; so 15 x 15 + 0 x 0 + 15 x 15 = 225 +0 +225 = 450.

Your Case 2: 10 Amp, 10 Amp, 20 Amp:
Sum of Amps squared = 10 x 10 + 10 x 10 + 20 x 20 = 100 + 100 + 400 = 600 Amps squared.

Therefore, the heating in Case 2 is 133.3% of the heating in Case 1.

You get exactly the same answer, but with more arithmetic, if you calculate the voltage drop in each conductor and multiply the voltage drop by the Amps in each conductor.