Comment by quickthrowman
4 days ago
4.5-5A at 240V = 9.72A at 120V
It’s the same level of power consumption. I’m not even sure what you’re asking at this point, to be honest.
4 days ago
4.5-5A at 240V = 9.72A at 120V
It’s the same level of power consumption. I’m not even sure what you’re asking at this point, to be honest.
You were talking about needing a second 240V 20A circuit, and you later backed that up by citing the spec sheet of 230V mini-split with a minimum circuit rating of 15A.
My argument was that you do not need such a circuit.
Technically you’re correct, a 12000 BTU minisplit only uses around 1000 watts while running which is just over 4A.
The breaker size being 20A 2P is a consequence of the NEC requiring you to size the wire based off the equipment nameplate rating of 15A, which is based off the full load amps of the motors inside the equipment.
Full load amps is the max amount of current a motor can draw at a specific voltage and is used for sizing wire and overcurrent protection for a piece of equipment. It doesn’t always match up the current a motor draws while it’s running normally. You take full load amps times 1.25 to get minimum circuit ampacity, which you use to size the conductors.
So while you are correct that a 240V 12000 BTU minisplit wont draw anywhere near 20A, the specific minisplit I looked at required a 20A breaker due to the minimum circuit ampacity being 15A. If the MCA was 12A, you could use a 15A breaker; an MCA of 8A would allow using a 10A breaker, and so on.
If you use fuses, you can size the overcurrent protection at 100%, breakers require 125% of the load for a continuous load. So you could use a 30A fusible disconnect switch fused at 15A for a unit with an MCA of 15A.
That's not the angle I'm taking. I'm not saying anything about what the mini-split actually uses. Give it the circuit that the nameplate asks for.
Instead I'm saying that particular minisplit is a lazy design and we can get a 12000 or higher BTU unit with a much smaller nameplate rating. Not only will it only need a single-pole breaker, the required circuit probably already exists.