# Breaker size



## VELOCI3 (Aug 15, 2019)

12A x 125%= 15A
20A bkr x 80%= 16A

Where did you see the 250%?


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## blueheels2 (Apr 22, 2009)

According to the manufacturer the LRA is 22.5. Seems like the 20 would be the best option. what say y'all


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## blueheels2 (Apr 22, 2009)

VELOCI3 said:


> 12A x 125%= 15A
> 20A bkr x 80%= 16A
> 
> Where did you see the 250%?
> ...




I must be confusing myself. We are supposed to use 240.38 to get the ampacity of the motor and then multiply by 250% (430.52). 125% of 12 amps is how we size the wire.


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## joe-nwt (Mar 28, 2019)

250% is max. I would try the 20A.


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## emtnut (Mar 1, 2015)

I'd go 15A breaker ... doubt it would trip ... tiny motor


If it's for a paying customer, I'd go 20A just in case of the remote possibility there was a call back for it tripping.


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## BillyMac59 (Sep 12, 2019)

You don't need to calculate the ampacity based on the horsepower rating. Code (Canadian anyways) says use the motor nameplate data. Overcurrent protection is therefore based on 12 amp.


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## JoeSparky (Mar 25, 2010)

I have an answer to your question. Cricket has decided not to respond to my PMs. I will not be contributing here until at least then. Happy to give an answer at certified badass electricians.


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## Wirenuting (Sep 12, 2010)

JoeSparky said:


> I have an answer to your question. Cricket has decided not to respond to my PMs. I will not be contributing here until at least then. Happy to give an answer at certified badass electricians.


This isn't an airport.
You don't need to announce your delay.


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## splatz (May 23, 2015)

I thought you sized the overloads based on the nameplate current rating but the breaker based on the horsepower using tables at the end of the chapter, for single phase motors it's table 430.248. Using that table, you use a FLC of 7.2A for a 1/3HP motor, 9.8A for a 1/2HP motor. 

For a 4/10 HP motor I am not sure if you're allowed to round up to 1/2HP or have to round down to 1/3HP. Now 430.110(C)(3) says for small motors not in 430.248 you can assume the FLC is 1/6 the LRC but I don't know if that would apply here, I think that's just for combination loads. 

Table 430.52 allows a maximum of 250% of FLC for inverse time breakers with single phase motors and you can round up to the next standard size. 

7.2A * 250% = 18A => 20A breaker

9.8A * 250% = 24.5A => 25A breaker


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## emtnut (Mar 1, 2015)

ValeoBill said:


> You don't need to calculate the ampacity based on the horsepower rating. Code (Canadian anyways) says use the motor nameplate data. Overcurrent protection is therefore based on 12 amp.


That is one of the rules that is different under NEC.


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## emtnut (Mar 1, 2015)

JoeSparky said:


> I have an answer to your question. Cricket has decided not to respond to my PMs. I will not be contributing here until at least then. Happy to give an answer at certified badass electricians.



I tried to get on, about 15 tries on the test, but I keep failing :sad:








.
.
:biggrin:
edit: Can you give me the answers Joe ? Or do I have to wait for Cricket to come back ??


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## JRaef (Mar 23, 2009)

*IF the motor has built-in thermal protection or you are using an overload relay*, then you can size the breaker at 250% max. But if the breaker IS the only protection for the motor, then it must be sized at a MAXIMUM of 115% of the FLA (125% if it is 1.15SF).


But in this case, that motor DOES have integral thermal OL protection. So you CAN go up to 250% of the nameplate FLA, i.e. 12 x 2.5 = 30A. Your CONDUCTORS however must be sized per the nearest applicable HP, in this case 1/2HP so 9.8A x 1.25 = 12.25A. 



Personally, I would use 12ga to the junction box and a 20A breaker, and never get a call back for the installation.


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