JDB3
Senior Member
- Location
- San Antonio, Texas
When figuring the 80% load of the wire for #12, is it correct to use the 25 amp rating?
25 amps X 80% = 20 amps.:?
25 amps X 80% = 20 amps.:?
80 %
- Thread starter JDB3
- Start date
- Status
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
In what context are you speaking of an 80% factor? Nothing says you can't load a wire to 100%.
JDB3
Senior Member
- Location
- San Antonio, Texas
In wiring a wall heater 4,000 watts @ 240 volts (actual voltage at site is 245 volts), would # 12 NM (romex) be acceptable? No central heat, so they may run continuous.
- Location
- Connecticut
- Occupation
- Engineer
No, #12 is not acceptable. You are required to treat the branch circuit load as continuous by 424.3(B). You need a branch circuit breaker that is no smaller than 25Amps. You can't put #12 on the 25A c/b per 240.4(D).
97catintenn
Senior Member
- Location
- Columbia, TN
Yes, but you should probably explain why and where you found it info as to the 125% for the OP.
- Location
- New Jersey
- Occupation
- Journeyman Electrician
kwired
Electron manager
- Location
- NE Nebraska
Seems to be a few threads of similar topic recently.
You are going about this the wrong way. You do not pick the overcurrent device then the conductor and then see if that works with the load.
First thing to do is determine what the load is, as well as if it is continuous or noncontinuous. You then determine what ampacity of conductor is needed.
210.19(A)(1) tells us that minimum ampacity shall be not less than the noncontinuous load plus 125 percent of the continuous load. There is similar worded requirement for feeder conductors as well as service conductors.
Once you have minimum ampacity of conductor you can select overcurrent protection. 240.4 says that conductors shall be protected against overcurrent in accordance with their ampacities specified in 310.15, unless otherwise permitted or required in 240.4(A) through (G).
If your conductor is already 125% of the load and you protect the conductor at its ampacity your overcurrent device will also be 125% of the load (or more if you were allowed to use next standard size higher).
You are going about this the wrong way. You do not pick the overcurrent device then the conductor and then see if that works with the load.
First thing to do is determine what the load is, as well as if it is continuous or noncontinuous. You then determine what ampacity of conductor is needed.
210.19(A)(1) tells us that minimum ampacity shall be not less than the noncontinuous load plus 125 percent of the continuous load. There is similar worded requirement for feeder conductors as well as service conductors.
Once you have minimum ampacity of conductor you can select overcurrent protection. 240.4 says that conductors shall be protected against overcurrent in accordance with their ampacities specified in 310.15, unless otherwise permitted or required in 240.4(A) through (G).
If your conductor is already 125% of the load and you protect the conductor at its ampacity your overcurrent device will also be 125% of the load (or more if you were allowed to use next standard size higher).
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