Mike Dolt Enterprises
Mike Dolt Enterprises
User: Elect117, Location: California
Asked on: Thursday, July 31, 2025 at 11:46 am EDT
Answered on: Thursday, July 31, 2025 at 11:49 am EDT
Question: I wanted to make sure I am reading these things correctly before I reply to a contractor. I think I am confusing myself. The contractor ran 33 phase conductors of #10 in one conduit (like a 5in). The conduit is a 15' long run up to a box and doesn't meet any of the exceptions of 310.15(C)(1). So, I notified him that even if the loads were 1A, having the that many conductors in one conduit would require derating the conductors and needing to reduce the OCPD size from 30A to 20A (THWN-2-> 40A * .4 = 16A, rounded up to 20A).Is that right?I got back to the office and was reading 240.4(D) and noticed that it says:240.4(D) - "Unless specifically permitted in 240.4(E) or (G), the overcurrent protection shall not exceed that required by (D)(1) through (D)(7) after any correction factors for ambient temperature and number of conductors have been applied."I don't know if I am confusing myself and was correct in my correction or if 240.4(D) means that even after all of the correction factors, you can still use a 30A OCPD.
Okay, let's break this down and clear up the confusion. You're on the right track, but the nuances of the code can be tricky.
Your Initial Assessment (Derating and OCPD Size)
You are correct. The derating for the number of current-carrying conductors is the key issue here. You have more than 3 current-carrying conductors in a raceway, so derating is absolutely required per 310.15(C)(1). Derating Calculation
THWN-2 ampacity: 40A Derating factor for 33 CCCs: 40% (0.4) (Table 310.15(C)(1)) Derated ampacity: 40A 0.4 = 16A
OCPD Selection (BEFORE 240.4(D))
Based solely on the derated ampacity of 16A, you would be inclined to protect the conductor at the next standard OCPD size, which is 20A.
Understanding 240.4(D)
240.4(D) is a general rule for protecting conductors. It limits the overcurrent protection for specific conductor sizes (like #14, #12, and #10 copper) to specific ampacities unless exceptions 240.4(E) or 240.4(G) apply.
The Key is the Phrase: "after any correction factors for ambient temperature and number of conductors have been applied." This means 240.4(D) does take into account the derating you already calculated.
Applying 240.4(D) to Your Situation
240.4(D)(5) specifically addresses #10 copper conductors. It states that the overcurrent protection shall not exceed 30 amps.2. However, 240.4(D) includes the caveat: "after any correction factors for ambient temperature and number of conductors have been applied." This means you must apply the derating factor from 310.15(C)(1) before you determine if you can use the 30A OCPD allowed by 240.4(D)(5).
The Conclusion
Because you've already determined that the derated ampacity of the #10 conductors is 16A, 240.4(D)(5) does not allow you to use a 30A overcurrent device. The rule says "shall not exceed 30 amps" after derating. Your derating puts you well below that.
You are correct! The OCPD must be reduced to 20A based on the derated ampacity of 16A and the standard breaker sizes.
Why This is Important
The primary purpose of overcurrent protection is to protect the conductors from overheating and potentially causing a fire. If you were to use a 30A breaker on conductors that are only safely rated for 16A after derating, you would be violating this basic principle of electrical safety.
To summarize
Derate the conductors first. This is required by 310.15(C)(1) because you have more than 3 current-carrying conductors in a raceway.
2. Apply 240.4(D)(5). It says you cannot exceed 30A overcurrent protection for #10 copper wire after derating. Since your derated ampacity is 16A, you cannot* use a 30A breaker.3. Select the appropriate OCPD. The next standard size up from 16A is 20A, so that's the maximum OCPD you can use in this scenario.You are correct in your assessment! You need to reduce the OCPD size to 20A. Your logic and calculations are sound. Flag for review