Mike Dolt Enterprises
Mike Dolt Enterprises
User: squaredan, Location: Pennsylvania
Asked on: Monday, April 18, 2011 at 3:46 pm EDT
Answered on: Tuesday, May 14, 2024 at 4:19 am EDT
Question: Hi,This is my first post on this forum andI hope I am in the right section for this question.. I have been studying for my master and I came across a question in the Mike Holt Exam Prep Book and I am not understanding why would not use Table 310.15B2a for Adjustment in this question.. The Question is" What size overcurrent protection device is required fro a 45A continuous load? the circuit is in a raceway with 14 current carrying conductors. (a) 45A (b) 50A (c) 60A (d) 70A "The answer in the book is 60AI understand we take 125% (210.20) of a continuous load 45A x 1.25= 56.25A and 240.4B lets us go to the next higher size 60A but then why wouldnt we take the Adjustment factor of table 310.15B2a of 14 current carrying conductors? I have been trying to find a code rule but not finding..Maybe it is wrong in the answer key??? Hope someone can help me on this one.. ThanksDan
The answer in the book is correct.Table 310.15(B)(2) applies to conductors in raceways, not to overcurrent devices protecting those conductors.
Table 310.15B2a is used to adjust the ampacity of conductors in raceways.
Rule 210.20 requires the overcurrent device to be rated not less than 125% of the continuous load.In this case, the continuous load is 45A, so the overcurrent device must be rated not less than 45A x 1.25 = 56.25A.
Table 240.4B shows that the next higher standard size overcurrent device is 60A.Therefore, the correct answer is (c) 60A. Flag for review