Dear Inspector,

I recently attended an IRC Combination Dwelling Inspector Exam Preparation course in Virginia sponsored by Northern Virginia ASHI. Some of the topics discussed were how to solve typical problems presented in the exams. I would like to share one of the typical electrical problems and share the process of producing the correct answer. The below will be based upon the 2003 International Residential Code, 4th printing.

Given:

Attic temperature 135 degrees (57°C) Copper Conductors Cable type NMC-B

Determine the ampacity of number 12 AWG conductors in a type NMC-B 12-2 w/g cable that is bundled with six similar cables run through an attic in Las Vegas, Nevada.

Note the question is not asking the maximum size overcurrent protection device - it is asking the maximum ampacity so table 3605.5.3, where you might go first, does not apply to this question.

First we must go to section 3605.4.4 to determine what temperature rating column to use when determining the ampacity of type NM cable conductors. We see in this section that for type NMC-B cable we use the 90°C column.

Then we locate table 3605.1 and in the 90° column for copper conductors we see that the allowable ampacity for number 12 is 30 amperes.

Next we must locate the ambient temperature correction factor due to the attic conditions. Again in the 90° copper conductor column we see the correction factor is .71 in the 56-60 ambient temperature row. We take 30 times .71 and the result is 21.3 amperes.

Finally, we must modify this result with the conductor proximity adjustment factor since there are cables bundled for distances greater than 2 feet (section 3605.3). The problem indicated "with six similar cables" indicating our total cables is seven. Since these are 12-2 conductors we count both the hot and neutral for a total of 14 conductors (both hot and neutral are current-carrying, we would also count both if it was 12-2 and a 240 volt circuit, both are current carrying. The only time you do not count the neutral in a condition like this is if the problem indicated 12/3 cable; the neutral only carries the difference in a 120-v multiwire circuit and the neutral is not used in a true 240-volt circuit). Looking at table 3605.3 we see the adjustment factor is .50 in the 10-20 conductors row. Taking our previous answer of 21.3 amperes times .50 we see our total ampacity is now 10.65, we could round this to 11 amperes.

Hope this helps us better understand use of the electrical tables in the IRC.

Talk next month,

Michael Casey
Kaplan Professional Schools
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