Quizzes on the Code

April 1, 2001
Conductors - Art. 310 What is the current-carrying ampacity of each of nine No. 12 THHN copper conductors installed in -in. EMT? (Assume all conductors are current-carrying.)a. 17Ab. 21Ac. 25Ad. 28A What is the current-carrying ampacity of each of six No. 14 THHN copper conductors routed through an ambient temperature of 105F?a. 17.4Ab. 18.6Ac. 19.8Ad. 21.4A What is the current-carrying ampacity of

Conductors - Art. 310

  1. What is the current-carrying ampacity of each of nine No. 12 THHN copper conductors installed in ¾-in. EMT? (Assume all conductors are current-carrying.)
    a. 17A
    b. 21A
    c. 25A
    d. 28A

  2. What is the current-carrying ampacity of each of six No. 14 THHN copper conductors routed through an ambient temperature of 105°F?
    a. 17.4A
    b. 18.6A
    c. 19.8A
    d. 21.4A

  3. What is the current-carrying ampacity of each of sixteen No. 12 THHN copper conductors, with diversity, located in the same raceway? (Assume only half of the conductors are current-carrying.)
    a. 17A
    b. 18A
    c. 20A
    d. 21A

  4. What is the current-carrying ampacity of each of twelve No. 12 THHN copper conductors, if routed in a raceway where there is a total of 16 current-carrying conductors?
    a. 17.2A
    b. 18.4A
    c. 19.6A
    d. 21.2A

answers and discussion

  1. (b), 21A. Cables or raceways enclosing four or more current-carrying conductors are required to be derated per Sec. 310-15(b)(2)(a). The NEC considers all ungrounded (phase) conductors to be current-carrying. Neutrals for Wye systems are considered current-carrying if the major portion of the load (more than 50%) consists of harmonic currents [Sec. 310-15(b)(4)(c)].

    Step 1: Find amperage of each conductor per Table 310-16. No. 12 THHN cu. = 30A.

    Step 2: Find derating factor per Table 310-15(b)(2)(a). 9 conductors requires a 70% derating factor.

    Step 3: Recalculate amperage, per Table 310-15(b)(2)(a). 30A × 70% = 21A.

    Solution: The ampacity of each conductor is 21A.

    Stallcup's Code Loop: Secs. 110-14(c), 310-15(b)(2)(a), and 310-15(b)(4)(c), and Tables 310-16 and 310-15(b)(2)(a).

  2. (a), 17.4A. See explanation for Questions 1 and the following:

    Conductors routed through ambient temperatures exceeding 86°F must be derated according to the correction factors listed in Table 310-16. The derating factors are listed in the ampacity correction factor at the bottom of Table 310-16. Select the derating factors based on the ambient temperature the conductors are exposed to.

    Step 1: Find amperage of conductors per Table 310-16. No. 14 THHN cu. = 25A

    Step 2: Find derating factor for multiple conductors per Table 310-15(b)(2)(a). 6 conductors requires an 80% derating factor.

    Step 3: Find derating factor for ambient temperature per Table 310-16. 105°F requires an 87% derating factor.

    Step 4: Recalculate amperage per Tables 310-15(b)(2)(a) and 310-16. 25A × 80% × 87% = 17.4A.

    Solution: The ampacity of each conductor is 17.4A.

    Stallcup's Code Loop: Secs. 110-14(c), 310-15(b)(2)(a), and 310-15(b)(4)(c), and Tables 310-16 and 310-15(b)(2)(a).

  3. (d), 21A. A 50% load diversity can be applied to a conductor ampacity when all the conductors are not loaded at the same time, if allowed by the authority having jurisdiction (AHJ).

    Step 1: Find amperage for eight current-carrying conductors per Table 310-16. No. 12 THHN cu. = 30A

    Step 2: Find derating factor per Table B-310-11 in Appendix B. 16 conductors requires a 70% derating factor.

    30A × 70% = 21A

    Solution: The ampacity of each conductor is 21A, however, only eight at any given time can have this ampacity rating.

    Stallcup's Code Loop: Sec. 110-14(c), and Tables 310-16 and B-310-11.

  4. (a), 17.2A. You can apply diversity to more than half of the conductors in a raceway or cable if they are limited to a certain calculated value.

    Step 1: Apply formula for 12 conductors per Tables B-310-11 and 310-16.

    A2 = √(0.5 × N/E) × A1, where A1= ampacity of conductor from Table 310-316 multiplied by the appropriate adjustment factor in Table B-310-11

    A2 = √(0.5 × 16/12) × 30A × 0.7
    A2 = 17.2A

    Solution: The ampacity of each conductor is 17.2A.

    Stallcup's Code Loop: Sec. 110-14(c), and Tables 310-16 and B-310-11.

Stallcup's Code Loops

Each quiz on the 1999 NEC is provided with Stallcup's Code Loops to show the direct correlation with other sections of the NEC. This allows the reader to review other sections of the NEC when determining if the correct answer to a problem is in one or multiple sections of the Code.

About the Author

James Stallcup Jr.

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