CSLB C-10 Electrical ContractorPlanning and EstimatingMedium

A commercial client requires a new 120V, 20A branch circuit for a series of general-use receptacles in an office space. The blueprints specify that the circuit will be wired with 12 AWG THHN conductors in EMT conduit. What is the minimum internal cross-sectional area (in square inches) required for the EMT conduit to properly accommodate these conductors, assuming a 40% fill limit and standard conductor areas?

  1. A0.166
  2. B0.133
  3. C0.0665
  4. D0.0532
Show answer & explanation

Correct answer: B. 0.133

For a 120V, 20A general-use receptacle circuit, there typically are two 12 AWG THHN ungrounded conductors, one 12 AWG THHN grounded conductor, and one 12 AWG bare equipment grounding conductor. That's a total of four 12 AWG conductors. According to CEC Chapter 9, Table 5, the approximate area for one 12 AWG THHN conductor is 0.0133 sq in. Total area occupied by conductors = 4 conductors * 0.0133 sq in/conductor = 0.0532 sq in. Since the conduit fill limit for 3 or more conductors is 40% (CEC Chapter 9, Table 1), the occupied area (0.0532 sq in) must be 40% or less of the conduit's internal area. Therefore, Minimum Conduit Area = Occupied Area / 0.40 = 0.0532 sq in / 0.40 = 0.133 sq in.

Why the other options are wrong

  • A. Incorrect. This value is derived from an incorrect calculation or fill percentage.
  • C. Incorrect. This value is derived from an incorrect calculation or fill percentage.
  • D. Incorrect. This is the total area occupied by the four 12 AWG conductors, not the required conduit area.

Conduit Sizing by Fill Percentage

Determining the minimum required internal area of a conduit to ensure the total cross-sectional area of all conductors installed does not exceed the maximum allowable fill percentage (e.g., 40% for 3+ conductors).

  • Prevents overheating of conductors and allows for easier pulling.
  • CEC Chapter 9, Table 1 specifies the fill percentages.
  • CEC Chapter 9, Table 5 lists conductor areas.

Memory trick: Conductor Area divided by Percent, tells the Conduit's size.

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