Box calculations are a key skill for electricians, and this free NEC electrical exam question focuses on determining the minimum cubic inch capacity for a box accommodating various conductors, devices, and fittings. Let’s break it down step by step to solve this problem.

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Question Breakdown

Question: Determine the minimum cubic inch capacity for a square junction box accommodating:

  • Three 16 AWG conductors
  • Two 14 AWG conductors
  • Three 14 AWG equipment grounding conductors
  • Two internal cable clamps
  • A single-gang device yoke connected to 14 AWG conductors
  • A 16 AWG pigtail

Required Capacity: Calculate the cubic inches required based on the NEC box fill rules.

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Step-by-Step Solution

Step 1: Identify All Items in the Box

To calculate the required cubic inch capacity, first, list everything that contributes to the box fill:

  • 3 x 16 AWG conductors
  • 2 x 14 AWG conductors
  • 3 x 14 AWG equipment grounding conductors
  • 2 x internal cable clamps
  • 1 x device yoke connected to 14 AWG conductors
  • 1 x 16 AWG pigtail (not applicable)

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Step 2: Determine the Box Fill Count for Each Item

Using NEC 314.16(B), determine the box fill contribution for each item:

  • 16 AWG Conductors: Each conductor counts as one unit of its size allowance:
    3 x 16 AWG = 3 units
  • 14 AWG Conductors: Each conductor counts as one unit of its size allowance:
    2 x 14 AWG = 2 units
  • Equipment Grounding Conductors: Count as one unit total for all grounding conductors:
    3 x 14 AWG = 1 unit
  • Internal Cable Clamps: Count as one unit total:
    2 x clamps = 1 unit
  • Device Yoke: Counts as two units of the largest conductor connected to it (14 AWG in this case):
    1 x yoke = 2 units
  • Pigtail: Does not count, as it originates and terminates within the box.

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Step 3: Calculate the Volume Allowance

Using NEC Table 314.16(B)(1), determine the volume allowances for each conductor size:

  • 16 AWG: Volume allowance = 1.75 in³
  • 14 AWG: Volume allowance = 2.00 in³

Now, calculate the total volume for each conductor type:

  • 3 x 1.75 in³ (16 AWG) = 5.25 in³
  • (2 + 1 + 1 + 2) x 2.00 in³ (14 AWG) = 12.00 in³

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Step 4: Calculate the Total Required Capacity

Add the calculated volumes together to determine the minimum box size:

Total Volume = Volume (16 AWG) + Volume (14 AWG)
Total Volume = 5.25 in³ + 12.00 in³
Total Volume = 17.25 in³

The minimum cubic inch capacity required for this junction box is 17.25 in³.

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Key Considerations for Junction Box Calculations

  1. Understand Box Fill Rules: NEC 314.16(B) outlines how each component contributes to box fill.
  2. Combine Grounding Conductors: All grounding conductors count as one unit, no matter how many are present.
  3. Device Yokes: Each yoke counts as two units of the largest conductor connected to it.
  4. Pigtails: Pigtails are excluded if they originate and terminate within the box.

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Common Mistakes to Avoid

  • Forgetting to Combine Grounding Conductors: Count all grounding conductors as one unit, not individually.
  • Excluding Clamps: Internal cable clamps contribute to box fill, so don’t overlook them.
  • Wrong Volume Allowance: Always refer to NEC Table 314.16(B)(1) for the correct volume allowance per conductor size.
  • Pigtail Miscalculation: Pigtails that do not leave the box are not included in the calculation.

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Why This Question Matters

Junction box capacity calculations ensure safe electrical installations, preventing overcrowding and overheating. Mastering these rules is crucial for electricians preparing for their NEC practice tests, especially for journeyman and master electrician exams.

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