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Topic: EGC count in the Jbox

User: Ponchik, Location: CA

Asked on: Friday, April 25, 2025 at 12:54 pm EDT
Answered on: Friday, April 25, 2025 at 12:54 pm EDT

Question: For box fill calculation, the first 4 counts as one based on the lasrgest size. After fourth one, each conductor counts as a quarter volume.so if I have (4) 10-2 cable and (3) 12-2 cable in the box, my EGC count is 1 3/4 of #10?TIA

You're very close, but there's a slight misunderstanding about how EGCs are counted. Here's the breakdown:1. Conductor Count

Each current-carrying and neutral conductor counts toward box fill.2.

Grounding Conductor Count

All equipment grounding conductors (EGCs) entering the box are counted together as a single conductor, based on the largest EGC entering the box.

Calculations for your scenario



(4) 10-2 Cables

Each 10-2 cable has two current-carrying conductors (hot and neutral) and one EGC (ground).


(3) 12-2 Cables

Each 12-2 cable has two current-carrying conductors (hot and neutral) and one EGC (ground).

Counting



Current-Carrying and Neutral Conductors

(4 cables
2 conductors/cable) + (3 cables 2 conductors/cable) = 8 + 6 = 14 conductors

EGCs

You have a total of 4 + 3 = 7 EGCs. Since the largest EGC is #10, all those EGCs together count as ONE #10 conductor.

So, to answer your question directly

The EGC count is

ONE #10 conductor
. NOT 1 3/4.

Important Considerations



Device Yokes

Device yokes (like switches or receptacles) also take up space in the box. Each yoke counts as twice the volume allowance for the largest conductor connected to the device.

Internal Clamps

Internal cable clamps (if present) count as one conductor, based on the largest conductor entering the box.


Conductor Size

You need to know the conductor sizes to determine the volume allowance for each conductor. The NEC provides tables (like NEC Table 314.16(B)) that list the volume allowance for different conductor sizes.

Box Size

You MUST make sure that the total volume of all conductors, devices, and clamps does not exceed the listed volume of the box.

Example Calculation (Hypothetical)

Let's assume you are using a metal 4"x1 1/2" box (21 cu in volume)1.

Current-Carrying Conductors

14 conductors2.

EGC

1 #103.

Device

1 device (example)


Total Conductor Count (#10 equivalent)

14 + 1 = 15 conductors.Based on NEC, one #10 conductor needs 2.25 Cu In. so 15 x 2.25 = 33.75 Cu in + 2 x 2.25 Cu in (assuming the device use a #10 conductor too)Total need 38.25 Cu In to house all wires. Since 4"x1 1/2" box have only 21 Cu in available, you should use a bigger box or use multiple gang box.

Disclaimer:
This is a simplified explanation. Always consult the National Electrical Code (NEC) and your local electrical codes for complete and accurate information. If you're not comfortable performing electrical work, hire a qualified electrician. Flag for review

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