Junction Boxes- Exceeding Fill Capacity- Doing the Math- Quincy Home Inspection

As a Wenatchee home inspector one item found frequently is junction boxes that have exceeded their fill capacity. Doing some calculations would ensure that the box capacity is not exceeded. Doing the math is all we need to do. But this is often overlooked.

All junction boxes have capacity associated with them in cubic inches.  Most non-metallic boxes on the home stores will have a Cubic Inch per junction box printed in them.
(For metal junction boxes capacities- link)

Each conductor (Wires- Hot (Black, red or others not white or green), Neutral (White) and Ground (bare or green) has an associated cubic inch by the AWG (American wire gauge).
14 Awg. = 2 Cu. In. per conductor,
12 Awg. = 2.25 Cu. In. per conductor,
10 Awg. = 2.5 Cu. In. per conductor,
8 Awg. = 3 Cu. In. per conductor
6 Awg. = 5 Cu. In. per current carrying conductor counted.

So it now time for the Math-
For this example we will use 12 Awg.  romex (NM) cable with two current carrying conductors. This is known as 12/2 cable which will have a black (hot), white (neutral) and bare wire (ground). So this is three wires total (proper terminology is conductors).

To figure out what capacity of box you need you will need to make a calculation based on the number of conductors, switches, receptacles and any clamps that will be in the junction box.

We will use 4 each 12/2 cables entering the box plus two receptacles for the calculation.

First we count all the current carrying conductors (8 each- 4 blacks and 4 whites).
Multiply by 2.25,  8  x 2.25 =18 cu. In.

Now all the grounds we will count them as one conductor (here we will have 4 grounds), 1 x 2.25 = 2.25

Next we must now account for the receptacles. Each receptacle (or switch if that was used) counts as 2 conductors. (For receptacles you use the largest conductor size in the box as the calculator, example if you had just 14 Awg conductors you use 2.00 cu in., if they were any 10 Awg you would use 2.5 cu in.)

So in this example we have 2 receptacles, so our calculation would be 2(2 receptacles) x 2(counts as 2 conductors) x 2.25 (our largest conductor size) = 9 cu in.

Now we total all these up- 18 + 2.25+ 9 = 29.25.  So we need a 30 cu in box as a minimum.

If you had any clamps, such as in a metal box, you would count those as one conductor using the same rule as the switch/receptacle.

Now we know the rules let’s not exceed the capacity, Okay!

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