  Find Local Contractor Call: 844-251-6308
No Obligation, Free Quotes
Home > Home Wiring USA > Definitions and Calculations > Box Fill Calculations (NEC 2002)

# Box Fill Calculations (NEC 2002)

By Warren Goodrich ## Toolbox

• Print
• Email
• Digg
• Del.icio.us
• Reddit
• Stumbleupon Share it
• Get help

If you are using conductors smaller than 4 awg. And all the same size conductors in the box use the following. We must first find the maximum numbers of conductors allowed in the box that you are using.

When you are calculating the total number of current carrying conductors in a box smaller than 4 awg conductors and all the same gauge and type insulation then, in addition to counting all actual current carrying conductors as one current carrying conductor for each conductor {any color of conductor except, green or bare}, you must add also as current carrying conductors as follows; NEC Article 314.16.B Example #1
An example of device boxes can be viewed by clicking on the picture icon to the left.

Each device (switch or receptacle) found in the box must be counted as 2 more current carrying conductors (NEC Article 314.16.4). Example #2
An example of a yoke of a device that must be counted as the 2 current carrying conductors may be viewed by clicking on the picture icon to the left.

## Counting Grounding Conductors, Clamps and Fittings

All grounding conductors in that box must be counted as a total to be one conductor count and considered as only one more current carrying conductor all together same as one conductor count. NEC Article 314.16.5 All clamps must be counted with the total number of clamps found in the box counted as only one more current carrying conductor. NEC Article 314.16.2 You must add one more current carrying conductor count if you have fittings such as studs or hickeys. Example #3
A hand drawn example of a hickey can be viewed by clicking on the picture icon to the left.

Keep in mind that you are not supposed to count conductors that neither enters your box nor leaves your box. [pigtails only that originated within the box and stays within the box, these are ignored].

If you are using a steel box that is listed in the NEC Chart 314.16.A, then the chart should give you the maximum current carrying conductors allowed in the steel box.

## Completing Calculations

Whether you are using a steel device box or a plastic device box a final result calculation of adding up all the number of current carrying conductors that you must count as being in that box, by your wiring design must be calculated. The total number of current carrying conductors that you designed to be in that box whether actual current carrying conductor or devices etc. listed above that also must be counted as current carrying conductor must then be compared to the cubic inch capacity of the device box you are calculating to fill to the maximum allowed in that box as required in NEC Article 314.16. The total number of current carrying conductors that you designed to be in that box must not exceed the total number of current carrying conductors allowed in that box comparing your cubic inch calculated to the actual cubic inch capacity of your device or lighting style box.

## Other Types of Calculation

You may also use the following method of calculation if you don’t have a Code book, are using a size of steel box that is not listed in chart NEC Article 314.16.A, or you are using a plastic or fiber box. If you are using a type of box that is not listed in the NEC Chart 314.16.A, then you must measure the inside dimensions of that box whether a device box or a lighting fixture style box or even a junction box that has all the same size conductors and the same type insulation and all conductors are smaller than 4 awg conductors, then multiply the height, times the length, times the depth of the box to find the cubic inch capacity of that box.

Now you must look in the NEC Chart 314.16.B to find the cubic inch required per conductor rated by the size of the conductor that you are using.

The Chart Mentioned Says That (NEC Article 314.16.B):

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.

## Calculating Current Carrying Conductors

If you are using all conductors of the same size in your box, then you must count the number of current carrying conductors [all colors including white but not counting green or bare] entering your box. Also do not count conductors that neither enters your box nor leaves your box. [pigtails only, these are ignored].

Now add to your number of current carrying conductor list, by counting all of the grounding conductors as one conductor [green or bare], no matter how many grounding conductors, just add the one current carrying conductor to your total number of current carrying conductor list. Remember that all grounding conductors [green or bare] must be counted as a total of one current carrying conductor. NEC Article 314.16.B.5

Clamps are also counted the same as grounding conductors, one current carrying conductor must be added for the total of all clamps found in the box. All clamps found within your box that are entering the box at least ½”, no matter how many, count as a total of one current carrying conductor, only, for all of these clamps. Now add this one conductor count to your total number of current carrying conductor list, if any of these clamps are present. NEC Article 314.16.B.2 A single gang plastic or fiber box will have no clamps to consider. They are exempt from a clamp requirement. NEC Article 314.17.C.Exception

Devices yokes must count as 2 conductors for each device. Count the number of devices {switches or receptacles}. Multiply the total number of devices times 2. The answer from multiplying the total number of devices by the 2 is the total number of current carrying conductors you must add to your total number of current carrying conductor list. NEC Article 314.16.B.4

Now this final total of your current carrying conductor list is the answer to the total number of current carrying conductors installed in your box.

## Calculating Cubic Inch Required

Now multiply the cubic inch required for the size of conductors in your box found in the NEC Chart 314.16.B and listed above in the previous paragraph telling how many cubic inch must be calculated per conductor by the answer that you found in your total number of current carrying conductor list. This is the total cubic inch required for all of the conductors, equipment, and devices in your box.

Compare this total cubic inch required to the total cubic inch capacity of your box. You must not exceed the capacity of your box with the total cubic inch required by your conductors, equipment, and devices that you installed in that box.

## Example Calculation

Example calculation of a box containing all the same size conductors in a specific box whether a device box, lighting style box, or junction box

You have a four gang plastic or fiber or steel device box that requires clamps as per the NEC. In this box you intend to install three switches, and one receptacle. Entering that box you have five Romex cables that are 12/2 with ground. You also have two of the three switches that are three way switches. Because of the three way switches you also have two 12/3 with ground Romex cables entering your box. Each cable has its own equipment grounding conductor and there are also four 12 awg pigtails that originate within the box and does not leave that box.

## How to Calculate the Example Above

To calculate the above example we would ignore the pigtail wires because they originate within the box and do not leave that box. We would then count all the equipment grounding conductors whether bare or with green insulation as only one current carrying conductor. We would count both the receptacle and the switches as two current carrying conductors each. We would then add one current carrying conductor for the combined total of all clamps within that box. All of the insulated conductors including the white wires must be considered as current carrying conductors but ignoring any green insulated conductors because they are already counted. We would use NEC Chart 314.16.B to perform our calculations. The calculation would be as follows:

1. Four switches and receptacles combined X 2 each = 8 current carrying conductors X 2.25 cu. in. = 18.00 cubic inch
2. All grounding conductors whether bare or green counts as 1 current carrying conductor X 2.25 cu. in. = 2.25 cubic inch
3. All clamps within that box counted as one current carrying conductor x 2.25 cu. in. = 2.25 cubic inch
4. Five 12/2 cables equal two current carrying conductors each equals 10 conductors x 2.25 cu. in. = 22.50 cubic inch
5. Two 12/3 cables equal three current carrying conductor each equals 6 conductors x 2.25 cu. in. = 13.50 cubic inch
6. Total cubic inch required by these conductors in this certain box = 58.50 cubic inches

## Final Result

Your device box is 8" wide x 4" tall x 2 ¾" deep. The total cubic inch of the inside of this box would be 88 cubic inches.

Therefore the above calculation when the total cubic inches are calculated as required for the conductors you installed in that certain box is 58.50 cubic inches and you have a total of 88 cubic inches inside that box so your are with the box fill maximum limit requirements of that certain box.

Conductors smaller than 4 awg. with different size conductors in the same box
If you are using different size conductors in the same box, you must perform the same conductor count and calculation as if using all of the same size conductors in your box as described above except the following concerns or added calculations.

You must separate each different size of current carrying conductor calculation into separate calculations for each size conductor multiplied times the assigned cubic inch required in the NEC Chart 314.16.B. You must also change the counts calculation for your devices, clamps, grounding conductors at a total of one, hickeys, and any other associated parts that are in that box that were mentioned in the above paragraph by calculating these items as if they were the same as if they were the largest conductor in the box, when you use the NEC Article 314.16.Bcalculations. Ignore any smaller conductors for the device and clamps etc. in your calculations section when it comes to these items mentioned above such as devices, clamps etc. They must be counted as if they were equal to the largest conductor found in that box that has different gauge conductors in that certain box . Only consider these equipment and devices etc. mentioned in the previous paragraph equal to the largest size conductor present in your box to calculate the current carrying conductors that you must add to your list for these equipment and devices. You must also change the one conductor count for the total number of grounding conductors using the largest grounding conductor size found in the box, when you use the NEC Chart 314.16.B.

Then add all of the cubic inch requirements from your list, all added together, for the total cubic inch required and compare that total cubic inch required to the cubic inch capacity of your box.

## Do Not Exceed Capacity

You must not exceed the total cubic inch capacity of the box compared to the actual total cubic inch required for the conductors, equipment, and devices etc. present in your box, as you have calculated.

## Four Gang Plastic, Fiber, or Steel Device Box That Requires Clamps

You have a four gang plastic or fiber or steel device box that requires clamps as per the NEC. In this box you intend to install three switches, and one receptacle. Entering that box you have three Romex cables that are 12/2 with ground. You also have two Romex cables that are 14/2 with ground You also have two of the three switches that are three way switches. Because of the three way switches you also have one 12/3 with ground Romex cable and one 14/3 with ground Romex cable entering your box. Each cable has its own equipment grounding conductor and there are also four 12 awg pigtails that originate within the box and does not leave that box.

## How to Calculate

To calculate the above example we would ignore the pigtail wires because they originate within the box and do not leave that box. We would then count all the equipment grounding conductors whether bare or with green insulation as only one current carrying conductor but we must calculate that one current carrying conductor as the largest conductor in the box being a 12 awg conductor. We would count both the receptacle and the switches as two current carrying conductors each again calculating these conductors used to count the current carrying conductors for those devices as 12 awg also being the largest conductor in the box no matter what size conductors are actually connected to those switches or receptacles [devices]. We would then add one current carrying conductor for the combined total of all clamps within that box. Again we must count that one conductor for the clamps to also be same as the largest conductor in that box. All of the insulated conductors including the white wires must be considered as current carrying conductors but ignoring any green insulated conductors because they are already counted. However we must count the 12 awg conductors as requiring 2.25 cubic inch per conductor and the 14 awg conductors as requiring 2 cubic inch per conductor. We would use NEC Chart 314.16.Bto perform our calculations. The calculation would be as follows:

## Example Calculation

Example calculation of a box containing different same size conductors in a specific box whether a device box, lighting style box, or junction box:

1. Four switches and receptacles combined X 2 conductors each = 8 conductors X 2.25 cu in. = 18.00 cubic inch
2. All grounding conductors whether bare or green counts as 1 current carrying conductor X 2.25 cu. in. = 2.25 cubic inch
3. All clamps within that box counted as one current carrying conductor x 2.25 cu. in. = 2.25 cubic inch
4. Three 12/2 cables equal two current carrying conductors each equals 6 conductors x 2.25 cu. in. = 13.50 cubic inch
5. Four one 12/3 cables equal three current carrying conductor each equals 3 conductors x 2.25 cu. in.= 6.75 cubic inch
6. Two 14/2 cables equal two current carrying conductors each equals conductors x 2.00 cu. in. = 8.00 cubic inch
7. One 14/3 cables equal three current carrying conductor each equals 3 conductors x 2.00 cu. in. = 6.00 cubic inch
8. Total cubic inch required by these conductors in this certain box = 56.75cubic inches

## Final Result

Your device box is 8"wide x 4" tall x 2 ¾" deep. The total cubic inch of the inside of this box would be 88 cubic inches.

Therefore the above calculation when the total cubic inches are calculated as required for the conductors you installed in that certain box is 56.75 cubic inches and you have a total of 88 cubic inches inside that box so your are with the box fill maximum limit requirements of that certain box.

IF YOU ARE USING CONDUCTORS 4 Awg. OR LARGER THEN REFER TO 314.28 OF THE NEC

## Calculating a Straight Pull

If you are installing conduits into a pull box that has a straight pull entering that box from one end of the box and leaving that box straight out the other end of that box then you have what is called a straight pull “pull”type box. To calculate a straight pull in a box concerning box fill when you are installing 4 awg conductors or larger in that box then you must calculates as follows;

## How to Calculate

If you have more than one conduit or just one conduit you would pick the largest conduit entering either end of that straight pull box. If you had a 4”conduit and a 2”conduit entering that straight pull box at one end and you had a 3”conduit and a 2”conduit entering that box from the other end of that same box then you would take the 4”diameter conduit to perform your calculation in sizing the minimum length of that straight pull box. You would ignore the two 2”conduits and also ignore the one 3”conduit. You would only use the largest conduit entering that straight pull box to perform your calculations. Then you would take that 4”conduit and multiply the diameter of that one conduit times 8 as required in NEC Article 314.28.A.1

## Example of Straight Pull Box Calculation An example matching the straight pull box and calculation below may be viewed by clicking on the picture icon to the left.

The answer to the calculation of the example straight pull box with a 3”and a 2”conduit entering one end plus a 3”and a 2 ½”conduit coming into the second end of this straight pull box you would discard both 2”conduits and also discard the one 3”conduit and perform you calculation using the one largest conduit entering that straight pull box ignoring all others at this time. The calculation would use the remaining 3”conduit. You would then multiply the 3”conduit times 8 as required in NEC Article 314.28.A.1.

## Final Result

The answer would be 24”. Therefore you box must be a minimum of 24”long between the to ends of that box that have straight pulls through that box.

The width of that box would only be required to be wide enough to install the 2 ½”and the 3”conduits in the same end of that box without running into the sides of that box. If you considered that the lock nuts would take up about ¼”to ½”more than the conduit itself you should measure the outside width of your two conduits and add about 2”leaving you room to install the lock nuts required to install and secure you conduits into the end of that box.

## Things to Remember When Calculating

You must remember that if conduits enter the opposite plane sides of that box while having straight pulls also going the other way through that box then you must again calculate those sides to find the distance required to be away from that opposite wall also.

If you have an angle pull anywhere in that box then you must refer to the rules of calculating the angle pull to discover the minimum distance to each opposite wall of that box and you no longer have just a straight pull box but an angle pull box.

If you have a straight pull and a “U”pull in the same box you may still calculate as a straight pull box, but you must also calculate the distance between the two conduits containing that “U”pull also multiplying that largest conduit of any “U”pull times 6 to find the minimum distance between conduits entering the same wall making that “U”pull. See below for further explanations of a “U”pull calculation.

## Example of Angle Pull Calculation

An example drawing matching the following calculation of an angle pull may be viewed by clicking on the picture icon to the left.

Calculating an Angle Pull with 4 Awg or Larger Conductors An example drawing matching the following calculation of an angle pull may be viewed by clicking on the picture icon to the left.

If you have boxes with conductors larger than 4 Awg. that are installed in a box, then you must calculate in a different manner. In short, a box with 4 Awg. or larger conductors with a “U”and / or an angle pull the box fill must be calculated by adding the diameter of all of the conduits on the same side of the box and in the same row for the first total. Then add each row installed on that same side of the box. Now pick the largest one row total, of each of the rows, and use that in your calculation. Ignore the other rows on that side of the box that is not in that same row. NEC Article 314.28.A.2 Now pick the largest conduit in that largest row and multiply that conduit diameter times 6. NEC Article 314.28.A.2 Now add the total diameters of all of the conduits in that same row to that times 6 multiplication answer. NEC Article 314.28.A.2 Ignore all of the other rows on that same side. This answer is the minimum distance in inches that the opposite wall of that box must be from that side that you calculated. NEC Article 314.28.A.2 Now calculate each side of the same box for the answer to the distance required to the opposite wall of the box from each side that you calculate. This is the size of box required after finding the distances required to each opposite walls of that box containing an angle or “U”pull to allow that box to contain the 4 Awg. or larger conductors. NEC Article 314.28.A.2

## Remember Minimum Distances

There also is a minimum distance allowed between two conduits on opposite wall containing that angle pull with one side of a box that contains a conductor going to a conduit on an adjacent side of the box in an angle pull. You must find the largest conduit of that certain angle pull and multiply it times 6 to find the minimum distance required to be between these to certain angle pull conduits containing the same conductors of that angle pull. Please be aware that you may have to calculate these distances between conduits with common angle pulls of the same wire many times just in one box. Example of Distance of Concern in an Angle Pull

To see an example of the distance of concern in an angle pull between to conduits of a common conductor in an angle pull, click on the picture icon to the left.

Calculating a U Pull with 4 awg or Larger Conductors

To calculate the distance required between to conduits entering the same side of a box in a “U”pull. You must take the largest of the two conduits containing a conductor of a “U”pull and multiply that conduit times 6. NEC Article 314.28.A.2 The answer to this multiplication would be the minimum distance between two conduits containing the same conductor in a “U”pull. NEC Article 314.28.A.2 Remember that even though you have a “U”pull you must still also calculate the distance from that wall where those two conduits enter that box to find the distance required to the opposite wall same as if it was an angle pull. Just refer to calculating the distance required to the opposite wall the same as if it was an angle pull as described in the previous section pertaining to angle pulls.

## Two Calculations Required

You must make two types of calculations in angle pulls and “U”pulls. One calculation to find the distance required to the opposite wall of each side that has a conduit entering that box and also you must make a second calculation to make sure you meet the minimum distance of the same set of conductors entering two conduits whether on the same wall or adjoining walls of that box.

Example of Calculating the Distance in a "U" Pull Pattern An example of calculating the distance required between two conduits entering the same side of a box thatcontain the same set of conductors in a “U”pull pattern - you may view an example of this calculation by clicking on the picture icon to the left.

Calculating an angle pull or “U”pull with 4 awg or larger conductors but these conductors are contained within a nonmetallic sheathed cable instead of a conduit

If you are using nonmetallic sheathed cables with 4 Awg. or larger in them then you must measure the widest point across the nonmetallic sheathed cable and treat this measurement as the diameter of a conduit in your calculation. The intent of the nonmetallic sheathed cable widest point is to find the size of conduit required to contain than nonmetallic sheathed cable. The calculations to discover the size of a box with 4 Awg. or larger conductors gets more complicated but it should give you a general idea on the subject.NEC Article 314.28.A.2

## Just For Fun

The following box is quite complicated with many conduits entering the same sides of the box

You may view this complicated angle pull box by clicking on the picture icon to the left.

## DON’T PANIC !

Lets just take this one thing at a time. The first thing you would want to do is add up all the diameters of all the conduits on each same wall calculating each wall separately as follows:

Step One
Wall "A" has three rows of conduits on that side wall. The total of each diameter conduit in each row are as follows:

• Top row Side "A" = 16 inches
• Second row Side "A" = 17 inches
• Bottom row of side "A" = 16 inches.

Comparing all the rows in total diameter of inches the second row with 17 inches would be the largest added diameter row. We would then discard the top and bottom rows and just concern your calculations with the largest total inches in the second row totaling 17 inches.

Step Two
Then we must find the largest conduit in that second row ignoring the top and bottom rows of conduits on side "A." Now take that largest conduit which is a 4" conduit in that second row of side "A" and multiply that 4" conduit times 6. This will give us a sub answer of 24". Then we must add the diameter of all the other conduits in that second row only but not counting that 4" conduit that we multiplied by the 6 again. The total diameters of the remaining conduits in that row would be 13". We must then add that 13" total of the remaining conduits in that specific second row to the answer we received in the times 6 multiplication, that would be 24. 13" plus 24" would be a total added together of 37". The answer on side "A" would then be 37" that side "A" must be from the opposite wall which would be side "C."

Calculating Side C
Now if there were conduits entering side “C”then we should have to calculate side “C”the same as you did side “A”to find which sides comparing side “A”to side “C”would be the most distance required to each opposite wall. We would then ignore the lesser distance in the comparison between sides “A”and side “C”when we discover the minimum distance required to the opposite walls of sides “A”and side “C”.

Calculating Side D