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# Box Fill Calculator

Enter the required parameters to precisely calculate â€œBox Fill" requirements for an electrical wiring box.

Largest conducting wire size

Are you using a box with internal clamps?

wires

Number of conducting wires

Will you need some support fittings?

devices

Number of devices

wires

Number of grounding conductors

Largest grounding wire size

Table of Content

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This box fill calculator precisely estimates the total box fill volumes for electrical utility boxes, prioritizing safety and electrical system reliability in various installations.

By using this tool, you can easily determine the perfect dimensions for your electrical boxes, ensuring they meet the safety standards of the National Electrical CodeÂ®.

## What Is Box Fill Calculation?

It is the calculation of the combined volume occupied by conducting wires, grounding wires, and other components in an electrical utility box.

The main objective of this box fill calculation is to prevent overcrowding in electrical boxes because an overloaded box can result in faults, arcing, or even pose a risk of fire.

## How To Calculate Box Fill?

Follow the below-mentioned steps to calculate box fill:

### Identify Components:Â

• Determine the number of conductors (wires) and their volume
• Recognize the device (switches, receptacles, etc.) that you have to install in the box
• Count any internal cable clamps, luminaire studs, or other fittings

### Assign Values:

• Every conductor is considered one
• All the devices have a specific value, mostly two

### Apply Multipliers:

• Multiply the total number of device mountings by the values specified in NEC Table 314.16(B)

### Calculate Box Fill:

• Put these values in the formulas to calculate the fill volume of each component and sum all the fill volumes to find the right box size

Now, let’s calculate the volume allowances for each component within an electrical box

### Conductor Fill:

$$\ A_w â€‹=\ n_wâ€‹$$

And

$$\ V_wâ€‹ =\ A_wâ€‹\times \ V_{largest\ conductorâ€‹}$$

Where

• $$\ A_wâ€‹\ is\ the\ conductor\ fill\ volume\ allowance$$
• $$\ n_wâ€‹\ represents\ the\ number\ of\ conducting\ wires$$
• $$\ V_wâ€‹\ is\ the\ conductor\ fill\ volume$$
• $$\ V_{largest\ conductor}â€‹\ is\ the\ free\ space\ we\ provide\ for\ the\ largest\ conductor$$

In the following table we have provided the conductor wire size and their allowance volumes. For ease, use our wire size calculator and optimize your wire size assessments.

 Size of Conductor (AWG) Volume Allowance Required Per Conductor (in.Â³) 18 1.5 16 1.75 14 2 12 2.25 10 2.5 8 3 6 5

Source Better Homes & Gardens

### Clamp Fill:

If you are using an electrical box with internal clamps, then:

$$\ A_c=\ 1$$

Otherwise (no clamps):

$$\ A_c=\ 0$$

$$\ V_câ€‹ =\ A_câ€‹\times \ V_{largest\ conductorâ€‹}$$

where

• $$\ A_c\ is\ the\ clamp\ fill\ volume\ allowance$$
• $$\ V_c\ is\ the\ clamp\ fill\ volume$$

### Support Fittings Fill:

If there are one or more luminaire studs or hickey inside the box:
$$\ A_s=\ 1$$If there are no luminaire studs or hickey inside the box:
$$\ A_s=\ 0$$$$\ V_sâ€‹ =\ A_sâ€‹\times \ V_{largest\ conductorâ€‹}$$

where

• $$\ A_s\ represents\ the\ support\ fittings\ fill\ volume\ allowance$$
• $$\ V_s\ is\ the\ support\ fittings\ volume$$

### Device or Equipment Fill:

$$\ A_d = 2 * n_d$$Where

• $$\ A_d\ is \ the\ representative\ of\ the\ device\ fill\ volume\ allowance$$
• $$\ n_d\ is\ the\ number\ of\ devices$$

$$\ V_d =\ A_d\times \ V_{largest\ conductor}$$

Where

• $$\ V_d\ is\ the\ device\ fill\ volume$$

### Equipment Grounding Conductor Fill:

If 1â€“4 grounding conductors that are entering in the electrical box,$$\ A_g=\ 1$$

If 5 or more:

$$\ A_g=\ 1+\dfrac{n_gâˆ’4}{4}$$
or
$$\ A_g=\dfrac{n_g}{4}$$

Where

• $$\ A_g\ is\ the\ representative\ of\ the\ grounding\ conductor\ fill\ volume\ allowance$$
• $$\ N_g\ is\ the\ total\ number\ of\ grounding\ conductors$$

In equation form, it can be expressed as:

$$\ V_g =\ A_g\times \ V_{largest\ ground\ Wire}$$

Where

• $$\ V_g\ shows\ the\ equipment\ grounding\ conductor\ fill\ volume$$
• $$\ A_g represents\ the\ grounding\ conductor\ fill\ volume\ allowance$$
• $$\ V_{largest\ ground\ Wire}\ is\ the\ free\ space$$

### The Total Box Fill Volume:

$$\ V_{total} =\ V_w+V_c+V_s+V_d+V_g$$

If the largest conductor is equal to the largest grounding wire size then find the total volume allowance and multiply it by the free space required for the largest conductor.

$$\ A_{total} =\ A_w+A_c+A_s+A_d+A_g$$

$$\ V_{total} =\ A_{total}\times \ V_{largest\ conductor}$$

Where

• $$\ V_{total}\ is\ the\ total\ box\ fill$$
• $$\ A_{total}\ represents\ the\ total\ volume\ allowance$$
• $$\ V_{largest\ ground\ Wire}\ is\ the\ free\ space\ as\ defined\ in\ Table\ 314.16(B)$$

## Example:

Suppose you have a box with 10 wires, where the largest conducting wire size is 12 AWG, the number of devices is 5, and there are 10 grounding conductors with the largest grounding wire size being 12 AWG, determine:

• Grounding wire volume
• Equipment grounding fill volume
• Total volume allowance needed
• Total box fill volume

Solution:

Given that:

Largest conducting wire size = 12 AWG = 2.25 cu inches
Number of conducting wires = 10
Devices = 5
Grounding conductors = 10
Largest Grounding conductor wire size = 12 AWG = 2.25 cu inches

$$\ A_w â€‹=\ n_wâ€‹$$

$$\ A_w â€‹=\ 10$$
Conductor Fill Volume:

$$\ V_wâ€‹ =\ A_wâ€‹\times \ V_{largest\ conductorâ€‹}$$
$$\ V_wâ€‹ =\ 10\times \ 2.25$$
$$\ V_wâ€‹ =\ 22.5$$

Device Fill Volume allowance:

$$\ A_d = 2 * n_d$$
$$\ A_d =\ 2\times \ 5$$
$$\ A_d =\ 10$$

Device Fill Volume:

$$\ V_d =\ A_d\times \ V_{largest\ conductor}$$
$$\ V_d =\ 10\times \ 2.25$$
$$\ V_d =\ 22.5$$

Equipment Grounding Fill Volume Allowance:

$$\ A_g=\dfrac{n_g}{4}$$
$$\ A_g=\dfrac{10}{4}$$
$$\ A_g=\ 2.5$$

Grounding Wire Volume:

$$\ Largest\ Grounding\ Wire\ Size =\ 12\ AWG$$
$$\ Grounding\ Wire\ Volume =\ 12\ AWG =\ 2.25\ cu\ inches$$

Equipment Grounding Fill Volume:

$$\ V_g =\ A_g\times \ V_{largest\ ground\ Wire}$$
$$\ V_g = 2.5*2.25$$
$$\ V_g =\ 5.625\ cu\ inches$$

Total Volume Allowance Needed:

$$\ A_{total} =\ A_w+A_c+A_s+A_d+A_g$$
$$\ A_{total} =\ 10+0+0+10+2.5$$
$$\ A_{total} =\ 22.5$$

Total Box Fill Volume:

$$\ V_{total} =\ V_w+V_c+V_s+V_d+V_g$$
$$\ V_{total} =\ 22.5+0+0+22.5+5.625$$
$$\ V_{total} =\ 50.625\ cu\ inches$$

Ensure precision by cross-checking your manual calculations with our electrical box fill calculator and optimize your results for a safe electrical system.

## The Importance of Electrical Box Fill Calculation:

The box fill calculations are very important in electrical installations for many reasons, including:

• Preventing Overcrowding: It prevents overheating, increased fire hazards, and potential damage to electrical components
• Compliance with Regulations: Helps to comply with the National Electrical Code (NEC) regulations, ensuring that the installations meet the necessary codes
• Enhancing System Reliability: Using the box fill calculations, you can form reliable electrical systems
• Safety: It’s the electricians’ main concern and can be achieved easily with the help of this calculation that they can perform manually or by using a box fill calculator.

## FAQâ€™s:

### How Many Wires Can I Put In An Electrical Box?

A standard single-gang box has 18 cubic inches of space, which can contain:

• 9 #14-gauge wires
• 8 #12-gauge wires
• 7 #10-gauge wires

These figures apply to boxes that house wires. If you have to install a box with devices, deduct two wires from the specified counts.

### Do Grounds Count In Box Fill?

Yes, the grounding wires are also included in calculating box fill volumes. According to the NFPA 70: National Electrical CodeÂ® 2020, a single volume allowance is specified for one to four equipment grounding conductors or equipment jumping conductors. Every additional grounding conductor needs an extra 1/4 volume allowance.

### What Is The NEC Code For Box Fill?

The box fill code, outlined in NEC 314.16 states that the boxes and conduit bodies must have the appropriate approved size to provide free space to all the enclosed conductors. The volume of the box calculated as per 314.16(A) should never be less than the fill calculation as determined in 314.16(B) in any case.