What is the minimum box volume required for two 12 AWG conductors and two 14 AWG conductors in a splice?

To determine the minimum box volume required for the conductors, it's important to understand how to calculate the volume based on the number and size of conductors.

According to the National Electrical Code (NEC), the volume allowance for conductors is calculated as follows:

• Each 12 AWG conductor requires a volume of 2.25 cubic inches.

• Each 14 AWG conductor requires a volume of 2.00 cubic inches.

• Additional volume must be added for any device or fitting in the box, which is not specifically indicated in this question but is typically included in calculations.

For this specific case:

• The volume for two 12 AWG conductors is:

( 2 \times 2.25 , \text{in.}^3 = 4.5 , \text{in.}^3 ).

• The volume for two 14 AWG conductors is:

( 2 \times 2.00 , \text{in.}^3 = 4.0 , \text{in.}^3 ).

Adding these volumes together gives:

( 4.5 , \text{in.}^3 + 4.0