Lesson 6



When calculating the allowable ampacity permitted on a conductor you must consider these factors.


·        The size of the conductor or CMA (circular mils area)

·        Conductor material

·        Conductor insulation

·        Temperature conductor is subjected to

·        Number or conductors installed in the raceway


As the circular mils area increases the higher the allowable ampacity that will flow.  The conductor material has a specific resistance per foot.  Copper has less resistance as compared to aluminum thus when using the same size wire and insulation, the conductor of copper will allow greater amount of current flow (I=E/R).  Insulating material also has an effect on dissipating heat faster.  If the insulation material is rated for a higher temperature then the higher the allowable ampacity permitted on the conductor. 


Since the currents listed in Table 310.15(B)(16) are for those conductors installed in a raceway with 3 current carrying conductors subjected to an ambient temperature of 86 degrees F, what happens when we increase the number of current carrying conductors and subject them to a higher ambient temperature?  The higher the temperature that the conductor is subjected to and if an increase in the amount of current carrying conductors, then the lower the amount of current that will be permitted on these conductors.  The demand factors in the NEC account for these factors (Bottom of T-310.15(B)(16), T-310.15(B)3(a)).  We simply look up our wire size and insulation ampacity and multiply by our demand factors.  For example if we are to install 6 #12 THHN copper conductors (all current carrying) in a run using 1/2 inch EMT through an area subjected to a temperature of 87 degrees F the permissible current on each conductor would be calculated as the following.



    30 amps T-310.15(B)(16)  Ampacity of 3 wires

X .96    T-310.15(B)(2)(a) Temperature Correction

  28.8 amps

X .80    T-310.15(B)(3)(a) Derating (More than 3 current carrying)

23.04 amps   Fuse at 20 amps acoording to 240.4(D)


Looking at the * in the bottom of Table 310.15(B)(16) says to see 240.4(D) which states not to exceed a 20 amp overcurrent device for a # 12 wire.


          When we are given the insulation type we almost always start derating our wires from the listed ampacity rating depending on the installation or general wiring method used (310.15(B)(3)).  Neutral conductors may be counted as a current carrying conductor if it meets the requirements of 310.15(B)(5).  Grounding or bonding conductors are not meant to carry current but shall only if a fault should occur.  These conductors shall not be current carrying.






          There are many types of boxes permitted for use.  The NEC code addresses the amount of conductors permitted in these boxes.  Boxes come in all different shapes and sizes.  A box is defined as width x thickness x depth.  If the unit of measurement is in the inch family then Inch x Inch x Inch = Cubic Inch.  The volume allowance for different size conductors occupying space in a box is given to us in T-314.16(B) is in that cubic inch capacity.  This is only applicable for wire sizes up to #6awg and if we have different conductors in the box.  If all conductors are the same size or smaller than #4awg (#6awg or smaller) we can use table 314.16(A) when sizing our metal box containing these conductors.  Non metallic boxes must have their volume allowances marked by the manufacturer in which the values given in Table 314.16(B) must be used.  This table 314.16(B) must also be used for different size wires in the same box.


          Using the charts what minimum size device box shall be used for a duplex receptacle using 2 nm cables each containing a # 14awg black, white, and an equipment grounding conductor with 1 internal clamp?




(1)               Black          2

(1)               White         2

(5)               EGC           1

(4)            Duplex Rec     2

(2)       Internal clamps    1

                         8 #14awg conductors  

Using T-314.16(A) choose = 3 x 2 x 3 ½


An alternate method is to calculate the cubic inch capacity of the conductors.


          Using T-314.16(B) for a # 14awg occupies 2 cubic inches.


#14 Black                   2 x 2cubic inches       =       4

#14 White                  2 x 2cubic inches       =       4

#14 EGC                    1 x 2cubic inches       =       2

#14 Duplex                 2 x 2cubic inches      =       4

#14 Internal clamps   1 x 2cubic inches         =      2


Choose a box with a cubic inch capacity of 16 cubic inches (3x2x3½).


          Device boxes installed shall be installed with the front edge of the box flush with the combustible surface or not greater than ¼ inch if installed with a noncombustible surface.  The area or opening around the box shall not exceed 1/8 inch from any edge of the box.  Boxes are designed to support different weights depending on the device or fixture to be attached.  These boxes if supporting 35lbs for a paddle fan and 50 lbs for a luminaire shall have their weight that the box is capable of supporting listed on the box.  If the box will be holding a lighting fixture above 50 lbs and the box does not have a marking, you must support the box independently.  A ceiling fan may also be supported independently if the paddle fan exceeds 70 lbs.  For paddle fans weighing 35-70lbs, the box must list the maximum weight it is capable of supporting and if it is suitable for that specific purpose.


          We have discussed when sizing boxes for conductors #6awg or smaller, but how do we size boxes when #4awg or larger conductors are present in the box?  The NEC article 314.28 addresses sizing these pull boxes.  For straight pulls we must multiply the size of the largest raceway entering the box by 8. 



            Length of pull box = Multiplier x Largest Raceway


            Width & Thickness = Large enough for Locknut and Bushing


In order to not exceed 360 degrees for a 4 inch conduit run we must install a pull box.  A pulling point or box for the installation of the conductors makes the pull easier.  The length on this pull box shall be at least:


8 x 4inches = 32 inches





6 x Largest Raceway + sum of other conduits= Distance to Opposite Wall


          For angle pulls, U pulls or splices, we take the largest raceway on the same wall and multiply by 6 and add the other raceways.  For example if a splice is to be made for # 2 awg conductors with three 2 inch conduits with a 4 inch conduit entering one side and all of them leaving on the adjacent side, we take the largest x multiply by 6 and add the rest. 


4 inch x 6 = 24 +2 + 2 + 2 = 30 inches


                   3oin x 30in box is required



          If we had different number of raceways or sizes on two walls, we would do this calculation twice. 


For example:

 On one wall we have conduits of one 4inch and three 2 inch conduits entering the box.

In this angle pull the wall where the conduits are leaving contain one 3 inch and two 2 inch raceways. 




The size of the box would be: 

          Dimension A = To the opposite wall






A = 6 x 4in = 24 + 2 + 2 + 2 = 30 inches


Dimension B = To opposite wall





B = 6 x 3in = 18 + 2 + 2 = 22inches





       A          X             B       =       PROPER SIZE BOX


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