2020 NEC FAQ's

[Dennis Alwon]

• Is pretwisting conductors required when using wirenuts?

This question is not answered in the NEC, it is a product listing question. Products are required to be used in accordance with their listing by 110.3(B).

Most (if not all) wirenuts are listed for use as follows: Pretwisting the conductors is not necessary, but not prohibited - as long as the conductors are twisted once the wirenut is in place, then the manufacturer's instructions have been followed and the code is satisfied. If the connection is not twisted, then there is a 110.3(B) violation when using most (if not all) brands of wirenuts.

That said, most electricians will agree that a pretwisted connection is the most professional method, as it allows the installer to see without a doubt that the connection is tight before installing the wirenut.

But pre-twisters and post-twisters are all welcome here.

 

[Dennis Alwon]

• Does a receptacle behind a fridge need to be a single receptacle?

This question has many different codes that inspire it, and they all address different aspects:

1. 210.52(B)(1) & (2) - If a duplex is installed, what's the other receptacle doing, if it's on the Small Appliance Branch Circuit (SABC)? Isn't this second receptacle behind the fridge prohibited?
2. 210.52(B)(1), exception 2 - If a duplex receptacle is installed, isn't this no longer an 'individual branch circuit', since there are two receptacles and one appliance?
3. 210.8(A)(6) - If the receptacle for the fridge is right next to the counter, and easily used from the countertop, does it need to be a single receptacle or GFCI protected?
4. 210.8(A)(7) - If the receptacle for the fridge is within 6' of a wet-bar sink, wouldn't it need to be a single receptacle or be GFCI protected?

I'll just take each as I've laid them out:

1. 210.52(B)(1) requires the receptacle outlets for walls, counters and refrigerators to be supplied from the SABCs. A receptacle outlet is an outlet box containing one or more receptacles; a duplex receptacle in an outlet box behind the fridge meets this requirement.
2. No: an individual branch circuit supplies only one appliance by the definition in Article 100. The definition does not discuss the number of receptacles supplied by the circuit, but the loads; therefore, a receptacle outlet supplying one appliance is an individual branch circuit.
3. No: A receptacle has to be installed over the countertop to serve the countertop (210.52(C)(5)). If it is installed behind the fridge, to serve the fridge, then 210.8(A)(6) does not apply to it. Further, there are no exceptions to 210.8(A)(6), all receptacles installed to serve the countertop are to be GFCI protected. Installing a single receptacle for a single appliance does not matter, as it does in basements and garages.
4. GFCI protection would be required whether the receptacle is a single or a duplex, as there are no exceptions for laundry, utility, and wet bar sinks as there are for basements and garages. Please note that a kitchen sink is not a wet bar sink by common interpretation, but wet bar sinks (or kitchen sinks) are not defined in the NEC. A refrigerator receptacle 1' from a sink in a kitchen does not require GFCI protection, but the same situation at a wet bar does.

Don't ask me to make sense of it, I'm just repeating what I believe.

 

[Dennis Alwon]

• Is sharing neutrals an acceptable practice? Is it code compliant?

Sharing neutrals is perfectly legal in most situations. The NEC term for several ungrounded conductors sharing a neutral is a "multiwire branch circuit." The definition is in Article 100. The key is, a multiwire branch circuit has a voltage between all the conductors of the circuit - two conductors on the "A" phase is not a multiwire branch circuit.

So, code compliance aside, is it a good idea? Many electricians do not approve of them, as disconnecting a neutral when one or more of the ungrounded conductors are live results in damage to equipment, and a shock hazard to the person working on the circuit.

The portion of the neutral that is connected to energized loads is carrying current, and can be lethal.

That said, 70% of members polled (here) use multiwire branch circuits (MWBCs), and proudly so. They are more efficient in terms of voltage drop, and in material used. Simply put, using one neutral for three line-to-neutral circuits in a commercial setting saves 66% of conductor over using dedicated grounded conductors for each circuit.

As for the potential for shock hazard, electricians need to be aware of the wiring method and disconnect and lock out all ungrounded conductors supplying a MWBC - just as they would if it were a simple two-wire circuit.

In the 2008 cycle, it appears that all handles supplying MWBCs will have to be tied together to prevent this hazard. Opinions abound about their use, but one thing is for certain; MWBCs continue to be used, and everyone should be aware of their benefits and drawbacks, for the safety of both the electrician and the equipment of the user.

Be sure to check out some applicable sections: 210.4, 501.40, 502.40, 505.21.

Related links:
Neutral on a multiwire branch circuit- started by ohhv May 28, 2008

 

[Dennis Alwon]

• Are anti-shorts (redheads) required for MC cable?

Not for MC.

NEMA has an engineering bulletin that explains that bushings are not required with MC.

Here it is, the original can be found on the web.

ENGINEERING DEPARTMENT
BULLETIN
No. 90

August 14, 2002

Use of Anti-Short Bushings for Terminating Type MC Cable
There has been much confusion within the Installation and Inspection communities regarding the
use of anti-short bushings for terminating Type MC cable. The confusion stems from the fact
that some MC cable manufacturers include anti-short bushings with their cable. The inclusion of
anti-short bushings with coils or reels of MC cable is based on historical practice relating to the
requirements of 320.40 of the NEC, which mandates the use of anti-short bushing or its
equivalent protection for Type AC Cable
Fittings used with Type MC Cable are required to be listed per 330.40 of the NEC. NEMA
supports the use of listed fittings for MC Cable. The design of these fittings may or may not
include an insulated throat however, they are required to be provided with a smooth, rounded end
stop so that the metal sheath of the cable will not pass through and the wires will not be damaged
in passing over the end stop. Whether or not an insulated throat is part of the listed product, these
listed MC fittings do not require an additional anti-short bushing. Anti-short bushings that may
be supplied by MC Cable manufacturers are for optional use by the installer, however they are
not required.


ROP #7-116 from the May 2001 Report on Proposals (ROP) for the 2002 NEC was a proposal
seeking to require anti-short bushings on all MC Cable termination installations.
The following is an excerpt from the Panel statement rejecting the proposal:

Anti-short bushings are not required for Type MC cable in accordance with the listing for
the product. The termination fittings approved for use with Type MC cables are designed
such that the wires will not come in contact with the cut edge of the armor; the throat of
the fitting is small enough to prevent contact with the armor. Type MC termination
fittings perform the same function for Type MC cable as Type AC terminations plus the
anti-short bushing do for Type AC cable.


NEMA supports the uniform adoption and enforcement of the NEC and recommends that local
Authorities Having Jurisdiction follow the requirements of NEC Section 330.40, Boxes and
Fittings for MC Cable. Section 330.40 requires that the fitting be listed, but does not mandate the
use of an anti-short bushing.

Distribution List:

Standards and Conformity Assessment Policy Committee
Codes and Standards Committee
NEMA Executive Staff

Edit for formatting

 

[Dennis Alwon]

• How do you insert a picture in the body of your message?

In order to post (insert) photos and images on this forum, the image must have a URL address in order to insert a photo.

How to post photos:

1. Open an account at Photobucket, Shutterfly or any other image hosting site. They are usually free.
Upload your images there.

2. Remember that the larger the image size, the longer it takes to upload and download. If need be, please resize your photos before uploading them.

3. When you want to post an image, right click on the image in your online hosting site (I am using Shutterfly), and you will see a dialogue box that looks like the one below. Now click on "Copy Image Address"

4. Now go to the thread you wish to post a photo in, and begin your response. When you get to where you want to insert an image, click on the icon between the link and the smiley face as shown below.

5. In the top left area of the new window that opens there is a link Icon.

Paste (ctrl + v) the image address in the window that opens. I already have an address posted in the image below. Now just choose insert and you are done.

If you want to upload an image from your computer then just go to step 4 and instead of choosing the link icon choose the one to the left of it and then select the file from your drive.

 

[Dennis Alwon]

American Disabilities Act
ADA Requirements for Switches and Receptacles

308.2 Forward Reach.

308.2.1 Unobstructed. Where a forward reach is unobstructed, the high forward reach shall be 48 inches (1220 mm) maximum and the low forward reach shall be 15 inches (380 mm) minimum above the finish floor or ground.

Figure 308.2.1 Unobstructed Forward Reach​

308.2.2 Obstructed High Reach. Where a high forward reach is over an obstruction, the clear floor space shall extend beneath the element for a distance not less than the required reach depth over the obstruction. The high forward reach shall be 48 inches (1220 mm) maximum where the reach depth is 20 inches (510 mm) maximum. Where the reach depth exceeds 20 inches (510 mm), the high forward reach shall be 44 inches (1120 mm) maximum and the reach depth shall be 25 inches (635 mm) maximum.

Figure 308.2.2 Obstructed High Forward Reach​

308.3 Side Reach.

308.3.1 Unobstructed. Where a clear floor or ground space allows a parallel approach to an element and the side reach is unobstructed, the high side reach shall be 48 inches (1220 mm) maximum and the low side reach shall be 15 inches (380 mm) minimum above the finish floor or ground.

EXCEPTIONS: 1. An obstruction shall be permitted between the clear floor or ground space and the element where the depth of the obstruction is 10 inches (255 mm) maximum.

2. Operable parts of fuel dispensers shall be permitted to be 54 inches (1370 mm) maximum measured from the surface of the vehicular way where fuel dispensers are installed on existing curbs.

Figure 308.3.1 Unobstructed Side Reach​

308.3.2 Obstructed High Reach. Where a clear floor or ground space allows a parallel approach to an element and the high side reach is over an obstruction, the height of the obstruction shall be 34 inches (865 mm) maximum and the depth of the obstruction shall be 24 inches (610 mm) maximum. The high side reach shall be 48 inches (1220 mm) maximum for a reach depth of 10 inches (255 mm) maximum. Where the reach depth exceeds 10 inches (255 mm), the high side reach shall be 46 inches (1170 mm) maximum for a reach depth of 24 inches (610 mm) maximum.

EXCEPTIONS: 1. The top of washing machines and clothes dryers shall be permitted to be 36 inches (915 mm) maximum above the finish floor.

2. Operable parts of fuel dispensers shall be permitted to be 54 inches (1370 mm) maximum measured from the surface of the vehicular way where fuel dispensers are installed on existing curbs.

Figure 308.3.2 Obstructed High Side Reach​

309 Operable Parts

 

[Dennis Alwon]

Charlie's Rule

Interpreting the NEC

It doesn't say what you think it says, nor what you remember it to have said, nor what you were told that it says, and certainly not what you want it to say. If by chance you are an instructor, it doesn't say what you have been saying, and if you?re an author, it doesn?t say what it?s intended to say.

Then what does it say? It says what it says. So if you want to know what it says, stop trying to remember what it says, don't ask anyone what is says and don?t think it says what you want it to say.

Go back and read it again and pay attention as though you were reading it for the first time. If you don?t like what it says, then get involved and try to change it. In the process, you might find out that what it says, it should be saying?

Mike Holt?s Comment: Most of the above was written by someone else, and I added my comments. If you think you have any suggestions to make this better, let me know. I would like to make a nice sticker to provide to the industry, naturally for free.

 

[Dennis Alwon]

How many nm cables can you install thru one hole

This section (334.80) is one of the most misinterpreted section in the NEC. Many inspectors say that only 2 cables are allowed but that is not what the Nec states. It says "when more than 2 cables are installed..." Let's see if I can clarify this for you.

This section is about heat that is generated thru the branch circuits or feeders. A safe installation is required so we need to allow a means for the cables to avoid overheating.

Table 310.15(C)(1) limits the number of current carrying conductor's that are bundled together. For instance, if you have 4-6 current carrying conductor's bundled then you have to de-rate the ampacity of the conductor's by 80%. From 7-9 current carrying conductor's you must de-rate 70%.

There are 2 factors we have to look at here.
1) Nm has 90C conductors inside the jacket so we can begin the de-rating from the 90C ampacity of the conductor- 110.14(C) allows this.
2) Section 334-80 limits the final ampacity of nm to 60C

Are you confused yet?

Let's look at an example.

There are 4 - 14/2 nm cables run thru a hole in the base plate of a stud wall. Table 310.16 (this table has been changed back to it's original section number) limits 14 awg copper to 25 amps at 90C. Since we have 4- cables we will have 8 current carrying conductor's. 7-9 current carrying conductor's must be de-rated at 70%

25*70% = 17.5 amps

This is still good to be installed on a 15 amp breaker. Thus we can have 4 cables thru a hole in that situation. The same holds true for 12-2 nm

30 * 70% = 21 amps

Thus a 20 amp breaker is still allowed since the 60 C rating of 12 awg is 20 amps.

The same rule applies if nm cable is run thru insulation but in either case at least 4- 12/2 or 14/2 cables can be run thru a hole. We are not limited to 2 cables. You can try the same calculation with 5 cables but you are then limited to 50% of the ampacity of the conductor.

For your convenience I quoted part of the section for you

Where more than two NM cables containing two or more current-carrying conductors are installed, without maintaining spacing between the cables, through the same opening in wood framing that is to be sealed with thermal insulation, caulk, or sealing foam, the ampacity of each conductor shall be adjusted in accordance with Table 310.15(C)(1) and the provisions of 310.14(A)(2), Exception, shall not apply.
Where more than two NM cables containing two or more current-carrying conductors are installed in contact with thermal insulation without maintaining spacing between cables, the ampacity of each conductor shall be adjusted in accordance with Table 310.15(C)(1).

Related Thread

More than 2 wires in a hole--- started by Olly - Jan 2020

 

[Dennis Alwon]

Understanding Table 220-55


The notes to this table are the most important part of the table and they must be read and understood. I will try and explain them in the following pages.

For household ranges that do not use the notes the process is quite simple. Just follow the table for the number of ranges and look up the answer or the demand percentage needed to calculate the load.

Column A is based on the percentage of the nameplate ratings of the cooking appliances that are less than 3 ½ kw. Use the nameplate rating of the appliance and multiply by the demand factor in Column A. For instance, 3 cooktops rated 3kw has a demand factor of 70% per table 220.55.

3 cooktops at 3 kw = 9 w total wattage. Using the 70% demand factor --- 9 kw *.7= 6.3 kw calculated load. Because of the diversity of the load the calculated load, in most cases, will be less than the nameplate ratings.

Column B is very much the same as Column A except that it is applies to ranges 3 ½ kw thru 8 ¾ kw.

Calculate the load for 20 ranges rated 8 kw.

20 ranges x 8 kw = 160kw but there is a demand factor for 20 ranges at 28% based on Column B for 20 ranges.

160 kw x .28 = 44.8 kw

Wow, 160 kw load is allowed to be connected at 44.8 kw. Why? Simply put it is because the ranges will never be on at the same time and with thermostats on ovens and range eyes the load will be much less than the sum of the nameplates.

Just remember that this section is based on the number of ranges and one must multiply the number of ranges by the kw of each unit to get the total kw then multiply by the demand factor.



Note 1

Over 12 kW through 27 kW ranges all of same rating. For ranges individually rated more than 12 kW but not more than 27 kW, the maximum demand in Column C shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kW.

Obviously, a 12 kw or a 27 kw range cannot be rated 8kw as Table 220.55 seems to show. This is where Note 1 comes into play.

Note 1 states that for any range larger than 12 kw we must increase the demand factor in Column C by 5% for each kw over 12.



Example 1

Let’s look at a 14kw range. A 14kw range is 2 kw larger than 12 kw (14kw-12kw). 5% for each kw over 12 kw means 5%x 2 = 10%. Take our demand factor 8kw (not 12kw) and multiple it by 10%.

8 kw X 10% = .8 kw. Now simply add it to the demand of 8kw.

8 kw +.8 kw = 8.8 kw.

8.8 kw= 8800 watts. Assuming 240v we can divide 8800/240= 36.67amps so a 40 amp circuit would be required for a 14 kw range.



Example 2

17 kw range

17 kw – 12 kw = 5 kw

5% for each kw over 12 kw = 5 x 5% = 25%

8 kw (demand) x 25% = 2 kw

8 kw + 2 kw = 10 kw

10000/240 = 41.7 amps or a 50 amp circuit.

Note 2

Over 8 3⁄4 kW through 27 kW ranges of unequal ratings. For ranges individually rated more than 8 3⁄4 kW and of different ratings, but none exceeding 27 kW, an average value of rating shall be calculated by adding together the ratings of all ranges to obtain the total connected load (using 12 kW for any range rated less than 12 kW) and dividing by the total number of ranges. Then the maximum demand in Column C shall be increased 5 percent for each kilowatt or major fraction thereof by which this average value exceeds 12 kW.

This one is a bit tricky because it hard to remember that any range rated less than 12 kw and greater than 8 ¾ kw must be used at 12 kw

Example 1

11 kw, 10 kw, 16 kw and 20 kw ranges. Since the 11 kw and 10 kw range are less than 12 kw they must be calculated at 12 kw per Note 2. What we have now is 2 - 12 kw ranges, a 16 kw and a 20 kw range. 12+12+16+20 = 60 kw and divide by 4 (number of ranges) = 60/4 = 15 kw. This 15 kw is an average of the range loads.

Table 220.55 shows a demand factor for 4 ranges at 17 kw. Note 2 states the same thing as Note 1 where the ranges are larger than 12 kw. Since the load is 15 kw the difference between 15 kw and 12 kw is 3 kw.

For each kw over 12 kw the load must be increased.

3 x 5%= 15%

15% x 17 kw = 2.55 kw

17 kw (demand load from table for 4 ranges) + 2.55 = 19.55 kw

Calculate the conductor size for the feeder for these ranges?

19,550/240 = 8.45 amps = 81 amps

At 75C a number 4 copper wire is needed.



Example 2

10 kw, 13 kw, 14 kw, 16 kw, 18 kw--- Again the 10 kw range must be used at 12 kw for this example.

12 + 13 + 14 + 16 + 18 = 73 kw/5 = 14.6 kw

From Table 220.55 -- 5 ranges at 12 kw ha a 20 kw demand. Now we have 14.6 kw as the average so 14.6 kw – 12 kw = 2.6 kw. Since .6 is a major fraction we must round up so now 2.6 kw is now 3 kw.

5% x 3 = 15%

15% x 20 kw (demand) = 3 kw

20 kw + 3 kw = 23 kw

Note 3

Over 1 3⁄4 kW through 8 3⁄4 kW. In lieu of the method provided in Column C, it shall be permissible to add the nameplate ratings of all household cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW and multiply the sum by the demand factors specified in Column A or Column B for the given number of appliances. Where the rating of cooking appliances falls under both Column A and Column B, the demand factors for each column shall be applied to the appliances for that column, and the results added together.

Example 1-- here we could use Column 3 but we will get a larger number so on tests you will probably always use Columns A or B

6 cooking appliances rated 4 kw, 5 kw, 6 kw, 7 kw, 2@ 8 kw. All fall in the range of over 1 ¾ kw thru 8 ¾ kw so we can use Note 3.

4+5+6+7+8+8 = 38 kw. In Column B for 6 appliances we have 43% demand factor so 38*.43 = 16.34 kw or 16 kw.

Example 2

6 cooking appliances but instead of them all being in Column B some will be in Column A…. 1 kw, 1 kw, 1.5 kw, 4 kw, 5kw, and 6kw.

Now we need to use Column A and B. We use Column A for the 2- 1kw and 1-1.5 kw. 1 + 1+ 1.5 = 3.5kw. For 3 appliances in Column A we have a demand factor of 70%. Thus 3.5 * .7 = 2.45 kw

For the 4kw, 5 kw, and 6 kw we must use Column B 4 +5 + 6 = 15 kw. For 3 appliances in Column B we have 55%.

15 *.55= 8.25 kw. Now we add them together…. 2.45 + 8.25 = 10.7 kw.




Note 4

Branch-Circuit Load. It shall be permissible to calculate the branch-circuit load for one range in accordance with Table 220.55. The branch circuit load for one wall-mounted oven or one counter-mounted cooking unit shall be the nameplate rating of the appliance. The branch-circuit load for a counter-mounted cooking unit and not more than two wall-mounted ovens, all supplied from a single branch circuit and located in the same room, shall be calculated by adding the nameplate rating of the individual appliances and treating this total as equivalent to one range.

The first sentence of note 4 says we can use Table 220.55—well duh—isn’t that what we have been doing… The second sentence states that we must use the nameplate rating for one oven or one counter mounted cooking unit (cooktop).

Example

An oven rated 5 kw will have a calculated load of 5kw. A cooktop rated 6 kw will have a calculated load of 6 kw. Pretty easy huh. But why isn’t there any demand allowance for these units? When an oven is turned on it will draw the rating of the unit so no demand is allowed. The cooktop does not have an oven to help with the diversity of the load so that unit must also be used at 100% also. It would be unusual to have all the burners going at the same time but it is possible.

The third sentence states that the branch circuit load for one cooktop and one or two wall ovens shall be calculated by adding the kw of all the units together and treating it as one unit.

Example

Calculate the load for one cooktop at 6 kw and 2 ovens rated 5 kw. As note 4 states we will add 6 + 5+ 5 = 16 kw.

From the table in column C we have a demand of 8 kw but we also have to use note 1. 16 kw is 4 kw greater than 12 kw so 5% x 4 = 20%

20% x 8 kw = 1.6 kw

1.6 kw + 8 kw = 9.6 kw

Here is an interesting fact.

16 kw range has the same branch circuit rating as a 9 kw range.

A 16 kw as we have just seen is calculated to be 9.6 kw. 9600/240 = 40 amps

A 9 kw range is rated 8 kw. 8000/240 = 33.3 amps or a 40 amp circuit

Let’s also look at 210.19(A)(3) which states for ranges of 83⁄4 kW or more rating, the minimum branch-circuit rating shall be 40 amperes

One last note is Note 5 which is self explanatory.

5. This table shall also apply to household cooking appliances rated over 1 3⁄4 kW and used in instructional programs





Last but not least is the calculation for ranges when we have a 3 phase service providing 208 volts.

An apartment has 10 ranges and is feed with a 3 phase 208 Volt system.

The first thing to do is figure out the max number of ranges between any 2 phases. There are 2 ways to do this but one is much simpler. You can draw it out on paper and count the number of ranges between phases but that can be difficult when you have 50 apartments. The other method is simply divide the number of ranges by 3 and if there is a remainder than add another ranges.

In the example above there are 10 ranges. As stated divide the 10 ranges by 3. 10/3= 3 with a remainder of 1 thus we have 4 ranges across 2 phases. If there were 11 ranges don’t get confused by the remainder of 2 because you would still only add 1 range for a total of 4 ranges across 2 phases.

Article 220.55 states Where two or more single-phase ranges are supplied by a 3-phase, 4-wire feeder or service, the total load shall be calculated on the basis of twice the maximum number connected

between any two phases.

There are 4 ranges and article 220.55 states to multiply by 2 and then look up that demand in Table 220.55. Column C shows 23 kw or 23,000 VA. This means there is a load of 11,500 per phase (23,000/2) or 34,500 VA for 3 phases (11,500/3). 34,500 va is the number to use when doing a load calculation for an entire apartment.