NEC Questions and Answers – Based on the 2011 NEC
September 2011 - Part 2 of 2
By Mike Holt for EC&M Magazine
Here’s the follow up to yesterday’s newsletter. This includes all of the answers to the questions sent, so you can see how you did.
Q1. Are lighting outlets in dwelling units allowed on the same circuit as receptacle outlets?
A1. An individual branch circuit is permitted to supply any load for which it’s rated, but in no case is the load permitted to exceed the branch-circuit ampere rating [210.23].
Branch circuits rated 15A or 20A supplying two or more outlets must only supply loads in accordance with 210.23(A), which says that a 15A or 20A branch circuit is permitted to supply lighting, equipment, or any combination of both.
Cord-and-plug-connected equipment not fastened in place, such as a drill press or table saw, must not have an ampere rating more than 80 percent of the branch-circuit rating [210.23(A)(1)].
Equipment fastened in place (other than luminaires) must not be rated more than 50 percent of the branch-circuit ampere rating if this circuit supplies luminaires, receptacles, or both [210.23(A)(2)].
Question: Can a whole house (central) vacuum motor rated 13A be installed on an existing 20A circuit that supplies more than one receptacle outlet?
Answer: No, an individual 15A or 20A branch circuit will be required.
Q2. What field labeling is required to identify circuits served by electrical equipment such as panelboards and motor control centers?
A2. All circuits, and circuit modifications, must be legibly identified as to their clear, evident, and specific purpose. Spare positions that contain unused overcurrent devices must also be identified. Identification must include sufficient detail to allow each circuit to be distinguished from all others, and the identification must be on a circuit directory located on the face or inside of the door of the panelboard. See 110.22 [408.4(A)].
All switchboards and panelboards supplied by a feeder in other than one- or two-family dwellings must be marked as to the device or equipment where the power supply originates [408.4(B)].
Q3. Does the NEC require that replacement receptacles be updated to provide GFCI protection?
A3. When existing receptacles are replaced in locations where GFCI protection is currently required, the replacement receptacles must be GFCI protected. This includes the replacement of receptacles in dwelling unit bathrooms, garages, outdoors, crawl spaces, unfinished basements, kitchen countertops, rooftops, or within 6 ft of laundry, utility, and wet bar sinks [406.4(D)(3)].
Q4. Is GFCI protection required for receptacles installed for coffee makers in non dwelling locations, such as a counter top with a sink in a conference room?
A4. GFCI protection is required for all 15A and 20A, 125V receptacles installed in commercial/industrial occupancies in the specific locations listed in 210.8(B), including receptacles installed within 6 ft of the outside edge of a sink [210.8(B)(5)].
Ex 1: In industrial laboratories, receptacles used to supply equipment where removal of power would introduce a greater hazard aren’t required to be GFCI protected.
Ex 2: Receptacles located in patient bed locations of general care or critical care areas of health care facilities aren’t required to be GFCI protected.
Q5. When a metal water pipe is used as a grounding electrode, is the concrete encased electrode satisfactory as the supplemental electrode, or is a ground rod still required?
A5. When an underground metal water pipe grounding electrode is present [250.52(A)(1)], it must be supplemented by one of the following electrodes [250.53 (D)(2)]:
• Metal frame of the building/structure electrode [250.52(A)(2)]
• Concrete-encased electrode [250.52(A)(3)]
• Ground ring electrode [250.52(A)(4)]
• Ground rod electrode meeting the requirements of 250.52(A)(5)
• Other listed electrodes [250.52(A)(6)]
• Metal underground systems, piping systems, or underground tanks [250.52(A)(8)]
The termination of the supplemental grounding electrode conductor must be to one of the following locations:
(1) Grounding electrode conductor
(2) Service neutral conductor
(3) Metal service raceway
(4) Service equipment enclosure
Ex: The supplemental electrode is permitted to be bonded to interior metal water piping located not more than 5 ft from the point of entrance to the building/structure [250.68(C)(1)].
Q6. When are systems required to be grounded?
A6. Systems operating below 50V aren’t required to be grounded or bonded in accordance with 250.30 unless the transformer’s primary supply is from [250.20(A)]:
- A 277V or 480V system.
- An ungrounded system.
For systems over 50V, the following systems must be grounded (connected to the earth) [250.20(B)]:
- Single-phase systems where the neutral conductor is used as a circuit conductor.
- Three-phase, wye-connected systems where the neutral conductor is used as a circuit conductor.
- Three-phase, high-leg delta-connected systems where the neutral conductor is used as a circuit conductor.
Q7. Is a supplemental grounding electrode always required?
A7. A single rod, pipe or plate electrode must be supplemented by an additional electrode that’s bonded to one of the following [250.53(A)(2)]:
(1) The single rod, pipe, or plate electrode
(2) The grounding electrode conductor of the single electrode
(3) The neutral service-entrance conductor
(4) The nonflexible grounded service raceway
(5) The service enclosure
Ex: If a single rod, pipe, or plate grounding electrode has an earth contact resistance of 25 ohms or less, the supplemental electrode isn’t required.
Q8. What are the requirements in the 2011 NEC for the marking of service equipment with the available fault current?
A8. Service equipment in other than dwelling units must be legibly field-marked with the maximum available fault current, including the date the fault current calculation was performed and be of sufficient durability to withstand the environment involved [110.24(A)].
When modifications to the electrical installation affect the maximum available fault current at the service, the maximum available fault current must be recalculated to ensure the service equipment ratings are sufficient for the maximum available fault current at the line terminals of the equipment. The required field marking(s) in 110.24(A) must be adjusted to reflect the new level of maximum available fault current [110.24(B)].
Ex: Field markings aren’t required for industrial installations where conditions of maintenance and supervision ensure that only qualified persons service the equipment.
Q9. How is the ampacity determined for NM-B Cable?
A9. Conductor ampacity is calculated on the 90°C insulation rating of the conductor in accordance with Table 310.15(B)(16), however the conductors must be sized to the termination temperature rating of 60°C [334.80].
Question: What size Type NM cable is required to supply a 9.60 kW, 240V, single-phase fixed space heater with a 3A blower motor? The terminals are rated 75°C.
(a) 2 AWG (b) 4 AWG (c) 6 AWG (d) 8 AWG
Answer: (b) 4 AWG
Step 1:Determine the total load in amperes:
I = VA/E
I = 9,600W/240V + 3A
I = 40A + 3A
I = 43A
Step 2: Conductor and Protection Size [424.3(B)]. Size the ungrounded conductors and overcurrent device at no less than 125 percent of the total heating load.
Conductor/Protection Size = Load x 1.25
Conductor/Protection Size = 43A x 1.25
Conductor/Protection Size = 53.75A
According to Table 310.15(B)(16), a 6 AWG conductor rated 55A at 60°C, protected with a 60A overcurrent device [240.6(A)].
If multiple Type NM cables pass through the same wood framing opening that’s to be sealed with thermal insulation, caulking, or sealing foam, the allowable ampacity of each conductor must be adjusted in accordance with Table 310.15(B)(3)(a) [334.80].
Q10. What are the acceptable wiring methods for PV circuits inside buildings/structures?
A10. PV system dc conductors (source and output) run inside a building/structure must be contained in a metal raceway, Type MC cable, or metal enclosure [690.31(E)].
Wiring methods cannot be located within 10 in. of the roof decking or sheathing, except where located directly below the roof surface that’s covered by PV modules and associated equipment and must be run perpendicular (90°) to the roof penetration point [690.31(E)(1)].
Note: The 10 in. from the roof decking/sheathing requirement is to prevent accidental contact with energized conductors from saws used by firefighters for roof ventilation during a structure fire.
FMC smaller than trade size ¾ or Type MC cable smaller than 1 in. in diameter installed across ceilings or floor joists must be protected by substantial guard strips that are at least as high as the wiring method. Where run exposed, other than within 6 ft of their connection to equipment, wiring methods must closely follow the building surface or be protected from physical damage by an approved means [690.31(E)(2)].
Wiring methods and enclosures containing PV source conductors must be marked with the wording “Photovoltaic Power Source” by labels or other approved permanent marking means [690.31(E)(3)]
The marking required by 690.31(E)(3) must be visible after installation and appear on every section of the wiring system separated by enclosures, walls, partitions, ceiling, or floors. Spacing between labels/marking must not be more than 10 ft and labels must be suitable for the environment [690.31(E)(4)].
Q11. How is the PV source circuit current calculated?
A11. The maximum PV source circuit current is calculated by multiplying the module nameplate short-circuit current rating (Isc) by 125 percent [690.8(A)(1)].
Example: What’s the maximum PV source circuit current for twenty-three series connected dc modules having a nameplate Isc of 11.80A
Maximum PV Source Circuit Current = Module Isc × 1.25
Maximum PV Source Circuit Current = 11.80A × 1.25
Maximum PV Source Circuit Current = 14.75A
The maximum PV output circuit current is equal to the sum of parallel PV source circuit currents as calculated in 690.8(A)(1) [690.8(A)(2)].
Example: What’s the maximum PV output circuit current for two strings, each containing twenty-three dc modules having a nameplate Isc of 11.80A?
Maximum PV Output Circuit Current = (Module Isc × 1.25)* × Number of Strings
Maximum PV Output Circuit Current = (11.80A × 1.25)* × 2
Maximum PV Output Circuit Current = (14.75A)* × 2
Maximum PV Output Circuit Current = 29.50A
*690.8(A)(1)
Q12. How is the conductor ampacity for PV circuits calculated?
A12. PV circuit conductors must be sized to the larger of 690.8(B)(2)(a) or 690.8(B)(2)(b).
Sizing according to 690.8(B)(2)(a): PV circuit conductors must be sized to carry not less than 125 percent of 690.8(A) current before the application of conductor ampacity correction [310.15(B)(2)(a) and 310.15(B)(3)(c)] and adjustment [310.15(B)(3)(a)].
Example—PV Source Circuit: What’s the minimum PV source circuit conductor ampacity before the application of conductor correction or adjustment for a string having a short-circuit current rating of 11.80A; assuming all terminals are rated 75°C?
Conductor Ampacity = (Module Isc × 1.25)* × 1.25
Conductor Ampacity = (11.80A × 1.25)* × 1.25
Conductor Ampacity = (14.75A)* × 1.25
Conductor Ampacity = 18.44A
Conductor Ampacity = 14 AWG rated 20A at 75ºC [Table 310.15(B)(16)]
*690.8(A)(1)
Example—PV Output Circuit: What’s the minimum PV output circuit conductor ampacity, before the application of conductor correction or adjustment, supplied by two PV source circuits, each having a short-circuit current rating of 11.80A; assuming all terminals are rated 75°C?
Conductor Ampacity = (Module Isc × 1.25 × Number of Source Circuits)* × 1.25
Conductor Ampacity = (11.80A × 1.25 × 2 strings)* × 1.25
Conductor Ampacity = 29.5A* × 1.25
Conductor Ampacity = 36.88A
Conductor Ampacity = 8 AWG rated 50A at 75ºC [Table 310.15(B)(16)]
*690.8(A)(2)
Compare this to sizing according to 690.8(B)(2)(b): Circuit conductors must be sized to carry 100% of 690.8(A) current after the application of conductor ampacity correction [310.15(B)(2)(a) and 310.15(B)(3)(c)] and adjustment [310.15(B)(3)(a)].
Example—PV Source Circuit: What’s the PV source circuit conductor ampacity after temperature correction for two current-carrying size 12 THWN-2 conductors in a circular raceway on the roof, ambient temperature is 90°F with 60°F temperature added in accordance with Table 310.15(B)(3)(c), supplying modules having a nameplate Isc rating of 11.80A?
Conductor Ampacity = Table 310.15(B)(16) Ampacity at 90°C Column x Temperature Correction
Temperature Correction = 0.58, Table 310.15(B)(2)(a) based on 150°F (ambient plus 60°F roof temperature adder)
Conductor Ampacity = 30A x 0.58
Conductor Ampacity = 17.40A, which has sufficient ampacity after correction and adjustment to supply 690.8(A)(1) calculated current of 14.75A (11.80A x 1.25).
Example—PV Output Circuit: What’s the PV output current ampacity after temperature correction for two current-carrying 8 size THWN-2 conductors in a circular raceway on the roof where the ambient temperature is 90°F supplied by two parallel PV source circuits, each having a Isc rating 11.80A?
Conductor Ampacity = Table 310.15(B)(16) Ampacity at 90°C Column x Temperature Correction
Temperature Correction = 0.58, Table 310.15(B)(2)(a) based on 150°F (ambient plus 60°F roof temperature adder)
Conductor Ampacity = 55A x 0.58 [Tables 310.15(B)(2)(a) and 310.15(B)(3)(c)]
Conductor Ampacity = 31.90A, which has sufficient ampacity after correction and adjustment to supply 690.8(A)(2) calculated current of 29.5A (11.80A x 1.25 x 2)
For more NEC Practice purchase Mike Holt's NEC Practice Questions book, Based on the 2011 NEC. www.MikeHolt.com/NEC
