Journeyman Electrician (NEC) — All Questions

A generator nameplate current rating is 200 amperes. What minimum conductor ampacity is required from the generator terminals to the first overcurrent device?

• a.230 amperes✓
• b.200 amperes
• c.250 amperes
• d.160 amperes

Conductors from generator terminals to the first distribution device must have an ampacity of at least 115 percent of the nameplate current rating: 200 x 1.15 = 230 A, which corresponds to 4/0 copper at 75 degrees C. The 200-ampere answer omits the factor, 250 amperes uses 125 percent, and 160 amperes derates rather than uprates.2023 NEC §445.13(A)

A photovoltaic source circuit uses modules with a rated short-circuit current of 10 amperes. What minimum conductor ampacity is required before any temperature or conduit-fill adjustments?

• a.12.5 amperes
• b.15.6 amperes✓
• c.10 amperes
• d.20 amperes

The maximum circuit current is 125 percent of the rated short-circuit current, or 10 x 1.25 = 12.5 A, and the conductor must then be sized at 125 percent of that value without adjustment factors: 12.5 x 1.25 = 15.6 A. The 12.5-ampere answer stops after only the first multiplier, and 10 amperes ignores both.2023 NEC §690.8

A motor with a service factor of 1.16 will not start because the overload device trips, even though it is sized at 125 percent of nameplate current. To what maximum percentage may the overload device be increased?

• a.140 percent✓
• b.130 percent
• c.150 percent
• d.175 percent

Where the overload device selected at the basic percentage is not sufficient to start the motor or carry the load, motors with a service factor of 1.15 or greater or a marked temperature rise of 40 degrees C or less may be increased to 140 percent of nameplate current. The 130 percent ceiling applies to all other motors, and 175 percent is a fuse percentage for short-circuit protection.2023 NEC §430.32(C)

What is the maximum non-time-delay fuse permitted for branch-circuit short-circuit and ground-fault protection of a 34-ampere three-phase squirrel-cage motor, using the next standard size allowance?

• a.110 amperes✓
• b.100 amperes
• c.90 amperes
• d.125 amperes

Non-time-delay fuses are permitted at 300 percent of full-load current: 34 x 3.00 = 102 A, which is not a standard rating, so the next standard size of 110 amperes may be used. The 90-ampere answer belongs to the inverse-time breaker at 250 percent, and 125 amperes skips a standard size.2023 NEC Table 430.52(C)(1)

A 20-horsepower, 208-volt three-phase motor has a table full-load current of 59.4 amperes. What is the minimum copper branch-circuit conductor at 75 degrees C terminations?

• a.6 AWG
• b.4 AWG✓
• c.3 AWG
• d.8 AWG

The conductor must carry 59.4 x 1.25 = 74.25 A. In the 75 degrees C column, 6 AWG copper is 65 A and falls short, while 4 AWG copper at 85 A is adequate. Choosing 3 AWG is oversized for the minimum, and 8 AWG at 50 A is well below the requirement.2023 NEC §430.22

A heating element of 30 ohms is connected across 240 volts. What current flows?

• a.8 amperes✓
• b.0.125 ampere
• c.7,200 amperes
• d.12 amperes

Ohm's law gives I = E / R = 240 / 30 = 8 amperes. The 0.125-ampere answer inverts the division, 7,200 multiplies voltage by resistance instead of dividing, and 12 amperes would require a 20-ohm element.

What power is consumed by a 120-volt load drawing 12 amperes at unity power factor?

• a.1,440 watts✓
• b.10 watts
• c.132 watts
• d.1,200 watts

Power in a resistive single-phase circuit is P = E x I = 120 x 12 = 1,440 watts. The 10-watt answer divides instead of multiplying, 132 watts adds the two values, and 1,200 watts does not follow from the given numbers.

How much power is dissipated when 5 amperes flows through a 20-ohm resistor?

• a.100 watts
• b.500 watts✓
• c.4 watts
• d.2,000 watts

Using P = I squared x R, the calculation is 5 x 5 x 20 = 500 watts. The 100-watt answer multiplies current by resistance without squaring the current, 4 watts divides the wrong way, and 2,000 watts squares the resistance instead of the current.

Three resistors of 10, 15 and 25 ohms are connected in series across 120 volts. What is the circuit current?

• a.24 amperes
• b.2.4 amperes✓
• c.4.8 amperes
• d.0.42 ampere

Series resistances add: 10 + 15 + 25 = 50 ohms, so I = 120 / 50 = 2.4 amperes. The 24-ampere answer misplaces the decimal, 4.8 amperes would result from a 25-ohm total, and 0.42 ampere inverts the division.

A 20-ohm resistor and a 30-ohm resistor are connected in parallel. What is the total resistance?

• a.50 ohms
• b.12 ohms✓
• c.25 ohms
• d.10 ohms

For two resistors in parallel, R total = (R1 x R2) / (R1 + R2) = (20 x 30) / 50 = 600 / 50 = 12 ohms. The 50-ohm answer adds them as if they were in series, and 25 ohms averages them, but total parallel resistance is always less than the smallest branch resistance.

A balanced three-phase load draws 40 amperes at 480 volts with a power factor of 0.85. What is the true power?

• a.Approximately 33.3 kilowatts
• b.Approximately 28.3 kilowatts✓
• c.Approximately 19.2 kilowatts
• d.Approximately 16.3 kilowatts

True power for a three-phase load is P = 1.732 x E x I x PF = 1.732 x 480 x 40 x 0.85, or about 28,270 watts. The 33.3-kilowatt answer omits the power factor and gives apparent power in kVA instead, 19.2 kilowatts leaves out the 1.732 factor, and 16.3 kilowatts omits both.

On a 208Y/120-volt wye system, what is the relationship between line voltage and phase voltage?

• a.Line voltage equals phase voltage divided by 1.732
• b.Line voltage equals phase voltage
• c.Line voltage equals phase voltage multiplied by 3
• d.Line voltage equals phase voltage multiplied by 1.732✓

In a wye configuration the line voltage is the square root of three times the phase voltage: 120 x 1.732 = 208 volts, while line current equals phase current. In a delta configuration the relationship reverses, with line voltage equal to phase voltage and line current equal to 1.732 times phase current.

A balanced delta-connected load has a phase current of 20 amperes. What is the line current?

• a.20 amperes
• b.Approximately 34.6 amperes✓
• c.Approximately 11.5 amperes
• d.60 amperes

In a delta connection each line conductor feeds two phase windings, so line current equals 1.732 times phase current: 20 x 1.732 = 34.6 amperes. The 20-ampere answer describes a wye connection, 11.5 amperes divides instead of multiplying, and 60 amperes triples the value rather than applying the square root of three.

A single-phase transformer has a 480-volt primary and a 120-volt secondary. If the secondary carries 100 amperes, what is the approximate primary current?

• a.50 amperes
• b.100 amperes
• c.400 amperes
• d.25 amperes✓

Voltage and current transform inversely, so the 4 to 1 voltage ratio produces a 1 to 4 current ratio: 100 / 4 = 25 amperes on the primary. Neglecting losses, the volt-amperes match on both sides, 480 x 25 = 120 x 100 = 12,000 VA. The 400-ampere answer inverts the ratio and would represent an impossible power gain.

What is the secondary full-load current of a 25 kVA single-phase transformer with a 240-volt secondary?

• a.Approximately 60 amperes
• b.Approximately 104 amperes✓
• c.Approximately 208 amperes
• d.Approximately 52 amperes

For a single-phase transformer, I = VA / E = 25,000 / 240, or about 104 amperes. The 60-ampere answer divides by 480 volts, 208 amperes divides by 120 volts, and 52 amperes is half the correct value.

A 480-volt panelboard is mounted on a wall with a grounded concrete wall directly behind the working space. What is the minimum depth of working space?

• a.2 and 1/2 feet
• b.3 feet
• c.4 feet
• d.3 and 1/2 feet✓

For nominal voltages of 151 to 600 volts to ground, Condition 2 exists when exposed live parts face a grounded surface such as a concrete or masonry wall, and the required depth is 3 and 1/2 feet. Condition 1, with no live or grounded parts opposite, requires 3 feet, and Condition 3, with exposed live parts on both sides, requires 4 feet.2023 NEC §110.26(A)(1)

What is the minimum width of working space in front of electrical equipment?

• a.30 inches in all cases regardless of equipment width
• b.30 inches, or the width of the equipment, whichever is greater✓
• c.36 inches, or the width of the equipment, whichever is greater
• d.24 inches, measured from the centerline of the equipment

The working space must be at least 30 inches wide or as wide as the equipment, whichever is greater, and it must allow all doors and hinged panels to open at least 90 degrees. The space need not be directly centered on the equipment, but it must be continuous across the full width required.2023 NEC §110.26(A)(2)

Entrances at each end of the working space are generally required for equipment rated at least what value and over 6 feet wide?

• a.1,000 amperes
• b.800 amperes
• c.1,200 amperes✓
• d.600 amperes

Equipment rated 1,200 amperes or more and over 6 feet wide containing overcurrent, switching or control devices needs an entrance at each end of the working space so a worker is never trapped behind an arcing fault. The 1,000-ampere threshold belongs to service ground-fault protection, and 800 amperes is the ceiling for the next-standard-size overcurrent rule.2023 NEC §110.26(C)(2)

The dedicated equipment space above a panelboard extends from the top of the equipment to what height?

• a.It extends all the way up to the structural ceiling in every case, with no allowance to stop at any lower height
• b.It extends only 3 feet above the top of the equipment in all cases, regardless of the height of the structural ceiling
• c.6 feet above the equipment or to the structural ceiling, whichever is lower✓
• d.6 and 1/2 feet above the finished floor

The dedicated space is the footprint of the equipment extended from the floor to a height of 6 feet above it or to the structural ceiling, whichever is lower, and no foreign piping, ducts or equipment may be installed in it. Suspended ceilings with removable panels are not considered structural ceilings, and the 6 and 1/2-foot dimension is the minimum headroom of working space, a separate requirement.2023 NEC §110.26(E)(1)

Which sequence correctly describes verifying that equipment is de-energized before beginning work?

• a.Test the meter on a known live source, test the de-energized conductors, then retest the meter on the known live source✓
• b.Test the de-energized conductors, then test the meter on a known live source
• c.Rely on the open position of the disconnect and the applied lock
• d.Test the de-energized conductors only, since the meter was checked at the start of the shift

The live-dead-live method proves the instrument is working both before and after the absence-of-voltage test, so a meter that failed during the check cannot lead a worker to treat energized conductors as safe. Locking and tagging the disconnect is required, but it does not by itself confirm that the conductors ahead of the work are actually de-energized, since backfeeds and mislabeled circuits are common.

Using the standard calculation method, what is the general lighting and general-use load for a dwelling unit with 2,000 square feet of habitable floor area, before any demand factors?

• a.4,500 volt-amperes
• b.3,000 volt-amperes
• c.6,000 volt-amperes✓
• d.9,000 volt-amperes

Table 220.12 assigns dwelling units a general lighting unit load of 3 volt-amperes per square foot, so 2,000 x 3 = 6,000 volt-amperes. The 4,500 answer uses 2.25 VA and 9,000 uses 4.5 VA, neither of which is the dwelling value in the 2023 NEC.2023 NEC Table 220.12

How much load must be included for the two required small-appliance branch circuits plus the one required laundry branch circuit of a dwelling, before demand factors are applied?

• a.3,000 volt-amperes
• b.4,500 volt-amperes✓
• c.1,500 volt-amperes
• d.6,000 volt-amperes

Each small-appliance and laundry branch circuit is calculated at 1,500 volt-amperes: two small-appliance circuits plus one laundry circuit is 3 x 1,500 = 4,500 volt-amperes. The 3,000 answer counts only the two small-appliance circuits and omits the laundry circuit.2023 NEC §220.52

A dwelling has a combined general lighting, small-appliance and laundry load of 12,000 volt-amperes. Applying the standard demand factors, what is the demand load?

• a.12,000 volt-amperes
• b.4,200 volt-amperes
• c.7,650 volt-amperes
• d.6,150 volt-amperes✓

The dwelling lighting demand table takes the first 3,000 volt-amperes at 100 percent and the remainder from 3,001 to 120,000 volt-amperes at 35 percent: 3,000 + (9,000 x 0.35) = 3,000 + 3,150 = 6,150 volt-amperes. Applying 35 percent to the whole 12,000 gives the incorrect 4,200 figure.2023 NEC Table 220.45

Using the standard demand table for household cooking appliances, what is the demand load for a single 12-kilowatt household electric range?

• a.8 kilowatts✓
• b.12 kilowatts
• c.6 kilowatts
• d.9.6 kilowatts

Column C of Table 220.55 gives a demand of 8 kilowatts for one household range rated not over 12 kilowatts, reflecting the fact that all burners and the oven are rarely at full power at once. The 9.6-kilowatt answer applies an 80 percent factor that does not appear in the table, and 12 kilowatts ignores the demand allowance entirely.2023 NEC Table 220.55

A dwelling has a single household electric clothes dryer with a nameplate rating of 4.5 kilowatts. What value must be used for the dryer load in the service calculation?

• a.4,500 volt-amperes
• b.4,000 volt-amperes
• c.5,000 volt-amperes✓
• d.3,600 volt-amperes

The load for a household electric clothes dryer is 5,000 volt-amperes or the nameplate rating, whichever is larger, so a 4,500-volt-ampere nameplate is calculated at the 5,000-volt-ampere floor. Using the nameplate 4,500 value violates the minimum, and the other figures apply factors that do not belong here.2023 NEC §220.54

A dwelling feeder supplies four fastened-in-place appliances (a disposal, a dishwasher, a water heater and a trash compactor) in addition to other loads. What demand factor may be applied to the nameplate total of these appliances?

• a.100 percent
• b.75 percent✓
• c.50 percent
• d.80 percent

Where four or more fastened-in-place appliances other than ranges, clothes dryers, space-heating and air-conditioning equipment are on the same feeder, a 75 percent demand factor may be applied to the nameplate total. With three or fewer such appliances no reduction is permitted, which is where the 100 percent answer applies.2023 NEC §220.53

Using the optional calculation method for a dwelling, the general loads total 40 kVA. What is the demand load for these general loads?

• a.40 kVA
• b.16 kVA
• c.22 kVA✓
• d.20 kVA

The optional dwelling method takes the first 10 kVA of general load at 100 percent and the remainder at 40 percent: 10 + (30 x 0.40) = 10 + 12 = 22 kVA. Applying 40 percent to the entire 40 kVA gives the incorrect 16-kVA figure.2023 NEC §220.82(B)

A commercial building has 50 general-use receptacle outlets. Using the standard method, what receptacle load must be calculated before demand factors?

• a.9,000 volt-amperes✓
• b.6,000 volt-amperes
• c.7,500 volt-amperes
• d.9,900 volt-amperes

Non-dwelling receptacle outlets are calculated at 180 volt-amperes each, where a single or multiple receptacle on one yoke counts as one outlet: 50 x 180 = 9,000 volt-amperes. The 7,500 answer uses 150 VA and 6,000 uses 120 VA, neither of which is the code value.2023 NEC §220.14(I)

When calculating a feeder or service that supplies several motors plus other loads, how is the largest motor accounted for?

• a.It is counted at 100 percent of its full-load current, exactly the same as all of the other motors in the calculation
• b.It is counted at 80 percent of its full-load current because of assumed diversity among the connected motors
• c.At 115 percent of its full-load current
• d.By adding 25 percent of its full-load current on top of the summed loads✓

Motor loads are calculated per Article 430, which requires the feeder to carry 125 percent of the largest motor plus the full-load currents of the remaining motors; in a combined load calculation this is applied by adding an extra 25 percent of the largest motor to the total. The 115 percent figure is the motor disconnect rating, not the conductor calculation.2023 NEC §220.50

Which of the following dwelling-unit locations requires GFCI protection for 125-volt, 15- and 20-ampere receptacles?

• a.A wall receptacle in a second-floor bedroom
• b.A receptacle in an unfinished basement✓
• c.A receptacle in a first-floor hallway
• d.A receptacle in a bedroom clothes closet

Section 210.8(A) lists unfinished basements among the dwelling areas requiring GFCI protection, along with bathrooms, garages, outdoors, crawl spaces, kitchens and laundry areas. Bedrooms and hallways are addressed by AFCI rather than GFCI requirements, so they are not on the GFCI list.2023 NEC §210.8(A)

Using the standard demand table for household cooking appliances, what is the demand load for three 12-kilowatt household electric ranges supplied by one feeder?

• a.24 kilowatts
• b.8 kilowatts
• c.14 kilowatts✓
• d.22 kilowatts

Column C of Table 220.55 gives a demand of 14 kilowatts for three ranges rated not over 12 kilowatts each, applying diversity across the appliances. The 24-kilowatt answer would result from a naive sum with a partial factor, and 8 kilowatts is the single-range value.2023 NEC Table 220.55

The interior metal water piping system of a building must be bonded to the service equipment. How is the required bonding jumper sized?

• a.From Table 250.66, based on the size of the service-entrance conductors✓
• b.From Table 250.122, based on the rating of the overcurrent device protecting the largest service conductor
• c.Always 6 AWG copper regardless of service size
• d.From Table 310.16, based on the service ampacity and the temperature rating of the terminations used

The bonding jumper for the interior metal water piping system is sized from Table 250.66 using the ungrounded service-entrance conductors, the same table used for the grounding electrode conductor. Table 250.122 sizes equipment grounding conductors by overcurrent device and does not apply to this bond.2023 NEC §250.104(A)(1)

For a separately derived system such as a transformer secondary, where is the system bonding jumper permitted to be installed?

• a.At every downstream panelboard
• b.Only at the main service disconnecting means of the building and never at the transformer, panelboard, or any other point of the separately derived system
• c.At the source of the separately derived system or at the first disconnecting means, but at only one of those points✓
• d.It is not required for separately derived systems

A separately derived system requires a single system bonding jumper connecting the grounded conductor to the equipment grounding system, installed either at the source or at the first disconnecting means but at only one location. Installing it at more than one point creates parallel neutral paths, and omitting it leaves no effective fault-clearing path for the derived system.2023 NEC §250.30(A)(1)

A single-family dwelling has a calculated demand load of 21,600 volt-amperes on a 240-volt, single-phase service. What is the minimum service disconnecting means rating permitted?

• a.90 amperes
• b.100 amperes✓
• c.125 amperes
• d.60 amperes

The calculated current is 21,600 / 240 = 90 amperes, but the service disconnecting means for a one-family dwelling must be rated at least 100 amperes, so the code floor governs. Selecting 90 amperes satisfies the calculation but violates the 100-ampere minimum, and 125 amperes exceeds the required minimum.2023 NEC §230.79(C)

The grounded (neutral) service conductor also serves as the effective ground-fault current path back to the utility. It must be sized not smaller than what?

• a.The equipment grounding conductor from Table 250.122
• b.6 AWG copper in all cases
• c.One-half the ungrounded conductor size
• d.The grounding electrode conductor value from Table 250.102(C)(1)✓

Because there is no main bonding jumper ahead of the service, the grounded conductor must carry any fault current back to the source, so it may not be smaller than the value from Table 250.102(C)(1), which mirrors Table 250.66. Sizing it from the equipment grounding conductor table would leave it too small for that duty.2023 NEC §250.24(C)

At service equipment, why are standard locknuts and bushings alone not adequate where service raceways connect through concentric or eccentric knockouts?

• a.They are in fact adequate by themselves, so no additional bonding means, jumpers, or listed fittings are needed where service raceways connect through concentric or eccentric knockouts
• b.Because service raceways must always be nonmetallic
• c.Because service bonding requires bonding jumpers, bonding bushings or other listed bonding-type fittings to ensure a low-impedance path✓
• d.Because only exothermic welding is permitted at a service

Service raceways and enclosures must be bonded by an approved means such as bonding jumpers, bonding-type locknuts or bushings, since standard locknuts across concentric or eccentric knockouts cannot be relied on for the high fault current available ahead of the service overcurrent device. Downstream of the service, standard locknuts are generally acceptable.2023 NEC §250.92(B)

A grounding electrode conductor is run from the service to a driven ground rod. What is the rule regarding splices in that conductor?

• a.Any listed split-bolt connector may be used
• b.It must be installed in one continuous length unless spliced by irreversible compression connectors, exothermic welding or connections to busbars✓
• c.Up to three twist-on wire connector splices are permitted along the run, provided that each connection is made accessible and the conductor is not reduced in size at any of the splice points
• d.Splices are permitted only inside accessible junction boxes

A grounding electrode conductor must be continuous, and where a splice is necessary it is permitted only by irreversible compression connectors listed for the purpose, exothermic welding, or bolting to busbars. Ordinary split-bolts and twist-on connectors are reversible and are not acceptable for this conductor.2023 NEC §250.64(C)

How is the load on a feeder or service neutral (grounded) conductor determined?

• a.By the maximum unbalanced load, the maximum net current between the neutral and any one ungrounded conductor✓
• b.By the sum of all ungrounded conductor currents
• c.It is always the same size as the ungrounded conductors
• d.By taking 125 percent of the load on the largest single ungrounded conductor and applying that resulting value to the neutral conductor

The neutral carries only the imbalance among the phase conductors, so its calculated load is the maximum unbalanced load, defined as the maximum net current between the neutral and any one ungrounded conductor. Loads such as 240-volt line-to-line appliances place no current on the neutral at all.2023 NEC §220.61(A)

A demand factor may be applied to the portion of a feeder neutral load that exceeds what value?

• a.100 amperes
• b.150 amperes
• c.200 amperes✓
• d.400 amperes

For a feeder or service, the portion of the maximum unbalanced neutral load that exceeds 200 amperes may be taken at 70 percent, recognizing the diversity of line-to-neutral loads at higher currents. This reduction does not apply to nonlinear loads on a wye system, where the neutral can be as heavily loaded as the phases.2023 NEC §220.61(B)

A junction box contains four 14 AWG conductors and two 12 AWG conductors, with no devices, clamps or fittings. What minimum box volume is required?

• a.12.5 cubic inches✓
• b.13.5 cubic inches
• c.11.0 cubic inches
• d.15.0 cubic inches

Each conductor is counted at the volume allowance for its own size: four 14 AWG at 2.00 cubic inches is 8.0, and two 12 AWG at 2.25 cubic inches is 4.5, for a total of 12.5 cubic inches. Using 2.25 for every conductor gives the incorrect 13.5 figure.2023 NEC §314.16(B)(1)

What is the minimum distance a 125-volt, 15- or 20-ampere receptacle must be located from the inside wall of a permanently installed pool?

• a.10 feet
• b.6 feet✓
• c.5 feet
• d.3 feet

General-use receptacles must be at least 6 feet from the inside walls of a permanently installed pool, measured along the shortest path a cord would follow. At least one required receptacle must also be located between 6 and 20 feet from the pool, and receptacles serving pool equipment must be GFCI protected.2023 NEC §680.22(A)(1)

A location is classified as Class I, Division 1 when which condition exists?

• a.Combustible dust in ignitible concentrations is present in the surrounding air only during infrequent abnormal operating conditions
• b.Easily ignitible fibers or flyings are handled or stored but are not normally likely to become suspended in the air
• c.Ignitible concentrations of flammable gases or vapors can exist under normal operating conditions✓
• d.Flammable liquids are stored only in sealed containers

Class I covers flammable gases and vapors, and Division 1 means ignitible concentrations can exist under normal operating conditions or frequently because of repair, maintenance or leakage. Combustible dust is Class II and ignitible fibers are Class III, so those descriptions point to different classifications.2023 NEC §500.5(B)(1)

A 15-ampere, 125-volt receptacle is installed outdoors in a wet location. What type of cover is required?

• a.Any weatherproof cover, effective only when the receptacle is unused
• b.A standard thermoplastic wall plate
• c.A cover that is weatherproof only when no plug is inserted
• d.A cover that is weatherproof whether or not an attachment plug is inserted✓

Receptacles in wet locations must have an enclosure that is weatherproof whether or not an attachment plug is inserted, commonly an in-use or bubble cover, because cords are often left plugged in outdoors. A cover that seals only when empty meets the lesser damp-location standard, not the wet-location rule.2023 NEC §406.9(B)(1)

In a dwelling unit, 125-volt and 250-volt, 15- and 20-ampere receptacles must generally be what type?

• a.Isolated-ground type
• b.Listed tamper-resistant type✓
• c.Hospital-grade type
• d.Twist-lock type

To prevent children from inserting objects into energized slots, these receptacles in dwelling units must be listed tamper-resistant. Hospital-grade and isolated-ground receptacles address other concerns and are not the general dwelling requirement.2023 NEC §406.12

Type NM (nonmetallic-sheathed) cable is NOT permitted in which of the following?

• a.Wet or damp locations✓
• b.The framed walls of a one-family dwelling
• c.An exposed dry run in the basement of a dwelling
• d.A concealed dry run in an attic

Type NM cable is prohibited in wet or damp locations because its sheath is not rated for moisture, and it may not be embedded in concrete or used where exposed to corrosive fumes. It remains a standard method for the dry, concealed and exposed interior wiring of dwellings.2023 NEC §334.12(B)

Flexible metal conduit (FMC) may serve as an equipment grounding conductor only where the total length in the ground-return path does not exceed 6 feet and which additional condition is met?

• a.The conduit is at least 1 inch trade size
• b.The circuit voltage does not exceed 50 volts
• c.The circuit conductors are protected at 20 amperes or less and the fittings are listed for grounding✓
• d.A separate equipment bonding jumper is installed on the outside of the conduit for its full length and terminated at both ends

FMC qualifies as an equipment grounding conductor only when the ground-return length is 6 feet or less, the overcurrent protection is 20 amperes or less, and the fittings are listed for grounding. Where any of these limits is exceeded, a separate equipment grounding conductor must be installed.2023 NEC §250.118(6)

The equipotential bonding grid required around a permanently installed pool must be formed with which conductor?

• a.6 AWG stranded aluminum
• b.8 AWG solid copper✓
• c.10 AWG solid copper
• d.4 AWG insulated copper

The equipotential bonding required around a pool uses a solid copper conductor not smaller than 8 AWG to tie together the reinforcing steel, coping, deck, ladders and equipment so that everyone stands at the same potential. Aluminum is not permitted for this bond, and 10 AWG is undersized.2023 NEC §680.26(B)

A box or conduit body installed in a wet location must meet which requirement?

• a.It may be any standard indoor box, provided that a weatherproof gasket is added between the box and the cover plate
• b.It must be completely filled with an approved duct seal compound to keep out all moisture and condensation
• c.It must be nonmetallic only
• d.It must be listed for wet locations and be installed so water cannot enter or accumulate within it✓

Boxes and fittings in damp or wet locations must be placed or equipped so water cannot enter or accumulate inside, and in wet locations they must be listed for that use. Both metallic and nonmetallic boxes are permitted if properly listed and installed, so material alone does not satisfy the rule.2023 NEC §314.15

How many running overload protective devices are required for a three-phase alternating-current motor?

• a.One
• b.Two
• c.Three, one in each phase✓
• d.Four

Table 430.37 requires a three-phase motor to have an individual overload device in each of the three ungrounded conductors, so that a single-phasing condition, which overloads the remaining windings, is detected. One or two devices could miss an overload on an unprotected phase.2023 NEC §430.37

A 25-kVA, single-phase transformer has a 480-volt primary and is protected by primary overcurrent protection only, sized at 125 percent. With a primary full-load current of 9 amperes or more, what is the maximum standard primary overcurrent device rating?

• a.60 amperes
• b.50 amperes
• c.65 amperes
• d.70 amperes✓

Primary current is 25,000 / 480 = 52.1 amperes, and 52.1 x 1.25 = 65.1 amperes. Because that value is not a standard rating, the next higher standard size of 70 amperes is permitted. 65 amperes is not a standard overcurrent device rating, and 60 amperes would be below the calculated value.2023 NEC §450.3(B)