Master Electrician (NEC) — All Questions

A one-family dwelling has 2,500 sq ft of habitable floor area. Using the standard method at 3 VA per sq ft, what is the general lighting and general-use receptacle load before demand factors?

• a.9,000 VA
• b.5,000 VA
• c.6,000 VA
• d.7,500 VA✓

2023 NEC 220.41 figures dwelling general lighting at 3 VA/sq ft: 2,500 x 3 = 7,500 VA, calculated before the 220.42 demand factors are applied.

A dwelling has 9,000 VA general lighting, 3,000 VA small-appliance, and 1,500 VA laundry. Applying the standard general-lighting demand (first 3,000 VA at 100%, remainder at 35%), what is the net general lighting load?

• a.6,675 VA✓
• b.7,275 VA
• c.13,500 VA
• d.4,725 VA

Sum = 9,000 + 3,000 + 1,500 = 13,500 VA. First 3,000 at 100% = 3,000; remaining 10,500 at 35% = 3,675. Net = 6,675 VA (2023 NEC 220.42).

Using Column C of the household-range demand table, what is the demand load for three 12-kW electric ranges served by one feeder?

• a.11,000 VA
• b.16,000 VA
• c.12,000 VA
• d.14,000 VA✓

2023 NEC Table 220.55, Column C, gives 14 kW for three ranges not over 12 kW: 14,000 VA.

A single household electric range is rated 14 kW. Using the 5%-per-kW-over-12 rule on the 8-kW Column C base, what is the demand load?

• a.10,000 VA
• b.9,600 VA
• c.8,800 VA✓
• d.8,000 VA

2023 NEC Table 220.55 Note 1: 14 kW is 2 kW over 12, so add 5% x 2 = 10% to the 8-kW base: 8,000 x 1.10 = 8,800 VA.

A dwelling feeder supplies four electric clothes dryers, each rated 5.5 kW. Applying the dryer demand factor, what is the demand load?

• a.22,000 VA✓
• b.27,500 VA
• c.20,000 VA
• d.16,500 VA

2023 NEC Table 220.54: 1-4 dryers are taken at 100%. 4 x 5,500 (each above the 5,000 VA minimum) = 22,000 VA.

A feeder supplies five electric clothes dryers, each rated 5 kW. Applying the dryer demand factor for five dryers, what is the demand load?

• a.21,250 VA✓
• b.25,000 VA
• c.23,750 VA
• d.18,750 VA

2023 NEC Table 220.54 lists 85% for five dryers: 5 x 5,000 x 0.85 = 21,250 VA (each dryer is figured at the 5,000 VA minimum).

A dwelling has five fastened-in-place appliances (dishwasher 1,200, disposal 900, water heater 4,500, compactor 1,000, wine cooler 600 VA) totaling 8,200 VA. What is the demand after the fastened-appliance factor?

• a.6,560 VA
• b.8,200 VA
• c.4,100 VA
• d.6,150 VA✓

2023 NEC 220.53 permits a 75% demand factor for four or more fastened-in-place appliances on the same feeder: 8,200 x 0.75 = 6,150 VA.

A dwelling's total calculated load is 27,600 VA on a 240-V, single-phase service. What is the calculated service current?

• a.100 A
• b.120 A
• c.115 A✓
• d.96 A

Current = volt-amperes / voltage: 27,600 / 240 = 115 A.

Under the optional dwelling method, the general loads (lighting, small-appliance, laundry, appliances, range) total 32,000 VA. Applying 220.82(B) (100% of the first 10 kVA, 40% of the remainder), what is the demand?

• a.12,800 VA
• b.18,800 VA✓
• c.22,000 VA
• d.32,000 VA

2023 NEC 220.82(B): 10,000 + 0.40 x 22,000 = 10,000 + 8,800 = 18,800 VA. The first 10 kVA stays at 100%.

Under the optional dwelling method a home has 5,000 VA of air-conditioning and 9,000 VA of central electric space heating (not run together). What single value is included for the heating/cooling load?

• a.5,000 VA
• b.5,850 VA✓
• c.9,000 VA
• d.14,000 VA

2023 NEC 220.82(C) takes the largest of the listed heating/cooling figures: A/C at 100% = 5,000 VA versus central heat at 65% = 5,850 VA. The larger, 5,850 VA, is used.

A 45-unit multifamily building has 720,000 VA of connected load and an applicable 220.84 demand factor of 26%. What is the calculated demand load?

• a.230,400 VA
• b.720,000 VA
• c.187,200 VA✓
• d.288,000 VA

2023 NEC 220.84 multiplies connected load by the Table 220.84 demand factor: 720,000 x 0.26 = 187,200 VA.

Three 12-kW household ranges on a feeder have a Column C demand of 14,000 VA. Using the range neutral allowance, what is the range contribution to the feeder neutral?

• a.8,400 VA
• b.11,000 VA
• c.14,000 VA
• d.9,800 VA✓

2023 NEC 220.61(B)(1) allows the feeder neutral for household ranges at 70% of the range demand: 14,000 x 0.70 = 9,800 VA.

A feeder neutral carries a maximum unbalanced load of 300 A. Applying the neutral demand allowance, what is the calculated neutral load?

• a.300 A
• b.250 A
• c.235 A
• d.270 A✓

2023 NEC 220.61(B)(2): 70% applies to the portion over 200 A: 200 + 0.70 x 100 = 200 + 70 = 270 A.

A 40-HP, 460-V, three-phase induction motor has a table full-load current of 52 A. What minimum branch-circuit conductor ampacity is required?

• a.52 A
• b.50 A
• c.65 A✓
• d.72 A

2023 NEC 430.22: motor branch-circuit conductors must be at least 125% of the table FLC: 52 x 1.25 = 65 A.

A 30-HP, 460-V, three-phase motor has a table full-load current of 40 A. What is the maximum inverse-time circuit breaker for branch-circuit short-circuit and ground-fault protection?

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

2023 NEC 430.52: inverse-time breaker at 250% of FLC = 40 x 2.50 = 100 A, which is a standard rating (240.6).

A 30-HP, 460-V, three-phase motor has a table full-load current of 40 A. What is the maximum time-delay (dual-element) fuse for branch-circuit short-circuit/ground-fault protection?

• a.80 A
• b.90 A
• c.60 A
• d.70 A✓

2023 NEC 430.52: time-delay fuses at 175% of FLC = 40 x 1.75 = 70 A, which is a standard fuse size.

A motor with a service factor of 1.15 has a nameplate full-load current of 40 A. What is the maximum overload protection at 125% of nameplate?

• a.46 A
• b.54 A
• c.44 A
• d.50 A✓

2023 NEC 430.32(A)(1): motors with a service factor of 1.15 or more may set overload at 125% of nameplate FLA: 40 x 1.25 = 50 A.

A feeder supplies a 50-HP (65 A) and a 30-HP (40 A) motor, both 460-V three-phase. What minimum feeder conductor ampacity is required?

• a.131.25 A
• b.121.25 A✓
• c.105 A
• d.115 A

2023 NEC 430.24: 125% of the largest motor FLC plus the sum of the others: (65 x 1.25) + 40 = 81.25 + 40 = 121.25 A.

What is the full-load primary current of a 112.5-kVA, 480-V, three-phase transformer?

• a.104 A
• b.117 A
• c.156 A
• d.135 A✓

Three-phase current = VA / (1.732 x V): 112,500 / (1.732 x 480) = 135 A.

What is the full-load secondary current of a 112.5-kVA, three-phase transformer with a 208-V secondary?

• a.541 A
• b.312 A✓
• c.270 A
• d.156 A

Secondary current = 112,500 / (1.732 x 208) = 312 A.

A 45-kVA, 480-V, three-phase transformer (primary FLC 54 A, over 9 A) has primary-only overcurrent protection at 125% maximum. What is the primary device rating (next higher standard permitted)?

• a.60 A
• b.80 A
• c.70 A✓
• d.90 A

2023 NEC Table 450.3(B): 125% of 54 A = 67.7 A; not a standard size, so the next higher standard device, 70 A, is permitted for primary-only protection.

A 150-kVA, 480-208Y/120-V transformer (secondary FLC 416 A) is protected on both sides. What is the maximum secondary overcurrent device at 125% (next higher standard permitted)?

• a.450 A
• b.600 A✓
• c.500 A
• d.520 A

2023 NEC Table 450.3(B): 416 x 1.25 = 520 A on the secondary; not a standard size, so the next higher standard rating, 600 A, is permitted.

A 120-V branch circuit carries 20 A over a one-way run of 100 ft using #10 Cu (CM = 10,380, K = 12.9). What is the approximate voltage drop?

• a.5.0 V✓
• b.6.2 V
• c.3.0 V
• d.2.5 V

Single-phase VD = (2 x K x I x L) / CM = (2 x 12.9 x 20 x 100) / 10,380 = 51,600 / 10,380 = 4.97 V.

A 480-V, three-phase feeder carries 60 A over 200 ft using #4 Cu (CM = 41,740, K = 12.9). What is the approximate voltage drop?

• a.3.2 V
• b.6.4 V✓
• c.4.8 V
• d.9.0 V

Three-phase VD = (1.732 x K x I x L) / CM = (1.732 x 12.9 x 60 x 200) / 41,740 = 268,100 / 41,740 = 6.42 V.

Using the previous feeder (6.42-V drop on a 480-V, three-phase feeder), what is the approximate percent voltage drop?

• a.3.0%
• b.1.3%✓
• c.0.7%
• d.2.7%

Percent VD = 6.42 / 480 = 0.0134 = 1.3%, within the 3% recommendation for a feeder.

A 240-V, single-phase circuit carries 24 A over a one-way run of 150 ft (K = 12.9). What is the smallest copper conductor that keeps voltage drop at or below 3% (7.2 V)?

• a.#10 AWG
• b.#8 AWG✓
• c.#12 AWG
• d.#6 AWG

Required CM = (2 x K x I x L) / Vdrop = (2 x 12.9 x 24 x 150) / 7.2 = 12,900 CM. #10 (10,380) is too small; #8 (16,510 CM) is the smallest that satisfies it.

Nine current-carrying #8 THHN conductors share one raceway. If the 90 C ampacity of #8 Cu is 55 A, what is the adjusted ampacity after the fill adjustment factor for 7-9 conductors?

• a.38.5 A✓
• b.44 A
• c.33 A
• d.55 A

2023 NEC Table 310.15(C)(1): 7 to 9 current-carrying conductors take a 70% factor: 55 x 0.70 = 38.5 A.

A #1/0 Cu THHN conductor has a 90 C ampacity of 170 A. In a 46 C ambient (correction factor 0.82), what is the corrected ampacity?

• a.125 A
• b.139 A✓
• c.170 A
• d.150 A

2023 NEC Table 310.15(B)(1): 170 x 0.82 = 139.4 A, about 139 A, before comparison to the termination temperature limit.

A #3 Cu THHN (90 C ampacity 110 A) runs with 8 current-carrying conductors (0.70 factor) in a 38 C ambient (0.91 factor). What is the adjusted-and-corrected ampacity?

• a.70 A✓
• b.77 A
• c.100 A
• d.85 A

Apply both factors to the 90 C ampacity: 110 x 0.70 x 0.91 = 70.1 A, about 70 A.

An office has a 10,800-VA general receptacle load. Applying the demand (first 10 kVA at 100%, remainder at 50%), what is the demand load?

• a.10,800 VA
• b.10,000 VA
• c.5,400 VA
• d.10,400 VA✓

2023 NEC 220.44 / Table 220.44: first 10,000 VA at 100% + remaining 800 VA at 50% = 10,000 + 400 = 10,400 VA.

A box contains five #12 AWG and three #10 AWG conductors and nothing else. Using the box-fill volume allowances, what minimum box volume is required?

• a.16.5 cu in
• b.20.0 cu in
• c.17.25 cu in
• d.18.75 cu in✓

2023 NEC Table 314.16(B): #12 = 2.25, #10 = 2.5 cu in. (5 x 2.25) + (3 x 2.5) = 11.25 + 7.5 = 18.75 cu in.

A device box has six #12 conductors, one #12 equipment grounding conductor, and one duplex receptacle. Using box-fill rules, what minimum box volume is required?

• a.22.5 cu in
• b.15.75 cu in
• c.20.25 cu in✓
• d.18.0 cu in

2023 NEC 314.16(B): six conductors = 6, all grounds count as one = 1, the device counts as two based on its largest conductor = 2. (6 + 1 + 2) x 2.25 = 20.25 cu in.

A motor feeder's largest branch-circuit short-circuit device is 110 A; the other motors on the feeder total 68 A of full-load current. What is the maximum feeder short-circuit/ground-fault device?

• a.150 A
• b.200 A
• c.178 A
• d.175 A✓

2023 NEC 430.62: the feeder device may not exceed the largest branch device (110 A) plus the other FLCs (68 A) = 178 A; the largest standard rating not exceeding 178 A is 175 A.

A commercial kitchen has eight kitchen-equipment units totaling 80 kW. Applying the demand factor for six or more units, what is the demand load?

• a.45 kW
• b.60 kW
• c.52 kW✓
• d.48 kW

2023 NEC Table 220.56 allows 65% for six or more units: 80 x 0.65 = 52 kW (not less than the sum of the two largest units).

In a commercial service calculation, the largest motor has a 65-A full-load current. How much is added for the largest-motor 25% factor?

• a.13 A
• b.81.25 A
• c.16.25 A✓
• d.65 A

2023 NEC 430.24 / 220.50: 25% of the largest motor FLC is added on top of the motor loads: 0.25 x 65 = 16.25 A.

An 800-A feeder overcurrent device supplies a 25-ft tap. Under the 25-ft feeder tap rule, what minimum tap-conductor ampacity is required?

• a.80 A
• b.160 A
• c.266.667 A✓
• d.200 A

2023 NEC 240.21(B)(2): the 25-ft tap ampacity must be at least 1/3 of the 800-A device: 800 / 3 = 267 A.

A 600-A feeder overcurrent device supplies a 10-ft tap. Under the 10-ft feeder tap rule, what minimum ampacity comes from the 1/10 provision?

• a.40 A
• b.60 A✓
• c.200 A
• d.133 A

2023 NEC 240.21(B)(1): the 10-ft tap ampacity must be at least 1/10 of the 600-A device: 600 / 10 = 60 A (and at least the load served).

A 112.5-kVA, 480-208Y/120-V transformer feeds a panel through 25-ft secondary conductors under 240.21(C)(6). What is the minimum secondary conductor ampacity?

• a.156 A
• b.260 A
• c.312 A✓
• d.390 A

Secondary FLC = 112,500 / (1.732 x 208) = 312 A. 2023 NEC 240.21(C)(6) requires the secondary conductors to have ampacity at least the transformer secondary current and terminate in a single OCPD not exceeding that ampacity.

A 150-kVA, 208-V, three-phase transformer has 2% impedance. What is the approximate available fault current at its secondary terminals (infinite primary)?

• a.8300 A
• b.10410 A
• c.20818 A✓
• d.15600 A

Secondary FLA = 150,000 / (1.732 x 208) = 416 A. Isc = FLA / %Z = 416 / 0.02 = 20,820 A.

A 300-kVA, 480-V, three-phase transformer has 5% impedance. What is the approximate available fault current at its secondary (infinite primary)?

• a.3600 A
• b.18040 A
• c.14432 A
• d.7217 A✓

Secondary FLA = 300,000 / (1.732 x 480) = 361 A. Isc = 361 / 0.05 = 7,217 A.

A feeder supplies a 100-A continuous load and a 50-A noncontinuous load. What minimum conductor ampacity is required?

• a.187.5 A
• b.175 A✓
• c.150 A
• d.200 A

2023 NEC 215.2(A)(1): 125% of the continuous load plus 100% of the noncontinuous load: (100 x 1.25) + 50 = 125 + 50 = 175 A.

A feeder supplies a 120-A continuous load and a 40-A noncontinuous load. What is the minimum standard overcurrent device rating?

• a.225 A
• b.190 A
• c.175 A
• d.200 A✓

215.3: device >= 125% continuous + noncontinuous = (120 x 1.25) + 40 = 190 A; the next standard size (240.6) is 200 A.

A feeder supplies six household clothes dryers, each rated 5 kW. Applying the dryer demand factor for six dryers, what is the demand load?

• a.18,000 VA
• b.25,500 VA
• c.22,500 VA✓
• d.30,000 VA

2023 NEC Table 220.54 lists 75% for six dryers: 6 x 5,000 x 0.75 = 22,500 VA.

A feeder supplies five household electric ranges, each rated 14 kW. Using Column C plus the over-12-kW adjustment, what is the demand load?

• a.20,000 VA
• b.22,000 VA✓
• c.24,000 VA
• d.25,000 VA

Table 220.55 Column C for five ranges = 20 kW; Note 1 adds 5% per kW over 12 (14 kW is 2 over): 20,000 x 1.10 = 22,000 VA.

A 208-V, three-phase service carries a calculated load of 90,000 VA. What is the calculated service current?

• a.250 A✓
• b.144 A
• c.216 A
• d.433 A

Three-phase current = 90,000 / (1.732 x 208) = 250 A.

A 480-V, three-phase feeder carries a calculated 150-A load. What transformer kVA does this correspond to?

• a.72 kVA
• b.104 kVA
• c.144 kVA
• d.125 kVA✓

kVA = (1.732 x V x I) / 1000 = (1.732 x 480 x 150) / 1000 = 124.7 kVA, about 125 kVA.

A box has internal cable clamps and contains eight #14 AWG conductors. Using box-fill rules, what minimum box volume is required?

• a.18.0 cu in✓
• b.22.5 cu in
• c.20.25 cu in
• d.16.0 cu in

2023 NEC 314.16(B): where clamps are present, add one conductor volume of the largest conductor. (8 conductors + 1 clamp allowance) x 2.0 (#14) = 9 x 2.0 = 18.0 cu in.

A #6 Cu THHN conductor (90 C ampacity 75 A) is run with twelve current-carrying conductors in one raceway. What is the adjusted ampacity?

• a.45 A
• b.37.5 A✓
• c.52.5 A
• d.26.25 A

2023 NEC Table 310.15(C)(1): 10 to 20 current-carrying conductors take a 50% factor: 75 x 0.50 = 37.5 A.

A retail show window is 30 ft long. Using the show-window unit load, what is the calculated show-window load?

• a.3,000 VA
• b.6,000 VA✓
• c.9,000 VA
• d.5,400 VA

2023 NEC 220.43(A) requires 200 VA per linear foot of show window: 30 x 200 = 6,000 VA.

An 8,000 sq ft office building is calculated for general lighting using the 2023 unit-load table. What is the general lighting load?

• a.24,000 VA
• b.32,000 VA
• c.16,000 VA
• d.28,000 VA✓

2023 NEC Table 220.42(A) lists 3.5 VA per sq ft for office/bank occupancies: 8,000 x 3.5 = 28,000 VA.