Master Electrician (NEC) — All Questions

A single-family dwelling has 2,000 sq ft of habitable floor area. Using the standard calculation method at 3 VA per sq ft, what is the general lighting and general-use receptacle load before any demand factors?

• a.3,000 VA
• b.6,000 VA✓
• c.4,000 VA
• d.8,000 VA

General lighting is figured at 3 VA per sq ft: 2,000 sq ft x 3 VA = 6,000 VA. This value is calculated before applying the general-lighting demand factors.

In the standard dwelling calculation, what is the minimum load to include for two required small-appliance branch circuits plus one laundry branch circuit, before any demand factor?

• a.1,500 VA
• b.4,500 VA✓
• c.6,000 VA
• d.3,000 VA

Each small-appliance and laundry circuit is figured at 1,500 VA. Two small-appliance circuits (2 x 1,500 = 3,000) plus one laundry circuit (1,500) equals 4,500 VA.

A dwelling has 7,200 VA general lighting, 3,000 VA small-appliance, and 1,500 VA laundry. Applying the standard demand factor (first 3,000 VA at 100%, remainder at 35%), what is the net general lighting load?

• a.4,095 VA
• b.11,700 VA
• c.7,200 VA
• d.6,045 VA✓

Sum = 7,200 + 3,000 + 1,500 = 11,700 VA. First 3,000 at 100% = 3,000; remaining 8,700 at 35% = 3,045. Net = 3,000 + 3,045 = 6,045 VA.

Using the demand table for household electric ranges, what is the demand load for one 12-kW electric range in a dwelling (Column C, one appliance)?

• a.9,600 VA
• b.12,000 VA
• c.6,000 VA
• d.8,000 VA✓

For a single household range not over 12 kW, the Column C demand is 8 kW = 8,000 VA. The 12-kW nameplate is reduced to an 8-kW calculated demand for one range.

What is the demand load for two 12-kW household electric ranges served from one feeder (range demand table, Column C, two appliances)?

• a.11,000 VA✓
• b.12,000 VA
• c.19,200 VA
• d.16,000 VA

Two ranges not over 12 kW have a Column C demand of 11 kW = 11,000 VA. This is less than the 24,000 VA connected total because of diversity.

A single household range is rated 16 kW. Using the 5% increase per kW over 12 kW rule on the 8-kW base, what is the demand load?

• a.9,600 VA✓
• b.16,000 VA
• c.10,400 VA
• d.8,000 VA

16 kW is 4 kW over 12 kW, so add 5% x 4 = 20% to the 8-kW base: 8,000 x 1.20 = 9,600 VA.

A dwelling has one 4-kW electric clothes dryer. What value must be used as the dryer demand load in the service calculation?

• a.5,500 VA
• b.4,500 VA
• c.4,000 VA
• d.5,000 VA✓

The dryer load is the nameplate rating or 5,000 VA, whichever is larger. Since 4,000 VA is below 5,000 VA, 5,000 VA must be used.2023 NEC 220.54

A dwelling has four fastened-in-place appliances (dishwasher, disposal, water heater, compactor) totaling 6,000 VA. What is the demand load after applying the fastened-appliance demand factor?

• a.3,000 VA
• b.4,500 VA✓
• c.6,000 VA
• d.7,500 VA

Where four or more fastened-in-place appliances are on the same feeder, a 75% demand factor may be applied: 6,000 x 0.75 = 4,500 VA.2023 NEC 220.53

A dwelling's total calculated load is 24,000 VA on a 240-V, single-phase service. What is the calculated service current?

• a.167 A
• b.100 A✓
• c.120 A
• d.83 A

Current equals volt-amperes divided by voltage: 24,000 VA / 240 V = 100 A.

Using the 75 C column, what is the smallest copper conductor rated to carry a 100-A service load?

• a.#2 AWG
• b.#4 AWG
• c.#3 AWG✓
• d.#1 AWG

At 75 C, #4 Cu is rated 85 A and #3 Cu is rated 100 A. The #3 AWG copper is the smallest conductor that meets the 100-A requirement.2023 NEC Table 310.16

For a 200-A dwelling service using the 83% conductor allowance, what is the smallest copper ungrounded conductor (75 C) permitted for the main power feeder?

• a.#1/0 AWG
• b.#3/0 AWG
• c.#2/0 AWG✓
• d.#4/0 AWG

200 A x 0.83 = 166 A required ampacity. At 75 C, #2/0 Cu is rated 175 A, which satisfies 166 A, so #2/0 is the smallest permitted.2023 NEC 310.12

An electric range has a Column C demand of 8,000 VA. Using the 70% neutral demand for ranges, what is the range contribution to the feeder neutral load?

• a.8,000 VA
• b.5,600 VA✓
• c.4,000 VA
• d.7,000 VA

The feeder neutral for a household range may be figured at 70% of the range demand: 8,000 x 0.70 = 5,600 VA.

From the motor full-load current table, what is the full-load current of a 25-HP, 480-V, three-phase induction motor?

• a.28 A
• b.42 A
• c.34 A✓
• d.40 A

The table value for a 25-HP, 460-480-V three-phase motor is 34 A. Table full-load current (FLC), not nameplate, is used for conductor and protection sizing.2023 NEC Table 430.250

A single continuous-duty motor has a table full-load current of 34 A. What minimum conductor ampacity is required for its branch circuit?

• a.50 A
• b.34 A
• c.40 A
• d.42.5 A✓

Motor branch-circuit conductors must be at least 125% of the FLC: 34 A x 1.25 = 42.5 A minimum ampacity.2023 NEC 430.22

A motor with a service factor of 1.15 has a nameplate full-load current of 34 A. What is the maximum overload protection setting at 125%?

• a.42.5 A✓
• b.45.9 A
• c.34 A
• d.39.1 A

For a motor with service factor 1.15 or greater, overload is set at up to 125% of nameplate FLA: 34 x 1.25 = 42.5 A.2023 NEC 430.32

For a motor with a 34-A full-load current, what is the maximum inverse-time circuit breaker for branch-circuit short-circuit and ground-fault protection (250%, then next higher standard size)?

• a.70 A
• b.100 A
• c.90 A✓
• d.80 A

Inverse-time breaker: 250% of FLC = 34 x 2.50 = 85 A. Where this does not correspond to a standard size, the next higher standard rating (90 A) is permitted.2023 NEC 430.52

A feeder supplies three motors with full-load currents of 34 A, 28 A, and 22 A. What minimum conductor ampacity is required for the feeder?

• a.110 A
• b.92.5 A✓
• c.105 A
• d.84 A

Feeder ampacity = 125% of the largest motor FLC plus the sum of the others: (34 x 1.25) + 28 + 22 = 42.5 + 50 = 92.5 A.2023 NEC 430.24

What is the full-load primary current of a 75-kVA, 480-V, three-phase transformer?

• a.72 A
• b.104 A
• c.90 A✓
• d.156 A

Three-phase current = VA / (1.732 x V): 75,000 / (1.732 x 480) = 90.2 A, about 90 A.

What is the full-load secondary current of a 75-kVA, three-phase transformer with a 208-V secondary?

• a.208 A✓
• b.180 A
• c.156 A
• d.90 A

Secondary current = 75,000 / (1.732 x 208) = 208.2 A, about 208 A.

A 75-kVA transformer has a 90-A primary current (over 9 A). With primary-only protection at 125% max, what is the primary overcurrent device rating (next higher standard permitted)?

• a.150 A
• b.100 A
• c.125 A✓
• d.110 A

125% of 90 A = 112.5 A. Where this does not match a standard rating, the next higher standard device (125 A) is permitted for primary-only protection.2023 NEC 450.3(B)

A 120-V branch circuit carries 16 A over a one-way run of 80 ft using #12 Cu (circular mils = 6,530, K = 12.9). What is the approximate voltage drop?

• a.5.1 V✓
• b.7.4 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 16 x 80) / 6,530 = 33,024 / 6,530 = 5.06 V.

Using the previous circuit (5.06-V drop on a 120-V circuit), what is the approximate percent voltage drop?

• a.3.0%
• b.6.0%
• c.4.2%✓
• d.2.1%

Percent VD = 5.06 / 120 = 0.042 = 4.2%, which exceeds the 3% recommended for branch circuits.

A 208-V three-phase feeder carries 40 A over 150 ft using #6 Cu (CM = 26,240, K = 12.9). What is the approximate voltage drop?

• a.3.0 V
• b.7.2 V
• c.9.0 V
• d.5.1 V✓

Three-phase VD = (1.732 x K x I x L) / CM = (1.732 x 12.9 x 40 x 150) / 26,240 = 134,050 / 26,240 = 5.11 V.

For a 112.5-kVA, 208-V, three-phase transformer with 2% impedance, what is the approximate available fault current at the secondary terminals (infinite primary)?

• a.15,617 A✓
• b.20,000 A
• c.6,250 A
• d.10,000 A

Secondary FLA = 112,500 / (1.732 x 208) = 312.3 A. Isc = FLA / %Z = 312.3 / 0.02 = 15,617 A.

A 45-kVA, 208-V, three-phase transformer has 2% impedance. What is the approximate available fault current at its secondary (infinite primary)?

• a.12,500 A
• b.3,120 A
• c.6,245 A✓
• d.9,000 A

Secondary FLA = 45,000 / (1.732 x 208) = 124.9 A. Isc = 124.9 / 0.02 = 6,245 A.

Six current-carrying THHN conductors share one raceway. If the 90 C ampacity of #6 Cu is 75 A, what is the adjusted ampacity after the fill adjustment factor for 4-6 conductors?

• a.55 A
• b.60 A✓
• c.52.5 A
• d.75 A

For 4 to 6 current-carrying conductors, apply an 80% adjustment factor: 75 A x 0.80 = 60 A.2023 NEC Table 310.15(C)(1)

A #3 Cu THHN conductor has a 90 C ampacity of 110 A. In a 50 C ambient (correction factor 0.82), what is the corrected ampacity?

• a.90 A✓
• b.82 A
• c.100 A
• d.110 A

Corrected ampacity = 110 A x 0.82 = 90.2 A, about 90 A, before comparing to the termination temperature limit.2023 NEC Table 310.15(B)(1)

A #2 Cu THHN (90 C ampacity 130 A) runs with 6 current-carrying conductors (0.80 factor) in a 40 C ambient (0.91 factor). What is the adjusted-and-corrected ampacity?

• a.130 A
• b.95 A✓
• c.104 A
• d.115 A

Apply both factors to the 90 C ampacity: 130 x 0.80 x 0.91 = 94.6 A, about 95 A.

An office has 45 general-use receptacle outlets. At 180 VA per outlet, what is the receptacle load before demand factors?

• a.5,400 VA
• b.4,500 VA
• c.9,000 VA
• d.8,100 VA✓

Each general-use receptacle outlet is figured at 180 VA: 45 x 180 = 8,100 VA.2023 NEC 220.14(I)

A commercial building has a 30,000-VA receptacle load. Applying the demand (first 10 kVA at 100%, remainder at 50%), what is the demand load?

• a.25,000 VA
• b.15,000 VA
• c.30,000 VA
• d.20,000 VA✓

First 10,000 VA at 100% = 10,000; remaining 20,000 at 50% = 10,000. Total demand = 20,000 VA.

A one-family dwelling has a total general load (lighting, small-appliance, laundry, fastened appliances, ranges) of 40,000 VA before HVAC. Using the optional method of 220.82(B), what is the demand for these general loads?

• a.22,000 VA✓
• b.16,000 VA
• c.40,000 VA
• d.28,000 VA

2023 NEC 220.82(B) takes 100% of the first 10 kVA plus 40% of the remainder: 10,000 + 0.40 x 30,000 = 10,000 + 12,000 = 22,000 VA. Trap: 16,000 VA applies 40% to the whole 40,000; the first 10 kVA must stay at 100%.2023 NEC 220.82(B)

Under the optional dwelling calculation, central electric space-heating load is included at what percent of nameplate?

• a.100%
• b.65%✓
• c.40%
• d.75%

2023 NEC 220.82(C)(3) permits central electric space heating at 65% of nameplate. Trap: the 40% figure (220.82(C)(5)) applies only to four or more separately controlled electric-heat units, not to central heat.2023 NEC 220.82(C)

A 10,000 sq ft office building is calculated for general lighting using the 2023 unit-load table for office/bank occupancies. What is the general lighting load?

• a.25,000 VA
• b.20,000 VA
• c.30,000 VA
• d.35,000 VA✓

2023 NEC Table 220.42(A) lists 3.5 VA per sq ft for banks and office buildings: 10,000 x 3.5 = 35,000 VA. Trap: 30,000 VA uses the 3 VA/sq ft dwelling/store value.2023 NEC Table 220.42(A)

A 6,000 sq ft retail store is calculated for general lighting using the 2023 unit-load table. What is the general lighting load?

• a.18,000 VA✓
• b.12,000 VA
• c.21,000 VA
• d.15,000 VA

2023 NEC Table 220.42(A) lists 3 VA per sq ft for stores: 6,000 x 3 = 18,000 VA. Trap: 21,000 VA mistakenly uses the 3.5 VA/sq ft office value.2023 NEC Table 220.42(A)

A commercial service supplies several motors; the largest has a 40-A full-load current. How much is added to the service calculation for the largest-motor factor?

• a.50 A
• b.40 A
• c.10 A✓
• d.8 A

2023 NEC 430.24 (with 220.50) requires 25% of the largest motor FLC be added on top of the motor loads: 0.25 x 40 = 10 A. Trap: 50 A adds 125% of the motor, double-counting the FLC that is already in the load.2023 NEC 430.24

A commercial kitchen has six kitchen-equipment units totaling 60 kW. Applying the demand factor for six or more units, what is the demand load?

• a.60 kW
• b.39 kW✓
• c.45 kW
• d.42 kW

2023 NEC Table 220.56 allows 65% for six or more units: 60 x 0.65 = 39 kW (result may not be less than the sum of the two largest units). Trap: 45 kW uses 75%, the four-appliance dwelling factor, not the commercial-kitchen table.2023 NEC 220.56

Transformer secondary conductors run 8 ft to a panelboard under the 10-ft secondary rule. Their ampacity must be not less than:

• a.10% of the primary overcurrent device rating
• b.the primary conductor ampacity
• c.the rating of the device or busbar they supply✓
• d.at least 300% of the transformer's rated secondary current

2023 NEC 240.21(C)(2) allows unprotected secondary conductors up to 10 ft if their ampacity is at least the calculated load and at least the rating of the device or busbar they terminate in. Trap: the 1/10 rule belongs to the 10-ft feeder tap 240.21(B)(1), not the transformer-secondary rule.2023 NEC 240.21(C)(2)

A 75-kVA, 480-208Y/120-V transformer feeds a panel through 20-ft secondary conductors under 240.21(C)(6). What is the minimum secondary conductor ampacity?

• a.208 A✓
• b.167 A
• c.260 A
• d.104 A

Secondary FLC = 75,000 / (1.732 x 208) = 208 A. 240.21(C)(6) requires the 20-ft secondary conductors to have ampacity at least equal to the transformer secondary current (208 A) and to terminate in a single OCPD not exceeding that ampacity. Trap: 260 A is 125% of secondary current, which would size the OCPD, not the minimum conductor.2023 NEC 240.21(C)(6)

A 400-A feeder overcurrent device supplies a 10-ft tap. Under the 10-ft feeder tap rule, what minimum tap-conductor ampacity comes from the 1/10 provision?

• a.200 A
• b.133 A
• c.13 A
• d.40 A✓

2023 NEC 240.21(B)(1) requires the 10-ft tap ampacity to be at least 1/10 of the 400-A device: 400 / 10 = 40 A (and at least the load served). Trap: 133 A is the 1/3 value that belongs to the 25-ft tap rule 240.21(B)(2).2023 NEC 240.21(B)(1)

A 600-A feeder overcurrent device supplies a 25-ft tap. Under the 25-ft feeder tap rule, what minimum tap-conductor ampacity is required?

• a.60 A
• b.200 A✓
• c.300 A
• d.150 A

2023 NEC 240.21(B)(2) requires the 25-ft tap ampacity to be at least 1/3 of the 600-A device: 600 / 3 = 200 A. Trap: 60 A applies the 1/10 rule, which is only for the 10-ft tap.2023 NEC 240.21(B)(2)

A 75-kVA transformer with a 208-A secondary (over 9 A) is protected on both the primary and the secondary. What is the maximum secondary overcurrent device (next higher standard permitted)?

• a.300 A✓
• b.260 A
• c.250 A
• d.225 A

Table 450.3(B) allows up to 125% on the secondary where both sides are protected: 208 x 1.25 = 260 A. Since 260 A is not a standard size, Note 1 permits the next higher standard rating, 300 A. Trap: 250 A is the next lower standard and would be too small to carry full secondary load.2023 NEC 450.3(B)

From the motor full-load current table, what is the full-load current of a 50-HP, 460-V, three-phase induction motor?

• a.52 A
• b.80 A
• c.65 A✓
• d.48 A

2023 NEC Table 430.250 lists 65 A for a 50-HP, 460-V three-phase motor. This table FLC (not nameplate) is used for branch-circuit and feeder sizing. Trap: 52 A is the 40-HP value.2023 NEC Table 430.250

A motor with a 34-A full-load current is protected by a non-time-delay (one-time) fuse. At 300%, what is the maximum branch-circuit short-circuit/ground-fault fuse (next standard size)?

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

Table 430.52 allows 300% for non-time-delay fuses: 34 x 3.00 = 102 A. Where this is not a standard size, 430.52(C)(1) Exception 1 permits the next higher standard fuse, 110 A. Trap: a time-delay (dual-element) fuse uses only 175% (60 A).2023 NEC 430.52

The optional calculation for a multifamily dwelling (220.84) may be used only where the building has at least how many dwelling units?

• a.3✓
• b.2
• c.5
• d.10

2023 NEC 220.84 applies to multifamily dwellings of three or more units that meet the stated conditions (single feeder per unit, electric cooking, and electric heat or air-conditioning). Trap: 5 confuses this with range-table diversity, not the unit-count threshold.2023 NEC 220.84

A feeder supplies five household electric ranges, each rated 12 kW. Using Column C of the range demand table, what is the demand load?

• a.17,000 VA
• b.25,000 VA
• c.20,000 VA✓
• d.21,000 VA

2023 NEC Table 220.55, Column C, gives 20 kW for five ranges not over 12 kW: 20,000 VA. Trap: 21,000 VA is the six-range value; 25,000 VA wrongly prorates 5 x 5 kW.2023 NEC Table 220.55

A service neutral carries a maximum unbalanced load of 250 A. Applying the neutral demand allowance, what is the calculated neutral load?

• a.250 A
• b.235 A✓
• c.225 A
• d.210 A

2023 NEC 220.61(B) permits 70% on the portion of neutral load over 200 A: 200 + 0.70 x 50 = 200 + 35 = 235 A. Trap: 250 A ignores the reduction; only the amount above 200 A is reduced.2023 NEC 220.61(B)

When more than two conductors are pulled into a single conduit, what maximum percent of the conduit's cross-sectional area may they occupy?

• a.53%
• b.31%
• c.60%
• d.40%✓

2023 NEC Chapter 9, Table 1: over two conductors is limited to 40% fill. Trap: 53% is the limit for one conductor and 31% for exactly two.2023 NEC Chapter 9 Table 1

A box contains eight #12 AWG conductors and nothing else. Using the box-fill volume allowance, what minimum box volume is required?

• a.16.0 cu in
• b.20.0 cu in
• c.18.0 cu in✓
• d.22.5 cu in

2023 NEC Table 314.16(B) assigns 2.25 cu in to each #12 conductor: 8 x 2.25 = 18.0 cu in. Trap: 16.0 uses the 2.0 cu in #14 value; 22.5 uses the 2.5 cu in #10 value.2023 NEC Table 314.16(B)

Under the optional method for adding load to an existing dwelling (220.83), after taking 100% of the first 8 kVA, the remaining other loads are calculated at what percent?

• a.40%✓
• b.35%
• c.50%
• d.65%

2023 NEC 220.83(B) takes 100% of the first 8 kVA and 40% of the remainder of the other load where additional air-conditioning or space-heating is being added. Trap: 35% is the general-lighting remainder factor from the standard method 220.42, not 220.83.2023 NEC 220.83(B)

A large multifamily building has 900,000 VA of connected load and an applicable 220.84 demand factor of 32%. What is the calculated demand load?

• a.450,000 VA
• b.900,000 VA
• c.288,000 VA✓
• d.360,000 VA

The optional multifamily method multiplies connected load by the Table 220.84 demand factor: 900,000 x 0.32 = 288,000 VA. Trap: 360,000 VA uses 40%; the demand factor falls as the number of units rises.2023 NEC 220.84

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.