1. A 120V, single-phase circuit supplies a 1,500W resistive space heater. Using the NEC 125% continuous load rule (Article 210.19), what is the minimum ampacity required for the branch-circuit conductors?
- A. 12.5 A
- B. 15.625 A ✓
- C. 20 A
- D. 25 A
Per NEC 210.19(A)(1), branch-circuit conductors for continuous loads must have an ampacity not less than 125% of the continuous load. Load current = 1,500W ÷ 120V = 12.5 A. Minimum ampacity = 12.5 A × 1.25 = 15.625 A. The conductor must be rated at least 15.625 A before rounding up to the next standard size.
2. A single-gang device box contains two 14/2 NM cables (each cable has two 14 AWG insulated conductors and one 14 AWG bare equipment grounding conductor) that enter through the box's internal cable clamps and terminate on a single-pole snap switch. No other conductors, fittings, or devices are present. Per NEC 314.16(B) and Table 314.16(B), what is the minimum box volume required?
- A. 12.0 cubic inches
- B. 14.0 cubic inches
- C. 16.0 cubic inches ✓
- D. 18.0 cubic inches
Per NEC 314.16(B), count the volume allowances: (1) Conductor fill – four 14 AWG insulated conductors = 4 allowances; (2) Clamp fill – one or more internal cable clamps = 1 allowance (based on the largest conductor, 14 AWG); (4) Device fill – the single-pole switch yoke = 2 allowances (double volume based on the largest conductor connected to it, 14 AWG); (5) Equipment grounding conductor fill – all EGCs together = 1 allowance based on the largest EGC (14 AWG). Total = 4 + 1 + 2 + 1 = 8 allowances. Table 314.16(B) assigns 2.00 cubic inches per 14 AWG conductor, so 8 × 2.00 = 16.0 cubic inches minimum. The 12.0 answer wrongly counts the switch as a single allowance and omits the clamps; 14.0 omits the clamp allowance; 18.0 wrongly counts each EGC separately.
3. A 240V, single-phase, 10 AWG copper conductor circuit is 150 feet long (one-way) and carries 24 amperes. Using the standard voltage drop formula (VD = 2 × K × I × D ÷ CM), where K = 12.9 for copper and CM for 10 AWG = 10,380, what is the approximate voltage drop?
- A. 4.5 V
- B. 7.1 V
- C. 8.9 V ✓
- D. 10.6 V
VD = (2 × K × I × D) ÷ CM = (2 × 12.9 × 24 × 150) ÷ 10,380. Numerator: 2 × 12.9 = 25.8; × 24 = 619.2; × 150 = 92,880. VD = 92,880 ÷ 10,380 ≈ 8.95 V ≈ 8.9 V. NEC 210.19(A) Informational Note recommends limiting voltage drop to 3% for branch circuits, which for 240V = 7.2 V, so this circuit exceeds the recommended limit. The formula uses 2× for single-phase two-wire circuits (round-trip distance).
4. According to NEC 220.52, what is the calculated load for the small-appliance branch circuits in a dwelling unit that has a kitchen, pantry, dining room, and breakfast room?
- A. 1,500 VA
- B. 3,000 VA ✓
- C. 4,500 VA
- D. 6,000 VA
NEC 220.52(A) requires a minimum of two 20-ampere small-appliance branch circuits for the kitchen, pantry, dining room, and breakfast room of a dwelling unit. Each circuit is calculated at 1,500 VA per NEC 220.52(A). Total = 2 circuits × 1,500 VA = 3,000 VA. These circuits are in addition to the general lighting load and must be included in the service or feeder calculation.
5. A residential electric range is rated 12 kW. Using NEC Table 220.55 Column C, what demand load should be used when calculating the service for a single-family dwelling?
- A. 8,000 W ✓
- B. 8,400 W
- C. 10,000 W
- D. 12,000 W
NEC Table 220.55, Column C applies to ranges rated over 8¾ kW through 12 kW. For one range in this rating, the demand load is 8,000 W (8 kW). This demand factor accounts for the fact that the range is rarely operated at full capacity. A 12 kW range does not require the full 12 kW to be used for calculation; the table allows 8 kW demand.
6. A single-family dwelling has 3,400 sq ft of floor area (calculated per NEC 220.5(C), excluding open porches, garages, and unfinished spaces not adaptable for future use). The dwelling has two 20-A small-appliance branch circuits and one laundry branch circuit. After applying the demand factors of NEC 220.42 to the general lighting, small-appliance, and laundry loads, what is the resulting demand load?
- A. 14,700 VA
- B. 7,095 VA ✓
- C. 5,520 VA
- D. 5,145 VA
Per Table 220.12, the general lighting load for a dwelling unit is 3 VA/sq ft: 3,400 × 3 = 10,200 VA. NEC 220.52(A) and (B) require 1,500 VA for each small-appliance circuit (2 × 1,500 = 3,000 VA) and 1,500 VA for the laundry circuit, and these loads are permitted to be included with the general lighting load when applying the demand factors of Table 220.42 (Table 220.42(A) in the 2023 NEC). Total = 10,200 + 3,000 + 1,500 = 14,700 VA. Applying the dwelling-unit demand factors: first 3,000 VA at 100% = 3,000 VA; remaining 11,700 VA at 35% = 4,095 VA; total demand = 7,095 VA. 14,700 VA ignores the demand factor; 5,520 VA omits the small-appliance and laundry loads; 5,145 VA incorrectly applies 35% to the entire load without the first 3,000 VA at 100%.
7. Four 10 AWG THHN copper conductors are installed in the same conduit. The ambient temperature is 86°F (30°C). Using NEC Table 310.16, what is the corrected ampacity of each 10 AWG THHN conductor after applying the conduit fill correction factor?
- A. 24 A
- B. 28 A
- C. 32 A ✓
- D. 40 A
Under NEC Table 310.16, 10 AWG copper THHN is taken from the 90°C column at 40 A. At 86°F (30°C), no ambient temperature correction is needed. With four current-carrying conductors in the same raceway, NEC 310.15(C)(1) requires an 80% adjustment factor for 4 through 6 conductors. Therefore, the adjusted ampacity is 40 A × 0.80 = 32 A. Terminal temperature limitations or the small-conductor rule may limit the allowable overcurrent protection in practice, but they do not change the adjusted conductor ampacity asked for here.
8. A 208V, 3-phase, 4-wire panel serves a balanced load of 30 kVA. What is the line current per phase?
- A. 48.1 A
- B. 72.2 A
- C. 83.3 A ✓
- D. 144.3 A
For a balanced 3-phase load: I = kVA × 1,000 ÷ (√3 × V_L-L) = 30,000 ÷ (1.732 × 208) = 30,000 ÷ 360.3 ≈ 83.3 A per phase. The √3 factor (1.732) is required for 3-phase calculations. Using single-phase formula (30,000 ÷ 208 = 144.2 A) or dividing by 3 instead of √3 are common errors.
9. According to NEC 430.22, a motor branch-circuit conductor must have an ampacity of at least what percentage of the motor's full-load current (FLC) for a continuous-duty motor?
- A. 100%
- B. 115%
- C. 125% ✓
- D. 150%
NEC 430.22(A) requires that branch-circuit conductors supplying a single motor used in a continuous duty application shall have an ampacity of not less than 125% of the motor's full-load current rating as determined by NEC 430.6(A)(1) (from NEC Tables 430.247–430.250). This is separate from the 125% continuous load rule of 210.19, but the result is the same multiplier for a different stated reason (motor starting and running characteristics).
10. A junction box contains the following 14 AWG conductors: 3 hot conductors entering, 3 neutral conductors entering, 2 equipment grounding conductors, and one single-pole switch (device). Per NEC 314.16(B), what is the minimum required box volume?
- A. 12.0 cu in
- B. 16.0 cu in
- C. 18.0 cu in ✓
- D. 20.0 cu in
Per NEC 314.16(B) and Table 314.16(B), each 14 AWG conductor allowance is 2.0 cubic inches. The 3 ungrounded conductors plus 3 grounded conductors count as 6 conductor equivalents. All equipment grounding conductors in the box together count as 1 conductor equivalent. One single-pole switch yoke counts as 2 conductor equivalents. Total conductor equivalents = 6 + 1 + 2 = 9; 9 × 2.0 cu in = 18.0 cu in.