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Electrical Exam Calculation Practice: Conduit Fill, Box Fill, Voltage Drop and Service Load

Exam-style calculation practice for US electrician licensing exams, based on the National Electrical Code (NFPA 70, 2023 edition). Each calculator shows its work: the formula, the NEC table it reads, and the numbers plugged in, so you can repeat the method on the exam without the tool.

Conduit fill (EMT, THHN)

Find the smallest EMT for a mix of THHN conductors, or check the percent fill of a trade size. The result lists each step with the Chapter 9 table it uses.

Each topic has 10 exam-style problems with worked solutions. Problems 1 and 2 are free below; the 12-Month Pass unlocks the other 8 in all four topics.

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Practice problems

Problem 1. Three 12 AWG THHN conductors run in 1/2 in EMT. What is the percent fill, and is it within the limit?

Worked solution
  1. Conductor areas: 3 x 12 AWG THHN @ 0.0133 sq in = 0.0399 sq in. Total = 0.0399 sq in. NEC Chapter 9, Table 5
  2. 3 conductors in the raceway, so the fill limit is 40% (1 conductor 53%, 2 conductors 31%, over 2 conductors 40%). NEC Chapter 9, Table 1
  3. Smallest EMT whose 40% area covers 0.0399 sq in: trade size 1/2 (0.304 x 0.4 = 0.1216 sq in). NEC Chapter 9, Table 4 (EMT)
  4. In 1/2 EMT (total area 0.304 sq in): 0.0399 / 0.304 = 13.13% fill, limit 40%: OK. NEC Chapter 9, Tables 1 and 4

Answer: 13.13%

Problem 2. How many 12 AWG THHN conductors are permitted in 3/4 in EMT?

Worked solution
  1. 40% of 3/4 EMT = 0.533 x 0.40 = 0.2132 sq in (more than 2 conductors). NEC Chapter 9, Tables 1 and 4 (EMT)
  2. One 12 AWG THHN = 0.0133 sq in. NEC Chapter 9, Table 5
  3. 0.2132 / 0.0133 = 16.03; a decimal of 0.8 or more rounds up, otherwise round down: 16 conductors. NEC Chapter 9, Note 7

Answer: 16

Problems 3 to 10 with worked solutions:

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Box fill

Add up the box fill allowances (conductors, grounds, clamps, studs, device yokes) and compare with the box volume.

Each topic has 10 exam-style problems with worked solutions. Problems 1 and 2 are free below; the 12-Month Pass unlocks the other 8 in all four topics.

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Practice problems

Problem 1. A 4 in x 1-1/2 in square box (21.0 cu in) has six 12 AWG insulated conductors, their equipment grounding conductors, and internal cable clamps. No devices. Does it pass?

Worked solution
  1. 6 x 12 AWG conductors x 2.25 cu in = 13.5 cu in. NEC 314.16(B)(1), Table 314.16(B)(1)
  2. 3 equipment grounding conductors: up to 4 count once, each one over 4 adds 1/4, based on the largest (12 AWG): 1 x 2.25 = 2.25 cu in. NEC 314.16(B)(5)
  3. Internal cable clamps (one or more) count once, based on the largest conductor (12 AWG): 2.25 cu in. NEC 314.16(B)(2)
  4. Required volume = 18 cu in; box volume = 21 cu in: OK. NEC 314.16(A), Table 314.16(A)

Answer: 18 cu in, passes

Problem 2. How many 14 AWG conductors are permitted in an empty 4 in x 2-1/8 in square box (30.3 cu in) with no clamps, grounds, or devices?

Worked solution
  1. Each 14 AWG conductor takes 2 cu in. NEC Table 314.16(B)(1)
  2. (30.3 – 0) / 2 = 15.15; round down: 15 conductors. NEC 314.16(A)

Answer: 15

Problems 3 to 10 with worked solutions:

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Voltage drop

Voltage drop and percent drop for a copper or aluminum run, plus the smallest size that meets your target drop. The long-form guide stays on our voltage drop calculator page.

Each topic has 10 exam-style problems with worked solutions. Problems 1 and 2 are free below; the 12-Month Pass unlocks the other 8 in all four topics.

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Practice problems

Problem 1. A 120 V single-phase circuit carries 16 A over 100 ft (one way) on 12 AWG copper. What are the voltage drop and percent drop?

Worked solution
  1. Single-phase: VD = 2 x K x I x L / CM, with L the one-way length. Exam formula; K from NEC Chapter 9, Table 8 resistance
  2. K = 12.9 for copper; 12 AWG = 6530 circular mils. NEC Chapter 9, Table 8
  3. VD = 2 x 12.9 x 16 A x 100 ft / 6530 = 6.32 V. Arithmetic
  4. Percent = 6.32 / 120 V x 100 = 5.27%. Recommended: 3% or less on a branch circuit, 5% or less feeder plus branch. NEC 210.19 and 215.2 informational notes (recommendation, not a requirement)

Answer: 6.32 V, 5.27%

Problem 2. A 240 V single-phase load draws 30 A, 150 ft away, on 10 AWG copper. What are the voltage drop and percent drop?

Worked solution
  1. Single-phase: VD = 2 x K x I x L / CM, with L the one-way length. Exam formula; K from NEC Chapter 9, Table 8 resistance
  2. K = 12.9 for copper; 10 AWG = 10380 circular mils. NEC Chapter 9, Table 8
  3. VD = 2 x 12.9 x 30 A x 150 ft / 10380 = 11.18 V. Arithmetic
  4. Percent = 11.18 / 240 V x 100 = 4.66%. Recommended: 3% or less on a branch circuit, 5% or less feeder plus branch. NEC 210.19 and 215.2 informational notes (recommendation, not a requirement)

Answer: 11.18 V, 4.66%

Problems 3 to 10 with worked solutions:

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Dwelling service load (optional method)

Optional method for a one-family dwelling at 120/240 V: general load with the 10 kVA / 40% demand, the larger of heating or cooling, then the service rating.

Each topic has 10 exam-style problems with worked solutions. Problems 1 and 2 are free below; the 12-Month Pass unlocks the other 8 in all four topics.

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Practice problems

Problem 1. A 1,500 sq ft house (120/240 V) has 2 small-appliance circuits, 1 laundry circuit, a 12 kW range, 5 kW dryer, 4.5 kW water heater, 1.2 kW dishwasher and 5 kVA of air conditioning. Using the optional method, what is the load and the minimum service rating?

Worked solution
  1. General lighting and receptacles: 3 VA x 1500 sq ft = 4500 VA. NEC 220.82(B)(1)
  2. Small-appliance circuits: 2 x 1,500 VA = 3000 VA; laundry: 1 x 1,500 VA = 1500 VA. NEC 220.82(B)(2); 210.11(C)(1), (C)(2)
  3. Appliances at nameplate: range 12000 VA + dryer 5000 VA + water heater 4500 VA + dishwasher 1200 VA = 22700 VA. NEC 220.82(B)(3)
  4. General load 31700 VA: first 10,000 VA at 100% + (31700 – 10,000) x 40% = 10,000 + 8680 = 18680 VA. NEC 220.82(B)
  5. Heating / cooling, take the largest: air conditioning 100% = 5000 VA. Use 5000 VA. NEC 220.82(C)(1)
  6. Total = 18680 + 5000 = 23680 VA; 23680 / 240 V = 98.67 A. NEC 220.82(A)
  7. Next standard rating at or above 98.67 A: 100 A. The minimum for a one-family dwelling is 100 A. NEC 240.6(A); 230.79(C)

Answer: 23680 VA, 98.67 A, 100 A service

Problem 2. A 2,000 sq ft house has 2 small-appliance circuits, 1 laundry circuit, a 10 kW range, 5.5 kW dryer, 4.5 kW water heater, no air conditioning, and 10 kW of central electric heat. Optional method: load and minimum service rating?

Worked solution
  1. General lighting and receptacles: 3 VA x 2000 sq ft = 6000 VA. NEC 220.82(B)(1)
  2. Small-appliance circuits: 2 x 1,500 VA = 3000 VA; laundry: 1 x 1,500 VA = 1500 VA. NEC 220.82(B)(2); 210.11(C)(1), (C)(2)
  3. Appliances at nameplate: range 10000 VA + dryer 5500 VA + water heater 4500 VA = 20000 VA. NEC 220.82(B)(3)
  4. General load 30500 VA: first 10,000 VA at 100% + (30500 – 10,000) x 40% = 10,000 + 8200 = 18200 VA. NEC 220.82(B)
  5. Heating / cooling, take the largest: central electric heat 65% = 6500 VA. Use 6500 VA. NEC 220.82(C), (C)(4)
  6. Total = 18200 + 6500 = 24700 VA; 24700 / 240 V = 102.92 A. NEC 220.82(A)
  7. Next standard rating at or above 102.92 A: 110 A. NEC 240.6(A)

Answer: 24700 VA, 102.92 A, 110 A service

Problems 3 to 10 with worked solutions:

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Sources and scope

Table values come from NFPA 70 (NEC) 2023: Chapter 9 Tables 1, 4, 5 and 8, Tables 314.16(A) and 314.16(B)(1), sections 314.16, 220.82, 230.79 and 240.6. They match the 2020 edition. Your state may enforce an older or amended edition, so check what your exam uses. The NEC is published by NFPA and can be read free online at nfpa.org/70. Problems and explanations are original and not affiliated with NFPA or any exam provider. Conduit fill covers EMT with THHN / THWN-2 only; voltage drop uses the exam constants K = 12.9 (copper) and 21.2 (aluminum) and does not replace an ampacity check.

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