Free Master Electrician Practice Test
This free master electrician practice test contains 38 original, unofficial questions with worked answers. It is focused practice—not a full-length exam—with electrical theory, reference-based calculations, and a separate 2026 National Electrical Code (NEC) section.
Question 1 of 38
Electrical theory · Three-phase current and power factor
A balanced, sinusoidal three-phase load uses 66.5 kW of real power at 480 V line-to-line with a 0.80 power factor. What is the line current, to the nearest ampere?
- A. 64 A
- B. 80 A
- C. 100 A
- D. 173 A
Show answer and explanation for Question 1
Answer: C. 100 A
Real power is P = √3 × V × I × PF, so I = P ÷ (√3 × V × PF) = 66,500 ÷ (1.732 × 480 × 0.80) ≈ 100 A. A (64 A) multiplies by the power factor instead of dividing by it. B (80 A) ignores the power factor. D (173 A) leaves out √3. This is operating current, not a conductor or breaker size.
Sources: U.S. DOE, Electrical Science Fundamentals Handbook, Vol. 3 (DOE-HDBK-1011/3-92), Module ES-09, pp. 20–21 (power in balanced three-phase loads).
Continue electrical calculations (Questions 2–10)
They test the math that sits under code questions: three-phase current, power, transformers, fault current, and voltage drop. Round only at the final step unless a question says otherwise. Answer each question before opening its explanation; keep your choices on paper to score the set below.
Question 2 of 38
Electrical theory · Power triangle
A linear load on a sinusoidal supply draws 36 kW of real power and 27 kvar of reactive power. What is its apparent power?
- A. 45 kVA
- B. 27 kVA
- C. 36 kVA
- D. 63 kVA
Show answer and explanation for Question 2
Answer: A. 45 kVA
Real and reactive power are the two legs of the power triangle, so S = √(P² + Q²) = √(36² + 27²) = 45 kVA. D (63 kVA) adds the legs as if they were in line. B and C each use only one leg. Bonus: the power factor here is 36 ÷ 45 = 0.80.
Sources: U.S. DOE, Electrical Science Fundamentals Handbook, Vol. 3 (DOE-HDBK-1011/3-92), Module ES-09, pp. 1–4 (power triangle; pf = P ÷ S).
Question 3 of 38
Electrical theory · Transformer efficiency (kW vs. kVA)
A transformer delivers 100 kVA to a load with a 0.90 power factor. Its real input power is 93 kW. What is its efficiency, to one decimal place?
- A. 90.0%
- B. 96.8%
- C. 93.0%
- D. 107.5%
Show answer and explanation for Question 3
Answer: B. 96.8%
Compare real power with real power. Output = 100 kVA × 0.90 = 90 kW. Efficiency = output ÷ input × 100 = 90 ÷ 93 × 100 = 96.8%. A reports the power factor, not efficiency. C just reuses the input number. D (107.5%) divides 100 kVA by 93 kW, mixing units and producing an impossible result above 100%.
Sources: U.S. DOE, Electrical Science Fundamentals Handbook, Vol. 3 (DOE-HDBK-1011/3-92), Module ES-09, p. 4 (power factor = real power ÷ apparent power); U.S. DOE, Electrical Science Fundamentals Handbook, Vol. 4 (DOE-HDBK-1011/4-92; June 1992), Module ES-13, p. 3, equation 13-3 (transformer efficiency).
Question 4 of 38
Electrical theory · Series-parallel circuits
A 48-volt DC source feeds a 2 Ω resistor in series with two 12 Ω resistors that are connected in parallel with each other. What current leaves the source?
- A. 1.85 A
- B. 4 A
- C. 8 A
- D. 6 A
Show answer and explanation for Question 4
Answer: D. 6 A
Two equal resistors in parallel: 12 ÷ 2 = 6 Ω. Add the series resistor: 2 + 6 = 8 Ω. Current = 48 ÷ 8 = 6 A. A treats all three resistors as series (26 Ω). B divides by one 12 Ω branch. C leaves out the 2 Ω resistor.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Sections ‘Ohm’s Law’ and ‘Resistance in series and parallel circuits’.
Question 5 of 38
Electrical theory · Resistive load power
A resistive heating element measures 24 Ω (treat it as constant). What power does it use at 208 V, to one decimal place?
- A. 1,802.7 W
- B. 8.7 W
- C. 2,400.0 W
- D. 4,992.0 W
Show answer and explanation for Question 5
Answer: A. 1,802.7 W
Combine Ohm's law and Watt's law: P = V² ÷ R = 208² ÷ 24 = 1,802.7 W. B (8.7) is the current in amperes (208 ÷ 24), not the power. C uses 240 V instead of the stated 208 V. D multiplies volts by ohms, which isn't a power formula.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Sections ‘Ohm’s Law’ and ‘Watt’s Law’; constant-resistance model.
Question 6 of 38
Electrical theory · Transformer current ratio
Three identical 24-volt coils, each drawing 10 A, are connected in parallel to a 240-to-24-volt single-phase control transformer. Ignoring losses and inrush, what is the primary current with all three coils energized?
- A. 30 A
- B. 3 A
- C. 300 A
- D. 1 A
Show answer and explanation for Question 6
Answer: B. 3 A
The secondary carries 3 × 10 = 30 A. Treat primary and secondary VA as equal: 24 V × 30 A = 720 VA, and 720 ÷ 240 V = 3 A. A copies the secondary current. C flips the ratio. D counts only one coil.
Sources: U.S. DOE, Electrical Science Fundamentals Handbook, Vol. 4 (DOE-HDBK-1011/4-92; June 1992), Module ES-13, p. 7, equations 13-5 and 13-6 (inverse voltage/current ratio); Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Resistance in series and parallel circuits’: branch currents sum.
Question 7 of 38
Electrical theory · Delta-wye transformer voltage
An ideal three-phase transformer bank has a delta primary fed at 480 V line-to-line and a wye secondary. Each primary winding has twice the turns of its secondary winding. What is the secondary line-to-line voltage, to one decimal place?
- A. 240.0 V
- B. 277.1 V
- C. 415.7 V
- D. 480.0 V
Show answer and explanation for Question 7
Answer: C. 415.7 V
In a delta, each winding sees the full line voltage: 480 V. A 2:1 turns ratio puts 240 V across each secondary winding. In a wye, line voltage = √3 × winding (phase) voltage = 1.732 × 240 = 415.7 V. A stops at the winding voltage. B is 480 ÷ √3, which skips the turns ratio. D ignores the transformer.
Sources: U.S. DOE, Electrical Science Fundamentals Handbook, Vol. 3 (DOE-HDBK-1011/3-92), Module ES-09, p. 20 (delta and wye line/phase relationships); U.S. DOE, Electrical Science Fundamentals Handbook, Vol. 4 (DOE-HDBK-1011/4-92; June 1992), Module ES-13, p. 5, equation 13-4 (voltage ratio equals turns ratio).
Question 8 of 38
Electrical theory · Transformer full-load current
What is the full-load secondary current of a 112.5 kVA, three-phase transformer with a 208Y/120-volt secondary?
- A. 312.3 A
- B. 540.9 A
- C. 468.8 A
- D. 135.3 A
Show answer and explanation for Question 8
Answer: A. 312.3 A
Three-phase: I = VA ÷ (V × 1.732) = 112,500 ÷ (208 × 1.732) ≈ 312.3 A; keep the denominator unrounded until the final step. B forgets √3. C uses single-phase 240 V math. D substitutes 480 V for the stated 208 V secondary.
Sources: U.S. DOE, Electrical Science Fundamentals Handbook, Vol. 3 (DOE-HDBK-1011/3-92), Module ES-09, pp. 20–21 (balanced three-phase apparent power).
Question 9 of 38
Electrical theory · Available fault current
A 50 kVA, single-phase, 120/240-volt transformer has 2% impedance. Assume an infinite primary source and ignore secondary conductor impedance. What is the approximate RMS symmetrical current for a bolted 240-volt line-to-line fault at its secondary terminals, to the nearest ampere?
- A. 4,167 A
- B. 10,417 A
- C. 5,208 A
- D. 208 A
Show answer and explanation for Question 9
Answer: B. 10,417 A
Full-load current at 240 V = 50,000 ÷ 240 A. Divide by the per-unit impedance: (50,000 ÷ 240) ÷ 0.02 ≈ 10,417 A. Use 0.02, not 2, and keep the unrounded full-load current until the final step. A multiplies by 20 (100 ÷ 5, a misread impedance). C calculates full-load current at 480 V before applying the 2% impedance. D is just the full-load current. This simplified estimate is not a complete protection or equipment-rating study.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Short circuit current calculations’: single-phase transformer terminal-current example; stated simplifying assumptions.
Question 10 of 38
Electrical theory · Three-phase voltage drop
A balanced 208-volt, three-phase load draws 50 A at unity power factor. The conductors run 200 ft one way and have a supplied resistance of 0.491 Ω per 1,000 ft. Ignoring reactance, what is the approximate line-to-line voltage drop as a percentage of 208 V, to one decimal place?
- A. 2.4%
- B. 8.5%
- C. 4.7%
- D. 4.1%
Show answer and explanation for Question 10
Answer: D. 4.1%
Three-phase: VD = 1.732 × R × I × L ÷ 1,000 = 1.732 × 0.491 × 50 × 200 ÷ 1,000 = 8.50 V. Percent = 8.50 ÷ 208 × 100 = 4.1%. C uses the single-phase multiplier 2 instead of 1.732. B reports the volts as a percentage. A (2.4%) divides the volts by 208 × 1.732 instead of by 208. The resistance value is supplied for this exercise.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Voltage drop calculations’: three-phase resistance formula and model limitations; Supplied exercise data, Question 10: resistance, current, length, voltage, and unity-power-factor assumptions.
Part 2: NEC calculations and reference practice (16 questions)
Each question identifies its source basis: 2023 tables, a method explained in the Minnesota Department of Labor and Industry (DLI) guide, or explicitly supplied worksheet data. Supplied-data questions test the calculation—not whether those inputs apply to every NEC edition or installation.
Question 11 of 38
Supplied motor data + 2023 ampacity table · Motor branch-circuit conductors
A 50 hp, 460-volt, three-phase, continuous-duty induction motor has a supplied full-load current (FLC) of 65 A for this exercise. Use 125% of that current for conductor ampacity. With three current-carrying THWN copper conductors, 75 °C terminations, 30 °C ambient, and no other adjustment, what is the smallest conductor below that meets the result?
- A. 4 AWG
- B. 6 AWG
- C. 3 AWG
- D. 2 AWG
Show answer and explanation for Question 11
Answer: A. 4 AWG
Required ampacity = 65 × 1.25 = 81.25 A. In the 2023 Table 310.16 copper 75 °C column, 6 AWG is 65 A and 4 AWG is 85 A, so 4 AWG is the smallest listed choice that meets 81.25 A. B (6 AWG) is what you would pick if you forgot the 125%. The 3 AWG and 2 AWG choices are larger than needed. The motor current is supplied; this question tests the percentage and ampacity-column selection, not recall of a motor table row.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Motor circuit conductor and overcurrent protection calculations’: 125% conductor method; HELUKABEL, Allowable Ampacity Tables, NFPA 70-2023, p. 1, left-hand Table 310.16, copper 75 °C column; Supplied exercise data, Question 11: 65 A motor current and installation conditions.
Question 12 of 38
Supplied protection worksheet · Motor inverse-time breaker
For this motor-protection worksheet, use a full-load current of 65 A and an inverse-time-breaker multiplier of 250%. The worksheet permits the next higher rating when the product falls between these listed ratings: 150, 175, 200, and 225 A. No further increase applies. What rating does the worksheet select?
- A. 150 A
- B. 175 A
- C. 200 A
- D. 225 A
Show answer and explanation for Question 12
Answer: B. 175 A
65 × 2.5 = 162.5 A. Under the stated next-higher-rating rule, select 175 A. A (150 A) is below the calculated value; C and D skip past the next rating. This is a supplied-rule calculation, not a statement that 250% or rounding up applies to every motor or protective device.
Sources: Supplied exercise data, Question 12: 65 A, 250%, the rating list, and the expressly supplied next-higher-rating condition.
Question 13 of 38
Motor calculation method (DLI guide) · Motor dual-element fuse
A 25 hp, 460-volt, three-phase motor has a supplied full-load current of 34 A. Use the DLI guide’s 175% dual-element (time-delay) fuse method for branch-circuit short-circuit and ground-fault protection. The exercise permits the next higher listed rating if needed, using 50, 60, 70, and 80 A. With no additional increase, what is the maximum rating selected?
- A. 50 A
- B. 85 A
- C. 70 A
- D. 60 A
Show answer and explanation for Question 13
Answer: D. 60 A
34 × 1.75 = 59.5 A. The next listed fuse rating is 60 A. B (85 A) is the unrounded 250% calculation for a different supplied multiplier, not the answer to this fuse exercise. A is below the selected maximum; the question does not establish whether a 50 A fuse would allow this motor to start. C goes past the next listed size.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Motor circuit conductor and overcurrent protection calculations’: dual-element fuse example, 175% and next-standard-rating steps; Supplied exercise data, Question 13: 34 A input and stated rating choices.
Question 14 of 38
Motor calculation method (DLI guide) · Nameplate vs. table current
For an ordinary, single-speed, general-purpose AC induction motor, with no special motor exception applicable, which of these is sized from the motor’s NAMEPLATE current rather than the NEC full-load current tables?
- A. Branch-circuit conductors
- B. The motor disconnect switch rating
- C. Separate overload protection
- D. Branch-circuit short-circuit and ground-fault protection
Show answer and explanation for Question 14
Answer: C. Separate overload protection
The DLI guide distinguishes table current for conductors, switch ratings, and short-circuit/ground-fault protection from nameplate current for separate overload protection. Mixing up the two is an easy way to miss motor questions. The ordinary-motor conditions in this question matter; this is not a rule for every motor arrangement.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Motor circuit conductor and overcurrent protection calculations’, discussion of 430.6(A)(1).
Question 15 of 38
Motor calculation method (DLI guide) · Feeder for several motors
A feeder supplies only three simultaneously operating, continuous-duty, 460-volt, three-phase motors: 20 hp, 10 hp, and 5 hp. For this exercise, their respective full-load currents are supplied as 27 A, 14 A, and 7.6 A. Using 125% of the largest motor current plus 100% of the others, what is the minimum feeder conductor ampacity?
- A. 48.6 A
- B. 60.75 A
- C. 55.35 A
- D. 33.75 A
Show answer and explanation for Question 15
Answer: C. 55.35 A
The DLI guide’s motor-feeder method uses 125% of the largest motor plus 100% of the others. 27 × 1.25 = 33.75; 33.75 + 14 + 7.6 = 55.35 A. A skips the 25% adder. B applies 125% to every motor. D stops after the largest motor.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Motor circuit conductor and overcurrent protection calculations’: feeder sum plus 25% of the largest motor; Supplied exercise data, Question 15: motor-current inputs and simultaneous continuous-duty operation.
Question 16 of 38
Supplied demand-factor worksheet · Dwelling general lighting demand
A worksheet for a 2,400 ft² one-family dwelling includes general lighting, two small-appliance circuits, and one laundry circuit. Use these supplied worksheet inputs: 3 VA per ft² for lighting; 1,500 VA for each small-appliance circuit; 1,500 VA for laundry; and a demand factor of 100% on the first 3,000 VA plus 35% on the remainder. What is the combined demand load for these loads only?
- A. 11,700 VA
- B. 6,045 VA
- C. 5,520 VA
- D. 3,045 VA
Show answer and explanation for Question 16
Answer: B. 6,045 VA
2,400 × 3 VA per ft² = 7,200 VA. Add 2 × 1,500 VA small-appliance and 1,500 VA laundry = 11,700 VA. Apply the supplied factors: first 3,000 VA at 100% = 3,000; remaining 8,700 × 35% = 3,045; total 6,045 VA. A skips the demand factors. C forgets the laundry circuit. D is only the 35% portion. This worksheet does not establish a complete dwelling service load or the rules for a particular NEC edition.
Sources: Supplied exercise data, Question 16: all unit loads, circuit counts, and demand factors are supplied exercise inputs.
Question 17 of 38
2023 NEC ampacity tables · Ambient correction plus bundling
Four current-carrying 6 AWG THHN copper conductors run together for more than 24 inches in one raceway through a space at 40 °C (104 °F). Terminations are rated 75 °C, and no other correction or exception applies. Starting from the 90 °C column of the 2023 NEC ampacity table, what is their adjusted ampacity?
- A. 75 A
- B. 68.25 A
- C. 60 A
- D. 54.6 A
Show answer and explanation for Question 17
Answer: D. 54.6 A
Table 310.16, 90 °C: 6 AWG = 75 A. Ambient correction for 90 °C insulation at 36–40 °C: 0.91. Adjustment for 4–6 current-carrying conductors: 80%. 75 × 0.91 × 0.80 = 54.6 A. B applies only temperature; C applies only bundling. Then compare against the 75 °C termination ampacity (65 A) and use the lower value for the circuit.
Sources: HELUKABEL, Allowable Ampacity Tables, NFPA 70-2023, p. 1, left-hand Table 310.16; p. 2, Tables 310.15(B)(1)(1) and 310.15(C)(1).
Question 18 of 38
2023 NEC ampacity tables · Counting current-carrying conductors
One raceway, longer than 24 inches, holds three separate three-phase, three-wire branch circuits of 10 AWG THHN copper plus one 10 AWG equipment grounding conductor. Use nine current-carrying conductors for adjustment; the equipment grounding conductor is not counted. All circuits run at the same time, ambient is 30 °C, terminations are rated 75 °C, and no other adjustment applies. Starting from the 2023 table’s 90 °C column, what is the adjusted ampacity of each circuit conductor?
- A. 40 A
- B. 32 A
- C. 28 A
- D. 20 A
Show answer and explanation for Question 18
Answer: C. 28 A
Use the nine current-carrying conductors specified in the question. Table 310.15(C)(1): 7–9 conductors = 70%. 10 AWG THHN at 90 °C = 40 A; 40 × 0.70 = 28 A. A applies no adjustment. B uses 80% by counting only 6. D uses 50% (10–20 conductors) by incorrectly treating the equipment grounding conductor as a tenth current-carrying conductor. This asks for adjusted ampacity, not a breaker selection.
Sources: HELUKABEL, Allowable Ampacity Tables, NFPA 70-2023, p. 1, left-hand Table 310.16; p. 2, Table 310.15(C)(1); Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Conductor ampacity calculations’: equipment grounding conductors excluded from current-carrying count.
Question 19 of 38
2023 NEC box fill · Box fill
A metal box has internal cable clamps and one duplex receptacle on a single yoke. Four 12 AWG insulated conductors enter and each terminates or passes through once; two 12 AWG equipment grounding conductors also enter. The largest conductors connected to the device are 12 AWG. There are no extra loops, support fittings, or other fill allowances. Under 2023 NEC 314.16(B), what is the minimum box volume?
- A. 13.5 in³
- B. 15.75 in³
- C. 18.0 in³
- D. 20.25 in³
Show answer and explanation for Question 19
Answer: C. 18.0 in³
Count volume allowances under 314.16(B): 4 conductors; 1 for all clamps; 1 for up to four EGCs; 2 for the device yoke. That's 8 × 2.25 in³ (12 AWG) = 18.0 in³. B misses the clamps. D counts each EGC separately. A misses the device's double allowance.
Sources: Minnesota DLI, Residential Electrical Inspection Checklist (2023 NEC), p. 5, item 046: 314.16 box-fill table and allowance notes.
Question 20 of 38
2023 NEC box fill · Box fill: many grounding conductors
Six 12 AWG equipment grounding conductors enter a box. Under 2023 NEC 314.16(B)(5), what volume allowance do they require together?
- A. 2.25 in³
- B. 3.375 in³
- C. 4.5 in³
- D. 13.5 in³
Show answer and explanation for Question 20
Answer: B. 3.375 in³
314.16(B)(5): up to four EGCs take one allowance (2.25 in³). Each additional EGC adds a quarter allowance: 2 × 0.5625 = 1.125 in³. Total 3.375 in³. C uses a half allowance per extra EGC rather than a quarter. A ignores the two additional conductors. D counts all six in full.
Sources: Minnesota DLI, Residential Electrical Inspection Checklist (2023 NEC), p. 5, item 046: 12 AWG = 2.25 in³; up to four grounding/bonding conductors count as one, then one-quarter per additional conductor.
Question 21 of 38
Supplied conduit-area worksheet · Conduit fill with mixed sizes
Three 6 AWG THWN and one 8 AWG THWN copper conductors will run in EMT longer than 24 inches. Use these supplied areas: each 6 AWG conductor = 0.0507 in²; the 8 AWG conductor = 0.0366 in². The permitted 40%-fill areas for the listed EMT sizes are ½ in.: 0.122 in²; ¾ in.: 0.213 in²; 1 in.: 0.346 in²; 1¼ in.: 0.598 in². What is the minimum listed trade size by fill?
- A. ½ in.
- B. ¾ in.
- C. 1 in.
- D. 1¼ in.
Show answer and explanation for Question 21
Answer: B. ¾ in.
Add the individual conductor areas: 0.0507 in² × 3 = 0.1521; add 0.0366 for a total of 0.1887 in². More than two conductors use the 40% fill column in this longer raceway. Compare the supplied permitted areas: ½ in. = 0.122 (too small); ¾ in. = 0.213 (fits). C and D are larger than needed. The areas are supplied; this question tests summing mixed sizes and choosing the first adequate raceway, not a cross-edition table lookup.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Raceway fill calculations’: mixed conductor sizes, individual conductor areas, and the 40% method; Supplied exercise data, Question 21: complete permitted-area comparison table.
Question 22 of 38
Electrical theory / supplied inputs · Single-phase voltage drop
A 240-volt, single-phase, 40 A load at unity power factor is 150 ft (one way) from the panel on 8 AWG copper. Use the supplied area of 16,510 circular mils and K = 12.9 Ω·cmil/ft, and neglect reactance. What is the approximate voltage drop, to one decimal place?
- A. 4.7 V
- B. 3.9 V
- C. 18.8 V
- D. 9.4 V
Show answer and explanation for Question 22
Answer: D. 9.4 V
VD = 2 × K × I × L ÷ CM = 2 × 12.9 × 40 × 150 ÷ 16,510 ≈ 9.4 V, about 3.9% of 240 V. A drops the 2 (current flows out and back). C doubles the 2. B is the percentage, not the volts. Some references use K = 12.8; read the value the question gives you.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Voltage drop calculations’: VD = 2KIL/CM, one-way length, and approximation limits; Supplied exercise data, Question 22: K = 12.9 Ω·cmil/ft, conductor area, load, and distance.
Question 23 of 38
Raceway method (DLI guide) · Raceway nipples
A conduit nipple 18 inches long connects two enclosures. What maximum percentage of its cross-sectional area may conductors fill?
- A. 31%
- B. 40%
- C. 53%
- D. 60%
Show answer and explanation for Question 23
Answer: D. 60%
The DLI guide’s raceway-fill section allows conduit nipples not longer than 24 inches to be filled to 60%. This 18-inch nipple meets that length condition. B (40%) is the guide’s normal limit for more than two conductors in a longer raceway; A and C are not the nipple allowance.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Raceway fill calculations’: conduit nipples not longer than 24 inches, 60% fill.
Question 24 of 38
Grounding method (DLI guide) · Shared equipment grounding conductor
One raceway holds three branch circuits protected at 20 A, 30 A, and 60 A. They share a single wire-type equipment grounding conductor. Which overcurrent device rating do you use to size it under Table 250.122?
- A. 110 A (the sum)
- B. About 37 A (the average)
- C. 60 A (the largest)
- D. 20 A (the smallest)
Show answer and explanation for Question 24
Answer: C. 60 A (the largest)
250.122(C): a single EGC run with multiple circuits is sized for the largest overcurrent device protecting conductors in that raceway or cable. The sum, average, and smallest device are not the basis specified by this rule. Use 60 A as the table-entry rating; this question does not ask for the conductor size.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), ‘Sample code question’, 250.122(C), largest overcurrent device for a shared EGC.
Question 25 of 38
2023 NEC small-conductor protection · Ampacity is not always the overcurrent limit
A general branch circuit uses 10 AWG copper with a 75 °C table ampacity of 35 A. No special small-conductor protection exception applies, and no temperature or conductor-count adjustment reduces its ampacity. Under the 2023 NEC small-conductor protection rule, what is the maximum overcurrent protection for this conductor?
- A. 20 A
- B. 30 A
- C. 35 A
- D. 40 A
Show answer and explanation for Question 25
Answer: B. 30 A
The general small-conductor protection limit for 10 AWG copper is 30 A, even though the 75 °C ampacity column lists 35 A. B is therefore the maximum under the stated conditions. A is below that maximum. C confuses table ampacity with the protection limit; D uses the 90 °C table ampacity and also exceeds the protection limit. Special motor and other exceptions are excluded by the question.
Sources: HELUKABEL, Allowable Ampacity Tables, NFPA 70-2023, p. 1, Table 310.16, 10 AWG copper row and small-conductor overcurrent-protection footnote; Minnesota DLI, Residential Electrical Inspection Checklist (2023 NEC), p. 2, item 014, 240.4 and general overcurrent-protection table; special-equipment caveat.
Question 26 of 38
Voltage-drop guidance (DLI guide) · Voltage-drop recommendation
The NEC's informational recommendation for combined feeder and branch-circuit voltage drop, for reasonable efficiency of operation, is not more than:
- A. 2%
- B. 3%
- C. 5%
- D. 10%
Show answer and explanation for Question 26
Answer: C. 5%
The recommendation is 5% total, with no more than 3% on either the feeder or the branch circuit. This general recommendation is not itself a mandatory limit; it does not override a separate applicable equipment or installation requirement. B (3%) is the per-part figure. A is below the stated combined recommendation, and D exceeds it.
Sources: Minnesota DLI, Electrical License Examination Guide (current guide; individual sections identified below), Section ‘Voltage drop calculations’: recommended 5% combined and 3% feeder/branch values.
Part 3: 2026 NEC changes (12 questions)
These questions use the 2026 NEC explanations in the government sources linked with each answer. They are separate from the 2023 table questions; do not treat the whole set as one state’s exam blueprint.
Question 27 of 38
2026 NEC · Finding load calculations
In the 2026 NEC, which article holds branch-circuit, feeder, and service load calculations?
- A. Article 100
- B. Article 110
- C. Article 120
- D. Article 220
Show answer and explanation for Question 27
Answer: C. Article 120
The 2026 edition moved load calculations from Article 220 to a new Article 120 in Chapter 1. D is where they lived through the 2023 edition. If your tabs or study notes still say 220, they're built for an older book.
Sources: City of Denison, TX, 2026 National Electrical Code Significant Changes, p. 1, Formatting Changes — Relocated Articles: Article 220 moved to Article 120.
Question 28 of 38
2026 NEC · Manufacturer instructions vs. the Code
Equipment installation instructions describe a setup that would not comply with an NEC requirement. Under 2026 NEC 110.3(B), what follows?
- A. The instructions override the NEC
- B. The installer may pick whichever is easier
- C. All of the instructions may be ignored
- D. The instructions can't justify a noncompliant installation
Show answer and explanation for Question 28
Answer: D. The instructions can't justify a noncompliant installation
The 2026 change says the manufacturer's instructions must result in an installation and use that complies with the Code. Instructions can be stricter, not weaker. A and B invent an override; C goes too far, since the rest of the instructions still apply.
Sources: Minnesota DLI, 2026 National Electrical Code FAQ, p. 1, item 1 (110.3(B)); City of Denison, TX, 2026 National Electrical Code Significant Changes, p. 2 (110.3(B)).
Question 29 of 38
2026 NEC · Splicing a grounding electrode conductor
Which splice uses the 2026 NEC 250.64(C)(2) allowance for listed grounding and bonding equipment?
- A. A fitting listed for grounding and bonding, installed where the splice stays accessible
- B. A twist-on connector concealed in a wall
- C. A twisted and taped joint
- D. An unlisted connector in an accessible spot
Show answer and explanation for Question 29
Answer: A. A fitting listed for grounding and bonding, installed where the splice stays accessible
The allowance needs both conditions: listed grounding and bonding equipment and an accessible location. B, C, and D each miss at least one. Exothermic welds and irreversible compression connectors remain permitted in accessible and non-accessible locations; they're separate methods.
Sources: Minnesota DLI, 2026 National Electrical Code FAQ, p. 4, item 11 (250.64(C)(2)); City of Denison, TX, 2026 National Electrical Code Significant Changes, p. 3 (250.64).
Question 30 of 38
2026 NEC · Garage GFCI
A new 120-volt, 20-ampere receptacle goes in the vehicle area of a dwelling's attached garage. The garage floor is above grade. Under 2026 NEC 210.8(A)(2), does being above grade remove the GFCI requirement?
- A. No; the elevation wording was removed
- B. Yes; garages need GFCI only at or below grade
- C. Yes; dry garages are exempt
- D. Only if the owner asks for it
Show answer and explanation for Question 30
Answer: A. No; the elevation wording was removed
The 2026 edition gives garages their own list item and removes the elevation wording. The above-grade vehicle area in this question still requires GFCI protection. B incorrectly treats the floor’s elevation as an exemption; C and D invent exemptions. This question does not establish how a particular earlier edition applied.
Sources: Minnesota DLI, 2026 National Electrical Code FAQ, p. 2, item 3, 210.8(A)(2).
Question 31 of 38
2026 NEC · Feeder surge protection
A new feeder panel serves an area used only as sleeping quarters in a fire station. Under 2026 NEC 215.18, can surge protection be left out because the area isn't a dwelling unit?
- A. Yes; only dwelling units are covered
- B. Yes; public buildings are excluded
- C. No; sleeping quarters in fire stations are covered
- D. Yes; sleeping areas never need it
Show answer and explanation for Question 31
Answer: C. No; sleeping quarters in fire stations are covered
2026 expanded the covered areas to include spaces used exclusively as sleeping quarters in fire, police, ambulance, rescue, ranger, and similar stations. The other options invent exclusions. This item tests coverage only, not SPD type or placement.
Sources: Minnesota DLI, 2026 National Electrical Code FAQ, p. 3, item 8 (215.18 covered areas, list item 5); City of Denison, TX, 2026 National Electrical Code Significant Changes, p. 3 (215.18).
Question 32 of 38
2026 NEC · Service disconnect location, one- and two-family
For a new two-family dwelling, where does 2026 NEC 230.70(A)(1) require the service disconnecting means?
- A. Anywhere inside, if the owner agrees
- B. In an indoor closet near the meter
- C. Somewhere remote that can't be seen from the dwelling
- D. A readily accessible outdoor location on or within sight of the dwelling
Show answer and explanation for Question 32
Answer: D. A readily accessible outdoor location on or within sight of the dwelling
2026 folded the former emergency-disconnect rule into 230.70: for one- and two-family dwellings, the service disconnect goes outdoors, readily accessible, on or within sight of the dwelling. Other buildings follow 230.70(A)(2), which still allows inside at the nearest point of entrance.
Sources: Minnesota DLI, 2026 National Electrical Code FAQ, p. 4, item 10 (230.70(A)(1) and (A)(2)).
Question 33 of 38
2026 NEC · Ceiling-fan outlet boxes
At a ceiling location in a dwelling living or sleeping area typical for a paddle fan, 2026 NEC 314.27(B)(2) requires the outlet box to meet one of two options. Which is one of them?
- A. A cover that looks like a fan-rated cover
- B. Drywall anchors instead of framing
- C. Any ordinary outlet box
- D. Installed to allow direct access through the box to framing that can support the fan, without removing the box
Show answer and explanation for Question 33
Answer: D. Installed to allow direct access through the box to framing that can support the fan, without removing the box
The two options are a box listed for ceiling-fan support, or a box installed so you can reach supporting framing through it without removing the box. A is only appearance; B and C don't provide listed or structural support.
Sources: Minnesota DLI, 2026 National Electrical Code FAQ, pp. 4–5, item 13 (314.27(B)(2)).
Question 34 of 38
2026 NEC · Damaged conductors
An evaluation finds that fire damage has made installed conductors unsuitable. Under 2026 NEC 300.4(C), what's required?
- A. Replace them
- B. Keep them if they still carry current
- C. Reduce the load instead
- D. Relabel the circuit and keep them
Show answer and explanation for Question 34
Answer: A. Replace them
New 2026 language says conductors and wiring methods made unsuitable by damage such as overheating, fire, corrosion, or water must be replaced. Still carrying current doesn't make a damaged conductor suitable. NEMA GD 1 and GD 2 give guidance on water- and fire-damaged equipment.
Sources: Minnesota DLI, 2026 National Electrical Code FAQ, p. 4, item 12 (300.4(C)); City of Denison, TX, 2026 National Electrical Code Significant Changes, p. 3 (300.4).
Question 35 of 38
2026 NEC · Wall space and fixed cabinets
When measuring dwelling wall space for receptacle spacing under 2026 NEC 210.52(A)(2), how is a fixed cabinet with a countertop treated?
- A. Counted as wall space
- B. Counted only if the countertop is wood
- C. Excluded from the wall-space measurement
- D. It removes all receptacle requirements in the room
Show answer and explanation for Question 35
Answer: C. Excluded from the wall-space measurement
2026 excludes any fixed cabinet from the wall-space measurement. Before, only cabinets without countertops or similar work surfaces were excluded. D goes too far: countertop receptacle rules still apply separately.
Sources: Minnesota DLI, 2026 National Electrical Code FAQ, p. 2, item 6 (210.52(A)(2)).
Question 36 of 38
2026 NEC · EV charging receptacles
A 50-ampere receptacle will supply electric vehicle supply equipment (EVSE). What does 2026 NEC 625.44 add?
- A. The receptacle must be listed for EVSE use
- B. Any receptacle with the matching blade pattern is fine
- C. Any 50 A-rated receptacle is fine
- D. The EVSE's own listing covers the receptacle
Show answer and explanation for Question 36
Answer: A. The receptacle must be listed for EVSE use
The new rule requires 30 A, 50 A, and 60 A receptacles used for EVSE to be listed for that use, because long continuous charging caused failures in ordinary receptacles. B and C substitute fit or rating for listing. D confuses one product's listing with another's.
Sources: City of Denison, TX, 2026 National Electrical Code Significant Changes, p. 3 (625.44).
Question 37 of 38
2026 NEC · High-leg identification
On a 2026 NEC high-leg delta system, what color does 110.5 require for the high-leg conductor?
- A. Green
- B. White
- C. Gray
- D. Orange
Show answer and explanation for Question 37
Answer: D. Orange
The 2026 significant-changes summary lists 110.5: high-leg conductors must be orange. Green, white, and gray aren't the high-leg marking under this rule.
Sources: City of Denison, TX, 2026 National Electrical Code Significant Changes, p. 2 (110.5).
Question 38 of 38
2026 NEC · Fuel dispenser emergency disconnect
A receptacle sits above a motor-fuel dispenser, inside its classified (hazardous) location. Under 2026 NEC 514.11, how does it relate to the emergency electrical disconnect?
- A. Receptacles are exempt
- B. It only needs a label
- C. The emergency disconnect must also disconnect it
- D. It may stay energized because it's above the dispenser
Show answer and explanation for Question 38
Answer: C. The emergency disconnect must also disconnect it
2026 clarified that the emergency disconnect must also de-energize receptacles located over or adjacent to dispensers within the classified location, so they can't become an ignition source during a spill or leak. The other options leave a live ignition source in place.
Sources: Minnesota DLI, 2026 National Electrical Code FAQ, p. 5, item 16 (514.11).
Check your score
Answer key: 1-C · 2-A · 3-B · 4-D · 5-A · 6-B · 7-C · 8-A · 9-B · 10-D · 11-A · 12-B · 13-D · 14-C · 15-C · 16-B · 17-D · 18-C · 19-C · 20-B · 21-B · 22-D · 23-D · 24-C · 25-B · 26-C · 27-C · 28-D · 29-A · 30-A · 31-C · 32-D · 33-D · 34-A · 35-C · 36-A · 37-D · 38-C.
Count how many you got right on your first try in each part. Exclude any question whose answer you opened before choosing. Your practice accuracy is correct answers ÷ attempted questions × 100; with no attempted questions, there is no percentage to report. Treat the result as feedback on these 38 questions only. It isn't an official score, and it doesn't predict whether you'll pass. Real exams use different questions and passing standards. These short parts are too small to measure a skill precisely, so use them to find topics, not to grade yourself.
For every miss, keep a quick error log:
| Question | What went wrong | What I'll redo |
|---|---|---|
| Example: 1 | Left out the power factor | Rewrite the formula with PF in the denominator, then rework without looking |
Then:
- Rework each missed calculation from a blank page. Write the formula, units, and each intermediate value.
- For each missed code question, find the cited provision in the matching edition by index, not by memory. For a supplied-data worksheet, redo the stated inputs and factors. On an open-book exam, speed of lookup is part of the skill.
- Retake only the missed questions a day or two later, and count that as a new attempt.
Which parts fit your exam?
Use the NEC edition named in your current exam bulletin for your test date. That isn't always the newest book, and it isn't always the edition your state enforces on job sites.
| Exam | Code edition on the exam | How to use this set | Official source |
|---|---|---|---|
| ICC R16 Master Electrician | 2023 NEC | Use theory and the explicitly labeled 2023 table questions; supplied-data exercises are calculation practice, not an R16 simulation. | ICC bulletin, R16 entry; ICC current exam list |
| ICC T16 Master Electrician | 2020 NEC | Theory and supplied-data arithmetic remain useful; this page does not supply a 2020 code-answer variant. | ICC current exam list, T16-N row |
| ICC G16 / F16 Master Electrician | 2017 / 2014 NEC | Theory and supplied-data arithmetic only; no older-edition code-answer variant is supplied here. | ICC current exam list, G16-N and F16-N rows |
| Texas Master Electrician (PSI for TDLR) | 2026 NEC for exams from September 1, 2026 | Use theory, supplied-data calculations, and the separate 2026 section; the bank is not a complete Texas outline simulation. | PSI Texas electrician bulletin, cover and Master Electrician sections |
| Minnesota Class A master | 2026 NEC in the current guide | Use theory, guide-based methods, supplied-data calculations, and the 2026 section; not a full Minnesota exam. | Minnesota DLI electrical license examination guide |
| Any other state or city exam | Named in that exam's bulletin | Use the same edition-matching process; no state-specific match is claimed. | Your licensing authority and its testing vendor |
A Texas note. TDLR’s exam page and PSI’s bulletin now identify the 2026 NEC exam reference. The bulletin states the September 1, 2026 effective date. This is an exam-reference date, not a statement about installation permits in every jurisdiction.
Studying for a 2026 exam? The 2026 edition moved load calculations from Article 220 to Article 120 (Denison’s significant-changes summary, p. 1). Question 16 supplies its own worksheet factors rather than claiming they establish the 2026 dwelling method. A 2023 table question and a 2026 changes question have different scopes; a correct answer here does not establish every requirement for a job-site design.
How three master electrician exams are set up
| Row label | ICC R16 Master Electrician | Texas Master Electrician | Minnesota Class A Master |
|---|---|---|---|
| Questions | 100, four-option multiple choice | Two portions: NEC Knowledge 75 (5 unscored) and Calculations 33 (3 unscored) | 80 |
| Time | 5 hours | 150 minutes + 170 minutes; unused time doesn't carry over | 5½ hours |
| Code book | Open book: 2023 NEC and Ugly's Electrical References (any edition) | Bring your own permitted soft-bound NEC; any soft-bound edition is allowed, but the questions reference 2026. Permanent publisher-made/provided tabs only; NEC Handbook not allowed | Soft-cover NEC supplied, without tabs; calculator supplied |
| Passing score | Consult the passing-score requirement for the specific exam in the ICC Exam Catalog | 70% on each portion | 70% |
| Fee | $120 through Pearson VUE | $78 for both portions; $78 retake | Not stated in the cited exam-format table; this does not mean the application is free |
| Retake | Wait 10 days after a failed attempt | Retake fee each attempt within your eligibility period | New application; eligible 30 days after the failure notice |
Sources: ICC bulletin — R16 entry and sections on passing requirements and retakes; PSI Texas bulletin — cover and printed pp. 8–11; Minnesota DLI guide, — reference-materials section and examination-length/scoring table.
Passing an exam doesn't license you by itself. ICC’s bulletin says plainly that it isn't a licensing agency; the city, county, or state that requires the exam decides who gets licensed and what else you need.
Five mistakes these questions are built around
- Nameplate vs. table current. The DLI guide’s ordinary-motor method distinguishes table current for conductors, switches, and short-circuit protection from nameplate current for separate overloads. Follow any stated exception, and use supplied values where a question provides them (Questions 11–15; DLI motor-calculation section).
- Next size up has limits. Use the next rating only when the applicable rule or the exercise explicitly permits it—not whenever a calculation falls between sizes (Questions 12 and 13).
- Derating factors multiply, and grounds don't count as current-carrying. Apply both the ambient correction and the conductor-count adjustment; do not replace one with the other (Questions 17 and 18; 2023 tables, p. 2).
- kW vs. kVA. Efficiency and power questions only work when you compare like with like (Questions 1–3).
- Read what the question assumes. Texas’s bulletin, pp. 8–10, tells its candidates to assume copper unless stated and not to apply code exceptions or optional calculation methods unless the question directs it. For another exam, follow that exam’s instructions rather than importing Texas’s rules.
Sources and verification
Written by the Castleport Test Prep Editorial Team. The named exam formats, published fees, reference-book policies, and code editions above were checked on September 28, 2026. Calculation results were independently recomputed; the source and scope of each question appear with its explanation.
The 2026 NEC explanations cite the Minnesota Department of Labor and Industry’s 2026 NEC FAQ and the City of Denison, Texas, 2026 NEC significant changes summary, not a review of the complete NFPA text. The labeled 2023 ampacity and box-fill questions use HELUKABEL’s 2023 table reproduction and Minnesota DLI’s 2023 inspection checklist. Supplied-data exercises state their inputs openly and do not claim those inputs apply in every edition.
Theory questions cite the U.S. Department of Energy’s Electrical Science handbooks (Volume 3 and Volume 4, June 1992) and the Minnesota examination guide. The older DOE publications support enduring electrical theory, not current installation-code requirements. The question scenarios and exercise values are independently authored. Source checking isn't a licensed professional review.
AI-assisted tools were used in developing and auditing this resource; the cited sources, not AI output, support its factual explanations. See our editorial methodology.
Castleport Test Prep is an independent exam prep publisher. We are not affiliated with, endorsed by, or approved by the International Code Council, PSI, the Texas Department of Licensing and Regulation, the Minnesota Department of Labor and Industry, or the National Fire Protection Association. Exam, credential, and code names are used to identify their subjects; trademarks belong to their respective owners.