Electrician Math Practice Test: 30 Questions + Math Warm-Up

This electrician math practice test has 30 original electrical calculations, each with a full worked answer, plus a 10-question no-calculator warm-up for apprenticeship applicants. This is unofficial practice using the assumptions in each question—not a full licensing exam or an installation-design guide.

A. Formulas and theory

A1 · Ohm's law
A 120-V heating element has a resistance of 16 Ω. What current does it draw?

A. 0.13 A
B. 136 A
C. 1,920 A
D. 7.5 A

Show answer and worked solution

Answer: D — 7.5 A
Work it: I = E ÷ R = 120 ÷ 16 = 7.5 A.
Why: You want current, so divide voltage by resistance. Check it: 7.5 A × 16 Ω = 120 V.
The other choices: A — Divides resistance by voltage — the formula flipped; B — Adds volts and ohms, which are different units; C — Multiplies voltage by resistance instead of dividing.
Reference: OpenStax College Physics 2e §20.2

Choose an answer before opening its solution. Applying for an apprenticeship? Go to the 10-question no-calculator warm-up.

A2 · Power and resistance
What is the resistance of a 1,500-W, 120-V heater at its rated voltage?

A. 0.08 Ω
B. 12.5 Ω
C. 9.6 Ω
D. 96 Ω

Show answer and worked solution

Answer: C — 9.6 Ω
Work it: R = E² ÷ P = 120² ÷ 1,500 = 14,400 ÷ 1,500 = 9.6 Ω.
Why: With voltage and power given, E² ÷ P goes straight to resistance. Cross-check: I = 1,500 ÷ 120 = 12.5 A, and 120 ÷ 12.5 = 9.6 Ω.
The other choices: A — Divides voltage by power (120 ÷ 1,500); B — That's the current in amperes (P ÷ E), not the resistance; D — Drops a zero from the wattage (14,400 ÷ 150).
Reference: OpenStax College Physics 2e §20.4

A3 · Series circuit
Resistors of 10 Ω, 20 Ω, and 30 Ω are connected in series across 120 V. What is the voltage across the 30-Ω resistor?

A. 60 V
B. 20 V
C. 40 V
D. 120 V

Show answer and worked solution

Answer: A — 60 V
Work it: R total = 10 + 20 + 30 = 60 Ω. I = 120 ÷ 60 = 2 A. Voltage across 30 Ω = 2 × 30 = 60 V.
Why: In series, the same current flows through every resistor, so the biggest resistor takes the biggest share of the voltage. The three drops (20 + 40 + 60) add back to 120 V.
The other choices: B — That's the drop across the 10-Ω resistor; C — That's the drop across the 20-Ω resistor; D — Full supply voltage across one resistor happens in a parallel circuit, not a series one.
Reference: OpenStax College Physics 2e §21.1

A4 · Series-parallel circuit
A 4-Ω resistor is in series with a parallel pair made of a 12-Ω and a 6-Ω resistor. A 64-V DC source feeds the whole network. What current does the source supply?

A. 2.9 A
B. 8 A
C. 4 A
D. 16 A

Show answer and worked solution

Answer: B — 8 A
Work it: Parallel pair = (12 × 6) ÷ (12 + 6) = 72 ÷ 18 = 4 Ω. Total = 4 + 4 = 8 Ω. I = 64 ÷ 8 = 8 A.
Why: Reduce the parallel pair to one equivalent resistance first, then add the series resistor.
The other choices: A — Treats all three resistors as series (64 ÷ 22); C — Ignores the 6-Ω branch (64 ÷ 16); D — Forgets the separate 4-Ω series resistor (64 ÷ 4).
Reference: OpenStax College Physics 2e §21.1

A5 · Three-phase current from kVA
A balanced 45-kVA load is supplied at 208 V line-to-line, three-phase. What is the line current? Round to the nearest 0.1 A.

A. 72.1 A
B. 216.3 A
C. 124.9 A
D. 374.7 A

Show answer and worked solution

Answer: C — 124.9 A
Work it: I = VA ÷ (E × √3) = 45,000 ÷ (208 × √3) ≈ 124.9 A.
Why: Three-phase current uses √3 (1.732) times the line-to-line voltage in the denominator.
The other choices: A — Divides by 3 instead of √3; B — Uses the single-phase formula (no √3); D — Multiplies by √3 instead of dividing.
Reference: DOE Fundamentals Handbook, Electrical Science Vol. 3, Module 9 (Basic AC Power), pp. 20–21

A6 · Three-phase current from kW and power factor
A balanced three-phase load uses 18 kW of real power at 208 V line-to-line and a 0.90 power factor. What is the line current? Round to the nearest 0.1 A.

A. 55.5 A
B. 32.1 A
C. 50.0 A
D. 96.2 A

Show answer and worked solution

Answer: A — 55.5 A
Work it: I = W ÷ (√3 × E × PF) = 18,000 ÷ (√3 × 208 × 0.90) ≈ 55.5 A.
Why: Watts are real power, so the power factor has to go in the denominator. A lower power factor means more current for the same kilowatts.
The other choices: B — Uses 3 instead of √3; C — Leaves out the power factor; D — Leaves out √3.
Reference: DOE Fundamentals Handbook, Electrical Science Vol. 3, Module 9 (Basic AC Power), pp. 20–21

A7 · Wye voltages
A balanced wye-connected system has 480 V line-to-line. What is the line-to-neutral voltage? Round to the nearest 0.1 V.

A. 277.1 V
B. 160.0 V
C. 480.0 V
D. 831.4 V

Show answer and worked solution

Answer: A — 277.1 V
Work it: V line-to-neutral = V line-to-line ÷ √3 = 480 ÷ √3 ≈ 277.1 V.
Why: In a wye system, line-to-neutral voltage is line-to-line voltage divided by √3 — which is why a 480Y/277-V system is labeled that way.
The other choices: B — Divides by 3 instead of √3; C — Makes no conversion; D — Multiplies by √3 instead of dividing.
Reference: DOE Fundamentals Handbook, Electrical Science Vol. 3, Module 9 (Basic AC Power), pp. 20–21

A8 · Transformer turns ratio
An ideal transformer has 600 primary turns and 150 secondary turns. The primary is supplied at 240 V AC. What is the secondary voltage?

A. 4 V
B. 90 V
C. 960 V
D. 60 V

Show answer and worked solution

Answer: D — 60 V
Work it: Vs = Vp × (Ns ÷ Np) = 240 × (150 ÷ 600) = 240 × 0.25 = 60 V.
Why: The secondary has one-quarter of the primary's turns, so an ideal transformer gives one-quarter of the voltage (a step-down).
The other choices: A — That's the turns ratio (600 ÷ 150) labeled as volts; B — Subtracts turns from volts; C — Uses the turns ratio upside down (a step-up).
Reference: OpenStax College Physics 2e §23.7, eq. 23.26

A9 · Efficiency
A motor delivers 10 hp of output while drawing 8,800 W of input. What is its efficiency? Use 1 hp ≈ 746 W and round to the nearest 0.1%.

A. 0.85%
B. 15.2%
C. 84.8%
D. 118.0%

Show answer and worked solution

Answer: C — 84.8%
Work it: Output ≈ 10 × 746 = 7,460 W. Efficiency = output ÷ input × 100 ≈ 7,460 ÷ 8,800 × 100 ≈ 84.8%.
Why: Convert horsepower to watts so both numbers use the same unit, then divide output by input.
The other choices: A — Writes the decimal 0.848 as a percent without multiplying by 100; B — That's the share of input not converted to shaft output ((8,800 − 7,460) ÷ 8,800), not the efficiency; D — Divides input by output. Efficiency can't be over 100%.
Reference: OpenStax College Physics 2e §7.6 (efficiency)

A10 · Voltage-drop percentage
A 240-V circuit measures 232.8 V at the load. What is the voltage drop as a percentage of the source voltage?

A. 3.1%
B. 7.2%
C. 97.0%
D. 3.0%

Show answer and worked solution

Answer: D — 3.0%
Work it: Drop = 240 − 232.8 = 7.2 V. Percent = 7.2 ÷ 240 × 100 = 3.0%.
Why: Percent voltage drop is measured against the source (supply) voltage.
The other choices: A — Divides by the load voltage (232.8) instead of the source voltage; B — That's the drop in volts, not percent; C — That's the percentage of voltage remaining at the load, not the drop.
Reference: Original calculation from the stated source and load voltages.

B. Branch-circuit and load calculations

Use the load factors stated in these exercises; the answers are calculation results, not complete service or circuit designs.

B1 · Continuous load — overcurrent device
A branch circuit supplies a 36-A continuous load and no noncontinuous load. For this exercise, size the overcurrent device at no less than 125% of that load and choose from 40-A, 45-A, or 50-A ratings. What is the smallest rating that meets this calculation?

A. 45 A
B. 36 A
C. 40 A
D. 50 A

Show answer and worked solution

Answer: A — 45 A
Work it: 36 × 1.25 = 45 A. 45 A is one of the supplied ratings.
Why: Apply the stated 125% factor, then select the smallest supplied rating at or above the result. This calculation alone does not check equipment or conductor protection limits.
The other choices: B — Forgets the 125% for continuous load; C — Rounds down below the required 45 A; D — Exceeds the calculated minimum; it isn't the smallest supplied rating that meets it.
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.

B2 · Receptacle load (non-dwelling)
In a commercial-building calculation exercise, how many general-purpose receptacle outlets (one mounting strap each) fit within a 15-A, 120-V branch-circuit load allowance, using 180 VA per strap? Assume no other loads and no continuous-load factor.

A. 8
B. 13
C. 15
D. 10

Show answer and worked solution

Answer: D — 10
Work it: Circuit capacity = 15 × 120 = 1,800 VA. 1,800 ÷ 180 = 10 receptacles.
Why: Use the 180-VA allowance supplied for each strap. Divide the circuit allowance by the allowance per strap; do not round a maximum count up.
The other choices: A — Applies an 80% limit (1,440 VA ÷ 180) that this counting method doesn't use; B — That's the answer for a 20-A circuit (2,400 ÷ 180 = 13.3, so 13); C — Divides 1,800 by 120 (volts) instead of 180 (VA per receptacle).
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.

B3 · Dwelling general lighting branch circuits
A 2,400 ft² dwelling needs how many 15-A, 120-V general lighting branch circuits at minimum in this calculation? Use 3 VA/ft² and 1,800 VA per circuit; do not include separately supplied appliance or laundry loads.

A. 3
B. 5
C. 7
D. 4

Show answer and worked solution

Answer: D — 4
Work it: 2,400 ft² × 3 VA/ft² = 7,200 VA. One 15-A, 120-V circuit = 1,800 VA. 7,200 ÷ 1,800 = 4 circuits.
Why: The branch-circuit count uses the stated 3 VA per square foot. Do not substitute the 2 VA/ft² used in the separate demand exercise below.
The other choices: A — Uses 2 VA/ft² instead of the stated 3 VA/ft² (4,800 ÷ 1,800 = 2.67, rounded up to 3); B — Loads each circuit to only 80% (1,440 VA); that limit isn't part of this count (7,200 ÷ 1,440 = 5); C — Adds 4,500 VA of appliance and laundry loads excluded from this question ((7,200 + 4,500) ÷ 1,800 = 6.5, rounded up).
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.; City of Austin, 2023-to-2026 NEC comparison, “Load Calculation,” PDF p. 2

B4 · Dwelling lighting demand — supplied factors
A 2,400 ft² dwelling has two small-appliance circuits and one laundry circuit. For this demand exercise, use 2 VA/ft² for general lighting, 1,500 VA for each small-appliance circuit, and 1,500 VA for the laundry circuit. Combine those loads, then take the first 3,000 VA at 100% and all the remainder at 35%. What is the resulting demand load?

A. 3,255 VA
B. 5,205 VA
C. 6,045 VA
D. 9,300 VA

Show answer and worked solution

Answer: B — 5,205 VA
Work it: General lighting = 2,400 × 2 = 4,800 VA. Add small-appliance (2 × 1,500 = 3,000) and laundry (1,500): 9,300 VA. Demand = 3,000 + (6,300 × 0.35) = 3,000 + 2,205 = 5,205 VA.
Why: The small-appliance and laundry loads are added to general lighting before the supplied demand factors are applied. This subtotal is not a complete dwelling service calculation.
The other choices: A — Applies 35% to the whole 9,300 VA; C — Uses 3 VA/ft² instead of the supplied 2 VA/ft² (7,200 + 4,500 = 11,700 VA; 3,000 + 8,700 × 0.35 = 6,045 VA). Right method, wrong input for this question; D — Skips the demand factor.
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.

B5 · Continuous plus noncontinuous load
A feeder supplies 40 A of continuous load and 20 A of noncontinuous load. Using 125% of the continuous load plus 100% of the noncontinuous load, what minimum conductor ampacity does this calculation produce before any correction or adjustment?

A. 60 A
B. 75 A
C. 87.5 A
D. 70 A

Show answer and worked solution

Answer: D — 70 A
Work it: (40 × 1.25) + 20 = 50 + 20 = 70 A.
Why: Only the continuous part gets the 125% factor; the noncontinuous part is taken at 100%.
The other choices: A — Leaves out the 125% for the continuous load; B — Applies 125% to the whole 60 A; C — Stacks an extra 125% on top of 70 A.
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.

B6 · Dryer load
A household electric dryer is rated 4,500 W at 240 V. For this load exercise, use the larger of 5,000 W and the nameplate rating, assume unity power factor, and add no further multiplier. What load does that represent in amperes? Round to the nearest 0.1 A.

A. 20.8 A
B. 18.8 A
C. 23.4 A
D. 26.0 A

Show answer and worked solution

Answer: A — 20.8 A
Work it: Use 5,000 W or the nameplate rating, whichever is larger: 5,000 W. 5,000 ÷ 240 ≈ 20.8 A.
Why: The supplied minimum is larger than the nameplate value. At the stated unity power factor, divide that wattage by voltage to find current.
The other choices: B — Uses the smaller 4,500-W nameplate — the classic trap; C — Adds 25% to the 4,500-W nameplate (5,625 ÷ 240); D — Adds 25% to 5,000 W (6,250 ÷ 240); this exercise specifies no extra multiplier.
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.; OpenStax College Physics 2e §20.4

B7 · Range demand
A single 12-kW household range operates at 240 V. For this demand exercise, use an 8-kW demand allowance and unity power factor. What is its demand load in amperes? Round to the nearest 0.1 A.

A. 25.0 A
B. 50.0 A
C. 66.7 A
D. 33.3 A

Show answer and worked solution

Answer: D — 33.3 A
Work it: Demand allowance = 8 kW. 8,000 ÷ 240 ≈ 33.3 A.
Why: Use the given 8-kW demand allowance, not the full 12-kW nameplate. Divide by the stated 240 V.
The other choices: A — Uses half the 12-kW nameplate (6,000 ÷ 240 = 25.0 A), not the supplied 8-kW demand; B — Uses the full 12-kW nameplate; C — Divides 8,000 VA by 120 V instead of 240 V.
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.; OpenStax College Physics 2e §20.4

C. Voltage drop and ampacity

C1 · Single-phase voltage drop
A 240-V, two-wire, single-phase, 30-A resistive load is 150 ft (one way) from the panel, fed with 10 AWG stranded copper. Use a conductor resistance of 1.24 Ω per 1,000 ft and neglect reactance. What is the approximate voltage drop? Round to the nearest 0.1 V.

A. 11.2 V
B. 5.6 V
C. 9.7 V
D. 22.3 V

Show answer and worked solution

Answer: A — 11.2 V
Work it: VD = 2 × L × R × I ÷ 1,000 = 2 × 150 × 1.24 × 30 ÷ 1,000 = 11.16 V (about 4.7% of 240 V).
Why: Current travels out on one conductor and back on the other, so single-phase voltage drop uses 2 × the one-way length.
The other choices: B — Forgets the return conductor (no 2); C — Uses the three-phase multiplier (1.732) on a single-phase circuit; D — Doubles the answer again.
Reference: OpenStax College Physics 2e §20.2; conductor resistance is a supplied input

C2 · Three-phase voltage drop
A balanced 208-V line-to-line, three-phase load draws 40 A per line and is 200 ft (one way) from the panel, fed with 8 AWG stranded copper. Use 0.778 Ω per 1,000 ft, unity power factor, and negligible reactance. What is the approximate line-to-line voltage drop? Round to the nearest 0.1 V.

A. 6.2 V
B. 12.4 V
C. 10.8 V
D. 32.3 V

Show answer and worked solution

Answer: C — 10.8 V
Work it: VD = √3 × L × R × I ÷ 1,000 = √3 × 200 × 0.778 × 40 ÷ 1,000 ≈ 10.8 V (about 5.2% of 208 V).
Why: For this balanced, resistive three-phase model, √3 replaces the 2 used for a two-wire single-phase circuit. L is still the one-way length.
The other choices: A — Leaves out the multiplier entirely; B — Uses the single-phase 2 instead of 1.732; D — Multiplies the length by 3. Use the one-way length once.
Reference: OpenStax College Physics 2e §20.2; DOE Electrical Science Vol. 3, Module 9, pp. 20–21; balanced three-phase relationships

C3 · Sizing a conductor for voltage drop
A 120-V, 16-A resistive load on a two-wire single-phase circuit is 100 ft (one way) away. Neglect reactance and use K = 12.9 Ω·cmil/ft for this copper-conductor model. Which listed conductor is the smallest that keeps voltage drop at or below the stated 3% target? Supplied circular-mil areas: 12 AWG = 6,530; 10 AWG = 10,380; 8 AWG = 16,510; 6 AWG = 26,240.

A. 12 AWG
B. 8 AWG
C. 10 AWG
D. 6 AWG

Show answer and worked solution

Answer: B — 8 AWG
Work it: Allowed drop = 0.03 × 120 = 3.6 V. CM = 2 × K × I × L ÷ VD = 2 × 12.9 × 16 × 100 ÷ 3.6 ≈ 11,467 cmil. 10 AWG (10,380) is too small, so the answer is 8 AWG (16,510).
Why: Pick the first size whose area is at least the calculated circular mils — never round down to the closest one. This checks the stated voltage-drop target only, not ampacity, protection, or complete installation compliance.
The other choices: A — Far below 11,467 cmil; C — Closest in size, but 10,380 is less than the 11,467 required; D — Works, but it's larger than the smallest size that meets 3%.
Reference: OpenStax College Physics 2e §20.3, resistance proportional to length divided by area; K and conductor areas are supplied inputs

C4 · Ambient temperature correction
Three current-carrying 10 AWG THHN copper conductors are in a raceway in a dry location at 43 °C ambient. Terminations are rated 75 °C. Use a 90 °C base ampacity of 40 A, a temperature factor of 0.87, and a 75 °C termination limit of 35 A. What corrected ampacity results after applying the factor and checking the termination limit, before separate overcurrent-protection limits?

A. 28.7 A
B. 35 A
C. 34.8 A
D. 40 A

Show answer and worked solution

Answer: C — 34.8 A
Work it: THHN is 90 °C insulation: Table 310.16 gives 40 A. Correction for 41–45 °C, 90 °C column = 0.87. 40 × 0.87 = 34.8 A. Termination check: 75 °C column = 35 A. The lower value, 34.8 A, governs.
Why: Apply the correction to the supplied 90 °C base, then keep the result at or below the termination limit. This is not permission to install a 34.8-A breaker: the cited 2023 table's small-conductor note limits 10 AWG copper protection to 30 A unless another code provision permits otherwise.
The other choices: A — Applies the 75 °C correction factor (0.82) to 35 A, rather than using the supplied 90 °C base and factor; B — Uses the termination limit without correcting for heat; D — Skips temperature correction.
Reference: HELUKABEL, NEC 2023 ampacity tables, PDF pp. 1–2

C5 · More than three current-carrying conductors
Six current-carrying 12 AWG THHN copper conductors share a raceway longer than 24 in. in a dry location at 30 °C ambient. Terminations are rated 75 °C. Use a 90 °C base ampacity of 30 A, an 80% conductor-count adjustment, and a 75 °C termination limit of 25 A. What adjusted ampacity results before separate overcurrent-protection limits?

A. 20 A
B. 24 A
C. 21 A
D. 25 A

Show answer and worked solution

Answer: B — 24 A
Work it: 90 °C column = 30 A. Adjustment for 4–6 conductors = 80%. 30 × 0.80 = 24 A. That's below the 75 °C limit of 25 A, so 24 A.
Why: Multiply the supplied base ampacity by 80%, then check the termination limit. This is not permission to install a 24-A breaker: the cited 2023 table's small-conductor note limits 12 AWG copper protection to 20 A unless another code provision permits otherwise.
The other choices: A — Uses the 75 °C value (25 A) × 80%; C — Uses the 70% factor for 7–9 conductors; D — Skips the adjustment.
Reference: HELUKABEL, NEC 2023 ampacity tables, PDF pp. 1–2

C6 · Correction and adjustment together
Nine current-carrying 8 AWG THHN copper conductors share a raceway longer than 24 in. in a dry location at 38 °C ambient. Terminations are rated 75 °C. Use a 90 °C base ampacity of 55 A, a temperature factor of 0.91, a 70% conductor-count adjustment, and a 75 °C termination limit of 50 A. What corrected and adjusted ampacity results? Round to the nearest 0.1 A.

A. 31.9 A
B. 35.0 A
C. 38.5 A
D. 50.0 A

Show answer and worked solution

Answer: B — 35.0 A
Work it: 90 °C column = 55 A. Temperature correction (36–40 °C, 90 °C column) = 0.91. Adjustment (7–9 conductors) = 70%. 55 × 0.91 × 0.70 = 35.035 A, or 35.0 A to the nearest 0.1 A. Below the 75 °C limit of 50 A.
Why: When both conditions apply, multiply by both factors.
The other choices: A — Starts from the 75 °C value (50 A) instead of the 90 °C insulation rating; C — Skips the temperature correction; D — That's the 75 °C termination limit with no derating.
Reference: HELUKABEL, NEC 2023 ampacity tables, PDF pp. 1–2

D. Motors, boxes and raceways

D1 · Motor branch-circuit conductors
A 10-hp, 460-V, three-phase, continuous-duty motor has a supplied table full-load current of 14 A. Apply the basic 125% conductor-sizing factor, with no special motor exception. What is the minimum branch-circuit conductor ampacity before correction, adjustment, and termination checks?

A. 17.5 A
B. 14.0 A
C. 21.0 A
D. 35.0 A

Show answer and worked solution

Answer: A — 17.5 A
Work it: 14 × 1.25 = 17.5 A.
Why: For the basic motor-conductor calculation specified here, use the supplied table full-load current, not a nameplate current, at 125%.
The other choices: B — Leaves out the 125%; C — Uses 150%; D — Uses the 250% breaker factor supplied in D2, not the 125% conductor factor in this question.
Reference: Minnesota DLI, Electrical license examination guide, “Motor branch circuit and feeder calculations”; 14 A is a supplied input

D2 · Motor short-circuit protection
A 3-hp, 230-V, single-phase motor has a supplied full-load current of 17 A. For this inverse-time-breaker calculation, multiply by 250%. If that result is between the supplied standard ratings of 40 A, 45 A, and 50 A, the exercise permits the next higher rating, but no additional increase. What is that rating?

A. 40 A
B. 45 A
C. 42.5 A
D. 50 A

Show answer and worked solution

Answer: B — 45 A
Work it: 17 × 2.50 = 42.5 A. Next higher standard size = 45 A.
Why: The stated calculation permits the next higher supplied rating when the percentage result falls between ratings. Do not carry this permission into a different application without its governing rule.
The other choices: A — Rounds down rather than following the stated next-higher-rating instruction; C — That's the calculated value, not one of the supplied standard ratings; D — Skips past the next standard size.
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.

D3 · Motor overload protection
A continuous-duty motor nameplate shows a full-load current of 15.6 A and a service factor of 1.15. For this overload exercise, use a maximum rating of 125% of the nameplate current. What is that rating?

A. 15.6 A
B. 17.9 A
C. 19.5 A
D. 21.8 A

Show answer and worked solution

Answer: C — 19.5 A
Work it: 15.6 × 1.25 = 19.5 A.
Why: This question specifies the nameplate current and a 125% multiplier. The 1.15 service factor is not the multiplier for this calculation.
The other choices: A — Leaves out the percentage; B — Uses 115% instead of the specified 125%; D — Uses 140%.
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.

D4 · Box fill
A 4-in. square metal box, 2-1/8 in. deep and marked 30.3 in³, holds two 12/2 NM cables with ground, one 12/3 NM cable with ground, one receptacle on a single yoke wired with 12 AWG, and internal cable clamps. Exactly three 12 AWG equipment grounding conductors enter the box; any pigtails remain entirely inside, and there are no other conductors or fittings. Using 2.25 in³ per 12 AWG allowance, what is the total required box-fill volume?

A. 20.25 in³
B. 24.75 in³
C. 22.5 in³
D. 29.25 in³

Show answer and worked solution

Answer: B — 24.75 in³
Work it: Insulated conductors: 2 + 2 + 3 = 7; their volume is 7 × 2.25 = 15.75 in³. Receptacle: 2 × 2.25 = 4.50 in³. Clamps: 1 × 2.25 = 2.25 in³. Grounds (three, so one allowance): 2.25 in³. Total = 24.75 in³ — it fits in 30.3 in³.
Why: Each counted item uses the 12 AWG allowance of 2.25 in³. This single receptacle yoke counts twice, the internal clamps together count once, and the three entering equipment grounding conductors together count once. This checks fill volume, not every installation requirement.
The other choices: A — Leaves out the receptacle; C — Counts the receptacle once instead of twice; D — Counts each ground wire separately.
Reference: Minnesota DLI, Electrical license examination guide, “Box fill calculations”; NEC 314.16(B) and Table 314.16(B)(1)

D5 · Raceway fill
Considering fill only, which is the smallest EMT (electrical metallic tubing) that can hold five 10 AWG THHN and three 12 AWG THHN conductors in a run longer than 24 in.? These eight include every conductor in the run. Supplied conductor areas: 10 AWG = 0.0211 in², 12 AWG = 0.0133 in². Supplied EMT 40% fill allowances: 1/2 in. = 0.122 in², 3/4 in. = 0.213 in², 1 in. = 0.346 in².

A. 1/2 in.
B. 1 in.
C. 1-1/4 in.
D. 3/4 in.

Show answer and worked solution

Answer: D — 3/4 in.
Work it: (5 × 0.0211) + (3 × 0.0133) = 0.1055 + 0.0399 = 0.1454 in². 1/2 in. (0.122) is too small; 3/4 in. (0.213) holds it.
Why: For this ordinary run with eight conductors, use the 40% fill allowances. Add every conductor's area, then find the first raceway size at or above the total. Ampacity and other installation limits are separate checks.
The other choices: A — 0.122 in² is less than 0.1454 in²; B — Works, but it's larger than necessary; C — Larger than necessary.
Reference: Minnesota DLI, Electrical license examination guide, “Conduit fill calculations”; areas supplied in the question

D6 · Straight-pull box
A pull box has 2-in. raceways entering opposite walls for a straight pull of 4 AWG and larger conductors. For this exercise, the distance between those walls must be at least 8 times the largest raceway trade size. What is that minimum distance?

A. 8 in.
B. 12 in.
C. 16 in.
D. 24 in.

Show answer and worked solution

Answer: C — 16 in.
Work it: 8 × 2 in. = 16 in.
Why: Apply the supplied 8-times rule to the largest raceway trade size. The question asks only for the dimension along this straight pull.
The other choices: A — Uses 4×; B — Uses the 6× angle-pull multiplier; D — Uses 12×.
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.

D7 · Angle-pull box
For an angle pull with 4 AWG and larger conductors, one wall has 2-in., 1-1/2-in., and 1-in. raceways in the same row. Use this exercise rule for the distance from those raceway entries to the opposite wall: 6 times the largest trade size plus the sum of the other trade sizes in that row. What is the minimum distance?

A. 12 in.
B. 14.5 in.
C. 16 in.
D. 27 in.

Show answer and worked solution

Answer: B — 14.5 in.
Work it: (6 × 2) + 1.5 + 1 = 14.5 in.
Why: Apply the supplied rule: 6 × the largest raceway plus the sum of the other raceways in the same row on that wall. This one dimension does not determine every box size or conductor-bending requirement.
The other choices: A — Forgets to add the other raceways; C — Uses the 8× straight-pull rule; D — Multiplies every raceway by 6.
Reference: Original calculation exercise using the factors and limits supplied in the question; see calculation scope.

Review your results

Count your correct answers in each section, and count any you skipped separately.

Review your results
Section	Questions	Your correct	Your skipped
A. Formulas and theory	10	___	___
B. Branch-circuit and load calculations	7	___	___
C. Voltage drop and ampacity	6	___	___
D. Motors, boxes and raceways	7	___	___
Total	30	___	___

Your total describes how you did on these 30 questions. It isn't a scaled score, and it can't tell you whether you'll pass a real exam. Six or seven questions per section is a small sample, so treat a low section score as a hint about where to look, not a verdict.

For a percentage, divide your number correct by 30 and multiply by 100. Keep skipped questions in that denominator. For example, 23 correct, 6 wrong, and 1 skipped is 23 ÷ 30 × 100 ≈ 76.7% of this set correct.

When you miss one, find the first line where your work differs from the worked answer. Check these steps:

The 25%. Apply 125% only where the question calls for it. D1 uses the supplied table current; D3 uses the nameplate current.
The multiplier. The two-wire voltage-drop model here uses 2; the balanced three-phase model uses √3 (about 1.732). Use the one-way length once.
The column. Use the stated insulation-column base and correction factors, then check the termination limit. Adjusted ampacity is not automatically the permitted breaker rating.
The rounding. For minimum-size questions, choose the smallest listed size that meets the result, not the nearest size below it. D2's next-rating instruction belongs to that exercise.
The input. A demand calculation done with the wrong VA/ft² can land on a wrong answer that's sitting right there as a choice (see B4).

Rework each miss the next day without looking at the solution. Then do a few more problems of the same type from your own study material with the code book open when the question requires a lookup.

Apprenticeship applicant? No-calculator math warm-up

The NECA-IBEW Electrical Training Center describes its Algebra and Functions section as 33 questions in 46 minutes, with no calculator (training center's aptitude-test instructions). That is the format published by this center, not a rule for every apprenticeship program; follow the instructions from the program you're applying to.

The 10 questions below are original math warm-ups, not a full aptitude test or a reading-comprehension section. Work them by hand. For official sample questions, see the electrical training ALLIANCE sample page.

E1 · Linear equation
Solve 3x − 7 = 20.

A. 4.3
B. 13
C. 9
D. 27

Show answer and worked solution

Answer: C — 9
Work it: 3x = 27, so x = 9.
Why: Undo the subtraction first (add 7 to both sides), then divide by 3.
The other choices: A — Subtracts 7 instead of adding (13 ÷ 3); B — Subtracts 7 from 20 and stops; D — Stops before dividing by 3.

E2 · Function evaluation
If f(x) = 2x² − 3, what is f(−2)?

A. −11
B. 11
C. 5
D. 13

Show answer and worked solution

Answer: C — 5
Work it: (−2)² = 4. 2 × 4 − 3 = 5.
Why: Square the input first — a negative number squared is positive.
The other choices: A — Treats (−2)² as −4; B — Adds 3 instead of subtracting (8 + 3); D — Squares 2x instead of x: (2 × −2)² − 3 = 13.

E3 · Fractions
3/8 + 5/12 = ?

A. 19/24
B. 1/24
C. 15/96
D. 8/20

Show answer and worked solution

Answer: A — 19/24
Work it: Common denominator 24: 9/24 + 10/24 = 19/24.
Why: Convert both fractions to the same denominator before adding the numerators.
The other choices: B — Subtracts instead of adding (10/24 − 9/24); C — Multiplies instead of adding; D — Adds tops and bottoms separately.

E4 · Percent
A 250-ft spool of wire is 38% used. How many feet remain?

A. 155 ft
B. 62 ft
C. 95 ft
D. 212 ft

Show answer and worked solution

Answer: A — 155 ft
Work it: Remaining = 62% of 250 = 0.62 × 250 = 155 ft.
Why: If 38% is used, 100% − 38% = 62% is left.
The other choices: B — Reports the percentage left (62%) as feet; C — That's the amount used; D — Subtracts 38 ft instead of 38%.

E5 · Rates and proportion
Five apprentices pull 1,200 ft of wire in 3 hours. At the same rate per person, how many feet do 8 apprentices pull in 4 hours?

A. 1,600 ft
B. 2,560 ft
C. 1,920 ft
D. 3,200 ft

Show answer and worked solution

Answer: B — 2,560 ft
Work it: Rate = 1,200 ÷ (5 × 3) = 80 ft per apprentice-hour. 80 × 8 × 4 = 2,560 ft.
Why: Find the rate for one person for one hour, then scale up by people and hours.
The other choices: A — Scales for more hours but not more people; C — Scales for more people but not more hours; D — Multiplies by the people but divides by the hours (1,200 × 8 ÷ 3).

E6 · Number pattern
What comes next: 3, 7, 15, 31, 63, …? The same rule—multiply by a fixed number, then add a fixed number—generates each new term.

A. 95
B. 126
C. 128
D. 127

Show answer and worked solution

Answer: D — 127
Work it: Each term = previous × 2 + 1. 63 × 2 + 1 = 127.
Why: The gaps (4, 8, 16, 32) double, so the next gap is 64: 63 + 64 = 127.
The other choices: A — Adds 32 again instead of 64; B — Doubles but forgets the +1; C — 128 is 2⁷. Every term here is one less than a power of 2, so the next is 2⁷ − 1.

E7 · Exponents
(2³)² × 2⁻⁴ = ?

A. 1/4
B. 2
C. 4
D. 64

Show answer and worked solution

Answer: C — 4
Work it: (2³)² = 2⁶. 2⁶ × 2⁻⁴ = 2⁶⁻⁴ = 2² = 4.
Why: A power of a power multiplies exponents; multiplying same-base powers adds exponents.
The other choices: A — Gets the sign of the final exponent wrong (2⁻²); B — Adds 3 + 2 for the first step (2⁵ × 2⁻⁴ = 2); D — Leaves out the 2⁻⁴ factor (2⁶ = 64).

E8 · Rearranging a formula
P = I²R. If P = 360 and I = 6, what is R?

A. 0.1
B. 10
C. 60
D. 2,160

Show answer and worked solution

Answer: B — 10
Work it: R = P ÷ I² = 360 ÷ 36 = 10.
Why: Isolate R by dividing both sides by I².
The other choices: A — Divides the wrong way (36 ÷ 360); C — Divides by I instead of I²; D — Multiplies P by I.

E9 · Inequality
Solve 4 − 2x > 10.

A. x > −3
B. x > 3
C. x < 3
D. x < −3

Show answer and worked solution

Answer: D — x < −3
Work it: −2x > 6. Divide by −2 and flip the sign: x < −3.
Why: Dividing or multiplying an inequality by a negative number reverses its direction.
The other choices: A — Forgets to flip the inequality sign; B — Loses the negative and forgets to flip; C — Loses the negative.

E10 · Quadratic equation
What is the positive solution of x² − 5x − 14 = 0?

A. 2
B. 7
C. 5
D. 14

Show answer and worked solution

Answer: B — 7
Work it: (x − 7)(x + 2) = 0, so x = 7 or x = −2. The positive solution is 7.
Why: Find two numbers that multiply to −14 and add to −5: −7 and +2.
The other choices: A — Drops the sign from the negative root (−2); C — That's the magnitude of the coefficient of x (−5), not a root; D — That's the magnitude of the constant term (−14), not a root.

Count your correct answers out of 10. As with the main set, that's feedback on these questions only.

Formulas and calculation inputs

E or V means voltage in volts; I means current in amperes; R means resistance in ohms (Ω); P means real power in watts. VA means volt-amperes, kVA means 1,000 VA, and PF means power factor. Use line-to-line voltage and line current for the balanced three-phase formulas. In the voltage-drop rows, L is one-way length in feet and R per 1,000 ft is the supplied conductor resistance.

Formulas and calculation inputs
Calculation	Formula or method	Scope here
Current, voltage, resistance	I = E ÷ R · E = I × R · R = E ÷ I	Resistive circuits in A1–A4
Power	P = E × I · P = I² × R · P = E² ÷ R	DC or a resistive AC load at unity power factor; AC values are root-mean-square (RMS) values
Series / parallel resistance	Series: R₁ + R₂ + … · Two in parallel: (R₁ × R₂) ÷ (R₁ + R₂)	A3–A4
Three-phase current	I = VA ÷ (E × √3) · I = W ÷ (E × √3 × PF)	Balanced three-phase loads; √3 ≈ 1.732
Wye voltages	Line-to-neutral = line-to-line ÷ √3	Balanced wye system
Transformer (ideal)	Vs ÷ Vp = Ns ÷ Np	AC transformer; p = primary, s = secondary, N = number of turns
Efficiency	Output ÷ input × 100%	Same power units; A9 supplies 1 hp ≈ 746 W
Continuous-load exercise	Continuous load × 125%, plus noncontinuous load × 100%	Factors expressly supplied in B1 and B5
Dwelling lighting load	Floor area × supplied VA/ft²	B3 counts circuits; B4 combines loads and applies its supplied demand tiers
Receptacle exercise	Circuit VA ÷ supplied VA per strap	B2 supplies 180 VA per strap and the loading assumptions
Dryer / range exercises	Selected demand watts ÷ voltage	B6 supplies a 5,000-W minimum; B7 supplies an 8-kW demand; both assume unity power factor
Voltage drop	Two-wire: 2 × L × R × I ÷ 1,000 · Balanced three-phase: √3 × L × R × I ÷ 1,000	Simplified resistive model with negligible reactance
Conductor area for voltage drop	CM = 2 × K × I × L ÷ VD	Two-wire model in C3; K = 12.9 Ω·cmil/ft is supplied, CM is area in circular mils, and VD is the allowed drop in volts
Ampacity	Table base × temperature correction × conductor-count adjustment, then compare with the termination limit	Supplied values in C4–C6, checked against the linked 2023 tables; overcurrent protection is a separate limit
Motor exercises	D1: supplied table current × 125% · D2: supplied current × 250%, then the stated next-rating rule · D3: nameplate current × 125%	Different tasks and explicitly stated assumptions, not a universal motor-sizing rule
Box fill	Add up allowances × volume per allowance	D4 uses 2.25 in³ per 12 AWG allowance
Raceway fill	Add conductor areas; compare with the supplied fill allowances	D5 checks fill only
Pull-box exercises	Straight: 8 × largest trade size · Angle: 6 × largest + others in the row	Supplied one-dimension rules in D6–D7, not complete box designs

Theory references: Ohm's law, power, series and parallel resistance, resistance and conductor area, transformer turns, efficiency, and DOE's balanced three-phase power relationships.

What these inputs mean: A supplied factor lets you practice the calculation without an off-page lookup. It does not establish that the same factor or simplified model applies to every installation. The 3% target in C3 is the target in that question, not a universal code limit. This set is not a complete 2023-to-2026 NEC crosswalk.

Edition and task still matter. For example, the City of Austin's comparison of the 2023 and 2026 NEC identifies a reduction from 3 to 2 VA/ft² for dwelling general lighting/receptacle load calculations while retaining 3 VA/ft² for the branch-circuit calculation (city comparison, “Load Calculation,” PDF p. 2). That distinction does not make B4 a complete dwelling service calculation or establish which edition governs your exam.

Which test are you preparing for?

A state or local electrician license exam. There's no single national electrician math test. Check your licensing authority's current instructions and candidate bulletin for the exact exam, NEC edition, allowed references, and calculator rules.

For example, Texas's current exam page and PSI's dated notice specify the 2026 NEC for exams from Sept. 1, 2026. PSI lists the Journeyman Electrician Calculations portion as 26 items (including 2 unscored) in 110 minutes, with 70% needed to pass, separate from the NEC Knowledge portion (PSI Texas electrician bulletin, cover and printed p. 13; TDLR exam information). Texas split the journeyman exam into those two portions starting March 11, 2025.

The bulletin's reference-material page also contains an older 2023-edition line; its effective-date notice and TDLR's current reference-material instructions both specify 2026. Do not treat that older line as the current exam-edition announcement (PSI, printed p. 8; TDLR, “Reference Materials”).

For fiscal year 2025, TDLR reported 1,301 passing results out of 6,328 on the Journeyman Calculations portion, a 20.56% pass rate (TDLR FY2025 exam statistics). Those are Texas exam results, not a count of individual people, and they don't describe any other state. This is a historical result, not a prediction for you or a benchmark for this practice set.

An apprenticeship aptitude test. The electrical training ALLIANCE's official sample covers algebra/functions and reading comprehension, rather than NEC calculations. Use the warm-up above for math practice and the official sample to see both subjects.

A class or a return to the trade. Section A is a good place to start. It needs no code book.

Common questions

Can I use a calculator? For this practice page, a calculator is useful for sections A–D; work E by hand. For an actual exam, check its candidate bulletin. Texas allows silent, non-programmable, battery-operated calculators without paper-tape printing or an alphabetic keyboard (PSI, “Security Procedures,” printed p. 4). The apprenticeship training center linked above says calculators aren't allowed for its aptitude test (center's instructions).

Do I need to memorize formulas? Use the formula table to review the electrical calculations, then solve again without looking. The algebra warm-ups explain their own steps. This set supplies its numerical factors; when practicing actual code lookups, knowing which table and edition to open is a separate skill.

Which NEC edition should I study? The one named by your licensing authority for your exam date—not automatically the newest edition or the edition printed on another practice page. The Texas example above is specific to Texas. The 2023 ampacity-table source used in C4–C6 is labeled as such; this page does not claim full alignment with a state's current exam blueprint.

Sources and how we checked this page

Texas exam format, NEC edition, and calculator rules: PSI Texas electrician candidate bulletin, cover and printed pp. 4, 8 and 13; TDLR electrician exam information.
Historical Texas exam results: TDLR Electrician Exam Statistics, Fiscal Year 2025, “Journeyman Calculations” row.
Selected ampacity values and factors: HELUKABEL, Allowable Ampacity Tables, NFPA 70: NEC 2023, PDF pp. 1–2, including the small-conductor protection note.
Box fill, conduit-fill method, and the basic motor-conductor method: Minnesota DLI, Electrical license examination guide, the named calculation sections. This is supporting instruction, not a claim that this page reproduces Minnesota's exam.
The dwelling 3-versus-2 VA/ft² distinction: City of Austin, comparison of 2023 and 2026 NEC changes, “Load Calculation.”
Circuit theory: the OpenStax sections linked beside the answers and above; U.S. DOE Fundamentals Handbook, Electrical Science Vol. 3, Module 9, pp. 20–21.
Aptitude-test context: NECA-IBEW Electrical Training Center and electrical training ALLIANCE sample questions.

Last checked Sept. 29, 2026: the cited Texas exam details and historical results, the cited aptitude-test descriptions, the selected technical source passages, and the arithmetic in all 40 answers. This wasn't a review of every state's licensing rules or the complete NEC.

How we made this page: These original questions were prepared with help from AI tools. Source passages and calculations were checked with AI assistance. The questions haven't been reviewed by a licensed electrician. See our editorial standards.

By the Castleport Test Prep Editorial Team

Castleport Test Prep is an independent exam prep publisher. We aren't affiliated with, endorsed by, or approved by NFPA, the Texas Department of Licensing and Regulation, PSI, the electrical training ALLIANCE, IBEW, NECA, or any other licensing authority or testing provider. NEC and National Electrical Code, and other exam and organization names, identify the subjects discussed; trademarks belong to their respective owners. These practice questions are original and aren't taken from any real exam. Using this page doesn't guarantee a score, a license, or a job.