Electrician Exam Formula Sheet

This electrician exam formula sheet covers electrical math and selected Code calculations for U.S. journeyman- and master-level study, with 18 worked examples. It's unofficial study material: your exam bulletin decides what you can bring into the room, and your Code edition controls the applicable rules.

Electrical math formulas

Ohm's law and power

E = volts · I = amps · R = ohms (Ω) · P = watts. Use these for DC circuits and purely resistive AC loads (heaters, incandescent lamps). Use RMS values for AC.

All values must describe the same load. For motors and other nonresistive AC loads, use the AC power formulas below. Volts × amps on those loads gives VA, not watts.

These formulas come from electrical theory, so they don't change between NEC editions. Use positive magnitudes and a formula whose denominator is not zero.

Source: DOE Electrical Science Handbook, Vol. 1, ES-01, pp. 14–17. Example: WX01.

Series and parallel resistance

These resistance checks assume two or more finite, positive resistors.

Source: DOE Vol. 1, ES-02, series and parallel circuits. Example: WX02.

AC power, power factor, and three-phase

VA = volt-amperes (apparent power) · W = watts (true power) · PF = power factor as a decimal · E in three-phase formulas = line-to-line voltage · I = line current. Use RMS voltage and current; the three-phase relationships below assume a balanced sinusoidal system.

• Power factor: PF = W ÷ VA (same as kW ÷ kVA). Power factor is not efficiency.
• kVA or kW? Multiply by 1,000 first: I = (kVA × 1,000) ÷ (E × 1.732).
• 1.732 is an approximation of √3. Handy products, rounded: 208 × √3 ≈ 360.3 · 240 × √3 ≈ 415.7 · 480 × √3 ≈ 831.4. Use the calculator’s √3 key when available.
• Balanced wye: line voltage = 1.732 × phase voltage; line current = phase current. Balanced delta: line voltage = phase voltage; line current = 1.732 × phase current.
• VA is apparent power, calculated directly from voltage and current. Don't divide a VA or kVA current calculation by PF. Don't invent a PF the problem didn't give. Treat PF as 1.0 only when the problem says the load is resistive or PF is 1. A current calculation that divides by PF requires PF greater than zero.

Sources: DOE Electrical Science Handbook, Vol. 3, module ES-09, pp. 2–4 and 19–20. Examples: WX03, WX04.

Transformers and energy

Here p means primary, s means secondary, and N means winding turns. The ideal turns ratio compares corresponding windings; it is not automatically the line-voltage ratio of a wye–delta transformer. Energy = kW × hours assumes constant power over that time.

Use the voltage of the side whose current you want. Voltage steps down, current steps up by the same ratio. A transformer's kVA is apparent power, so no PF goes in these current formulas. Rated current tells you what the transformer carries, not what fuse to install. That's a Code rule (see Transformer overcurrent protection).

Sources: DOE Electrical Science Handbook, Vol. 4, module ES-13; EIA, Measuring electricity. Example: WX05.

Motor math (theory)

746 W is the exact value of "electric horsepower." Mechanical horsepower is approximately 745.7 W, so use whatever the question gives (NIST SP 811, Appendix B.9). Horsepower is shaft output, so divide by efficiency to get electrical input.

Exam trap: these formulas calculate operating current from the supplied output, efficiency, voltage, and PF. A Code sizing question normally uses the applicable table current for motor conductors and short-circuit/ground-fault protection; overload protection uses nameplate current (see Motor circuits).

Frequency is the supply frequency in hertz; speed is revolutions per minute. Pole count means poles, not pole pairs. Efficiency and PF are separate, positive decimal inputs.

Sources: DOE Vol. 4, module ES-12, p. 6 (synchronous speed and slip); DOE, Determining Electric Motor Load and Efficiency, “Input Power Measurements” and “The Slip Method.” Example: WX06.

Voltage drop

• L is the one-way length in feet. The 2 accounts for the two-wire loop; √3 is the balanced three-phase line relationship. Don't double the distance too.
• K = 12.9 for copper and 21.2 for aluminum are common approximate 75 °C study values, with K in Ω·cmil/ft. They can be derived from NEC Chapter 9, Table 8. The table lists 1,000 kcmil copper at 0.0129 Ω per 1,000 ft at 75 °C (aluminum: 0.0212). Multiply by 1,000,000 circular mils and divide by 1,000 ft and you get 12.9. K for a single wire size can differ slightly: 8 AWG copper works out to about 12.8, which is the value Minnesota's exam guide uses in its example. If the question gives you a K, use it.
• CM = circular mils, from NEC Chapter 9, Table 8:

• 3% of common voltages: 120 V → 3.6 V · 208 V → 6.24 V · 240 V → 7.2 V · 480 V → 14.4 V.
• Is 3% always a Code requirement? The usual 3% branch-circuit and 5% feeder-plus-branch-circuit guidance is in Informational Notes, not a blanket mandatory limit. Section 90.5(C) distinguishes those notes from requirements. A specific circuit rule or the problem’s stated limit can still control the answer. On an exam, use the limit the question gives you.
• These are resistance-only study formulas for the stated two-wire or balanced three-phase model. A more general AC voltage-drop calculation can require reactance and power factor, not just conductor resistance. Here R is resistance in Ω per 1,000 ft, I is amperes, and the percentage uses the corresponding source voltage—line-to-line for three-phase.

Sources: Minnesota DLI calculation guide, “Voltage Drop”; Schneider Electric, voltage drop in steady load conditions.

Examples: WX07, WX08, WX09.

AC theory

Use hertz for f, seconds for T, and ohms for R, XL, XC, and Z. Here L is inductance, not the run length used in voltage drop. These are ideal, steady-state sinusoidal relationships; peak is not peak-to-peak.

Convert first: 1 mH = 0.001 H, 1 µF = 0.000001 F. The impedance formula is for series circuits only; parallel circuits need a different method.

Source: DOE Vol. 3, ES-07 p. 5; ES-08 pp. 1, 6, 14, 19. Example: WX10.

Units and percentages

Capital M is mega; lowercase m is milli (NIST, SI prefixes).

• p% of x = x × (p ÷ 100) — 125% of 20 A = 25 A
• x increased by p% = x × (1 + p ÷ 100) — 20 A increased by 125% = 45 A
• p% of a rating = rating × (p ÷ 100) — 80% of 20 A = 16 A; whether an 80% limit applies is a separate Code question.

When a Code rule says "not less than 125% of," multiply by 1.25.

Example: WX11.

NEC calculation formulas

Code calculations need both a formula and the right rule for the question. The 2023 references below are labeled separately from selected 2026 changes; this is not an edition-by-edition replacement for the Code. The 2026 NEC moved load calculations from Article 220 to Article 120, but do not assume every section number or rule stayed the same. Use the edition your bulletin names.

Sources: NFPA’s explanation of the 2026 load-calculation changes; NFPA 70 edition access.

Continuous loads and breaker sizing

• Continuous load: maximum current expected to last 3 hours or more.
• Overcurrent device: rating ≥ (continuous load × 1.25) + noncontinuous load (2023 NEC 210.20(A) branch circuits; 215.3 feeders). The listed 100%-rated assembly exception has different conditions; do not apply the 125% shortcut to it automatically.
• Conductor: the usual pre-adjustment minimum is (continuous load × 1.25) + noncontinuous load. Separately, the ampacity remaining after applicable adjustment and correction must carry the actual maximum load; termination limits still apply (2023 NEC 210.19(A), 215.2(A)).
• Maximum entirely continuous load under the ordinary 125% rule: breaker rating × 0.80. It's the same rule turned around, because 1 ÷ 1.25 = 0.80. This is not a universal 80% limit for every circuit or assembly.
• Next size up: 2023 NEC 240.4(B) requires all its conditions: the conductors are not part of a branch circuit supplying more than one receptacle for portable cord-and-plug-connected loads; their ampacity falls between standard device ratings; and the next higher rating does not exceed 800 A. This does not erase small-conductor limits or the special rules for particular equipment. Standard ratings are in Table 240.6(A).

Source: 2023 NEC text, Articles 210, 215 and 240, printed pp. 70–84, 70–91, and 70–123–124. Example: WX11.

Ampacity adjustment and correction

allowable ampacity = table ampacity × temperature correction factor × adjustment factor

• Ampacities: Table 310.16. Ambient temperature correction: 310.15(B). Adjustment factors above: Table 310.15(C)(1).
• You can start from the conductor's insulation rating (for example, the 90 °C column for THHN) when adjusting. The final answer can't exceed what the terminations allow. Use the equipment marking and the applicable temperature, conductor-size, and equipment conditions in NEC 110.14(C); 90 °C insulation does not automatically mean 90 °C terminals.

The conductor count and any exception must come from the problem and the applicable rule, not from counting every wire as current-carrying.

Sources: licensed 2023 NEC ampacity-table extract, Tables 310.16 and 310.15(C)(1); Minnesota DLI guide, “Ambient Temperature Correction” and “Adjustment Factors.” Example: WX12.

Box fill

required box volume = number of volume allowances × volume per conductor

That shortcut applies when the allowances use one conductor size. For mixed sizes, calculate each size’s allowances and add their volumes.

How to count (NEC 314.16(B)); the reference text, printed pp. 70–175–176, explains the conductor, clamp, support-fitting, and yoke allowances. The newer grounding-conductor rule is described separately below:

• Each conductor that enters and is spliced or terminated in the box: 1 each. A conductor passing straight through: 1.
• All internal cable clamps together: 1, sized by the largest conductor.
• Luminaire studs and hickeys: 1 for each type, sized by the largest conductor in the box.
• Each yoke or strap containing one or more devices (receptacle, switch): 2, sized by the largest conductor connected to a device on that yoke or strap.
• Equipment grounding conductors entering the box: up to four count as 1 allowance. Each one beyond four adds ¼. Use your edition’s 314.16(B)(5) to determine the conductor-size basis, especially with mixed sizes. The extra-quarter rule was introduced in 2020; the earlier rule counted the ordinary group as one. WX13 uses only 12 AWG conductors, so its size basis is unambiguous.
• Pigtails that originate and terminate entirely inside the box do not add conductor-fill allowances. Do not assume all small fixture wires are exempt.

Box volumes are in 314.16(A). Do not round a required minimum volume down.

The 14–10 AWG values are also listed in the current DLI checklist; the 8 AWG reference allowance is shown in Table 314.16(B), 2017 text, printed p. 70–176. Use the table in your assigned edition.

Sources: Minnesota DLI 2026 inspection checklist, item 046; DLI examination guide, “Box Fill.” Example: WX13.

Conduit fill

total conductor area (Chapter 9, Table 5) ≤ allowed area for the raceway and trade size (Chapter 9, Table 4)

The short-nipple allowance applies to nipples between boxes, cabinets, or similar enclosures; apply the other relevant table notes as well.

When every conductor is the same size and insulation type, the Informative Annex C tables give the maximum count directly. Count physical conductor area, including equipment grounding conductors; this is not the current-carrying-conductor count used for ampacity adjustment. Passing a fill calculation does not establish ampacity.

Sources: NEC Chapter 9, Table 1 and its notes, 2017 reference text, printed p. 70–679; Minnesota’s current examination guide, “Conduit Fill,” for the 40% and short-nipple 60% calculations. Use the table and notes in your assigned edition. Example: WX14.

Motor circuits

There is no single instantaneous-trip percentage for every AC motor. Short-circuit/ground-fault protection and overload protection are separate calculations. In WX15, the motor currents, protection percentages, and next-standard-size instruction are supplied, so the exercise tests the calculation rather than an unstated table lookup.

Source: Minnesota DLI examination guide, “Motor Calculations” and the branch-circuit and feeder examples. Example: WX15.

Transformer overcurrent protection

• Rated current: see Transformers and energy.
• Protection calculation: rated current × the percentage for the applicable table row. Primary-only protection, primary-and-secondary protection, current range, voltage class, and the table notes are not interchangeable.
• For the 125% primary-only case specified in WX16, calculate 125% of rated primary current, then use the next higher standard rating only because the exercise expressly permits it. This is not a universal 125% rule for every transformer.

For an actual Code question, select the applicable row and notes in Table 450.3(B) of the assigned edition. Transformer-winding protection does not by itself settle secondary-conductor protection; see 240.4(F) and 240.21(C).

Sources: Minnesota DLI guide, “Transformer Calculations,” for rated current; 2023 NEC, 240.4(F), printed p. 70–123, for the separate secondary-conductor issue. Example: WX16.

Dwelling load calculation (standard method)

This table is a 2023 NEC reference for selected dwelling feeder and service calculations, not a complete service calculation.

The 2023 optional method is a different calculation: its eligible general-load component uses the first 10 kVA at 100% and the remainder at 40%, with heating and air-conditioning loads handled separately. Section 220.82 also sets eligibility conditions, including a qualifying three-wire supply and a 100 A or larger service or feeder. Do not mix that method into the standard-method rows above.

Selected 2026 changes: load calculations move to Article 120. General lighting and receptacle load for dwelling feeders and services becomes 2 VA per sq ft in 120.41, while the minimum general-purpose branch-circuit calculation remains 3 VA per sq ft under 120.13. The 1,500 VA small-appliance and laundry provisions appear in 120.52. Do not use the 2 VA figure to calculate the minimum number of general-purpose branch circuits.

Sources: 2023 NEC, 220.41–220.82, printed pp. 70–94–98; NFPA’s 2026 dwelling-load explanation; NFPA’s 2026 Handbook correction page, replacement p. 103, containing 120.52–120.54. Examples: WX17, WX18.

Rounding and exam instructions

• Fractions of an amp in the relevant load calculation: less than 0.5 may be dropped under 2023 NEC 220.5(B). North Carolina's exam handbook spells it out: 99.5 A rounds to 100 A, 99.4 A to 99 A. This is not permission to round a minimum box volume, conductor area, or required ampacity down.
• Nominal voltages such as 120, 120/240, 208Y/120, 240, 480Y/277, and 480 V are used for calculations unless the question says otherwise (2023 NEC 220.5(A)).
• Exceptions and optional methods: bulletins disagree. Texas's bulletin (p. 8) says to ignore exceptions and optional calculation methods unless the question directs you to them. North Carolina's handbook (p. 7) says to apply exceptions and alternatives unless the question excludes them. Read your own bulletin's instructions.
• Copper unless stated: Texas's bulletin tells candidates to treat every conductor as copper unless the question says otherwise.

Which formula should I use?

Check your answer's unit last. Amps, watts, VA, ohms, and kWh aren't interchangeable.

Worked examples

These 18 examples are original, unofficial study calculations, not real exam questions. Carry full calculator precision and round only at the end; round a required minimum upward when necessary to avoid understating it. The motor- and transformer-protection exercises supply their protection rules explicitly.

WX01: Heater current, power, and energy

Given: A 32 Ω resistive heating element on 240 V RMS runs at constant power for 90 minutes.

• I = E ÷ R = 240 ÷ 32 = 7.5 A
• P = E² ÷ R = 240² ÷ 32 = 1,800 W = 1.8 kW
• 90 min = 1.5 h → 1.8 kW × 1.5 h = 2.7 kWh

Answer: 7.5 A, 1,800 W, 2.7 kWh. Watch out: 1,800 × 90 isn't kWh. Convert both power and time to kW and hours first.

Source: DOE Vol. 1, ES-01; EIA, Measuring electricity.

WX02: Three resistors in parallel

Given: 18 Ω, 36 Ω, and 12 Ω in parallel on 72 V DC.

• 1/R = 1/18 + 1/36 + 1/12 = 2/36 + 1/36 + 3/36 = 6/36 → R = 6 Ω
• I = 72 ÷ 6 = 12 A

Check: Branch currents are 4 A + 2 A + 6 A = 12 A, and 6 Ω is smaller than the smallest branch (12 Ω). Adding the resistances (66 Ω) would treat it as a series circuit.

Source: DOE Vol. 1, ES-02, parallel circuits.

WX03: VA is not watts

Given: A single-phase load draws 12 A at 240 V RMS with PF 0.80.

• VA = 240 × 12 = 2,880 VA
• W = 240 × 12 × 0.80 = 2,304 W

Answer: 2,880 VA; 2,304 W. The gap between them comes from power factor, not efficiency. Nothing about losses is given here.

Source: DOE Vol. 3, ES-09, pp. 2–4.

WX04: Three-phase current from kVA

Given: A balanced sinusoidal 36 kVA three-phase load at 208 V line-to-line, PF 0.90. Find current to one decimal place and real power in kW.

• I = 36,000 ÷ (208 × √3) ≈ 99.926 A → 99.9 A
• true power = 36 kVA × 0.90 = 32.4 kW

Answer: 99.9 A, or 100 A if the question requests rounding to the nearest whole amp; 32.4 kW. Don't divide the current by 0.90: current is being calculated from apparent power, not real watts.

Source: DOE Vol. 3, ES-09, pp. 19–20.

WX05: Transformer primary and secondary current

Given: An ideal single-phase 15 kVA transformer, 480 V primary, 240 V secondary.

• Ip = 15,000 ÷ 480 = 31.25 A
• Is = 15,000 ÷ 240 = 62.5 A
• Turns ratio = 480 ÷ 240 = 2:1

Answer: 31.25 A primary, 62.5 A secondary, 2:1. The higher-voltage side carries less current.

Source: DOE Vol. 4, ES-13, transformer ratios.

WX06: Motor electrical input and operating current

Given: A motor delivers 7.5 hp at 90% efficiency and 0.82 PF on a balanced sinusoidal 460 V line-to-line three-phase supply. Use 746 W per hp.

• output = 7.5 × 746 = 5,595 W
• input = 5,595 ÷ 0.90 = 6,216.666… W
• I = (5,595 ÷ 0.90) ÷ (460 × √3 × 0.82) ≈ 9.52 A

Answer: 6.22 kW input, 9.52 A. Enter efficiency as 0.90, not 90. This is calculated operating current; an ordinary Code conductor-sizing question for this motor would use the applicable Table 430.250 value instead.

Source: DOE, Determining Electric Motor Load and Efficiency, “Input Power Measurements”.

WX07: Single-phase voltage drop, then resize

Given: A 240 V two-wire single-phase resistance-model circuit, 30 A, 150 ft one way, 10 AWG copper (10,380 cmil), K = 12.9. Compare with a 3% voltage-drop target and then try 8 AWG (16,510 cmil).

• VD = (2 × 12.9 × 30 × 150) ÷ 10,380 = 116,100 ÷ 10,380 ≈ 11.18 V
• % VD = (116,100 ÷ 10,380) ÷ 240 × 100 ≈ 4.66%
• Try 8 AWG (16,510 cmil): VD = 116,100 ÷ 16,510 ≈ 7.03 V; (116,100 ÷ 16,510) ÷ 240 × 100 ≈ 2.93%

Answer: 10 AWG drops 4.66%, over a 3% target; 8 AWG drops 2.93%. The 150 ft is used once. The 2 handles the return path. This comparison resolves voltage drop only, not the full conductor-sizing question.

Source: Minnesota DLI guide, “Voltage Drop”; the K, areas, and target are supplied exercise inputs.

WX08: Three-phase voltage drop

Given: A balanced three-phase resistance-model circuit at 480 V line-to-line, 50 A, 200 ft one way, 6 AWG copper (26,240 cmil), K = 12.9.

• VD = (√3 × 12.9 × 50 × 200) ÷ 26,240 ≈ 8.52 V
• % VD = [(√3 × 12.9 × 50 × 200) ÷ 26,240] ÷ 480 × 100 ≈ 1.77%

Answer: 8.52 V, 1.77%. Using 2 instead of 1.732 would overstate the drop by about 15%.

Source: Minnesota DLI guide, “Voltage Drop”; the K and area are supplied exercise inputs.

WX09: Smallest conductor for a voltage-drop target

Given: A 120 V two-wire resistance-model circuit, 16 A, 100 ft one way, copper with K = 12.9, maximum 3% drop. Choose from the conductor areas in this sheet.

• allowed VD = 120 × 0.03 = 3.6 V
• CM = (2 × 12.9 × 16 × 100) ÷ 3.6 = 41,280 ÷ 3.6 = 11,466.666… cmil

Answer: 8 AWG. 10 AWG has only 10,380 cmil, so it's too small; pick the first listed size at or above 11,466.666… cmil (11,467 cmil when stated as a whole-number minimum). Ampacity still has to be checked separately. This formula only answers the voltage-drop part.

Source: Minnesota DLI guide, “Voltage Drop”; the K, conductor choices, and target are supplied exercise inputs.

WX10: Convert microfarads before finding reactance

Given: An ideal 47 µF capacitor on a sinusoidal 60 Hz supply.

• C = 47 ÷ 1,000,000 = 0.000047 F
• XC = 1 ÷ (2 × π × 60 × 0.000047) ≈ 56.4 Ω

Answer: 56.4 Ω. Plugging in 47 as farads gives an answer a million times too small.

Source: DOE Vol. 3, ES-08, p. 6; NIST SI prefixes.

WX11: A continuous load and the 125% rule

Given: A 4,500 W, 240 V resistive storage water heater. The problem specifies a continuous load, the ordinary 125% overcurrent-device rule, no 100%-rated assembly exception, and no additional load or manufacturer-imposed sizing condition. Find the minimum standard device rating under those conditions.

• I = 4,500 ÷ 240 = 18.75 A
• 18.75 × 1.25 = 23.4375 A → next standard rating at or above that minimum: 25 A

Answer: 25 A minimum standard overcurrent-device rating under the stated conditions; conductor sizing is a separate check. "125% of" means × 1.25. Multiplying by 2.25 ("increased by 125%") would give a wrong 42.19 A.

Source: 2023 NEC, 210.20(A) and Table 240.6(A); continuous classification is supplied by this problem.

WX12: Ampacity adjustment

Given: A conductor with a table ampacity of 30 A shares a raceway with five other current-carrying conductors (six total). Ambient correction factor: 1.00. The problem states that the adjustment table applies and no conductor-count exception applies.

• 30 × 1.00 × 0.80 = 24 A

Answer: 24 A. Six conductors falls in the 4–6 row (80%). Then confirm the termination temperature limit doesn't set a lower number.

Source: 2023 NEC licensed table extract, Table 310.15(C)(1).

WX13: Box fill with more than four grounds

Given: A junction box with internal clamps and five 12/2 cables with 12 AWG grounds (every wire spliced in the box). No devices or other fittings. Use the up-to-four-grounds-plus-quarters rule.

• Insulated conductors: 5 cables × 2 = 10
• Grounds: 5 → first four = 1, fifth = ¼ → 1.25
• Clamps: 1
• Total: 10 + 1.25 + 1 = 12.25 allowances × 2.25 cu in. = 27.5625 cu in.

Answer: at least 27.5625 cu in., or 27.57 cu in. when expressing the minimum to two decimal places. Do not round this minimum down to 27.56. Under the earlier ordinary one-allowance grounding rule, this example would give 27.0 cu in. Use the rule in your edition.

Source: Minnesota DLI guide, “Box Fill”; 2026 DLI checklist, item 046.

WX14: Conduit fill

Given: Nine insulated conductors, each 0.0133 sq in. including insulation (from the problem). The raceway's total internal area is 0.533 sq in. (from the problem). Use the 40% fill rule, not the short-nipple allowance.

• conductor area = 9 × 0.0133 = 0.1197 sq in.
• allowed at 40% = 0.533 × 0.40 = 0.2132 sq in.

Answer: It fits the stated fill limit (0.1197 ≤ 0.2132). Nine conductors means "over 2," so the 40% limit applies. This does not answer the separate ampacity-adjustment question.

Source: Minnesota DLI guide, “Conduit Fill”; physical areas and the applicable percentage are supplied.

WX15: Motor branch circuit and feeder

Given: A continuous-duty three-phase motor with a table full-load current of 28 A (from the problem). A feeder also supplies two other continuous-duty motors of 16 A each and no other loads. For this exercise, use 125% for the branch conductors, 125% of the largest motor plus the others for the feeder, 250% for the inverse-time breaker, and 175% for the dual-element fuse. The problem permits the next higher standard rating when the protection calculation falls between ratings; do not apply a starting-current exception.

• Branch conductors: 28 × 1.25 = 35 A
• Inverse-time breaker, maximum: 28 × 2.50 = 70 A (a standard size)
• Dual-element fuse, maximum: 28 × 1.75 = 49 A → next standard size 50 A
• Feeder: (28 × 1.25) + 16 + 16 = 67 A

Answer: 35 A minimum branch-conductor ampacity; 70 A breaker or 50 A fuse maximum under the supplied protection rules; 67 A minimum feeder ampacity before any separate adjustment, correction, or termination check. Overloads would use the nameplate current, not 28 A.

Source: Minnesota DLI guide, “Motor Calculations”; 2023 standard device ratings, Table 240.6(A). The exercise supplies the protection factors and rounding permission.

WX16: Transformer primary-only protection

Given: A balanced 75 kVA three-phase transformer, 480 V line-to-line primary, 208Y/120 V secondary. The problem specifies the primary-only 125% case and permits the next higher standard rating when 125% is not a standard rating. Find rated line currents and that primary-device limit; do not size secondary conductors.

• Ip = 75,000 ÷ (480 × √3) ≈ 90.2 A
• Is = 75,000 ÷ (208 × √3) ≈ 208.2 A
• Primary-only protection: [75,000 ÷ (480 × √3)] × 1.25 ≈ 112.8 A → next higher standard rating 125 A

Answer: 90.2 A primary, 208.2 A secondary, 125 A maximum primary device under the supplied rule. This result does not establish secondary-conductor protection.

Source: Minnesota DLI guide, “Transformer Calculations”; 2023 standard device ratings, Table 240.6(A). The exercise supplies the 125% case and rounding permission.

WX17: Dwelling general lighting demand, 2023 vs. 2026

Given: A 2,000 sq ft house with two small-appliance circuits and one laundry circuit. Compare the 2023 and 2026 dwelling feeder/service lighting unit loads. For this comparison, the problem supplies 1,500 VA per small-appliance or laundry circuit and demand factors of 100% for the first 3,000 VA and 35% for the remainder. Calculate this component only, not the full service load.

Same house, 700 VA apart. Use your exam's edition. These figures include only the lighting, small-appliance, and laundry component; they are not the house’s total service demand.

Source: 2023 NEC, 220.41, 220.45 and 220.52; NFPA’s 2026 dwelling unit-load explanation; the demand factors are supplied for the comparison.

WX18: Range over 12 kW

Given: One 15 kW household range, 2023 NEC standard method, using Table 220.55 Column C and Note 1.

• 15 kW is 3 kW over 12 → 3 × 5% = 15% increase
• 8 kW × 1.15 = 9.2 kW

Answer: 9.2 kW of calculated demand (9,200 VA for this range calculation). The 5% increments increase the 8 kW table value, not the 15 kW nameplate rating.

Source: 2023 NEC, Table 220.55 and Note 1, printed p. 70–97.

Can you bring a formula sheet into the electrician exam?

Do not assume an open-book exam permits a separate formula sheet. The reference rules below are program-specific. What varies is whether you can write formulas inside your own code book before exam day. Some exams allow it, some ban it, and one bulletin contradicts itself. Your exact candidate bulletin and any clarification from the licensing authority or testing provider control.

What to do:

1. Open your bulletin and search it for "notes," "tabs," "reference," and "paper."
2. If your rules permit the relevant notes, copy formulas into the permitted locations in your code book before test day, in the form the bulletin requires (for example, ink, or highlighted pencil in North Carolina). Resolve a contradictory instruction before relying on notes.
3. If notes aren't allowed, practice recalling the electrical math formulas and locating the references your exam supplies or permits. Do not assume every formula on this page will be available in the exam room.

Calculators: The Texas bulletin, printed p. 4, specifies a silent, battery-operated, non-programmable calculator without an alphabetic keyboard. North Carolina, p. 3, and Minnesota, p. 4, provide one. Follow the rule for your own examination.

British Columbia candidates: SkilledTradesBC publishes separate formula sheets for different exam levels. Its Common Core Level 1–2 sheet says formulas are already included in the exam's reference materials, so candidates shouldn't bring that sheet. Pick your exam level on the Construction Electrician page.

Which NEC edition's numbers should you use?

The one your bulletin names, and that may not match what's enforced on job sites. North Carolina's July 2025 handbook lists the 2020 NEC. Texas’s examination page specifies the 2026 NEC, effective September 1, 2026. Minnesota's guide lists the 2026 NEC. Those are examination-reference examples, not a state-by-state field-adoption list.

Match the rule to your exam date and exact license, not just the newest Code book available. The selected 2023/2026 comparison in WX17 does not make every other Code example on this page a 2026 calculation.

How to study with this sheet

• Learn the math before the Code. Calculation questions can be a whole section. Texas's journeyman exam has a separate Calculations portion with 26 items—24 scored and two unscored—and 110 minutes, and you have to pass that portion separately (Texas bulletin, pp. 12–14).
• Work the examples without looking, then check each step.
• Practice by topic: Motor calculation practice questions: 32 worked problems. Keep the assumptions and Code edition stated with each problem.
• Broader prep: master electrician exam prep.

You can also repeat WX01–WX18 on this page: cover the answer, write down the formula and units, then compare your steps. A missed calculation tells you what to revisit; this sheet does not produce an official score or a pass prediction.

Sources

By Castleport Test Prep Editorial Team. Last verified September 30, 2026: the linked examination-reference rules, the specific Code and technical passages cited on this page, and all 18 worked calculations. Sources were checked September 29–30; the cited document editions are identified rather than relabeled as new publications.

Exam bulletins:

• Texas: PSI/TDLR Electrician Candidate Information Bulletin, updated Sept. 15, 2026; TDLR electrician exam page
• North Carolina: NCBEEC Examination Information Handbook, July 2025, pp. 3, 7
• Massachusetts: PSI Examination Candidate Information Bulletin, effective Nov. 1, 2024, p. 2 Q11 and pp. 6–7; its conflicting note instructions are described above
• Minnesota: DLI Electrical License Examination Guide, reference-material rules and electrical-calculation sections
• ICC: What should I bring to the test site?, updated Nov. 10, 2025; exact program rules also apply
• British Columbia: SkilledTradesBC Construction Electrician resources and the Common Core Level 1–2 formula sheet, July 2022

Code and technical references:

• NFPA 70 edition access; the cited 2023 NEC excerpt supports the specified Article 210, 215, 220, and 240 passages, not every later article or the full 2026 Code
• NFPA’s 2026 dwelling-load explanation and 2026 Handbook correction page, replacement p. 103
• Licensed 2023 ampacity-table extract, Tables 310.16 and 310.15(C)(1)
• Minnesota’s 2026 inspection checklist, box-fill item 046; the 2020 equipment-grounding allowance change
• NEC Table 8 conductor-property reference, a historical table scan; material, stranding, coating, and 75 °C conditions are specified beside the values
• 2017 NEC reference text, Chapter 9 Table 1 and notes, for the explicitly identified conduit-fill reference
• DOE Electrical Science Handbook, Vol. 1, Vol. 3, and Vol. 4 (DOE-HDBK-1011-92; 1992 theory references, not installation codes)
• DOE, Determining Electric Motor Load and Efficiency, input-power and slip-method sections
• NIST SP 811, Appendix B.9 (horsepower) and NIST SI prefixes
• EIA: Measuring electricity
• Schneider Electric: Voltage drop in steady load conditions

The worked examples are original, and their arithmetic was recalculated for this page. They are not official exam questions. Checking sources is not a licensed professional review, and this sheet is not installation guidance.

Castleport Test Prep is an independent exam prep publisher. We are not affiliated with, endorsed by, or approved by NFPA, PSI, Pearson VUE, the International Code Council, SkilledTradesBC, or any state licensing board. Exam and credential names are used for identification, and trademarks belong to their respective owners.