Electrician Exam Math: Complete Formula Guide with Worked Examples

Calculation questions separate the candidates who pass from those who don't. Most journeyman and master exams are 20–40% math — and on an open-book exam, speed matters as much as accuracy. You can't afford to re-derive a formula under pressure.

This guide gives you every formula in exam-ready form, with the worked examples that build the speed and pattern recognition you need.

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The Formula Foundation: Four Quantities You Must Know

Every electrical calculation involves some combination of these four quantities:

Master the relationships between these four and you can solve nearly any basic electrical math problem.

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Ohm's Law

E = I × R

The three forms:

The Ohm's Law Triangle: Draw a triangle with E on top, I on the bottom-left, R on the bottom-right. Cover the unknown — what remains is the formula.

Worked Examples

Example 1: A circuit has a resistance of 24Ω and carries 5A. What is the voltage?

• E = I × R = 5 × 24 = 120V

Example 2: A 240V circuit has a resistance of 40Ω. What is the current?

• I = E ÷ R = 240 ÷ 40 = 6A

Example 3: A 120V circuit draws 8A. What is the resistance?

• R = E ÷ I = 120 ÷ 8 = 15Ω

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The Power Formulas

Power (P) in watts relates to voltage, current, and resistance through three equivalent formulas:

The Power Wheel

The full set of 12 Ohm's Law / Power formulas is sometimes called the "Power Wheel":

        E
   _____|_____
  |           |
 I×R        P/I
  |           |
  |___________| 
        I
   _____|_____
  |           |
 E/R       P/E
  |___________|
        R
   _____|_____
  |           |
 E/I       P/I²
  |___________|
        P
   _____|_____
  |           |
 E×I       I²R
  |___________|

For the exam, the most useful forms to memorize:

• P = E × I (most common)
• I = P ÷ E (finding current from a wattage and voltage)
• E = P ÷ I (finding voltage from wattage and current)

Worked Examples

Example 1: A 240V water heater draws 20A. What is its wattage?

• P = E × I = 240 × 20 = 4,800W (4.8 kW)

Example 2: A 1,500W space heater operates on 120V. What current does it draw?

• I = P ÷ E = 1,500 ÷ 120 = 12.5A

Example 3: A 10Ω resistor carries 4A. What power does it dissipate?

• P = I² × R = 4² × 10 = 16 × 10 = 160W

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Series and Parallel Circuits

Series Circuits

In a series circuit, current is the same everywhere. Voltages and resistances add.

• R_total = R₁ + R₂ + R₃ + …
• E_total = E₁ + E₂ + E₃ + …
• I is the same in every part of the circuit

Example: Three resistors — 4Ω, 6Ω, and 10Ω — are connected in series across 120V.

• R_total = 4 + 6 + 10 = 20Ω
• I = E ÷ R = 120 ÷ 20 = 6A (same through all three)
• Voltage across the 10Ω: E = I × R = 6 × 10 = 60V

Parallel Circuits

In a parallel circuit, voltage is the same across every branch. Currents and conductances add.

• 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + …
• E is the same across every branch
• I_total = I₁ + I₂ + I₃ + …

Shortcut for two equal resistors in parallel: R_total = R ÷ 2

Shortcut for any two resistors in parallel: R_total = (R₁ × R₂) ÷ (R₁ + R₂)

Example: Two 60Ω resistors in parallel across 120V.

• R_total = 60 ÷ 2 = 30Ω
• I_total = E ÷ R = 120 ÷ 30 = 4A
• Each branch draws: 120 ÷ 60 = 2A (branches add: 2 + 2 = 4A ✓)

Exam tip: Parallel circuit resistance is always less than the smallest individual resistor. If your answer is higher than the smallest resistor, something went wrong.

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Power Factor

Power factor (PF) is the ratio of true power (watts, W) to apparent power (volt-amperes, VA):

PF = W ÷ VA

The three power quantities:

Key conversions:

• W = VA × PF
• VA = W ÷ PF
• kVA = kW ÷ PF

Example: A motor draws 15A at 240V and has a power factor of 0.85. What is the true power?

• VA = E × I = 240 × 15 = 3,600 VA
• W = VA × PF = 3,600 × 0.85 = 3,060W (3.06 kW)

Exam tip: If no power factor is stated, assume PF = 1.0 (unity). The exam will always tell you when PF matters.

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Single-Phase vs. Three-Phase Power

Single-Phase Power

• Apparent power: VA = E × I
• True power: W = E × I × PF
• Current: I = VA ÷ E (or I = W ÷ (E × PF))

Three-Phase Power

Three-phase circuits use the multiplier 1.732 (√3) to account for the phase relationships:

• Apparent power: VA = 1.732 × E × I
• True power: W = 1.732 × E × I × PF
• Current: I = VA ÷ (1.732 × E)

Exam tip: 1.732 is √3. It only appears in three-phase formulas — never in single-phase. If you see "three-phase" in the problem, 1.732 is almost certainly needed.

Three-Phase Worked Examples

Example 1: A 480V, 3-phase motor draws 30A at PF = 0.90. What is the true power in kW?

• W = 1.732 × 480 × 30 × 0.90
• W = 1.732 × 480 × 30 × 0.90 = 1.732 × 12,960 = 22,447W
• True power ≈ 22.4 kW

Example 2: A 3-phase, 208V panel has a 45 kVA load. What is the line current?

• I = (kVA × 1,000) ÷ (1.732 × E)
• I = (45,000) ÷ (1.732 × 208)
• I = 45,000 ÷ 360.3 = 124.9A → 125A

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Voltage Drop

Voltage drop is one of the most tested calculations on master electrician exams — and one of the most commonly fumbled. The NEC recommends voltage drop not exceed 3% for branch circuits and 5% for the combined feeder and branch circuit (NEC 210.19 Informational Note, 215.2 Informational Note).

The Voltage Drop Formula

Single-phase (or DC):

VD = (2 × K × I × L) ÷ CM

Three-phase:

VD = (1.732 × K × I × L) ÷ CM

Where:

• K = resistivity constant (12.9 for copper, 21.2 for aluminum)
• I = load current in amps
• L = one-way conductor length in feet
• CM = circular mils of the conductor (from NEC Chapter 9, Table 8)

Common Conductor Circular Mil Values (Table 8)

Voltage Drop Worked Example

Example: A 120V single-phase circuit uses 12 AWG copper wire and runs 75 feet (one way) to a 16A load. What is the voltage drop? Does it comply with the NEC recommendation?

• K = 12.9 (copper)
• I = 16A
• L = 75 ft (one-way)
• CM = 6,530 (12 AWG from Table 8)

VD = (2 × 12.9 × 16 × 75) ÷ 6,530
VD = (30,960) ÷ 6,530
VD = 4.74V

VD% = (4.74 ÷ 120) × 100 = 3.95%

Result: Exceeds the NEC 3% recommendation for branch circuits. To comply, upgrade to 10 AWG (CM = 10,380):

VD = (2 × 12.9 × 16 × 75) ÷ 10,380 = 30,960 ÷ 10,380 = 2.98V (2.48%) ✓

"Reverse" Voltage Drop: Finding the Minimum Conductor Size

Sometimes the exam asks: what is the minimum conductor size to limit voltage drop to 3%?

Formula rearranged for CM:

CM = (2 × K × I × L) ÷ VD_max

Example: What is the minimum conductor size for a 120V, 20A single-phase circuit, 100 feet one-way, limiting VD to 3%?

• VD_max = 120 × 0.03 = 3.6V
• CM = (2 × 12.9 × 20 × 100) ÷ 3.6
• CM = 51,600 ÷ 3.6 = 14,333 CM

From Table 8: 10 AWG = 10,380 (too small), 8 AWG = 16,510 ✓

Minimum conductor: 8 AWG

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Transformer Calculations

Turns Ratio

The voltage ratio equals the turns ratio:

E_primary / E_secondary = N_primary / N_secondary

A transformer that steps 480V down to 120V has a turns ratio of 480:120 = 4:1.

Primary and Secondary Current

For a single-phase transformer, VA is constant (ignoring losses):

E_primary × I_primary = E_secondary × I_secondary = VA

Current is inversely proportional to voltage — step down voltage, step up current.

Example: A 10 kVA, 240V/120V single-phase transformer is fully loaded. What is the secondary current?

• I_secondary = VA ÷ E_secondary = 10,000 ÷ 120 = 83.3A
• I_primary = VA ÷ E_primary = 10,000 ÷ 240 = 41.7A

Three-Phase Transformer Current

I = kVA × 1,000 ÷ (1.732 × E)

Example: A 75 kVA, 480V/208V three-phase transformer. What is the secondary full-load current?

• I = (75,000) ÷ (1.732 × 208)
• I = 75,000 ÷ 360.3 = 208.2A → 208A

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Horsepower to Watts Conversion

Motors are rated in horsepower (HP) but electrical calculations use watts or amps. The conversion:

1 HP = 746 watts

Example: A 5 HP motor at 240V single-phase, PF = 1.0 (ideal). What current does it draw?

• P = 5 × 746 = 3,730W
• I = P ÷ (E × PF) = 3,730 ÷ 240 = 15.5A

Exam tip: In practice, use the NEC motor tables (Table 430.248 for single-phase, 430.250 for three-phase) rather than converting HP to watts — the NEC tables account for efficiency and are required for sizing conductors and protection. The HP → watts conversion is useful for quick estimates and physics problems.

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Quick Reference: All Formulas in One Place

Ohm's Law

Power

Power Factor

Three-Phase

Voltage Drop

K values: Copper = 12.9 | Aluminum = 21.2
NEC recommendation: 3% max VD for branch circuits; 5% combined feeder + branch

Transformers

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Common Exam Math Mistakes

1. Using total conductor length instead of one-way length in the voltage drop formula — the "2 ×" factor already accounts for the return conductor, so L is always the one-way distance
2. Forgetting 1.732 on three-phase problems — if the problem says three-phase, 1.732 is almost certainly in the formula
3. Confusing VA and watts — always check whether the problem gives you VA (apparent power) or watts (true power) before applying formulas
4. Not converting HP to watts when needed — 1 HP = 746W, not 1,000W
5. Using the wrong K value — copper is 12.9, aluminum is 21.2; always check the conductor material
6. Skipping power factor when given — if the problem specifies a PF, it must be used; ignoring it gives a wrong answer

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Ready to Practice Electrical Math?

GetLicenseReady's practice questions include calculation problems for every formula type on this page — timed, with step-by-step solutions and NEC references.

Related reading: NEC Table 310.16 — Wire Sizing and Ampacity | NEC Article 220 — Load Calculations | NEC Article 430 — Motors | NEC Chapter 9 — Conduit Fill Calculations