Journeyman Electrician Exam Calculations

Quick Facts

How to Approach Any Calculation

Every exam calculation can be broken into three steps. First, identify which part of the code governs the question. Second, pull the starting values from the right table. Third, apply any corrections, multipliers or demand factors in the order the code gives them. Write down each step. When you check your work, you will see where you went wrong.

The examples below use the 2023 NEC. The section numbers are given so you can follow along in your book.

Ohm’s Law and Power

These formulas show up as simple questions and inside harder ones.

• E = I × R: voltage equals current times resistance.
• P = E × I: power equals voltage times current.
• P = I² × R and P = E² ÷ R: useful variations.

Example. A 240 V water heater element has a resistance of 12 Ω. Current = 240 ÷ 12 = 20 A. Power = 240 × 20 = 4,800 W.

Example. A 1,500 W heater on 120 V draws 1,500 ÷ 120 = 12.5 A.

For three-phase loads, add √3 (about 1.732): P = E × I × 1.732 (for unity power factor), so I = P ÷ (E × 1.732). A 30 kW three-phase heater at 208 V draws 30,000 ÷ (208 × 1.732) ≈ 83.3 A.

Conductor Ampacity with Corrections

Steps:

1. Choose the conductor’s starting ampacity from Table 310.16, using the column for the insulation temperature rating.
2. Apply the ambient temperature correction from Table 310.15(B)(1) if the ambient is not 30°C (86°F).
3. Apply the adjustment factor from Table 310.15(C)(1) if there are more than three current-carrying conductors in the raceway or cable.
4. Check that the result is not higher than the termination limit in 110.14(C) allows.

Example. Six current-carrying 8 AWG THHN copper conductors share a conduit in a 38°C ambient. THHN is rated 90°C, so start with the 90°C value: 55 A. Ambient correction at 36–40°C for 90°C insulation: 0.91. Adjustment for 4 to 6 conductors: 80%. Result: 55 × 0.91 × 0.80 = 40.04 A. Then compare with the termination limit: the 75°C value for 8 AWG copper is 50 A, and the 60°C value is 40 A. The corrected value of about 40 A is the controlling ampacity.

Using the 90°C column for corrections and then checking against the termination column is allowed by 110.14(C) and 310.15. It is one of the most common question types.

Box Fill

Steps from 314.16(B):

1. Count one allowance for each conductor that starts outside the box and ends or is spliced in it. A conductor that passes through without a splice also counts as one.
2. Count one allowance for all internal cable clamps together, based on the largest conductor.
3. Count one allowance for all support fittings such as fixture studs.
4. Count two allowances for each device yoke, based on the largest conductor connected to the device.
5. Count one allowance for all equipment grounding conductors together, based on the largest.
6. Multiply the total by the volume per conductor in Table 314.16(B). If sizes are mixed, use each size’s volume.

Example. A box holds three 12/2 NM cables with grounds, entering through internal clamps, and one receptacle. Conductors: 3 cables × 2 insulated conductors = 6. Clamps: 1. Device yoke: 2. Grounds: 1. Total: 10 allowances. Volume: 10 × 2.25 in³ = 22.5 in³.

Conduit Fill

Steps:

1. Look up each conductor’s area in Chapter 9, Table 5.
2. Add the areas.
3. Use Chapter 9, Table 4 to find the raceway with enough area at the allowed fill percentage: 40% for three or more conductors.

If all conductors are the same size and type, Annex C gives the number directly, which saves time. Remember that a nipple 24 in or shorter may be filled to 60% (Note 4 to the Chapter 9 tables).

Dwelling Load Calculations

The standard method for a one-family dwelling follows Article 220 Parts III and IV (the optional method in Part IV is a separate calculation). The core values below have not changed between the 2020 and 2023 editions, although several sections were renumbered in 2023.

Example: a 2,000 ft² house.

1. General lighting: 2,000 ft² × 3 VA = 6,000 VA.
2. Small-appliance circuits: 2 × 1,500 VA = 3,000 VA.
3. Laundry circuit: 1,500 VA.
4. Subtotal: 10,500 VA.
5. Apply the general lighting demand factors: the first 3,000 VA at 100% = 3,000 VA. The remaining 7,500 VA at 35% = 2,625 VA. Net = 5,625 VA.
6. Add the range: one 12 kW range is 8 kW under Table 220.55 Column C = 8,000 VA.
7. Add the dryer: 5,000 W minimum or the nameplate, whichever is larger (220.54). Assume 5,000 VA.
8. Add fixed appliances, applying 75% if there are four or more of them (220.53), and the larger of the heating or air conditioning load (220.60).

Then divide the total VA by 240 V to get the service current. For a total of 18,625 VA before heating and cooling, 18,625 ÷ 240 ≈ 77.6 A. The minimum service for a one-family dwelling is still 100 A under 230.79(C).

Motor Calculations

Steps for a single motor on a branch circuit:

1. Find the full-load current (FLC) from Table 430.248 (single-phase) or 430.250 (three-phase). Do not use the nameplate for this step (430.6(A)(1)).
2. Size conductors at 125% of FLC (430.22).
3. Size the short-circuit and ground-fault protective device using Table 430.52. If the result is not a standard size, 430.52(C)(1) Exception 1 allows the next higher standard size.
4. Size overloads from the nameplate current using 430.32.

Example. A 10 hp, 460 V, three-phase motor. FLC from Table 430.250: 14 A. Conductors: 14 × 1.25 = 17.5 A, so 12 AWG copper (20 A in the 60°C column) meets the ampacity requirement. Inverse time breaker: 14 × 2.50 = 35 A, which is a standard size. Dual-element fuse: 14 × 1.75 = 24.5 A, so the next standard size is 25 A.

Voltage Drop

The NEC recommends, in informational notes to 210.19 and 215.2, a maximum of 3% voltage drop on branch circuits or feeders and 5% overall. These are recommendations rather than requirements in most cases, but exam questions often ask you to calculate them.

Single-phase formula: VD = (2 × K × I × L) ÷ CM, where K is about 12.9 for copper, I is current, L is the one-way length in feet and CM is the conductor area in circular mils from Chapter 9, Table 8.

Example. A 120 V, 16 A load is 100 ft from the panel on 12 AWG copper (6,530 CM). VD = (2 × 12.9 × 16 × 100) ÷ 6,530 ≈ 6.3 V. That is about 5.3% of 120 V, above the 3% recommendation, so a larger conductor would be advisable. With 10 AWG (10,380 CM), VD ≈ 4.0 V, about 3.3%. With 8 AWG (16,510 CM), VD ≈ 2.5 V, about 2.1%.

Grounding Conductor Sizing

These are table lookups rather than calculations, but they often appear in calculation sections.

• Grounding electrode conductor: Table 250.66, based on the size of the largest service-entrance conductor. For 2/0 copper service conductors, the minimum is 4 AWG copper. The conductor to a rod electrode never needs to be larger than 6 AWG copper (250.66(A)).
• Equipment grounding conductor: Table 250.122, based on the rating of the overcurrent device. For a 60 A circuit, 10 AWG copper. For 200 A, 6 AWG copper. If the ungrounded conductors are increased for voltage drop, the equipment grounding conductor must be increased in proportion (250.122(B)).

Practice the Steps

Speed comes from repetition. Work each type of calculation until you can do it without notes, then take our Journeyman Electrician Practice Test (NEC 2023), which mixes calculation and lookup questions with a code reference for every answer. If you are new to the code book, start with our Journeyman Electrician NEC Study Guide.

Common Mistakes

• Forgetting the 125% multiplier on continuous loads and motor conductors.
• Using the wrong ampacity column: check the insulation rating and the termination limit.
• Applying demand factors to the wrong loads: lighting demand factors apply to the lighting, small-appliance and laundry loads, not the range or dryer.
• Using the one-way length twice: the single-phase voltage drop formula already includes the factor of 2.
• Rounding too early: carry decimals until the final step.

Where to Go Next

• EPA 608 Apprentice Exemption