Electrician Prep

Texas Electrician Exam Calculations: The Complete Breakdown

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Texas Electrician Exam Calculations: The Complete Breakdown

The Calculations portion of the Texas journeyman electrician exam has a pass rate of just 20.56%. That makes it the single hardest part of the entire licensing process. You get 26 questions and 110 minutes, and you need to score 70% or higher to pass.

The good news is that the calculation types are well-defined. There are no surprise formulas or mystery topics. Every question falls into one of six categories, and this guide walks you through each one with the formulas, NEC references, and step-by-step examples you need.

The Six Calculation Categories

Based on the TDLR content outline, here is what the Calculations portion covers:

| Category | % of Calculations | Approx. Questions | |---|---|---| | Conductor Sizing and Overcurrent Protection | 30% | ~8 | | Load Calculations | 27% | ~7 | | Motor Calculations | 15% | ~4 | | Voltage Drop | 12% | ~3 | | Conduit and Box Fill | 10% | ~2-3 | | General Electrical Calculations | 6% | ~1-2 |

Conductor sizing and load calculations together account for 57% of the section. If you master just those two categories, you are more than halfway to passing.

Category 1: Conductor Sizing and Overcurrent Protection (30%)

This is the most heavily tested category. You need to select the correct conductor size based on the load, installation conditions, and NEC rules.

The Core Process

  1. Determine the load current (in amps)
  2. For continuous loads, multiply by 125% (per NEC 210.20(A))
  3. Look up the base ampacity in NEC Table 310.16
  4. Apply temperature correction factors if ambient temperature is not 30C (86F)
  5. Apply adjustment factors if more than three current-carrying conductors are in the raceway
  6. Verify the conductor size meets the requirements based on terminal temperature ratings (110.14(C))
  7. Select the overcurrent device from standard sizes in 240.6(A)

Key Tables

Table 310.16 is the main ampacity table. It lists conductor ampacities for insulated conductors rated up to 2000V in a raceway, cable, or directly buried. The table has columns for 60C, 75C, and 90C conductor insulation types.

Temperature correction factors are found in Table 310.15(B)(1). For example, if the ambient temperature is 40C (104F), the correction factor for 75C rated conductors is 0.88. You multiply the base ampacity by this factor.

Adjustment factors for bundled conductors are in Table 310.15(C)(1). For 4 to 6 current-carrying conductors in a raceway, you multiply the base ampacity by 0.80. For 7 to 9 conductors, it is 0.70.

Example Problem

A 50A continuous load uses THWN-2 copper conductors (rated 90C) installed in conduit with 6 current-carrying conductors. Ambient temperature is 40C. What is the minimum conductor size?

Step 1: Continuous load adjustment: 50A x 1.25 = 62.5A

Step 2: Find the correction and adjustment factors:

  • Temperature correction for 90C conductor at 40C ambient: 0.91
  • Adjustment for 6 conductors in raceway: 0.80

Step 3: Calculate the required ampacity before derating: 62.5A / (0.91 x 0.80) = 62.5 / 0.728 = 85.85A

Step 4: Look up in Table 310.16, 90C column: #4 AWG THWN-2 copper has an ampacity of 95A (which exceeds 85.85A)

Step 5: Check terminal temperature rating. For circuits over 100A, you can use the 75C column. For 100A or less, most equipment terminations are rated 60C or 75C. #4 AWG in the 75C column is 85A, which exceeds 62.5A. The conductor passes.

Minimum conductor size: #4 AWG copper THWN-2

Overcurrent Protection

Once you have the conductor size, you need to select the overcurrent device. The general rule from 240.4 is that the overcurrent device cannot exceed the ampacity of the conductor. However, 240.4(B) allows the "next size up" standard overcurrent device when the ampacity of the conductor does not match a standard size, provided the circuit is 800A or less.

Standard OCPD sizes per 240.6(A): 15, 20, 25, 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 110, 125, 150, 175, 200, 225, 250, 300, 350, 400, 450, 500, 600...

Practice conductor sizing with our free calculator

Category 2: Load Calculations (27%)

Load calculations determine the minimum size of feeders and service entrance equipment. The exam focuses heavily on dwelling unit calculations using the Standard Calculation Method (NEC Article 220, Parts III and IV).

Standard Dwelling Unit Load Calculation Steps

Step 1: General Lighting and Receptacle Load

  • Multiply the dwelling unit square footage by 3 VA per square foot (Table 220.12)
  • This covers general lighting and general-use receptacle outlets

Step 2: Small Appliance Branch Circuits

  • Add 1,500 VA for each 20A small appliance branch circuit
  • Minimum of two circuits required (210.11(C)(1))
  • Total: at least 3,000 VA

Step 3: Laundry Branch Circuit

  • Add at least 1,500 VA for the laundry branch circuit (220.53)

Step 4: Apply Demand Factors (Table 220.42)

  • First 3,000 VA at 100%
  • Remaining VA from 3,001 to 120,000 at 35%

Step 5: Fixed Appliance Loads

  • Add the nameplate rating of each fixed appliance (water heater, dishwasher, disposal, etc.)
  • If four or more fixed appliances (other than range, dryer, AC, or heating), apply 75% demand factor

Step 6: Dryer Load (Table 220.54)

  • Use nameplate rating or 5,000 VA, whichever is larger

Step 7: Cooking Equipment (Table 220.55)

  • For a single range not exceeding 12 kW, use a demand of 8 kW
  • For larger ranges or multiple ranges, use the columns in Table 220.55

Step 8: Heating or Air Conditioning (220.60)

  • Use the larger of the heating load or cooling load (not both)
  • Do not apply demand factors to the larger load

Step 9: Total and Size the Service

  • Add all loads together
  • Divide by voltage (240V for single-phase) to get amperage
  • Select the service size from standard sizes

Example Problem

A 2,000 sq ft dwelling has a 5 kW dryer, a 12 kW range, a 4,500 VA water heater, a 1,200 VA dishwasher, a 900 VA disposal, a 5-ton (60A) AC unit, and a 10 kW electric furnace. Two small appliance circuits and one laundry circuit.

General lighting: 2,000 x 3 = 6,000 VA Small appliance: 2 x 1,500 = 3,000 VA Laundry: 1,500 VA Subtotal: 10,500 VA

Apply Table 220.42: 3,000 at 100% = 3,000 + 7,500 at 35% = 2,625. Total = 5,625 VA

Fixed appliances (3 items): 4,500 + 1,200 + 900 = 6,600 VA (fewer than 4, so no 75% demand factor applies)

Dryer: 5,000 VA (Table 220.54) Range: 8,000 VA (Table 220.55, Column A for single range not over 12 kW) Heating vs. AC: AC = 60A x 240V = 14,400 VA; Furnace = 10,000 VA. Use the larger: 14,400 VA

Total: 5,625 + 6,600 + 5,000 + 8,000 + 14,400 = 39,625 VA Amperage: 39,625 / 240V = 165.1A Service size: 175A or 200A (next standard size)

Use our load calculation tool to practice

Category 3: Motor Calculations (15%)

Motor calculations follow a specific process that differs from standard circuit sizing. The key rule: always use the NEC table values for full-load current, never the motor nameplate.

Motor Branch Circuit Sizing

Step 1: Find FLC from NEC tables

  • Table 430.248: Single-phase motors
  • Table 430.250: Three-phase motors

Step 2: Size branch-circuit conductors at 125% of FLC (430.22)

Step 3: Size branch-circuit overcurrent protection per Table 430.52

  • Standard fuse: 300% of FLC
  • Inverse time breaker: 250% of FLC
  • If the calculated value does not match a standard size, round up to the next standard size

Step 4: Size overload protection at 115% of nameplate FLA for standard motors (430.32)

Example Problem

A 25 HP, 3-phase, 460V squirrel cage motor with FLA of 32A on the nameplate. Size the branch circuit conductors and inverse time circuit breaker.

Step 1: Table 430.250 FLC for 25 HP, 460V, 3-phase = 34A

Step 2: Conductor: 34A x 1.25 = 42.5A. From Table 310.16 (75C column): #8 AWG copper (50A)

Step 3: OCPD: 34A x 2.50 = 85A. Next standard size = 90A inverse time breaker

Step 4: Overload: Nameplate FLA x 1.15 = 32 x 1.15 = 36.8A

Category 4: Voltage Drop (12%)

Voltage drop is not a mandatory NEC requirement but a recommendation. The NEC suggests that voltage drop should not exceed 3% for branch circuits and 5% total for feeders plus branch circuits combined (210.19(A) Informational Note No. 4).

The Formulas

Single-phase: VD = (2 x K x I x D) / CM

Three-phase: VD = (1.732 x K x I x D) / CM

Where:

  • VD = voltage drop in volts
  • K = resistivity constant (12.9 for copper, 21.2 for aluminum)
  • I = load current in amps
  • D = one-way distance in feet
  • CM = conductor area in circular mils (from NEC Chapter 9, Table 8)

Percentage voltage drop: (VD / Source Voltage) x 100

Example Problem

A 120V single-phase circuit supplies a 16A load through #12 AWG copper conductors over a distance of 100 feet. What is the voltage drop?

CM for #12 AWG: 6,530 (from Chapter 9, Table 8) VD = (2 x 12.9 x 16 x 100) / 6,530 = 41,280 / 6,530 = 6.32V Percentage: 6.32 / 120 x 100 = 5.27%

This exceeds the recommended 3% for a branch circuit. You would need to increase the conductor size.

Use our voltage drop calculator

Category 5: Conduit and Box Fill (10%)

Conduit Fill

NEC Chapter 9, Table 1 sets the maximum fill percentages:

  • 1 conductor: 53% of conduit area
  • 2 conductors: 31% of conduit area
  • 3 or more conductors: 40% of conduit area

The process:

  1. Find the cross-sectional area of each conductor from Chapter 9, Table 5
  2. Add up the total area of all conductors
  3. Find the conduit size from Chapter 9, Table 4 that provides enough area at the correct fill percentage

Box Fill (Article 314.16)

Each component in a box counts for a certain volume based on the largest conductor connected:

| Component | Volume Allowance | |---|---| | Each conductor entering the box | 1x the volume from Table 314.16(B) | | All internal clamps (combined) | 1x | | Each support fitting | 1x | | Each device or equipment (yoke) | 2x | | All equipment grounding conductors (combined) | 1x |

For #14 AWG, the volume per conductor is 2.0 cubic inches. For #12 AWG, it is 2.25 cubic inches.

Example Problem

A box contains four #12 AWG conductors, two #12 AWG equipment grounding conductors, one cable clamp, and one single-gang device. What is the minimum box volume?

  • 4 conductors: 4 x 2.25 = 9.0 cu in
  • Grounding conductors (all count as 1): 1 x 2.25 = 2.25 cu in
  • Clamp (all count as 1): 1 x 2.25 = 2.25 cu in
  • Device (counts as 2): 2 x 2.25 = 4.50 cu in
  • Total: 18.0 cu in

Practice box fill with our calculator

Category 6: General Electrical Calculations (6%)

This category covers basic electrical theory calculations:

Ohm's Law: E = I x R (Voltage = Current x Resistance)

Power formulas:

  • P = E x I (Watts = Volts x Amps)
  • P = I^2 x R
  • P = E^2 / R

Three-phase power: P = 1.732 x E x I x PF

VA calculations: VA = Volts x Amps (for single-phase), VA = 1.732 x Volts x Amps (for three-phase)

These questions tend to be more straightforward than the NEC-specific calculations. If you are comfortable with basic electrical math, this is a place to pick up easy points.

Time Management Strategy

You have 110 minutes for 26 questions, which is about 4 minutes and 15 seconds per question. That sounds generous, but multi-step calculations can eat time quickly.

Quick wins first. Scan through all 26 questions and answer the ones you can solve in under 2 minutes first. Ohm's Law problems, simple conductor lookups, and single-step calculations should take less than a minute each.

Flag and skip. If a question requires a full dwelling unit load calculation, flag it and come back to it after you have answered the faster questions. These multi-step problems can take 8 to 12 minutes each.

Use your codebook efficiently. Most time is lost not on calculations but on looking up values. Tab your NEC at Tables 310.16, 220.42, 220.54, 220.55, 430.248, 430.250, 430.52, and Chapter 9 Tables 1, 4, 5, and 8 before exam day.

Double-check units. Many wrong answers come from unit errors, like using VA when the question asks for amps, or forgetting to convert kW to VA.

How to Practice

Do not just read about calculations. Work through them by hand with your NEC codebook open, exactly as you would on exam day.

Start with single-concept problems. Practice conductor sizing problems until you can do them without hesitation. Then move to load calculations, then motors, then voltage drop.

Progress to multi-step problems. Real exam questions often combine concepts: size the conductor for a motor circuit that runs through conduit in a high-temperature environment. You need to apply motor rules, temperature correction, and conduit adjustment all in one problem.

Take timed practice tests. Set a timer for 110 minutes and work through 26 calculation problems. The time pressure changes everything.

Take our free practice test to see where you stand on calculations.

Sources:

  1. Texas Department of Licensing and Regulation (TDLR) - Electrician Exam Information
  2. TDLR - Electrician Exam Statistics, Fiscal Year 2025
  3. National Fire Protection Association - NFPA 70: National Electrical Code, 2023 Edition
  4. PSI Exams - TDLR Electrician Exam Test Candidate Instructions
  5. Electrical Contractor Magazine - Building the Foundation: Key Changes to Articles 90-130 of the 2026 NEC