Texas journeyman exam · 2026 NEC
Calculations on the Texas journeyman exam
The Calculations portion of the Texas journeyman exam is 26 multiple-choice questions in 110 minutes, and they come from six families: box fill, conduit fill, voltage drop, motor circuits, services and three-phase. Each family is a short method plus one or two figures you read from your book. Below, each is explained as a method with one worked example in our own numbers; wherever a step needs a figure the book prints, the example uses a made-up figure for the arithmetic, not the book’s — read the real one in your code book.
Box fill
The method
Count everything in the box the way Article 314 counts it: each conductor, plus the allowances it gives for clamps, fittings, devices and grounding conductors. Multiply that count by the volume Article 314 allows for the conductor size, and compare it with the volume of the box.
Worked example
A box marked 30 cubic inches. Your count comes to 9. Say the allowance for the conductor size is 2.6 cubic inches (a made-up figure for the arithmetic, not the book’s — read the real one in your code book).
- 9 × 2.6 = 23.4 cubic inches needed.
- 23.4 is less than 30.
The box is big enough, with 6.6 cubic inches to spare.
Conduit fill
The method
Add up the cross-sectional area of every conductor, from the Chapter 9 table for its insulation type and size. Compare the total with the area the raceway may be filled to for that number of conductors, from the Chapter 9 table for the raceway type and trade size.
Worked example
Ten conductors of one size, each 0.0300 square inches, in a raceway whose allowed fill area for that count is 0.37 square inches (both made-up figures for the arithmetic, not the book’s — read the real ones in your code book).
- 10 × 0.0300 = 0.300 square inches.
- 0.300 is less than 0.37.
The ten conductors fit.
Voltage drop
The method
This is Ohm’s law over the length of the run. For a single-phase circuit the current goes out and back, so the drop is 2 × one-way length × current × resistance per foot. The resistance per 1,000 feet for each size and metal is in Chapter 9; the question tells you the target or asks for the drop itself.
Worked example
A 120-volt circuit, 16 amps, 90 feet one way. Say the conductor’s resistance is 0.27 ohms per 1,000 feet (a made-up figure for the arithmetic, not the book’s — read the real one in your code book).
- 2 × 90 × 16 × 0.27 ÷ 1,000 = 0.778 volts.
- 0.778 ÷ 120 = 0.65%.
The drop is about 0.78 volts, 0.65% of the supply.
Motor circuits
The method
Size a motor circuit from the full-load current in the Article 430 motor tables, not the nameplate (the nameplate sets the overloads). The branch-circuit conductors carry a percentage of that current that Article 430 sets; pick the conductor from the Article 310 ampacity table; size the short-circuit and ground-fault device from Article 430’s table of maximum settings by device type.
Worked example
Say the motor table gives 21.4 amps for this motor, and say the conductor percentage were 130% (both made-up figures for the arithmetic, not the book’s — read the real ones in your code book).
- 21.4 × 1.30 = 27.8 amps.
The branch-circuit conductors need an ampacity of at least 27.8 amps — then read the real figures in Article 430 and redo it.
Services
The method
A dwelling service is a load calculation: the general lighting load from the floor area, the small-appliance and laundry loads, the demand factors that reduce the total, then the fixed appliances, the range and the larger of heating or cooling. In the 2026 edition these steps are in Article 120. Divide the final volt-amperes by the voltage and choose the next standard size your book lists.
Worked example
Say the calculation, after demand factors, comes to 40,800 volt-amperes on a 120/240-volt service (our number, not a worked dwelling).
- 40,800 ÷ 240 = 170 amps.
The service needs a rating of at least 170 amps; choose the next standard rating your book lists.
Three-phase
The method
For a balanced three-phase load, current = power ÷ (1.732 × line voltage), where 1.732 is the square root of 3. Three-phase voltage drop uses 1.732 in place of the 2 in the single-phase formula: 1.732 × one-way length × current × resistance per foot.
Worked example
A balanced 30-kilowatt load at 480 volts, three-phase.
- 1.732 × 480 = about 831 volts.
- 30,000 ÷ 831 = 36.1 amps.
Each line carries about 36 amps.
Working them on the day
110 minutes for 26 questions is about four minutes each. Write the method, look up the one or two figures, and check the answer against the choices before you move on.
A calculator is allowed if it is silent, battery-powered, non-programmable, with no paper tape and no alphabet keyboard.