Texas Journeyman Calculation Check: 10 Worked Problems

New to electrical work in Texas? Choose your next step.

Start with no experience Find a local route Apprentice-license steps Free aptitude practice

Report an issue on this page

If something on this page is unclear, wrong, or broken, you can tell me here.

Your email (so I can reply) What problem did you run into? I am a human, not a bot

Leave this empty

Try ten original calculation questions from our Texas Journeyman Electrician Exam Prep workbook (2026 NEC). Keep the conditions and units beside your working. Use 1.732 for √3 in this check.

You can check every answer and read the working without giving us an email address. The score summarizes these ten questions; it does not predict a licensing-exam result.

1. A single-phase load draws 10 amperes at 240 volts, and a wattmeter reads 1,920 watts. What is the power factor?

A. 0.80B. 0.85C. 1.25D. 0.75

Worked answer and wrong turns

Correct value: 0.80. Apparent power is 240 x 10 = 2,400 VA, and power factor is true power divided by apparent power: 1,920 / 2,400 = 0.80. 1.25 inverts the ratio, and a power factor can never be above 1.

Source Electrical theory (power factor = watts / volt-amperes). No Code lookup.

Source: Texas Journeyman workbook, Chapter 1, Question 8. Options have been reordered for this check; the values and correct answer are unchanged.

2. A 480Y/277-volt, three-phase feeder supplies only 50 kVA of lighting, balanced across the phases, and the lighting is a continuous load. The feeder overcurrent device is not part of an assembly listed for operation at 100% of its rating, and no other exception applies. Before rounding to a standard size, what is the minimum device rating?

A. 130.2 AB. 60.1 AC. 75.2 AD. 104.2 A

Worked answer and wrong turns

Correct value: 75.2 A. 50,000 VA ÷ 831.36 = 60.14 A of line current, and 1.25 × 60.14 = 75.2 A for the continuous lighting.

Worked Calculation

Given

Feeder: 480Y/277 V, three-phase, 50 kVA balanced lighting
The lighting is a continuous load
Overcurrent device not part of an assembly listed for operation at 100% of its rating; no other exception applies

Find

Minimum feeder overcurrent device rating before rounding (A)

Formula

Device ≥ 1.25 × VA ÷ (1.732 × 480 V)

Substitute

Line current = 50,000 / (1.732 × 480) = 50,000 / 831.36 = 60.14 A
Continuous load at 125%: 1.25 × 60.14
Minimum device rating = 75.2 A

Result

75.2 A

Check

The question asks for the value before rounding to a standard size.

Code path NEC 215.5 (was 215.3 in 2023)

60.1 A leaves out the 125% factor.
104.2 A uses the single-phase formula: 50,000 / 480 = 104.2.
130.2 A applies 125% to the single-phase figure: 1.25 × 104.2 = 130.2.

Source: Texas Journeyman workbook, Chapter 13, Question 250. Options have been reordered for this check; the values and correct answer are unchanged.

3. Field-installed feeder tap conductors, not over 10 ft long, leave the enclosure in which the tap is made. The feeder is protected at 400 amperes, the tap ends in a 30-ampere circuit breaker, and the calculated load on the tap is 24 amperes. What minimum ampacity must the tap conductors have?

A. 30 AB. 40 AC. 133 AD. 24 A

Worked answer and wrong turns

Correct value: 40 A. The one-tenth limit is 400 ÷ 10 = 40 A, which is larger than the 24 A load and the 30 A breaker, so 40 A governs.

Worked Calculation

Given

Field-installed feeder tap, not over 10 ft, leaving the enclosure where the tap is made
Feeder protection: 400 A
Tap terminates in a 30 A breaker
Calculated load on the tap: 24 A

Find

Minimum tap conductor ampacity (A)

Formula

Ampacity ≥ largest of calculated load, terminating device, and feeder OCPD ÷ 10

Substitute

Limit 1, the calculated load: at least 24 A
Limit 2, the device the tap terminates in: at least 30 A
Limit 3, one-tenth of the feeder device rating because the field-installed tap leaves its enclosure: 400 / 10 = 40 A
The largest limit governs: 40 A

Result

40 A

Code path NEC 240.21(B)(1)

24 A meets only the calculated-load limit.
30 A meets the load and terminating-device limits but not the one-tenth limit.
133 A is the one-third ratio from the 25-ft tap rule: 400 / 3 = 133.

Source: Texas Journeyman workbook, Chapter 13, Question 251. Options have been reordered for this check; the values and correct answer are unchanged.

4. A feeder conductor has a base ampacity of 115 amperes. In an ambient temperature where the correction factor is 0.82, what is the corrected ampacity?

A. 115.0 AB. 140.2 AC. 82.0 AD. 94.3 A

Worked answer and wrong turns

Correct value: 94.3 A. 115 A × 0.82 = 94.3 A after temperature correction.

Worked Calculation

Given

Base ampacity: 115 A
Temperature correction factor: 0.82

Find

Temperature-corrected ampacity (A)

Formula

Corrected ampacity = base ampacity × correction factor

Substitute

Corrected = base × factor
115 × 0.82 = 94.3 A

Result

94.3 A

Code path NEC 310.15(B)

82.0 A uses 100 A as the base: 100 × 0.82.
115 A ignores the correction factor.
140.2 A divides by the factor instead of multiplying: 115 / 0.82.

Source: Texas Journeyman workbook, Chapter 13, Question 252. Options have been reordered for this check; the values and correct answer are unchanged.

5. Six current-carrying conductors share a raceway, giving an adjustment factor of 0.80. If each conductor has a base ampacity of 130 amperes, what is the adjusted ampacity?

A. 130 AB. 96 AC. 104 AD. 117 A

Worked answer and wrong turns

Correct value: 104 A. With six current-carrying conductors in the raceway, 130 A × 0.80 = 104 A.

Worked Calculation

Given

Base ampacity: 130 A
Adjustment factor (six current-carrying conductors): 0.80

Find

Ampacity adjusted for six conductors (A)

Formula

Adjusted ampacity = base ampacity × adjustment factor

Substitute

Adjusted = base × factor
130 × 0.80 = 104 A

Result

104 A

Code path NEC 310.15(C)(1)

96 A uses 120 A as the base: 120 × 0.80.
117 A uses a factor of 0.90: 130 × 0.90.
130 A ignores the adjustment factor.

Source: Texas Journeyman workbook, Chapter 13, Question 253. Options have been reordered for this check; the values and correct answer are unchanged.

6. A feeder conductor with a base ampacity of 175 amperes is installed where the temperature correction factor is 0.88 and the conductor adjustment factor is 0.80. What is the final ampacity?

A. 123.2 AB. 140.0 AC. 154.0 AD. 112.0 A

Worked answer and wrong turns

Correct value: 123.2 A. 175 A × 0.88 = 154.0 A, and 154.0 × 0.80 = 123.2 A once both factors are applied.

Worked Calculation

Given

Base ampacity: 175 A
Temperature correction: 0.88
Adjustment: 0.80

Find

Final feeder ampacity with both factors (A)

Formula

Ampacity = base × temperature correction × adjustment

Substitute

Both factors apply: 175 × 0.88 × 0.80
175 × 0.88 = 154.0
154.0 × 0.80 = 123.2 A

Result

123.2 A

Code path NEC 310.15(B) and 310.15(C)(1)

112.0 A applies the 0.80 adjustment twice: 175 × 0.80 × 0.80.
140.0 A applies only the adjustment: 175 × 0.80.
154.0 A applies only the temperature correction: 175 × 0.88.

Source: Texas Journeyman workbook, Chapter 13, Question 254. Options have been reordered for this check; the values and correct answer are unchanged.

7. A single-phase feeder carries 80 amperes a one-way distance of 200 feet using conductors of 52,620 circular mils. Using K = 12.9 ohm-cmil/ft and VD = 2 x K x I x L / CM, what is the voltage drop?

A. 11.77 VB. 15.69 VC. 3.92 VD. 7.84 V

Worked answer and wrong turns

Correct value: 7.84 V. 2 × 12.9 × 80 × 200 = 412,800, and 412,800 ÷ 52,620 = 7.84 V.

Worked Calculation

Given

Current: 80 A, single-phase
One-way length: 200 ft
K = 12.9
Conductor area: 52,620 circular mils

Find

Single-phase feeder voltage drop (V)

Formula

VD = 2 × K × I × L ÷ CM

Substitute

VD = 2 × K × I × L / CM
2 × 12.9 × 80 × 200 = 412,800
412,800 / 52,620 = 7.8449 V

Result

7.84 V

Check

7.8449 V is shown as 7.84 V.

Code path NEC 215.4(A)(2) Informational Note No. 2

3.92 V leaves out the factor 2 for the return conductor: 206,400 / 52,620.
11.77 V uses a factor of 3 instead of 2: 3 × 12.9 × 80 × 200 / 52,620 = 11.77.
15.69 V uses a factor of 4 instead of 2: 4 × 12.9 × 80 × 200 / 52,620 = 15.69.

Source: Texas Journeyman workbook, Chapter 13, Question 255. Options have been reordered for this check; the values and correct answer are unchanged.

8. A feeder operating at 240 volts has a voltage drop of 7.84 volts. What is the percentage of voltage drop?

A. 1.63%B. 3.27%C. 6.53%D. 7.84%

Worked answer and wrong turns

Correct value: 3.27%. 7.84 V ÷ 240 V = 0.0327, and 0.0327 × 100 = 3.27%.

Worked Calculation

Given

Voltage drop: 7.84 V
System voltage: 240 V

Find

Voltage drop as a percent of 240 V (%)

Formula

VD% = VD ÷ system voltage × 100

Substitute

Percent = drop / voltage × 100
7.84 / 240 = 0.0327
0.0327 × 100 = 3.27%

Result

3.27%

Check

3.267% rounds to 3.27%.

Code path NEC 215.4(A)(2) Informational Note No. 2 (feeder voltage drop)

1.63% divides by 480 V: 7.84 / 480.
6.53% divides by 120 V: 7.84 / 120.
7.84% treats the volts as a percentage.

Source: Texas Journeyman workbook, Chapter 13, Question 256. Options have been reordered for this check; the values and correct answer are unchanged.

9. A three-phase, 480-volt feeder supplies a balanced load of 120,000 VA. What is the feeder load in amperes?

A. 144 AB. 160 AC. 250 AD. 288 A

Worked answer and wrong turns

Correct value: 144 A. 120,000 VA ÷ 831.36 = 144.34 A, which rounds to 144 A.

Worked Calculation

Given

Load: 120,000 VA
Three-phase, 480 V

Find

Three-phase feeder load current (A)

Formula

I = VA ÷ (1.732 × 480 V)

Substitute

I = VA / (1.732 × V)
1.732 × 480 = 831.36
120,000 / 831.36 = 144.34 A

Result

144 A

Check

Keep 1.732 x 480 = 831.36 unrounded; 144.34 A rounds to 144 A.

Source Electrical theory (balanced three-phase current = VA / (1.732 x V)). No Code lookup.

160 A does not match the calculation: 160 × 831.36 = 133,018 VA, more than the 120,000 VA load.
250 A leaves out the square root of 3: 120,000 / 480.
288 A doubles the correct result: 144 × 2.

Source: Texas Journeyman workbook, Chapter 13, Question 257. Options have been reordered for this check; the values and correct answer are unchanged.

10. A 120/240-volt, single-phase, 3-wire feeder supplies only linear line-to-neutral loads. One ungrounded conductor carries 90 amperes and the other carries 65 amperes, both at the same lagging power factor. What is the actual neutral current for these simultaneous loads?

A. 90 AB. 155 AC. 25 AD. 65 A

Worked answer and wrong turns

Correct value: 25 A. With the stated linear loads at the same phase angle, the neutral carries the difference: 90 A − 65 A = 25 A.

Worked Calculation

Given

Line A: 90 A
Line B: 65 A
Linear line-to-neutral loads, same lagging power factor (same phase angle)

Find

Actual neutral current for the stated loads (A)

Formula

Neutral = difference between the two ungrounded conductor currents

Substitute

Neutral carries the unbalance: 90 – 65 = 25 A

Result

25 A

Code path Electrical theory: actual neutral-current phasor difference

65 A is the smaller line current.
90 A is the larger line current.
155 A adds the lines: 90 + 65.

Source: Texas Journeyman workbook, Chapter 13, Question 258. Options have been reordered for this check; the values and correct answer are unchanged.

Your ten-question check:

Choose one next step

Open the worked answer for a missed question, name the operation or condition you missed, and redo that question before starting a new set.

See the Journeyman workbook, or choose a different exam book. Paperback and Kindle are separate purchases; our workbook is not the NEC itself.

Keep your review organized

Open a free fillable study plan, mistake log and reference checklist directly. No email or account is required. The email study series is not active.

Open the three free study tools

These are study examples under the conditions shown. The NEC paths refer to the workbook’s 2026 edition. Actual installation decisions require all applicable conditions and local requirements.