Texas Journeyman Electrical Practice Exam Pt 2
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The second half of the Texas Journeyman Electrician Exam focuses on electrical calculations, a section where many candidates struggle. This portion contains 26 multiple-choice questions to be completed in 110 minutes, with a required 70% passing score. Combined with your documented 7,000+ hours of supervised experience, mastering the Calculations portion is essential to securing your Texas Journeyman license once you reach 8,000 hours.
The Calculations portion tests your ability to apply formulas and NEC rules to real-world electrical scenarios. Topics include service and feeder load calculations, branch circuit sizing, motor and transformer calculations, conduit and conductor fill, overcurrent protection, and special occupancies. Every question assumes copper conductors unless otherwise stated, so precision and familiarity with the 2023 National Electrical Code® (NEC) are critical. Time management matters here — calculation errors or second-guessing can quickly eat away at your 110-minute window.
Dakota Prep prepares you with Texas journeyman calculation practice exams, problem-solving drills, and NEC-aligned exercises. With over 3,000 electrician exam practice questions, you’ll build confidence in service sizing, load calculations, and formula application under pressure. Start your free Texas journeyman electrician practice exam today and gain the skills you need to pass the Calculations portion on your first try.
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NEC 2017: 450.3(B), Table 450.3(B)NEC 2020: 450.3(B), Table 450.3(B)NEC 2023: 450.3(B), Table 450.3(B)Step 1: Calculate primary current: I = 30000 / (√3 * 480) = 36.1 A. Step 2: Calculate secondary current: I = 30000 / (√3 * 208) = 83.3 A. Step 3: Determine primary protection: 36.1 A * 2.5 = 90.3 A (250% for primary with secondary protection). Step 4: Determine secondary protection: 83.3 A * 1.25 = 104.1 A (125% for secondary > 9A).
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NEC 2017: 450.3(B), Table 450.3(B)NEC 2020: 450.3(B), Table 450.3(B)NEC 2023: 450.3(B), Table 450.3(B)Step 1: Calculate primary current: I = 30000 / (√3 * 480) = 36.1 A. Step 2: Calculate secondary current: I = 30000 / (√3 * 208) = 83.3 A. Step 3: Determine primary protection: 36.1 A * 2.5 = 90.3 A (250% for primary with secondary protection). Step 4: Determine secondary protection: 83.3 A * 1.25 = 104.1 A (125% for secondary > 9A).
What is the minimum size 60°C (140°F) rated copper conductors required to supply an AC transformer arc welder (non motor generator) with a 50 ampere-rated primary current and a 60 percent duty-cycle?
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NEC 2017: Table 630.11(A), Table 310.15(B)(16)NEC 2020: Table 630.11(A), Table 310.16NEC 2023: Table 630.11(A), Table 310.16Step 1: Determine the effective primary currentFor a welder with a duty cycle other than 100%, we need to calculate the effective primary current using the formula from NEC Article 630.11(A):Effective current = Primary current × √(duty cycle)Effective current = 50 A × √(0.60) = 38.73 AStep 2: Apply the NEC sizing requirementNEC 630.11(A) requires conductors that supply one or more welders to have an ampacity of at least 100% of the effective primary current.Required ampacity = 38.73 AStep 3: Select the conductor sizeUsing the 60°C column of NEC Table 310.16 for copper conductors, we need to find the smallest conductor size that has an ampacity of at least 38.73 A.The 8 AWG copper conductor has an ampacity of 40 A at 60°C, which is the next size up that meets our requirement.
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NEC 2017: Table 630.11(A), Table 310.15(B)(16)NEC 2020: Table 630.11(A), Table 310.16NEC 2023: Table 630.11(A), Table 310.16Step 1: Determine the effective primary currentFor a welder with a duty cycle other than 100%, we need to calculate the effective primary current using the formula from NEC Article 630.11(A):Effective current = Primary current × √(duty cycle)Effective current = 50 A × √(0.60) = 38.73 AStep 2: Apply the NEC sizing requirementNEC 630.11(A) requires conductors that supply one or more welders to have an ampacity of at least 100% of the effective primary current.Required ampacity = 38.73 AStep 3: Select the conductor sizeUsing the 60°C column of NEC Table 310.16 for copper conductors, we need to find the smallest conductor size that has an ampacity of at least 38.73 A.The 8 AWG copper conductor has an ampacity of 40 A at 60°C, which is the next size up that meets our requirement.
What is the minimum electrical service size required for a 2,000 sq ft residence with general lighting, 6 fastened-in-place kitchen appliances (2,000 VA each), one 12 kW range, 3 small appliance circuits (1,500 VA each), one laundry circuit (1,500 VA), an 11,000 VA electric furnace, a 13,000 VA central air conditioner, and a 7,000 VA EV charging station?
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NEC 2017: Table 220.12, 220.42, 220.52, 220.53, 220.55, 220.60NEC 2020: 220.11, 220.14(J), 220.42, 220.52, 220.53, 220.55, 220.60NEC 2023: 220.5(C), 220.41, 220.42, 220.52, 220.53, 220.55, 220.60Step 1: Calculate all individual loads
Step 2: General lighting = 2,000 sq ft × 3 VA = 6,000 VA
Step 3: Kitchen appliances = 6 × 2,000 VA = 12,000 VA
Step 4: Range = 12 kW = 12,000 VA
Step 5: Small appliance circuits = 3 × 1,500 VA = 4,500 VA
Step 6: Laundry circuit = 1,500 VA
Step 7: Electric furnace = 11,000 VA
Step 8: Central air conditioning = 13,000 VA
Step 9: EV charging station = 7,000 VA, but per NEC minimum requirement, use 7,200 VA
Step 10: Group the general lighting, small appliances, and laundry loads: 6,000 VA + 4,500 VA + 1,500 VA = 12,000 VA
Step 11: Apply NEC demand factors to lighting group: first 3,000 VA at 100% plus remainder at 35%: 3,000 VA + (9,000 VA × 0.35) = 6,150 VA
Step 12: Apply 75% demand factor to kitchen appliances (more than four appliances): 12,000 VA × 0.75 = 9,000 VA
Step 13: For range rated at 12 kW, use NEC base demand of 8,000 VA
Step 14: For non-coincident loads (furnace and AC), select the larger: 13,000 VA
Step 15: Sum all adjusted loads: 6,150 VA + 9,000 VA + 8,000 VA + 13,000 VA + 7,200 VA = 43,350 VA
Step 16: Convert to amperes at 240V: 43,350 VA ÷ 240V = 180.6 A
Step 17: According to NEC 240.6, the next standard service size above 180.6 A is 200 A
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NEC 2017: Table 220.12, 220.42, 220.52, 220.53, 220.55, 220.60NEC 2020: 220.11, 220.14(J), 220.42, 220.52, 220.53, 220.55, 220.60NEC 2023: 220.5(C), 220.41, 220.42, 220.52, 220.53, 220.55, 220.60Step 1: Calculate all individual loads
Step 2: General lighting = 2,000 sq ft × 3 VA = 6,000 VA
Step 3: Kitchen appliances = 6 × 2,000 VA = 12,000 VA
Step 4: Range = 12 kW = 12,000 VA
Step 5: Small appliance circuits = 3 × 1,500 VA = 4,500 VA
Step 6: Laundry circuit = 1,500 VA
Step 7: Electric furnace = 11,000 VA
Step 8: Central air conditioning = 13,000 VA
Step 9: EV charging station = 7,000 VA, but per NEC minimum requirement, use 7,200 VA
Step 10: Group the general lighting, small appliances, and laundry loads: 6,000 VA + 4,500 VA + 1,500 VA = 12,000 VA
Step 11: Apply NEC demand factors to lighting group: first 3,000 VA at 100% plus remainder at 35%: 3,000 VA + (9,000 VA × 0.35) = 6,150 VA
Step 12: Apply 75% demand factor to kitchen appliances (more than four appliances): 12,000 VA × 0.75 = 9,000 VA
Step 13: For range rated at 12 kW, use NEC base demand of 8,000 VA
Step 14: For non-coincident loads (furnace and AC), select the larger: 13,000 VA
Step 15: Sum all adjusted loads: 6,150 VA + 9,000 VA + 8,000 VA + 13,000 VA + 7,200 VA = 43,350 VA
Step 16: Convert to amperes at 240V: 43,350 VA ÷ 240V = 180.6 A
Step 17: According to NEC 240.6, the next standard service size above 180.6 A is 200 A
What is the ampacity of 500 kcmil copper conductors with THWN insulation installed in a wet location with eight current-carrying conductors in a 75-foot raceway at an ambient temperature of 125°F?
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NEC 2017: Table 310.15(B)(16), Table 310.15(B)(2)(a), Table 310.15(B)(3)(a)NEC 2020: Table 310.16, Table 310.15(B)(1)(1), Table 310.15(C)(1)NEC 2023: Table 310.16, Table 310.15(B)(1)(1), Table 310.15(C)(1)#500 kcmil THWN copper ampacity before derating = 380 amperes380 amperes × .67 (temp. correction) × .7 (adj. factor) = 178.22 amperes
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NEC 2017: Table 310.15(B)(16), Table 310.15(B)(2)(a), Table 310.15(B)(3)(a)NEC 2020: Table 310.16, Table 310.15(B)(1)(1), Table 310.15(C)(1)NEC 2023: Table 310.16, Table 310.15(B)(1)(1), Table 310.15(C)(1)#500 kcmil THWN copper ampacity before derating = 380 amperes380 amperes × .67 (temp. correction) × .7 (adj. factor) = 178.22 amperes
Three continuous-duty motors rated at 30 hp, 25 hp, and 20 hp are powered by a 230-volt, three-phase system. The feeder conductor requires a _____ ampacity.
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NEC 2017: 430.24, Table 430.250NEC 2020: 430.24, Table 430.250NEC 2023: 430.24, Table 430.250From Table 430.250, the FLC for a 30 hp, 230V motor is 80A, for a 25 hp, 230V motor is 68A, and for a 20 hp, 230V motor is 54A. We apply 125% to the largest motor (80A * 1.25 = 100A) and add the FLC of the other two motors (68A + 54A = 122A). The total is 100A + 122A = 222A. This question tests the ability to calculate ampacity for multiple large motors of different sizes.
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NEC 2017: 430.24, Table 430.250NEC 2020: 430.24, Table 430.250NEC 2023: 430.24, Table 430.250From Table 430.250, the FLC for a 30 hp, 230V motor is 80A, for a 25 hp, 230V motor is 68A, and for a 20 hp, 230V motor is 54A. We apply 125% to the largest motor (80A * 1.25 = 100A) and add the FLC of the other two motors (68A + 54A = 122A). The total is 100A + 122A = 222A. This question tests the ability to calculate ampacity for multiple large motors of different sizes.
If a heat pump rated at 40 A, 240 V, 1-phase was installed 70 m (229.66 ft) from its supply, the minimum size of XHHW conductor with a k-factor of 12.9 that would not allow more than a 3 percent voltage drop is __________
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NEC 2017: Chapter 9, Table 8NEC 2020: Chapter 9, Table 8NEC 2023: Chapter 9, Table 8Explanation: To find the minimum size of the conductor, we need to calculate the circular mil area (CM) required to limit the voltage drop to 3%. The formula for CM is:CM = (2 x K x I x D) / VD permittedWhere K is the resistivity constant for the conductor material (XHHW), I is the current (40 A), D is the distance (70 m), and VD permitted is the allowed voltage drop (3% of 240 V = 7.2 V).First, we need to convert the distance from meters to feet: 70 m * 3.281 ft/m = 229.67 ft.Next, we can use the resistivity constant for copper (K = 12.9) as a reference, since the question does not specify the conductor material.CM = (2 x 12.9 x 40 A x 229.67 ft) / 7.2 V = 32,919 CMNow, we can reference Chapter 9, Table 8 in the NEC codebook to find the conductor size with a circular mil area equal to or greater than 32,919 CM. According to the table, a No. 4 AWG conductor has a circular mil area of 41,740 CM, which is sufficient to limit the voltage drop to 3%.
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NEC 2017: Chapter 9, Table 8NEC 2020: Chapter 9, Table 8NEC 2023: Chapter 9, Table 8Explanation: To find the minimum size of the conductor, we need to calculate the circular mil area (CM) required to limit the voltage drop to 3%. The formula for CM is:CM = (2 x K x I x D) / VD permittedWhere K is the resistivity constant for the conductor material (XHHW), I is the current (40 A), D is the distance (70 m), and VD permitted is the allowed voltage drop (3% of 240 V = 7.2 V).First, we need to convert the distance from meters to feet: 70 m * 3.281 ft/m = 229.67 ft.Next, we can use the resistivity constant for copper (K = 12.9) as a reference, since the question does not specify the conductor material.CM = (2 x 12.9 x 40 A x 229.67 ft) / 7.2 V = 32,919 CMNow, we can reference Chapter 9, Table 8 in the NEC codebook to find the conductor size with a circular mil area equal to or greater than 32,919 CM. According to the table, a No. 4 AWG conductor has a circular mil area of 41,740 CM, which is sufficient to limit the voltage drop to 3%.
Maximum number of No. 10 AWG XHHW conductors that can be pulled into a run of 1 1/2 (41) EMT is __________.
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NEC 2017: Chapter 9, Table 4NEC 2020: Chapter 9, Table 4NEC 2023: Chapter 9, Table 4Step 1: Find the maximum area for conductors within the conduitMaximum percent of area of Trade size 1 1/2 (41) EMT conduit that can be used for conductor fill = 40 percent; Chapter 9 Table 1Cross-sectional Area of 41 EMT conduit at 40 percent = 0.814 in² (526 mm²); Chapter 9 Article 358Step 2: Find the area of a single No. 10 AWG XHHW conductorCross-sectional Area of No. 10 AWG XHHW = 0.0243 in² (15.68 mm²); Chapter 9 Table 5Step 3: Calculate the maximum number of conductorsMaximum number of conductors = 0.814 in² / 0.0243 in² = 33.49
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NEC 2017: Chapter 9, Table 4NEC 2020: Chapter 9, Table 4NEC 2023: Chapter 9, Table 4Step 1: Find the maximum area for conductors within the conduitMaximum percent of area of Trade size 1 1/2 (41) EMT conduit that can be used for conductor fill = 40 percent; Chapter 9 Table 1Cross-sectional Area of 41 EMT conduit at 40 percent = 0.814 in² (526 mm²); Chapter 9 Article 358Step 2: Find the area of a single No. 10 AWG XHHW conductorCross-sectional Area of No. 10 AWG XHHW = 0.0243 in² (15.68 mm²); Chapter 9 Table 5Step 3: Calculate the maximum number of conductorsMaximum number of conductors = 0.814 in² / 0.0243 in² = 33.49
Calculate the smallest standard transformer size (in kVA) needed to power a 3-phase, 4-wire, 120/208 V load drawing 282 A.
The correct answer is
Step 1: Calculate the power of the load.P = √3 * V * I * PFP = √3 * 208 V * 282 A * 1 (assuming PF = 1)P = 101,612.7 VA ≈ 101.61 kVAStep 2: Choose the next standard transformer size.The next standard size up from 101.61 kVA is 112.5 kVA.
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Step 1: Calculate the power of the load.P = √3 * V * I * PFP = √3 * 208 V * 282 A * 1 (assuming PF = 1)P = 101,612.7 VA ≈ 101.61 kVAStep 2: Choose the next standard transformer size.The next standard size up from 101.61 kVA is 112.5 kVA.
A 1-phase commercial store service at 120/240 V, that has an ammeter reading of 170 A and a watt- meter reading of 36,000 W, has __________ power factor
The correct answer is
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What is the minimum service size required for a 1,500 sq ft dwelling with general lighting, 4 kitchen appliances (2,500 VA each), one 14 kW range, 2 small appliance circuits (1,500 VA each), one laundry circuit (1,500 VA), a 9,000 VA electric heating system, an 11,000 VA air conditioning unit, and a 4,000 VA dedicated home office circuit?
The correct answer is
NEC 2017: Table 220.12, 220.42, 220.52, 220.53, 220.55, 220.60NEC 2020: 220.11, 220.14(J), 220.42, 220.52, 220.53, 220.55, 220.60NEC 2023: 220.5(C), 220.41, 220.42, 220.52, 220.53, 220.55, 220.60Step 1: Calculate the individual loads
Step 2: General lighting = 1,500 sq ft × 3 VA = 4,500 VA
Step 3: Kitchen appliances = 4 × 2,500 VA = 10,000 VA
Step 4: Range = 14 kW = 14,000 VA
Step 5: Small appliance circuits = 2 × 1,500 VA = 3,000 VA
Step 6: Laundry circuit = 1,500 VA
Step 7: Electric heating system = 9,000 VA
Step 8: Air conditioning unit = 11,000 VA
Step 9: Home office circuit = 4,000 VA
Step 10: Group the general lighting, small appliances, and laundry loads: 4,500 VA + 3,000 VA + 1,500 VA = 9,000 VA
Step 11: Apply NEC demand factors to the lighting/small appliance/laundry group: first 3,000 VA at 100% plus remainder at 35%: 3,000 VA + (6,000 VA × 0.35) = 5,100 VA
Step 12: Apply 75% demand factor to kitchen appliances: 10,000 VA × 0.75 = 7,500 VA
Step 13: Calculate range demand using 8,000 VA base plus 5% per kW over 12 kW: 8,000 VA × (1 + 0.10) = 8,800 VA
Step 14: For non-coincident loads (heating, AC, home office), select the largest: 11,000 VA
Step 15: Sum all adjusted loads: 5,100 VA + 7,500 VA + 8,800 VA + 11,000 VA = 32,400 VA
Step 16: Convert to amperes at 240V: 32,400 VA ÷ 240V = 135 A
Step 17: According to NEC 240.6, the next standard service size above 135 A is 150 A
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NEC 2017: Table 220.12, 220.42, 220.52, 220.53, 220.55, 220.60NEC 2020: 220.11, 220.14(J), 220.42, 220.52, 220.53, 220.55, 220.60NEC 2023: 220.5(C), 220.41, 220.42, 220.52, 220.53, 220.55, 220.60Step 1: Calculate the individual loads
Step 2: General lighting = 1,500 sq ft × 3 VA = 4,500 VA
Step 3: Kitchen appliances = 4 × 2,500 VA = 10,000 VA
Step 4: Range = 14 kW = 14,000 VA
Step 5: Small appliance circuits = 2 × 1,500 VA = 3,000 VA
Step 6: Laundry circuit = 1,500 VA
Step 7: Electric heating system = 9,000 VA
Step 8: Air conditioning unit = 11,000 VA
Step 9: Home office circuit = 4,000 VA
Step 10: Group the general lighting, small appliances, and laundry loads: 4,500 VA + 3,000 VA + 1,500 VA = 9,000 VA
Step 11: Apply NEC demand factors to the lighting/small appliance/laundry group: first 3,000 VA at 100% plus remainder at 35%: 3,000 VA + (6,000 VA × 0.35) = 5,100 VA
Step 12: Apply 75% demand factor to kitchen appliances: 10,000 VA × 0.75 = 7,500 VA
Step 13: Calculate range demand using 8,000 VA base plus 5% per kW over 12 kW: 8,000 VA × (1 + 0.10) = 8,800 VA
Step 14: For non-coincident loads (heating, AC, home office), select the largest: 11,000 VA
Step 15: Sum all adjusted loads: 5,100 VA + 7,500 VA + 8,800 VA + 11,000 VA = 32,400 VA
Step 16: Convert to amperes at 240V: 32,400 VA ÷ 240V = 135 A
Step 17: According to NEC 240.6, the next standard service size above 135 A is 150 A
