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.
Eight No. 10 AWG XHHW aluminum conductors, when installed in a run of EMT, and protected by circuit breakers have a maximum ampacity rating of __________.
The correct answer is
NEC 2017: Table 310.15(B)(3)(a)NEC 2020: Table 310.15(C)(1)NEC 2023: Table 310.15(C)(1)
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NEC 2017: Table 310.15(B)(3)(a)NEC 2020: Table 310.15(C)(1)NEC 2023: Table 310.15(C)(1)
A receptacle located 1600 mm (5.25 ft) away from a hot tub can be protected by __________
Using an isolating transformer
A Class A ground fault circuit interrupter
An independent bare ground extended back to the system ground
Cannot be installed in this location
The correct answer is
NEC 2017: 680.32NEC 2020: 680.32NEC 2023: 680.32
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NEC 2017: 680.32NEC 2020: 680.32NEC 2023: 680.32
In the motor branch circuit of a 30 hp autotransformer-type controlled motor, separate fuses or an inverse time circuit breaker must be rated or set at not more than ________ percent of the motor full-load current for a general-use switch to be permitted as the disconnecting means.
The correct answer is
NEC 2017: 430.109(D)(3)NEC 2020: 430.109(D)(3)NEC 2023: 430.109(D)(3)Step 1: Verify that 30 hp is within the specified range of over 2 hp up to and including 100 hp. Step 2: Review the three provisions, focusing on provision (3). Step 3: Identify that the code specifies "not more than 150 percent of the motor full-load current" for the rating or setting of separate fuses or an inverse time circuit breaker. Step 4: Confirm that 150% is the correct percentage stated in the code.
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NEC 2017: 430.109(D)(3)NEC 2020: 430.109(D)(3)NEC 2023: 430.109(D)(3)Step 1: Verify that 30 hp is within the specified range of over 2 hp up to and including 100 hp. Step 2: Review the three provisions, focusing on provision (3). Step 3: Identify that the code specifies "not more than 150 percent of the motor full-load current" for the rating or setting of separate fuses or an inverse time circuit breaker. Step 4: Confirm that 150% is the correct percentage stated in the code.
What is the minimum total volt-amperes (VA) that must be included in the calculation for the required small-appliance and laundry circuit loads when determining the demand load for a one-family dwelling?
The correct answer is
NEC 2017: 210.11(C)(1) & (2), 220.52(A) & (B)NEC 2020: 210.11(C)(1) & (2), 220.52(A) & (B)NEC 2023: 210.11(C)(1) & (2), 220.52(A) & (B)2-small appliance circuits @ 1,500 VA each = 3,000 VA1-laundry circuit @ 1,500 VA each = 1,500 VATotal = 4,500 VAPlease note that in ND, it requires 3 small appliance circuits.
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NEC 2017: 210.11(C)(1) & (2), 220.52(A) & (B)NEC 2020: 210.11(C)(1) & (2), 220.52(A) & (B)NEC 2023: 210.11(C)(1) & (2), 220.52(A) & (B)2-small appliance circuits @ 1,500 VA each = 3,000 VA1-laundry circuit @ 1,500 VA each = 1,500 VATotal = 4,500 VAPlease note that in ND, it requires 3 small appliance circuits.
What is the minimum size requirement for a copper grounding electrode conductor connected to the concrete-encased building steel used as the grounding electrode system, if an AC service is supplied with four (4) parallel sets of size 500 kcmil aluminum conductors, and the grounding electrode conductor doesn't connect to another type of electrode that requires a larger size conductor?
The correct answer is
NEC 2017: Table 250.66, 250.66(B)NEC 2020: Table 250.66, 250.66(B)NEC 2023: Table 250.66, 250.66(B)
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NEC 2017: Table 250.66, 250.66(B)NEC 2020: Table 250.66, 250.66(B)NEC 2023: Table 250.66, 250.66(B)
Maximum number of No. 10 AWG XHHW conductors that can be pulled into a run of 1 1/2 (41) EMT is __________.
The correct answer is
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
Which of the following areas in a dwelling are required to have all 120-volt, single-phase, 15- and 20-ampere branch circuits supplying outlets or devices protected by a listed arc-fault circuit interrupter (AFCI)?
The correct answer is
NEC 2017: 210.12(A)NEC 2020: 210.12(A)NEC 2023: 210.12(A)
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NEC 2017: 210.12(A)NEC 2020: 210.12(A)NEC 2023: 210.12(A)
A maximum of _____ size 12/2 AWG with ground Type NM cables can be installed in an outlet box with a total volume of 18 cubic inches if the box contains a duplex receptacle outlet.
The correct answer is
NEC 2017: Table 314.16(B), 314.16(B)(4)NEC 2020: Table 314.16(B), 314.16(B)(4)NEC 2023: Table 314.16(B), 314.16(B)(4)Step 1: Determine box capacityCalculate how many individual #12 wires the box can hold:Box volume ÷ Volume per #12 wire = Total wire capacity18 cu. in. ÷ 2.25 cu. in. = 8 wiresStep 2: Account for deviceSubtract the wires used by the duplex receptacle. 314.16(B)(4) states that a double volume allowance on the largest conductor connected to the deviceTotal wire capacity - Device wires = Remaining wire capacity8 wires - 2 wires = 6 wiresStep 3: Identify cable compositionNote the number of wires in each 12/2 NM cable:12/2 with ground NM cable contains 3 wires where 1 is groundingStep 4: Calculate grounding allowanceUp to four equipment grounding conductors enter a box, a single volume allowance shall be made based on the largest equipment grounding conductor entering the box. 6 wires - 1 wires = 5 wiresStep 5: Calculate conductor allowanceDivide remaining capacity by conductors per cable:Remaining wire capacity ÷ conductors per cable = Number of cables allowed5 wires ÷ 2 wires = 2.5 cables -> 2 cables
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NEC 2017: Table 314.16(B), 314.16(B)(4)NEC 2020: Table 314.16(B), 314.16(B)(4)NEC 2023: Table 314.16(B), 314.16(B)(4)Step 1: Determine box capacityCalculate how many individual #12 wires the box can hold:Box volume ÷ Volume per #12 wire = Total wire capacity18 cu. in. ÷ 2.25 cu. in. = 8 wiresStep 2: Account for deviceSubtract the wires used by the duplex receptacle. 314.16(B)(4) states that a double volume allowance on the largest conductor connected to the deviceTotal wire capacity - Device wires = Remaining wire capacity8 wires - 2 wires = 6 wiresStep 3: Identify cable compositionNote the number of wires in each 12/2 NM cable:12/2 with ground NM cable contains 3 wires where 1 is groundingStep 4: Calculate grounding allowanceUp to four equipment grounding conductors enter a box, a single volume allowance shall be made based on the largest equipment grounding conductor entering the box. 6 wires - 1 wires = 5 wiresStep 5: Calculate conductor allowanceDivide remaining capacity by conductors per cable:Remaining wire capacity ÷ conductors per cable = Number of cables allowed5 wires ÷ 2 wires = 2.5 cables -> 2 cables
To enclose the conductors of a Solar PV System between the point of entry into the building and the first readily accessible disconnecting means, __________ can be used.
The correct answer is
NEC 2017: 690.31(G)NEC 2020: 690.31(B)NEC 2023: 690.31(B)
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NEC 2017: 690.31(G)NEC 2020: 690.31(B)NEC 2023: 690.31(B)
A 20 hp, 208-volt, three-phase, Design B motor in a woodworking shop frequently trips its time-delay (dual-element) fuse during startup. The maximum permitted fuse rating to resolve this issue is _____ amperes.
The correct answer is
NEC 2017: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6NEC 2020: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6NEC 2023: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6Step 1: Find the FLC for a 20 hp, 208-volt, three-phase Design B motor in Table 430.250. The FLC is 59.4 amperes. Step 2: Calculate 225% of FLC: 59.4 × 225% = 133.65 amperes. Step 3: The next lower standard fuse size is 125 amperes, which doesn't exceed 225% of FLC. Step 4: Therefore, the maximum permitted fuse rating is 125 amperes.
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NEC 2017: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6NEC 2020: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6NEC 2023: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6Step 1: Find the FLC for a 20 hp, 208-volt, three-phase Design B motor in Table 430.250. The FLC is 59.4 amperes. Step 2: Calculate 225% of FLC: 59.4 × 225% = 133.65 amperes. Step 3: The next lower standard fuse size is 125 amperes, which doesn't exceed 225% of FLC. Step 4: Therefore, the maximum permitted fuse rating is 125 amperes.
