California Journeyman Electrical Practice Exam
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The California General Electrician certification exam is a critical step for electricians ready to advance their careers. This closed-book test includes 100 multiple-choice questions over 4 hours and 30 minutes, with a required 70% passing score. Eligibility comes through work experience (8,000 hours), completion of an apprenticeship program, or other state-approved pathways. Passing the exam earns you the official California Electrical Certification Card, valid for three years.
The exam content is heavily based on the National Electrical Code (NEC) and measures both theoretical knowledge and real-world application. Test topics include electrical installation requirements, conductor and conduit sizing, load calculations, equipment grounding and bonding, overcurrent protection, and safety regulations. Many questions require fast NEC codebook navigation, so learning to reference articles efficiently is as important as memorizing formulas.
Dakota Prep helps you prepare with California journeyman practice tests, NEC code questions, and timed exam simulations. With over 3,000 electrician exam practice questions built to mirror the state test, you’ll gain the confidence to handle calculation problems, code lookups, and tricky wording. Start today with a free California journeyman electrician practice exam and set yourself up to pass on your first attempt.
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)
That's correct! Way to go
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: 680.32
NEC 2020: 680.32
NEC 2023: 680.32
That's correct! Way to go
NEC 2017: 680.32
NEC 2020: 680.32
NEC 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.
That's correct! Way to go
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 VA
That's correct! Way to go
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 VA
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)
That's correct! Way to go
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 4
NEC 2020: Chapter 9, Table 4
NEC 2023: Chapter 9, Table 4
Step 1: Find the maximum area for conductors within the conduit
Maximum percent of area of Trade size 1 1/2 (41) EMT conduit that can be used for conductor fill = 40 percent; Chapter 9 Table 1
Cross-sectional Area of 41 EMT conduit at 40 percent = 0.814 in² (526 mm²); Chapter 9 Article 358
Step 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 5
Step 3: Calculate the maximum number of conductors
Maximum number of conductors = 0.814 in² / 0.0243 in² = 33.49
That's correct! Way to go
NEC 2017: Chapter 9, Table 4
NEC 2020: Chapter 9, Table 4
NEC 2023: Chapter 9, Table 4
Step 1: Find the maximum area for conductors within the conduit
Maximum percent of area of Trade size 1 1/2 (41) EMT conduit that can be used for conductor fill = 40 percent; Chapter 9 Table 1
Cross-sectional Area of 41 EMT conduit at 40 percent = 0.814 in² (526 mm²); Chapter 9 Article 358
Step 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 5
Step 3: Calculate the maximum number of conductors
Maximum 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)
That's correct! Way to go
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 capacity
Calculate how many individual #12 wires the box can hold:
Box volume ÷ Volume per #12 wire = Total wire capacity
18 cu. in. ÷ 2.25 cu. in. = 8 wires
Step 2: Account for device
Subtract the wires used by the duplex receptacle.
314.16(B)(4) states that a double volume allowance on the largest conductor connected to the device
Total wire capacity - Device wires = Remaining wire capacity
8 wires - 2 wires = 6 wires
Step 3: Identify cable composition
Note the number of wires in each 12/2 NM cable:
12/2 with ground NM cable contains 3 wires where 1 is grounding
Step 4: Calculate grounding allowance
Up 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 wires
Step 5: Calculate conductor allowance
Divide remaining capacity by conductors per cable:
Remaining wire capacity ÷ conductors per cable = Number of cables allowed
5 wires ÷ 2 wires = 2.5 cables -> 2 cables
That's correct! Way to go
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 capacity
Calculate how many individual #12 wires the box can hold:
Box volume ÷ Volume per #12 wire = Total wire capacity
18 cu. in. ÷ 2.25 cu. in. = 8 wires
Step 2: Account for device
Subtract the wires used by the duplex receptacle.
314.16(B)(4) states that a double volume allowance on the largest conductor connected to the device
Total wire capacity - Device wires = Remaining wire capacity
8 wires - 2 wires = 6 wires
Step 3: Identify cable composition
Note the number of wires in each 12/2 NM cable:
12/2 with ground NM cable contains 3 wires where 1 is grounding
Step 4: Calculate grounding allowance
Up 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 wires
Step 5: Calculate conductor allowance
Divide remaining capacity by conductors per cable:
Remaining wire capacity ÷ conductors per cable = Number of cables allowed
5 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)
That's correct! Way to go
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), 240.6
NEC 2020: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6
NEC 2023: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6
Step 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.
That's correct! Way to go
NEC 2017: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6
NEC 2020: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6
NEC 2023: Table 430.250, 430.52(C)(1), 430.52(C)(1)(a)(2), 240.6
Step 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.
