Michigan Journeyman Electrical Practice Exam
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The Michigan Journeyman Electrician exam is a critical step for electricians advancing their careers in the state. To qualify, you must have 8,000 hours of supervised electrical work experience over a minimum of 4 years, verified with notarized employer documentation. Your experience must cover electrical construction, wiring, and equipment maintenance under a licensed Michigan journeyman or master electrician.
The exam itself consists of 80 multiple-choice questions completed in 2.5 hours, with a required 75% passing score. Content areas include grounding and bonding, overcurrent protection, wiring methods, services and feeders, motors, load calculations, special occupancies, lighting, appliances, and electrical theory. You’ll also be tested on Michigan-specific laws and rules, making it essential to master both the 2023 NEC and the Michigan Electrical Code Rules Part 8. Efficient code navigation and strong calculation skills are key to success.
Dakota Prep prepares you with Michigan journeyman practice tests, NEC-aligned drills, and state-specific exam simulations. With over 3,000 electrician exam practice questions, you’ll build speed, accuracy, and confidence in every exam area. Start with a free Michigan journeyman electrician practice exam today and take the first step toward passing your licensing test on the 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)
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.
