The journeyman electrical practice exam represents a crucial milestone for aspiring electricians, typically taken after completing 4-6 years of apprenticeship and accumulating 8,000 hours of experience. This NEC practice exam thoroughly evaluates both theoretical knowledge and practical applications through a comprehensive electrical journeyman practice test that ensures candidates are prepared for independent work.
The National Electrical Code (NEC) prep test forms the cornerstone of the examination. The NEC code practice test comprises the largest portion of electrical practice questions, where candidates must demonstrate mastery of code requirements. This includes installation standards, circuit calculations, grounding procedures, and equipment specifications. Success on the electrical journeyman practice exam requires quick navigation of the codebook, as many questions on the NEC practice exam require direct code consultation.
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The correct answer is
NEC 2017: 680.10
NEC 2020: 680.10
NEC 2023: 680.10
That's correct! Way to go
NEC 2017: 680.10
NEC 2020: 680.10
NEC 2023: 680.10
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.
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)
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
The correct answer is
NEC 2017: 690.31(A)
NEC 2020: 690.31(A)
NEC 2023: 690.31(A)
That's correct! Way to go
NEC 2017: 690.31(A)
NEC 2020: 690.31(A)
NEC 2023: 690.31(A)
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