USVI Journeyman Electrician Licensing Exam 2026 – 190
Comprehensive Practice Questions with Answers & Detailed
Rationales, 2023 NEC (NFPA 70), Electrical Wiring, Circuit
Calculations, Grounding and Bonding, Overcurrent Protection,
Motors, Electrical Equipment, Safety Standards, Troubleshooting
Scenarios & USVI DLCA Licensing Exam Preparation
DOMAIN I: ELECTRICAL THEORY & FUNDAMENTALS (Questions 1–20)
Question 1
A journeyman electrician is troubleshooting a 120V single-phase circuit with a resistive load.
The measured current is 12 amperes. What is the resistance of the load?
A. 0.10 ohms
B. 10 ohms
C. 1,440 ohms
D. 120 ohms
Correct Answer: B
Explanation: Using Ohm's Law (R = V ÷ I), resistance equals 120V ÷ 12A = 10 ohms.
Question 2
An electrical apprentice asks why AC is used for power distribution instead of DC. Which of
the following is the primary advantage of AC?
A. AC produces more heat per ampere
B. AC voltage can be easily transformed
,C. AC requires larger conductors than DC
D. AC flows only in one direction
Correct Answer: B
Explanation: AC voltage can be stepped up or down using transformers, enabling efficient
long-distance transmission.
Question 3
A circuit has a power factor of 0.80. If the apparent power is 15 kVA, what is the true power
consumed by the load?
A. 12 kW
B. 18.75 kW
C. 15 kW
D. 0.053 kW
Correct Answer: A
Explanation: True power = Apparent power × Power factor = 15 kVA × 0.80 = 12 kW.
Question 4
In a parallel circuit with three branches, each branch has a resistance of 30 ohms. What is
the total resistance of the circuit?
A. 90 ohms
B. 30 ohms
C. 10 ohms
D. 15 ohms
,Correct Answer: C
Explanation: For parallel resistances of equal value: R_total = R ÷ n = 30 ÷ 3 = 10 ohms.
Question 5
A 240V single-phase feeder supplies a load of 48 amperes. What size copper conductor
(THHN, 75°C) is required at minimum, disregarding voltage drop?
A. 8 AWG
B. 6 AWG
C. 4 AWG
D. 10 AWG
Correct Answer: B
Explanation: NEC Table 310.16: 6 AWG copper THHN at 75°C has an ampacity of 65A,
sufficient for 48A.
Question 6
Which of the following best defines the difference between a "short circuit" and a "ground
fault"?
A. A short circuit involves a hot-to-neutral fault; a ground fault involves a hot-to-ground fault
B. A short circuit is a high-resistance path; a ground fault is a low-resistance path
C. A short circuit only occurs in DC systems
D. A ground fault is always less dangerous than a short circuit
Correct Answer: A
, Explanation: A short circuit is an unintended hot-to-neutral or hot-to-hot connection; a
ground fault is unintended contact between an ungrounded conductor and ground.
Question 7
A 3-phase, 480V motor draws 25 amperes per phase at a power factor of 0.85. What is the
input power in kW?
A. 12.24 kW
B. 17.65 kW
C. 21.20 kW
D. 10.40 kW
Correct Answer: B
Explanation: P = √3 × V × I × PF = 1.732 × 480 × 25 × 0.85 = 17,650W or 17.65 kW.
Question 8
What is the primary reason for grounding the neutral conductor at the service point?
A. To reduce conductor ampacity requirements
B. To stabilize line-to-neutral voltage and provide a fault current path
C. To increase the available fault current
D. To eliminate the need for equipment grounding conductors
Correct Answer: B
Explanation: Grounding the neutral stabilizes voltage to ground and creates a low-
impedance path for fault current to facilitate overcurrent device operation.
Comprehensive Practice Questions with Answers & Detailed
Rationales, 2023 NEC (NFPA 70), Electrical Wiring, Circuit
Calculations, Grounding and Bonding, Overcurrent Protection,
Motors, Electrical Equipment, Safety Standards, Troubleshooting
Scenarios & USVI DLCA Licensing Exam Preparation
DOMAIN I: ELECTRICAL THEORY & FUNDAMENTALS (Questions 1–20)
Question 1
A journeyman electrician is troubleshooting a 120V single-phase circuit with a resistive load.
The measured current is 12 amperes. What is the resistance of the load?
A. 0.10 ohms
B. 10 ohms
C. 1,440 ohms
D. 120 ohms
Correct Answer: B
Explanation: Using Ohm's Law (R = V ÷ I), resistance equals 120V ÷ 12A = 10 ohms.
Question 2
An electrical apprentice asks why AC is used for power distribution instead of DC. Which of
the following is the primary advantage of AC?
A. AC produces more heat per ampere
B. AC voltage can be easily transformed
,C. AC requires larger conductors than DC
D. AC flows only in one direction
Correct Answer: B
Explanation: AC voltage can be stepped up or down using transformers, enabling efficient
long-distance transmission.
Question 3
A circuit has a power factor of 0.80. If the apparent power is 15 kVA, what is the true power
consumed by the load?
A. 12 kW
B. 18.75 kW
C. 15 kW
D. 0.053 kW
Correct Answer: A
Explanation: True power = Apparent power × Power factor = 15 kVA × 0.80 = 12 kW.
Question 4
In a parallel circuit with three branches, each branch has a resistance of 30 ohms. What is
the total resistance of the circuit?
A. 90 ohms
B. 30 ohms
C. 10 ohms
D. 15 ohms
,Correct Answer: C
Explanation: For parallel resistances of equal value: R_total = R ÷ n = 30 ÷ 3 = 10 ohms.
Question 5
A 240V single-phase feeder supplies a load of 48 amperes. What size copper conductor
(THHN, 75°C) is required at minimum, disregarding voltage drop?
A. 8 AWG
B. 6 AWG
C. 4 AWG
D. 10 AWG
Correct Answer: B
Explanation: NEC Table 310.16: 6 AWG copper THHN at 75°C has an ampacity of 65A,
sufficient for 48A.
Question 6
Which of the following best defines the difference between a "short circuit" and a "ground
fault"?
A. A short circuit involves a hot-to-neutral fault; a ground fault involves a hot-to-ground fault
B. A short circuit is a high-resistance path; a ground fault is a low-resistance path
C. A short circuit only occurs in DC systems
D. A ground fault is always less dangerous than a short circuit
Correct Answer: A
, Explanation: A short circuit is an unintended hot-to-neutral or hot-to-hot connection; a
ground fault is unintended contact between an ungrounded conductor and ground.
Question 7
A 3-phase, 480V motor draws 25 amperes per phase at a power factor of 0.85. What is the
input power in kW?
A. 12.24 kW
B. 17.65 kW
C. 21.20 kW
D. 10.40 kW
Correct Answer: B
Explanation: P = √3 × V × I × PF = 1.732 × 480 × 25 × 0.85 = 17,650W or 17.65 kW.
Question 8
What is the primary reason for grounding the neutral conductor at the service point?
A. To reduce conductor ampacity requirements
B. To stabilize line-to-neutral voltage and provide a fault current path
C. To increase the available fault current
D. To eliminate the need for equipment grounding conductors
Correct Answer: B
Explanation: Grounding the neutral stabilizes voltage to ground and creates a low-
impedance path for fault current to facilitate overcurrent device operation.
