Hawaii Journeyman Electrician Exam Prep 2026 | 200
Practice Questions with Answers & Rationales | Electrical
Code Study Guide
SECTION A: GENERAL ELECTRICAL THEORY, OHM’S LAW, AND SAFETY
1. A 240V single-phase circuit supplies a resistive load of 12 ohms. What is the current?
A. 10 A
B. 20 A
C. 28.8 A
D. 2,880 A
Answer: B
Explanation: I = V/R = 240V / 12Ω = 20 A.
2. Three 15Ω resistors are connected in parallel. What is the total resistance?
A. 45 Ω
B. 15 Ω
C. 5 Ω
D. 1.5 Ω
Answer: C
Explanation: 1/Rt = 1/15 + 1/15 + 1/15 = 3/15 = 1/5 → Rt = 5 Ω.
3. A 5 hp, 230V single-phase motor has a nameplate full-load current of 28 A. What is the minimum
branch circuit conductor ampacity before applying any adjustment factors?
A. 28 A
B. 35 A
C. 42 A
D. 56 A
Answer: B
,Explanation: NEC 430.22(A): Conductors for a single motor must have ampacity at least 125% of
FLC = 28 × 1.25 = 35 A.
4. A 120V branch circuit feeds a continuous load of 1,200 VA and a non‑continuous load of 800
VA. What is the minimum branch circuit rating?
A. 15 A
B. 20 A
C. 25 A
D. 30 A
Answer: B
Explanation: Continuous load = 1,200/120 = 10 A × 1.25 = 12.5 A; plus non‑continuous = 800/120
= 6.67 A; total = 19.17 A → next standard size 20 A.
5. A worker receives an electric shock that causes the heart to beat chaotically (ventricular
fibrillation). What current level typically causes this?
A. 1–5 mA
B. 10–20 mA
C. 100–200 mA
D. 2–5 A
Answer: C
Explanation: Ventricular fibrillation occurs at 100–200 mA across the heart. Lower currents may
cause “let‑go” threshold.
6. What is the minimum working clearance in front of a 480V panelboard with exposed live parts
on one side and grounded parts on the other?
A. 3 ft
B. 3.5 ft
C. 4 ft
D. 5 ft
Answer: B
,Explanation: NEC 110.26(A)(1) Condition 2: 151–600V requires 3.5 ft clearance when exposed
live parts on one side and grounded on the other.
7. A 240V feeder supplies a continuous load of 80 A and a non‑continuous load of 30 A. What is
the minimum conductor ampacity?
A. 110 A
B. 120 A
C. 130 A
D. 140 A
Answer: C
Explanation: NEC 215.2(A)(1): (80 × 1.25) + 30 = 100 + 30 = 130 A.
8. A 3‑phase, 208V motor draws 25 A at 0.85 power factor. What is the real power in kW?
A. 5.3 kW
B. 7.7 kW
C. 9.2 kW
D. 12.4 kW
Answer: B
Explanation: P(kW) = (V × I × √3 × PF)/1000 = (208 × 25 × 1.732 × 0.85)/1000 = 7,660/1000 =
7.66 kW.
9. A 10 AWG copper conductor with THHN insulation has a 90°C ampacity of 40 A. If the ambient
temperature is 45°C, what is the corrected ampacity?
A. 32.8 A
B. 34.8 A
C. 36.4 A
D. 40 A
Answer: B
Explanation: NEC Table 310.15(B)(2)(a): Correction factor for 90°C conductor at 45°C = 0.87. 40
× 0.87 = 34.8 A.
, 10. A 120V, 15 A general‑use receptacle is installed in a dwelling bedroom. What is the maximum
load permitted on that receptacle if other loads are on the same branch circuit?
A. 12 A
B. 15 A
C. 20 A
D. 10 A
Answer: A
Explanation: NEC 210.23(A)(1): A 15 A branch circuit shall not supply a total load exceeding 12
A for lighting or cord‑and‑plug appliances.
11. Three resistors of 10 Ω, 20 Ω, and 30 Ω are connected in series across 120V. What is the voltage
drop across the 20 Ω resistor?
A. 20 V
B. 40 V
C. 60 V
D. 80 V
Answer: B
Explanation: Total R = 60 Ω; I = 120/60 = 2 A; V = 2 × 20 = 40 V.
12. A 25 kVA, 480V‑120/208V three‑phase transformer has a primary overcurrent device set at 40
A. Is this allowed for a supervised location?
A. Yes, because primary protection can be up to 125% of primary FLC
B. No, because primary protection must be exactly 100% of FLC
C. Yes, but only if secondary protection is also provided
D. No, because 40 A is too low
Answer: A
Explanation: Primary FLC = 25,000/(480×1.732) = 30.1 A; 125% = 37.6 A, next standard size 40
A is permitted.
Practice Questions with Answers & Rationales | Electrical
Code Study Guide
SECTION A: GENERAL ELECTRICAL THEORY, OHM’S LAW, AND SAFETY
1. A 240V single-phase circuit supplies a resistive load of 12 ohms. What is the current?
A. 10 A
B. 20 A
C. 28.8 A
D. 2,880 A
Answer: B
Explanation: I = V/R = 240V / 12Ω = 20 A.
2. Three 15Ω resistors are connected in parallel. What is the total resistance?
A. 45 Ω
B. 15 Ω
C. 5 Ω
D. 1.5 Ω
Answer: C
Explanation: 1/Rt = 1/15 + 1/15 + 1/15 = 3/15 = 1/5 → Rt = 5 Ω.
3. A 5 hp, 230V single-phase motor has a nameplate full-load current of 28 A. What is the minimum
branch circuit conductor ampacity before applying any adjustment factors?
A. 28 A
B. 35 A
C. 42 A
D. 56 A
Answer: B
,Explanation: NEC 430.22(A): Conductors for a single motor must have ampacity at least 125% of
FLC = 28 × 1.25 = 35 A.
4. A 120V branch circuit feeds a continuous load of 1,200 VA and a non‑continuous load of 800
VA. What is the minimum branch circuit rating?
A. 15 A
B. 20 A
C. 25 A
D. 30 A
Answer: B
Explanation: Continuous load = 1,200/120 = 10 A × 1.25 = 12.5 A; plus non‑continuous = 800/120
= 6.67 A; total = 19.17 A → next standard size 20 A.
5. A worker receives an electric shock that causes the heart to beat chaotically (ventricular
fibrillation). What current level typically causes this?
A. 1–5 mA
B. 10–20 mA
C. 100–200 mA
D. 2–5 A
Answer: C
Explanation: Ventricular fibrillation occurs at 100–200 mA across the heart. Lower currents may
cause “let‑go” threshold.
6. What is the minimum working clearance in front of a 480V panelboard with exposed live parts
on one side and grounded parts on the other?
A. 3 ft
B. 3.5 ft
C. 4 ft
D. 5 ft
Answer: B
,Explanation: NEC 110.26(A)(1) Condition 2: 151–600V requires 3.5 ft clearance when exposed
live parts on one side and grounded on the other.
7. A 240V feeder supplies a continuous load of 80 A and a non‑continuous load of 30 A. What is
the minimum conductor ampacity?
A. 110 A
B. 120 A
C. 130 A
D. 140 A
Answer: C
Explanation: NEC 215.2(A)(1): (80 × 1.25) + 30 = 100 + 30 = 130 A.
8. A 3‑phase, 208V motor draws 25 A at 0.85 power factor. What is the real power in kW?
A. 5.3 kW
B. 7.7 kW
C. 9.2 kW
D. 12.4 kW
Answer: B
Explanation: P(kW) = (V × I × √3 × PF)/1000 = (208 × 25 × 1.732 × 0.85)/1000 = 7,660/1000 =
7.66 kW.
9. A 10 AWG copper conductor with THHN insulation has a 90°C ampacity of 40 A. If the ambient
temperature is 45°C, what is the corrected ampacity?
A. 32.8 A
B. 34.8 A
C. 36.4 A
D. 40 A
Answer: B
Explanation: NEC Table 310.15(B)(2)(a): Correction factor for 90°C conductor at 45°C = 0.87. 40
× 0.87 = 34.8 A.
, 10. A 120V, 15 A general‑use receptacle is installed in a dwelling bedroom. What is the maximum
load permitted on that receptacle if other loads are on the same branch circuit?
A. 12 A
B. 15 A
C. 20 A
D. 10 A
Answer: A
Explanation: NEC 210.23(A)(1): A 15 A branch circuit shall not supply a total load exceeding 12
A for lighting or cord‑and‑plug appliances.
11. Three resistors of 10 Ω, 20 Ω, and 30 Ω are connected in series across 120V. What is the voltage
drop across the 20 Ω resistor?
A. 20 V
B. 40 V
C. 60 V
D. 80 V
Answer: B
Explanation: Total R = 60 Ω; I = 120/60 = 2 A; V = 2 × 20 = 40 V.
12. A 25 kVA, 480V‑120/208V three‑phase transformer has a primary overcurrent device set at 40
A. Is this allowed for a supervised location?
A. Yes, because primary protection can be up to 125% of primary FLC
B. No, because primary protection must be exactly 100% of FLC
C. Yes, but only if secondary protection is also provided
D. No, because 40 A is too low
Answer: A
Explanation: Primary FLC = 25,000/(480×1.732) = 30.1 A; 125% = 37.6 A, next standard size 40
A is permitted.
