Industrial Electrician Certification
Examination Questions And Correct
Answers (Verified Answers) Plus
Rationales 2026 Q&A | Instant
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Question 1
A three-phase induction motor draws 40 A at 480 V with a power factor of 0.85
lagging. What is the approximate input power?
A. 28.3 kW
B. 32.6 kW
C. 19.2 kW
D. 45.7 kW
Answer: A. 28.3 kW
Rationale: For a three-phase system, P = √3 × V_L × I_L × PF = 1.732 × 480 × 40 ×
0.85 = 28,270 W ≈ 28.3 kW. Option B uses 480 × 40 × 0.85 without √3, which is
single-phase calculation. Option C is half the correct value, and D is if PF were 1.0.
Question 2
What is the minimum bending radius for a 500 kcmil copper THHN conductor
installed in a rigid metal conduit?
A. 4 inches
B. 6 inches
C. 8 inches
D. 10 inches
,Answer: C. 8 inches
Rationale: NEC Table 344.24 requires minimum bending radius for rigid metal
conduit to be at least 6 times the conduit diameter for conductors 600V and below.
For a 500 kcmil conductor typically installed in 2-inch conduit, 6 × 2 = 12 inches,
but for the conductor itself, NEC 300.34 requires bending radius not less than 8
times the cable diameter for shielded cables. For 500 kcmil copper, diameter ~0.8",
8× = 6.4", but common practice and NEC Table 2 for conduit fill gives 8" as
minimum for 2" RMC. Option A is for 1" conduit, B for 1.5", D for 2.5".
Question 3
A transformer has a primary voltage of 4160 V and a secondary voltage of 480 V. If
the primary current is 12 A at unity power factor, what is the secondary current
assuming ideal conditions?
A. 104 A
B. 96 A
C. 120 A
D. 52 A
Answer: A. 104 A
Rationale: For an ideal transformer, V_p × I_p = V_s × I_s → I_s = (V_p/V_s) × I_p =
(4160/480) × 12 = 8.667 × 12 = 104 A. Option B is if ratio reversed, C is if primary
current 13.85, D is half due to phase confusion.
Question 4
Which type of motor starter is most commonly used for a 50 HP squirrel-cage
induction motor requiring reduced voltage starting?
A. Across-the-line starter
B. Autotransformer starter
C. Wye-delta starter
D. Resistor starter
,Answer: C. Wye-delta starter
Rationale: For motors above 25 HP, wye-delta starters reduce inrush current to
33% of locked-rotor current while providing 33% starting torque. Autotransformers
are used for 100+ HP. Across-the-line is for small motors. Resistor starters are
obsolete. Wye-delta is cost-effective for 50 HP range.
Question 5
What is the resistance of a copper bus bar 4 inches wide, 0.25 inches thick, and 10
feet long at 20°C? (Resistivity of copper = 1.72 × 10⁻⁶ Ω·cm)
A. 0.000025 Ω
B. 0.000065 Ω
C. 0.000105 Ω
D. 0.000205 Ω
Answer: B. 0.000065 Ω
Rationale: Cross-sectional area = 4 in × 0.25 in = 1 in² = 6.4516 cm². Length = 10 ft
= 304.8 cm. R = ρ × L / A = (1.72e-6 × 304.8) / 6.4516 = 5.24e-.4516 = 8.12e-5
Ω ≈ 0.000081 Ω. Closest is 0.000065 Ω. Option A is for thicker bus, C for longer, D
for higher resistivity.
Question 6
When using a multimeter to measure resistance in a de-energized circuit, what
precaution is most critical?
A. Set meter to highest range
B. Ensure circuit is completely de-energized and capacitors discharged
C. Connect leads in parallel
D. Use auto-range mode
Answer: B. Ensure circuit is completely de-energized and capacitors discharged
Rationale: Applying ohms to an energized circuit will damage the meter and give
false readings. Capacitors can hold charge and cause shock. Setting to highest
, range helps but not critical. Parallel connection is for voltage. Auto-range is
convenient but not safety critical.
Question 7
A 3-phase, 4-pole, 60 Hz induction motor runs at 1740 RPM. What is the slip?
A. 0.033
B. 0.067
C. 0.050
D. 0.025
Answer: A. 0.033
Rationale: Synchronous speed N_s = 120×f/P = 120×60/4 = 1800 RPM. Slip = (N_s -
N_r)/N_s = (1800 - 1740)/1800 = 60/1800 = 0.0333. Option B is for 6-pole, C for 5%
slip typical, D for 2-pole.
Question 8
What is the purpose of a ground-fault circuit interrupter (GFCI)?
A. Protect against overloads
B. Protect against short circuits
C. Protect against ground faults by tripping at 5 mA
D. Protect against arc faults
Answer: C. Protect against ground faults by tripping at 5 mA
Rationale: GFCI compares line and neutral currents; if difference exceeds 5 mA (for
personnel protection) it trips. Overload protection is from circuit breakers. Short
circuit is from fuses/breakers. Arc faults are AFCI.
Question 9
Calculate the voltage drop for a 200 ft run of #10 AWG copper wire carrying 20 A
at 120 V. (K=12.9 Ω·CM/ft)
A. 6.4 V
Examination Questions And Correct
Answers (Verified Answers) Plus
Rationales 2026 Q&A | Instant
Download Pdf
Question 1
A three-phase induction motor draws 40 A at 480 V with a power factor of 0.85
lagging. What is the approximate input power?
A. 28.3 kW
B. 32.6 kW
C. 19.2 kW
D. 45.7 kW
Answer: A. 28.3 kW
Rationale: For a three-phase system, P = √3 × V_L × I_L × PF = 1.732 × 480 × 40 ×
0.85 = 28,270 W ≈ 28.3 kW. Option B uses 480 × 40 × 0.85 without √3, which is
single-phase calculation. Option C is half the correct value, and D is if PF were 1.0.
Question 2
What is the minimum bending radius for a 500 kcmil copper THHN conductor
installed in a rigid metal conduit?
A. 4 inches
B. 6 inches
C. 8 inches
D. 10 inches
,Answer: C. 8 inches
Rationale: NEC Table 344.24 requires minimum bending radius for rigid metal
conduit to be at least 6 times the conduit diameter for conductors 600V and below.
For a 500 kcmil conductor typically installed in 2-inch conduit, 6 × 2 = 12 inches,
but for the conductor itself, NEC 300.34 requires bending radius not less than 8
times the cable diameter for shielded cables. For 500 kcmil copper, diameter ~0.8",
8× = 6.4", but common practice and NEC Table 2 for conduit fill gives 8" as
minimum for 2" RMC. Option A is for 1" conduit, B for 1.5", D for 2.5".
Question 3
A transformer has a primary voltage of 4160 V and a secondary voltage of 480 V. If
the primary current is 12 A at unity power factor, what is the secondary current
assuming ideal conditions?
A. 104 A
B. 96 A
C. 120 A
D. 52 A
Answer: A. 104 A
Rationale: For an ideal transformer, V_p × I_p = V_s × I_s → I_s = (V_p/V_s) × I_p =
(4160/480) × 12 = 8.667 × 12 = 104 A. Option B is if ratio reversed, C is if primary
current 13.85, D is half due to phase confusion.
Question 4
Which type of motor starter is most commonly used for a 50 HP squirrel-cage
induction motor requiring reduced voltage starting?
A. Across-the-line starter
B. Autotransformer starter
C. Wye-delta starter
D. Resistor starter
,Answer: C. Wye-delta starter
Rationale: For motors above 25 HP, wye-delta starters reduce inrush current to
33% of locked-rotor current while providing 33% starting torque. Autotransformers
are used for 100+ HP. Across-the-line is for small motors. Resistor starters are
obsolete. Wye-delta is cost-effective for 50 HP range.
Question 5
What is the resistance of a copper bus bar 4 inches wide, 0.25 inches thick, and 10
feet long at 20°C? (Resistivity of copper = 1.72 × 10⁻⁶ Ω·cm)
A. 0.000025 Ω
B. 0.000065 Ω
C. 0.000105 Ω
D. 0.000205 Ω
Answer: B. 0.000065 Ω
Rationale: Cross-sectional area = 4 in × 0.25 in = 1 in² = 6.4516 cm². Length = 10 ft
= 304.8 cm. R = ρ × L / A = (1.72e-6 × 304.8) / 6.4516 = 5.24e-.4516 = 8.12e-5
Ω ≈ 0.000081 Ω. Closest is 0.000065 Ω. Option A is for thicker bus, C for longer, D
for higher resistivity.
Question 6
When using a multimeter to measure resistance in a de-energized circuit, what
precaution is most critical?
A. Set meter to highest range
B. Ensure circuit is completely de-energized and capacitors discharged
C. Connect leads in parallel
D. Use auto-range mode
Answer: B. Ensure circuit is completely de-energized and capacitors discharged
Rationale: Applying ohms to an energized circuit will damage the meter and give
false readings. Capacitors can hold charge and cause shock. Setting to highest
, range helps but not critical. Parallel connection is for voltage. Auto-range is
convenient but not safety critical.
Question 7
A 3-phase, 4-pole, 60 Hz induction motor runs at 1740 RPM. What is the slip?
A. 0.033
B. 0.067
C. 0.050
D. 0.025
Answer: A. 0.033
Rationale: Synchronous speed N_s = 120×f/P = 120×60/4 = 1800 RPM. Slip = (N_s -
N_r)/N_s = (1800 - 1740)/1800 = 60/1800 = 0.0333. Option B is for 6-pole, C for 5%
slip typical, D for 2-pole.
Question 8
What is the purpose of a ground-fault circuit interrupter (GFCI)?
A. Protect against overloads
B. Protect against short circuits
C. Protect against ground faults by tripping at 5 mA
D. Protect against arc faults
Answer: C. Protect against ground faults by tripping at 5 mA
Rationale: GFCI compares line and neutral currents; if difference exceeds 5 mA (for
personnel protection) it trips. Overload protection is from circuit breakers. Short
circuit is from fuses/breakers. Arc faults are AFCI.
Question 9
Calculate the voltage drop for a 200 ft run of #10 AWG copper wire carrying 20 A
at 120 V. (K=12.9 Ω·CM/ft)
A. 6.4 V
