NEIEP 400 Final Exam Version 2 - 2026/2027
| Questions & 100% Correct Verified Answers
| Grade A | Elevator Industry Exam
DOMAIN 1 – Advanced Electrical Theory & Power Distribution (20 Q)
Q1. A 30 HP, 460 V, 3-phase motor nameplate shows 35 A FLA, SF 1.15. Per NEC Article
620.61(B), the max inverse-time breaker size for the branch-circuit
short-circuit/ground-fault protective device is:
A. 35 A × 1.25 = 43.75 A → 45 A
B. 35 A × 1.15 × 1.25 = 50.3 A → 55 A
C. 35 A × 2.5 = 87.5 A → 90 A
D. Table 430.250 30 HP = 40 A × 2.5 = 100 A
Correct Answer: C
Rationale: NEC 620.61(B) permits up to 250 % of name-plate FLA for elevator motor
inverse-time breakers. 35 A × 2.5 = 87.5 A → next standard 90 A. Option A uses general
125 % rule; D uses table FLC instead of name-plate.
Q2. A wye-connected 460 V, 3-phase supply feeds a delta-connected motor. The motor’s
phase voltage is approximately:
A. 266 V
B. 460 V
,C. 798 V
D. 230 V
Correct Answer: B
Rationale: In a delta load each winding is placed line-to-line; therefore phase voltage
equals line voltage (460 V). Option A is line-to-neutral voltage of supply (460/√3).
Q3. A solid-state brake coil is rated 180 V DC and 2.5 A. The existing step-down
transformer is 460 V primary, 120 V secondary. Which buck-boost connection will
achieve the required coil voltage?
A. 120 V secondary in subtractive (buck) mode with 60 V boost module
B. 120 V secondary in additive (boost) mode with 60 V module → 180 V
C. Use 230 V secondary only
D. Connect coil across 460 V line
Correct Answer: B
Rationale: Boost connection adds module voltage to secondary: 120 V + 60 V = 180 V
exact rating. Subtractive (A) would yield 60 V; 460 V (D) exceeds coil rating.
Q4. A power-quality meter on a 480 V elevator branch reads 5th harmonic voltage THD =
8 %. The most likely source and its primary effect is:
A. VFD switching → increased RMS current & transformer heating
B. Capacitor bank → resonance only
C. Incandescent lamps → flicker
,D. Encoder wiring → position drift
Correct Answer: A
Rationale: Elevator VFDs generate harmonics (5th, 7th) causing extra copper & core
losses. Capacitors (B) may resonate but are not the source; lamps (C) not significant;
encoder (D) unrelated.
Q5. A 25 A, 460 V, 3-phase motor branch circuit (75 °C copper THHN) requires an
equipment grounding conductor per NEC 250.122. The minimum size is:
A. 12 AWG
B. 10 AWG
C. 8 AWG
D. 14 AWG
Correct Answer: B
Rationale: NEC 250.122 table: OCPD ≤ 60 A → 10 AWG EGC. 25 A circuit typically
protected by 40-60 A breaker; 12 AWG (A) allowed only if ≤ 20 A OCPD.
Q6. A clamp meter on one phase of a balanced 30 HP, 460 V motor reads 38 A under
load. The expected approximate input power is:
A. √3 × 460 V × 38 A = 30.3 kW (ignore power factor)
B. 460 V × 38 A = 17.5 kW
C. 30 HP × 746 W = 22.4 kW (output)
D. 30.3 kW / 0.9 PF = 33.6 kVA
, Correct Answer: A
Rationale: Three-phase power (line values) = √3 × V_L × I_L = 1.732 × 460 × 38 ≈ 30.3
kW (real if PF=1). Option B omits √3; C is mechanical output; D adds PF division
unnecessarily.
Q7. An insulation-resistance test on a 460 V motor winding reads 2.5 MΩ at 20 °C.
According to IEEE 43, the minimum acceptable value for an AC motor is:
A. 1 MΩ
B. 2 MΩ
C. 5 MΩ
D. 100 MΩ
Correct Answer: B
Rationale: IEEE 43 recommends minimum 2 MΩ for most AC machines < 1 kV. Values <
2 MΩ warrant investigation; > 5 MΩ is excellent but 2 MΩ is pass/fail threshold.
Q8. A power-factor meter reads 0.72 lagging for a motor load. To improve PF to 0.95,
the required capacitor kVAR equals:
A. kW × (tan φ₁ – tan φ₂)
B. kVA × (PF₁ – PF₂)
C. kW / PF₁ – kW / PF₂
D. √3 × V × I × sin θ
Correct Answer: A
| Questions & 100% Correct Verified Answers
| Grade A | Elevator Industry Exam
DOMAIN 1 – Advanced Electrical Theory & Power Distribution (20 Q)
Q1. A 30 HP, 460 V, 3-phase motor nameplate shows 35 A FLA, SF 1.15. Per NEC Article
620.61(B), the max inverse-time breaker size for the branch-circuit
short-circuit/ground-fault protective device is:
A. 35 A × 1.25 = 43.75 A → 45 A
B. 35 A × 1.15 × 1.25 = 50.3 A → 55 A
C. 35 A × 2.5 = 87.5 A → 90 A
D. Table 430.250 30 HP = 40 A × 2.5 = 100 A
Correct Answer: C
Rationale: NEC 620.61(B) permits up to 250 % of name-plate FLA for elevator motor
inverse-time breakers. 35 A × 2.5 = 87.5 A → next standard 90 A. Option A uses general
125 % rule; D uses table FLC instead of name-plate.
Q2. A wye-connected 460 V, 3-phase supply feeds a delta-connected motor. The motor’s
phase voltage is approximately:
A. 266 V
B. 460 V
,C. 798 V
D. 230 V
Correct Answer: B
Rationale: In a delta load each winding is placed line-to-line; therefore phase voltage
equals line voltage (460 V). Option A is line-to-neutral voltage of supply (460/√3).
Q3. A solid-state brake coil is rated 180 V DC and 2.5 A. The existing step-down
transformer is 460 V primary, 120 V secondary. Which buck-boost connection will
achieve the required coil voltage?
A. 120 V secondary in subtractive (buck) mode with 60 V boost module
B. 120 V secondary in additive (boost) mode with 60 V module → 180 V
C. Use 230 V secondary only
D. Connect coil across 460 V line
Correct Answer: B
Rationale: Boost connection adds module voltage to secondary: 120 V + 60 V = 180 V
exact rating. Subtractive (A) would yield 60 V; 460 V (D) exceeds coil rating.
Q4. A power-quality meter on a 480 V elevator branch reads 5th harmonic voltage THD =
8 %. The most likely source and its primary effect is:
A. VFD switching → increased RMS current & transformer heating
B. Capacitor bank → resonance only
C. Incandescent lamps → flicker
,D. Encoder wiring → position drift
Correct Answer: A
Rationale: Elevator VFDs generate harmonics (5th, 7th) causing extra copper & core
losses. Capacitors (B) may resonate but are not the source; lamps (C) not significant;
encoder (D) unrelated.
Q5. A 25 A, 460 V, 3-phase motor branch circuit (75 °C copper THHN) requires an
equipment grounding conductor per NEC 250.122. The minimum size is:
A. 12 AWG
B. 10 AWG
C. 8 AWG
D. 14 AWG
Correct Answer: B
Rationale: NEC 250.122 table: OCPD ≤ 60 A → 10 AWG EGC. 25 A circuit typically
protected by 40-60 A breaker; 12 AWG (A) allowed only if ≤ 20 A OCPD.
Q6. A clamp meter on one phase of a balanced 30 HP, 460 V motor reads 38 A under
load. The expected approximate input power is:
A. √3 × 460 V × 38 A = 30.3 kW (ignore power factor)
B. 460 V × 38 A = 17.5 kW
C. 30 HP × 746 W = 22.4 kW (output)
D. 30.3 kW / 0.9 PF = 33.6 kVA
, Correct Answer: A
Rationale: Three-phase power (line values) = √3 × V_L × I_L = 1.732 × 460 × 38 ≈ 30.3
kW (real if PF=1). Option B omits √3; C is mechanical output; D adds PF division
unnecessarily.
Q7. An insulation-resistance test on a 460 V motor winding reads 2.5 MΩ at 20 °C.
According to IEEE 43, the minimum acceptable value for an AC motor is:
A. 1 MΩ
B. 2 MΩ
C. 5 MΩ
D. 100 MΩ
Correct Answer: B
Rationale: IEEE 43 recommends minimum 2 MΩ for most AC machines < 1 kV. Values <
2 MΩ warrant investigation; > 5 MΩ is excellent but 2 MΩ is pass/fail threshold.
Q8. A power-factor meter reads 0.72 lagging for a motor load. To improve PF to 0.95,
the required capacitor kVAR equals:
A. kW × (tan φ₁ – tan φ₂)
B. kVA × (PF₁ – PF₂)
C. kW / PF₁ – kW / PF₂
D. √3 × V × I × sin θ
Correct Answer: A
