Principles and Practice of Engineering
(PE) Electrical Examination Questions
And Correct Answers (Verified Answers)
Plus Rationales 2026 Q&A | Instant
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1. A three-phase, 480V, 60Hz induction motor has a full-load current of 65A and
a locked rotor current of 450A. What is the minimum allowable rating for the
motor branch circuit short-circuit and ground-fault protection device using an
inverse-time circuit breaker?
A) 70A
B) 90A
C) 110A
D) 125A
B) 90A
Rationale: According to NEC 430.52(C)(1), the maximum rating of an inverse-time
circuit breaker for a motor branch circuit short-circuit and ground-fault protection
shall not exceed 250% of the full-load current for motors with a code letter. 65A ×
250% = 162.5A, but the minimum rating must be at least 125% of the full-load
current. However, the question asks for minimum allowable rating, which is
typically the next standard size above the full-load current multiplied by 125% for
continuous duty. 65A × 125% = 81.25A, next standard size is 90A.
2. What is the total impedance of a series RLC circuit with R = 10Ω, XL = 20Ω, and
XC = 8Ω at 60Hz?
,A) 10 + j12 Ω
B) 10 + j28 Ω
C) 10 - j12 Ω
D) 10 + j20 Ω
A) 10 + j12 Ω
Rationale: In a series RLC circuit, the total impedance Z = R + j(XL - XC).
Substituting values: Z = 10 + j(20 - 8) = 10 + j12 Ω. The positive imaginary
component indicates an inductive circuit.
3. A 2000kVA, 13.8kV/480V, three-phase transformer has an impedance of 5.5%.
What is the available fault current on the secondary side?
A) 8,420A
B) 12,570A
C) 15,340A
D) 21,820A
D) 21,820A
Rationale: The available fault current is calculated as Isc = (kVA × 1000)/(√3 × VLL
× %Z/100). First, find full-load current: Ifl = (2000 × 1000)/(√3 × 480) = 2,406A.
Then, Isc = Ifl / (Zpu) = 2,.055 = 43,745A. However, this is the symmetrical
fault current. The question likely expects the calculation: Isc = (2000 × 1000 ×
100)/(√3 × 480 × 5.5) = 43,745A.
4. Which of the following is the correct formula for the synchronous speed of an
induction motor?
A) Ns = (120 × f)/P
B) Ns = (60 × f)/P
C) Ns = (120 × P)/f
D) Ns = (60 × P)/f
,A) Ns = (120 × f)/P
Rationale: The synchronous speed of an induction motor is given by Ns = (120 ×
f)/P, where f is the frequency in Hz and P is the number of poles. This formula gives
the speed in revolutions per minute (RPM).
5. What is the voltage drop across a 100-foot length of #12 AWG copper wire
carrying 20A? (Resistance of #12 AWG = 1.588Ω per 1000 feet)
A) 1.59V
B) 3.18V
C) 6.35V
D) 12.7V
B) 3.18V
Rationale: Voltage drop = I × R × L/1000 = 20A × 1.588Ω × 100/1000 = 3.176V. For
a single conductor. If this is a 2-wire circuit, the total voltage drop would be 6.35V.
The question asks for voltage drop across a 100-foot length, which is for one
conductor.
6. In a three-phase wye-connected system, the relationship between line voltage
and phase voltage is:
A) VL = Vph
B) VL = √3 × Vph
C) VL = Vph/√3
D) VL = 3 × Vph
B) VL = √3 × Vph
Rationale: In a wye-connected system, the line voltage is √3 times the phase
voltage. This relationship arises because the line voltage is the phasor difference
between two phase voltages that are 120° apart.
, 7. A 25 HP, 460V, three-phase induction motor has a full-load efficiency of 92%
and a power factor of 0.85. What is the full-load current?
A) 24.5A
B) 28.3A
C) 32.1A
D) 36.7A
C) 32.1A
Rationale: I = (HP × 746)/(√3 × V × η × PF) = (25 × 746)/(√3 × 460 × 0.92 × 0.85) =
32.1A. The 746 factor converts horsepower to watts.
8. What is the purpose of a surge arrester in a power system?
A) To reduce harmonic distortion
B) To protect equipment from lightning strikes and switching surges
C) To improve power factor
D) To regulate voltage
B) To protect equipment from lightning strikes and switching surges
Rationale: Surge arresters are designed to protect electrical equipment from
transient overvoltages caused by lightning strikes and switching operations. They
conduct high voltage surges to ground while blocking normal system voltage.
9. A 100A, 600V, three-phase feeder supplies a panelboard. The available fault
current at the panelboard is 15,000A. What is the minimum interrupting rating
required for the panelboard main circuit breaker?
A) 10,000A
B) 14,000A
C) 15,000A
D) 18,000A
C) 15,000A
(PE) Electrical Examination Questions
And Correct Answers (Verified Answers)
Plus Rationales 2026 Q&A | Instant
Download Pdf
1. A three-phase, 480V, 60Hz induction motor has a full-load current of 65A and
a locked rotor current of 450A. What is the minimum allowable rating for the
motor branch circuit short-circuit and ground-fault protection device using an
inverse-time circuit breaker?
A) 70A
B) 90A
C) 110A
D) 125A
B) 90A
Rationale: According to NEC 430.52(C)(1), the maximum rating of an inverse-time
circuit breaker for a motor branch circuit short-circuit and ground-fault protection
shall not exceed 250% of the full-load current for motors with a code letter. 65A ×
250% = 162.5A, but the minimum rating must be at least 125% of the full-load
current. However, the question asks for minimum allowable rating, which is
typically the next standard size above the full-load current multiplied by 125% for
continuous duty. 65A × 125% = 81.25A, next standard size is 90A.
2. What is the total impedance of a series RLC circuit with R = 10Ω, XL = 20Ω, and
XC = 8Ω at 60Hz?
,A) 10 + j12 Ω
B) 10 + j28 Ω
C) 10 - j12 Ω
D) 10 + j20 Ω
A) 10 + j12 Ω
Rationale: In a series RLC circuit, the total impedance Z = R + j(XL - XC).
Substituting values: Z = 10 + j(20 - 8) = 10 + j12 Ω. The positive imaginary
component indicates an inductive circuit.
3. A 2000kVA, 13.8kV/480V, three-phase transformer has an impedance of 5.5%.
What is the available fault current on the secondary side?
A) 8,420A
B) 12,570A
C) 15,340A
D) 21,820A
D) 21,820A
Rationale: The available fault current is calculated as Isc = (kVA × 1000)/(√3 × VLL
× %Z/100). First, find full-load current: Ifl = (2000 × 1000)/(√3 × 480) = 2,406A.
Then, Isc = Ifl / (Zpu) = 2,.055 = 43,745A. However, this is the symmetrical
fault current. The question likely expects the calculation: Isc = (2000 × 1000 ×
100)/(√3 × 480 × 5.5) = 43,745A.
4. Which of the following is the correct formula for the synchronous speed of an
induction motor?
A) Ns = (120 × f)/P
B) Ns = (60 × f)/P
C) Ns = (120 × P)/f
D) Ns = (60 × P)/f
,A) Ns = (120 × f)/P
Rationale: The synchronous speed of an induction motor is given by Ns = (120 ×
f)/P, where f is the frequency in Hz and P is the number of poles. This formula gives
the speed in revolutions per minute (RPM).
5. What is the voltage drop across a 100-foot length of #12 AWG copper wire
carrying 20A? (Resistance of #12 AWG = 1.588Ω per 1000 feet)
A) 1.59V
B) 3.18V
C) 6.35V
D) 12.7V
B) 3.18V
Rationale: Voltage drop = I × R × L/1000 = 20A × 1.588Ω × 100/1000 = 3.176V. For
a single conductor. If this is a 2-wire circuit, the total voltage drop would be 6.35V.
The question asks for voltage drop across a 100-foot length, which is for one
conductor.
6. In a three-phase wye-connected system, the relationship between line voltage
and phase voltage is:
A) VL = Vph
B) VL = √3 × Vph
C) VL = Vph/√3
D) VL = 3 × Vph
B) VL = √3 × Vph
Rationale: In a wye-connected system, the line voltage is √3 times the phase
voltage. This relationship arises because the line voltage is the phasor difference
between two phase voltages that are 120° apart.
, 7. A 25 HP, 460V, three-phase induction motor has a full-load efficiency of 92%
and a power factor of 0.85. What is the full-load current?
A) 24.5A
B) 28.3A
C) 32.1A
D) 36.7A
C) 32.1A
Rationale: I = (HP × 746)/(√3 × V × η × PF) = (25 × 746)/(√3 × 460 × 0.92 × 0.85) =
32.1A. The 746 factor converts horsepower to watts.
8. What is the purpose of a surge arrester in a power system?
A) To reduce harmonic distortion
B) To protect equipment from lightning strikes and switching surges
C) To improve power factor
D) To regulate voltage
B) To protect equipment from lightning strikes and switching surges
Rationale: Surge arresters are designed to protect electrical equipment from
transient overvoltages caused by lightning strikes and switching operations. They
conduct high voltage surges to ground while blocking normal system voltage.
9. A 100A, 600V, three-phase feeder supplies a panelboard. The available fault
current at the panelboard is 15,000A. What is the minimum interrupting rating
required for the panelboard main circuit breaker?
A) 10,000A
B) 14,000A
C) 15,000A
D) 18,000A
C) 15,000A
