Electrical Power Systems Exam 2026/2027 |
80 Practice Questions with Answers | CT
Ratios, NFPA 70E, Arc Flash, Grounding,
Three-Phase, Harmonics & Renewable
Energy | Pass PE, FE & Journeyman
Exams
Description:
Master the 2026/2027 Electrical Power Systems exam with 80 original questions and
detailed explanations. Covers CT/PT ratios, NFPA 70E approach boundaries, arc flash,
grounding, three-phase calculations, harmonics, renewable energy, and NEC 2026
updates. Ideal for PE Power, FE Electrical, journeyman, and university finals.
Download the complete 2026/2027 exam paper now and boost your pass rate.
,Electrical Engineering Exam 2026/2027: 80 Questions & Answers
Question 1
A current transformer (CT) has a rated primary current of 600 A and a rated secondary current
of 5 A. If the primary current is 450 A, what is the secondary current under ideal conditions?
A) 3.25 A
B) 3.75 A
C) 4.25 A
D) 4.75 A
Answer: B
Explanation: The CT ratio is calculated as Ipri_rated / Isec_rated = = 120. Under
ideal conditions, Isec = Ipri / (CT ratio) = = 3.75 A. This relationship assumes the
transformer operates within its linear region and no saturation effects are present.
Question 2
A potential transformer (PT) with a ratio of 200:1 is connected to a 13.8 kV primary system.
What secondary voltage would be expected under ideal operating conditions?
A) 60 V
B) 69 V
C) 115 V
D) 120 V
Answer: B
Explanation: Using the ideal PT relationship Vsec = Vpri / (PT ratio), we calculate 13,800 V
/ 200 = 69 V. This represents the standard secondary voltage commonly used for metering
applications in medium voltage systems.
Question 3
A current transformer has a rated secondary current of 5 A and a rated secondary burden
voltage of 10 V. Calculate the CT burden in volt-amperes (VA).
A) 25 VA
B) 50 VA
C) 75 VA
D) 100 VA
Answer: B
Explanation: CT burden is calculated as VA = Vsec × Isec = 10 V × 5 A = 50 VA. This
represents the maximum load that can be connected to the secondary winding while
, maintaining specified accuracy class performance.
Question 4
A meter reads 125.3 A while the true current is 122.8 A. What is the percentage error of the
meter?
A) 1.84%
B) 2.04%
C) 2.24%
D) 2.44%
Answer: B
Explanation: Using the meter error formula %Error = (Measured − True) / True × 100% =
(125.3 − 122.8) / 122.8 × 100% = 2..8 × 100% = 2.036% ≈ 2.04%. The positive error
indicates the meter reads higher than the actual value.
Question 5
Two ammeters measure the same current. Ammeter A reads 98.2 A and Ammeter B reads
101.6 A. What is the percent difference between the two measurements?
A) 2.82%
B) 3.12%
C) 3.40%
D) 3.68%
Answer: C
Explanation: Percent difference is calculated as %Diff = |A−B| / ((A+B)/2) × 100% = |98.2 −
101.6| / ((98.2 + 101.6)/2) × 100% = 3.4 / (199.8/2) × 100% = 3..9 × 100% = 3.403% ≈
3.40%.
Question 6
During an insulation resistance test, a technician applies 1000 V DC and measures a leakage
current of 2.5 μA. What is the insulation resistance in megohms?
A) 250 MΩ
B) 400 MΩ
C) 500 MΩ
D) 600 MΩ
Answer: B
Explanation: Using Ohm's law for insulation testing, R = Vtest / Ileak = 1000 V / (2.5 × 10⁻⁶
A) = 400 × 10⁶ Ω = 400 MΩ. This value would generally be considered acceptable for most
medium voltage equipment.
80 Practice Questions with Answers | CT
Ratios, NFPA 70E, Arc Flash, Grounding,
Three-Phase, Harmonics & Renewable
Energy | Pass PE, FE & Journeyman
Exams
Description:
Master the 2026/2027 Electrical Power Systems exam with 80 original questions and
detailed explanations. Covers CT/PT ratios, NFPA 70E approach boundaries, arc flash,
grounding, three-phase calculations, harmonics, renewable energy, and NEC 2026
updates. Ideal for PE Power, FE Electrical, journeyman, and university finals.
Download the complete 2026/2027 exam paper now and boost your pass rate.
,Electrical Engineering Exam 2026/2027: 80 Questions & Answers
Question 1
A current transformer (CT) has a rated primary current of 600 A and a rated secondary current
of 5 A. If the primary current is 450 A, what is the secondary current under ideal conditions?
A) 3.25 A
B) 3.75 A
C) 4.25 A
D) 4.75 A
Answer: B
Explanation: The CT ratio is calculated as Ipri_rated / Isec_rated = = 120. Under
ideal conditions, Isec = Ipri / (CT ratio) = = 3.75 A. This relationship assumes the
transformer operates within its linear region and no saturation effects are present.
Question 2
A potential transformer (PT) with a ratio of 200:1 is connected to a 13.8 kV primary system.
What secondary voltage would be expected under ideal operating conditions?
A) 60 V
B) 69 V
C) 115 V
D) 120 V
Answer: B
Explanation: Using the ideal PT relationship Vsec = Vpri / (PT ratio), we calculate 13,800 V
/ 200 = 69 V. This represents the standard secondary voltage commonly used for metering
applications in medium voltage systems.
Question 3
A current transformer has a rated secondary current of 5 A and a rated secondary burden
voltage of 10 V. Calculate the CT burden in volt-amperes (VA).
A) 25 VA
B) 50 VA
C) 75 VA
D) 100 VA
Answer: B
Explanation: CT burden is calculated as VA = Vsec × Isec = 10 V × 5 A = 50 VA. This
represents the maximum load that can be connected to the secondary winding while
, maintaining specified accuracy class performance.
Question 4
A meter reads 125.3 A while the true current is 122.8 A. What is the percentage error of the
meter?
A) 1.84%
B) 2.04%
C) 2.24%
D) 2.44%
Answer: B
Explanation: Using the meter error formula %Error = (Measured − True) / True × 100% =
(125.3 − 122.8) / 122.8 × 100% = 2..8 × 100% = 2.036% ≈ 2.04%. The positive error
indicates the meter reads higher than the actual value.
Question 5
Two ammeters measure the same current. Ammeter A reads 98.2 A and Ammeter B reads
101.6 A. What is the percent difference between the two measurements?
A) 2.82%
B) 3.12%
C) 3.40%
D) 3.68%
Answer: C
Explanation: Percent difference is calculated as %Diff = |A−B| / ((A+B)/2) × 100% = |98.2 −
101.6| / ((98.2 + 101.6)/2) × 100% = 3.4 / (199.8/2) × 100% = 3..9 × 100% = 3.403% ≈
3.40%.
Question 6
During an insulation resistance test, a technician applies 1000 V DC and measures a leakage
current of 2.5 μA. What is the insulation resistance in megohms?
A) 250 MΩ
B) 400 MΩ
C) 500 MΩ
D) 600 MΩ
Answer: B
Explanation: Using Ohm's law for insulation testing, R = Vtest / Ileak = 1000 V / (2.5 × 10⁻⁶
A) = 400 × 10⁶ Ω = 400 MΩ. This value would generally be considered acceptable for most
medium voltage equipment.
