Free PE Power Practice Test
Practice test: 80 questions
Here are 80 original practice questions for the NCEES PE Electrical and Computer: Power exam, matched to the October 2025 specification and to the code editions the exam supplies — including the 2020 NEC, not 2023. Every question has a worked solution, there's no sign-up, and these are unofficial Castleport questions, so a score here is practice feedback, not an NCEES score.
Choose an answer, then check it or reveal the solution. These are unofficial practice questions.
Question 1 of 80: Transformer secondary fault
PEP-01 · Transmission and Distribution Analysis · Fault current analysis · Choose one answer
A 1,500 kVA, 480Y/277 V transformer has 5.75% impedance. Assume an infinite (zero-impedance) utility source and ignore motor contribution. What is the bolted three-phase fault current at the secondary terminals?
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Answer: A. 31.4 kA
Why: Full-load current = 1,500,000/(√3 × 480) = 1,804 A. With an infinite source, I_sc = I_FL / Z_pu = 1,804/0.0575 = 31,378 A ≈ 31.4 kA. This is the conservative upper bound; real source impedance lowers it.
Why the other choices miss: 1.80 kA is the full-load current. 18.1 kA uses 3 instead of √3. 54.3 kA leaves out √3 entirely.
Source: Southern Illinois University ET 332b, Lesson 10: transformer regulation, efficiency, short-circuit current — Slide 3 (regulation); slides 11–12 (rated current, % impedance); slides 26–27 (Isc); slide 29 (efficiency)
Question 2 of 80: Continuous load OCPD
PEP-02 · Electrical Safety · Wiring methods and installations · Choose one answer
A branch circuit serves 36 A of continuous load and 8 A of noncontinuous load. The breaker is a standard (not 100%-rated) device. What is the smallest standard breaker rating permitted by NEC 210.20(A)?
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Answer: B. 60 A
Why: Required rating = 125% of continuous + 100% of noncontinuous = 1.25 × 36 + 8 = 53 A. The next standard rating at or above 53 A is 60 A.
Why the other choices miss: 44 A forgets the 125% factor. 50 A is below the 53 A minimum. 55 A isn't a standard breaker rating.
Source: EC&M: Code Q&A — Branch Circuit Overcurrent Protection (Holt) — 210.20(A); EEPower: Protecting Motor Branch Circuits from Short Circuits and Ground Faults (Mari, 2024) — Table 430.52 percentages; 240.6 10 A rating added in 2023
Question 3 of 80: Line current from kVA
PEP-03 · Circuit Analysis · Three-phase circuits · Enter a number
A balanced 150 kVA load is served at 480 V, three-phase. What is the line current? Enter amperes.
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Answer: 180.4 A (accepted range 178.6–182.2 A)
Why: I = S / (√3 × V_LL) = 150,000 / (√3 × 480) = 180.4 A.
Common mistakes: Dividing by 480 alone gives 312.5 A (single-phase formula). Dividing by 3 × 480 gives 104.2 A.
Source: MIT OCW 6.061 course notes, Ch. 3: Polyphase networks — Line-line voltages section; Sec. 6 three-phase transformers; Sec. 8.2
Question 4 of 80: Motor branch short-circuit protection
PEP-04 · Protection · Overcurrent protection · Choose one answer
A squirrel-cage motor has a full-load current of 34 A from NEC Table 430.250. It is protected by an inverse-time breaker for branch-circuit short-circuit and ground-fault protection. Using the Table 430.52 percentage and the permission to go to the next higher standard rating, what is the maximum breaker rating?
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Answer: C. 90 A
Why: For squirrel-cage motors, Table 430.52 allows an inverse-time breaker up to 250% of FLC: 2.5 × 34 = 85 A. 85 A isn't a standard rating, and the NEC permits the next higher standard rating: 90 A. (Overload protection is a separate requirement.)
Why the other choices miss: 85 A is the calculated value, not a standard rating. 80 A rounds down unnecessarily. 100 A skips past the next standard size.
Source: EEPower: Protecting Motor Branch Circuits from Short Circuits and Ground Faults (Mari, 2024) — Table 430.52 percentages; 240.6 10 A rating added in 2023
Question 5 of 80: CT ratio
PEP-05 · Measurement and Instrumentation · Instrument transformers and metering · Choose one answer
A relay is fed from an 800:5 current transformer. During normal load the relay reads 3.2 A on its CT input. What is the primary current?
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Answer: D. 512 A
Why: An 800:5 CT has a ratio of 800/5 = 160. Primary current = 3.2 A × 160 = 512 A.
Why the other choices miss: 296 A divides by √3 for no reason (a CT ratio is a straight current ratio). 887 A multiplies by √3. 2,560 A multiplies by 800 and forgets the 5 A secondary rating.
Source: Southern Illinois University ET 332b, Lesson 11: transformer nameplate data and connections — Slide 19 (delta–wye, 30° shift, third-harmonic trapping); slide 20 (open delta); slide 22 (CT ratio); U.S. Bureau of Reclamation FIST Vol. 3-8: Protective Relays and Associated Circuits (rev. 9/2021) — Sec. 5.2 p. 17; Sec. 7 p. 21; Sec. 8.3 p. 24
Question 6 of 80: Full-load current
PEP-06 · Electric Power Devices · Transformers · Choose one answer
A single-phase 50 kVA, 7,200–240 V transformer is fully loaded. What is the secondary current?
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Answer: A. 208 A
Why: I = S/V = 50,000/240 = 208 A.
Why the other choices miss: 6.9 A is the primary current (50,000/7,200). 120 A divides by √3 — that's for three-phase. 417 A divides by 120 V.
Source: Southern Illinois University ET 332b, Lesson 10: transformer regulation, efficiency, short-circuit current — Slide 3 (regulation); slides 11–12 (rated current, % impedance); slides 26–27 (Isc); slide 29 (efficiency)
Question 7 of 80: Bonding a lightning protection system
PEP-07 · General Applications · Lightning protection · Choose one answer
A building has a lightning protection system with its own ground terminals. Under the NEC, how must that system relate to the building's grounding electrode system?
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Answer: B. It must be bonded to the building or structure grounding electrode system
Why: NEC 250.106 requires the lightning protection system grounding to be bonded to the building or structure grounding electrode system. Bonding keeps the two systems at nearly the same potential during a strike, which reduces the chance of a dangerous flashover between them.
Why the other choices miss: Isolation is the old intuition this rule corrects. The lightning protection electrodes don't replace the required grounding electrode system. The bonding rule is in the NEC's grounding article (250.106), so it's fair game on an exam that supplies the 2020 NEC.
Source: EC&M Code Q&A: Bonding the lightning protection system (Holt) — 250.106
Question 8 of 80: Slip
PEP-08 · Rotating Machines · Machine types and applications · Enter a number
A 4-pole, 60 Hz induction motor runs at 1,750 rpm under load. What is its slip? Enter a percentage.
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Answer: 2.78% (accepted range 2.73–2.83%)
Why: Synchronous speed n_s = 120f/P = 120 × 60/4 = 1,800 rpm. Slip = (1,800 − 1,750)/1,800 = 0.0278 = 2.78%.
Common mistakes: Dividing by 1,750 instead of 1,800 gives 2.86%.
Source: MIT OCW 6.061 course notes, Ch. 10: Induction machines — Sec. 2–4 (slip, speed); Sec. 6.2 (volts/hertz control); Sec. 6.3–6.4 (torque ∝ |V|²)
Question 9 of 80: Six-pulse bridge output
PEP-09 · Power Electronic Circuits and Control Devices · Power electronics · Enter a number
An ideal three-phase, six-pulse diode bridge is fed from 480 V line-to-line. Neglecting source inductance, what is the average dc output voltage? Enter volts.
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Answer: 648.2 V (accepted range 641.7–654.7 V)
Why: V_d0 = (3√2/π) × V_LL = 1.35 × 480 = 648 V.
Common mistakes: 0.9 × 480 = 432 V uses the single-phase bridge factor. √2 × 480 = 679 V is the peak line-to-line voltage, not the average.
Source: Linköping University TSTE19 Power Electronics, Lecture 3 (rectifiers) — Slide 17 (single-phase bridge); slide 27 (three-phase bridge); slide 35 (commutation drop)
Question 10 of 80: Adding utility source impedance
PEP-10 · Transmission and Distribution Analysis · Fault current analysis · Choose one answer
The same 1,500 kVA, 5.75% transformer is fed from a utility source with 500 MVA of available three-phase fault capacity. Treat both impedances as pure reactances. What is the secondary three-phase fault current now?
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Answer: C. 29.8 kA
Why: On the transformer's 1.5 MVA base, the utility impedance is 1.5/500 = 0.003 pu. Total = 0.0575 + 0.003 = 0.0605 pu. I_sc = 1,804/0.0605 = 29.8 kA.
Why the other choices miss: 31.4 kA ignores the source. 601 kA is the utility's capacity expressed at 480 V with no transformer. 7.0 kA mixes bases (adding 0.2 pu from a 100 MVA base without converting).
Source: Southern Illinois University ET 332b, Lesson 10: transformer regulation, efficiency, short-circuit current — Slide 3 (regulation); slides 11–12 (rated current, % impedance); slides 26–27 (Isc); slide 29 (efficiency); University of Utah ECE 3600 — Per-unit notes (Stolp) — pp. 2–3 (base impedance, base current, change of base)
Question 11 of 80: Ampacity adjustment (2020 numbering)
PEP-11 · Electrical Safety · Wiring methods and installations · Choose one answer
Four current-carrying conductors run in one raceway. Each has a Table 310.16 ampacity of 50 A, and no ambient temperature correction applies. What is the adjusted ampacity under the 2020 NEC?
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Answer: D. 40 A
Why: For four to six current-carrying conductors, the adjustment factor is 80%. 50 A × 0.80 = 40 A. In the 2020 NEC, this table is Table 310.15(C)(1) — earlier editions numbered it 310.15(B)(3)(a), so search by the 2020 number on exam day.
Why the other choices miss: 50 A ignores bundling. 35 A applies the 70% factor that starts at seven conductors. 45 A is a 90% factor that isn't in the table.
Source: EC&M: Code Q&A — Conductor Ampacity Adjustment Requirements (Holt, 2021; 2020 NEC) — Table 310.15(C)(1) examples
Question 12 of 80: Delta load line current
PEP-12 · Circuit Analysis · Three-phase circuits · Choose one answer
A balanced delta-connected heater has 30 Ω of resistance in each leg and is supplied at 480 V line-to-line. What is the line current?
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Answer: A. 27.7 A
Why: In delta, each leg sees full line voltage: phase current = 480/30 = 16.0 A. Line current in a balanced delta = √3 × phase current = 27.7 A.
Why the other choices miss: 16.0 A is the leg (phase) current. 48.0 A is 3 × 16. 9.2 A divides by √3 instead of multiplying.
Source: MIT OCW 6.061 course notes, Ch. 3: Polyphase networks — Line-line voltages section; Sec. 6 three-phase transformers; Sec. 8.2
Question 13 of 80: Smallest standard breaker (2020)
PEP-13 · Protection · Overcurrent protection · Choose one answer
Under the 2020 NEC, what is the smallest standard ampere rating for inverse-time circuit breakers in 240.6(A)?
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Answer: B. 15 A
Why: In the 2020 edition, the standard inverse-time breaker ratings start at 15 A. A 10 A standard rating for fuses and inverse-time breakers was added in the 2023 NEC — a newer-edition change the exam's 2020 code doesn't include.
Why the other choices miss: 10 A is the 2023 answer, which is wrong for an exam that supplies the 2020 NEC. 5 A and 20 A aren't the lowest 2020 breaker rating.
Source: EEPower: Protecting Motor Branch Circuits from Short Circuits and Ground Faults (Mari, 2024) — Table 430.52 percentages; 240.6 10 A rating added in 2023
Question 14 of 80: Two-wattmeter power factor
PEP-14 · Measurement and Instrumentation · Instrument transformers and metering · Choose one answer
Two wattmeters measure a balanced three-phase, three-wire load. W1 reads 8.0 kW and W2 reads 3.0 kW. What is the load power factor?
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Answer: C. 0.79
Why: Total P = W1 + W2 = 11.0 kW. For a balanced load, Q = √3(W1 − W2) = √3 × 5.0 = 8.66 kvar. tan θ = 8.66/11.0 = 0.787, so θ = 38.2° and PF = cos 38.2° = 0.79.
Why the other choices miss: 0.45 is (W1 − W2)/(W1 + W2), which is neither tan θ nor PF. 0.38 is just W2/W1. 0.91 comes from leaving out the √3 in the tan θ formula.
Source: NJIT ECE 342 Lab 2: Three-phase power measurements — Two-wattmeter method and N−1 rule
Question 15 of 80: Open-delta capacity
PEP-15 · Electric Power Devices · Transformers · Choose one answer
A closed-delta bank of three 50 kVA single-phase transformers loses one unit and runs open-delta. What is the maximum balanced three-phase load without overloading either remaining transformer?
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Answer: D. 86.6 kVA
Why: In open delta, each remaining transformer carries the full line current at line-to-line voltage, so each unit limits the line current to I = 50 kVA / V_LL. A balanced load is S = √3 × V_LL × I = √3 × 50 = 86.6 kVA. That's 57.7% of the original 150 kVA bank, or 86.6% of the two units' combined nameplate.
Why the other choices miss: 100 kVA adds the two nameplates, but the open-delta currents aren't in phase with the windings' voltages, so you can't use both fully. 75 kVA and 150 kVA aren't supported by the circuit.
Source: Southern Illinois University ET 332b, Lesson 11: transformer nameplate data and connections — Slide 19 (delta–wye, 30° shift, third-harmonic trapping); slide 20 (open delta); slide 22 (CT ratio)
Question 16 of 80: SPD type on the line side
PEP-16 · General Applications · Surge protection · Choose one answer
Under 2020 NEC Article 242, which type of surge-protective device (SPD) is permitted on the supply (line) side of the service disconnect?
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Answer: A. Type 1
Why: Type 1 SPDs are permanently connected devices that are normally installed on the supply side of the service disconnect. Type 2 goes on the load side of the service disconnect, Type 3 on the load side of branch-circuit protection, and Type 4 devices are components inside equipment installed by the manufacturer. Article 242 is itself a 2020 change: it replaced the old Articles 280 and 285.
Why the other choices miss: Type 2 is the most common panel-mounted SPD, which makes it a tempting wrong answer, but it belongs on the load side of the service disconnect.
Source: EC&M: NEC Requirements for Overvoltage Protection (Holt, 2023; 2020 NEC) — 242.12–242.18 SPD types; EC&M: 2020 National Electrical Code Changes — 220.12; 230.67; Article 242
Question 17 of 80: Machine characteristics
PEP-17 · Rotating Machines · Machine types and applications · Match each item
Match each machine or operating condition to its characteristic.
Items:
- Overexcited synchronous motor
- Synchronous condenser
- Wound-rotor induction motor
- Induction motor in normal motoring
Choices:
- A. Synchronous machine with no mechanical load, used for power factor correction
- B. External rotor resistance raises starting torque and lowers starting current
- C. Rotor turns slightly below synchronous speed (positive slip)
- D. Supplies reactive power to the system's inductive loads
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Answer: Overexcited synchronous motor → D. Supplies reactive power to the system's inductive loads; Synchronous condenser → A. Synchronous machine with no mechanical load, used for power factor correction; Wound-rotor induction motor → B. External rotor resistance raises starting torque and lowers starting current; Induction motor in normal motoring → C. Rotor turns slightly below synchronous speed (positive slip)
Why: Overexciting a synchronous motor makes it deliver reactive power, which is why it can correct plant power factor; a synchronous condenser does only that job. A wound rotor brings the rotor circuit out through slip rings so external resistance can shape starting. An induction motor needs slip to induce rotor current, so it runs below synchronous speed when motoring.
Common mistakes: It's easy to swap the two synchronous rows: both supply reactive power, but only the condenser has no shaft load.
Source: Southern Illinois University ET 332b, Lesson 19: power factor correction using synchronous motors — Slide 4; All About Circuits, Lessons in Electric Circuits Vol. II, Ch. 13: Wound-rotor induction motors — Starting and speed control; MIT OCW 6.061 course notes, Ch. 10: Induction machines — Sec. 2–4 (slip, speed); Sec. 6.2 (volts/hertz control); Sec. 6.3–6.4 (torque ∝ |V|²)
Question 18 of 80: Single-phase bridge output
PEP-18 · Power Electronic Circuits and Control Devices · Power electronics · Choose one answer
An ideal single-phase full-bridge diode rectifier is fed from 240 V rms. What is the average dc output voltage?
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Answer: B. 216 V
Why: V_d0 = (2√2/π) × V_s ≈ 0.9 × 240 = 216 V.
Why the other choices miss: 108 V is the half-wave value (0.45 × V_s). 240 V confuses rms with average. 339 V is the peak.
Source: Linköping University TSTE19 Power Electronics, Lecture 3 (rectifiers) — Slide 17 (single-phase bridge); slide 27 (three-phase bridge); slide 35 (commutation drop)
Question 19 of 80: Three-phase feeder drop
PEP-19 · Transmission and Distribution Analysis · Voltage drop · Choose one answer
A 480 V three-phase feeder carries 200 A at 0.85 PF lagging over 500 ft (one-way). Conductor R = 0.05 Ω and X = 0.04 Ω per 1,000 ft per conductor. Using the approximation VD = √3 · I · (R cos θ + X sin θ) · L, what is the line-to-line voltage drop?
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Answer: C. 11.0 V
Why: For L = 500 ft: R = 0.025 Ω, X = 0.020 Ω. sin θ = 0.527. R cos θ + X sin θ = 0.02125 + 0.01054 = 0.03179 Ω. VD = 1.732 × 200 × 0.03179 = 11.0 V (2.3% of 480 V).
Why the other choices miss: 6.4 V leaves out √3. 12.7 V uses the single-phase factor of 2. 19.1 V multiplies by 3.
Source: MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation)
Question 20 of 80: Working space depth
PEP-20 · Electrical Safety · Wiring methods and installations · Choose one answer
A 480Y/277 V panelboard (277 V to ground) will be installed facing a concrete block wall, with exposed live parts on one side of the working space. What is the minimum depth of working space under NEC Table 110.26(A)(1)?
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Answer: D. 3.5 ft
Why: 277 V to ground falls in the 151–600 V column. Live parts on one side with grounded parts on the other is Condition 2, and concrete, brick, or tile walls count as grounded. Condition 2 at 151–600 V requires 3.5 ft.
Why the other choices miss: 3 ft is Condition 1 (or any condition at 150 V or less to ground). 4 ft is Condition 3 (live parts on both sides). 6.5 ft is the headroom requirement, not depth.
Source: IAEI Magazine: Working Space for Electrical Equipment (Hunter, 2013) — Table 110.26(A)(1) conditions
Question 21 of 80: Real power, wye load
PEP-21 · Circuit Analysis · Three-phase circuits · Choose one answer
A balanced wye load has an impedance of 9.6 + j7.2 Ω per phase and is supplied at 208 V line-to-line. What is the total real power?
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Answer: A. 2.88 kW
Why: Phase voltage = 208/√3 = 120.1 V. |Z| = √(9.6² + 7.2²) = 12.0 Ω, so I = 120.1/12.0 = 10.0 A. P = 3 I² R = 3 × 10.0² × 9.6 = 2,880 W ≈ 2.88 kW.
Why the other choices miss: 3.60 is the apparent power in kVA. 2.16 is the reactive power in kvar. 8.65 kW uses 208 V as the phase voltage of a wye load.
Source: MIT OCW 6.061 course notes, Ch. 3: Polyphase networks — Line-line voltages section; Sec. 6 three-phase transformers; Sec. 8.2; MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation)
Question 22 of 80: Service ground-fault protection
PEP-22 · Protection · Overcurrent protection · Select all that apply
A new 480Y/277 V solidly grounded service has a 1,200 A service disconnect. Which statements about ground-fault protection of equipment under NEC 230.95 are correct? Select all that apply.
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Answer: A, B and C
Why: 230.95 applies to solidly grounded wye services of more than 150 V to ground with service disconnects rated 1,000 A or more. A 480Y/277 V system is 277 V to ground, so this 1,200 A disconnect needs ground-fault protection with a maximum setting of 1,200 A and no more than 1 second of delay at 3,000 A or more.
Why the other choices miss: A 208Y/120 V system is only 120 V to ground, so 230.95 doesn't apply.
Source: IAEI Magazine: Performance testing requirements for ground-fault protection equipment — 230.95 settings; U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 5 — ¶5-1d(4) (motoring generator); ¶5-2c (differential); ¶5-4a(4) (service ground-fault protection)
Question 23 of 80: Working on CT secondaries
PEP-23 · Measurement and Instrumentation · Instrument transformers and metering · Select all that apply
A technician must remove a panel meter from the secondary circuit of an in-service current transformer. Which practices are correct? Select all that apply.
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Answer: A and B
Why: An energized CT keeps driving its secondary current. If that circuit is opened, the CT tries to push current through an open circuit and can develop very high voltage that damages insulation and endangers people. Reclamation's relay manual says CT secondaries must never be open-circuited with the primary energized, recommends shorting switches or blocks, and says CT and PT secondary circuits are grounded at only one point.
Why the other choices miss: Option C is wrong because the hazard exists whenever the primary is energized, not just during faults. Option D is wrong because a fuse is a deliberate way to open the circuit — exactly what you must avoid.
Source: U.S. Bureau of Reclamation FIST Vol. 3-8: Protective Relays and Associated Circuits (rev. 9/2021) — Sec. 5.2 p. 17; Sec. 7 p. 21; Sec. 8.3 p. 24
Question 24 of 80: Voltage regulation
PEP-24 · Electric Power Devices · Transformers · Enter a number
A transformer's secondary reads 246 V at no load and 240 V at rated load. What is its percent voltage regulation? Enter a percentage.
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Answer: 2.5% (accepted range 2.45–2.55%)
Why: %VR = (V_NL − V_FL)/V_FL × 100 = (246 − 240)/240 × 100 = 2.5%.
Common mistakes: Dividing by the no-load voltage gives 2.44%.
Source: Southern Illinois University ET 332b, Lesson 10: transformer regulation, efficiency, short-circuit current — Slide 3 (regulation); slides 11–12 (rated current, % impedance); slides 26–27 (Isc); slide 29 (efficiency)
Question 25 of 80: Dwelling-unit service SPD (new in 2020)
PEP-25 · General Applications · Surge protection · Select all that apply
Which statements about NEC 2020 section 230.67 are correct? Select all that apply.
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Answer: A and B
Why: 230.67 is a 2020 addition: every service supplying dwelling units needs an SPD, and it must be Type 1 or Type 2, located at or next to the service (with an allowance for the next panelboard downstream).
Why the other choices miss: Type 3 devices sit at the utilization equipment and don't satisfy a service-level requirement. The rule has no lightning-flash-density condition.
Source: EC&M: 2020 National Electrical Code Changes — 220.12; 230.67; Article 242
Question 26 of 80: Synchronous motor speed
PEP-26 · Rotating Machines · Machine types and applications · Choose one answer
A 6-pole synchronous motor runs from a 60 Hz supply within its rating. What is its speed?
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Answer: B. 1,200 rpm
Why: A synchronous motor runs at synchronous speed regardless of load: n = 120f/P = 120 × 60/6 = 1,200 rpm.
Why the other choices miss: 1,164 rpm applies an induction-motor slip a synchronous motor doesn't have. 1,800 and 3,600 rpm are 4- and 2-pole speeds.
Source: MIT OCW 6.061 course notes, Ch. 10: Induction machines — Sec. 2–4 (slip, speed); Sec. 6.2 (volts/hertz control); Sec. 6.3–6.4 (torque ∝ |V|²)
Question 27 of 80: VFD volts per hertz
PEP-27 · Power Electronic Circuits and Control Devices · Power electronics · Choose one answer
A 460 V, 60 Hz induction motor runs on a drive using constant volts-per-hertz control. What output voltage should the drive apply at 45 Hz?
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Answer: C. 345 V
Why: Constant V/Hz keeps air-gap flux roughly constant below base speed. 460/60 = 7.67 V/Hz, so at 45 Hz: 7.67 × 45 = 345 V.
Why the other choices miss: Holding 460 V at 45 Hz overfluxes the motor. 613 V scales the wrong way. 259 V scales by the square of the frequency ratio.
Source: MIT OCW 6.061 course notes, Ch. 10: Induction machines — Sec. 2–4 (slip, speed); Sec. 6.2 (volts/hertz control); Sec. 6.3–6.4 (torque ∝ |V|²)
Question 28 of 80: Shunt capacitor voltage support
PEP-28 · Transmission and Distribution Analysis · Voltage regulation and support · Choose one answer
A long radial feeder serves a large lagging (inductive) load, and voltage at the far end is low. What is the main effect of adding shunt capacitors at the load end?
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Answer: D. They supply the load's reactive power locally, reducing reactive current through the feeder and raising the load-end voltage
Why: Shunt capacitors supply reactive power at the load, so less reactive current flows through the feeder's series impedance. Less current through that impedance means less drop, so the load-end voltage rises. That's why shunt compensation is used for voltage support as well as power factor.
Why the other choices miss: Capacitors don't add real power. Shunt devices don't change the line's series reactance (series capacitors do, by reducing it). Reducing reactive current does change voltage drop.
Source: MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation); Southern Illinois University ET 332b, Lesson 19: power factor correction using synchronous motors — Slide 4
Question 29 of 80: Interrupting rating
PEP-29 · Electrical Safety · Wiring methods and installations · Select all that apply
The available fault current at the line terminals of a new 480 V main breaker is 31.4 kA. Which interrupting ratings satisfy NEC 110.9? Select all that apply.
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Answer: C, D and E
Why: 110.9 requires an interrupting rating at least equal to the fault current available at the device's line terminals. Any rating of 31.4 kA or more qualifies: 35, 42, and 65 kA.
Why the other choices miss: 22 and 25 kA are below the available fault current. Don't confuse interrupting rating (110.9, for devices that interrupt fault current) with short-circuit current rating (110.10, for equipment).
Source: EC&M: General Requirements of the NEC (Holt, 2020) — 110.9 vs 110.10
Question 30 of 80: Open-phase sequence currents
PEP-30 · Circuit Analysis · Symmetrical components · Choose one answer
Phase C of a wye-connected feeder opens. With ABC sequence, the currents are Ia = 100∠0° A, Ib = 100∠−120° A, and Ic = 0. What is the magnitude of the negative-sequence current?
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Answer: A. 33.3 A
Why: With a = 1∠120°: I2 = (Ia + a²Ib + aIc)/3. a²Ib = 100∠(240° − 120°) = 100∠120°. Ia + a²Ib = 100∠0° + 100∠120° = 100∠60°, so I2 = 33.3∠60° A. (For comparison, I1 = 66.7∠0° A and I0 = 33.3∠−60° A.)
Why the other choices miss: 0 A assumes the currents are still balanced. 66.7 A is the positive-sequence magnitude. 100 A is a phase current, not a sequence component.
Source: MIT OCW 6.061 course notes, Ch. 4: Introduction to symmetrical components — Sec. 2 (Eqs. 4–7); Sec. 7.1 single line-to-ground fault
Question 31 of 80: Differential relay on a through fault
PEP-31 · Protection · Protective relaying · Choose one answer
A bus differential relay (device 87) has CTs on every circuit connected to the bus. A fault occurs on an outgoing feeder beyond that feeder's CT. How should the relay respond?
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Answer: B. It should not operate — current entering the zone equals current leaving it
Why: Differential protection compares the current entering the protected zone with the current leaving it. For an external (through) fault, the fault current enters on the source circuits and leaves on the faulted feeder, so the phasor sum is close to zero and the relay restrains. Only a fault inside the zone produces differential current.
Why the other choices miss: Magnitude alone doesn't trip a differential relay. Time-delayed backup for this fault belongs to the feeder's own overcurrent protection and upstream relays. Fault type doesn't change the zone logic.
Source: U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 5 — ¶5-1d(4) (motoring generator); ¶5-2c (differential); ¶5-4a(4) (service ground-fault protection); U.S. Army TM 5-811-14, Appendix E: Partial relay device numbers list — Device numbers 21–87
Question 32 of 80: Polarization index
PEP-32 · Measurement and Instrumentation · Insulation testing · Choose one answer
An insulation resistance test on a Class B motor stator winding reads 150 MΩ at 1 minute and 240 MΩ at 10 minutes. Which interpretation is best?
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Answer: C. PI = 1.6; below the 2.0 recommended minimum for Class B insulation, so the result warrants investigation
Why: Polarization index = R(10 min) / R(1 min) = 240/150 = 1.6. Reclamation FIST 3-1, citing IEEE 43, gives recommended minimum PI values of 1.5 for Class A and 2.0 for Classes B and F. A PI of 1.6 is therefore below the recommended minimum for Class B insulation; the test result should be investigated rather than treated as acceptable on PI alone.
Why the other choices miss: 1.0 is not the benchmark — a PI near 1 means almost no absorption current, which is a warning sign. 0.63 inverts the ratio. 90 is a difference, not a ratio.
Source: U.S. Bureau of Reclamation FIST Vol. 3-1: Testing Solid Insulation of Electrical Equipment — ¶9 and ¶14 (temperature correction); ¶12 p. 5 (polarization index)
Question 33 of 80: Capacitor at reduced voltage
PEP-33 · Electric Power Devices · Capacitors · Choose one answer
A 50 kvar capacitor bank rated 480 V is operated at 460 V. What reactive power does it supply?
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Answer: D. 45.9 kvar
Why: Capacitor output Q = V²/X_C, so it scales with voltage squared: 50 × (460/480)² = 45.9 kvar.
Why the other choices miss: 47.9 kvar scales linearly. 52.2 kvar inverts the ratio. 50.0 kvar ignores the lower voltage.
Source: MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation)
Question 34 of 80: Point-source illuminance
PEP-34 · General Applications · Illumination/lighting · Choose one answer
A small luminaire is mounted 4 m above a horizontal work surface. Point P on the surface is 3 m horizontally from the spot directly under the luminaire. The luminous intensity toward P is 2,000 cd. What is the horizontal illuminance at P?
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Answer: A. 64 lx
Why: Use E = I·cos θ / d². The distance to P is √(4² + 3²) = 5 m, and cos θ = 4/5 = 0.8 (θ measured from vertical). E = 2,000 × 0.8 / 25 = 64 lx.
Why the other choices miss: 80 lx skips the cosine term (that's illuminance on a surface facing the light). 125 lx uses the 4 m height as the distance. 222 lx uses the 3 m offset as the distance.
Source: MIT OCW 4.401 Lecture 11 (lighting): inverse-square law for a point source — Slide: flux, illuminance, intensity
Question 35 of 80: Autotransformer vs primary impedance
PEP-35 · Rotating Machines · Motor starting · Choose one answer
Two reduced-voltage starters are compared on the same motor, each set to apply 65% of line voltage at start: an autotransformer starter and a primary-reactor starter. Which draws less current from the supply line during starting?
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Answer: B. The autotransformer starter (about 42% of full-voltage line current vs about 65%)
Why: Both apply 65% voltage to the motor, so motor current and torque are about the same. But an autotransformer steps current down on its line side: line current is about 0.65² ≈ 42% of the full-voltage value, while a series reactor passes the motor's current straight through (about 65%). That's why the autotransformer gives the most torque per line ampere.
Why the other choices miss: Equal motor voltage does not mean equal line current. Reactor losses aren't what distinguishes the two.
Source: Schneider Electric Digest: Reduced-voltage starters — description of all types — Autotransformer, wye-delta, part-winding sections; EC&M: Reduced-Voltage Starters — Choosing the Best Type (Hinman, 1998) — Autotransformer vs primary-impedance line current
Question 36 of 80: Commutation (overlap) voltage drop
PEP-36 · Power Electronic Circuits and Control Devices · Power electronics · Choose one answer
A six-pulse diode bridge is fed from 480 V line-to-line, 60 Hz, through a source inductance of 0.5 mH per phase. It delivers a steady 100 A dc. What is the average output voltage including commutation drop?
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Answer: C. 630 V
Why: V_d = 1.35 V_LL − (3/π) ω L_s I_d. Commutation drop = (3/π) × 377 × 0.0005 × 100 = 18.0 V. V_d = 648.2 − 18.0 = 630 V.
Why the other choices miss: 648 V ignores source inductance. 612 V subtracts the drop twice. 666 V adds it.
Source: Linköping University TSTE19 Power Electronics, Lecture 3 (rectifiers) — Slide 17 (single-phase bridge); slide 27 (three-phase bridge); slide 35 (commutation drop)
Question 37 of 80: Capacitor kvar to reach 0.95
PEP-37 · Transmission and Distribution Analysis · Power factor correction · Choose one answer
A 500 kW load runs at 0.75 PF lagging. How much capacitor kvar raises it to 0.95 PF lagging?
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Answer: D. 277 kvar
Why: Q1 = 500 × tan(cos⁻¹ 0.75) = 441 kvar. Q2 = 500 × tan(cos⁻¹ 0.95) = 164 kvar. Capacitor = 441 − 164 = 277 kvar.
Why the other choices miss: 441 is the original reactive power. 164 is the remaining reactive power. 140 is the drop in kVA, not kvar.
Source: MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation)
Question 38 of 80: Class I, Division 1
PEP-38 · Electrical Safety · Hazardous locations · Select all that apply
Which conditions make a location Class I, Division 1? Select all that apply.
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Answer: A, B and C
Why: Division 1 means the hazard can be present in normal operation, frequently during maintenance or leakage, or at the same moment equipment failure creates an ignition source. These three conditions mirror the NEC Article 500 definitions, which OSHA reproduces in 29 CFR 1910.399.
Why the other choices miss: Closed containers that leak only on rupture, and areas kept safe by mechanical ventilation, are Division 2 conditions — hazardous only under abnormal circumstances.
Source: OSHA 29 CFR 1910.399 — Definitions (hazardous locations) — Class I Division 1/2; Class I Zone 0/1/2
Question 39 of 80: Single line-to-ground fault
PEP-39 · Circuit Analysis · Symmetrical components · Choose one answer
At a solidly grounded bus, the prefault voltage is 1.0 pu and the Thevenin sequence impedances are Z1 = Z2 = j0.15 pu and Z0 = j0.10 pu. What is the magnitude of the fault current for a bolted single line-to-ground fault?
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Answer: A. 7.5 pu
Why: For a single line-to-ground fault the sequence networks are in series, so I1 = I2 = I0 = E/(Z1 + Z2 + Z0) = 1.0/0.40 = 2.5 pu. The phase fault current is Ia = 3 I0 = 7.5 pu.
Why the other choices miss: 2.5 pu is the sequence current; you must multiply by 3. 6.67 pu is the three-phase fault current (1/0.15). 10.0 pu leaves Z0 out of the denominator.
Source: MIT OCW 6.061 course notes, Ch. 4: Introduction to symmetrical components — Sec. 2 (Eqs. 4–7); Sec. 7.1 single line-to-ground fault
Question 40 of 80: Distance relay fault location
PEP-40 · Protection · Protective relaying · Choose one answer
A distance relay (21) uses a 600:5 CT and a VT ratio of 1,000:1. During a bolted fault, it measures 2.0 Ω on its secondary side. The line's positive-sequence impedance is 0.8 Ω per mile (primary). About how far away is the fault?
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Answer: B. 20.8 mi
Why: Convert secondary ohms to primary: Z_pri = Z_sec × (VT ratio / CT ratio) = 2.0 × (1,000/120) = 16.7 Ω. Distance = 16.7/0.8 = 20.8 miles.
Why the other choices miss: 2.5 mi divides secondary ohms by the per-mile impedance without converting. 16.7 is the primary impedance in ohms, not miles. 300 mi inverts the ratio.
Source: U.S. Army TM 5-811-14, Appendix E: Partial relay device numbers list — Device numbers 21–87
Question 41 of 80: Temperature and insulation resistance
PEP-41 · Measurement and Instrumentation · Insulation testing · Choose one answer
You trend insulation resistance on the same transformer each year. Why should each reading be corrected to a common base temperature before comparing them?
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Answer: C. Insulation resistance changes strongly with temperature, so uncorrected readings taken at different temperatures can hide or fake a trend
Why: Insulation resistance varies a great deal with temperature (it falls as insulation warms), so Reclamation's FIST 3-1 corrects readings to a 40 °C base with a material-specific coefficient before comparing them. It also notes the correction is only approximate, which is another reason to test at similar temperatures when you can.
Why the other choices miss: The test set's output voltage isn't the issue. Correction matters for insulation resistance trending specifically. Resistance falls, not rises, as temperature increases — so a hot reading looks worse, not better, than the same insulation tested cold.
Source: U.S. Bureau of Reclamation FIST Vol. 3-1: Testing Solid Insulation of Electrical Equipment — ¶9 and ¶14 (temperature correction); ¶12 p. 5 (polarization index)
Question 42 of 80: Battery runtime
PEP-42 · Electric Power Devices · Electrical energy storage · Choose one answer
A 100 kWh battery system is limited to 90% depth of discharge. Its inverter is 95% efficient. How long can it carry a constant 20 kW ac load?
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Answer: D. 4.28 h
Why: Usable dc energy = 100 × 0.90 = 90 kWh. Delivered ac energy = 90 × 0.95 = 85.5 kWh. Runtime = 85.5/20 = 4.28 h.
Why the other choices miss: 5.00 h uses nameplate energy. 4.50 h ignores inverter losses. 3.85 h applies the 90% limit twice.
Source: Worked calculation from the definitions given in the question.
Question 43 of 80: 2020 NEC office lighting unit load
PEP-43 · General Applications · Illumination/lighting · Choose one answer
An office building has 20,000 ft² of floor area. Using the office unit load in 2020 NEC Table 220.12, which already includes the 125% continuous-load factor specified by the table note, what minimum general lighting load results?
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Answer: A. 26.0 kVA
Why: The 2020 NEC office unit load is 1.3 VA/ft², and the table note states that its unit loads already include the 125% continuous-load multiplier. So 1.3 × 20,000 = 26,000 VA = 26.0 kVA; do not multiply that table result by 1.25 again.
Why the other choices miss: 70.0 kVA uses the old 3.5 VA/ft² office value from the 2017 NEC. 60.0 kVA uses 3 VA/ft², which is not the 2020 office-table value. 32.5 kVA incorrectly applies another 125% multiplier to a Table 220.12 value that already includes it.
Source: EC&M: Understanding Load Calculations and the 2020 NEC (Vigstol, 2019) — Table 220.12 office value, 2017 vs 2020; EC&M: 2020 National Electrical Code Changes — 220.12; 230.67; Article 242
Question 44 of 80: Rank starters by starting torque
PEP-44 · Rotating Machines · Motor starting · Put in order
Put these reduced-voltage starting methods in order from highest to lowest starting torque, as a percentage of full-voltage starting torque.
- A. Wye-delta
- B. Autotransformer, 80% tap
- C. Autotransformer, 50% tap
- D. Autotransformer, 65% tap
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Answer: B. Autotransformer, 80% tap → D. Autotransformer, 65% tap → A. Wye-delta → C. Autotransformer, 50% tap
Why: Autotransformer starting torque scales with the square of the tap: 80% → 64%, 65% → 42%, 50% → 25%. Wye-delta reduces both inrush and torque to about 33% of the delta-connected values. Order: 80% tap (64%) → 65% tap (42%) → wye-delta (33%) → 50% tap (25%).
Common mistakes: The common slip is treating tap percentage as torque percentage. A 50% tap gives only 25% torque, which drops it below wye-delta.
Source: Schneider Electric Digest: Reduced-voltage starters — description of all types — Autotransformer, wye-delta, part-winding sections
Question 45 of 80: Evaluate a ladder rung
PEP-45 · Power Electronic Circuits and Control Devices · Relays, switches, Boolean and ladder logic · Choose one answer
A ladder rung has two branches in parallel driving output Y. Branch 1: normally open contacts X1 and X2 in series. Branch 2: one normally closed contact X3. The inputs are X1 = 1, X2 = 0, and X3 = 1. What is Y?
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Answer: B. Y = 0
Why: Series contacts form an AND and parallel branches form an OR; a normally closed contact passes power only when its input is 0. Y = (X1 · X2) + X3′ = (1 · 0) + 0 = 0, so Y is off.
Why the other choices miss: X1 alone can't complete branch 1 because X2 is open. X3 = 1 opens the normally closed contact, so branch 2 is open too. A seal-in contact isn't needed to evaluate this rung.
Source: EC&M: The Basics of Ladder Logic (Fehr, 2003) — Series = AND, parallel = OR, inverted inputs
Question 46 of 80: Current THD
PEP-46 · Transmission and Distribution Analysis · Power quality · Choose one answer
A drive's input current contains a 200 A fundamental, 30 A of 5th harmonic, 20 A of 7th, and 10 A of 11th. Other harmonics are negligible. What is the current THD?
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Answer: C. 18.7%
Why: THD = √(I5² + I7² + I11²)/I1 = √(900 + 400 + 100)/200 = 37.4/200 = 18.7%.
Why the other choices miss: 15.0% counts only the 5th harmonic. 30.0% adds harmonics arithmetically instead of root-sum-square. 14.4% divides by the sum of all the currents instead of the fundamental.
Source: Eaton: How do you calculate THD and TDD? (Harmonics FAQ) — THD calculation
Question 47 of 80: Equipment approval and marking
PEP-47 · Electrical Safety · Hazardous locations · Choose one answer
Which statement about equipment in Class I hazardous locations is correct?
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Answer: D. Equipment approved for a Division 1 location may be installed in a Division 2 location of the same class and group
Why: Division 1 equipment meets the stricter requirement, so it can be used in Division 2 of the same class and group. The reverse is not true. Equipment must also be marked for class, group, and operating temperature, and that temperature marking may not exceed the ignition temperature of the gas or vapor present.
Why the other choices miss: Appearance doesn't confer an approval. A sealed enclosure doesn't excuse a surface temperature above the ignition temperature. Groups matter because different gases need different protection.
Source: OSHA 29 CFR 1910.307 — Hazardous (classified) locations — (c)(2)(ii) temperature marking; (e) Division 1 equipment in Division 2
Question 48 of 80: Change of base
PEP-48 · Circuit Analysis · Per-unit system · Choose one answer
A generator's reactance is 0.15 pu on its own ratings of 50 MVA and 13.8 kV. What is the reactance on a 100 MVA, 13.2 kV system base?
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Answer: A. 0.328 pu
Why: Z_new = Z_old × (S_new/S_old) × (V_old/V_new)² = 0.15 × (100/50) × (13.8/13.2)² = 0.15 × 2 × 1.093 = 0.328 pu.
Why the other choices miss: 0.300 ignores the voltage change. 0.275 inverts the voltage ratio. 0.075 inverts the MVA ratio.
Source: University of Utah ECE 3600 — Per-unit notes (Stolp) — pp. 2–3 (base impedance, base current, change of base)
Question 49 of 80: Instantaneous pickup vs inrush
PEP-49 · Protection · Protective relaying · Choose one answer
A 2,000 kVA, 12.47 kV transformer has primary instantaneous (50) relays. For coordination, model magnetizing inrush as 12 × full-load current for 0.1 s. Which of these instantaneous pickup settings (primary amperes) avoids tripping on inrush?
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Answer: B. 1,400 A
Why: FLA = 2,000,000/(√3 × 12,470) = 92.6 A. Inrush point = 12 × 92.6 = 1,111 A at 0.1 s. An instantaneous element operates with no intentional delay, so its pickup must be above 1,111 A. Only 1,400 A qualifies.
Why the other choices miss: 300, 750, and 1,000 A are all below the inrush point and would trip when the transformer is energized.
Source: U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 4 — ¶4-2d(2) (relay coordination interval); ¶4-6c (transformer inrush)
Question 50 of 80: Fall-of-potential probe spacing
PEP-50 · Measurement and Instrumentation · Ground resistance testing · Enter a number
A fall-of-potential test is run on a single ground rod in uniform soil. The current probe is placed 50 m from the rod. Using the 62% method, how far from the rod should the potential probe be placed? Enter the distance in meters.
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Answer: 30.9 m (accepted range 29.9–31.9 m)
Why: Place the potential probe on a straight line between the electrode and the current probe, about 61.8% of the way out: 0.618 × 50 m ≈ 30.9 m (about 31 m). The 62% figure assumes uniform soil and enough spacing that the electrode and current probe don't interact.
Common mistakes: A common mistake is measuring 62% from the current probe instead of from the electrode under test (that gives 19 m).
Source: Metrel application note: Earth resistance measurement and the 62% rule — Probe placement and assumptions
Question 51 of 80: PV module temperature derating
PEP-51 · Electric Power Devices · Alternative power generation · Choose one answer
A PV module is rated 400 W at Standard Test Conditions. Its maximum-power temperature coefficient is −0.35%/°C. At 1,000 W/m² irradiance and a cell temperature of 60 °C, what output do you expect?
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Answer: C. 351 W
Why: STC is 1,000 W/m², AM1.5, and 25 °C cell temperature. ΔT = 60 − 25 = 35 °C. P = 400 × (1 − 0.0035 × 35) = 400 × 0.8775 = 351 W.
Why the other choices miss: 400 W ignores temperature. 379 W uses only a 15 °C rise. 316 W doubles the coefficient.
Source: Penn State AE 868: Interpreting a PV manufacturer datasheet — Standard Test Conditions definition
Question 52 of 80: Load factor
PEP-52 · General Applications · Energy management and demand calculations · Enter a number
A plant uses 238,080 kWh in a 31-day month. Its peak 15-minute demand that month is 480 kW. What is the monthly load factor? Enter a percentage.
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Answer: 66.7% (accepted range 66.2–67.2%)
Why: Load factor = average load ÷ peak load. Hours in the month = 31 × 24 = 744 h. Average load = 238,080 kWh ÷ 744 h = 320 kW. Load factor = 320 ÷ 480 = 0.667 = 66.7%.
Common mistakes: Using a 30-day month gives 68.9%, a small but real error. Dividing peak by average (1.5) inverts the ratio.
Source: HOMER Pro user manual: Load factor — Definition
Question 53 of 80: Voltage dip and starting torque
PEP-53 · Rotating Machines · Motor starting · Choose one answer
During starting, a large induction motor's terminal voltage dips to 90% of rated. Approximately what percentage of its rated-voltage locked-rotor torque does it develop?
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Answer: D. 81%
Why: Induction motor torque varies with the square of applied voltage: 0.90² = 0.81, so about 81%.
Why the other choices miss: 90% assumes a linear relationship. 73% cubes the ratio. 100% ignores the dip.
Source: MIT OCW 6.061 course notes, Ch. 10: Induction machines — Sec. 2–4 (slip, speed); Sec. 6.2 (volts/hertz control); Sec. 6.3–6.4 (torque ∝ |V|²)
Question 54 of 80: Relay device numbers
PEP-54 · Power Electronic Circuits and Control Devices · Relays, switches, Boolean and ladder logic · Match each item
Match each ANSI/IEEE device number to its function.
Items:
- 27
- 50
- 51
- 86
Choices:
- A. Instantaneous overcurrent relay
- B. AC time overcurrent relay
- C. Lockout relay
- D. Undervoltage relay
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Answer: 27 → D. Undervoltage relay; 50 → A. Instantaneous overcurrent relay; 51 → B. AC time overcurrent relay; 86 → C. Lockout relay
Why: These are standard device function numbers used on one-line diagrams and relay schemes: 27 undervoltage, 50 instantaneous overcurrent, 51 ac time overcurrent, 86 lockout (hand-reset auxiliary that trips and blocks closing).
Common mistakes: The usual mix-up is 50 vs 51: 50 has no intentional delay, while 51 follows an inverse time curve.
Source: U.S. Army TM 5-811-14, Appendix E: Partial relay device numbers list — Device numbers 21–87
Question 55 of 80: Delta–wye characteristics
PEP-55 · Transmission and Distribution Analysis · Transformer connections · Select all that apply
Which statements about a standard delta–wye transformer bank are correct? Select all that apply.
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Answer: A, B and C
Why: A delta–wye bank shifts voltages by 30° between the windings, the delta circulates third-harmonic currents instead of passing them to the supply, and the wye side creates a new neutral point for 120/208 V or 277/480 V four-wire service.
Why the other choices miss: Because of that 30° shift, a delta–wye bank can't be paralleled with a wye–wye or delta–delta bank — the secondary voltages wouldn't line up in phase.
Source: Southern Illinois University ET 332b, Lesson 11: transformer nameplate data and connections — Slide 19 (delta–wye, 30° shift, third-harmonic trapping); slide 20 (open delta); slide 22 (CT ratio); MIT OCW 6.061 course notes, Ch. 3: Polyphase networks — Line-line voltages section; Sec. 6 three-phase transformers; Sec. 8.2
Question 56 of 80: Hospital essential electrical system
PEP-56 · Electrical Safety · Special occupancies and systems · Match each item
Match each hospital load to the branch of the essential electrical system that normally serves it.
Items:
- Exit signs and egress lighting
- Receptacles and task lighting in patient care areas
- Medical gas equipment and operating room HVAC
Choices:
- A. Critical branch
- B. Equipment branch
- C. Life safety branch
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Answer: Exit signs and egress lighting → C. Life safety branch; Receptacles and task lighting in patient care areas → A. Critical branch; Medical gas equipment and operating room HVAC → B. Equipment branch
Why: Under NEC Article 517, a hospital's essential electrical system is split into three branches. The life safety branch keeps exits usable (egress lighting, exit signs, fire alarm). The critical branch serves direct patient care — mainly receptacles and lighting in patient care areas. The equipment branch serves major equipment such as medical gas systems and operating room HVAC.
Common mistakes: A common error is putting operating room HVAC on the critical branch because it's critical to surgery; it is equipment-branch load.
Source: EC&M: Essential Electrical Systems at Health Care Facilities (Biason) — Branch descriptions
Question 57 of 80: Per-unit line impedance
PEP-57 · Circuit Analysis · Per-unit system · Choose one answer
A 138 kV transmission line has a series impedance of 50 Ω. What is its per-unit impedance on a 100 MVA, 138 kV base?
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Answer: A. 0.263 pu
Why: Z_base = V_base² / S_base = 138² / 100 = 190.4 Ω (line-to-line kV and three-phase MVA). Z_pu = 50 / 190.4 = 0.263 pu.
Why the other choices miss: 0.788 pu uses the line-to-neutral voltage with three-phase MVA. 3.81 pu inverts the ratio. 0.0263 pu is a decimal-place slip.
Source: University of Utah ECE 3600 — Per-unit notes (Stolp) — pp. 2–3 (base impedance, base current, change of base)
Question 58 of 80: Reverse power on a generator
PEP-58 · Protection · Protective relaying · Choose one answer
A generator's prime mover loses its input while the breaker stays closed. The machine starts drawing real power from the system and running as a motor. Which relay function is applied to detect this?
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Answer: B. 32 — directional power relay
Why: When a generator loses prime-mover input, it motors — drawing real power from the system. A directional power (32) relay detects the reversal of real power flow.
Why the other choices miss: Voltage and frequency can stay normal while a generator motors on a stiff system. A differential relay sees no internal fault.
Source: U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 5 — ¶5-1d(4) (motoring generator); ¶5-2c (differential); ¶5-4a(4) (service ground-fault protection); U.S. Army TM 5-811-14, Appendix E: Partial relay device numbers list — Device numbers 21–87
Question 59 of 80: Current probe spacing
PEP-59 · Measurement and Instrumentation · Ground resistance testing · Choose one answer
You plan a fall-of-potential test on a substation ground grid whose largest dimension (its diagonal) is 20 m. Using the practical rule that the current-probe distance should be at least five times the size of the earthing system, what is the minimum distance from the grid to the current probe?
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Answer: C. 100 m
Why: 5 × 20 m = 100 m. Large grids need long leads; the 62% method only works when the electrode under test and the current probe are far enough apart to act like separate hemispherical electrodes.
Why the other choices miss: 20 m and 40 m are too close, so the probes' voltage gradients overlap and the reading is too low. 62 m mixes up the potential-probe percentage with the current-probe spacing.
Source: Metrel application note: Earth resistance measurement and the 62% rule — Probe placement and assumptions
Question 60 of 80: Wind speed and power
PEP-60 · Electric Power Devices · Alternative power generation · Choose one answer
Average wind speed at a turbine site rises from 6 m/s to 9 m/s. Ignoring changes in efficiency and air density, by what factor does the power available in the wind increase?
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Answer: D. 3.38
Why: Power in the wind is ½ρAv³, so it scales with the cube of speed: (9/6)³ = 1.5³ = 3.38.
Why the other choices miss: 1.50 is linear. 2.25 is squared. 5.06 is the fourth power. (Separately, a turbine can never capture more than the Betz limit of 16/27 of that power.)
Source: Queen Mary University of London — Power in the wind (course notes) — Power ∝ v³; Betz limit
Question 61 of 80: Demand factor
PEP-61 · General Applications · Energy management and demand calculations · Choose one answer
A building has 500 kVA of connected load. Its measured maximum demand is 325 kVA. Which value is the demand factor?
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Answer: A. 0.65
Why: Demand factor = maximum demand ÷ total connected load = 325 ÷ 500 = 0.65. It's a dimensionless ratio, normally 1 or less.
Why the other choices miss: 1.54 inverts the ratio. 0.35 is 1 − 0.65. 65 kVA confuses a ratio with a load.
Source: EC&M: Ensuring Accuracy in Demand Factors with the NEC (Kuether, 2025) — Article 100 demand factor definition
Question 62 of 80: Real power transfer
PEP-62 · Transmission and Distribution Analysis · Power flow · Enter a number
Two buses are connected by a lossless line with X = 0.5 pu. Both voltages are 1.0 pu, and the angle between them is 20°. What real power flows across the line? Enter per-unit.
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Answer: 0.684 pu (accepted range 0.677–0.691 pu)
Why: P = V1 V2 sin δ / X = (1.0)(1.0) sin 20° / 0.5 = 0.342/0.5 = 0.684 pu.
Common mistakes: Using cos 20° gives 1.88 pu. Forgetting to divide by X gives 0.342 pu.
Source: MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation)
Question 63 of 80: Which code governs a utility substation
PEP-63 · Electrical Safety · Special occupancies and systems · Choose one answer
An electric utility is designing a new distribution substation that will be under its exclusive control. Among the codes NCEES supplies for this exam, which primarily governs the safety rules for installing this station?
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Answer: B. 2017 NESC (ANSI C2), Part 1
Why: The NESC covers the installation, operation, and maintenance of electric supply and communication lines and associated utility equipment. In the 2017 NESC, Part 1 is the safety rules for electric supply stations and equipment (Part 2 overhead lines, Part 3 underground lines, Part 4 work rules). Utility-controlled installations like this are outside the NEC's scope.
Why the other choices miss: The NEC covers premises wiring. NFPA 70E covers safe work practices, not installation design. NFPA 497 is a recommended practice for classifying hazardous areas.
Source: IEEE SA — National Electrical Safety Code (NESC) overview — Scope; 2017 National Electrical Safety Code — table of contents (library record) — Parts 1–4; Mike Holt, Illustrated Guide to Changes to the NEC (sample pages): 90.2(B)(5) electric utilities — 90.2(B)(5)
Question 64 of 80: Line-to-line from line-to-neutral
PEP-64 · Circuit Analysis · Phasor diagrams · Choose one answer
A balanced wye source has ABC (positive) phase sequence and Van = 277∠0° V. What is Vab?
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Answer: C. 480∠+30° V
Why: For positive sequence, line-to-line voltages are √3 larger than line-to-neutral voltages and lead them by 30°. Vab = √3 × 277∠(0° + 30°) = 480∠30° V.
Why the other choices miss: 480∠−30° is the result for ACB (negative) sequence. 277∠30° forgets the √3. 480∠0° forgets the phase shift.
Source: MIT OCW 6.061 course notes, Ch. 3: Polyphase networks — Line-line voltages section; Sec. 6 three-phase transformers; Sec. 8.2
Question 65 of 80: Current-limiting fuses
PEP-65 · Protection · Protective devices · Choose one answer
What makes a current-limiting fuse 'current limiting'?
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Answer: D. Within its current-limiting range, it opens in less than one-half cycle, before fault current reaches its available peak
Why: Current-limiting fuses act so fast that, within their current-limiting range, they open in less than half a cycle. That keeps the let-through current below the peak the circuit could otherwise deliver, which reduces thermal and magnetic stress on downstream equipment.
Why the other choices miss: Time delay is a separate characteristic (dual-element fuses). Every fuse opens on sustained overload, which isn't what 'current limiting' means. Fuses are replaced, not reset.
Source: U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 3 — ¶3-3e (current-limiting fuses); ¶3-7 (automatic reclosing)
Question 66 of 80: Open- and short-circuit tests
PEP-66 · Electric Power Devices · Testing · Select all that apply
Which statements about transformer open-circuit and short-circuit tests are correct? Select all that apply.
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Answer: A and B
Why: The open-circuit test applies rated voltage with one winding open, so the input power is essentially core loss. The short-circuit test circulates rated current through a shorted winding at reduced voltage, so the input power is essentially copper loss. Together they give the losses used in the efficiency formula.
Why the other choices miss: The open-circuit test is at rated voltage, not rated current. Rated voltage on a short-circuited transformer would produce a fault, not a test.
Source: Southern Illinois University ET 332b, Lesson 10: transformer regulation, efficiency, short-circuit current — Slide 3 (regulation); slides 11–12 (rated current, % impedance); slides 26–27 (Isc); slide 29 (efficiency)
Question 67 of 80: Equipment grounding conductor size
PEP-67 · General Applications · Grounding · Choose one answer
A feeder is protected by a 100 A circuit breaker. Using NEC Table 250.122, what is the minimum copper equipment grounding conductor (assume the ungrounded conductors were not upsized for voltage drop)?
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Answer: A. 8 AWG
Why: Table 250.122 sizes the equipment grounding conductor from the rating of the overcurrent device ahead of it, not from the conductor ampacity. For 100 A, the minimum is 8 AWG copper (6 AWG aluminum).
Why the other choices miss: 10 AWG is the 60 A row. 6 AWG is the 200 A row. Matching the phase conductors is not the rule (though 250.122(B) does require a proportional increase if the phase conductors are upsized for voltage drop).
Source: ExpertCE: Sizing grounding conductors — NEC 250.122 — Table 250.122 rows
Question 68 of 80: Steady-state stability limit
PEP-68 · Transmission and Distribution Analysis · Power system stability · Choose one answer
For the same line (V1 = V2 = 1.0 pu, X = 0.5 pu), the transfer must rise to 1.5 pu. What is the steady-state power angle, and what is the theoretical maximum transfer?
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Answer: B. 48.6°; 2.0 pu
Why: P_max = V1 V2 / X = 1.0/0.5 = 2.0 pu, reached at δ = 90°. For 1.5 pu: sin δ = 1.5/2.0 = 0.75, so δ = 48.6°. The closer δ gets to 90°, the less margin remains before the systems pull out of step.
Why the other choices miss: 30.0° solves sin δ = 0.5. 41.4° uses cos⁻¹ instead of sin⁻¹. The maximum depends on the voltages and reactance, not on the present loading.
Source: MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation)
Question 69 of 80: Shock vs arc-flash boundaries
PEP-69 · Electrical Safety · Shock and burns · Select all that apply
Which of these are shock protection boundaries under NFPA 70E? Select all that apply.
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Answer: A and B
Why: NFPA 70E uses two shock protection boundaries: the limited approach boundary (inside it a shock hazard exists) and the restricted approach boundary (inside it the likelihood of shock is highest; only qualified workers may cross). The arc flash boundary is a burn boundary, and it can fall inside or outside the shock boundaries.
Why the other choices miss: The arc flash boundary protects against thermal injury, not shock. NEC working space is an installation clearance, not a 70E approach boundary.
Source: OSHA Fact Sheet FS-4474 (11/2024): Establishing Boundaries Around Arc Flash Hazards — Boundary definitions; footnote 3
Question 70 of 80: Identify Vbc on a phasor diagram
PEP-70 · Circuit Analysis · Phasor diagrams · Select a spot on the diagram
The diagram shows four line-to-line voltage phasors (480 V each) drawn with Van = 277∠0° as the reference, ABC sequence. Select the phasor that represents Vbc.
Diagram labels:
- P — Phasor P at +30°
- Q — Phasor Q at −90°
- R — Phasor R at +150°
- S — Phasor S at −30°
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Answer: Q (Phasor Q at −90°)
Why: Vab = 480∠30°. In ABC sequence each line-to-line voltage lags the previous one by 120°, so Vbc = 480∠(30° − 120°) = 480∠−90° (phasor Q) and Vca = 480∠150° (phasor R).
Common mistakes: P is Vab. R is Vca. S (−30°) is where Vab would sit in ACB sequence.
Source: MIT OCW 6.061 course notes, Ch. 3: Polyphase networks — Line-line voltages section; Sec. 6 three-phase transformers; Sec. 8.2
Question 71 of 80: Reclosers
PEP-71 · Protection · Protective devices · Choose one answer
Why are automatic reclosers widely used on overhead distribution circuits but not on circuits feeding cables or transformers?
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Answer: C. Most overhead faults are temporary and clear when the line is briefly de-energized, while cable and transformer faults usually are not temporary
Why: Most overhead distribution faults are temporary (the S&C bulletin cites more than 80%), such as a branch brushing a line. Opening and reclosing restores service without a sustained outage. Faults in cable and transformer loads are usually permanent, so reclosing into them only adds damage — which is why TM 5-811-14 applies reclosers to aerial systems.
Why the other choices miss: None of the other options is the deciding factor. What matters is whether a fault is likely to be temporary: reclosing restores service after a temporary fault, but reclosing into a permanent cable or transformer fault just adds damage.
Source: S&C Electric, TripSaver II Cutout-Mounted Recloser, Descriptive Bulletin 461-32 — pp. 1–2; U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 3 — ¶3-3e (current-limiting fuses); ¶3-7 (automatic reclosing)
Question 72 of 80: Efficiency at half load
PEP-72 · Electric Power Devices · Testing · Choose one answer
Tests on a 500 kVA transformer give a core loss of 1.2 kW and a full-load copper loss of 4.8 kW. What is its efficiency at half load and 0.9 power factor?
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Answer: D. 98.9%
Why: Output = 0.5 × 500 × 0.9 = 225 kW. Copper loss scales with current squared: 4.8 × 0.5² = 1.2 kW. Core loss stays 1.2 kW. η = 225/(225 + 1.2 + 1.2) = 98.9%.
Why the other choices miss: 97.4% uses full-load copper loss at half load. 98.7% is the full-load efficiency. 99.5% leaves out core loss.
Source: Southern Illinois University ET 332b, Lesson 10: transformer regulation, efficiency, short-circuit current — Slide 3 (regulation); slides 11–12 (rated current, % impedance); slides 26–27 (Isc); slide 29 (efficiency)
Question 73 of 80: Single rod electrode
PEP-73 · General Applications · Grounding · Choose one answer
A service has a single driven ground rod as its only grounding electrode, and no resistance test has been done. What does NEC 250.53 require?
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Answer: A. A supplemental electrode, installed not less than 6 ft from the rod, unless the single rod is shown to be 25 Ω or less
Why: A rod electrode must be supplemented by an additional electrode unless the single rod has a resistance to earth of 25 Ω or less. The supplemental electrode must be at least 6 ft away. Without a test showing 25 Ω or less, you install the second electrode — and two rods is the maximum the rule requires.
Why the other choices miss: One rod isn't automatically enough. Rods closer than 6 ft overlap their resistance areas and gain little. The NEC doesn't require driving rods until a 5 Ω target is met.
Source: EC&M: Installing a Grounding Electrode Using Ground Rods (Holt, 2017) — 250.53(A)(2)–(3)
Question 74 of 80: Arc flash boundary threshold
PEP-74 · Electrical Safety · Shock and burns · Choose one answer
NFPA 70E defines the arc flash boundary as the distance at which the incident energy equals:
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Answer: B. 1.2 cal/cm² (about 5.0 J/cm²)
Why: The arc flash boundary is where incident energy falls to 1.2 cal/cm² (5.0 J/cm²) — the level at which an unprotected worker could receive a second-degree burn.
Why the other choices miss: The other values are often seen on arc-rated clothing and labels, but they are not the boundary definition.
Source: OSHA Fact Sheet FS-4474 (11/2024): Establishing Boundaries Around Arc Flash Hazards — Boundary definitions; footnote 3
Question 75 of 80: Complex power
PEP-75 · Circuit Analysis · Single-phase circuits · Choose one answer
A 240 V single-phase source supplies a load impedance of 8 + j6 Ω. What are the real power and power factor?
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Answer: C. 4,608 W at 0.80 lagging
Why: |Z| = 10 Ω, so I = 240/10 = 24 A. S = V I* = 240 × 24∠36.87° = 5,760∠36.87° VA = 4,608 W + j3,456 var. PF = 4,608/5,760 = 0.80, lagging because the load is inductive (+j6).
Why the other choices miss: 5,760 is the apparent power in VA. 3,456 is the reactive power. An inductive load has a lagging, not leading, power factor.
Source: MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation)
Question 76 of 80: Relay coordination time interval
PEP-76 · Protection · Coordination · Choose one answer
For the maximum fault on a feeder, the feeder relay operates in 0.25 s. Using a 0.3 s coordination time interval between relays in series, what is the minimum operating time for the upstream relay at that same fault current?
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Answer: D. 0.55 s
Why: Upstream time ≥ downstream time + CTI = 0.25 + 0.30 = 0.55 s. TM 5-811-14 gives 0.3–0.4 s as the usual interval between relays in series.
Why the other choices miss: 0.30 s is the interval alone. 0.45 s uses too small a margin. 1.00 s works but isn't the minimum and slows fault clearing.
Source: U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 4 — ¶4-2d(2) (relay coordination interval); ¶4-6c (transformer inrush)
Question 77 of 80: Rubber gloves and arc flash (70E-2021)
PEP-77 · Electrical Safety · Shock and burns · Choose one answer
A qualified electrician wears rubber insulating gloves with leather protectors for a task inside both the restricted approach boundary and the arc flash boundary. What did the 2021 edition of NFPA 70E change about how those gloves are treated?
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Answer: A. It recognizes rubber insulating gloves with leather protectors as providing arc flash protection in addition to shock protection
Why: The 2021 edition states that rubber insulating gloves with leather protectors provide arc flash protection in addition to shock protection. The gloves still must satisfy the shock-protection requirements for the voltage and task.
Why the other choices miss: Leather protectors are part of the recognized combination, not prohibited. Requiring separate arc-rated gloves over them is the pre-2021 assumption this change addresses. Shock protection inside the restricted approach boundary still applies.
Source: IRINFO: Significant Changes in the 2021 Edition of NFPA 70E (Nelson, Feb. 2021) — Table 130.5(C) likelihood change
Question 78 of 80: Capacitor for unity PF
PEP-78 · Circuit Analysis · Single-phase circuits · Choose one answer
A 120 V, 60 Hz single-phase load draws 10 A at 0.60 power factor lagging. What parallel capacitance raises the power factor to 1.0?
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Answer: B. 176.8 μF
Why: S = 120 × 10 = 1,200 VA. P = 720 W, so Q = √(1,200² − 720²) = 960 var. The capacitor must supply 960 var: X_C = V²/Q = 14,400/960 = 15 Ω. C = 1/(2π × 60 × 15) = 176.8 μF.
Why the other choices miss: 132.6 μF sizes the capacitor from P instead of Q. 221.0 μF uses S. 44.2 μF uses 240 V instead of 120 V in X_C = V²/Q.
Source: MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation)
Question 79 of 80: Which device should open first
PEP-79 · Protection · Coordination · Select a spot on the diagram
A radial one-line diagram shows, from source to load: the service main breaker (M), a feeder breaker (F), a panelboard branch breaker (B), and a load. A bolted fault occurs on the branch circuit between B and the load. In a selectively coordinated system, select the device that should open.
Diagram labels:
- M — Service main breaker
- F — Feeder breaker
- B — Branch breaker
- L — Load
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Answer: B (Branch breaker)
Why: Selective coordination localizes an overcurrent to the circuit or equipment affected. The branch breaker nearest the fault opens; the feeder and main stay closed and serve as backup only if B fails.
Common mistakes: Opening F or M would drop healthy circuits. The load isn't a protective device.
Source: IAEI Magazine: Selective Coordination for Critical Systems (Neeser, 2014) — Article 100 definition
Question 80 of 80: DC feeder voltage drop
PEP-80 · Circuit Analysis · Direct current circuits · Choose one answer
A 125 V dc station battery feeds a 100 A load 200 ft away (one-way). Conductor resistance is 0.2 Ω per 1,000 ft. What are the voltage drop and percent drop?
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Answer: C. 8.0 V, 6.4%
Why: A dc circuit has a go and a return conductor: 2 × 200 ft = 400 ft. R = 400 × 0.0002 Ω/ft = 0.08 Ω. Drop = 100 A × 0.08 Ω = 8.0 V, which is 8.0/125 = 6.4%.
Why the other choices miss: 4.0 V counts only one conductor. 16 V doubles the length twice. 8.0 V at 3.2% divides by 250 V.
Source: MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks — Sec. 7 (complex power, power factor); Sec. 9–10 (power flow through a reactance, shunt compensation)
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Score yourself by knowledge area
Count your correct answers in each row. An item counts only if it's fully right — every correct box for "select all" questions, every pair for matching, and the full sequence for ordering. That matches how NCEES scores its alternative item types: right or wrong, with no partial credit (NCEES CBT page).
| Knowledge area | NCEES range on the real exam (of 80) | Questions in this set | Question numbers |
|---|---|---|---|
| 1. Measurement and Instrumentation | 6–9 | 7 | 5, 14, 23, 32, 41, 50, 59 |
| 2. General Applications | 8–12 | 9 | 7, 16, 25, 34, 43, 52, 61, 67, 73 |
| 3. Electrical Safety | 10–15 | 11 | 2, 11, 20, 29, 38, 47, 56, 63, 69, 74, 77 |
| 4. Circuit Analysis | 10–15 | 12 | 3, 12, 21, 30, 39, 48, 57, 64, 70, 75, 78, 80 |
| 5. Power Electronic Circuits and Control Devices | 5–8 | 6 | 9, 18, 27, 36, 45, 54 |
| 6. Rotating Machines | 5–8 | 6 | 8, 17, 26, 35, 44, 53 |
| 7. Electric Power Devices | 8–12 | 9 | 6, 15, 24, 33, 42, 51, 60, 66, 72 |
| 8. Transmission and Distribution Analysis | 8–12 | 9 | 1, 10, 19, 28, 37, 46, 55, 62, 68 |
| 9. Protection | 10–15 | 11 | 4, 13, 22, 31, 40, 49, 58, 65, 71, 76, 79 |
| Total | 80 | 80 |
Our counts sit inside every NCEES range, but they are our practice design. NCEES publishes ranges, not the exact mix on your exam form (PE Power specification). With 6–12 questions per area here, treat an area score as a pointer to what to study next, not a measurement.
Show / hide the complete answer key
Answer key
1: A · 2: B · 3: 180.4 A · 4: C · 5: D · 6: A · 7: B · 8: 2.78% · 9: 648.2 V · 10: C
11: D · 12: A · 13: B · 14: C · 15: D · 16: A · 17: 1→D, 2→A, 3→B, 4→C · 18: B · 19: C · 20: D
21: A · 22: A+B+C · 23: A+B · 24: 2.5% · 25: A+B · 26: B · 27: C · 28: D · 29: C+D+E · 30: A
31: B · 32: C · 33: D · 34: A · 35: B · 36: C · 37: D · 38: A+B+C · 39: A · 40: B
41: C · 42: D · 43: A · 44: BDAC · 45: B · 46: C · 47: D · 48: A · 49: B · 50: 30.9 m
51: C · 52: 66.7% · 53: D · 54: 27→D, 50→A, 51→B, 86→C · 55: A+B+C · 56: 1→C, 2→A, 3→B · 57: A · 58: B · 59: C · 60: D
61: A · 62: 0.684 pu · 63: B · 64: C · 65: D · 66: A+B · 67: A · 68: B · 69: A+B · 70: Q
71: C · 72: D · 73: A · 74: B · 75: C · 76: D · 77: A · 78: B · 79: B · 80: C
What your practice score means
Use it to find weak areas and repeat mistakes. It isn't an NCEES score or a pass prediction. NCEES reports the real exam as pass/fail, converts raw scores to a scaled score, and doesn't publish the passing score because it varies slightly with difficulty (NCEES Examinee Guide, p. 14). These 80 questions haven't been equated to an NCEES form, so this set has no validated passing percentage.
One rule that does carry over: NCEES doesn't deduct points for wrong answers, so never leave a question blank on exam day (Examinee Guide, p. 14).
Check your code editions before you study
The exam gives you the NCEES PE Power Reference Handbook plus these standards on screen, and it scores code answers against the listed edition only: solutions based on other editions get no credit (PE Power specification, codes and standards page). If you work from a newer code book at your job, these are the differences most likely to cost you points.
| Standard | Edition supplied on the exam | Why the edition matters |
|---|---|---|
| NFPA 70 (NEC) | 2020 | Office lighting unit load is 1.3 VA/ft², and the 2020 table note includes the 125% continuous-load multiplier in that unit load (the 2017 NEC used 3.5 VA/ft²) (EC&M). Article 242 replaced Articles 280 and 285, and 230.67 added surge protection for dwelling services (EC&M). The bundling table is Table 310.15(C)(1) (EC&M). A 10 A standard breaker rating doesn't exist until the 2023 edition (EEPower). |
| NFPA 70E | 2021 | The 2021 edition recognizes rubber insulating gloves with leather protectors as arc flash protection as well as shock protection, and Annex D now points to IEEE 1584-2018 for arc flash calculations (IRINFO). |
| ANSI C2 (NESC) | 2017 | Organized in four parts: supply stations, overhead lines, underground lines, and work rules (2017 NESC contents). |
| NFPA 30B | 2023 | Aerosol products — listed on the exam. |
| NFPA 497 | 2021 | Classifying flammable gas and vapor locations — listed on the exam. |
| NFPA 499 | 2021 | Classifying combustible dust locations — listed on the exam. |
On exam day the standards open one chapter at a time, and Ctrl+F doesn't work — you search with the search box beside the reference (specification; Examinee Guide, p. 10). Practice finding the 2020 section numbers, not just the rules. If NCEES posts a new Power specification or codes list for your test date, use that list instead of this table.
PE Power exam at a glance
| Item | What NCEES says |
|---|---|
| Questions | 80 (NCEES) |
| Time | 9-hour appointment: 2-minute nondisclosure agreement, 8-minute tutorial, 8-hour exam, 50-minute scheduled break (NCEES) |
| Pace | 8 hours ÷ 80 questions = 6 minutes per question on average (our arithmetic) |
| When and where | Computer-based, year-round, at Pearson VUE test centers (NCEES) |
| Question types | Multiple choice plus alternative item types: multiple correct, point and click, drag and drop, and fill in the blank. No partial credit (NCEES CBT) |
| Units | SI and US Customary (specification) |
| References | Closed book; electronic Reference Handbook and listed standards provided on screen (specification) |
| Calculator | One NCEES-approved model: Casio fx-115 and fx-991 models; HP 33s and HP 35s only; TI-30X and TI-36X models (NCEES exams page) |
| Fee | $400 to NCEES; your licensing board may charge separately (NCEES) |
| Results | Usually 7–10 days (NCEES) |
| Pass rates (July 2026 update) | First-time takers: 62% of 1,344. Repeat takers: 42% of 721 (NCEES) |
What to do with your misses
A missed question is only useful if you know why you missed it. For each one, write down the question number and which of these went wrong:
- Concept: you didn't know the principle (for example, that open-delta capacity is √3 × one unit, not two units).
- Setup: right idea, wrong equation or base (a per-unit base mix-up, single-phase factor on a three-phase drop).
- Code edition or lookup: you knew a rule but from the wrong edition, or couldn't find it fast.
- Units or arithmetic: √3 slips, kVA vs kW, percent vs per-unit.
- Misread: you answered a different question than the one asked.
Then group misses by knowledge area using the table above. Relearn the method behind the biggest group, and practice on new problems in that area before you come back to these — repeating the same item mostly tests memory. For broader PE exam prep guidance and mixed practice, use our PE exam prep hub.
Quick answers
Is the PE Power exam open book? No. It's closed book with an electronic reference: NCEES supplies the Reference Handbook and the listed standards on screen, and you can't bring your own (NCEES).
Which NEC edition is on the exam? The 2020 NEC (NFPA 70-2020), under the specification effective beginning October 2025 (specification).
Does NCEES have its own practice exam? Yes. NCEES says its PE prep materials, including a practice exam, are in the exam-prep area of your MyNCEES account (NCEES; MyNCEES exam prep). It's the only practice written by the exam owner, so it's worth taking alongside free sets like this one.
Should I guess if I'm stuck? Yes. Wrong answers aren't penalized (Examinee Guide, p. 14). Flag it, pick your best answer, and move on.
What if I don't pass? You'll get a diagnostic report by knowledge area. You can retake once per testing window and no more than three times in 12 months, and some boards are stricter (Examinee Guide, pp. 5 and 14). Details: NCEES exam retake policy and NCEES exam results.
Who decides whether I can sit for the exam? Your state licensing board. Register through NCEES, but check your board's requirements first (NCEES board directory; our NCEES exam registration guide).
Sources, verification, and independence
By the Castleport Test Prep Editorial Team. Last verified October 8, 2026: PE Power exam format, the October 2025 specification and knowledge-area ranges, the supplied codes and editions, NCEES item types, calculator list, Examinee Guide policies, and the July 2026 pass rates were checked against the NCEES pages listed below.
How we made these questions: each question is original and mapped to the published NCEES specification, with the technical or code reference used for the underlying rule listed directly under the item. During the October 8, 2026 audit, the calculation keys, item counts, answer-key synchronization, current NCEES exam facts, and higher-risk code-edition claims were rechecked. The questions have not been reviewed by a licensed professional engineer. Several code questions rely on published explanations of the 2020 NEC and NFPA 70E rather than the full code text; the source line under each question names the reference used.
Official NCEES sources
- NCEES — PE Electrical and Computer exam page (Power section; Scoring & Pass Rates)
- NCEES — PE Electrical and Computer: Power CBT Exam Specifications (effective October 2025)
- NCEES — Computer-based testing: alternative item types
- NCEES Examinee Guide (May 2026)
- NCEES — Exams page (calculator policy)
Technical and code references used in the questions
- Southern Illinois University ET 332b, Lesson 10: transformer regulation, efficiency, short-circuit current
- EC&M: Code Q&A — Branch Circuit Overcurrent Protection (Holt)
- EEPower: Protecting Motor Branch Circuits from Short Circuits and Ground Faults (Mari, 2024)
- MIT OCW 6.061 course notes, Ch. 3: Polyphase networks
- Southern Illinois University ET 332b, Lesson 11: transformer nameplate data and connections
- U.S. Bureau of Reclamation FIST Vol. 3-8: Protective Relays and Associated Circuits (rev. 9/2021)
- EC&M Code Q&A: Bonding the lightning protection system (Holt)
- MIT OCW 6.061 course notes, Ch. 10: Induction machines
- Linköping University TSTE19 Power Electronics, Lecture 3 (rectifiers)
- University of Utah ECE 3600 — Per-unit notes (Stolp)
- EC&M: Code Q&A — Conductor Ampacity Adjustment Requirements (Holt, 2021; 2020 NEC)
- NJIT ECE 342 Lab 2: Three-phase power measurements
- EC&M: NEC Requirements for Overvoltage Protection (Holt, 2023; 2020 NEC)
- EC&M: 2020 National Electrical Code Changes
- Southern Illinois University ET 332b, Lesson 19: power factor correction using synchronous motors
- All About Circuits, Lessons in Electric Circuits Vol. II, Ch. 13: Wound-rotor induction motors
- MIT OCW 6.061 course notes, Ch. 2: AC power flow in linear networks
- IAEI Magazine: Working Space for Electrical Equipment (Hunter, 2013)
- IAEI Magazine: Performance testing requirements for ground-fault protection equipment
- U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 5
- EC&M: General Requirements of the NEC (Holt, 2020)
- MIT OCW 6.061 course notes, Ch. 4: Introduction to symmetrical components
- U.S. Army TM 5-811-14, Appendix E: Partial relay device numbers list
- U.S. Bureau of Reclamation FIST Vol. 3-1: Testing Solid Insulation of Electrical Equipment
- MIT OCW 4.401 Lecture 11 (lighting): inverse-square law for a point source
- Schneider Electric Digest: Reduced-voltage starters — description of all types
- EC&M: Reduced-Voltage Starters — Choosing the Best Type (Hinman, 1998)
- OSHA 29 CFR 1910.399 — Definitions (hazardous locations)
- EC&M: Understanding Load Calculations and the 2020 NEC (Vigstol, 2019)
- EC&M: The Basics of Ladder Logic (Fehr, 2003)
- Eaton: How do you calculate THD and TDD? (Harmonics FAQ)
- OSHA 29 CFR 1910.307 — Hazardous (classified) locations
- U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 4
- Metrel application note: Earth resistance measurement and the 62% rule
- Penn State AE 868: Interpreting a PV manufacturer datasheet
- HOMER Pro user manual: Load factor
- EC&M: Essential Electrical Systems at Health Care Facilities (Biason)
- Queen Mary University of London — Power in the wind (course notes)
- EC&M: Ensuring Accuracy in Demand Factors with the NEC (Kuether, 2025)
- IEEE SA — National Electrical Safety Code (NESC) overview
- 2017 National Electrical Safety Code — table of contents (library record)
- Mike Holt, Illustrated Guide to Changes to the NEC (sample pages): 90.2(B)(5) electric utilities
- U.S. Army TM 5-811-14 Coordinated Power Systems Protection, Ch. 3
- ExpertCE: Sizing grounding conductors — NEC 250.122
- OSHA Fact Sheet FS-4474 (11/2024): Establishing Boundaries Around Arc Flash Hazards
- S&C Electric, TripSaver II Cutout-Mounted Recloser, Descriptive Bulletin 461-32
- EC&M: Installing a Grounding Electrode Using Ground Rods (Holt, 2017)
- IRINFO: Significant Changes in the 2021 Edition of NFPA 70E (Nelson, Feb. 2021)
- IAEI Magazine: Selective Coordination for Critical Systems (Neeser, 2014)
Castleport Test Prep is an independent exam prep publisher, not affiliated with, endorsed by, or approved by NCEES. These are original practice questions, not NCEES exam questions. Exam and credential names identify the subject of this resource; trademarks belong to their respective owners. Practice results don't guarantee a passing score or licensure.