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PE Power exam: what the problems actually look like

In NCEES Squared 2025 (FY2024-25), PE Power had the lowest first-time pass rate of any PE depth most engineers take. The mathematics is not the reason; most of it is algebra you did in your second year. The reason is that four plausible numbers sit in front of you, three of them are mistakes you could genuinely make, and some of what is being tested is not in the handbook at all. Here are three practice problems, worked in full.

· by the FE → PE Prep team

80

Questions

9 hours

Appointment

9

Knowledge areas

6

Codes supplied

The anatomy of a PE Power question

Four things are true of nearly every question on this exam, and together they explain why engineers who know the material still fail it.

  • It says “most nearly.” The options are close. You cannot eliminate three of them by inspection, which means you have to actually do the problem.
  • The wrong answers are engineered. A good distractor is not a random number. It is the answer you get from one specific, plausible error. A clean-looking result is not evidence you are right; every option in these three problems is somebody’s honest mistake.
  • Some of it is not in the handbook. NCEES supplies the PE Electrical and Computer: Power Reference Handbook and six codes, and the specification examines practice that lives in those codes and in IEEE standards. Formula recall is not the whole job; knowing the convention is.
  • The clock is tighter than the appointment suggests. The 9-hour appointment holds 8 hours of exam time; the rest is the nondisclosure agreement, an 8-minute tutorial and a 50-minute scheduled break. That leaves 6 minutes per question, and it has to cover reading, a handbook or code lookup, and checking.

Three practice problems

These are three of our free PE Power questions, figures included. Try each one before you open the solution. When you do open it, read the distractor notes, not just the answer.

1. Full-load line current of a three-phase motor

The most routine calculation in power engineering, and it is on the exam precisely because it is routine. Nothing here is difficult. The question is whether you keep three separate factors straight while tired.

  1. 1
    medium
    A. Three-phase circuits
    A three-phase induction motor delivers 50 hp at full load. At this load its efficiency is 91% and its power factor is 0.86 lagging. The motor is supplied at 480 V (line-to-line). The full-load line current is most nearly:
    Three-phase induction motor fed by three line conductors A, B and C, with the line-to-line voltage marked between lines A and B and the line current on line A3-phase supplyABCMVLL = 480 VIL50 hpη = 91%pf = 0.86 lag

    Figure — drag or pinch to move around

    1. 44.9 A
    2. 49.3 A
    3. 52.2 A
    4. 57.3 A
    Show the answer & worked solution

    Answer: D.57.3 A

    Shaft output: Pout=50×746 W=37.3P_{out}=50\times746\text{ W}=37.3 kW. Electrical input: Pin=Pout/η=37.3/0.91=40.99P_{in}=P_{out}/\eta=37.3/0.91=40.99 kW. The handbook's induction-motor power flow (§4.2.5) gives Pin=3 VTILcos⁡θP_{in}=\sqrt3\,V_TI_L\cos\theta, so IL=40.99×1033(480)(0.86)=57.3I_L=\dfrac{40.99\times10^3}{\sqrt3(480)(0.86)}=57.3 A. Distractors: 52.2 A leaves out the efficiency, taking the shaft output as the electrical input; 49.3 A leaves out the power factor, treating the input kW as kVA; and 44.9 A divides the shaft output by 3 VT\sqrt3\,V_T with neither the efficiency nor the power factor.
What the distractors teach

Every wrong answer drops a factor. Skip the efficiency and treat shaft output as electrical input: 52.2 A. Skip the power factor and treat real power as apparent power: 49.3 A. Skip both and divide the shaft output by √3 times the line voltage alone: 44.9 A. All three are clean-looking numbers, and none of them is a slip you would notice by staring at your answer.

2. Power factor after switching in a capacitor bank

A capacitor bank problem looks like subtraction, and that is the trap. Reactive power comes off the reactive leg only, the real power does not move, and the load's kvar has to come from its kW and power factor, not from a kVA the question never gives you.

  1. 2
    medium
    C. Power factor correction
    A plant load draws 600 kW at 0.72 lagging power factor. A 250 kvar, three-phase capacitor bank is then switched onto the plant bus. The resulting power factor is most nearly:
    Figure for this questionUtilityPlant busQC bankP, PF1 lag

    Figure — drag or pinch to move around

    1. 0.771
    2. 0.877
    3. 0.919
    4. 0.964
    Show the answer & worked solution

    Answer: B.0.877

    A shunt bank changes only the reactive power and leaves the real power fixed, so the handbook's correction relation (§5.1.3), ΔPreactive=Preal(tan⁡Φ1−tan⁡Φ2)\Delta P_{reactive}=P_{real}(\tan\Phi_1-\tan\Phi_2), holds with PrealP_{real} constant. The load's reactive power is Q1=Ptan⁡Φ1=600tan⁡(cos⁡−10.72)=600×0.9639=578.3Q_1=P\tan\Phi_1=600\tan(\cos^{-1}0.72)=600\times0.9639=578.3 kvar. The bank supplies ΔPreactive=250\Delta P_{reactive}=250 kvar, leaving Q2=578.3−250=328.3Q_2=578.3-250=328.3 kvar, so tan⁡Φ2=328.3/600\tan\Phi_2=328.3/600 and PFnew=cos⁡(tan⁡−1328.3600)=0.877PF_{new}=\cos\left(\tan^{-1}\dfrac{328.3}{600}\right)=0.877 lagging. The distractors take the load's kvar as Psin⁡Φ1=416.4P\sin\Phi_1=416.4 instead of Ptan⁡Φ1P\tan\Phi_1 (0.964); hold the original 833.3 kVA fixed and take sin⁡Φ2=Q2/S1\sin\Phi_2=Q_2/S_1, although a shunt bank holds the kW, not the kVA, constant (0.919); and subtract only one phase's share of the bank, QC/3Q_C/3 (0.771).
What the distractors teach

Three different ways to mishandle the power triangle, each producing a believable power factor. Take the load's reactive power as P·sin Φ instead of P·tan Φ, putting the 600 kW where Q = S·sin Φ needs kVA, and you get 0.964. Keep the original 833 kVA and take sin Φ2 = Q2/S1, as if the bank only tilted the same triangle, and you get 0.919; a shunt bank holds the kW constant, not the kVA, so the new triangle is smaller. Subtract only one phase's share of the three-phase bank, 83 kvar instead of 250, and you get 0.771. Done right, the load's 578 kvar drops to 328 kvar against the same 600 kW, and the power factor is 0.877.

3. Zone 1 reach of a distance relay

This one is different in kind, and it is the reason this article exists. Distance relaying is named explicitly in the October 2025 specification. The PE Electrical and Computer: Power Reference Handbook lists the distance relay only as device number 21 and gives no reach settings. The 80% Zone 1 convention comes from IEEE C37 relaying practice, and you are expected to know it.

  1. 3
    medium
    B. Protective relaying (differential, distance, undervoltage, pilot)
    A 138 kV line has a positive-sequence impedance of 56 Ω56\ \Omega primary. Its distance (21) relay is supplied by 3000:5 CTs and by 80,500:115 V VTs connected line-to-ground. With Zone 1 set to 80% of the line, the Zone 1 reach in secondary (relay) ohms is most nearly:
    Figure for this questionBus S3000:580,500:115 V21tripreach = ? Ω (secondary)ZL = 56 Ω (primary)Zone 1 = 80%Bus R

    Figure — drag or pinch to move around

    1. 38.4 Ω
    2. 44.8 Ω
    3. 48 Ω
    4. 66.5 Ω
    Show the answer & worked solution

    Answer: A.38.4 Ω

    CTR =3000/5=600=3000/5=600 and VTR =80500/115=700=80500/115=700 (the VT's own rating — it is connected line-to-ground), so Zsec=Zpri×CTRVTR=56×0.8571=48 ΩZ_{sec}=Z_{pri}\times\dfrac{CTR}{VTR}=56\times0.8571=48\ \Omega. Zone 1 =0.8×48=38.4 Ω=0.8\times48=38.4\ \Omega secondary — deliberately short of the remote bus so measurement errors cannot cause an instantaneous overreach trip for the next line's faults. The other options set Zone 1 to the full line, leaving out the 80% margin (48 Ω); leave the reach in primary ohms, never referring it to the relay (44.8 Ω); and divide the 80,500 V VT primary by 3\sqrt3 as if it were a line-to-line rating (66.5 Ω).
What the distractors teach

Forget the 80% margin and set Zone 1 to the whole line: 48 Ω. Leave the reach in primary ohms, 80% of the 56 Ω line, and you get 44.8 Ω. Divide the 80,500 V VT rating by √3 as if it were line-to-line and you get 66.5 Ω, but these VTs are connected line-to-ground, so their own rating is already the phase voltage. The 80% is not arbitrary padding. It exists so that CT, VT and line-data errors cannot make an instantaneous zone overreach into the next line. Knowing why is what stops you picking 48.

The six codes NCEES hands you — and the catch

This is where PE Power differs sharply from PE Mechanical HVAC, which supplies only a handbook. Here the specification lists six standards, supplied inside the exam software as searchable PDFs, at these revision years:

Codes and standards supplied in the PE Power exam, per the NCEES October 2025 specification
StandardTitle
NFPA 70-2020National Electrical Code (NEC)
ANSI C2-2017National Electrical Safety Code (NESC)
NFPA 70E-2021Electrical Safety in the Workplace
NFPA 30B-2023Manufacture and Storage of Aerosol Products
NFPA 497-2021Classification of hazardous locations: gases and vapors
NFPA 499-2021Classification of hazardous locations: combustible dusts

Two things in that list cost people points. The revision years are scored: NCEES states that solutions based on other standards do not receive credit, so an answer worked from a newer NEC is wrong even when it is right in your office. And the standards are presented one chapter at a time: only a single chapter can be open and searched at once, so a lookup you have not rehearsed is expensive in a way that practicing with a paper code book never shows you.

Where the questions come from

NCEES publishes a question range for each knowledge area in its exam specification. It is the most useful study document you are probably not reading, because it tells you where your hours belong, and for this exam it contains a genuine surprise.

PE Electrical and Computer: Power knowledge areas and question counts, per the NCEES October 2025 specification
Knowledge areaQuestions
Electrical SafetyWiring methods, hazardous locations, special occupancies, shock and burns. Tied for the largest block, and it is mostly code work, not calculation.10–15
Circuit AnalysisThree-phase, symmetrical components, per-unit, phasors, single-phase and DC. The arithmetic backbone.10–15
ProtectionOvercurrent, protective relaying, devices and coordination. The area most engineers under-study relative to its weight.10–15
General ApplicationsLightning and surge protection, illumination, energy management, grounding.8–12
Electric Power DevicesTransformers, capacitors, storage, PV and wind, and testing.8–12
Transmission and Distribution AnalysisVoltage drop and regulation, power factor, power quality, fault current, transformer connections, power flow, stability.8–12
Measurement and InstrumentationInstrument transformers and metering, insulation testing, ground-resistance testing.6–9
Power Electronic Circuits and Control DevicesConverters, inverter-based resources and VFDs; relays, switches, Boolean and ladder logic.5–8
Rotating MachinesMachine types and applications, motor starting. Smaller than its reputation suggests.5–8

Read the top of that table again. Electrical Safety is tied with Circuit Analysis and Protection for the largest block on the exam, and it is almost entirely code work. Rotating Machines, which is what most people picture when they hear “power exam,” is one of the two smallest. Study time allocated by intuition tends to get this exactly backwards.

For the pass-rate picture alongside this, the companion guide covers it: how hard is the PE Power exam?

Common questions

What do PE Power exam questions actually look like?
Mostly multiple choice with four options, usually phrased 'most nearly', and the options sit close enough that you cannot eliminate them by inspection. Many carry a figure: a one-line diagram, a ladder-logic rung, a phasor diagram, a relay characteristic. NCEES gives you 8 hours of exam time for 80 questions, inside a 9-hour appointment that also holds the nondisclosure agreement, an 8-minute tutorial and a 50-minute scheduled break. The break does not come out of your exam time, so you have 6 minutes per question on average.
Which references are supplied in the PE Power exam?
NCEES specifies the exam as closed book with an electronic reference. You get the PE Electrical and Computer: Power Reference Handbook plus six codes and standards as searchable PDFs: the 2020 NEC, the 2017 NESC, NFPA 70E-2021, NFPA 30B-2023, NFPA 497-2021 and NFPA 499-2021. Two details matter on exam day: the standards are provided as individual chapters and only one chapter can be open and searched at a time, and NCEES scores standards questions against those revision years specifically, so an answer based on a newer NEC does not receive credit.
Is everything on the PE Power exam in the reference handbook?
No, and that is one of the harder things to discover late. NCEES says the handbook leaves out the theories, conversions, formulas and definitions examinees are expected to know, and the specification also examines practice that lives in the codes and in IEEE standards: protective relaying conventions, coordination, hazardous-location classification. The distance-relay problem in this article is a clean example: neither the primary-to-secondary ohms conversion nor the 80% Zone 1 convention is printed in the handbook.
How many questions are on each PE Power topic?
The October 2025 specification allocates 10-15 questions each to Electrical Safety, Circuit Analysis and Protection; 8-12 each to General Applications, Electric Power Devices, and Transmission and Distribution Analysis; 6-9 to Measurement and Instrumentation; and 5-8 each to Power Electronic Circuits and Control Devices and to Rotating Machines. The surprise for most candidates is that Electrical Safety is tied for the largest block while Rotating Machines is one of the smallest.
Does the PE Power exam use SI or US customary units?
Both. The specification states plainly that the exam uses both the International System of units and the US Customary System, so unit handling is part of what is being tested rather than a formatting detail. Practicing in only one system leaves a gap the exam will find.

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Ten free PE Power problems, the motor problem above among them, each with the figure and the full worked solution. No signup.

Format, knowledge areas and supplied standards are from the NCEES PE Electrical and Computer: Power specification effective October 2025. Specifications and fees change over time. Confirm current specifics at ncees.org. Independent study resource; not affiliated with, endorsed by, or sponsored by NCEES.