Free PE Power practice questions
Five sample questions from the EngiByte question bank, each with a worked solution. Try each one before you read the answer. These are original practice questions, not NCEES questions.
Question 1
A balanced three-phase load draws 100 A per line at 480 V line-to-line with a power factor of 0.9. What real power is it consuming? (P = √3 × V_LL × I_L × pf)
- 43.2 kW
- 74.8 kW
- 83.1 kW
- 129.6 kW
Answer: B. P = 1.732 × 480 × 100 × 0.9 ≈ 74.8 kW. The distractors are the classic slips: 43.2 kW forgets the √3 (treating it as single-phase), 83.1 kW forgets the power factor, and 129.6 kW multiplies by 3 instead of √3, the most common three-phase error. The √3 arises because line-to-line voltage is √3 times the phase voltage in a wye system; internalize that and three-phase power problems become one-line arithmetic.
Question 2
An industrial plant draws 1,000 kW at 0.80 lagging power factor. Approximately how much capacitive reactive power (kvar) must be added to correct the power factor to 0.95?
- ~170 kvar
- ~420 kvar
- ~750 kvar
- ~1,080 kvar
Answer: B. At 0.80 pf, the load's reactive power is Q = P·tan(cos⁻¹0.80) = 1,000 × 0.75 = 750 kvar. At 0.95 pf it would be 1,000 × 0.329 ≈ 329 kvar. The capacitors must supply the difference: 750 − 329 ≈ 421 kvar.
Question 3
A load draws 80 kW of real power and 60 kvar of reactive power. What is its apparent power in kVA?
- 20 kVA
- 100 kVA
- 140 kVA
- 4,800 kVA
Answer: B. Real and reactive power add like the legs of a right triangle: S = √(80² + 60²) = √(6,400 + 3,600) = 100 kVA, the classic 3-4-5 triangle scaled up.
Question 4
A 10 MVA transformer has 8% impedance and is fed from a source strong enough to treat as infinite. Approximately what fault level (short-circuit MVA) is available at its secondary terminals for a three-phase fault?
- 0.8 MVA
- 12.5 MVA
- 125 MVA
- 1,250 MVA
Answer: C. With an infinite source and nominal prefault voltage, the transformer limits the three-phase short-circuit level to 10 MVA/0.08 = 125 MVA. Finite source and cable series impedances reduce that source contribution. Other connected sources, prefault voltage, impedance tolerances, and DC offset must also be considered when selecting breaker duty.
Question 5
A three-phase induction motor has 4 poles and runs on a 60 Hz supply. What full-load speed would you expect on its nameplate?
- ~1,200 rpm
- ~1,750 rpm
- Exactly 1,800 rpm
- ~3,600 rpm
Answer: B. Synchronous speed is 120f/p = 120 × 60 / 4 = 1,800 rpm, but an induction motor must run slightly slower than the rotating field, the 'slip' is what induces rotor current and produces torque. Typical full-load slip of 2-3% gives roughly 1,750 rpm. Exactly 1,800 rpm is the trap: at synchronous speed an induction motor produces zero torque. The other options are the synchronous speeds of 6-pole (1,200) and 2-pole (3,600) machines.
Want more? The practice exam page describes the full mocks, and the cheat sheet lists the formulas used above.
Frequently asked questions
Are these real PE exam questions?
No. They are original practice questions written for EngiByte and are not NCEES questions.
Are the solutions explained?
Yes. Each answer shows the calculation and the common mistake the wrong options represent.
Where can I get more practice questions?
The EngiByte app has a diagnostic, a daily sprint and four timed mock exams, linked below.