Power factor correction practice problems, single and three-phase

Invented AC power problems for electrical apprentices: the power triangle, capacitor kVAr and µF, and star and delta line and phase current. Press Check to mark each answer, or open the working.

Educational practice only. Not for sizing real capacitor banks, cables or protective devices.

Set

Rules, signs and rounding

      Answers are saved on this device as you type.

      The formulas, in the order you use them

      • Power triangle: S² = P² + Q², power factor = cos φ = P ÷ S, Q = P × tan φ.
      • Single-phase: S = V × I and I = P ÷ (V × cos φ).
      • Correction keeps P: QC = Q₁ − Q₂ = P (tan φ₁ − tan φ₂).
      • Capacitance: C = QC ÷ (2π f V²), with QC in var and V across the capacitor.
      • Star: Vph = VL ÷ √3, IL = Iph. Delta: Vph = VL, IL = √3 × Iph.
      • Three-phase power: P = √3 × VL × IL × cos φ, and IL = P ÷ (√3 × VL × cos φ).

      The course menu sets the voltages, the frequency, kVAr or kVAR, φ or θ, star or wye, and the rounding rule. Its syllabus is named under Rules.

      An example from set EXPF24

      A single-phase motor takes 4.1 kW from 230 V, 50 Hz at a power factor of 0.6 lagging, to be raised to 0.92. tan φ₁ = 0.8 ÷ 0.6 = 1.3333, so Q₁ = 4.1 × 1.3333 = 5.47 kVAr. tan φ₂ = 0.426, so Q₂ = 1.75 kVAr. The capacitor supplies QC = 5.47 − 1.75 = 3.72 kVAr, and C = 3720 ÷ (2π × 50 × 230²) = 224 µF. The current falls from 29.7 A to 19.4 A. Keeping S the same instead of P gives Q₂ = 2.68 kVAr, a common slip. Open set EXPF24

      Questions

      How do you work out the capacitor for power factor correction?

      Find Q₁ = P × tan φ₁ and Q₂ = P × tan φ₂. The capacitor supplies the difference, QC = Q₁ − Q₂. Then C = QC ÷ (2π f V²), with QC in var and V the voltage across the capacitor. Multiply by 10⁶ for microfarads.

      Why does P stay the same after correction?

      An ideal capacitor takes no real power. It only supplies reactive power, so the load still does the same work while Q and S fall. Keeping S the same instead of P is a common slip.

      What is the difference between kW, kVA and kVAr?

      kW is real power, the part that does work. kVAr is reactive power, which flows back and forth in the inductance. kVA is apparent power, V × I. They form a right-angled triangle: kVA² = kW² + kVAr², and power factor = kW ÷ kVA.

      Which voltage goes across three-phase correction capacitors?

      In delta each capacitor sits between two lines, so it has the line voltage. In star it has the phase voltage, VL ÷ √3. Each capacitor supplies a third of the total kVAr, so star capacitors need three times the capacitance of delta ones.

      Why is three-phase power √3 × line voltage × line current × cos φ?

      Total power is 3 × Vph × Iph × cos φ. In star Vph = VL ÷ √3 and Iph = IL; in delta Vph = VL and Iph = IL ÷ √3. Either way 3 ÷ √3 = √3 is left, so P = √3 VL IL cos φ.

      Can I use this to size a capacitor bank for a real motor?

      No. Every load here is invented, and there is no box for your own load. Real correction needs the measured load, harmonics, switching, discharge resistors, ratings and protection, which these exercises leave out.

      Sources, read 2 November 2026: T. R. Kuphaldt, Lessons in Electric Circuits, vol. II, Calculating Power Factor and Practical Power Factor Correction (the All About Circuits textbook); City & Guilds 2365-03 qualification handbook, Unit 302; DHET Electrical Trade Theory N2 and N3 syllabi; NIMI Electrician 2nd year trade theory; BIS IS 12360; NCCER Electrical curriculum.