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Circuit Analysis: AC, Linear Circuits and Transients

A Y-connected load with an impedance of Z = 8 + j6 Ω per phase, as in figure (a), is connected to a balanced three-phase source with a line voltage of 200 V RMS. Which of the following is closest to the total three-phase active power?

Y (star) and Δ (delta) connections of a three-phase load (a) Y a Z b Z c Z N Ia Vp = VL/√3 (b) Δ a b c Z Z Z Ia Iab IL = √3 × Ip
Y (star) and Δ (delta) connections of a three-phase load · View in diagrams
  1. AAbout 1.1 kW
  2. B3.2 kW
  3. C9.6 kW
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Answer: B. 3.2 kW

Method 1 (one phase times 3): the phase voltage of the Y is 200/√3 ≈ 115.5 V and |Z| = √(8² + 6²) = 10 Ω, so the phase current (= line current) is about 11.55 A. The active power of one phase is I²R = 11.55² × 8 ≈ 1067 W, and for three phases 3 × 1067 ≈ 3200 W = 3.2 kW. Method 2 (three-phase power formula): the power factor is cos φ = 8/10 = 0.8, so P = √3 × VL × IL × cos φ = √3 × 200 × 11.55 × 0.8 ≈ 3.2 kW. Method 3 (simplify the formula): P = 3 × (VL/√3)² × R/|Z|² = VL² × R/|Z|² = 200² × 8/100 = 3200 W, which can be calculated without √3 at all.

  • A. About 1.1 kW is the power of one phase. The total for three phases is three times that.
  • B. Correct. P = √3 VL IL cos φ ≈ 3.2 kW.
  • C. 9.6 kW is calculated by applying the 200 V line voltage directly to each phase, which corresponds to the power of the same load connected in Δ.
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