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kW, kVA, kVAR in parallel impedances

Short answer

Branch 1: I₁=10A, 1.2kW, 1.6kVAR lag, 2kVA. Branch 2: I₂=8.94A, 0.8kW, 1.6kVAR lead, 1.789kVA. Total P=2kW, Q=0, S=2kVA, power factor=1 unity.

A voltage of 200∠53° V is applied across impedances in parallel. The value of impedances are (12 + j16)Ω and (10–j20)Ω. Determine the kW, kVA and kVAR in each branch and the power factor of the whole circuit.

The magnitude of the voltage is 200 V. Its phase angle (53°) does not affect the powers, which depend only on the magnitudes and the impedances. (The same problem in another paper gives 200∠0° V, with identical results.)

Z₁ = 12 + j16 = 20∠53.13° Ω, Z₂ = 10 − j20 = 22.36∠−63.43° Ω

Branch 1

  • I₁ = 200/20 = 10 A (lagging 53.13°)
  • kVA: S₁ = VI₁ = 200 × 10 = 2000 VA = 2.0 kVA
  • kW: P₁ = I₁²R = 100 × 12 = 1200 W = 1.2 kW
  • kVAR: Q₁ = I₁²X = 100 × 16 = 1600 VAR = 1.6 kVAR (lagging)
  • pf₁ = 12/20 = 0.6 lagging

Branch 2

  • I₂ = 200/22.36 = 8.944 A (leading 63.43°)
  • kVA: S₂ = 200 × 8.944 = 1789 VA = 1.789 kVA
  • kW: P₂ = I₂²R = 80 × 10 = 800 W = 0.8 kW
  • kVAR: Q₂ = I₂²X = 80 × 20 = 1600 VAR = 1.6 kVAR (leading)
  • pf₂ = 10/22.36 = 0.447 leading

Whole circuit

  • P = 1200 + 800 = 2000 W
  • Q = 1600 (lag) − 1600 (lead) = 0 VAR
  • S = √(2000² + 0²) = 2000 VA = 2 kVA

Total current I = S/V = 10 A.

Power factor = P/S = 2000/2000 = 1 (unity)

The lagging reactive power of the first branch is cancelled exactly by the leading reactive power of the second.

✓ Answer checked by Claude Opus 5.5 (high effort)

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