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Delta load 20∠60° power and wattmeter readings

Short answer

Each phase impedance 20∠60° Ω draws 4840 W; total power 14.52 kW. Wattmeter readings: W1 = 14.52 kW, W2 = 0 W (cos φ = 0.5).

Each phase of a 3-phase, delta-connected load consists of an impedance Z = 20∠60° ohm. The line voltage is 440 V at 50 Hz. Compute the power consumed by each phase impedance and the total power. What will be the readings of the two wattmeters connected?

Each phase impedance Z = 20∠60° Ω, delta connected, V_L = 440 V, 50 Hz.

Phase current

In delta V_ph = V_L = 440 V.

I_ph = 440/20 = 22 A, lagging by 60° (cos φ = 0.5)

Power per phase

P_ph = V_ph I_ph cos φ = 440 × 22 × 0.5 = 4840 W (= I²R = 22² × 10)

Total power

P = 3 × 4840 = 14 520 W = 14.52 kW

Wattmeter readings

Line current I_L = √3 I_ph = 1.732 × 22 = 38.1 A, φ = 60°.

  • W₁ = V_L I_L cos(30° − φ) = 440 × 38.1 × cos(−30°) = 440 × 38.1 × 0.866 = 14 520 W
  • W₂ = V_L I_L cos(30° + φ) = 440 × 38.1 × cos 90° = 0 W

Sum = 14.52 kW, equal to the total power. At power factor 0.5 one wattmeter reads zero and the other reads the whole power.

✓ Answer checked by Claude Opus 5.5 (high effort)

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