Two wattmeter method proof

Statement: In a three-wire three-phase system, the sum of the readings of two wattmeters equals the total power consumed, for a balanced or an unbalanced load.

Connection

The current coil (CC) of W₁ is in line B and that of W₂ in line R. The pressure coils (PC) of both are connected to line Y, the common line.

 R ──[ W2: CC ]───────────┬── load
        PC ──────┐        │
 Y ──────────────┴────────┼── load
        PC ──────┐        │
 B ──[ W1: CC ]───────────┴── load

General proof

Instantaneous power of a star-connected load (or its star equivalent):

p = v_R i_R + v_Y i_Y + v_B i_B

For a three-wire system i_R + i_Y + i_B = 0, so i_Y = −(i_R + i_B). Substituting:

p = v_R i_R + v_B i_B − v_Y (i_R + i_B) = (v_R − v_Y) i_R + (v_B − v_Y) i_B = v_RY i_R + v_BY i_B

• W₂ reads the average of v_RY i_R
• W₁ reads the average of v_BY i_B

So the total power is P = W₁ + W₂.

Balanced load: phasor proof

Take RYB sequence with V_RN = V_ph∠0°, V_YN = V_ph∠−120°, V_BN = V_ph∠120°. The load current lags its phase voltage by φ. Let V_L be the line voltage and I_L the line current.

• V_RY = V_RN − V_YN = V_L∠30°; I_R = I_L∠−φ. Angle between them = 30° + φ. W₂ = V_L I_L cos(30° + φ)
• V_BY = V_BN − V_YN = V_L∠90°; I_B = I_L∠(120° − φ). Angle between them = 30° − φ. W₁ = V_L I_L cos(30° − φ)

Sum:

W₁ + W₂ = V_L I_L [cos(30° − φ) + cos(30° + φ)] = V_L I_L × 2 cos 30° cos φ = V_L I_L × 2 × (√3/2) × cos φ = √3 V_L I_L cos φ

This is the total three-phase power. Hence proved.

Further results

• W₁ − W₂ = V_L I_L [cos(30° − φ) − cos(30° + φ)] = V_L I_L sin φ
• tan φ = √3 (W₁ − W₂)/(W₁ + W₂)
• At unity pf (φ = 0) both read equal. At pf 0.5 (φ = 60°) W₂ = 0. For pf below 0.5, W₂ reads negative.