Justify the two wattmeter method

Statement: The power of a three-wire three-phase load can be measured with two wattmeters, because the sum of their readings equals the total power.

Justification

Connect the current coils of the two wattmeters in lines R and B, and join both pressure coils to the third line Y.

Total instantaneous power: p = v_R i_R + v_Y i_Y + v_B i_B

Since i_R + i_Y + i_B = 0 (three-wire system), i_Y = −(i_R + i_B), so

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

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

Hence P = W₁ + W₂. This is true for star or delta and for balanced or unbalanced loads.

For a balanced load with lagging angle φ:

• W₁ = V_L I_L cos(30° − φ)
• W₂ = V_L I_L cos(30° + φ)

W₁ + W₂ = V_L I_L × 2 cos 30° cos φ = √3 V_L I_L cos φ, the total three-phase power.

Two wattmeters are used instead of three because the neutral point does not need to be accessible (so it works for delta loads and for star loads without a neutral), and one meter is saved. For a balanced load it also gives the power factor: tan φ = √3 (W₁ − W₂)/(W₁ + W₂).