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Two wattmeter method connection diagram

Short answer

Current coils are placed in lines R and B, with pressure coils connected between their own lines and the third line Y. The sum of readings gives total three-phase power: W1 = V_L I_L cos(30°–φ), W2 = V_L I_L cos(30°+φ), and P = W1+W2 = √3 V_L I_L cos φ, while tan φ = √3 (W1–W2)/(W1+W2).

Explain with connection diagram the measurement of 3-φ power using two wattmeters.

Principle: The total power in a three-wire three-phase circuit (star or delta, balanced or unbalanced) can be measured with two wattmeters. The algebraic sum of their readings is the total power.

Connection diagram

 R ──[ W2: CC ]───────────┬──► load
        PC ──────┐        │
 Y ──────────────┴────────┼──► load
        PC ──────┐        │
 B ──[ W1: CC ]───────────┴──► load
  • The current coils (CC) are in series with two of the lines (R and B).
  • One end of each pressure coil (PC) is connected to the line of its own current coil, and the other ends of both are joined to the third line (Y).
  • Each wattmeter measures the average of (line-to-Y voltage) × (current in its own line).

Readings (balanced load, lagging power factor cos φ)

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

Total power P = W₁ + W₂ = √3 V_L I_L cos φ

Also: tan φ = √3 (W₁ − W₂)/(W₁ + W₂), so the power factor of the load can be found from the readings. Reactive power Q = √3 (W₁ − W₂).

Notes

  • At unity power factor both wattmeters read the same.
  • At cos φ = 0.5, one wattmeter reads zero and the other reads the total power.
  • For cos φ < 0.5, one wattmeter reads negative. Its pressure-coil connection is reversed to get a deflection, and that reading is subtracted.
  • It cannot be used for a four-wire system with a neutral current, where three wattmeters are needed.

✓ Answer checked by Claude Opus 5.5 (high effort)

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