--- title: "Two wattmeter method connection diagram" subject: "Basic Electrical Engineering" chapter: 8 key: "semester_1/basic_electrical_engineering:two-wattmeter-connection-diagram" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/two-wattmeter-connection-diagram site: BCT Engineering --- # Two wattmeter method connection diagram Basic Electrical Engineering · Chapter 8: Three-Phase Circuit Analysis - Asked in: 2072 Chaitra, 4 marks; 2070 Chaitra, 4 marks (asked 2 times) - Video: [Three Phase Circuit Tutorial - 1 Wattmeter 2 Wattmeter Methods - Solved Problems](https://www.youtube.com/watch?v=da5otX5VmEs) — EzEd Channel - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/two-wattmeter-connection-diagram - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8 ## Question Explain with connection diagram the measurement of 3-φ power using two wattmeters. ## 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). ## Answer **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 ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the two-wattmeter connection, how are the pressure coils joined? A. Both to the neutral B. One end of each to its own line, the other ends joined to the third line C. Both across the same two lines D. In series with the load Answer: B. Each PC goes between its own line and the third (common) line. (Key point: Connection) 2. Which relation gives the load power factor angle from the readings? A. tan φ = √3 (W1 - W2)/(W1 + W2) B. tan φ = (W1 + W2)/(W1 - W2) C. cos φ = W1/W2 D. tan φ = (W1 - W2)/√3 (W1 + W2) Answer: A. tan φ = √3 (W1 - W2)/(W1 + W2) for a balanced load. (Key point: Power factor from readings) 3. At power factor 0.5 in a balanced load, what do the wattmeters read? A. Both equal B. One reads zero and the other the total power C. One reads negative D. Both read zero Answer: B. At φ = 60°, cos(30° + 60°) = 0, so one meter reads zero. (Key point: Special case) 4. When is the two-wattmeter method not sufficient? A. Balanced delta load B. Unbalanced star load without neutral C. Four-wire system with neutral current D. Lagging power factor Answer: C. With a neutral carrying current, i_R + i_Y + i_B is not zero, so three wattmeters are needed. 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