--- title: "Justify the two wattmeter method" subject: "Basic Electrical Engineering" chapter: 8 key: "semester_1/basic_electrical_engineering:two-wattmeter-justify-use" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/two-wattmeter-justify-use site: BCT Engineering --- # Justify the two wattmeter method Basic Electrical Engineering · Chapter 8: Three-Phase Circuit Analysis - Asked in: 2067 Asoj, 4 marks - Video: [2 measurement of three phase power using two wattmeter method](https://www.youtube.com/watch?v=11DtY0iu_Fo) — Divya NS - 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-justify-use - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8 ## Question Justify the use of two wattmeter to calculate the power of 3 φ load. ## Short answer In a three-wire system, iR+iY+iB=0, so instantaneous power p = (vR-vY)iR + (vB-vY)iB = vRY iR + vBY iB. Two wattmeters with current coils in lines R and B and pressure coils to Y read the averages of these terms, their sum gives total three-phase power, valid for unbalanced loads without neutral. ## Answer **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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the third phase's power be eliminated? A. The sum of line currents is zero in a three-wire system B. The load is purely resistive C. V_Y is zero D. The supply is DC Answer: A. i_Y = -(i_R + i_B) for three-wire systems. (Key point: Condition) 2. What is W1 + W2 for a balanced load? A. V_L I_L cos φ B. √3 V_L I_L cos φ C. 3 V_L I_L cos φ D. V_L I_L sin φ Answer: B. 2 cos 30° cos φ = √3 cos φ. (Key point: Sum) 3. Why are only two wattmeters needed instead of three? A. The third phase has no power B. The third line is used as the common point, so no neutral access is needed C. The meters are more accurate D. Cost is the only issue Answer: B. One meter is saved because the pressure coils share the third line. (Key point: Why two) ## Related in Basic Electrical Engineering - [Two wattmeter method for unbalanced load](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Atwo-wattmeter-unbalanced-load-sketch) — Basic Electrical Engineering, chapter 8 - [Delta load 20∠60° power and wattmeter readings](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-load-20-60-440v-two-wattmeter) — Basic Electrical Engineering, chapter 8 - [Two wattmeter method proof](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Atwo-wattmeter-method-proof) — Basic Electrical Engineering, chapter 8, asked 3× - [Two wattmeter method connection diagram](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Atwo-wattmeter-connection-diagram) — Basic Electrical Engineering, chapter 8, asked 2× - [Neutral current and power, 380 V unbalanced load](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aunbalanced-star-neutral-current-380v) — Basic Electrical Engineering, chapter 8 - [Find R for maximum power to load](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amax-power-load-4-ohm-find-r) — Basic Electrical Engineering, chapter 3 ## Related in other subjects - [Two-wattmeter method star load](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Atwo-wattmeter-star-load) — Basic Electrical and Electronics Engineering, chapter 1 - [Justify: series motor power; shunt motor for traction](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ajustify-series-motor-power-shunt-traction) — Utilization of Electrical Energy, chapter 4 - [Justify not self starting; two starting methods](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Amotor-starting-two-methods) — Electric Machines II, chapter 2 - [Justify higher voltage/more circuits for more power](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Ajustify-higher-voltage-more-circuits) — Transmission and Distribution System Design, chapter 2