---
title: "Two wattmeter method proof"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:two-wattmeter-method-proof"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/two-wattmeter-method-proof
site: BCT Engineering
---

# Two wattmeter method proof

Basic Electrical Engineering · Chapter 8: Three-Phase Circuit Analysis

- Asked in: 2074 Asoj, 6 marks; 2073 Chaitra, 6 marks; 2069 Asar, 8 marks (asked 3 times)
- Video: [Two wattmeter method of 3-phase power measurement, Electrical concepts with MCQ.](https://www.youtube.com/watch?v=E4fJTNTXO1k) — €lectrical_¢oncepts
- 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-method-proof
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

Prove that sum of the readings of two wattmeters is equal to the total three phase power in measurement of power of 3-phase circuit by 2 wattmeter method.

## Short answer

In a three-wire system, instantaneous power p = v_RY i_R + v_BY i_B. Two wattmeters measure the averages of these products, so total power P = W1 + W2. For balanced load, W1 = V_L I_L cos(30°–φ), W2 = V_L I_L cos(30°+φ), sum = √3 V_L I_L cos φ.

## Answer

**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.

```text
 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which condition lets the total power be written as the sum of two wattmeter readings?
   A. i_R + i_Y + i_B = 0 in a three-wire system
   B. The load must be balanced
   C. The supply must be four-wire
   D. The power factor must be unity
   Answer: A. With i_Y = -(i_R + i_B), the third term can be eliminated, which holds for any three-wire load. (Key point: Condition used)

2. After substituting i_Y, the instantaneous power becomes:
   A. v_R i_R + v_B i_B
   B. v_RY i_R + v_BY i_B
   C. v_RY i_Y + v_BY i_B
   D. v_R i_R - v_B i_B
   Answer: B. p = (v_R - v_Y) i_R + (v_B - v_Y) i_B = v_RY i_R + v_BY i_B. (Key point: Resulting expression)

3. For a balanced load with lagging φ, the two readings are:
   A. V_L I_L cos(30° - φ) and V_L I_L cos(30° + φ)
   B. V_L I_L cos φ and V_L I_L sin φ
   C. V_L I_L cos(60° - φ) and cos(60° + φ)
   D. V_L I_L sin(30° - φ) and sin(30° + φ)
   Answer: A. W1 = V_L I_L cos(30° - φ) and W2 = V_L I_L cos(30° + φ). (Key point: Balanced readings)

4. Using cos A + cos B = 2 cos((A+B)/2) cos((A-B)/2), the sum W1 + W2 equals:
   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: C. The sum is V_L I_L x 2 cos 30° cos φ = √3 V_L I_L cos φ. (Key point: Sum of readings)

## Related in Basic Electrical Engineering

- [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 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
- [Total power of series-parallel circuit](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-total-power-265uf-19mh) — Basic Electrical Engineering, chapter 6
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- [Justify the two wattmeter method](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Atwo-wattmeter-justify-use) — Basic Electrical Engineering, chapter 8
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