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

# Two wattmeter method for unbalanced load

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

- Asked in: 2068 Shrawan, 8 marks
- Video: [Is it possible to measure three-phase power in an unbalanced load by the two-wattmeter method?](https://www.youtube.com/watch?v=XBdnDZdma7U) — GRADplus
- 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-unbalanced-load-sketch
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

Describe the measurement of power in a unbalanced three phase load by two wattmeter method with neat sketch of circuit diagram.

## Short answer

Connect current coils of two wattmeters in two lines and pressure coils from those lines to the third line. Total power is the algebraic sum of the two readings, W1+W2, because instantaneous power p = vRY iR + vBY iB in a three-wire system, valid for any unbalanced load without needing the neutral point.

## Answer

**Principle:** For any three-wire three-phase load, whether balanced or unbalanced, star or delta, the total power is the sum of the readings of two wattmeters.

#### Circuit diagram

```text
 R ──[ W1: CC ]──────────────┬──► Z_R ─┐
        PC ──────┐           │         │
                 │           │         ● N / Δ
 Y ──────────────┴───────────┼──► Z_Y ─┤
        PC ──────┐           │         │
 B ──[ W2: CC ]──────────────┴──► Z_B ─┘
```

- The current coils (CC) of W₁ and W₂ are in lines R and B.
- The pressure coils (PC) of both are connected between their own line and the third line (Y).

#### Principle for an unbalanced load
Instantaneous power p = v_R i_R + v_Y i_Y + v_B i_B. In a three-wire system i_R + i_Y + i_B = 0, so i_Y = −i_R − i_B. Then

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

- W₁ reads the average of v_RY i_R = V_RY I_R cos(angle between V_RY and I_R)
- W₂ reads the average of v_BY i_B = V_BY I_B cos(angle between V_BY and I_B)

**P = W₁ + W₂**

This does not depend on any balance. No assumption about load equality is used. For an unbalanced load the two readings are generally unequal, and the simple formulas cos(30° ± φ) no longer apply, because the phase angles and the line currents differ.

#### Procedure
1. Connect the wattmeters as shown and read W₁ and W₂.
2. If a wattmeter deflects backwards (power factor below 0.5 in a balanced case, or a strongly reactive phase), reverse its pressure-coil leads and treat the reading as negative.
3. Total power = W₁ + W₂ (algebraic sum).

#### Limits
- It works only for a three-wire system. With a neutral wire and unbalanced load, three wattmeters are needed.
- For a balanced load, tan φ = √3 (W₁ − W₂)/(W₁ + W₂) gives the power factor, but this relation is not valid for unbalanced loads.

## Check yourself

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

1. Does the two-wattmeter method require a balanced load?
   A. Yes, it works only for balanced loads
   B. No, it works for any three-wire load, balanced or unbalanced
   C. Only for star
   D. Only for delta
   Answer: B. The proof uses only i_R + i_Y + i_B = 0, not balance. (Key point: Validity)

2. Where are the current coils and pressure coils placed?
   A. CCs in lines R and B; PCs between their own line and line Y
   B. CCs in the neutral
   C. PCs across the load
   D. CCs in all three lines
   Answer: A. CCs in two lines, PCs joined to the third line. (Key point: Connection)

3. For an unbalanced load, the formula cos(30° ± φ):
   A. Still applies
   B. No longer applies, since angles and currents differ
   C. Applies only for delta
   D. Gives the reactive power
   Answer: B. The readings are V_RY I_R cos(angle) and V_BY I_B cos(angle) with their actual angles. (Key point: Formulas)

4. If a wattmeter deflects backwards, what is done?
   A. Ignore it
   B. Reverse its pressure coil leads and take the reading as negative
   C. Swap the current coil only and add the reading
   D. Replace with three wattmeters
   Answer: B. The reversed reading is subtracted: P = W1 + W2 algebraically. (Key point: Negative reading)

## Related in Basic Electrical Engineering

- [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
- [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
- [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
- [Neutral current and power, unbalanced star](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aunbalanced-star-neutral-current-415v) — Basic Electrical Engineering, chapter 8

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