Two-wattmeter method
Three Phase Systems · Lesson 16 · Intermediate
Purpose
Apply the two-wattmeter method to three-wire systems and interpret positive, zero, or negative readings safely.
Learning objectives
- Describe connections conceptually
- Sum readings for active power
- Derive balanced-load reactive information
- Handle negative readings correctly
Two wattmeters can determine total active power in a three-phase three-wire system because Kirchhoff current law removes one independent line current. The method can measure balanced or unbalanced active power, but PF-angle formulas add a balance assumption.
Core theory
Each wattmeter current element is placed in a different line and its voltage element referenced to the remaining line according to the instrument method. Total active power is W1 + W2 with algebraic signs.
For a balanced sinusoidal load, Q = √3(W1 − W2) under the declared connection/sign convention and tanφ = √3(W1 − W2)/(W1 + W2). Swapping meter identities reverses the Q sign.
At PF below 0.5 one reading becomes negative; at PF 0.5 one becomes zero. Reverse a connection only according to the instrument procedure and retain the reading's negative algebraic sign.
| Term | Meaning | Symbol | Unit |
|---|---|---|---|
| Wattmeter | Instrument measuring active power from voltage and current channels | Not applicable | Not applicable |
| Algebraic sum | Sum retaining positive and negative signs | Not applicable | Not applicable |
| Three-wire system | Three line conductors with no load neutral current path | Not applicable | Not applicable |
P = W1 + W2; tanφ = √3(W1 − W2)/(W1 + W2)The angle relation requires a balanced sinusoidal load and declared connection convention.
W for P; ratio dimensionlessAssumptions: Balanced sinusoidal load; W1 = 8 kW, W2 = 2 kW.
- Power: P = 8 + 2 = 10 kW.
- Angle: tanφ = √3(8−2)/(8+2) = 1.039.
- PF: φ ≈ 46.1°, so PF = cosφ ≈ 0.693.
Total active power is 10 kW and power-factor magnitude is approximately 0.693.
Reasonableness check: Both readings are positive, consistent with PF above 0.5.
- Discarding a negative reading
- Using the PF formula on unbalanced loads
- Changing connections while energised
Where this appears in practice
The method is used in meters, commissioning, laboratories, and diagnosis of three-wire loads.
Knowledge check
How is total active power found when one wattmeter reads negative?
Add the readings algebraically. Do not replace the negative value with its magnitude.
Answer: Add the readings algebraically. Do not replace the negative value with its magnitude.
Practical exercise
Calculate P and PF magnitude for three fictional reading pairs, including one negative reading.
Summary
- Two readings give total active power
- Signs matter
- PF derivation assumes balance
Sources and review
- IEC 60038: IEC standard voltages: IEC; 2009+A1:2021; International
- IEC 61921: Low-voltage power-factor-correction banks: IEC; 2017; International
- Harmonics and Power Quality Analysis webinar Q&A: IET; Current online guidance; United Kingdom
Editorial review date: 2026-08-22. Professional electrical review is pending.