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Circuits, electronics and noise

Balanced reactive power from two wattmeters

Total reactive power of a balanced three-phase load from the two wattmeter readings, with W₁ = V I cos(30° − φ) and W₂ = V I cos(30° + φ), for lagging φ positive.

Ref
ELN-44
Rev
v1
Field
Circuits, electronics and noise
Result
Q [var]

For a balanced three-phase load the two wattmeter readings are W₁ = V I cos(30° − φ) and W₂ = V I cos(30° + φ), with V the line voltage, I the line current and φ the load angle (lagging positive). Subtracting them gives W₁ − W₂ = V I sin φ, and the total reactive power is Q = √3 (W₁ − W₂). Swapping the meters changes the sign, so the connection must be known. This applies to balanced loads only.

Total reactive power of a balanced three-phase load from the two wattmeter readings, with W₁ = V I cos(30° − φ) and W₂ = V I cos(30° + φ), for lagging φ positive. Swapping the meters reverses the sign.

Calculator

Solve for
var
W
−10,00010,000
W
−10,00010,000
  • Total real power, W₁ + W₂3,000 W
  • Apparent power, √(P² + Q²)3,464 VA
  • Load angle φ30 °

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Reactive power (var) current value 1,732 var
W1Wattmeter 1 readingKey driver
−19,10015,600
W2Wattmeter 2 reading
−13,90020,800
  • With readings of 2,000 W and 1,000 W the load takes 3,000 W and 1,732 var, at a load angle of 30°.
  • Equal readings mean a unity power factor, with no reactive power, and equal and opposite readings mean a purely reactive load.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (var)Local stepChange in result
W1 Wattmeter 1 reading−10,000 to 10,000 W−19,05015,59034,640+400 W+40 %
W2 Wattmeter 2 reading−10,000 to 10,000 W−13,86020,78034,640+400 W−40 %

Sweep of Wattmeter 1 reading

Wattmeter 1 reading (W)Reactive power (var)
−10,000−19,050
−7,500−14,720
−5,000−10,390
−2,500−6,062
0−1,732
2,5002,598
5,0006,928
7,50011,260
10,00015,590

Sweep one variable

Reactive power (var)Hover, touch, or use the arrow keys on the chart
−20k−10k010k −10k−50000500010k Wattmeter 1 reading (W) now 1,732

Variables and how each one contributes

Q

Reactive power

Result · var

What the equation gives you. Everything else feeds it.

W1

Wattmeter 1 reading

W · typical −10,000 to 10,000

The reading of wattmeter 1, in W. Raises Q in direct proportion, with a factor of √3. Raising it by 10 % (to 2,200) moves the result by +20 %, from 1,732 to 2,078 var.

W2

Wattmeter 2 reading

W · typical −10,000 to 10,000

The reading of wattmeter 2, in W. Lowers Q in direct proportion, with a factor of √3. Raising it by 10 % (to 1,100) moves the result by −10 %, from 1,732 to 1,559 var.

Worked example with the values above

  1. Difference: W₁ − W₂ = 2,000 − 1,000 = 1,000 W.
  2. Multiply by √3 = 1.7321.
  3. Q = 1,732.1 var.

Assumptions and limits

  • Sinusoidal steady state with RMS values, and a balanced three-phase load where stated.
  • Accurately connected meters in the standard two-wattmeter arrangement.

Common mistakes

  • Using peak voltage where RMS is needed.
  • Ignoring the sign of one wattmeter reading, which can be negative for power factors below 0.5.
  • Using line-to-line voltage where the phase voltage is needed, or the reverse, a factor of √3.

Statistics and uncertainty

Meter errors, supply voltage fluctuation and unbalance all move these results. Wattmeter readings carry the meter’s accuracy class, and for the balanced-load formulas any unbalance gives an error, so check the readings against the total power.

Where it is used

  • Power-factor correction in plants and workshops.
  • Measuring power in three-phase motors and checking delta loads and cables.

Maths you need

  • RMS values: the equivalent steady value of an AC quantity
  • Real, reactive and apparent power: P, Q and S
  • Three-phase systems: line and phase quantities differ by √3

Related formulas

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