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Circuits, electronics and noise
Complex power from voltage and current phasors
Real power P from RMS voltage and current phasors given in rectangular parts, with the current entering the load’s positive terminal.
- Ref
- ELN-42
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- P [W]
With RMS phasors V = Vr + jVi and I = Ir + jIi, the complex power is S = V I* = P + jQ. Real power is P = Vr Ir + Vi Ii, and reactive power is Q = Vi Ir − Vr Ii, positive for an inductive load (current lagging). The apparent power is |S| = |V||I| and the power factor is P/|S|. The calculator returns P; the others are in the derived rows.
Real power P from RMS voltage and current phasors given in rectangular parts, with the current entering the load’s positive terminal. The reactive power Q (positive for an inductive load) and the apparent power are in the derived rows.
Calculator
- Reactive power Q400 var
- Apparent power |S|500 VA
- Power factor, P/|S|0.6
- Phase angle between voltage and current53.13 °
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.
- The load takes P = 300 W and Q = 400 var from an apparent power of 500 VA.
- The power factor is 0.6; a positive Q means the current lags the voltage, as in a motor.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Ir Current, real part | −50 to 50 A | −5,000 | 5,000 | 10,000 | +2 A | +66.7 % |
| Vi Voltage, imaginary part | −500 to 500 V | −1,700 | 2,300 | 4,000 | +20 V | −26.7 % |
| Vr Voltage, real part | −500 to 500 V | −1,500 | 1,500 | 3,000 | +20 V | +20 % |
| Ii Current, imaginary part | −50 to 50 A | 300 | 300 | 0 | +2 A | 0 % |
Sweep of Current, real part
| Current, real part (A) | Real power (W) |
|---|---|
| −50 | −5,000 |
| −37.5 | −3,750 |
| −25 | −2,500 |
| −12.5 | −1,250 |
| 0 | 0 |
| 12.5 | 1,250 |
| 25 | 2,500 |
| 37.5 | 3,750 |
| 50 | 5,000 |
Sweep one variable
Variables and how each one contributes
Real power
What the equation gives you. Everything else feeds it.
Voltage, real part
The real part of the voltage phasor, in V. Raises the real power in proportion to Ir. Raising it by 10 % (to 110) moves the result by +10 %, from 300 to 330 W.
Voltage, imaginary part
The imaginary part of the voltage phasor, in V. Raises the real power in proportion to Ii. Raising it by 10 % (to 0) moves the result by +0 %, from 300 to 300 W.
Current, real part
The real part of the current phasor, in A. Raises the real power in proportion to Vr. Raising it by 10 % (to 3.3) moves the result by +10 %, from 300 to 330 W.
Current, imaginary part
The imaginary part of the current phasor, in A. Raises the real power in proportion to Vi. Raising it by 10 % (to −4.4) moves the result by +0 %, from 300 to 300 W.
Worked example with the values above
- Real parts: Vr Ir = 100 × 3 = 300 W.
- Imaginary parts: Vi Ii = 0 × −4 = 0 W.
- P = 300 + 0 = 300 W.
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 Vr Ir − Vi Ii, which is the real part of the plain product V I and not of V I*.
- Reversing the sign of Q by taking the wrong sign of the current angle.
- Using peak phasors where RMS phasors are needed, which doubles the power.
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