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The watts are for real power and the VA for reactive power, but isn´t the same thing at the end? Why does not use only watts or only vars?

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  • $\begingroup$ There's some good answers below, but there's also a bit of semantics here. I rather like when you say Newton-meters it usually refers to torque not energy. When you say Joules (which has the same units) you definitely mean energy. So whenever you hear volt-amps, it should prompt you to think about whether the load is reactive or not (i.e. are the AC current and voltage exactly in phase or not) $\endgroup$ – Roger Wood Mar 20 at 22:23
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Complex Power is composed of a Real component (dissipated power in Watts) and an Imaginary component (Reactive stored energy in Volt-Amperes Reactive (VARs)). The total is combined as a vector quantity (Volt-Amperes = VA).

A common quantity to be concerned with is Power Factor (PF). PF is the cosine of the relative phase angle between the AC Voltage waveform and the AC Current waveform at the load terminals. In a purely resistive load, the AC voltage and current waveforms are IN PHASE (no reactive component exists for a Real load). In this ideal case, the PF is 1.0 because the cosine of the relative phase angle, 0 degrees, equals 1.00. PF ranges from 0.00 to 1.00 depending on the relative phase angle. A PF value of 0.80 is commonly used to estimate (quickly) the total amount of energy (real and reactive) being delivered to the load terminals to perform work.

Actual work is performed by the Real component in Watts. Reactive energy does not perform work and is thus wasted energy. PF correction is achieved by adding a reactive component to the load impedance to reduce the magnitude of the relative phase angle between the voltage and current waveforms. This reduces the VARs and increases the Watts by rotating the VA vector closer to the dissipated component of Power along the horizontal real axis. This reduces wasted energy and increases actual work at the load for the same amount of energy generated at the power station (generator).

For an actual real-world load, which is composed of a Real component (dissipated power in Watts) and a Reactive component (Stored energy in VARS), the PF is the ratio of the Real component (Watts) to the magnitude of the Total (VA). In this case, the PF will be less than the ideal value of 1.00 because the magnitude of the relative phase angle (between the AC voltage and current waveforms) is greater than 0 degrees.

This can be easily visualized using a "Power Triangle." In the Power Triangle, the Real dissipated power (Watts) is drawn along the positive horizontal (X-axis). The Reactive component (VARs) is drawn along the vertical (Y-axis). The hypotenuse of the right triangle represents the total vector magnitude in Volt-Amperes (VA). If the load impedance is inductive, the Reactive component is drawn +90 degrees relative to the Real component along the positive Y-axis. Similarly, if the load impedance is capacitive, the Reactive component is drawn -90 degrees relative to the Real component along the negative Y-axis.

This discussion can easily be found in most Sophomore/introductory-level textbooks for Electrical Engineering principles.

I hope this clarifies the reader's question. Thanks, and be well.

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You pay the electricity company for both the "pure" (resistive, real) watt components AND the reactive (capacitive or inductive, imaginary) watt components in the power that enters your building through the meter box.

This means it is very important to trim or null the inductive component out by adding the right amount of capacitance, or vice versa, so the only portion of the power you will be billed for is the real component that is performing useful work for you in your plant.

In this sense, real and reactive power are, at the end of the day, completely different things.

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  • $\begingroup$ Its a valid idea, but I was meaning from the physics view, it would be the same either one unit or other. Its more or less like a pleonasm. $\endgroup$ – kong001 Mar 20 at 1:27
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    $\begingroup$ At least in Germany, power meters in private households measure real power only and the customer is billed accordingly. Only if the energy company detects an unusually large phase angle will they start billing for apparent power. This is of importance in many industrial operations where e.g. large inductances need to be compensated in consequence. $\endgroup$ – Nephente Mar 20 at 8:05
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Residential electricity meters measure real power in watts not reactive power in Volt-Amps. The former is the average value of the instantaneous voltage times current. The latter is the rms voltage times the rms current. If the sinusoidal voltage and current are exactly in phase then the two measures are identical. If the sinusoidal voltage and current are exactly 90 degrees out of phase then the power in watts is zero but the volt-amps are unchanged. Only real power can do work or provide heat. Capacitive loads or (especially) inductive loads like motors can cause the voltage and current to be out of phase.
Power companies are only allowed to charge residences for real power. The situation is different for large commercial enterprises who may have to pay a surcharge for a poor power factor. This can motivate them to make their load less reactive (e.g. by adding a large capacitor across a large motor)

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Yes, $W$ and $VA$ are the same. Similarly, you could measure pressure in $Pa$ or $Nm^{-2}$. $Hz$ and $Bq$ are both $s^{-1}$ but one is usual for frequency and the other for the rate of radioactive decays. The usage is just conventional. If you wanted to totally pure and consistent then don't use any derived units, just base ones. So, $kg m^2 s^{-3}$ than $W$.

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