In this article, I will discuss the exact answer for the question which most of you are not aware. Before presenting my answer, I would like to give the answers which most of the electrical engineer thinks that they are correct and this answer even presented by many standard books also that is we use conjugate to match the convention of reactive power polarity. But this is not complete answer.
Lets discuss the point -
In case of Inductor, the active power, P = 0 because the instantaneous power of inductor is like symmetric sinusoid as shown in fig
and the reactive power, Qind = +VIsinΦ
In case of capacitor also the active power, P = 0 because the instantaneous power of capacitor is also like symmetric sinusoid but opposite to inductor as shown in fig
and the reactive power, Qcap = -VIsinΦ
From the above figures, the instantaneous power of inductor and capacitor are opposite to each other i.e anti-phase.
For both cases, the average power is zero but there is an exchange of power. So the maximum value will call it as reactive power.
As the waveform says both are anti-phase that's why we have taken Qind as +ve and Qcap as -ve or vice-versa. But as standard convention we take Qind as +ve (Qind = +VIsinΦ) because most of the loads are inductive in nature and very less loads are capacitive.
So the key point is
Qind = +VIsinΦ
and Qcap = -VIsinΦ
Lets check -
If we don't take conjugate, then Power S= VI
i. In this case we are considering for lagging load or inductive load or R-L load and taking voltage phasor as reference phasor.
S = Vã„¥0⁰ x Iã„¥-Φ
= VIㄥ-Φ
= VIcosΦ - jVIsinΦ
where S = P + jQ
and here P = VIcosΦ and Q = - VIsinΦ
Here Q is negative(-ve) but we should get Q as
positive (+ve) or equal to Qind. That's why to match the convention we take conjugate.
If S = VI*
= VIㄥ+Φ
= VIcosΦ + jVIsinΦ
From the above figures, the instantaneous power of inductor and capacitor are opposite to each other i.e anti-phase.
For both cases, the average power is zero but there is an exchange of power. So the maximum value will call it as reactive power.
As the waveform says both are anti-phase that's why we have taken Qind as +ve and Qcap as -ve or vice-versa. But as standard convention we take Qind as +ve (Qind = +VIsinΦ) because most of the loads are inductive in nature and very less loads are capacitive.
So the key point is
Qind = +VIsinΦ
and Qcap = -VIsinΦ
Lets check -
If we don't take conjugate, then Power S= VI
i. In this case we are considering for lagging load or inductive load or R-L load and taking voltage phasor as reference phasor.
S = Vã„¥0⁰ x Iã„¥-Φ
= VIㄥ-Φ
= VIcosΦ - jVIsinΦ
where S = P + jQ
and here P = VIcosΦ and Q = - VIsinΦ
Here Q is negative(-ve) but we should get Q as
positive (+ve) or equal to Qind. That's why to match the convention we take conjugate.
If S = VI*
= VIㄥ+Φ
= VIcosΦ + jVIsinΦ
Now the Q is positive or the convention of Q is matched with Qind .
ii. In this case we are considering for leading load or capacitive load or R-C load and taking voltage phasor as reference phasor.
S = Vã„¥0⁰ x Iã„¥+Φ = VIã„¥+Φ = VIcosΦ + jVIsinΦ
where S = P + jQ
and here P = VIcosΦ and Q = + VIsinΦ
Here Q is positive(+ve) but we should get Q as
negative(-ve) or equal to Qcap. That's why again to match the convention we take conjugate.
If S = VI*
= VIㄥ-Φ
= VIcosΦ - jVIsinΦ
Now the Q is negative or the convention of Q is matched with Qcap.
That means there is a coherence between the convention what we have taken on the mathematics what we have followed.
So, this is the answer normally given by any text book or any other websites, but this is not complete answer. Now i am extending little bit.
Suppose, we are taking a reference phasor and considering the voltage phasor and current phasor for lagging load.
With respect to reference phasor, I is lagging by an angle β and V is leading by angle α .
∴ power factor angle = α + β
If we calculate power without taking conjugate, then S = VI
S = V∠α x I∠-β
= VIㄥ(α-β)
= VIcos(α-β) + jVIsin(α-β)
The angle is supposed to be (α+β) but in the calculation the angle is (α-β). So again wrong thing. That' why again to have a match, we take conjugate of the current.
So, S = VI*
S = V∠α x I∠β
If we calculate power without taking conjugate, S = VI, where V➝VㄥΦ and Iã„¥0
= VIㄥ(α+β)
= VIcos(α+β) + jVIsin(α+β)
Now the convention of Q is matched.
Lets look at another example where we have taken current as reference phasor for lagging load.
S = VIㄥ+Φ S = VIcosΦ + jVIsinΦ
Here Q is positive and the convention of Q is matched with Qind.
S = VI* ➠ If we take voltage as reference or reference other than the current.
S = VI directly ➠ If we take current as reference.
Here Q is positive and the convention of Q is matched with Qind.
Conclusion :
∴ The answer what is concluded so far is to have coherence between the convention and mathematics, we use conjugate.S = VI* ➠ If we take voltage as reference or reference other than the current.
S = VI directly ➠ If we take current as reference.
But most of the cases we take voltage as reference because current keeps on changing with respect to load. That's why we mostly use S = VI*.
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