Why Capacitor is Generator of Reactive Power while Inductor an Absorber? Understanding Reactive Power

Consider an Inductive circuit and assume the instantaneous voltage be

v = VmSinωt

Then the current will lag behind the voltage by some angle Ø and the instantaneous current will be as

i = ImSin(ωt-Ø)

Now to understand the concept of Reactive Power, we will calculate the instantaneous power. The instantaneous power will be multiplication of instantaneous voltage and current as given below.

p = vi

= VmSinωt x ImSin(ωt-Ø)

= VmIm[Sinωt x Sin(ωt-Ø) ]

= (VmIm/2) [2 x Sinωt x Sin(ωt-Ø)]

From the Trigonometrical Formula,

2SinAxSinB = Cos (A-B) – Cos(A+B)

p = (VmIm/2) [CosØ – Cos(2ωt – Ø) ]

= VI[CosØ – Cos(2ωt – Ø)]

Where V = RMS value of voltage

I = RMS value of current

Thus  instantaneous power is

 p = VICosØ – VICos(2ωt – Ø)] = VICosØ(1- Cos2ωt) – VISinØ Sin2ωt
It is quite clear from the expression of instantaneous power is that, it is made of two parts one is VICosØ(1- Cos2ωt) and another one is VISinØ Sin2ωt.
• The first part VICosØ(1- Cos2ωt) pulsates around the same average power VICosØ but never goes negative as VICosØ(1- Cos2ωt) can be zero at most. Thus we see that average power due to this part of instantaneous power is some finite value VICosØ to which we call Active Power.
• Second part of instantaneous power i.e. VISinØ Sin2ωt pulsates from -VISinØ to VISinØ and therefore the average power will become zero. This means that this part of instantaneous power only travels back and forth without doing nay work. We call this component is known as Reactive Power.

Thus instantaneous power can be written as

p = P(1- Cos2ωt) -QSin2ωt

Mind that both P and Q has the same unit Watt but to show that Q is Reactive Power, it is expressed in terms of VAR.

Now, observe that SinØ will be negative for Capacitor and hence

Q = Negative for Capacitor.

Which means that Capacitor is not consuming Reactive Power rather it supplies Reactive Power and hence Generator of Reactive Power.

For Inductor, SinØ = Positive, therefore

Q = Positive, which implies that an Inductor consumes Reactive Power.

To conclude, it is better to say that a Capacitor is supplying lagging current rather than taking leading current.

Thank you!

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