_{m}Sinωt

_{m}Sin(ωt-Ø)

_{m}Sinωt x I

_{m}Sin(ωt-Ø)

_{m}I

_{m}[Sinωt x Sin(ωt-Ø) ]

_{m}I

_{m}/2) [2 x Sinωt x Sin(ωt-Ø)]

_{m}I

_{m}/2) [CosØ – Cos(2ωt – Ø) ]

p = VICosØ – VICos(2ωt – Ø)] = VICosØ(1- Cos2ωt) – 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!

Good day, i have a question. What will happen to the generators in parallel if one generator is running with capacitive pf -kvar and the other one with reactive power factor +kvar. Many thanks for an answer.