Dear Tej,

Thank you for sharing your question with me.Your question is an elementary and fundamental question. In classical electrical engineering, we have circuit elements and electrical sources. According to the variation with time, the sources are either direct current or alternating current sources.The electrical circuit elements are the resistors, the capacitors and inductors. According to their nature, the resistors dissipates electrical energy by converting it to heat, while the capacitors and inductors store electrical energy. The capacitor stores electrical energy in its electric field and the inductors in its magnetic field. therefore they behave reactively in the electrical circuits and are termed reactive elements.

In an AC circuit with a sinusoidal electrical sources, one introduced the concept of the impedance where the impedance of a resistor =R, the impedance of the capacitor is XC=1/jwC and the the impedance of the inductor isXL=jwL. With any combination of these elements the impedance Z= R+jX where X is the reactance. Applying an AC voltage V on the impedance Z, an AC current I passes through the resistance such that I=V/Z. Since Z is complex it can be expressed in magnitude Zm and phase phi. Consequently, the current will have also a magnitude and phase. Im=V/Zm and the phase is = -phi. Hence , the concept of the phasor is introduced. If the impedance is reactive the current lags behind the input voltage and if it is capacitive it will llead the the voltage. the power dissipated in the Z is Pa =0.5 V Im cos phi and the reactive power which is the energy storage rate in the reactive elements Pr=0.5 V Im sin phi.

Now let us discuss some special cases:

The impedance is purely inductive, phi = -90 degrees and consequently the active power is zero and the total power will be reactive. The same holds for a pure capacitive load. If Z is purely resistive , the reactive power will be zero and only there is

active power.

Where is the problem?

Most of the loads on the public electrical utility network are inductive. ON the other side, the ideal electrical loads must be pure resistive. We see from the equation of the active power that it is proportional to the power factor cos phi. Since cos phi will be highest at phi equals zero and cos phi =1. Therefore to supply an inductive load with the same active energy the required volt ampere must be increased by 1/ cos phi.

Therefore the voltage source must pump a generation power Pa/cos phi. Accordingly the cost of the power generation increases with the same factor. In addition during the power transmission on the transmission line from the power station to a distant load more power will be dissipated in the cables. Also, larger cable cross sections will be required and hence larger cost of the transmission network.

What is the remedy?

The remedy is to compensate the inductive reactance by a capacive reactance by adding capacitor to specific network nodes. This increases the power factor and improves the power network.

The power generator can be considered as a voltage source with an electromotive force and a source impedance which in the case of the utility supply very small. The power factors are defined for loads not for power sources

I hope i covered the main points in your questions.

Mr Abdelhalim Zekry, thank you very much for spending your valuable time for me. thank you sir

Let us recollect the Basic Triangles in Electrical Circuits-Voltage Drop Triangle,KVA Trianngle,Current Triangle (Real Power Componnent or Wattful Component,Reactive Component or Wattless Component and actual Current Phasor).

In my view these age old Triangles will explain the suject clearly regarding neutralization of Inductive KVAR or Improvement of Power Factor or Drawing of Phasor Diagrams for Circuits or Machines showing the Voltage Drops across different Components

The Inductance and Capacitance Elements,wherever they are present in an Electric Circuit, Cause only Wattless Component of Current because Self Induced Emf across Inductance Vl = L * d (I) / dt. and Current through a Capacitor

Ic = C * d(V) / dt. and V or I of the Circuit is Sinusoidal. The Power consumed by these elements are in quadrature to Real Power and is called Reactive Power Q. The Real Power consumed by them is Zero( as P = V * I * Cos(Phi) and Phi = 90 degrees for these Reactive Elements).

Also refer to the discussions being held on Reactive Power in R.G.

The Generator has to supply both Real and Reactive Powers,total called Apparent Power S = P + jQ. The Sign can be + or - for inductive Load,depending on the convention you use for S(stated by me in R..G. earlier)

Capacitive KVAR being of Leading Power Factor will neutralise the Inductive KVAR which is of Lagging Power Factor to that extent and reduces the the Nett Apparent Power in KVA that has to be Produced by the Generator.

Also refer to the Answer given above by Prof. Abdelhalim Zekry.

P.S.

Mr Pisupati Subramanyam ., thank you for your explanation sir.

There are two more Triangles which are Fundamental - the Impedance Triangle and Admittance Triangle.Both help in finding the Impedance Angle and help in calculation of A.C. quantities in Polar Form.

These Triangles form the Basis for the above three Triangles.

P.S.

In the case of A.C. Supply they acquire Real Power in one half Cycle and dissipate in the other half cycle. Net Real Power is Zero. Reactive Current Component causes the Voltage Drop.

See the Power and Current Waveforms for the energy storing elents in my book ,"Basic Concepts of Electrical Engineering"