Analysis On Power Factor

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The electrical energy is almost exclusively generated, transmitted and distributed in the form of alternating current. Therefore, the question of power factor immediately comes into picture. Most of the loads (e.g. induction motors, arc lamps) are inductive in nature and hence have low lagging power factor. The low power factor is highly undesirable as it causes an increase in current, resulting in additional losses of active power in all the elements of power system from power station generator down to the utilization devices. In order to ensure most favorable conditions for a supply system from engineering and economical standpoint, it is important to have power factor as close to unity as possible.
The cosine of angle between voltage and current in an a.c. circuit is known as
Power factor.

In an a.c. circuit, there is generally a phase difference φ between voltage and current. The term cosφ is called the power factor of the circuit. If the circuit is inductive, the current lags behind the voltage and the power factor is referred to as lagging. However, in a capacitive circuit, current leads the voltage and power factor is said to be leading.
Fig: 1.1

Consider an inductive circuit taking a lagging current I from sup- ply voltage V; the angle of lag being φ.  The phasor diagram of the circuit is shown in Fig. 1.1
The circuit current I can be resolved into two perpendicular components, namely;
     a)  Icos φ in phase with V
b)I sin φ 90° out of phase with V
The component I cosφ  is known as active or wattful component, Fig. 1.1 whereas component I sinφ is called the reactive or wattless component. The reactive component is a measure of the power factor. If the reactive component is small, the phase angle φ is small and hence power factor cosφ will be high. Therefore, a circuit having small reactive current (i.e., I sinφ) will have high power factor and vice-versa. It may be noted that value of power factor can never be more than unity.
1)It is a usual practice to attach the word ‘lagging’ or ‘leading’ with the numerical value of power factor to signify whether the current lags or leads the voltage. Thus if the circuit has a p.f. of 0.5 and the current lags the voltage, we generally write p.f. as 0.5 lagging.
2)Sometimes power factor is expressed as a percentage. Thus 0.8 lagging power factor may be expressed as 80% lagging.
The improvement of power factor is very important for both consumers and generating stations as discussed below:
1)For Consumers: A consumer has to pay electricity charges for his maximum demand in kVA plus the units consumed. If the consumer improves the power factor, then there is a reduction in his maximum kVA demand and consequently there will be annual saving due to maximum demand charges. Although power factor improvement involves extra annual expenditure on account of p.f. correction equipment, yet improvement of p.f. to a proper value results in the net annual saving for the consumer.
2)For generating stations: A generating station is as much concerned with power factor improvement as the consumer. The generators in a power station are rated in kVA but the useful output depends upon kW output. As station output is kW= kVA x cosφ , therefore, number of units supplied by it depends upon the power factor. The greater the power factor of the generating station, the higher is the kWh it delivers to the system. This leads to the conclusion that improved power factor increases the earning capacity of the power station.
If a consumer improves the power factor, there is reduction in his maximum kVA demand and hence there will be annual saving over the maximum demand charges. However, when power factor is improved, it involves capital investment on the power factor correction equipment. The consumer will incur expenditure every year in the shape of annual interest and depreciation on the investment made over the p.f. correction equipment. Therefore, the net annual saving will be equal to the annual saving in maximum demand charges minus annual expenditure incurred on p.f. correction equipment.
The value to which the power factor should be improved so as to have maximum net annual saving is known as the most Economical Power Factor.

Consider a consumer taking a peak load of PkW at a power factor of cos φ1and charged at a rate of Rs x per kVA of maximum demand per annum. Suppose the consumer improves the power factor to cos φ2 by installing p.f. correction equipment. Let expenditure incurred on the p.f. correction equipment be Rs y per kVAR per  annum. The power triangle at the original p.f. cos φ1 Is OAB andfor the improved p.f. cos φ2, it is OAC [See Fig. 1.2].

Fig: 1.2
kVA max. demand at cos φ1 , kVA1 = P/cos φ1 = P sec φ1
kVA max. demand at cos φ2 , kVA2 = P/cos φ2 = P sec φ2
Annual saving in maximum demand charges
                                                    = Rs x (kVA1 kVA2)
                                                    = Rs x (P sec φ1 P sec φ2)
                                                    = Rs x P(sec φ1 sec φ2)
Reactive power at cos φ1 , kVAR1 = P tan φ1
Reactive power at cos φ2’ kVAR2 = P tan φ2
Leading kVAR taken by p.f. correction equipment
                                                    = P (tan φ1  tan φ2)
Annual cost of p.f. correction equipment
                                                   = Rs Pv (tan φ1  tan φ2)
Net annual saving, S = exp. (i)  exp. (ii)
                                                   = xP (sec φ1_ sec φ2)   yP (tan φ1   tan φ1)
In this expression, only is variable while all other quantities are fixed. Therefore, the net annual saving will be maximum if differentiation of above expression w.r:t. φ1 is zero i.e.
                                                d/dφ2   (S) = 0                                                             
or   d/dφ2   [xP (sec φ1  sec φ2 )  yP (tan φ1 tan φ2) = 0                          
or   d/dφ(xP sec φ1)  d/dφ(xP sec φ2)  d/dφ(yP tan φ1) + yP (tan φ2) = 0
or   0  xP sec φ2 tan φ2  0 + yP sec2 φ2 ) = 0
or                                 x tan φ2 + y sec φ2 = 0
or                                                       tan φ2 = y/x sec φ2
or                                                         sin φ2 = y/x
\Most economical power factor, Cos φ2Öi  sin2 φ2 = Öi  (y/ x)2
It may be noted that the most economical power factor (cos φ2 ) depends upon the relative costs of supply and p.f. correction equipment but is independent of the original p.f. cos φ1.
The analysis of power factor can also be made in terms of power drawn by the a.c. circuit. If each side of the current triangle oab of Fig. 1.3 is multiplied by voltage V, then we get the power triangle OAB shown in Fig. 1.3 where
OA = VI cos φ and represents the active power in watts or kW
AB = VI sin φ and represents the reactive power in VAR or kVAR
OB = VI and represents the apparent power in VA or kVA
The following points may be noted form the power triangle:
Fig: 1.3

1)  VI Sinφ The apparent power in an ac. circuit has two components viz,
active and reactive power at right angles to each other.
                            OB2 = OA2+AB2
or   (apparent power)2 = (active power)2 + (reactive power)2
or                    (kVA)2 = (kW)2 + (kVAR)2
2)  Power factor, cos φ = OA⁄OB= active power apparent power =KW ⁄ KVA
Thus the power factor of a circuit may also be defined as the ratio of active power to the apparent power. This is a perfectly general definition and can be applied to all cases, whatever be the waveform.
3)  The lagging reactive power is responsible for the low power factor. It is clear from the power triangle that smaller the reactive power component, the higher is the power factor of the circuit.
                      kVAR = kVA sin φ = cos φ sin φ
                      kVAR = kWtanφ
4)  For leading currents, the power triangle becomes reversed. This fact provides a key to the power factor improvement. If a device taking leading reactive power (e.g. capacitor) is connected in parallel with the load, then the lagging reactive power of the load will be partly neutralised, thus improving the power factor of the load.
5)  The power factor of a circuit can be defined in one of the following three ways:
a)Power factor = cos φ = cosine of angle between V and I
b)Power factor = R/Z= Resistance ⁄ Impedance
c)Power factor = VI cos φ ⁄ VI= Active power ∕ Apparent Power
6)  The reactive power is neither consumed in the circuit nor it does any useful work. It merely flows back and forth in both directions in the circuit. A wattmeter does not measure reactive power.
Let us illustrate the power relations in an ac. circuit with an example. Suppose a
circuit draws a current of 10 A at a voltage of 200 V and its p.f. is 0.8 lagging. Then,
                    Apparent power = VI = 200 x 10 = 2000 VA
                    Active power     = VI cos φ = 200 x 10 x 0.8 = 1600 W
                    Reactive power = VI sin φ = 200 x 10 x 0.6 = 1200 VAR
The circuit receives an apparent power of 2000 VA and is able to convert only 1600 watts into active power. The reactive power is 1200 VAR and does no useful work. It merely flows into and out of the circuit periodically. In fact, reactive power is a liability on the source because the source has to supply the additional current (i.e., I sinφ).
The power factor of an AC electric power system is defined as the ratio of the real power to the apparent power, and is a number between 0 and 1 (frequently expressed as a percentage, e.g. 0.5 pf = 50% pf). Real power is the capacity of the circuit for performing work in a particular time. Apparent power is the product of the current and voltage of the circuit. Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power can be greater than the real power.
In an electric power system, a load with low power factor draws more current than a load with a high power factor, for the same amount of useful power transferred. The higher currents increase the energy lost in the distribution system, and require larger wires and other equipment. Because of the costs of larger equipment and wasted energy, electrical utilities will usually charge a higher cost to industrial or commercial customers where there is a low power factor.
Linear loads with low power factor (such as induction motors) can be corrected with a passive network of capacitors or inductors. Non-linear loads, such as rectifiers, distort the current drawn from the system. In such cases, active power factor correction is used to counteract the distortion and raise power factor. The devices for correction of power factor may be at a central substation, or spread out over a distribution system, or built into power using equipment.
Instantaneous and average power calculated from AC voltage and current with a unity power factor (φ=0, cosφ=1)

Instantaneous and average power calculated from AC voltage and current with a zero power factor (φ=90, cosφ=0)

Instantaneous and average power calculated from AC voltage and current with a lagging power factor (φ=45, cosφ=0.71)
In a purely resistive AC circuit, voltage and current waveforms are in step (or in phase), changing polarity at the same instant in each cycle. Where reactive loads are present, such as with capacitors or inductors, energy storage in the loads result in a time difference between the current and voltage waveforms. This stored energy returns to the source and is not available to do work at the load. Thus, a circuit with a low power factor will have higher currents to transfer a given quantity of real power than a circuit with a high power factor. A linear load does not change the shape of the waveform of the current, but may change the relative timing (phase) between voltage and current.
Circuits containing purely resistive heating elements (filament lamps, strip heaters, cooking stoves, etc.) have a power factor of 1.0. Circuits containing inductive or capacitive elements (lamp ballasts, motors, etc.) often have a power factor below 1.0.


AC power flow has the three components: real power (P), measured in watts (W); apparent power (S), measured in volt-amperes (VA); and reactive power (Q), measured in reactive volt-amperes (VAR).
The power factor is defined as:
In the case of a perfectly sinusoidal waveform, P, Q and S can be expressed as vectors that form a vector triangle such that:
If φ is the phase angle between the current and voltage, then the power factor is equal to, and
Since the units are consistent, the power factor is by definition a dimensionless number between 0 and 1. When power factor is equal to 0, the energy flow is entirely reactive, and stored energy in the load returns to the source on each cycle. When the power factor is 1, all the energy supplied by the source is consumed by the load. Power factors are usually stated as "leading" or "lagging" to show the sign of the phase angle, where leading indicates a negative sign.
If a purely resistive load is connected to a power supply, current and voltage will change polarity in step, the power factor will be unity (1), and the electrical energy flows in a single direction across the network in each cycle. Inductive loads such as transformers and motors (any type of wound coil) consume reactive power with current waveform lagging the voltage. Capacitive loads such as capacitor banks or buried cable generate reactive power with current phase leading the voltage. Both types of loads will absorb energy during part of the AC cycle, which is stored in the device's magnetic or electric field, only to return this energy back to the source during the rest of the cycle.
For example, to get 1 kW of real power, if the power factor is unity, 1 kVA of apparent power needs to be transferred (1 kW ÷ 1 = 1 kVA). At low values of power factor, more apparent power needs to be transferred to get the same real power. To get 1 kW of real power at 0.2 power factor, 5 kVA of apparent power needs to be transferred (1 kW ÷ 0.2 = 5 kVA). This apparent power must be produced and transmitted to the load in the conventional fashion, and is subject to the usual distributed losses in the production and transmission processes.


Electrical loads consuming alternating current power consume both real power and reactive power. The vector sum of real and reactive power is the apparent power. The presence of reactive power causes the real power to be less than the apparent power, and so, the electric load has a power factor of less than 1.
The reactive power increases the current flowing between the power source and the load, which increases the power losses through transmission and distribution lines. This results in additional costs for power companies. Reactive power can require the use of wiring, switches, circuit breakers, transformers and transmission lines with higher current capacities. Therefore, power companies require their customers, especially those with large loads, to maintain their power factors above a specified amount (usually 0.90 or higher) or be subject to additional charges. Electricity utilities measure reactive power used by high demand customers and charge higher rates accordingly. Some consumers install power factor correction schemes at their factories to cut down on these higher costs.
It is often desirable to adjust the power factor of a system to near 1.0. This power factor correction is achieved by switching in or out banks of inductors or capacitors. For example the inductive effect of motor loads may be offset by locally connected capacitors. When reactive elements supply or absorb reactive power near the load, the apparent power is reduced.
Power factor correction may be applied by an electrical power transmission utility to improve the stability and efficiency of the transmission network. Correction equipment may be installed by individual electrical customers to reduce the costs charged to them by their electricity supplier. A high power factor is generally desirable in a transmission system to reduce transmission losses and improve voltage regulation at the load.
Power factor correction brings the power factor of an AC power circuit closer to 1 by supplying reactive power of opposite sign, adding capacitors or inductors which act to cancel the inductive or capacitive effects of the load, respectively. For example, the inductive effect of motor loads may be offset by locally connected capacitors. If a load had a capacitive value, inductors (also known as reactors in this context) are connected to correct the power factor. In the electricity industry, inductors are said to consume reactive power and capacitors are said to supply it, even though the reactive power is actually just moving back and forth on each AC cycle.
The reactive elements can create voltage fluctuations and harmonic noise when switched on or off. They will supply or sink reactive power regardless of whether there is a corresponding load operating nearby, increasing the system's no-load losses. In a worst case, reactive elements can interact with the system and with each other to create resonant conditions, resulting in system instability and severe over voltage fluctuations. As such, reactive elements cannot simply be applied at will, and power factor correction is normally subject to engineering analysis.
An automatic power factor correction unit is used to improve power factor. A power factor correction unit usually consists of a number of capacitors that are switched by means of contactors. These contactors are controlled by a regulator that measures power factor in an electrical network. To be able to measure 'power factor', the regulator uses a CT (Current transformer) to measure the current in one phase.
Depending on the load and power factor of the network, the power factor controller will switch the necessary blocks of capacitors in steps to make sure the power factor stays above 0.9 or other selected values (usually demanded by the energy supplier).
  1. Reactive Power Control Relay
  2. Network connection points
  3. Slow-blow Fuses
  4. Inrush Limiting Contactors
  5. Capacitors (single-phase or three-phase units, delta-connection)
  6. Transformer Suitable voltage transformation to suit control power (contactors, ventilation)
Instead of using a set of switched capacitors, an unloaded synchronous motor can supply reactive power. The reactive power drawn by the synchronous motor is a function of its field excitation. This is referred to as a synchronous condenser. It is started and connected to the electrical network. It operates at full leading power factor and puts VARs onto the network as required to support a system’s voltage or to maintain the system power factor at a specified level. The condenser’s installation and operation are identical to large electric motors. Its principal advantage is the ease with which the amount of correction can be adjusted; it behaves like an electrically variable capacitor. Unlike capacitors, the amount of reactive power supplied is proportional to voltage, not the square of voltage; this improves voltage stability on large networks. Synchronous condensors are often used in connection with high voltage direct current transmission projects or in large industrial plants such as steel mills.


2.1 Power in resistive Load

Consider a circuit for a single-phase AC power system, where a 120 volt, 60 Hz AC voltage source is delivering power to a resistive load: (Figure below)


Fig: 2.1
Ac Source drives a purely resistive load.


In this example, the current to the load would be 2 amps, RMS. The power dissipated at the load would be 240 watts. Because this load is purely resistive (no reactance), the current is in phase with the voltage, and calculations look similar to that in an equivalent DC circuit. If we were to plot the voltage, current, and power waveforms for this circuit, it would look like Figure below.
Fig: 2.2                                                                                                     Current is in phase with voltage in a resistive circuit.
Note that the waveform for power is always positive, never negative for this resistive circuit. This means that power is always being dissipated by the resistive load, and never returned to the source as it is with reactive loads. If the source were a mechanical generator, it would take 240 watts worth of mechanical energy (about 1/3 horsepower) to turn the shaft.
Also note that the waveform for power is not at the same frequency as the voltage or current! Rather, its frequency is double that of either the voltage or current waveforms. This different frequency prohibits our expression of power in an AC circuit using the same complex (rectangular or polar) notation as used for voltage, current, and impedance, because this form of mathematical symbolism implies unchanging phase relationships. When frequencies are not the same, phase relationships constantly change.
As strange as it may seem, the best way to proceed with AC power calculations is to use scalar notation, and to handle any relevant phase relationships with trigonometry.
2.2 Power in reactive Load
Consider a circuit for a single-phase AC power system, where a 120 volt, 60 Hz AC voltage source is delivering power to a reactive AC circuits: (Figure below)
For comparison, let's consider a simple AC circuit with a purely reactive load in Figure below.
Fig: 2.3
AC circuit with a purely reactive (inductive) load.

Fig: 2.4
Power is not dissipated in a purely reactive load. Though it is alternately absorbed from and returned to the source.
Note that the power alternates equally between cycles of positive and negative. (Figure above) This means that power is being alternately absorbed from and returned to the source. If the source were a mechanical generator, it would take (practically) no net mechanical energy to turn the shaft, because no power would be used by the load. The generator shaft would be easy to spin, and the inductor would not become warm as a resistor would.
Now, let's consider an AC circuit with a load consisting of both inductance and resistance in Figure below.

Fig: 2.5
AC circuit with both reactance and resistance.

At a frequency of 60 Hz, the 160 mill henry of inductance gives us 60.319 Ω of inductive reactance. This reactance combines with the 60 Ω of resistance to form a total load impedance of 60 + j60.319 Ω, or 85.078 Ω ∠ 45.152o. If we're not concerned with phase angles (which we're not at this point), we may calculate current in the circuit by taking the polar magnitude of the voltage source (120 volts) and dividing it by the polar magnitude of the impedance (85.078 Ω). With a power supply voltage of 120 volts RMS, our load current is 1.410 amps. This is the figure an RMS ammeter would indicate if connected in series with the resistor and inductor.
We already know that reactive components dissipate zero power, as they equally absorb power from, and return power to, the rest of the circuit. Therefore, any inductive reactance in this load will likewise dissipate zero power. The only thing left to dissipate power here is the resistive portion of the load impedance. If we look at the waveform plot of voltage, current, and total power for this circuit, we see how this combination works in Figure below.

Fig: 2.6
A combined resistive/reactive circuit dissipates more power than it returns to the source. The reactance dissipates no power; though, the resistor does.
As with any reactive circuit, the power alternates between positive and negative instantaneous values over time. In a purely reactive circuit that alternation between positive and negative power is equally divided, resulting in a net power dissipation of zero. However, in circuits with mixed resistance and reactance like this one, the power waveform will still alternate between positive and negative, but the amount of positive power will exceed the amount of negative power. In other words, the combined inductive/resistive load will consume more power than it returns back to the source.
Looking at the waveform plot for power, it should be evident that the wave spends more time on the positive side of the center line than on the negative, indicating that there is more power absorbed by the load than there is returned to the circuit. What little returning of power that occurs is due to the reactance; the imbalance of positive versus negative power is due to the resistance as it dissipates energy outside of the circuit (usually in the form of heat). If the source were a mechanical generator, the amount of mechanical energy needed to turn the shaft would be the amount of power averaged between the positive and negative power cycles.
Mathematically representing power in an AC circuit is a challenge, because the power wave isn't at the same frequency as voltage or current. Furthermore, the phase angle for power means something quite different from the phase angle for either voltage or current. Whereas the angle for voltage or current represents a relative shift in timing between two waves, the phase angle for power represents a ratio between power dissipated and power returned. Because of this way in which AC power differs from AC voltage or current, it is actually easier to arrive at figures for power by calculating with scalar quantities of voltage, current, resistance, and reactance than it is to try to derive it from vector, or complex quantities of voltage, current, and impedance that we've worked with so far.

  • In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
  • In a purely reactive circuit, no circuit power is dissipated by the load(s). Rather, power is alternately absorbed from and returned to the AC source. Voltage and current are 90o out of phase with each other.
  • In a circuit consisting of resistance and reactance mixed, there will be more power dissipated by the load(s) than returned, but some power will definitely be dissipated and some will merely be absorbed and returned. Voltage and current in such a circuit will be out of phase by a value somewhere between 0o and 90o.

2.3 True, Reactive, and Apparent power

We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power. This “phantom power” is called reactive power, and it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts. The mathematical symbol for reactive power is (unfortunately) the capital letter Q. The actual amount of power being used, or dissipated, in a circuit is called true power, and it is measured in watts (symbolized by the capital letter P, as always). The combination of reactive power and true power is called apparent power, and it is the product of a circuit's voltage and current, without reference to phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S.
As a rule, true power is a function of a circuit's dissipative elements, usually resistances (R). Reactive power is a function of a circuit's reactance (X). Apparent power is a function of a circuit's total impedance (Z). Since we're dealing with scalar quantities for power calculation, any complex starting quantities such as voltage, current, and impedance must be represented by their polar magnitudes, not by real or imaginary rectangular components. For instance, if I'm calculating true power from current and resistance, I must use the polar magnitude for current, and not merely the “real” or “imaginary” portion of the current. If I'm calculating apparent power from voltage and impedance, both of these formerly complex quantities must be reduced to their polar magnitudes for the scalar arithmetic.
There are several power equations relating the three types of power to resistance, reactance, and impedance (all using scalar quantities)
                       P= true power, P= I2R, P= E2/R
                       Measured in units of watts
                       Q= Reactive power, Q= I2X , Q= E2/X
                       Measured in units of volt amps-Reactive (VAR)
                       S= Apparent power, S= I2Z , S= E2/Z, S=IE
                       Measured in units of volt amps (VA)
Please note that there are two equations each for the calculation of true and reactive power. There are three equations available for the calculation of apparent power, P=IE being useful only for that purpose. Examine the following circuits and see how these three types of power interrelate for: a purely resistive load in Figure below, a purely reactive load in Figure below, and a resistive/reactive load in Figure below.
Resistive load only:                                                                                                                                                                
Fig: 2.7                                                   
True power, reactive power, and apparent power for a purely resistive load.
Reactive load only:

Fig: 2.8
True power, reactive power, and apparent power for a purely reactive load.

Resistive/reactive load:
Fig: 2.9
True power, reactive power, and apparent power for a resistive/reactive load.
These three types of power — true, reactive, and apparent — relate to one another in trigonometric form. We call this the power triangle

Fig: 2.10
Power triangle relating apparent power to true power and reactive power.
Using the laws of trigonometry, we can solve for the length of any side (amount of any type of power), given the lengths of the other two sides, or the length of one side and an angle.
The low power factor is mainly due to the fact that most of the power loads are inductive and, therefore, take lagging currents. In order to improve the power factor, some device taking leading power should be connected in parallel with the load. One of such devices can be a capacitor. The capacitor draws a leading current and partly or completely neutralises the lagging reactive component of load current. This raises the power factor of the load.
                  (i)                                          (ii)                                        (iii)
Fig: 3.1
To illustrate the power factor improvement by a capacitor, consider a single phase load taking lagging current I at a power factor cos φ1 as shown in Fig. 3.1
The capacitor C is connected in parallel with the load. The capacitor draws current IC which leads the supply voltage by 90°. The resulting line current I΄ is the phasor sum of I and IC and its angle lag is φ2 as shown in the phasor diagram of Fig. 3.1 (iii). It is clear that φ2 is less than φ1, so that cos φ2 is greater than cos φ1. Hence, the power factor of the load is improved. The following points worth noting:
I.The circuit current I΄after p.f. correction is less than the original circuit current I.
II. The active or wattful component remains the same before and after p.f. correction because only the lagging reactive component is reduced by the capacitor.
                                              Icos φ1 = I΄cos φ2
III.The lagging reactive component is reduced after p.f. improvement and is equal to the difference between lagging reactive component of load (I sin φ1) and capacitor current (IC) i.e.,
                                             I΄sin φ2 = I sin φ1 IC
IV.As                                I cos φ1 = I΄ cos φ2
                                       VI cos φ1 = V I΄  cos φ2            [Multiplying by V]
Therefore, active power (kW) remains unchanged due to power factor improvement.
V.                                  I΄ sin φ2 = I sin φ1  I
                                V I΄ sin φ2 = VI sin φ1  VIC              [Multiplying by V]
Net kVAR after p.f. correction = Lagging kVAR before p.f. correction  leading kVAR of equipment.
Normally, the power factor of the whole load on a large generating station is in the region of 0.8 to 0.9. However, sometimes it is lower and in such cases it is generally desirable to ke special steps to improve the power factor. This can be achieved by the following equipment:
  1. Static capacitors.
  2. Synchronous condenser.
  3. phase advancers.
The power factor can be improved by connecting capacitors in parallel with the equipment operating at lagging power factor. The capacitor (generally known as static capacitor) draws a leading current and partly or completely neutralises the lagging reactive component of load current. This raises the power factor of the load. For three-phase loads, the capacitors can be connected in delta or star as shown in Fig.  Static capacitors are invariably used for power factor improvement in factories.

Fig: 3.2

  • They have low losses.
  • They require little maintenance as there are no rotating parts.
  • They can be easily installed as they are light and require no foundation.
  • They can work under ordinary atmospheric conditions.
  • They have short service life ranging from 8 to 10 years.
  • They are easily damaged if the voltage exceeds the rated value.
  • Once the capacitors are damaged, their repair is uneconomical.
Static capacitor banks are intended to increase bandwidth and stability of power transmission lines, to stabilize voltage in load carrying components, to decrease power loss, to improve quality of distribution and reactive power compensation.
Capacitor banks can be operated separate units and as part of static thyristor compensators, filtering/compensating devices and in connection with shunting reactors, and with shunt reactors, controlled reactors and vacuum-reactor groups.
Technical Data:
  • Nominal Capacity, MVAr: 5 – 200
  • Nominal Voltage, kV: 6, 10, 35, 110
  • Neutral Terminal: isolated or dead-grounded
  • Enclosure: in accordance to GOST 15150-69 У1, УХЛ1
Other capacity and voltage options are available by customer's request

Fig: 3.3
A synchronous motor takes a leading current when over-excited and, therefore, behaves as a capacitor. An over-excited synchronous motor running on no load is known as Synchronous condenser. When such a machine is connected in parallel with the supply, it takes a leading current which partly neutralises the lagging reactive component of the load. Thus the power factor is improved.
Fig shows the power factor improvement by synchronous condenser method. The 3φ load takes current IL at low lagging power factor cos φL. The synchronous condenser takes a current Im which leads the voltage by an angle φm.. The resultant current I is the phasor sum of Im and IL and lags behind the voltage by an angle φ. It is clear that φ is less than φL so that cos φ is greater than cos φL. Thus the power factor is increased from cos φL to cos φ . Synchronous condensers are generally used at major bulk supply substations for power factor improvement.


Fig: 3.4

  • By varying the field excitation, the magnitude of current drawn by the motor can be changed by any amount. This helps in achieving step less control of power factor.
  • The motor windings have high thermal stability to circuit current .
  • The fauls can be removed easily.
  • There are considerable losses in the motor.
  •  The maintenance cost is high.
  • It produces noise.
  • Except in sizes above 500 kVA, the cost is greater than that of static capacitors of the same rating.
  • As a synchronous motor has no self-starting torque, therefore, auxiliary equipment has to be provided for this purpose.


Phase advancers are used to improve the power factor of induction motors. The low  power factor of an induction motor is due to the fact that stator winding draws exciting current which lags behind the supply voltage by 90º . If the exciting ampere turns can be provided from some other a.c. source, then a stator winding will be relieved of exciting current and a power factor of the motor can be improved. This job accomplished by the phase advancer which is simply an ac. exciter. The phase advancer is mounted on the same shaft as the main motor and is connected in the rotor circuit of the motor. It provides exciting ampere turns to the rotor circuit at slip frequency. By providing more ampere turns than required. the induction motor can be made to operate on leading power factor like an over-excited synchronous motor.
Phase advancers have two principal advantages. Firstly. As the exciting ampere turns are supplied at slip frequency, therefore, lagging kVAR drawn by the motor are considerably reduced. Secondly, phase advancer can be conveniently used where the use of synchronous motors is unadmissible. However, the major disadvantage of phase advancers is that they are not economical for motors beIow 200 H.P.
In electrical engineering, a synchronous condenser (sometimes synchronous compensator) is a synchronous motor that is not attached to any driven equipment. Its field is controlled by a voltage regulator to either generate or absorb reactive power as needed to support a system’s voltage or to maintain the system power factor at a specified level. The condenser’s installation and operation are identical to large electric motors.
Increasing the device's field excitation results in its furnishing magnetizing power (kilovars) to the system. Its principal advantage is the ease with which the amount of correction can be adjusted. The energy stored in the rotor of the machine can also help stabilize a power system during short circuits or rapidly fluctuating loads such as electric arc furnaces. Large installations of synchronous condensers are sometimes used in association with high-voltage direct current converter stations to supply reactive power.
Synchronous motors are used principally in large power applications because of their high operating efficiency, reliability, controllable power factor, and relatively low sensitivity to voltage dips. They are constant-speed machines with applications in mills, refineries, power plants, and the like, to drive pumps, compressors, fans, pulverizers, and other large loads, and to assist in power-factor correction. One very interesting .application is the 44-MW, b-ky, 60-Hz, 50-pole, 144 r/min synchronous motors used to drive the Queen Elizabeth 11 passenger ship. A solid-state volts/hertz drive circuit provides speed control through frequency adjustment.
Synchronous machines designed specifically for power-factor control have no external shafts and are called synchronous condensers. They “float” on the bus, supplying reactive power to the system. The direction of the reactive power, and hence the power factor of the system, is adjusted by changing the field excitation of the machine.
Unless otherwise specified, when discussing the behavior of individual motors and/or generators, it will be assumed that the machine is connected to a power source of unlimited capacity and zero impedance, called an infinite bus. The terminal voltage and frequency of the infinite bus remain constant and are unaffected by any power drawn from or supplied to the infinite bus. Large power systems in the United States and other highly industrialized countries may be considered to approximate those of the infinite bus
When starting, the rotor is allowed to accelerate to its maximum speed as an induction motor using its built-in damper windings.2 At this speed the slip is very small and the rotating flux of the stator moves very slowly relative to the revolving rotor. Direct current is then applied to the magnet windings, forming alternate north and south poles that “lock” in rotational synchronism with the corresponding opposite poles of the rotating flux, the slip is zero, all induction-motor action ceases, and the machine operates at synchronous speed. The synchronous speed, as developed in Section 4.5 in Chapter 4, is expressed in terms of the stator frequency and the number of poles:
ns=  r/min                               (1)
If the rotor magnets are energized before the machine reaches its maximum speed as an induction motor, the rotor may not synchronize and severe vibration will occur; every time a pole of the rotating stator flux passes a rotor pole, alternate attraction and repulsion occurs. Such out-of-step operation, called pole slipping, also causes cyclic current surges and torque pulses at slip frequency in the armature windings.
The adverse effects of frequent starting, or operating with unbalanced three-phase line voltages, are the same as for induction motors.
A simplified circuit diagram for rotor and stator connections is shown in Figure 4.1 A varistor, or resistor and switch, connected across the field leads during locked rotor and acceleration, prevents a high induced emf in the field windings, and the induced current in the circuit formed by the field windings and external resistor provides additional induction-motor torque.
The circuit for a brushless excitation system is shown in Figure 8.6. A frequency- sensitive solid-state control circuit monitors the frequency of the emf induced in the motor field winding by the rotating flux of the stator. The frequency of the emf is the same as that in the squirrel-cage winding and is a function of the frequency of the applied stator voltage and the slip. That is,
   fs = sfs                                       (2)
At locked rotor s = 1 .0, causing the rotor frequency to equal the applied stator frequency. As the rotor accelerates, the slip becomes smaller, and the frequency decreases.
When the frequency-sensitive circuit observes the “high frequency” generated in the rotor coils at locked rotor, and during high-slip acceleration, it closes SCR-2 and opens SCR- I; opening SCR- 1 blocks the exciter current, and closing SCR-2 connects the discharge resistor across the field windings.


Fig: 4.1
Simplified circuit diagrams for rotor and stator connections of a synchronous motor.
At near-synchronous speed, the rotor frequency is very low (approaching zero), and the control circuit opens closes SCR-l, opening SCR-2 disconnects the discharge resistor, and admits current to the field windings. The solid-state switching is programmed to close SCR-l at an instant that will ensure that the rotor poles will be facing opposite polarity, thus preventing pole slipping.

Fig: 4.2
Circuit diagram of a brushless excitation system for a synchronous machine. (Courtesy Dresser Rand, Electric Machinery)
Synchronous motors without built-in starting components are not energized until they have been accelerated to approximately synchronous speed by an auxiliary motor or turbine. No load is connected to a synchronous motor during the acceleration period. When near synchronous speed, the magnet windings are energized, and the field current adjusted to obtain an induced stator voltage approximately equal to the supply voltage. The circuit breaker connecting the stator to the three-phase supply is dosed when the induced stator voltage is in phase with the supply voltage. This procedure, called paralleling.
Reversing a Synchronous Motor
The direction of rotation of a synchronous motor is determined by its starting direction, as initiated by induction-motor action. Thus, to reverse the direction of a three-phase synchronous motor, it is necessary to first stop the motor and then reverse the phase sequence of the three-phase connections at the stator. Reversing the current to the field windings will not affect the direction of rotation.
Although the rotor of a synchronous motor rotates in synchronism with the rotating flux of the stator. Increases in shaft load cause the rotor magnets to change their angular position with respect to the rotating flux. This displacement angle can be seen viewing the rotor with a strobe light synchronized with the stator frequency.
Figure 4.3 shows a “strobe view” of a salient-pole motor operating at synchronous speed with no load on the shaft. The direction of rotation is counterclockwise (CCW). As the machine is loaded, the rotor changes its relative position with respect to the rotating flux of the stator. Lagging behind it by angle & Angle & expressed in electrical degrees, is called the power angle, load angle, or torque angle. The broken line on the South Pole in Figure 8.7 shows the position of the rotating magnets relative to the rotating flux of the stator for a representative load.
A synchronous motor operates at the same average speed for all values of load from no load to peak load. When the load on a synchronous motor is increased, the motor slows down just enough to allow the rotor to change its angular position in relation to the rotating flux of the stator, and then goes back to synchronous speed. Similarly, when the load is removed, it accelerates Just enough to cause the rotor to decrease its angle of lag in relation to the rotating flux, and then goes back to synchronous speed. When the peak load that the machine can handle is exceeded, the rotor pulls out of synchronism.
The torque developed by all synchronous motors has two components: a reluctance-torque component and a magnet-torque component. The reluctance-torque component is due to the normal characteristic of magnetic materials in a magnetic field to align themselves so that the reluctance of the magnetic circuit is a minimum


Fig: 4.3
Strobe view of a salient-pole motor operating at synchronous speed Shaft. (Courtesy Dresser Rand, Electric Machinery) The magnet-torque component is due to the magnetic attraction between the field poles (magnets) on the rotor and the corresponding opposite poles of the rotaring stator flux.
The analysis of synchronous-motor behavior in this chapter is based on magnet torque alone. This is justified for cylindrical-rotor motors because surface provides an insignificant reluctance torque. It is also justified motors operating from 50 percent rated load to above 100 percent rated load with power factors ranging from unity to leading; the reluctance torque significantly smaller than the magnet torque. The performance salient-pole motors.
The air-gap flux in a synchronous motor includes a rotating field flux φf due to DC current (If) in the rotating magnets, and a rotating armature flux. Called armature reaction flux (φar), caused by the three-phase armature currents in the tator windings. The magnitude and phase angle of the armature-reaction flux is a function of the magnitude and phase angle of the armature current. Both the magnet flux and the armature reaction flux rotate in the same direction, and at synchronous speed with respect to the armature windings.
The rotating fluxes φf and φar generates speed voltages in the stator conductor, as visualized in Fig 4.4 for one armature phase of a three-phase synchronous motor.








Fig: 4.4
Separate circuits showing emfs generated by component magnetic fields for one phase of a synchronous motor: (a) due to rotating magnets: (b) due to rotating armature- reaction flux motor. Although shown as separate diagrams for ease of visualization, both φf   and φar rotate within the same three-phase stator winding.
The rotor magnets sweeping the stator conductors, as shown in Figure 8.8(a), generate a speed-voltage, called a counter-emf (cemf) or excitation voltage, that acts in opposition to the applied voltage. The speed-voltage is proportional to the field flux and the speed of rotation. Expressed in equation form:
Ef = ns φE f       ————————– (3)
Expressing the flux in terms of magnetomotive force (mmf) and the reluctance of the magnetic circuit,
Ef = ns  K                                      (4)
Where:                 E, = excitation voltage/phase (V)
                             ns = synchronous speed (r/min)
                             φf = pole flux (Wb)
                             Nf= number of turns of conductor in field coil
                             If = DC field current (A)
                             = reluctance of magnetic circuit (A-t/Wb)
                             Kf= constant
The speed of a synchronous motor is equal to the speed of the rotating therefore constant for a given frequency. Thus, as indicated in Eq. (i), the excitation voltage is a function of the field current alone. Note, however, that because of magnetic saturation effects, the reluctance of the magnetic circuit is not constant. Hence φf and Ef are not proportional to If.
The cemf plays a significant role in synchronous-motor operation. And as will be      explained later, may be less than, equal to, or greater than the applied stator voltage. Its adjustment, through changes in field current, is also used to correct system power factor.
4.3.2 ARMATURE-REACTION VOLTAGE                                                                                                                                                                                                  The rotating armature-reaction flux sweeping the stator conductors, as visualized in Fig 4.4(b), generates a speed voltage called the armature-reaction voltage. The armature-reaction speed voltage, expressed in terms of armature-reaction flux is
Ear =  ns φar  Ka  ————————– (5)
Where:                         Ear = armature-reaction voltage (V)
                                     φar = armature-reaction flux (Wb)
                                     ns   = synchronous speed (r/min)
                                     Ka = constant
Neglecting the effects of magnetic saturation, the armature-reaction flux is proportional to the armature current. Hence, the armature-reaction voltage may be conveniently expressed in terms of the armature current and a fictitious armature reaction. Thus,
Ear =  Ia . J Xar ————————– (6)
where:                         Xar = armature-reaction reactance (Ω/phase).
The equivalent-circuit model for one armature phase of a cylindrical rotor synchronous motor is shown in Figure 4.5(a). All values are per phase. Applying Kirchhoff’s voltage law to Figure 4.5(a),


                                 Into Eq.




Fig: 4.5
(a) Equivalent-circuit model for one phase of a synchronous-motor armature; (b) hasor diagram corresponding to the equivalent-circuit model in (a).
where:                         Ra = armature resistance (Ω /phase)
                                    Xt = armature leakage reactance (Ω /phase)
                                    Xs = synchronous reactance (Ω l/phase’)
                                    Zs = synchronous impedance (Ω /phase)
                                   VT = applied voltage/phase (V)
A phasor diagram showing the component phasors and the tip-to-tai1 determination of Vt is shown in Figure 4.4(b). The phase angle δ of the excitation voltage Fig 4.4(b) is equal to the torque angle in Figure 4.3. The torque angle is also called the load angle or power angle.
Except for very small machines, the armature resistance of a synchronous motor is
relatively insignificant compared to its synchronous reactance, enabling Eq(10) be approximated as
VT= Ef + Ia JXs ————————– (11)
The equivalent-circuit and phasor diagram corresponding to Eq.(11) Figure 4.6 and are normally used for the analysis of synchronous-motor behavior as the motor responds to changes in load and/or changes in field excitation.
From the geometry of the phasor diagram,
Ia Xs cos θi = -Ef sin δ                         (12)
Multiplying through by Vand rearranging terms,
VTIa cos θi sin δ                 (13)
Since the left side of Eq. (13) is an expression for active power input, the magnet power/phase developed by the synchronous motor can be expressed as
Pin.Iφ= VTIa cos θi                                (14)
Pin.Iφ=  sin δ                             (15)
Thus, for a three-phase synchronous motor,
Pin = 3 * VTIa cos θi                             (16)
Pin =  sin δ                               (17)


Fig: 4.6
(a) equivalent-circuit (b) phasor diagram for a synchronous motor, assuming armature resistance is negligible.
Equation (15), called the synchronous-machine power equation, expresses the magnet power/phase developed by a cylindrical-rotor motor, in terms of its excitation voltage and power angle. Assuming a constant source voltage and constant frequency, Eqs. (14) and (15) may be expressed as proportionalities that are very useful in synchronous-machine analysis:
P α Ia cos θ                                        (18)
P α Ef sin δ                                        (19)
The effect that changes in shaft load have on armature current, power angle and power factor are illustrated in Figure 4.7 the applied stator voltage, frequency, and field excitation are assumed constant. The heavy lines indicate the initial load conditions, and the light lines indicate the new steady-state conditions that correspond to doubling the shaft load. Note: In accordance with Eqs. (18) and (19) doubling the shaft load doubles both Ia cos θi  and Ef sin δ. When adjusting phasor diagrams to show new steady-state conditions, the line of action of the new IaJXs phasor must be perpendicular to the new Ia phasor. Furthermore, as shown in Figure 4.7. if the excitation is not changed, increasing the shaft load causes the locus of the Ef phasor to describe a circular arc, increasing its phase angle with increasing shaft load. Note also that an increase in shaft load is also accompanied by a decrease in θi resulting in an increase in power factor.


Fig: 4.7
Phasor diagram showing effect of changes in shaft load on armature current, power angle, and power factor of a synchronous motor.
As additional load is placed on the machine, the rotor continues to increase its lag relative to the rotating flux, thereby increasing both the angle of lag of the cemf phasor and the magnitude of the stator current. During all this loading, however, except for the transient conditions whereby the rotor assumes a new position in relation to the rotating flux, the average speed of the machine does not change. Finally, with increased loading, a point is reached at which a further increase in fails to cause a corresponding increase in motor torque, and the rotor pulls out of synchronism. The point of maximum torque occurs at a power angle of approximately 90° for a cylindrical-rotor machine, as is evidenced in Eq. (17). The critical value of torque that causes a synchronous motor to pull out of synchronism is called the pull-out torque.
Intuitively we can expect that increasing the strength of the magnets will increase the magnetic attraction, and thereby cause the rotor magnets to have a closer alignment with the corresponding opposite poles of the rotating stator flux; the result is a smaller power angle. Proof of this behavior can be seen in Eq. (17). Assuming a constant shaft load, the steady-state value of Ef sin δ must be constant. A step increase in Ef will cause a transient increase in Ef sin δ. and the rotor will accelerate. As the rotor changes its angular position, δ decreases until Ef sin δ has the same steady-state value as be fore, at which time the rotor is again operating at synchronous speed angular position of the rotor magnets relative to the rotating flux of the a fraction of a second.
The effect of changes in field excitation on armature current, pc power factor of a synchronous motor operating with a constant shaft load, from a constant voltage, constant frequency supply, is illustrated in Figure – (19), for a constant shaft load,


Fig: 4.8

Phasor diagram showing the effect of changes in field excitation on armature current power angle, and power factor of a synchronous motor
Ef1 sin δ1 = Ef2 sin δ2 = Ef3 sin δ3 = Ef sin δ                       (20)
This is shown in Figure 4.8 where the locus of the tip of the E1 phasor is a straight line parallel to the Vt phasor. Similarly, from Eq. (18), for a constant shaft load.
Ia1 cos θi1 =  Ia2 cos θi2 = Ia3 cos θi3 = Ia cos θi                      (21)
This is shown in Figure 4.8, where the locus of the tip of the Ia phasor is a line perpendicular to the Vt phasor.
Note that increasing the excitation from Ef1 to Ef3 in Figure 4.8 caused the angle of the current phasor (and hence the power factor) to go from lagging to leading. The value of field excitation that results in unity power factor is called normal excitation. Excitation greater than normal is called over excitation, and excitation
less than normal is called under excitation. Furthermore, as indicated in Figure 4.8. when operating in the overexcited mode,  ‌‌ Ef  >  Vt  .


Curves of armature current vs. field current or armature current vs. excitation voltage are called V curves, and are shown in Figure 4.9. for representative synchronous- motor loads. The curves are related to the phasor diagram in Figure 4.8. and illustrate the effect of different values of field excitation on armature current and power factor for specific shaft loads. Note that increases in shaft load require increases in field excitation in order to maintain unity power factor.
The locus of the left endpoints of the V curves in Figure 4.9 represents the stability limit (δ =  90°). Any reduction in excitation below the stability limit for a particular load will cause the rotor to pull out of synchronism.
The constant-load V curves shown in Figure 4.9 may be plotted from laboratory data (Ia vs If) as If is varied, or graphically by plotting  Ia  vs  Ef  from a family of phasor diagrams such as that shown in Figure 8.12, or from the following mathematical expression for the V curves6:
Ia= . [ E2+ V2T – 2  ] 0.5                 (22)
Equation () is based on the geometry of the phasor diagram and assumes Ra S negligible. Note: If the developed torque is less than the shaft load plus wind age and friction, instability will occur, and the expression under the square root sign will be negative.
The family of V curves shown in Figure 4.9 represents computer plots of Eq. (22), assuming a three-phase, 40-hp, 220-V, 60-Hz synchronous motor with a synchronous reactance of 1.27 (Ω /phase).


Fig: 4.9
Family of representative V curve for a Synchronous motor

Fig: 5.1

  • Variac
  • Ammeter (DC)
  • Ammeter (AC)
  • Variable register AC supply voltage
  • DC supply voltage
  • Tacometer
  • Connector
When supply voltage= 315V then r.p.m = 1120 rpm
When supply voltage= 345V then r.p.m = 1230 rpm
When supply voltage= 430V then r.p.m = 1500 rpm
          N= 120f⁄P                        here
            = 120×50⁄4                    f= 50 Hz     
            =1500 r.p.m                    P= 4


Fig: 5.2

  • When excitation voltage vf = 0 then the armature current= 0.78 A
  • When excitation voltage vf = 50 then the armature current= 0.52 A
  • When excitation voltage vf =100 then the armature current= 0.2 A
  • When excitation voltage vf =125 then the armature current= Unity
  • When excitation voltage vf = 150 then the armature current= 0.15A
  • When excitation voltage vf = 200 then the armature current= 0.8 A
  • When excitation voltage vf = 220 then the armature current= 1.2A
As the load on a synchronous motor increases, the armature (stator) current Ia increases regardless of excitation. For under and over excited motors, the power factor (p.f.) tends to approach 1 with increase in load. The change in power factor is greater than the change in Ia with increase in load.
The magnitude of armature current varies with excitation. The current has large value both for low and high values of excitation. In between, it has minimum value corresponding to a certain excitation. The variations of I with excitation are known as V curves because of their shape.
For the same output load, the armature current varies over a wide range and so causes the power factor also to vary accordingly. When over-excited, the motor runs with leading power factor and with lagging power factor when under-excited. In between, the power factor is unity. The minimum armature current corresponds to unity power factor.


An over-excited synchronous motor has a leading power factor. This makes it useful for power factor correction of industrial loads. Both transformers and induction motors draw lagging currents from the line. On light loads, the power drawn by induction motors has a large reactive component and the power factor has a very low value. The current flowing to supply reactive power creates losses in the power system. In an industrial plant, synchronous motors can be used to supply some of the reactive power required by induction motors. This improves the plant power factor and reduces supply current.
A synchronous condenser provides step-less automatic power factor correction with the ability to produce up to 150% additional MVARs. The system produces no switching transients and is not affected by system electrical harmonics (some harmonics can even be absorbed by synchronous condensers). They will not produce excessive voltage levels and are not susceptible to electrical resonances. Because of the rotating inertia of the condenser, it can provide limited voltage support during short power outages.
The use of rotating synchronous condensers was common through the 1950s. They remain an alternative (or a supplement) to capacitors for power factor correction because of problems that have been experienced with harmonics causing capacitor overheating and catastrophic failures. Synchronous condensers are also very good for supporting voltage. The reactive power produced by a capacitor bank is in direct proportion to the square of its terminal voltage, where a synchronous condenser's reactive power declines less rapidly, and can be adjusted to compensate for falling terminal voltage. This reactive power improves voltage regulation in situations such as starting large motors, or where power must travel long distances from where it is generated to where it is used, as is the case with power wheeling (distribution of electric power from one geographical location to another within an electric power distribution system).
Synchronous Condensers may also be referred to as Dynamic Power Factor Correction systems. These machines can prove very effective when advanced controls are utilized. A PLC based controller with pf controller and regulator will allow the system to be set to meet a given power factor or can be set to produce a specified amount of KVAR. In both cases, the output of the machine obviously cannot exceed its capability.



As was mentioned before, the angle of this “power triangle” graphically indicates the ratio between the amount of dissipated (or consumed) power and the amount of absorbed/returned power. It also happens to be the same angle as that of the circuit's impedance in polar form. When expressed as a fraction, this ratio between true power and apparent power is called the power factor for this circuit. Because true power and apparent power form the adjacent and hypotenuse sides of a right triangle, respectively, the power factor ratio is also equal to the cosine of that phase angle. Using values from the last example circuit:

It should be noted that power factor, like all ratio measurements, is a unitless quantity.
For the purely resistive circuit, the power factor is 1 (perfect), because the reactive power equals zero. Here, the power triangle would look like a horizontal line, because the opposite (reactive power) side would have zero length.
For the purely inductive circuit, the power factor is zero, because true power equals zero. Here, the power triangle would look like a vertical line, because the adjacent (true power) side would have zero length.
The same could be said for a purely capacitive circuit. If there are no dissipative (resistive) components in the circuit, then the true power must be equal to zero, making any power in the circuit purely reactive. The power triangle for a purely capacitive circuit would again be a vertical line (pointing down instead of up as it was for the purely inductive circuit).
Power factor can be an important aspect to consider in an AC circuit, because any power factor less than 1 means that the circuit's wiring has to carry more current than what would be necessary with zero reactance in the circuit to deliver the same amount of (true) power to the resistive load. If our last example circuit had been purely resistive, we would have been able to deliver a full 169.256 watts to the load with the same 1.410 amps of current, rather than the mere 119.365 watts that it is presently dissipating with that same current quantity. The poor power factor makes for an inefficient power delivery system.
Poor power factor can be corrected, paradoxically, by adding another load to the circuit drawing an equal and opposite amount of reactive power, to cancel out the effects of the load's inductive reactance. Inductive reactance can only be canceled by capacitive reactance, so we have to add a capacitor in parallel to our example circuit as the additional load. The effect of these two opposing reactances in parallel is to bring the circuit's total impedance equal to its total resistance (to make the impedance phase angle equal, or at least closer, to zero).
Since we know that the (uncorrected) reactive power is 119.998 VAR (inductive), we need to calculate the correct capacitor size to produce the same quantity of (capacitive) reactive power. Since this capacitor will be directly in parallel with the source (of known voltage), we'll use the power formula which starts from voltage and reactance:

Let's use a rounded capacitor value of 22 µF and see what happens to our circuit: (Figure below)
Parallel capacitor corrects lagging power factor of inductive load. V2 and node numbers: 0, 1, 2, and 3 are SPICE related, and may be ignored for the moment.

The power factor for the circuit, overall, has been substantially improved. The main current has been decreased from 1.41 amps to 994.7 milliamps, while the power dissipated at the load resistor remains unchanged at 119.365 watts. The power factor is much closer to being 1:

Since the impedance angle is still a positive number, we know that the circuit, overall, is still more inductive than it is capacitive. If our power factor correction efforts had been perfectly on-target, we would have arrived at an impedance angle of exactly zero, or purely resistive. If we had added too large of a capacitor in parallel, we would have ended up with an impedance angle that was negative, indicating that the circuit was more capacitive than inductive.
A SPICE simulation of the circuit of (Figure above) shows total voltage and total current are nearly in phase. The SPICE circuit file has a zero volt voltage-source (V2) in series with the capacitor so that the capacitor current may be measured. The start time of 200 msec ( instead of 0) in the transient analysis statement allows the DC conditions to stabilize before collecting data. See SPICE listing “pf.cir power factor”.

           pf.cir power factor
           V1 1 0 sin(0 170 60)
           C1 1 3 22uF
           v2 3 0 0
           L1 1 2 160mH
           R1 2 0 60
           # resolution stop start
           tran 1m     200m 160m
The Nutmeg plot of the various currents with respect to the applied voltage Vtotal is shown in (Figure below). The reference is Vtotal, to which all other measurements are compared. This is because the applied voltage, Vtotal, appears across the parallel branches of the circuit. There is no single current common to all components. We can compare those currents to Vtotal.

Zero phase angle due to in-phase Vtotal and Itotal . The lagging IL with respect to Vtotal is corrected by a leading IC .
Note that the total current (Itotal) is in phase with the applied voltage (Vtotal), indicating a phase angle of near zero. This is no coincidence. Note that the lagging current, IL of the inductor would have caused the total current to have a lagging phase somewhere between (Itotal) and IL. However, the leading capacitor current, IC, compensates for the lagging inductor current. The result is a total current phase-angle somewhere between the inductor and capacitor currents. Moreover, that total current (Itotal) was forced to be in-phase with the total applied voltage (Vtotal), by the calculation of an appropriate capacitor value.
Since the total voltage and current are in phase, the product of these two waveforms, power, will always be positive throughout a 60 Hz cycle, real power as in Figure above. Had the phase-angle not been corrected to zero (PF=1), the product would have been negative where positive portions of one waveform overlapped negative portions of the other as in Figure above. Negative power is fed back to the generator. It cannot be sold; though, it does waste power in the resistance of electric lines between load and generator. The parallel capacitor corrects this problem.
Note that reduction of line losses applies to the lines from the generator to the point where the power factor correction capacitor is applied. In other words, there is still circulating current between the capacitor and the inductive load. This is not normally a problem because the power factor correction is applied close to the offending load, like an induction motor.
It should be noted that too much capacitance in an AC circuit will result in a low power factor just as well as too much inductance. You must be careful not to over-correct when adding capacitance to an AC circuit. You must also be very careful to use the proper capacitors for the job (rated adequately for power system voltages and the occasional voltage spike from lightning strikes, for continuous AC service, and capable of handling the expected levels of current).
If a circuit is predominantly inductive, we say that its power factor is lagging (because the current wave for the circuit lags behind the applied voltage wave). Conversely, if a circuit is predominantly capacitive, we say that its power factor is leading. Thus, our example circuit started out with a power factor of 0.705 lagging, and was corrected to a power factor of 0.999 lagging.

  • Poor power factor in an AC circuit may be “corrected”, or re-established at a value close to 1, by adding a parallel reactance opposite the effect of the load's reactance. If the load's reactance is inductive in nature (which is almost always will be), parallel capacitance is what is needed to correct poor power factor.


When the need arises to correct for poor power factor in an AC power system, you probably won't have the luxury of knowing the load's exact inductance in henrys to use for your calculations. You may be fortunate enough to have an instrument called a power factor meter to tell you what the power factor is (a number between 0 and 1), and the apparent power (which can be figured by taking a voltmeter reading in volts and multiplying by an ammeter reading in amps). In less favorable circumstances you may have to use an oscilloscope to compare voltage and current waveforms, measuring phase shift in degrees and calculating power factor by the cosine of that phase shift.
Most likely, you will have access to a wattmeter for measuring true power, whose reading you can compare against a calculation of apparent power (from multiplying total voltage and total current measurements). From the values of true and apparent power, you can determine reactive power and power factor. Let's do an example problem to see how this works: (Figure below)

Fig: 6.3
Wattmeter reads true power; product of voltmeter and ammeter readings yields appearant power.
First, we need to calculate the apparent power in kVA. We can do this by multiplying load voltage by load current:

As we can see, 2.308 kVA is a much larger figure than 1.5 kW, which tells us that the power factor in this circuit is rather poor (substantially less than 1). Now, we figure the power factor of this load by dividing the true power by the apparent power:

Using this value for power factor, we can draw a power triangle, and from that determine the reactive power of this load: (Figure below)


Fig: 6.4
Reactive power may be calculated from true power and appearant power.
To determine the unknown (reactive power) triangle quantity, we use the Pythagorean Theorem “backwards,” given the length of the hypotenuse (apparent power) and the length of the adjacent side (true power):


If this load is an electric motor, or most any other industrial AC load, it will have a lagging (inductive) power factor, which means that we'll have to correct for it with a capacitor of appropriate size, wired in parallel. Now that we know the amount of reactive power (1.754 kVAR), we can calculate the size of capacitor needed to counteract its effects:

Rounding this answer off to 80 µF, we can place that size of capacitor in the circuit and calculate the results: (Figure below)


Fig: 6.5
Parallel capacitor corrects lagging (inductive) load.
An 80 µF capacitor will have a capacitive reactance of 33.157 Ω, giving a current of 7.238 amps, and a corresponding reactive power of 1.737 kVAR (for the capacitor only). Since the capacitor's current is 180o out of phase from the the load's inductive contribution to current draw, the capacitor's reactive power will directly subtract from the load's reactive power, resulting in:

This correction, of course, will not change the amount of true power consumed by the load, but it will result in a substantial reduction of apparent power, and of the total current drawn from the 240 Volt source: (Figure below)



Fig: 6.6
Power triangle before and after capacitor correction.
The new apparent power can be found from the true and new reactive power values, using the standard form of the Pythagorean Theorem:

This gives a corrected power factor of (1.5kW / 1.5009 kVA), or 0.99994, and a new total current of (1.50009 kVA / 240 Volts), or 6.25 amps, a substantial improvement over the uncorrected value of 9.615 amps! This lower total current will translate to less heat losses in the circuit wiring, meaning greater system efficiency (less power wasted). 
With respect to Fluorescent and High intensity Discharge Lamps.
All Discharge Lamps, such as Fluorescent Lamps, High Pressure Mercury Vapour Lamps, Sodium Lamps, Metal Halide Lamps, etc., require ballasts (chokes) or transformers for their operation. These devices are inductive in nature. When a Discharge Lamp is switched on, it draws Apparent Power from the mains. This Apparent Power (VA) has two components; one is the Active Power (W) actually being consumed by the lamp for illuminating it, and the other is the Reactive Power (VAr) feeding the electromagnetic circuit of the control gear.
Power factor is the ratio of Active Power (W) to the Apparent Power (VA) (Figure 1).

Power factor is also denoted by p.f. or cos Ø . Typical p.f. of ballasts is from 0.35 to 0.5.
As mentioned above, the inductive components, such as ballasts, draw Reactive Power (VAr) from the mains. It lags behind the Active Power (W) by 90o (Figure 1). A capacitor, if connected across the mains, will also draw Reactive Power [VAr(c)], but it leads the Active Power (W) by 90o. The direction of the capacitive Reactive Power [VAr(c)] is opposite to the direction of the inductive Reactive Power (VAr) (Figure 6.7).


Fig: 6.7
If a capacitor is connected in parallel with an inductive load, it will draw capacitive Reactive Power [VAr(c)]. The effective Reactive Power drawn by the circuit will reduce to the extent of the capacitive Reactive Power [VAr(c)], resulting in reduction of Apparent Power from VA to VA1. The phase angle between the Active Power and the new Apparent Power VA1 will also reduce from Ø to Ø1 (Fig: 6.8). Thus the power factor will increase from cos Ø to cos Ø1.New
p.f. = cos Ø1 =

By selecting a capacitor of an appropriate value, the power factor can be corrected to 1. However, in practice, the power factor is improved to fall between 0.9 and 0.95.
                                                                                                                                                                                                                Fig: 6.8                                                                                                                                                                       


If your electricity bill contains a power factor penalty or if you are charged for peak load measured in kVA, power factor correction can help you reduce such surcharges. Look at your electricity bill to determine if such charges are relevant to your business.
If you are paying power factor charges, it will probably be worthwhile investing in power factor correction equipment. Often your investment in such equipment can be recovered in a 2-3 years and from then on will provide ongoing savings to you business.
The electrical power used by motors and fluorescent tubes, has two components; active power and reactive power. The active power is converted into useful work – turning motors or producing light- but the reactive component is only used to “energize” the magnetic or electrostatic properties of the equipment. Reactive power still has to be produced by the generator and transmitted to your business over the electricity network. Producing and transmitting the reactive power creates costs for the electricity companies which are passed on to you as power factor penalties or surcharges.


  Power factor diagram without factor correction

The Power Factor is, roughly speaking, a measure of what proportion of the total power supplied to your business is actually used by the equipment. The power factor value can vary between just above 0 and 1. The aim is keep the power factor as close as possible to 1, as power factor penalties generally cut in if it drops below 0.85.


  Diagram with power factor correction

Fig: 6.10
To effectively reduce your power factor, you will need to obtain a technical assessment and design parameters for the power factor correcting capacitors to ensure they match the requirements of your equipment. Your energy supplier may be able to assist you in supplying this technical assistance, but there are also technical consultants available who will have the expertise.
Why not consider having a complete energy audit done for you business, which also addresses your power correction requirements? Generally the cost savings from energy efficiency actions identified in the audit will more than pay for the cost of the audit.


  • If you use equipment containing electric motors, or florescent lights extensively, you may have a problem with your power factor.
  • Check your bill for power factor penalties or peak load measured in kVA.
  • A poor power factor will result in additional electricity costs to your business.
  • Obtain expert technical advice to determine how your power factor can be corrected and install correction equipment if required.
  • Consider also conducting a full energy audit, as these generally identify cost effective energy savings opportunities.
The main advantages of the Power Factor Correction are:
1. The electrical load on the Utility is reduced, thereby allowing the Utility to supply the surplus power to other consumers, without increasing its generation capacity.
2. Most of the Utilities impose low power factor penalties. By correcting the power factor, this penalty can be avoided.
3. High power factor reduces the load currents. Therefore, a considerable saving is made in the hardware cost, such as cables, switchgear, substation transformers, etc.
Synchronous motors load the power line with a leading power factor. This is often usefull in cancelling out the more commonly encountered lagging power factor caused by induction motors and other inductive loads. Originally, large industrial synchronous motors came into wide use because of this ability to correct the lagging power factor of induction motors.
This leading power factor can be exaggerated by removing the mechanical load and over exciting the field of the synchronous motor. Such a device is known as a synchronous condenser. Furthermore, the leading power factor can be adjusted by varying the field excitation. This makes it possible to nearly cancel an arbitrary lagging power factor to unity by paralleling the lagging load with a synchronous motor. A synchronous condenser is operated in a borderline condition between a motor and a generator with no mechanical load to fulfill this function. It can compensate either a leading or lagging power factor, by absorbing or supplying reactive power to the line. This enhances power line voltage regulation.
In industrial application where both synchronous motors and induction motors are used. The synchronous motor is usually operated at a leading power factor to compensate for the lagging power factor of the induction motors. In those cases where a synchronous motor is operated whithout load, merely for the perpose of improving the power factor of a system, the machine is called a synchronous condenser.machines specifically designed for this application are built whithout external shafts.                                    
The electrical power used by motors and fluorescent tubes, has two components; active power and reactive power. The active power is converted into useful work – turning motors or producing light- but the reactive component is only used to “energize” the magnetic or electrostatic properties of the equipment. Reactive power still has to be produced by the generator and transmitted to your business over the electricity network. Producing and transmitting the reactive power creates costs for the electricity companies which are passed on to you as power factor penalties or surcharges.
  1. Charles I Hubert . ELECTRICAL MACHINE  theory ,operation,  
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  1. V.K Mehata . Rohit Mehata. PRINCIPAL OF POWER SYSTEM
  1. Jhon j.Gainger, William D. Stevenson.JR POWER SYSTEM ANALYSES
  1. Tony R Kuphaldp. Under the terms & condition of the Design science License POWER ELECTRIC CIRCUIT (Copy Right)
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