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    Apparent Power: Components and Physical nterpretat ionAlexander E. Emanuel

    Worcester Polytechnic Institute, Worcester, MA 01609

    Abstract - This study explains the basic attributesas well as the physical properties of the apparent pow=S . The apparent power is divided in active and nonactivepowers. I t is shown that all forms of nonactivepower havea common form of manifestation, parasitic oscillations ofenergy among electrical components and electromecani-cal, electrochemical, or electrothermal systems. The mainconclusion of this paper is summarized in a recommenda-tion to incorporate all the nonactive powers, except thefundamental one, in one entity.

    K e y Words Power Definitions. Harmonics. PowerQuality.

    1. INTRODUCTIONSingle-phase sinusoidal circuit theory gives for the

    instantaneous power of a load characterized by the ac-tive power P = VIcusB and leading power factor PF =cosB = P / S , he well known expression:

    p = P - s4 2 w t - ) (1)The power loss dissipated in the feeder dedicated to theabove load is:

    Two basic properties of the apparent powerS re revealedby the equations (1) and (2) :I 1I . In single-phase and sinusoidal situations the

    apparent power i s the amplitude of the instanta-neous power oscillations.

    The frequency of these oscillations is w/1c = 2f, .e.twice th e power systems frequency.

    II . If the load voltage is kept constant, the variablepower loss in the feeder i s proportional with theapparent power squared.

    IThe universally accepted resolution of the apparent

    power in active power P and reactive power Q = V I sin Byields:

    S2= P2+Q2 (3)n o m (2) and (3) results:

    Q2AP = r-(1+ F)V2 (4)This expression shows that minimum feeder losses are o btained when the load reactive power Q = 0. An otherfundamental fact learned from (4) is that when the loadvoltage V and the current I are kept constant the loadactive power P is maximized when Q = 0. Mathemat-ically this can be readily observed if we rewrite (2) nd(3)differently:

    andQ2=--V2APr

    SThe fact that V and I are kept constant means the samething as keeping AP nd S constants. All these observa-tions lead to a third important property of S:

    III. The apparent power of a load or cluster ofloads supplied b y a wmm on feeder, i s the maxi-mum active power that can be transmitted throughthe feeder, while keeping the load voltage and thefeeder losses constant. This definition can be ex-tended to a voltage or current source: The apparentpower of a souire i s the mazimum active powei-that can be supplied b y the source while keepingita output voltage and internal losses constant.

    0-7803-5105-3/98/$10.00 0 1998IEEE

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    It is this definition of S that leads to the notion of Secondly, the very definition ofS in three phase systems,utilization factor or, as t is better known, the power f a e is still controversial(l,2,3]and a universal agreement willtor: depend on the acceptance of nonactive powers definitions.

    There are four elementary forms of nonactive powers:F = P IS (5 )Substituting in (5) the mean values of P and S taken fora time interval T, esults:

    Thi s expression gives a clearer meaning to PF, the powerfactor quantifies feeder utilization; W p s the actual ener-gy delivered to the load and Ws s the maximum possibleenergy that could be delivered to the load, during thetime T,while keeping the energy lost in the feeder con-stant, i.e:

    Aw =[ ~ d t constantProperties I1and 111, aswell as he PF nterpretation holdtrue for any type of load, linear or nonlinear, and for anytype of circuit topology, including polyphase balanced orunbalanced.

    When a load transformer is included in the circuit(Fig.l), the to tal losses are:

    I. Displacement AngleThis phenomenon was explained and named in 1888 bythe ac circuit pioneers "displacement of current phase"[4,5}. Assuming a single-phase load with sinusoidal volt-age and current waves:

    v = JZVsinwt (8 )i = f i I s in (wt+e)

    = i = vIase - vicos(2wt + e)(9)

    the resulting instantaneous power is(10)

    If the load is a resistance, B = 0, and the instantaneouspower p, and energy zu are:

    p = VI(1- cos2wt)W = sin 2wt

    vz s2 vzAPe d 2 +- r - + -R V R This s the case ofPF = 1,at any moment p 1 , i.e. the(7 ) energy rate of flow can not be negative. This is a remark-able property, since in spite of th e alterna ting voltage andw(vd he load.

    where r = F + p+r s ~nd R is the mamething branch currentw8ves, he energy flow to-resistance. Th e relation A P ersus9 emains nearly lin-ear. The error caused by rs , 1s is not significant, morenear'y constant when alineto-line voltage V and a balanced threephase load.over, the traIISformer (the fixed lose)^ remain In the (lo), when 8 # 0, during eachvaries. In (7) it wM half- cycle p is positive and negative, hence the energy wflowsn both ktions. For example, duringk x - 0 controls the loadunbalance. In this case! he instantaneous power is:

    = P P + P u (18)where p p = (2+p)VZ is the instantaneous active powercomponent and

    pu = (1- P)VZcos2wtis the unbalanced instantaneous power.

    W e n P = 1, the load is balanced and p = p p = 3VZ.In this c- the instantanems power is is constant, nooverall energy oscillations occurs. This is the most mtable property of polyphase systems, it translates into aconstant electromagneticmotor torque and a nil neutralcurrent. It is true that for unity power factor each linecaries he intrinsical power VZ[l--cos(2wt--p,,~)], (whereP 0 . k = 0, 1200, tl%P), as shown in Fig.4. The oscillat-ing terms of these three instantaneous powers canceleachother, i.e. at a certain moment on one ine the. oscillatingpower flows from th e source o th e load and returns viathe other twolies. In each phase, however, the intrinsi-cal active powers are positive an d the total instantaneouspower is constant.

    When the load is unbalanced, P # 1, an oscillatingterm p u is created, Figd, and in the transferred energyequation two terms appear:

    wT = W p +WUwhere w p = Pt is the useful energy and

    is the nonactive term caused by the load unbalance. Inthis case no energy is stored in the load. It can be provedtha t an unbalanced load converts part of the energy de-livered by the positive sequence components in negativeenergy carried by negative and zero sequence powers. Theinteraction between the alternator positive sequence volt-ages and the nonpositive sequence currents, generated bythe unbalanced load, is causing the nonactive energy os-cillations.

    The unbalanced loads have aPF < 1,even if the loadhas no L or C omponents. This fact can be explainedasfollows: Let us assume an equivalent three-phase system,perfectly balanced and with PF=1, that delivers the sameactive power 3VZ, as he initial unbalanced system, whilemaintaining the same load voltage V. F'" (18) results:

    P = (2 + p)vz = 3vzewhere 2 + P zI , =-is the equivalent current flowing in the lines supplyingthe equivalent load. The line power loss in the equivalentlines is: AP,= 3 t ~ z 3rz2(-)' + P3In the unbalanced system flowsa neutral current with therms value:and assuming that the neutral and the line conductorshave the same resistance t , esults th at the power loss inthe unbalanced system is:AP = r[2z2+P'I' + I - P)'z2] = (3+ 2 ~ ' ~ ) r ~ 'One can easily ascertain that AP > AP,. Calculationsyield: 5AP -AP' = -(1- /3)2rZ23This result demonstrates that for an unbalanced load,even purely resistive, PF < 1.

    z, = (1 - p)z

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    The reader should also receive the concept that thepolyphase system is viewed by the utility engineerasoneentity, a single electromagnetic waveguide meant t o tr?fer the electric energy to the polyphase load. This claimrelies on the fact that best system utilization takes placewhen the efficiency is maximized.

    IV. Current/Voltage DistortionIn networks with sinusoidal voltage sources nonlinear

    loads cause the direet distortion of currents and the indi-rect distortion of the voltage impressed across the loads.The flow of harmonic currents causes additional losses inconductors, reducing feeders utilization. The nonactivepowers associated with the flow of harmonics can be d escri bed assuming a sinusoidal voltage, v = d V s i n u t ,supplying a nonlinear load simulated by an array of har-monic currents:

    The harmonic order h is an integer or a noninteger num-ber. The instantaneous power has two terms:

    P = P m + P D Jwhere pp 1 is the fundamental instantaneous power:

    pp1 = PI+PIcos2wt +Q1sin2wtWe recognize a fundamental active power PI= VI1c o s 8 i land a fundamental reactive power Q 1 = VZ1 sin Oi lThe econd term of the instantaneous power is due t oharmonic currents and has the expression:

    p D I = XDIh sinut in(hwt+ a i h ) (20)hflwhere D l h = VZ,. The instantaneous power caused byeach harmonic current and the sinusoidal voltage v is adouble oscillation at the frequencies(hf )f, that will nottransfer energy. In Fig.6 are presented typical waveformscharacteristic to the interaction of a 60 Hz voltage witha 300 Hz harmonic current. The energy oscillogram, inspite of its unusual waveform, has the expected nil meanvalue.

    The nonactive powers caused by distortion are onlythe terms D I h . Similar conclusionswill be =ached whenone considers a current source i = 4Z si nw t supplyingan array of harmonic voltage sources connected in series,(an other hypothetical representation of a nonlinear load).

    In this case the nonactive powers caused by the voltagedistortion will have the expressionD V h = zvh. In practiceboth current and voltage waveforms are distorted. In thiscase the instantaneous power has four terms:

    ~ = ~ ~ + P D I + P D V + P H v l i l + v l i H + V H i l+ V H iHThe terms p o i and ~ D V re nonactive powers , p o l isgiven in (20) and

    p D v = sin(wt+ i l ) sin(hwt+eul)h f l

    where D V h = &Vh. The instantaneous power p~ is dueto interaction among harmonic currents and voltages:pH = x p h { 1 cos[2(hwt+0,h)l) +xQhsin 2 ( u t +o Uh )

    h#1 h#1+E~V~Z.i n ( w t+em) sin(nwt +@in)For this case we observe the harmonic active and re-

    m.nm#n

    active powers of order h:p h = v h l h ca(&heth)Q h = vhzh sin(&h&h)

    and terms D, = VmZ, caused by the cross products ofcurrent and voltage harmonics of different orders. Theapparent power of this load is:

    s = %"&"

    The only active powers are the fundamental PI nd thetotal harmonic active power P H =&,.Ph. All othersterms are nonactive powers, and each elementary term istied to an oscillation of energy without net transfer to thebad. The power factor for this situation is:

    Evidently this analysishas proved the existence of a hostof elementary nonactive powers. Now two question shouldbe addressed: First; Does the physical mechanism, thatgoverns nonactiveenergies oscillations differ from one typeof nonactive ascillation t o the other? Second; Are thereany reasons to group separately some nonactive powersfiom others, when the purpose of monitoring the nonac-tive powers is for commercial use? Specifically, when thegoals are limited to electric energy measurement, evalu-ation of power quality and the fair recovery of investedcapital in mitigating equipment.

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    3. N ON A C TI V E EN ER GY OS C I LLA TI ON S :P H Y S I C A L N A T U R EIn practical power networks the four modes of no nw

    tive energy generation are intertwined. The interactionsbetween the four modes may lead to an enormous num-ber of elementary nonactive powers. For example if anonlinear load is mnnected in parallel with a modulat-ed load, then the subharmonicscaused by the modulatedload will cause oscillation of energy with the harmonicsvoltage produced by the nonlinear bad and vice-versa. Jnorder to reach solutions that allow a pragmatic approachto the nonactive power quantification, it is crucial to ob-tain a better insight into the physical mechanisms thatcontrol the flow of nonactive energy.

    An electromechanical system that will help to fathomsome physical aspects related to the nonactive energiesgeneration is depicted in Fig.7a. Thiisingle-phase circuitis supplied by an ideal alternator that produces the si-nusoidal voltage vc . Th e load consists of a resistance R,a reactance X = ( I / w C )- w L and a separately exciteddc motor. The motor is assumed ideal, i.e i t is losslessand linear, its back EMF e = KIFRM. The motor drivesa frictionless flywheel with the moment of inertia J M .The connection alternator load has the resistance r and anegligible impedance. The load voltage, U = fiVsinwtis chosen as reference for the phasor diagram given inFig.7b.Following are the equations that govern the motor behav-ior:

    v = KIFRM (21)

    where K s the motor constant and T s the electromag-netic torque.Substitution of (22) in (21) gives:

    f iv sinw t = KIF J S i M d =- i M dJM CMwhere CM= J M / ( K I F ) ~s an equivalent capacitance thatcan replace the motor - JM system. If JM is replaced bya torsion bar, then an nductance becomes he equivalentcomponent. A combination of torsional bar and flywheelis equivalent with a pardel L M , CMcircuit.

    For steady-state conditions the motor current is:i M = &IMcoswt; IM VWCM

    The remaining linear loads have an active current:in = JZZnsinwt; I n = V I R

    and the reactive current:ix = f i ~ xo s w t ; IX = V/X

    The total current is:i = &Isin(wt + e); I = & + zX + M >*

    Tbe alternator voltage is:vc = A v C sin(wt+ e)

    V: = v2+ T Z ) ~ 2r1 COS@p = vci = P -Scos(2wt+y)

    P = vGzos 6; 6= e -

    where from Fig.% results:(23)

    The instantaneous power delivered by the alternator is:(24)

    where= 6+2= e+

    tan6 = (~ZsinB)/(V+rIcos8)and S = VcI.

    To understand the flow of instantaneous powers onemust know the expressions of each components power.The instantaneous electrical power produced by the al-ternator equals the mechanical power received:

    The mechanical power has a constant term P,, suppliedby the prime mover and a time variable term caused byvariation in the angular velocityRc. Assuminga two-polealternator, its steady-state angular velocity is:

    f l c = w + A qThe termAwt Q w and it is a periodic fluctuation with aminute peak-ta-peak value. The mean value of Rc s:

    Substitution of (24) in (25)gives:

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    This equation gives:P = P ,

    Jc&mZ, == S~os(2wt 7)dtandIntegration of (26) yields

    (26)

    where Ki is the integration constant equal with the aver-age value of < >= w 2, hence

    = w2 + -sin(%t +7 )wJc= W +AwJ2 W +~ w ( A w ~ )

    Fkom here results

    The kinetic energy stored in JG is:1 1 S5Jcfl; = -Jew:+- in(2wt+7 )2 2w (27)

    As expected, the kinetic energy has two terms: A con-stant one and a fluctuating one. The significance of thesecond term becomes clear if when the energy deliveredby the voltagesourcevc is calculated. Rom (24) is found:

    Comparing (27) and (28) it becomes evident that all theoscillationsof energy, active and nonactive,are supportedby the reservoir of energy stored in JG. It is slso clearthat in t he system sketched in Fig.?a the apparent powerS, delivered by the alternator, has a clear physical mean-ing; S is indeed the amplitude of the oscillationsof power,be it mechanical or electrical.

    The motor angular velocity is:

    and the power delivered by the motor is:

    The instantaneous powers caused by the remaining com-ponents are readily found:

    is the power supplied to the combined reactance X,V2RR=Pn-PRw2wt; PR=-=r12

    is the power flowing through the resistance R andPA = r12- 1 2 c o s ( 2 w t +26)

    is the power flowing through r .The total instantaneous power is:

    P = P A + P R +PX +PW= PR+ r12- PR+ r12cos26)co s 2wt

    + ( QM + Qx + l'sin 2e) sin 2wt= P - cos(2wt + e+ E )

    The apparent power S computed from these equations is:S2= (P R +r12 os 2e)2+ (QM +Q X + r12sin 2e)2

    Since PR= VI ose and QM+Qx = VI in0, results:s2= VI)^ + (r1212+2 r ~ 1 ~OSe

    This expression is identical with the apparent power o btained from9 ( V I C ) ~sing equation (23).This resultconfirms the validity of the model used and summarizedin Fig.8 where the actual energy paths are depicted. Thesolid arrows indicate the unidirectional flowof useful ener-gy. The mechanical energy delivered by the prime moveris converted in electric energy and delivered to r and R.The dashed arrows show the flow of intrinsical oscillatingenergy between the active components r and R and JG.The dotted arrows trace the flow of nonactive energy thatoscillates between the inertive components C, , M andJC.

    The same method can be extended to a nonlinear net-work,Fig.9s. The voltage Ud across the diode d, Fig.Sd,has two components:

    wheresinwt& =-ZV2

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    The analysis of the circuit becomes simpler if the diodeis replaced by the two fictitious voltage sources v& andV H , Fig.9b. The current i has also two distinctive com-ponents:

    i = i1+ i HThe fundamental current il is governed by the equation:

    v =U& +Ri,and the nonfundamental curreat i~ by

    UH = RiH

    The source delivers the active power P = V2/(2R) .Halfgoes to v& and half to R. The active power delivered toz)dl is converted in harmonic active power th at includes adc component 2V2/n2R and the term

    cos(2hwt)v 1=+ c h2 - )2h=x4,..

    Tbe dentity- 1 a 2 1

    helped to obtain (29). Thisvery equation (29) shows thatv supplies:

    p = vi = v ( i l+ i ~ )w&il+ v&iH +Ri2+RiliHThe voltage source U& provides:

    p& = V& i = V& i1+ W&iHThe nonfundamental voltagesoufce UH injects:

    p~ = VHi = VHil+ V H ~ HPower entering resistor R is:

    p n = R(il + H)2= fi: +R i i +2 R i l i ~= RG + H ~ H 2 v ~ i 1 .There are three intrinsical components carrying the BG-

    tive energy: Instantaneous active power supplied by thesource U:

    vz2RpA=%2+v&il = - ( 1 - m 2 d )

    power entering U&:V*4Rv&il = -(I -cos&)

    and power delivered by UH :6:+014

    RM = VHiH =

    (29)-

    from UH, .e the diode d that acts like a frequency con-verter, it receives power at fundamental frequency andconverts it in dc an even harmonics. The flow paths forthe active energies is shown in Fig.9e.

    The nonactiveoscillations of energy are shown in Fig.9f.They take place between the diode and th e source U , ow-ever, part of it involves resistance R. Th e nonactive poweroscillations supported by the source are:

    1sinw t co shwt-[sinwtv2 + h2-k2.4. .

    Just like in the previous example all the oscillations ofenergy are sustained by the kinetic energy stored in J c .

    iTR

    The studied sys tem has the active power:P = V2/2R=VI12

    the apparent power:

    s= Jz- = /-and th e power factor is:

    PF = P / S = l / f iThe studied circuit has no inertive components, the loadis purely resistive, however, the insertion of the diode, afrequency converter, i.e. a harmonic generator, causes

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    oscillations of nonactive energy and as a result, PFc1.A last example to be analyzed is the threephase sys-

    tem shown in Fig.10. The alternator supplies two loads:First load is a purely inductive and balanced unit supplied by a reactive component ZL. The second load ispurely resistive but unbalanced just like the load used todemonstrate the flow of unbalanced instantaneous power.The line currents are:

    i, = sin w t - J z ~ ~ w s w t$6 = fizsin(wt - 1200)- i ~ L c o s ( u t 1200)ic = &Z sin(wt+ 1200)- i ~ Lc os (w t +1200)

    causing the instantaneous power:P = P o + % ) b + P c '

    PVZ(1- cos 2wt) - 2VZL sin 2wt+VZ[1- cos(2wt+ 120)]- ~ V Z Lin(2ut+ 120')+VI[1- cos(%t - ZO')] - 2 v Z ~in(2wt- 20')

    As it was shown in the previous section when P=O,p = 3VZ. If P#O results:p = (2+P )VZ- (1 -p)Vzcw2w t

    In both cases ZL has no impact on the instantaneouspower expression. The feeder losses, however axe are af-fected by ZL. The three inductances L, ause reactivepower flow, however, the oscillations of energy that takeplace between each three- phase source and the co rmsponding L, re 120' out of phase. This is causing non-active energy to flow from the a lternator to the inductiveload on one phase and to return via the other two phas-es. Moreover, the sum of the three instantaneous non=tive powers is nil as long as the load is balanced. Thisnonactive power, caused by displacement angle can notcause mechanical oscillations. The same conclusions aretrue for the oscillations associated with the flow of a0tive power. The instantaneous power oscillation due top u = (p - 1)VZcos2wtwill be sustained by the kineticenergy stored in Jc . The flow of the instantaneous ener-gies is summarized in Fig.lOb to 10e.

    It issignificant for this study to realize that if a Stein-metz compensator isconnected n parallel with the unbal-anced load, Fig.lOf, the unbalanced power pu is cancelledby the nonactive powercaused by the inertive componentsCs, s, N nd LN.This shows again that there are nodifferences between the physical natures of WQ and wu.

    4. CONCLUSIONSThis study proves that all forms of nonactive po-

    wers stem from energy manifestations that have a com-mon mark: Energy oscillations between different sources,sources and loads or loads and loads. The net energytransfer linked with all the nonactive powers is nil.Due to the unique significance of the fundamental powers,SIPIand Q1, that comes from the basic fact that electricenergy is a product expected to be generated, deliveredand bought in the form of a 60 or 50 Hz electromagneticfield, it is useful to separate[6,7,8] the apparent power Sin fundamental SI and the nonfundamentalSN apparentpowers:

    The term SN lumps all the nonactive powers, it con-tains also a minute amount of harmonic active power,PH.The harmonic active power rarely exceeds 0.005P1, andis most often wasted in motors and other equipment ad-ditional losses, hence it is a polluting component too. Ina first approximation an industrial nonlinear load[8] canbe evaluated from the measurements of PI, QI nd SN.The further subdivision of SN n other components pro-vides information on the required dynamic compensatoror static filter capacity and level of current and voltagedistortion.

    s2= s: +s;

    5. REFERENCESP.S.Filipski, ,"Apparent Power -A MisleadingQuan-tity in the Non-Sinusoidal Power Theory", Euro-pean 'bansactions on Electrical Power Engineering,ETEP V01.3,~No.1, Jan/Febr 1993 pp 21-26P.S. Filipski, "Polyphase Apparent Power and Pow-er Factor under Distorted Waveform Conditions",IEEE 'Itans. on Power Delivery, vo1.6, 1991, ppP.S. Filipski, R. Arseneau, "Definition and Mea-surement of Apparent Power Under Distorted Wave-form Conditions", IEEE Tutorial Course 90 EH0327-PWR, pp 37-42 and in the Proceedings of the3rd Intnl. Conf. on Harmonics in Power Systems,ICHPS 111,Sept. 1988, Nashville, Indiana, pp 127-31O.B. Shallenberger, "The Energy of AlternatingCurrents", The Electrical World, March 3, 1888,p.114W. tanley, "Phenomena of Retardation in the In-duction Coil", AIEE, Vol.V, No.4, Jan. 1888, p.97-108

    1161-65

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    [SI A.E. Emanuel, "on the Assessment of HarmonicPollution", IEEE 'Ifansactions on Power Delivery,V01.10, No.3, July 1995, pp.169898[?'IEEE Working Group on Nonsinusoidal Situations,"Practical Definitions for Power Systems with Non-sinusoidal Waveforms and Unbalanced Loads: A-IsovDiscussion",IEEE aansac tions on Power Delivery,Vol.11, No.1, Jan. 1996, pp.79-101

    [8] J.H.C. Pretorius, "Modelling of Distorted Electri-calPowerand its PracticalCompensation in Indus-trial Plant", D.Ing. Thesis,Rand Afrikaans Uni-versity, December 1997. (J.D van W y k and P.H.Swart supervisors)

    D v ( i ) v m

    iI

    (b)-+L (4 E V(3) so00. I

    Fig.1 Feeder and Load; (a) One-Line Diagram;(b )Equiden t Single-phase Circuit.

    iii

    Fig.3 Modulation; (a) Voltage and Current Waveforms;(b) Instantaneous Power; (c) Energies.

    , P

    c V I

    C V I

    Fig.2 SigIe-Phase Sinusoidal Conditions;(a) Voltage and Cumnt Waveforms;(b) Instantaneous Power; (c ) Energies.

    Fig.4 Threephase System with Unity Power factor.Instantaneous Power. VI = 100W.

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    200

    0

    P

    Fig.S Unbalanced Thre-Phase System.VI = 100W, = 0.25;(a) Instantaneous Powers: p o , pb, pc and the Total(b) A ctive Power P, otal Instantaneous Power p andInstantaneous Nonactive Power p u .

    ~ n t h x h " ower p = a + b +Pc;

    W

    Fig.6 Instantaneous Power Caused by a HarmonicCurrent and F'undamental Voltage;(a) Voltage and Current Waveforms;(b) I nstantaneo us Power p ; (c ) Energy Waveform.

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    Fig.7 Linear S ystem with ElectromechanicalLoads;(a) Systems Schematic; (b) Phasor Diagram.

    Fig.8 Energy Flow in a Linear ElectromechanicalSystem.

    Fig.9 Energy Flow in a Nonlinear Circuit;(a) Schem atic; (b) Equivalent Circuit: (c) Waveforms;(d) Diode Voltage; (e) Flow of Ac tive Energy(Unidirectional)( f ) Flow of N onactive Energy.

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    a

    Fa.10 Unbalanced Three-Phaae, our-WireSystem;(a) Circuit Diagram; (b) Active Energy Flow;(c ) Active Energy Oscillations;(d) Nonactive Energy Oxillations due to t;(e) Nonactive Energy Oscillation s due to Unbalance;(f) Steinmetz Compensator.

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