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ECA Chapter 3

Bm3352 sinusoidal state analysis

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83 views31 pages

ECA Chapter 3

Bm3352 sinusoidal state analysis

Uploaded by

Vijay Prakash
Copyright
© © All Rights Reserved
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chapter - a) Sinusoidal Steady-State Analysis {1 Introduction lnpevious chapter we studied about the circuits in which the current remained constant. In this {dager we willtudy about the circuits in which the current vary ina repetitive manner. Such types efeicitsare known as altemating current circuits Alternating current is abbreviated as ac hence ‘cut with alternating cu wn as ac circuit. An altemating current is one that periodically reverses its direction, Some examples for aumating current is shown in Fig.4.1. Note that here the instantaneous value is plotted versus time, indall these plots are known as waveforms, Wot, wy a PV\ > e « o) 4123 thee) 9 hae « ® Fig. (a,b) ac waveforms (pulsating de the instantaneous values of current changes between two values by reversing sec. Therefore the current is ac. Sit the waveform shown in Fig.4.1(b)is ince italso reversing the polarity periodically. But in Fig.4.1(c) the magnitude ‘These types of waveforms are often referred nother example for pulsating de. OO Yr Siraoidal Steady-State Analysis 4.3 If is flux linked with the coil at any angle @, $= NBA cos 8, Let 9= or, then, induced emf (y= 4 ()-=t=-4 oncos a) = NBA osin at when t=, th pln te cis pro te edhe e=V,sin ot 44 Electric Circuit Analysis ‘Sinusoidal Steady-State Analysis 4.5 when a= 24 jortant terms and definitions =¥, sin (28) = Vor vs aprere . c=? jm: The graph or lot o tained by plotting the |. ay renee rrent/voltage against a base of : 4.22 Pacey hate ot canbe represented LY | fit ‘Mathematically the sinusoidal waveform 7 i= [, sin a nn quency, which sequal othe “The“o termin 5 (4.2)refrsto angular ordi fet mad of forever insressed by =25 yy, rate in radians pr second, The waveform repeat “ti increased by T, we get ise. This va of 7-2 for which the waveform repeat itself is known as time perio Ss - — ™— ‘The reciprocal ofthe time period is known as frequency defined by ; Fig4.5(a) Fie) requency 1 fee ofS ‘The number of cycles that a waveform complete in one cycle is known as frequency. It is measured in Hz. ‘This quantity corresponds o the number of oscillations per second and i eras relation between the frequency and radian frequency is given by a a Suppose ifthe frequency ofthe oscillation is #50 Hz, then the period of the wave! 1 Taip 7-20 mec, andthe radian frequency =25/'=100 rad/see ‘The lifference between hasavaloe tt 0 oho ete wens sown in Fig isthe send ‘waveform can be represented ag. W2"°f0Rm is having a zero value, Note thatthe Fig Peakvatue: The maximum value positive or negative, attained by a waveform is called peak value. (AR- Bla)sinx +(LAer+ RB)cosex =P cosa ‘Comparing LHS and RHS we get ol’, RY, 0 Bae, Ase ™ RoE The forced response =e si Rhg (= prin EO 4.4 Phasor Representation ‘A phasors a mathematical representation ofan ac quantity in polar form. T ‘or current in polar form, first we write peak value of the voltage or ‘angle Gsssocited with that voltage or current (some authors use ‘examples of representing sinusoids in phasor form. Sinusoidal form Phasor notation (0) =17 sin(300e +30") 17230°7 M()=15 sino 15ZOv 4) =0.01sin (1071 90%) 0.012-90° ¥ N repre ioe hat the frequency of sinusoidal waveform doe not appear in its phasor ‘When analysing ac circuits — rculs using phasors, we assume that all voltages and currents have D®™% Sinusoidal Steady Stee Analysis 4.13 ye: 44 wae the following voltages with phasor : v0) =100c08(at +659) ¥, +80") y, in(cox -80°) solion @ v0) = 100c0s (or +65°) (= 100sin or +65° + 99°) =100sin(at +1559) @ ¥,(1) =500sin(or +80°) V,, = 500; = 80° =V, =500280° w ¥4(0)=25sin(wt -80°) ¥,=25, 0=-9P =, =252-80° Example: 4.5 Represent the following voltages with phasor 0 v(0 =-Ssin (we +10") @ v(t) =—S0c0s (aw -70°) Solution 0 y(t) =-Ssin(ov +10°) 4.14 Blectie Circuit Analysis. = Ssin(or +190°) or 52-170" @ =; =502200°F Represented the following voltages with phasors ¥0 = 100 sin(or + 20°) ¥40 = 10 cos(ar - 30°) ANS: V, = 10260° ¥(0 =-25 sin(oe + 160°) ANS: V,=252-20° Fa aun an 4.5. The Complex Forcing Function Letus consider the sinusoidal volage ¥()= 7, cos(o1 +6)... In phasor form the and vector can be expressed as 1LO = 1 oe tn the same way we can represent the phasor VV LO RO orn Muitply V with &™ yields Ve" avd ay fen ANS: ¥, = 100220° v4 otto. Fig 4.20 ‘Sinusoidal Steady-Stare Analysis 4.15 rom Buler’s theorem we have Re[Fe™ | = 1, costa + 0) wrnnnnne(424) where ¥, =I and ssnns(428) Vis phasor representati Lets consider two sinusoidal signals wih same fequncy given by Yem{or +4. a 1 ,cos(at + 8,). The addition of to sinusoids sequal 17, 008(0" +8) + pq cos(r +8). snu(h26) In phasor form Ki = M20, and ¥, = V0, Wehave ¥,0s(0r +4) +¥,,co(or +4) = Reh") +Rel Ye]... (427) ste ADB) (4.29) 0) ry 208( 04 +8) + Vy 605(0H + 8) = 7, 608(00 +) coesonrnnnnnen(432) with V2 Way + VN O= LIAR, ne (432) From Eq. (431) we find that the sum of two sinusoids with same frequency equals another Simsoid at that frequeney. Hiweapplya signal f(t) = 4,cosax + 4, sinay to acireuit then te forced response in any branch sequal to Y(t) = Bost + B, sina... ‘sing the relation sinat = cos(or - 90°) the Eq. (4.33) can be expressed using a single sinusoid. (4.33) sn (A34) ‘Simsoidal SteadyState Analysis 4.17 4.16: Electric Crewt Analysis 4.6 Complex Algebra - “Phas aed ‘Acomplex number ‘has j wel 5 x_n “nemesis GAA) traction of complex numbers must be jon and su a st be performed in rectangular form. To add ex numbers add real part and imaginary part separately. Similarly subtraction can be oF py subtracting the real and imaginary parts separately 46.1. Polar form c=Mze Here Mis magnitude ofthe complex number and isthe an “0-2 ‘between the line joining the origin andthe point and positive real —_ tagiude of beconplesnanber'C Fo clledasabsoute vate, (5-4)+ j6+2) ‘and is designated by a Fora complex number 2= 1+ jy the polar form representati eran ‘Temulipication can be carried out in both polar torm and rectangular form. Let us consider two Z=MLB crenne plex numbers where Me FF on and Paes = 5h th) tan 2 ———— vel =m -9n)+ (-#=-1) aThecompex umber +jy ao ca bepreseted in rignometic frm singe Re(=.2,) =~ lm a w presen polar form the multiplication canbe carried out 6059+ 00 os ee ay Pomel hr oes es Ifwe denote x=reos@ andy =rs It on db xt yyere a1 then Wenne " 219 LO, #8, nnn AS) rPay Noe thatthe magnitude of Z, and Z, are multiplied and phase 6, and 6 are added. 4.18 Blectrie Cirewit Anais Example: 46 (Convert the following complex numbers to polar form @3+A Gi)-4+,A — Giii)3-4 (iv)- 4-3. Solution: la sexty=34p. The polar form |Z) 20 2j-VF oF 424 “Here the x and yare postive then the poins are lie inthe frst quadrant. Thats 0 < 09 i +3" = 1649 = 25 <5 ‘Here the x is negative and. ‘Positive then the points lies is the second quadrant. TH is90 <0 180" Inthis case 6 =tan (4) 1 Second quadrant Simoidal SteadyState Anais 4.19 Heres positive and yis negative then the (+ pont es inthe fourth quadrant. <0<0° Inthis case 8 =216.87° (or) Polar form is given by SLNG87T (or) $2 -143.13° ‘Sirasoidal Steady-State Analy 4.20. Blectrie Circuit Analysis teady-State Analysis 4.21 ttt (4.50) n Ea (4.4) we et Example: 47 Bee Conver owing rom ino e ve © sz | TAS £2499 rn B9(635) we get 5724999 25 se required rectangular form is given by * xtiyeS4s (40), pyslon of complex number Letus consider two numbers 2, =3, 4/4: Now Ath (4.51) | Ath 5 | Tedivision of complex number canbe performed complex number inthe OsTIx=y. sm simnsnsrnnsnnsscnny (gy, itinaor. Already we know that when a complex number by its complex conjugate ~~ " theresutant is a real number. Therefore multiply and divide the numerator and denominator by ‘Substitute Eq. (4.48) in Eq. (4.47) we get, complex conjugate of the denominator. 2+ (057727=25 2433 hat tJ — HA) at att J mH ath at | In polar form Lets consider two complex numbers in polar form wai (42) Feamp tan 45°22 48 x Mutiply the following numbers. PF FPY natn ao) 4230,5220° 4.22. Blectric Cieuit Analysis 4.230°,5220° = 5x 430° + 20° = 20250° a 129083240" 1290%3.240° =1% 390" + 40° =32130° Example: 4.9 Example: 4.10 100230" + 1002 -99° ‘Convert the above polar form into rectangular form ) 10434 2-72 1o.z20° 3290° Convert the rectangular form into polar form Soe oo PA SAR ci eser-ic0 ‘Convert the reetangular form into polar form 10s ja 10.77221.80"_ 2-f2 282-4 10+ 3 1044Z16.69° _ 10.44 16 690 433.698 Sod 7 36052—33 6 3.605 =2.896250.38" ‘The numerator and denominator are in polar form. hence 10220" 10199 gor 22-70" $290" 7g rr 424 Blectrie Cirent Analysis ‘Sirsoidat Stead)-State Analysis 4.25 4.7, Complex Plane | Foor ockwise direction by 90°, wy, The eter jin complex number =1-yisequalto VI. In Vector —s| 49 Phasor Relationship for R, Land C smathematies, xs known as real stand yis known Sher operaiod — consi . eetlamementandyistnown Heron 7 Ins secon we obtain simple relationship betwee te vage andcurent phasors ores, ‘are also known as in phase (active) and quadrature (reactive) Vorert| inductors and capacitors in term of impedance. components respectively. (=) ftavcw voor Penick n dager eee Pcemercochva does ie vector Vis dvb operate, YV =, serene rae Oe te he compe care int fo jena etre itn comarclokwioe decker gh Fig23 eee me ye cones S00" (0) = Lease + in When hops aplic counterclockwise from the reference ana, nn Y BeFetisV=-/¥, The vor? ~ Vicosan + j¥, sina = R[I, cova + jt;sinet] SC. resistor then the caret Wea (y= eons) = 2oos(100+ 20°) 492 Inductor Incase ofan inductor the voltage v() ae VN Peper tear caret oportnalty tH Forexample if v(t) = 100s(100« + 20°) is applied toa Sinusoidal Steady-State Analysis 4.27 v + ab oh ee be piedomsin (frequen domain |, ear Voltage and current relation fran at inductor Fig29 Fig 4.28 rom Eq. (4.60) V,26, = @L 1,20 + 90° 20, = 20,+9°. 20, = £0,- 9" From the above equation we can find thatthe inductor current lags the voltage by 90°. 493 Capacitor + ne = arcapacitor the voltage current relationship is given by : =c® acy TOY TE, Fo CF men (a6) “ST” Tie o o 1d ody fee ree = jo cv") sen ( 465) = jocy,e* 1,20, = jo CV,20, i VLB, = FE LO nnn 66) Inphasor form ve G61, From Eq. (4.67) we cat voltage is 1 From Ba, (4.66) we have "Pinal phasor current by fistor az 7 vr 428 Electric Circuit Anais o£ 0? vr co reget My i the voltage by 90°. ‘which says that the current leads ‘5 | ‘ocorcrand vgs enti tine ema an feteney domain ay Fig431, 4.10 Impedance In previous scion we got the follwing relatos between phasor volIAge and phasor apy resistor, inductor and capacitor Forresistoy | = Rl Forinductor | W= jl ws ' 1 Foroapactor Y= Ted ‘where Z is called impedance which impedance conveyste ‘notion that 2 “impedes” phasor current /, just as resistance R resists instantaneous current i To distinguish the impedance of resistor, capacitor and inductor we add subseriptstoZ. Tha we have the following equations, GJ Z=R+iX cere ‘Sinusoidal SteadyState Analysis 429 IY carn ere ismagnite 2 is angle of, na. (4.81) the impedance is given in polar form in rectangular form, lis generally denoted phere R = Re(Z)is known a resistance and “xe In(z) is known as reactance ‘The magnitude of Z = ARH conn (4.83) and phase angle aX 8, = AN cnn BA) Forresistance the impedance is pure resistive and reactance is zero, whereas for capacitor sakindvtor the impedances are purely reactive and resistance is zero (485) semen 86) Fig A32 Impedance dagram ()RLCireait O)RCCreuit 4.30, Electric Circuit Amalysis ow \) yh Current if ¥ = 20c0s(1000K + 30%) is, is applied for the element (i) erates =F 4.12 Admittance . The reciprocal of impedance is represented BY ee 30°); @ = 1000 rad /see ] : ‘Since 2s a complex number, Yisalsoa complex number. Hence Yin rectangular fama, be represented by ‘here G is known as conductan . i) 2° SE Fooosionie® a? i = 0.2cos(10004 + 120°) Example: 4.13 Rex . a ‘The impedance and adnitane of crit elements are given in table be 706 ” Element Impedance Admittance | Resistance a 1g 35/4. (0.12- j0.19) R Inductance Jol ale Jal Comite | ot "poe ene eeec Tec | vr 432. Blecirie Cirouit Analysis Simsoldal Steady-State Analysis 4.33 pLSerles clroult : 4 ii Rk Mg === Example: 4.14 it inductance Lhas the same a att sit shown in v ns with i imit RL a Ate fequeney «1000 reser il of and L? “ey, cot ge tnd i 200 resistor in series wth a2 indo om The = oe in jeformand the voltage V,Fig4.34 RL seriescireait, Fig. 435 Phasor diagram of problem este etance leads the current ‘aeRL sere ica — \ b= Ba) | ptt i lant of Vand V,, therefore leads the current by 0 which is less wie a oats RL circuit is shown in Fig 435. ae Cis wt Ppa ettal impedance of i. . Fig433 e Z=2,+ . = Re je 7 ' ZaR+y loan “The magnitude of impedance | a * - fee eee _20= jo wtocrinmaeerZi9 cae 2 Oa tan b= tan" Ss Te “a 7 “ "Fos sa 0+ ta therefore Z=(|20=VR ea ECan, auaing rl pas we ‘The impedance diagram of RL circuit is showmin Fig. 36. - et I?the applied voltage v = V,co8(or + 6), then the current | : p= Maseslot 8) = 4a" = 1600 = @ = 20rad see aoe = 1,,008(04 +8) wn aoe i wok oo ose? 2000 ) (4.103) =L-St™ _ 200 _ J ¢ easy 413 Phasor Diagram Reample: 415 Sern inoue roi ton st see | ey, iL, RC and RLC series coennn ¥Y about relation between current and voltage ou = 150c0s($00) Acminscicitconitof R = 100 and = 20nd! andepliod vole» = s0=n(500), tthe curent the voltage across esistanc, he volage across indutance nd the Ps em wit respect to applied voltage 7 T is 434 Blectrie Crit Anais Sinusoidal Steady-Sate Analysis 4.35 : le se impedance diagram for RC ics shown in Fig. 4, rive pplied voltage is ¥ = V,cosor V,,c0s(ot + 6) then the curent ea sre i (ED) oc = 1,c0s(or +8) is-45°, where Ie si 1. (4.112) 4132 RC Series circuit fe Fe 1 O= tan no (4.113) Trample: 4.16 Consider an RC series iret with R = 100 nd C= 37 Te applied vole isgiven ty = 50c05(10000%) . Calculate the current V,, Vand the phase angle of current with respect __ api volage. ing the caren yan egh 8 esr than 0 Te pr diagram of RC circuits shown in Fig. 439. an) Slaton: ze Te applied voltage in polar formis 500° R= 100 y he ' ce y ‘The magnitude ofthe impedance is ae Me ‘and the angle of impedance is ott 4.36, Electric Cent Analyst Simsoidal Steady-State Analysis 437 $020°___ 4,472226.56° edances in Series 1 Tr182 = 2656 ie 26,56]10 = 44.72.226.56° or acircuit in which the impedances ZZ, ‘connected in series. Let the phasor volage urrent are V and I respectively. Since KYL -yoltage phasor we write VKH tint Vy is for the 4.13.3 RLC Series Circuit 0 Consider a series RLC circuit shown in Fig.4.42. The voltage V, is inphase with the current, the voltage V, leads the current 90° and the voltage V. lags the current by 90°, The total impedance in the circuit is vie “Belge Ly AZAD AD tot By ce AM) ‘The voltage across impedance Z, is V, = IZ = Rs j(ot-2e x | 7 = R+i(K,-X%, Whe, From Eq, (4.115) we can find thatthe reactance ispositiveif X, > X, and negativeif X, < X_.If X, > X, the circuit behaves like an RL series circuit ‘and current-lags the applied volige by an angle 9. If X, < X¢ thecireuit behaves like an RC series circuit and current leas the applied voltage by angle g. The ‘Phasor diagram for both cases are shown in Fig 4.43, ‘The voltage across series impedances divi + 415 Impedances in Parallel and phase angle is tan“! x —%- R 1 i o- seen ean a pst Ss J Hayy >> | The impedance diagram for zl shown in Fig 4.44 forboth cases are aK my | ® Figs . ofz| BO fr 4.38. letric Crit Analysis ‘Sinusoidal Steady-State Analysis 4.39 BK Z, = jol = j(2)() = jan ith he above values of impedances and phasors redraw the circuitas shown in Fig. 4.32 and. © ex aply Kitch: as oF network theorem to sole for desied branch cues and vag poample: 417 ‘Aseries RL circuit has L = 0.014 and an impedance of 11.180. With a sinusoidal voltage applied, the current lags the voltage by 26.565°. Calculate ‘a and R. Solution: Given 4.16 AC Circuit Analysis Gives Inthis section we study about the method used to solve R= 200) < om ‘current, voltage and power in any AC circuit. Ifthe a Z=VR +L = 118= \(0.020) +(0.01) 0° @ = $00 rad /sec- Hrample: 4.18 ‘A pure inductance of L = 0.1#1 passes a current i= 10cos1000r amperes. What is the voltage ‘cross the element 7 Solution: Given T=1020° © = 2 rad /see The impedances in the circuit can be @ = 1000rad /see “alulated as 440. Bc Cre Anat Example: 4.19 ee wpe in Ther Simscide Stady-Sate Anais 441 Aresistor 7 ed voltage % = 1000x000 +30") os. te ee ~ plo ite capacitance? Solution: Given R=50 =e +37) i ‘The current leads the voltage by 50° X= Se x . ie rc aprast >> oss ol x,=Leans oR = ns? Given eer C= 16x10°F our roo ‘Te impedance diagram and phasor diagram is shown in Fig.49. Example: 4.20 alee ements currents ‘Apurecapacitance C= lQuF passes current = 10sin2000tamperes. Find the voli, a area Cenaceae = ‘the element. w= i(0) = 100s (10001 ~ 140°) Seti Sedation: T= 1020"; @ = 2000 rad /sec Goee v(t) = 100sin(10007 + 150°) = = 500 it (10001 - 140") = 10sin( 1000 - 140° + 90°) = 10sin(1000 - So") InPolar form vy, 0°(502 ~ 90°) = 5002 - 90° ee mage ¥(0) = 500sin(2000" - 90°) volts el pieann 2-4-1020 ‘revit elements ifthe applied voltage and curents are v(t) = SOsin (zon +6" = (939+ 3.42)0 in (20001 +959 i iagan. Find and draw the impedance diagram, phasor 442. lecrie Cre Anayt . Simesotdl SteadyState Analysis 443 ample: 4.23 uency will th Aeris cit has R= 1000 an =H: At eeney eet voltage by 30°? Solution: Fora series RC circuit “(4 vow) WR, ion: 20uF and ¢ = 30° : Given. R = 100; C = 2041 find +sina ~ an{ 458) of 48) “1 1 ) Zi Z veo Seat 1 in(ct + 90°) + 100sin(or + 30°) © Go0x 203107 x1an30") = 100[2sin(ar + 60°)cos30] Pe = 173.2sin(ct + 60°) 13.2 f= 13783 He v,, = 132 = imary, i as (voltmeter reading) Exam ole: 4.24 ‘An RL series circuit has L=21mH. Ata frequency of SOHz current lags the voltage yyy, Find, ‘wo impedances Z, and Z, are connected in series with syoluge source V= 50.4, Find the voltage across each soi Solan: inpdance and draw the voltage phasor. one {Jas a= a"() 45° Sition: Figaso For an RL series circuit ; 350 ) a 7s = 36162—17.19 Figs = 36.162 -17.1°F 53.139 = 1838236057 Euample: 4.27 Deen the circuit of Fig.4.52 there is a voltage drop across j5 reactance of 16,67.2—45°¥. ine 2, 444 Blectric Circuit Anais Solution: Given 1x, =16612-457 teste anced 7 5 04S" «520° 2 3Rae 202, =182,=00+/90 2, =18- (10+ j8) = 6-J92 Example: 4.27 If @ = S00 rad/sec and 1, = 254° inthe circuit shown in Fig.4.53. Find 7(9, Solution: Given © = 500rad see Z, = iX, = jol = j(500)(2010") = loa 1, = 25240 Y, = A,X, = 2524 10290" = 252130° Lette current through 100 is 7 Wehave pele Oe 70 21-0820) = 2.795.266.56° Sirol Seeds Anas 445 y= [2.79526656" io 425230" $95 139» =16.7+ 19.15 = 2SAL48.9°Y ) = 25.Asin(5001 + 48.99) one 428 xelt the impedance Z, inthe series cic, mestan " 00 . b souton Vv =10020° - > 1=252~45° Peas 10020° 2757 — aye = OAS = (28.28 + 28.28) Z,=(0-j9a 2, = (28.28 + j2828) - (10- js) = (18.28 + j33.28)0 ample: 4.29 ‘Three impedances Z, = (10+ j10)Gx Z, = j60 and Z, = 80 are connected in series to taunknown voltage source V. Find J and Vif the voltage drop across Z, is 21.08.18.43°Y. Soon: V, = 21.08218.43%2, = 80 Hy UOBABAS «9 635 18.48°A z 3 Z,=2,+2,+2, = (10+ /10) + (-/16)+(8)= 18-60. vem, = [2.35218.43°]18 - 36] = 5020? I= Simsoidal Steady Site Anais 447 445 Blesrc Creu Anat | ga tii nr 8 sho in Example: 430 nd determine the circuit constang, fo el combination of (~ 625) diagram a | apeparal ‘nd 3009 canbe replaced by ‘Constroct the phasor and impedance following voltage and cure. _¢ (625) = 150 sins + SI = si s00"-25°4 Ae = (243.823 Now In pol form Z,, = (600 4800) In polar form = (234.146 +. 23 117) = (477969 + j175.468)0 sample: 4.32 stor, 10H inductor and a capacitor Care connected in parallel. Find the dance at @ = 100 rad see if C = 20uF the magnitude ofthe impedance when C= 1QuF Ifthe magnitude ofthe impedance is 125 at © = 100rad/sec, find C. (@) At what values of o the magnitude of the impedance equal to 1000 if C= 20,F Selon: 12579 mH X, =o ko77650 Exampic: 431 = (100)(10) = 1000 Find 2, if = 800 rad Ise forth shown in Fi Keak ees ie src inFig4.56, © @C 00x 20x10 Solin | 1 1 1 The reactance pall PR jee tsacince X= ae ' | "8 jel “tad “sear h es eet ~ 300 * ooo * Sag FO 00 Tracie ott gp = 0.005 + 2%. non = (8000.6) Fig4s6 =480 ere , 448. lecricChrexitAnabyss C= l0nF oc = 1003x1010" =! Om 0 1, jooot Lt = 300 * 71000 (©) Given [2] = 125 ce? f= hens + j100c - 0001) (¢h) = (ors + (uae non" =C=745uF © &,|=100 wz 1 1 y* =p fgg 1220810 2 (om - (oan +[2nx10*o 2.) ty 1) Wx10%o- 275x107 ( We al = 75x10 22x10" @- d= 366410? OT 86x10 > 2x10%e! 866x107 129 866x107 + fi o 2x 2x10" 86510") — 4x2 10% x (1) Simacidel Steady-State Analysis 449 0.0866, = ee = 0 = 444 rad /sec prample: 433 ‘An impedance of 342+ j4kQ ‘18 Connected jr impedance aka eam a of 62 — j6k2 ‘Solution: Total Impedance = (32+ j4kQ) + (640) J6KQ) = (9k2~ jako) current my V2, _50(3k + ja votage sees = Fi = OE a eases VZ, _ 50(6k - 6k) 22 OI) sooine-marey Example: 4.34 ‘Two impedances Z, = 120230°0 and Z, = 4060°0 connected to an ac supply. Ifthe catent through Z, is 80. 30° mA determine the current through Z, Solution: Given (802 - 30°)120230° LaF = 2402 ~ 60° md 4.50. Electric Cireit Anat Example: 4.35 ; Find Zand Y,, of the two branch pars na 2, = (212) 747, = 9.1 + j6.176 = 115232.485° el cireuitand compute CUTER Ch bag oun roo) z<(6+ 709 3 22, soar) = 0.086952 ~ 32.485° = 0.0733 ~ j0.0467 Simcidal SteadyState Analysis 4.51 X, = wb = $00(001) = 5 ‘the impedance Z, = 50 2, 22,42, where Z, is the impedance of the remaining two elements Z,=(8-J9)-j5=5- 0 R=50 Figo 1 | X. = Gg == C= 200uF puample: 437 athe uit shown in Fig 4.63 find the two branch currents and the total current. construct the phasor diagram, \ r : Solution: t . v 10a me cea eames weg fw fe To rscas “BOL BAIS Z, = L450 nai ZZ___ (10220°)(20245*) ¥ _ 1000 . =o arene Ls ay eae 24 Tez, ~ WLI + 20S 200 265° , = 3334 333 = 4712454 ~arasaar = BSED 1oozase * : = 13321194 ee TTagae = P84 A three clement series circuit contains one inductance Z, B ! ne indu = 0.01H. The applied voltage a! 1, = Y= 100245" _jozasea TIRE cent ae shown nthe phasor diagram, If «= 500 rad/see what are the ott 7" 02 : p= Le = OLAS sa ‘Solution: n= 7 * 20245" From the phasor diagram V =1002 ~ 45° fase Not | Pample: 438 - a 1=14.142- 99° eal voltmeter placed across 50 resistor shown inFig.65 eke sa zt, ig ‘ads 25 volts. What is ammeter reading ? sa TTL — ase Solutions oy smmmcan a vac Be Pr combination of 108 seniors ca be Lt Tom the data We ha Placed by a single resistor 5. joo tt he single resist Feats From the data we have 1432 Electric Circuit Analysis ‘Simutcdal Steady-State Analysis 4.53 ‘we know ammeter reading is 7 Example: 439 Determine curent in 30 resistor. Find the vale of Jif, = 2292 Selation: 4.54 Electric Circuit Analysis ‘Sinusoidal Steady Si 439 RMS Value of Cartent and Vag dna 435 sistance R carrying ao carving a curren it) which nthe circuits (f= pay \? hich is periodic with period 7. Let the 14 nnn A139) 4.140) Ife eet aries ade current /then the power p=RP. nnn (4.41) Comparing Eq (4136) andE9.(4.137) we can sy thatany nage, eee that any periodic cent) produces same Pe fiom which average value of zeroing t= This value of Fis known root mean square value [RMS value: The rms value ofa periodic current (voltage) is equal tothe de curent (voltage) that would deliver the same average power toa resistance. Thatis ¥, Where 7 =“ and I= am 1 ‘FF ie the rms values of the phasors V and J respectively. Tk 2 ™ = ‘Similarly the rms value ofthe voltage is given by ‘alu 600 is known as power factor. 4.18 Average Power ‘The average power P is also defined as .erms value is given by 4 1 p=—_ Toner # Te ll on(doe) which “Soren ie average rat of energy transfer over and, gee 7 vi", eR 4.20 Apparent Power (Complex Power) and Power Factor on(or +2) 1,cos(an +8) p= “idacos(e, -8) = Fal £05(8, ~ 8) = M1e04(,-4) = Moose smote ene tained set eset fowe ctor indo bs againg power fc In capacive ‘then the power factor is said to be leading power factor. : Let us considera tworterminal circuit wit i sider a two-terminal circuit with rms voltage and current phasors V,., 8 Yom = Poul 4, 1 = n|Ze, ‘we define complex power as sey, =W where J," is the complex conju vs Conjugate ofthe rms curent phasor. The real part of Sis" = Re(S) = Re(V1) = Vremwe Simoida Su Sees teady-State Analysis. 4.357 rea by im(S) = In(v7) = ino aoa isknown 25 reactive power wnt complex power Si l80 known as apparent poner Me gverage reactive and SPPAFEN overs as exe ag P=rR Q=rx sary pa ofS 421 Power Tri Copia presentation of complex power ihnown a power sample: 4.42 ‘voltage source V = 2402 - 30*V has three parallel impedances Z, = 25210°Q, 2, =102 ~ 60°02. and Z, = 1590°Q. Find complete power information, Solution: 2402 ~30° abe. =9.62-40° hog sae ~* v 2402-30" 5 oY 2 MOE sacar hee 0c 2402-30 67-120" 13290" 4.58 Electric Cirewit Analysis: s, = 2990 49883 §, = 3840 = 5149 - j7483 Teal el power ‘Total reactive power” Example: 4483 Atwolement series ciruitof R = 502 and X, = 100 hasan effective applica vig, 100 Determine the power triangle. Solution: »/\om Given _ | P=FR=(8.944)' 5 = 4007 ee Q = FX = (8.944)'10 = 80074 lagging S= FZ = (8.944) 11.18 = 894.4VA R power factor pf = Example: 4.44 Ifthe complex power Sin branch 2 is 1SOOVA, wi Find eee ome Sin anc 2s 1500/4, wha willbe the indication on teense Solution: | Impedance in branch 2is = V4 6 = Va complex power =H, = 150074 S= 12, = 1300"4 P rigs P= 1500 _ no = 236 1, = 14952 - 6343° Sil Sastre 459 ‘otal current Power factor = 7 = 0.614 Example: 4.45 ‘The impedance of a circuit is Z = 6+ j80 and an applied phasor volage V = 50245", determine the power triangle, The different approaches to obtain the power triangle are 1 P = FR=(5)'6 = 25(6) = 150" 460. Electric Cirewt Anais 2 seW= Q = Visind = (250)s0 porwr factor = cos 53.13°=0.6 HEB 3. S=V0 = 50245 528. 13° = 250253.13" = 150+ j200 P = Re(S)= 1507 Q = Im(S) = 200var lagging pf =cos®= cos $3.13° = 0.6 lagging iF 7 Pay Fig4st Example: 4.46 | ‘Aseries circuit of R=10ohnis and X.=1Sohms hasan applied phasor voltage V=50 Find the complete power and power triangle Solution: P_7es2aw Vv =502-907 Z =10— ISM = 18.0282 -5631°2 ¥___s0Z-9 ea 1a a Re en BIBL 33.604 S = = (502 -90")(2.773233.69°) = 138.6732 -56.31° Figa = 76923 ~ 115.383 P= 16923 O-115.383var leading pf = 60s $6.31 = 0.5547 leading Example: 4.47 In @ circuit the applied voltage v= 100sin(ox +20°) and a resultin {= Ssin(ot ~ 6%) amperes, determine the power triangle, Solution: 100 v=|— N20, 1 = 5, (3) sek wo V=TON20% 1 = 3.5352 — 69° Sista Sead site Anas * eh ‘the complex power $= yy" = (70.71.2299) 352609 = 250280: ) = Bale P2462 real power P= G4" resctive power O = 246.2varlaggng S=250v4 power factor = c0s 80° = 0.1736 Example: “as Given the circuit of Fig.4.83 determine the, Power tangle - | solution: td From Fig.4.83 pea current pa Fi Fig489 cal SVE = (252-90°)3 5362135" = 884z45° = 625 + j62.5VA crszsvar Thereal power P= 6250 cone Reactive power Q = 88.4var lagging ‘Complex power S = 88.4VA Power factor = cos 45° = 0.707 eee Short Questions and Answers 1 What is the impedance of an inductance? Z, = job =f 2. Whatis the impedance of s capacitor? where

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