Homework9 - Let 3'8 z =[J)= is x1 J=S J8=4 14 V" 8 j8 1...

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Unformatted text preview: Let 3 '8 - _- - z = [J )= is x1 J=S+J8=4+14 V " 8+j8 1+} 1—; 2 Next =12£G°= 2100 4+2? 4+{8+_}'8) Finally [12£O°}{1.3421—26.6°]* S=—=T.2+j3.6 VA ,3 I =1.3424’—26.6° A FILE—4 Repreaeu’r the circuit in The frequencyr domain and label the nude voltages: l 152 .l— The mode equaticua are: Séfl=h+ -={:r : \-'1[1—j]+jT1=lCI 1‘1 — 1: 2 12 TE— 113] +¢=D : V1(a.25+_;')+13(2]=fl 1:2 — ‘1 ‘3 3'2 —_}'2 1 Using MATLAE: then H II 00 | | Id Ch RI ._. ha 53“ RD CI .. ,1} "Kc-w the complex power can be calculated as, l I a o I [—[—)V1] (26572—1269 ) [—flzaéy 3 3 3 5: —-=— = c.3339: -— VA 2 2 J 39 Finally 3 3 S=P+jQ=jE 2: 23:0. 9:51-1:53 FILE—5 125.1 j2Dfl v = 50,5120” v rma o ’J 0 ° =w=fl=2aaaa1a°a 16+j12 lflfifififi?” I s=n" = (saauaoxasa—aatafl): 125 asaatfl=1aa+fi5 VA FILE—11 Represent the circuit in the £requeney denlain a3 I H armrcc mud Deublntg the anglea cf vilit) increa sea the angle cf vilit) by 7’5“ {c} increases the angle 0f the load current by TV. {a} dcea not change the impedance. {b} doe-s net change the complex power. FILE—12 Rapreaau’r tlla circuit in The iraquency damain in 1240': U SCI :1: 2 6.61 1.93 _ I[m)=[%_i]] =[1.lfl+jfl.33)3 =(l.1D—jfl.33)=1.159'~”-'5T A Using KVL give-3. ' I 1 139m '15? _ RHiL—C =W=10-4¢'=1U+J3§1 2 , _g ' 5“ 3:10;! 1 Alld 3——=3 :5 _=5 :1: P11.5—13 Acoscor‘u' L Analysis using Mathcad [cxl l_5_3.111cd): Enter the parameters of the voltage source: A ;= 13 m ;= 3 Enter the Average and Reactive Power delivered to the RL circuit: p ;= 3 Q ;= .5 The complex power delivered to the RL circuit is: S := P + j -Q The impedance seen by the voltage source is: z ;= i 2-3 . 1111(2) Calculate the required values of Ft and L R := Re(Z] L := R = 5.16 L = 2.15 (II) The mesh current is: I := g The complex power delivered by the source is: Sv := SV 2 3 + 6i . . . _ i- I-R} The complex power delivered to the resistor Is. Sr := — Sr 2 8 2 The complex power delivered to the inductor is: 31 := St = 5i VerifvSv=Sr+Sl: Sr+31=8+6i Sv=8+6i P1 1 .6—7 (Using all 11115 Tallies} (a) W = 220[?.6] =16”? VA P 131? = — = =.T88 pf VI [GTE a=cos'1pr = 30.0“ :5 Q = VIsii13=1036 VAR (b) To reatore tlie pf to 1.0. a capacitor is required to eliminate Q by introducing —Q. then FE 2WD 2 1030=—= { ‘3 :> XC=4TQ X X f‘ C = = 56.5;11: (0X = (3773(4?) (c) P = VI 0055' where 49 = 0° then 1317' = 2201 I = 6.0A for corrected pf 3Note I = 16A for nncorrectedpf. P11.6—8 First load: 31 = F +19 = P(1+ jtan(coa_1(.6)))= 500(1+ j tan 511°] = 500+ 300? kVA Second load: 51 = 400 + jfiOO EVA Total: 5 = 51+S2 =900+j126? kVA Edam =P+jP tan (cos—1{.90))=900+j436 VA 3 6—. Q :3de F}: c, From the "rector diagram: 5 mm = S + jQ . Therefore “5—. 5 dcait‘ni. . _ P I ‘too 900+ f436 = 900+ 512?? + Q => Q = —841VAR 1: ' i V 3 3 ' I * =—_}'S4l i=2 = =fl=j11392>2=—j1139=— “I Z —j341 —j341 STTC Finally. C l = 2.21] ,uF = (1189)(3T7) P11.6—9 (a) s = P + jQ = P+ th011(eo<1—1pf)=1000 + 3'10001011(005_10.3)=1000 + 1750 VA S 1000+'?50 . Let V1210010° 1'711115.Tl1e11I =_—=—J=10+_]T.5 : I=lfl—_}'T.5 A ‘ L 100z0° 7 D ‘ =& = 8136.90 = 0.4+;‘40 V ZL =_L _ D I 11311—369 TL =[0.4 + 100011.014) + 111(1) = (12.0 + 10.0)(10 — 3'15) = 200.500 1: b I: ( ) For111ax111111111pm1=ertransfer. we require [6.4+ j4.8)| = Z1 | 22m = 1 . YL+YMW l 1 l . —_=YL +11”!w :Ym =—_——_=;0.15 5 (15.4—14.3) 0.4—;40 0.4+;40 Then Em = —j6.67’ f2 50 we need 0 capaei‘ror given by 1 _ 1 _ _ = 6.61 :> C = — = 1'50 11F mC (6.6T)(200) P11.IS—10 (48$0”)(1.0T5£—33.3°]* _ , (a) S =f = Eiglifigfifl = 20.2T+_; 16 VA 25 2 6945+ '13.39 2 lififiifififi“ (13} VI = *1: (—Jt) =[q—)= ESETJESJ" V I [1.0105— 303°) [1.010530%] ‘r’ 2001;251D Zl =—1= —= 26.33£53.4°=12+jl4 =12 +j6(4] Q I l.07’6£—33.3° R1=12§1011dL1=4H (130 = en's—1(056): 50°. |s| =i — fl= 3.10:1 100 pf _ 0.50 , 2(3253g56“) ¥ 1.3 =—*=15.3751?.7° 1- (lflfifi—fifij") Y3 15.372117” =—=14.23£56°=3+j11.33=3 +_;'6(1.97) Q 23 =— 1 1.D?6£—33.3° R3: SflaildL3=2H. PllJ—l Use superposition since we have Two different fi‘equency sources. Pint consider the dc SOUI'CE ((0:0): 12 11:14 ]=12 A 12+2 133:112 R=(12)3(2)=233 w 12 fl Next. consider The ac source (to = 20 rad-"5): —_}'60 _- 3-: 12: 9.167 f5} flizlléfim J _ +2+j4 5 (12—;5) Then I: (2)=—“3?m=125 w :— P3: 2 Now using power superposition P=P§+P§ =ESS+125=413 TN PllJ—l Use superposition since we have two different frequency sources. First consider co = 2000 rad-"s SOILI‘CE: Current division yields 8 I I 11:5 :2 =% 453.5991 Wit" 39 —+3 5 Then |11|43 Pi _ j = 20 w 5 —£—171.9° A fl Then Now using power superposition P = 1:; + P; = 22 W P11.8—4 Zt=4DD+j80CI 11 and EL =2000 || —j1000=400—j80{) .o . 3 . . . . Since EL = E.t the average power dehvered to the load 1s 111axn1111111 and cannot be increased by adjusting The value of the capacitance. The voltage across The 2000 E! resistor is = 2.5 —;5 = 5.5955“ 15 So s s '3 P: 1 =7-'.Eén1"lfi.r 2000 is The average power delivered to The 2000 Q resistor. PILS—S Notice That Zr.11ot ZL. is being adjusted .W'hen 2.1 is fixed. Then The average power delivered to The load is maximized by choosing ZL = Zfl‘. In contrast. when ZL is fixed. then The average power delivered to the load is maximized by 111i11h11izi11g the real part of Zr. In this case. choose R = 0. Since no average power is dissipated by capacitors or inductors. all of The average power provided by source is delivered To the load. P11.9—5 :tfifl 28.0 +! watt! 9 t J59 -|l— ii— I1 I2 Mesh aquatic-113: 10=-;511+[_;'9(11 —Ig)+;313] D: 2312 +[J-n5I2 +j3(11—Ig]]-[19(Il-Ig)+f3lz] 14 —jfi Il_1+::r —_,t'l'5 18+}? I2 {1| Solving the 1116-511 aquatic-11$. tug. 115mg MATLAE. yielda 01' 11:2.62i—T2" A and I1 =D.53£{]° A than V = JEWEL—13]I+j312 = jEtII—jtil2 = 13 5143“ V Finally 1-1!) = 23CDS{BUI+IUC} V P11.9—6 (t) '1 12 fl —* IE=U:>II=IDAO°A:>1'1{U)=IUA 10 Go A .2 2 L 0 w: L1 :1] (0} = (0.5))(10) :15 J (b) '1 12 9 £3 9 I2 Megh equations: 3612-13114: :9 11:212 11: 10100;; ::> IE: 510°A Then w=éLli33{U}+éLli13{U)—Mi1(U}i3(Dj = £01309): +£(1.2)(5)2 —(0.5) (103(5) = G (C) L [7+j5112—j311=0 I2 =3.25 .4494" A mflflfl (3(3) = 3.25cos(5r+49.4°) A 15(9) = 2.12 A Finally 1r1«'=%([ill.3)(lfll)2 +£(1.2)(2.132436)(10)(2.12)=5.0 J P11.£|‘—'er Mesh aquatic-113: —TT + jSIl + jiflIl—IE}—j611+j6{Il—Il}+ jSIl = U 313+j5fll— Ilj—j511= D 5101?ng yields I2 = (1.54 527°} I1 [Ki-13)"? I1(_jl ll = 171" Then T '2- Z =I—I= 8.2+j' = 3.4 £14051 1 FILE-S The tail voltages are given by ‘1=J51y—j2Uy—L)—I4Iz=j4Iy—j3h T2 =j4(11—Il]—lel+Jl-EIE = j211 —j313 v3 =j812 —j4Il+j2(Il—IE)=—j211+j613 The 1116-311 equations are 5 [1+1] + 5 (11 49+}; =1MD“ Cambining and solving yield-3 ll+j6 1C! —6— '4 '1] 5D + '1“) Il=+ = —j =1.2ifl.23°A 1+J5 45—34 SID +j33 —fi—j4 S+j3 Finally V=—j5 I2 = 6J1] fi—SQJ’E“ A :> v(r}=ficoa[lr — 397°} V P11.1fl—4 EDD , 3 T , - r wfi.L= =39 2 n=——fimmfi=m4mw : mexfizmzwm (5)1 3+2 P11.10—5 32 ' - Z = ,0 + “mil: 7’!- n" 320 9 Maximum} power Transfer requires ij _ 320 . , = 1601;?! and . = 30 JmL j H; 50 n = 2. Then 53L = 640 kg so 540 103 = —"‘_ =6.4H 103 P11.10—6 1 2:3[2+6)=Zfl A- 16,I_0v 2+2 2 I1 I2 _" "‘_ " 1 1 l..-'-.| I _ __ 2— Jim—2 9 16,113 v o “a: . 1:2 v ISL_=—I,=1I1=i1610 =32540°A . - 2 2 2 1 2 +3 Then 1‘400 2 = ‘ —3.?540°Q 1 10° _ I.» ix; P11.1~[I—'ar . 1, 12—311 2 4 HA; ' 6 6 1 V Il=——L=——1 l ' 12 V Z: 1:45 I1" P11.1{I—8 Source and line 10!.) Refrigerator ZL=_2 5 =e.3+;e.13r2 1[ze(1e+fl.54}]_3.1423° lU+lO+jT.54 25 = 88 k‘W-"home Therefore. 529 k‘N are required for air-L hemea. Pll.lfl—9 {a} '1 (ml gm.) —} {— I‘ I ' ' 15 £1 5 E I 45 o I . . is o 12 it I I Coil voltages: v1 =j16 11 T3 = 3'12 II Ii-Iesh equations: 3 I1 +11—5145“ :13 —12 12 —v2 = :1 Substitute the coil voltages into the mesh equations and do some algebra: 311 +116 I1 = 5545“ => I1 = 0235—134“ 1211+j1211=|3| 2: I1 =0 v2 = —12 12 = 0 lb} szm) s“; I 12 11 I Coil voltages: v1 = 31511 +j312 1':=j1212+jt3[1 Mesh equations: 311+11—51459=o —1212—1'z=fl Substitute the coil voltages into the mesh equations and do some algebra: (C) 311+{J'1fiI1+j3131|=5145° 12I1+(J'12I1+J'311:I=C| 12+j12 3 . I =— I =— '—1 I 1 3.3 1 2H J: I3“ ' [3+jlfi-H E {j—l]+_}'3 iI: =5£45° 2} I2 =U.133fi—l413 K 2' _ 1': = —12 II = 1.5515139" I255} 15 '. BEE . II Idea] . Coil voltages and euneuts': 1'1=—m T. 3.55 * Il=_3.5511 15 Mesh equations: 3 11+1]—5./_”45“ = D —13 I2 — 1'2 = 5 Substimte into the second mesh equation and do acme algebra: ’ 15 ‘3 3.55. F 15 2 _—Il I=—‘*1 1 II —12 ' . 3.55 5 15 1.3.5,. '2' xi; 311+13| 3 I1 = 53453 33.55,, => I1 = 0.203145” T: —12] =—12 1 ——I 1 3.55 K 15 ]_12[1€J‘II . 3.55 121'} ' 13.203145” = 2.33145“ PILIU—lfl '3 CI II=L=L=HWA lflD-J'TS {1+;‘3}+[4—j3] 1'] =(4—j3]2£fl°=1{}1—36.9° v ...
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Homework9 - Let 3'8 z =[J)= is x1 J=S J8=4 14 V" 8 j8 1...

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