HW8s - Plfl.3—3 r11 120m “a 23 “:25 lfl'J =...

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Unformatted text preview: Plfl.3—3 r11 _ 120m: “a 23 “:25: lfl'J} = 300cns{12flflz{2x mfl+55°y =3605{2.43r+55°} = I500 llz a 2.4xx[lgu ]=432° 2:» f{2xlfl'3}=3flfl m5[432°| 55”}:3flfl casflZT’}: —18fl.5 mA :5: PlU.4—1 Lfl+Rl=—v, : gum: =—4flflm53flflr :1: Jr a“: T11“! = Am53flflr+fisin30m then sz—EUDAsin3flflr+3flfl3cas3flflt . Substituting and equating coefficients gives —3DflA+IEflB = D } A = 41.46 3uu3+120A=—4uu B ‘Hj Then em = —fl.4660530fl!—I.1551113Dflr=1.24cosfi30flr—GB‘“) A FUELS—2 354—45" 0 3 5L+3LST 4— 3+ =51+ ELSE” 4— '3+—L—36_S?° [ I 5325—3130] L f 5 I = 5£+31.S?°[4.4E—j3_36} = 5£+EI.87°(5.fi£—36.ET°}| = zsz+45°= MJ'E ”1445 Plus—4 (62:13:?) [-4+.;‘3+23~“5)=—12_1—.;21_3 :3 az—lllandéz—EIS Plfl.fi—2 4msEEV9 4H Apply KCL at node a: —"'4m52" +4.25iv+:=fl 1 :1? Apply K‘H'L tn the right mesh: 4£+4ii—v=fl :5 v=4f+4iiL d: u'lr After :1me algebra: a” d Fr+SEF+5i=4cps 2:“ Now use .I'= I,“ Rekfllnfl} and 4 ms 2! = 4Rp{€2t} ta write all :1? 2 [In Rte-{emflfifl + 5 fig" Re :E”‘”*;] + 5 [In Re{g"”""'}] = 4 Rep”: Re {37: [1,, 4““? + 5 % [In E‘MI] + 5 [In HW’E = Re{4 a“: Re {—4 24W” (319’ +5 (42 94”?!“ ejz']+5 31"!” 312'} = HEW 32'} _4 9491”” (fig 349;”)+5 449:, = 4 4 _ 4 _ 4 —4+5[_;'2}+5 14:10 14.05534 In a” = = {1.3935 —34° 5.1:] =u_593 (3:351:21 45°} A PIES—2 r _ a Z: '* —&= —EDflDz—155“fl =4532+2113_;= 1e |_;'mf. -1 2x1|_3 z195° 50R=4532£1 and i=fl= 2'13 :1: 2x113“ =1.flfimI] Plfl.9-l 2, --~.. .r“ 5?- -—~; g' I: .5 '+ -'$ 1. H I J J 1001'2? {all z,=3+_;'4 = 5353.10 s2 and 22:11—13 = 3.5 5—45a 52 {b} Total impedance = :2:,+z2 =3+j4+8—_;'3=1|—_f4= l1_?z—2GI.D°£2 (c) =1flfl£fl° _ IUD lflfl = gum => :{:}=3_55 m5{1250i+20.fl°}fi [ _ 11-12 11.?1—200 1].? Plug—3 l: l l + l 2 = wan: 413° +fl_54DS/_”I m1 = {—0.349— _ffl_65?]+[—fl_fl94 ”0.532) ={—0.349—u094]+_;'{—u.65?+0532} = —fl_443 —j (J. I25 : H.4fiflél5'fi‘“ :[:}= 45¢} 005112! +196“) mA PllJJJ—S _f15=_f{21-?96}[3-ID'5]| lzefflv I: 12. =D.43/_’—3T°A 20+_;15 {{a‘}: D.4Ecus{21-T96r—3T°j A I I50 £0” 16D éfl" _ W m —_;1325 + 3011+; 31? = 0.53 45.5“" A i{.‘} = 0.53cnsi120zr‘+5.9“} A {b} 1 _ mums _ 1604')“ _ {—.;'|99l{300+.f25 III _ 2565—599" —;199+3uu+;251 =fl.625£59.9° A rm=u525 ms (Eflflrzr+59.9°] A Plums ”IDS": JD 1 n The mesh equations are: (113— 3'] Il +U} 1.2 +|3|l3 =10 .Ji- [L_..|F[2 + J13. = D 01] +.;'12+U—.I} l3 =flfl Solving these mesh equations using Crmner‘s rule yields: {Io—Jr) It] a j 0 j u _,r 10 [l—j) _ so —_;'20 1|_ 23.334'I15m z: i[!j=8.38ms[103i+?7_5°) A. - .i Plfl.11—4 #209 Find V“: v“ = (5 540°] —ED +_;so so +jSU—j20 _ Q :[5 £400] EUJEZ 21.9 lflflzfififlfil“ = 4-54—21? v 11720 9 Find It: — '20 30+ '30 z =M=234419° 52 —_;'2u + an + 130 The Thevenin equivalent is ...
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