20 ox] Fial the Morten’ ~eyuivr ckt. te the leet of terminels @ and b, Selutens 1) Remove the external circuitry aad apply a shoot okt. 2) Solve fer Ty which it the current throwgh ‘rhe short oky Sinte 2 Sources are present use Super posiHen theorem: | kill Current source Creplace uith open ckt) . + 7 Poke; GA vesester fs Shoefed oof ‘ta $ = Meh VW tty, a aie 2. Kill voltage source Cheplace with Shorf ekh) By CPR: UT, Iv, 2a) ie =-BA. 3, Com bine # < bin te get Ly Int Ty, ware(-8) 26 6.354 cowbimeed —...ex / con bent eer 3.) kill Sources ant solve for Ry (replece Current seures wit open ckt ) Ureplate voltese Source with Short ck h) Rye tenes = CC) _ 2 B42] Gate £) Redraw errgien! ck, Ck. in a with — Morfon Eguy' ay 10, 4384) “ ~ Becton Fga'. Ck? Motes In Schematic Ly was shown as being forced cut ef terwi'nal “a” Since the computed Ty value wes negative , the civection of the source [$s drawn to feree Ccuyren? out of termine! ~b”. Current 56(MAKimum lower _Iransten _ ‘"eerein A load will rece‘ue macimam power rem a linear bilateral de network when it totel resistive Value is exactly equal to the Thevenin resistance of the network as “teen “ by the lead. makiinum Pow er transfer 7pnmaor terre Complex Numbers + Hirt He eg taitC tO a Converscon Between Forms. Ce 58Math Gperutieous uith Com plex Mur bers Given: C,= A, +78, nel Ey = Ayy gp Ba Addition C,)+Cy = (A +4) 4 9 (B+) Subtraction: t,-% = C4,-42) + 7 (B-B,) and Subtraction are eatvest in A dkdi' tien rechangular form, Malte plication ancl Division are easiest in Polaw form. os c Ce, Nites |i really 1IZ0 57- rnesors A Phesor fs. Complee namber which can be wed to represent a Sinasatdal Ac wave form. Vn Sin (ut +0) 7 \ Te Domain Value Expression tov Si nasoid Vn SinCut+e) 27V, 26 ee Phason eae Devain ex| Gives: f2 60H . ‘ - cae . y= 50 . (37 730°) i “, MW, 730 sin (2776 40°) Finl Ci Sola tions By Kuh ~ewtay +1, 70 Livi th, +My, Convent fo Poser Porsins mn, = 50 (02) L 30° = F.35 L490" 30.6/-4 417.68 % Re tonvets peak to RMS Wy > 30 (6.707) £60" = BIA LZ b0% = 10.6149 18:37 = Vy By Vat = (joertt-ar) + p(i759 + 18.37) = : eer ee eeeare eel Ej = YUAR + ZICOE = FUT LT? ON be tine A0Om yu VEC SET) Sen (3778 417°) © 17. sen (377 HHI? 60 ( )Tinpoclunce ancl the Phator fag ram Resistive Elements rer purely resistive elewent the current and voltage are in phee. Zo where @ is the Tupedance of a res/stor, < Ohns Lav for Chasors A Phaser is oa Is a rotating Vector with & magnitucfe and phase ousle sociated weth i7, 7 7 &xf Find the Current A, + ; Sketch 4 «md a FR Ar zpoo sin & Solutio! Solution’ 1) First convert MW to Chasor Ferm: = 100 Sinwt =P Ta 100 20° QA pfly ohms Laws nowt > Vv o Ty. 10 L0 ag f0 ark R slo . Z)Cencut pack to Me bomans al dn 2OSinwt Amps.: ex 435, tsinlintsio%) Find WW, Sketch 4 and a oo... Gu ae hk) Conver “a to es or. - Aa4sin(wt+io?) <7 Ls 420° Anys. 2.) Apaly dhm's Law 7 WaT Ey = (4L30)(2. 20°) = 6.230" volts, da) Convert Dack to Hime elotmarh : w= EG sin(utt30") — velts A Phaser = Diagram is pht th the compl plane showing the magnitude and phese relationships of creat qaanh ties, For the above ceample the Phaser Di'agr am would pe: Motes Lo awd Vo are jn plese since if's a ores'shye chrewt 62Inductive Reactance current induchive element the For a purely lags the voltaye by 40°. of an inchuctor Mee anh s wl In rectangular Ferm ; Ohn's Law fan Inductors i ef Az 5 sin(wtt30) Find voltage A, sai ] Sketeh W aad “A, eng she 1) Convert « and 2, te poser Form: Ae Soin (ut 30°) = T= 530° Ziska OM = GZ" 2) Apply obns ler $V TEs (52°) ) (4 Lae!) = 30L" Hy Cent hack Jo Hime demain? Anz 20 Sin (wt +120" ae SI think of hasord as rotating cow / T lags V Ting Dowen ey 63 gerh phasor Dioyvam by 40°Capacitive Weaclrance Fer a pure ly capaci tive lowes the voltage by 9° a Z, is the Impedance a capacitor aie ie “we Clement the current rectangular for =s> = -o a ? ull gee we Ohmi lee for Capactters Es V = ne La 7 P = : Tee = 6 Sin (wt~6o°) Find voltage 4 Sketeh An and v 1) Convecl A anol Zz, te phaser fore Fe GL Fax, Aw’. 05 LZ : 7 OFLA A A) Apply Owm's Can i Vat Es (626) (0 5 Zo) Vz3 Z (~6o +e) = 34-90" 1) eoavct dak te Hae domains An = 3 sin (wrt-l50") I leads V by 40° + ; : t ca Vine Doman Gragh Fa feeThe compler puantibies Corresponding fo the impedance of Ry Ly and C elements Hn be graphect in the Cemy ley pire. 7 XL 240" a ake For any Con Tourabeon (series, parollel, series-parailel, er. the angle associated with the tote/ img ecdance Ts the angle by which the source current lag ¢ the applied vo|tege. For incluctive networks Gy wll be fiehive, For Capacitive networks 6, will be negative. Series Configuration For Twmpadances in Series the total Inpedance iS: > toi osGE &x/ herb —_ Retr Z0 Sola trons 2.24 ao a) Apply series B vale: Praw the Tnpedance disgram, Find — Fotel Impecdame 27 } L) Convert Impedance te ghee form ; ZL" Be Byte teere 6290" Convert to rectangalar z,> 4 +87 form conveet back to Polar form: Beas ef Bea kee 7 Heep % 2,2 6L0° > 2) Aprly Serv & rales Zp 2 A1te a £63.43° te wld: Dettrmine total Z, Draw ee 1) Convert to phuier form! a By = 1040" > B3F 12.Z-%0 7,2 6S +102 40+ |2L-90 convert fo rect. form be sold: FG +log ~ng = b-2y Convert back te pole Lem: Sy = G38 a Zo Nb 43 G, neg 3 2% capachve 2 Avagran.For the Series AC Cirewt the current 15 the Same threwh each element. +V¥- Phosor Ss - Equations: : {2 Te E@&) Ei . a « V, Se tee tee 2 rt 8,43 = ieT2, , Yet By fy KVL; -E4¥V+%y20 Eavi tv so ® v Divider Rules me E(S By ‘oltage Prvide: “ 2 Zh The Power Supplied to the Crrearf ssi Ps EX cos 0, Where: E is Yoltaye in Ams Lots current fn Rms O, fs the phere angle detnesn Fand T is also the angle of The Powen Factor Fp ie: Fp = Cos Oy so PRET FpR-L Circult sa jan Convent to Phasers : ee ltt sihwt —y Es lH Zo" © BX Convert te RMS Fe 4ie(207) Zoo's loo Lo" VOlts Ring Find tel Su pedance » ees eaeas ° 2,7 Egt B= 320 +4L40° converting te rect forms Bar 3+ Ze a converting te polar for mz B,5 on L313" By Omni Lane 7 Tnpedence qe E . 100° Diagram z. 5 Zsa? : : < 6,3 5313 os + 53.137 Ve = 60v. a 2 : ° Ves TE, = (mo Zs9)(4L 90") 80L 3687" han, Cheek by Keyes i -E + Vy V2? 4 EF eth = bo Za S3s* +60 LIE! Converting to rect form: Es 36-g48 + G4 +A 48 Es tee +70 = JooLo" v. kos € Pheso- Dire 63ex) wee Cow tinued Tote! Powen delivered fo Rb ckty peste ty tee ot FCI are eftective (KMS) valses P, x (lev.)(RoA) cos 581° = 2ee0 ws C6) = [Roo Ww Power dissipated by Resistor + Pe= I°R = (204) "(3a) = tools) = 1200, Tote! Power ditsipated by RL ekh: Pr = Pre Pi = Vd tos Op t+ VT tos & x (60)(29) cos 0" + (60) (20) cos 70° = l2oow. + 06 = 200 Ww, T wote: [Inductor dissipates Bere power Power Factor: Fp = co Oy > Cos 53.18 © 06 Lagging l, Fe Legging becaue T Tey; E fn oy indubive ch eo Since PEI Fp Ales Pie P called “Real Ponce sn wats P EL ET called “VA “or Apparest Power Als Fy= KR 2 EE CHa ar mE REE eed CosO= ET wa” eR 2 G7R~C Cirewt en/ _ Reta Xba Convert te Phesues: 70 oh 427.07 S10 (mbt 918) >To ta take Convert to RMS; R= 07 shaleet4 5319" : i : Fe 7076.07) Lud = FA L533" kas Find tote limp eefance / Zit Wet Bex CLo re L 10° te rect forms BR G-Zh LK pele forms Bye 10-8 Z~ 53s? Conver ting converting to By Ohui Laws i E=I2,* (saZsun') (0.0 Zour) sov. Z.0° Yt Tep= (sa 254) (604 0’) “ Tae 3ov, £53.13" % = (sa L508 ia Zo) a eae Check by KVb: ~E+VatVer0 3 Fe Uty = 30. LEM + 40 Z- 36.87 Convert to rect. form: Ea 18 +724 + 3a -fa4 = 50 4/0 E2 50L0° 4 checks ie Tn pecance Diagram Ope ~FbNg? I Convert to time clomarn: Bn S0Z0" > Cn YRS Sin wt = 70-7 sin wl Kes . (ut +5343) = RHR sin(uct+ 503 ) Vat BoZsrni? a Mya VE 30 bin ) = 36.07 )2 96.56 sin Curb 3697 J Ves to L060? 9 We = Aa 40 Sin (ut