Electrical Circuit Analysis 1, Study notes of Electrical Circuit Analysis

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Kassin, Shacly Sec tion 4 \- O lane's Jaw —————— DB Senew — Peratled 4: Swe bole! - Element Squevel Unit * No \eege Goure —C>}— Volk Cv\ + * Current Source —G)— Arperve CAY we Resisten le \ olhw CN) f2\ Sek Ln. _Xbeoe £ Piha + looo x \oo00 % \oo0o ka “ieee A ——> wA g — + \o00 + \ooo Ve LTR Le J XS ob Volk Awe P= V Tos erent CA) elo 4 Power volk (wv) Cwak bY cy) re ke Words! - @) oleltveved (voltage 5 = No troge Supplied = Nol tage : ouce _ No \tone Con umred = Neb cis Si pated Cia eri stere, ——> _ Peover Con Sumed = Power At ssi pated Cin Resistene Gy © =e NES Nin R s\ Cevies QRetistane Pleo \ eae NE Im Series Connection , x i ord Current ig the Same. eS Total a = a = aly = aw V+ Xeeq = , FEA), 2 AQ - Qpecial Gse - Voltage Divider :- Ra Pa (Sy As- Eid the Fyuivalent Resistonae for he Fo Mou ing Cie fet bs~ Sk l6 len Vane ISoktL hokn bokar b =a Qok pn roe 2Sk&e 6kn Gols - —> Req - Gox Yo =Q2yukn2 —> Req | 60 +o Ls) Joka lok Sec Ub, lo, Req Series 1 =16 +lo+ 24 = SOkN asknrv wa b -K, Qok rn —>Rey = Ses SO ae ag oN g \So4+ So se aS Rey, = ASt3tSt+I0 ‘ac, [ ey « bo wal Exawple dt- We Wurverts Th BD ta the @ Ciel ove uA BAAS ral OS) Ty bY Power dissipated in each JZeristor c) Va AV Clee ake power delivered by the larren & Sourle if equal +29 power absorbed IA all oNker eleweunt- Solution t - SP oStuwme. Gorreke: in eacls boanek K pat polevities a Clording te He Qorrent: aSsuwed. Te | eee eee a ee es eee ae 4 O72 as a + dhe a Oo s- 2 am 44 oe, = Bal A— kvlLOYd: _,¢-6) ee Vs - \Y (|= 6 | Ny 2-\sav] be 4 Oo nh Rae KAYO : HT, + (F+v2 e4VT, - 44h = C= 6) COE Cay = 24, =0 \ Ta= 2a | a+ = a3 ig sA ——— eel ead 3A or KVLC3% 0 ANy + 24T2 + GAbey ty =e (6) S =0 a\SA + 24 03) + mek be egyel 49 Yervo- GE Se eine) = avy remery -6 = T) 4 2 Te-gaA Power fupplied in Gz | ae -|S2 x 745 126g w Th only oe SeurG@ = V4 Jy Pa = 23) 0 W Pye tage Scourge Vir= Sete 4 [ Power PS ab for Leal by Ve \tese SouvG , beGute the fe the =tve sec nsline4 LOT sien Caccen'd Aivection \S of Sapo Vagrecel of — Flowing Out feoen Na Maze Spey] Prortege = Nx T= &@ %-4 = ~Z20W r me, Ly -ve is fhe indi GtHon TNE bet Gane = =the ae ee a rt Ta is Supposed be ‘5 ae ie is ab Sorbed be opposite divection Sleek 4 Ouet AL Node Voltage Ue thool Question 4:- Write the nodal equations of the networks Cheown below, when find +e node Nolteger - eS ate & o> ———} \- sA = bya Sa & —= Neb Sree 1.- Fir now of eenbel noder C& Etenhel vroder\ Step at- Reslave One of the eStenteal nodes or a Ntherentie nook - Ste P@i- Wik node equshonrs , wha write node eavorion oMume aff Crrrentse goin Vv ovk of the WNoole- (kel) ST=0 @ each moodle Ao Kool ail poole Noiteyer ond all Curreets in Gack branch. Geer ve Yiode CAX+ - busty litany nodal equation br AY —S+wWNa-e (i A Va - N@ =o 2 ) 2 Na S+4++ S|+ ve(- 4] = 5 accety OD Node CR): - Ne- Na a RA. + V@-9 = 9 Q 4 ve FHL EME SE * A1-E © Go\ham {AY Linu [tortoud\ 4 on) OAL Ae Mode S, | t- Va a 8+ OFF V > noole No \tet ers — Ve o-38s N it Super-Node = Section o) —s We Gi iba Cuper Noole branch !- wher there is only a Vo\tepe Cour le WW the rane between a Shwe CWenrwel\ pnodlen: N ty . 4 Aan 2 —> a mm <-z QA -— rt aS —> Note i- in order +o Colve 4 Super- Noole Yethed, you Cennoe take Vi or NQ OS yeferen@ noolé. oole| enolysis t ex Wool i V,-o . ole Civ — ) ie > +i 20 ir . No- 7 Wode (2d: ~- =) ae A= $ Vase 252 -~o——(\al hy cdo! ing A AY ee ve “v4 % Ns. -3 _ at if (@) + 4 o Vo- Cour ego tba! bo VY 4 VX 4 a8 - © And Os we eno \ Vie Noe Iv | Ln Solve “ x4 A 4 Ciwu Sree ea t= a =| V,= 4 Melt ‘ No = z Vole 2 That - When we Solve fr o Quper— node Yoren ol Seats - weote Qiper noc eqs Hea Eelord:-— oaodld +the noole eqe Kons of hode 1 A 2 ond write $6 in one queen, | vin Qu per — node HA lafwcail, A+ M | Va, Va -23 a .. L Moola!) yvoele (2) | Secten, © Hh Aegis Steps: ee _ : . Assume Current in eacl \cop ole mene on @ach resistence (4 ~) _ Apply” KYL each nicep- _ = su Ex 4, ree — _ 62. ~ so ee +O, ay 2) —— = __ ty Find ae Phew each resistor er Hes ‘Rocky 455 pwwwwr- - - ie 8 40+ 6T, + 30,1) 2020 ‘ a _49T,-3, = 30 Sci ¥ kv 2s 20 43/T,-T]J +4 T4020 . Song [Ti Bh *T- ST, =-S0 —>|2) ' Ts 1A Te 663.0 ; : Tz 1-7, = C66) 3.434 E - (Rowe sed) Se chon Mew Anely Gs Sees!- —_ 2 AggSume Mivvent in eaclk loop. ClooP Cur ver | wed Current) pus Poteriby on cook yeaisiente C+ -% sapely kvl" in each loop Frowele 4 Use the wet —-lervent wetheoad +o hiro Vo in the Cir otk Chants Seet- AStume lurrenrkE in eccet loop [, oEbure ing loop Curreets in clockoice divec Hon | as Cole. SS Ving t £2 22 Breultencouslys — Dechion A: : —— —___Supectte a : —— he “When, Cutten | ae 2 Meshes 5 We are te loach, een ong —_____ the Cucrert source. & ther | ie es — — iS _Keown_as_Supermes\s Some. steps fo cHesh : x Assume Gucreat_in each loop x Pik Polatty Sone ne a a a a er wv | —____—————— “We thowe Current Source lbelineca 2_loags ————a oes\. » —— oe eae atta a a a