Finding Thevenin and Norton Equivalents in Electrical Circuits, Study notes of Business Economics

Download Finding Thevenin and Norton Equivalents in Electrical Circuits and more Study notes Business Economics in PDF only on Docsity! Thevenin’s and Norton’s Equivalent Circuit Tutorial. (by Kim, Eung) Thevenin's Theorem states that we can replace entire network by an equivalent circuit that contains only an independent voltage source in series with an impedance (resistor) such that the current-voltage relationship at the load is unchanged. Norton's Thereom is identical to Thevenin's Theorem except that the equivalent circuit is an independent current source in parallel with an impedance (resistor). Therefore, the Norton equivalent circuit is a source transformation of the Thevenin equivalent circuit. Original Circuit Norton‘s Equivalent Circuit Thevenin’s Equivanlent Circuit Int e They are erchangeabl 1 How to find Thevenin's Equivalent Circuit? If the circuit contains You should do Resistors and independent sources 1) Connect an open circuit between a and b. 2) Find the voltage across the open circuit which is Voc. Voc = Vth. 3) Deactivate the independent sources. Voltage source open circuit Current source short circuit 4) Find Rth by circuit resistance reduction Resistors and dependent sources or independent shorces 1) Connect an open circuit between a and b. 2) Find the voltage across the open circuit which is Voc. Voc = Vth. If there are both dependent and independent sources. 3) Connect a short circuit between a and b. 4) Determine the current between a and b. 5) Rth = Voc / Iab If there are only dependent sources. 3) Connect 1 Ampere current source flowing from terminal b to a. It = 1 [A] 4) Then Rth = Voc / It = Voc / 1 Note: When there are only dependent sources, the equivalent network is merely RTh, that is, no current or voltage sources. 2 FP . of vallelamrecfed We cam seperife the vesisters juct> = Sore groups fy Yesicters . a per > C 4 (14. ) Vib = CSN SNS i | | | ° (445 _ se 2, Vab = B+ | + (44) + ( SISNSS oD *~ Carved] Divider i loher we Nave multiple resistors L ls Ty valle| conedfio , the Soure ft *e Current will be divided ito eack poralle/ branch accersling fe KCl. Is = ti + J2ot1e cam be souplified as since paralle| —tenneéted Ye Sistors voffase ACKOSS one swale resist as CRNAs 143)» Ce each resisto- (S the some MS V. we al / Vv a= RS Jen ithe+ l= > He? hy ae vw therefore 4 Be = at ak \ ev = Vai mt F> \ Ta= a, ° pe Is arn Cae eek ok B.* Be Bs can be Thus, the cuweuf Posing, thigh Ye resistor Rn found as h=%= = ms 1 Bn” 1, ” Br " Gemea> % In general, We com yepect} che currext divider principle by Jn= Js - Gn (? is Cockcfauce ‘\ ie ZG i By, JZ €xanp/e 7 ) oA By and Br is Seves-tunedted tae the cumerf Source, therefore te Currect Floiag, acess Ri ond fia is just the sane as /oA. Hwee Be md fia Ave parale/-anecfedl , the cuwesf wil be Sotledl noe Two Sromches . is= a Gp® fo a” Node x Ef we sef the minus. termina a a < PV (4) best 1 . me the voltage Sourte *S ground, then A the velfage of Node-K /S Vas , b Peply KCL te “Noole- x. Vas-/0 , ab pg mo /0 4o » Vab= — BU = Vea. there fre Theverins Civcutt is Ron= (220 the polarity af the voltose source [Ss Yeversed sinte Thevenin volfase Source Ts mMtuus value . As you know , the impedence ™ theveum's Cirule is the Sane os the iw peclowee wa Nod’ Grealt, So, (Ff ee feud Shot-cinutt camer across ab at © istfead af opou~cireult voltage acnss ab, yor can ‘find Norton's Equivabout Civonlt. . node—y. /oSe 4a. Lets séf the vttasxe of node “4 as Vy. Apply KeL to ned em Ae + Va tat Bo fo. 4o al ——e a 2 Vs -8.3.V Fasnceta 3 Flim ta: Tne Ves —83V = -2.0954 42 450 ||, example >. 6X wa Jor O Find opor—circult voltage actoss ab. Qa o> JO SL reste au be ignore sme Cs al ba mee Neale a cs oper . Apply KVL, 6i -2A4+ 64-20 =O ~AzRA Therefore Vep= 6A = (AV ® Fad short-cicult (uncut acess ab. bes Using 4wo mesh cawesfS , we have / —20¢b6u—2i tb Ad =O bUe-d + loiz=e \ A = Are At From these three equations » we obtain A= 18. A = Jeb @ Fem Vab (oper-cirmle ond Dap (shot—ciroie), find Bah Vab te = vob 2S (360 Am Lab (20 f 136 ~ therefore Thevevin's Eqvtvalosf Commit JS s a ee fad Rens [3.60° Norfon’s Eypriolert Creme, ws example #>. o Sine tle Cinult hos no Fadopardet Bite, K=o when we Coanedf an spor Cir fo ab, therefore Vab=o wd dab = © Lope) Cclose >. So, we can vat use Rena Var lite example 3) Jab So, we connect [A test Cawet Ssune qo ab. Then we com sae Ren= Vab 1A (A let& sef cho minus node f the voltage Soae aS growrd for veferevice - Apply KCL to note A. | Naty - 2 Veb oy = zB + Tp | =o Vab \ ond as Me ly SS ‘ Vab Theetoe. feos of ZS> + ab —/=o 2 Ub lo », & > Reps Mba S28 | 13