Question

kennesaw.view.usgedu uit Analysis I XLS Group H3 Surme Bb Collaborate Sessions ats > Labs > Lab 5 > Lab-5 2101-03 - Collabot < Task 1. Thévenin Analysis a. Analytical Calculations: Determine the resulting Thévenin equivalent for circuit 1 by calculating the Voc (using Voltage Divider Rule or Nodal Analysis), Isc (using Mesh Analysis) and Rth (using Delta-Wye conversion with voltage source deactivated). You must show detailed step-by-step calculations with relevant circuit diagrams. Given: R1 = 1 k-2, R2 = 2.2 k12, R3 = 820 12, R4 = 220 22, R6 = 470 12, RL = 330 12. Write your final answers in Table 1 of the Datasheet and attach the calculations in the Appendix section. Without calculations, no points will be awarded for Table 1. R1 R2 R3 + I source v load SRL ? w R4 V1

Answer 1

820.2 220.2 3300 RL 4102 Rég 820. Voc Br. Qu 1920.2 w Noc : To-fend open Circuí voltage 1ka 22ke (Voc) Thevenins voltage (276), open circuit the Load terminals Bv. and find open caten voltage acroGo that terminals. Apply kve in Loop 1 * To find voc :- =) -8 + 2,9 (is-() +0.22) +0.47) Új =0. AM ike. + 2.89 (41)-2-2012) = 8 -(1) > Apply kve in Loop 2. 2.2be →1.62 +0.89 29+ 2.912-41)=0. +4.02) n = 2.2in i = 1.8272+2+ la Qubotiteste in Equation U 4702 9.89 (1.8272+27) lg - 2.2 lg =8, = ₂ i1 4.744902405 MA Apply krL in 3rd Loop * -(2023711 -0.82) (y + VOC =0. cloth = 60.22) +10.82119 = 3.143184 vots hoad terminals and * To find Iobc : Short Circuit the fend Chort circuit Current through it - Apply kve in Loop (1). QIg o Bokel → -8+ 2.2 (Ig-I3) + 0. 2262 0.22 (J.-Ig) +0.4+1 = PI 21. YIGC → 2.89 (I 1) -0.29 €2) -2.213 = 8.-(1) Fo.aoke From Circuit Ig toc =2.596 +12mA. ƏYoc unke Bv. Cscanned With Fk Scanner

Loop . → Apply krt in 0.29 ( I9-17) + 0.82619-13) =0, 0.8113 => 1.04 I9 -0.2211 0.-12) → Apply KrL en Loop => 1X J3 + 0.82C 13-12.) + (9.9) (13-11) -0, 4.0213-0.82 Ig 2.214 =0,-39 By Golving (1) (2) & (3) IL 6.8168 MA I2 = 5.2235 mA = ISC 4.796 MA I3 E the hoad * To find Rith open Círcuit) Circuit terminalo atst bear by de activating the Independent sources C voltage Gource with Ghort Circuít and current Bource coith open 2.2 kb, 0.2252 and 0.utke ake Bo Convert them to Forming A 0.82 ke d - Connection AWW с 2.2ke B I RACC Rith A A be M ho RA 0.22ke - BAB BBC BB 0.4 B B B C 2.2X0.47 = 0.354485ke RA AAB. BBC BABT BBct BAC 9.2 +0.47 +0.22 RB = 2.24 0.22 -0.16 44740 skoz 2.89 ko -0.035778546ke 0.22x0.47 2.89 1ka AMW 1.357785522 0.987479 ke 10.3517ko M 0.826 MMA 0.16747kel Rth. *th 0.0 35777785 AM 0.0357 el ke Rth 3. hth = 0.607575 kn CS$canned with Cam Singer 604.5752