Question

#1. Use Thevenin's Theorem to find the voltage V, in the circuit below. 3 Ω 2Ω Ο Ο + 50 V 6Ω Σ 0.5υ, & 10 Ω ob

Answer 1

Sola. for ai a giren akti 3 Il 212 (22 +0.5V) Sovet 41-12) + a 10 o.svx b As we have, depenele dent some UTH RTH = Isc

apply are KrL hee get in hop 0 -311-6(11-32) +50 = 0 - 3 11 -6 Il +612+50? 911-612 = - 10 Apply Kul in loop @ - 2 I 2 -10 (I2 + O'JVX) +6C IF I2) =o But UX = 6 (11-12) -212- -10(Iz to.5x6 (11-12 12))+6 (11-12)=0 - 2 12 -10(I2+3(11-12)) + 6(11-12) 20 -212 -10 12-30 11 + 30 I2 +611-61220 - 24 Il +1212 co .: 241, -1212 10 2 .:uth=10(I2 +0.5 ux =10(12+0.5 vx) soleig egno, o we get 11--SpA, 123 -10 NTA = 10(-140 +0.546 (15+)) *h = 10 (5) soo .

Noth 166.67 V Now short ckt 'ab' Therefore 10 s ceill des couded (52 +0.5 Ux) 로프루 T so ISC 22 -311-6 (Ir-Il) +50 20 911-612 = 70 - 2I2 + 6 (11-12) =0 - 2 I2 +611-612 =0 GIl .8 Iz =0 - 4 Solvry we geti 12 11 = 1 00 9 CA / 2 - A -25 3 Isc = I2 + 0.8VX = 25 **+05*6 (4-6) ISC A = 16.67 A ..RH = uth Rn こ 500/3, 501 3 S1032

thee fron, Therinins equiratut clef as RTH + 1 + UTH 18+ Vo = TH=166.620

Multimeter-XMM1 XMM1 16.667 A A V dB R1 R3 Set... O! 30 20. XMM2 11 10.5 Mho R2 602 R4 100 V1 -50 V HB Multimeter-XMM2 Х 166.667 V A V 2 dB Set...

Note- please find me in comment box if you find any difficulty. Please note that in solution mistakenly I have written Rth=Vth/Rth which is Rth=Vth/Isc. As load resistor is not given, output voltage is nothing but thevenins equivalent voltage across load. Simulation circuit is attached, showing same result. Good luck