Question

4. Determine the Thevenin and Norton equivalents (Rth, RN, Vrh, and In) for the circuit shown below at terminals a-b. 4 A 49 82 w 2 31022 w 48 V

Answer 1

given Circuit The vinin ab FA voltaze acron open Circuit Voltage at a,b= Vab b 42 nodul Analysin at v 38.2 10-1 42 - 4+ at V-V₂ 4 8v sk 4 VO . (4+4) - vz(+1)=4 2 V, - Vg = 16 no al Anolyan V. Vzovit V2 - 48 V2 2 ($+$J-US= 6 (J-v(-6 V2 3/₂ - 2y +48 2 2 -V₂ = 16 (+2 - 2/4 of 3V2 = 42 2 V2 =64 ie Va=V₂ = 32 v Vy = 32V V6= 10 X 4 [vc voltage acrom ion] V6=40v -3 v Varv = 32-40 Van= Vaba a Vatna -8v

Rih and Ra from ab seen same Registance was dix charged. ie Vo Hase Source shorted when Circuit Current Aource opened RID, RN Wy b CU (4+43/18 sian pour 4RE = 8118 8.2 = 42 RN S htio 142 Rin= Rth= RNE b 14 m2

Circuit as The vienin Rih 142 Pui valent below 1402 M 8V 6 fig 1 to get The vehin to apply Source transformation equivalent Nortor's Circuit Rin Source Thinsformation With Rib =RN Van IN Rh Nortorns equivalent In 149 14 الها