Question

For the circuit show below, use superposition to solve for V 6 ΜΩ 40 μΑ 1 ΜΩ + V - 75 μA (1) X4 MΩ 200 kΩ

Answer 1

given circuit 404A ime 6m2 L.WMT- +V- LWW 200ķh 754A ► 24 m2 E3v Find V By Superposition theorem Superposition theorem :- In any Linean, active, Bilateral Blue consisting of no. of Energy sources and Resistances. the effect produced in any element when all sources act at a time equals to the sum of effect poo duced in same elemen! when each source considered individually,

NOTE while Applying superposition theorem each Sub circuit as only one independent source the other independent sources De energised. where Ideal voltage source short circuited IDeal current source Oper circuited, Dependent sources can't be suppressed. * Voltage's have specific polarities, * currents have unique Directions.

por: Considered 3 volts only Im2 6m2 w tv WW 2008-th . white Z - 3y whug 1. IM 2 _ MWWW kook-2 ty- 3V z whit iMh WWW ty 10.9me 5 n o 3v1.4m2, 6mm 200k 2 are in series te=100.2 + 4ooke 1. = 10m2 +0.2m2 Rez 10.2me

By voltage Division Rule: reoltage Across ime is im. =-3x - : 102M-2 tim.ch =-3x II 12myce -3 V = -0.267 volte

Step-2 considered 75.44 only imar Gyer Two w MM 75.46 2006. 4 m2 ime 6m. 200*2ty- 54 me 54A 26-2me 15MA Q 4 + L V imar

By cuorent Division Rule. Te = 7540 x- 4me 4m2 +6-2..2 time = TEMA X 4mot. 11.2 más I = 15x4 HA 1102 T = 26.78 el A i voltage A cross 112 V = I x 1m. = 26.78 4A Ximen = 26.78 x 10 x 18 V = 26.78 volts

Step-3 considered 404A only Thi thoto whug Mooit v 200K who olur. throh + A - how to 2005-2 whil TV - 4m2 YouA O E 200k 3 ome

у 1, 3, мя ЗООК- фоцит lome By curgent Division Rule Т. = 40 съК- x 2 2005-2 +1 М 2 том. - 4-оя хо» м-п- - О» 2 м-, + Јmn - Tби-- Фолін Х О•& var. 11. М. - Чох. ЦА 19 Т= 0.71 1A Voltage Across ims is '

ie V= I xime 6 = 0.71X10 X10 V = 0.71 volts 6 By Superposition theorem Voltage heroes ime is sum) All three cases - voltages V = -0.967 + 96.78 +0.71 V = 27.223 volts " Resulted Answen is V = 27.223 volts