--Kirchhoff's voltage law i.e. the sum of voltages in the
loop/mesh is always '0'
--The superposition theorem states that for any linear bilateral
circuit with more than 1 independent source the voltage or current
of the element can find out with the sum of current or voltage due
to individual element
--Thevenin's Theorem states that “Any linear circuit containing
several voltages and resistances can be replaced by just one single
voltage in series with a single resistance connected across the
load“
I for the given circuit using nodal analysis Luhmo Š olmu 0 The for given node circuit voltage is contain we tuo nodes. can corite KCL as. ule hu = 14 + 34 -0, for node voltage no. we can write kol as a M: +22+6) + y-v-4 = 0 . 1 substituting in (1. we get 4N = 14ts 14 3 2 + 1 Sy = 151. [~= 6v). Scanned with CamScanner
1,- ( 2 mA 12 ) 1-2 no 4+2 MO 4 . ALMA (h omma I-2 mesh considering Mi we can have Kve equation as. & Hifat I 12- Pixs - (11-29x1 + (1^2 - 10*1 = 12 - 31, I O !1 1, = 12 + I 3. considewing mesh Ma we can have kul equation as. + 2x1 + I 2 = (1 -2 -110*1 0 ( -2) X1 + ! Le AI - Il = 2 - substituting from (1) we get Haf - 12+2 = 2,! .31=6 ! 1. = 2mA Scanned with CamScanner
& 4kr considering be Inaesne the . 12 v the supply to be active and other sources current could be conside as open circuit. considering Loop L. 1 तसं 12 + 4 to, t 2 10,=0, (70,-4 mA), source to be active and other sowees to be Considering ama in acine. Lis t using current division rell. weget. 902 = 2X alta - 28% PO2 = 4 mA and otuus to be inachine active Considering A Sowee to be resistan here the 2kR & 4ku parallel with or desistance herce а саме – ишћа fteno Вот очало - Yetike - hence logo or Pos Doi +102 + Deoz - 2 tý +0, 2A. os opposite direction of cs Scanned withegative direction shows opport! CamScanner Assum prion
(4) To Il 10 find find t the the vinin equivalent resistance istance twitch off witch off all all the the resistone Sources, Muu un {4kn. Rot-4tre, Rua = 4.k . to and and voltage . the voltage theninin find load resistance is opencircited accross the load resistane! lee we Cånd the Vog l i yam A in & 12 KCL at node With . using using ke2 at node Vi Vin Vihti 2 +6 = 0 4 4 we have considering kee at node ni we have .. via Vith a 6-? so, ū - Substituting in ③ we get, "= 4th+32 uth - Ven +32 = -9 4 +lJL NAM -4 v. Nth = Nth -JA cooth -:110 - Nha Scanned vkith CamScanner