Question

(b) Assume that you have been asked to find V. in the circuit of Figure Q2(b). (1) Between mesh and nodal analysis technique, which method would you recommend? Explain why. [2 marks] (11) Use your recommended method of analysis to find Vo. [13 marks]

Answer 1

Node Analysis Mesh Analysis * Mesh analysis * Planer and Planer network Non Planer network * Used KEL * used KUL Procedure of Mesh amolysis I Procedure of Node Analysis 1) Indentified total no mesh . Tatal no of node 1) Assign the current durection Wassign the voltage of each mesh 11 develop KVL equation for mode is token as each mesh referenc node W) by solved KVL equation the referenc- node find loop current as potential should Seat node, one of to ground | 11 develop KCL equation Iw) by solved KEL equation find node voltage Below circuit used Mesh Analysis because You two Joop current know Node analysis not used here because one node voltage know (11 used Mesh Analysis Is, 12 LOA 852 Bj2 m 8j 15 gov! T-jar Ig 11 1, 0 10 40°A Isa loop KUL -15/90 +5 (I, - I 2) - jz (I - Ių)

b e (5-2;) I, -5 12 +j214 = 15/90º =j15 0 loop ② kVL (8-5j) 12 +5 (1,-1,) +j261,-13) = 0 - 5 I, + (13-3;) I 2 - j2 Iz = 0 ④ Joop (11) Iz = -Is, =-12 20° = - 12 Joop current negative sign source current Is, because oppositive direction Joop 4 I4 = - Isa=-10 (40' = (-7.66-642) same condition above Iz put equation 1 Iu put equation

(5-21) I, -51, +j21-766-6.421 ) = j15 15-2j) I, -512 = (-12.85 +30-32) W -51, +(13-3j)I₂ - j2 (-12) = - 51, +(13-3;) I, = -24 W equation (1) multiply by 5 equation 1 multiply by (5-2;) 5 (5-3)2,- 251, = 5 (-12-35+30:32) -5(5/21)1, +(3-3j) (5-4) 1,= (24))(5-21) 1, [ 13-3;) (5–2j) -25] = 5(12.85+30-321) +(-24;)(5-21) 1, [34-41] =(-112-25 +316) In = (-1-80-1-243) A Iz value put equation (5-2)),-5 (-1.80-1:24;) = -12.85+30:32] (5-2) 1,=421-85 +24-120) I =(-5,43 +266 j) A Vo = 5 (I, - Iz) 15/(-5,43+2-66j) - (-1-80 -1.24) 51-3.63 +3 19;) V = 5 (5.32/132.94° IV. = (26.63 2132.94%) v Vo