Question

Practice Problem 3.8 By inspection, obtain the node-voltage equations for the circuit in Fig. 3.28 Answer: 3A 50 w U2 5 192 2A 1.25 -0.2 -1 0 -02 0.2 0 0 -1 0 1.25 -0.25 0 0 -0.25 1.25 2003 + 2A Figure 3.28 For Practice Prob. 3.8. Copricht OMG-2007 Solution: We have five meshes, so the resistance matrix is 5 by S. The diagonal esc FI 80 DOO 000 F3

Answer 1

Given circuit * is below given Чл. Vy 193 3 VA 2 A Uz 512 2A 2on Nodal the analysis: solve by using at oj Apply Nodal Analysis 19 26,- U2 11-13 =0 + 20 5 V, + 40,-442 + 200,- 2013 = 20 2013 = 0 [Divide by 20) 25 V, -402 0,212 - Oil 13 =0 1.25 V, > CS Scanned with CamScanner

at Node va) Apply Nodal anolysis 42-0 5 = 0 5 12-21 = 5 - 5 -0.2 V, +0.2 Va = 13 at No de Apply Nodal analysis 123-11 뺌 +3 + 13-14 ។ 4 (03-0,1 +12 + V3-4 = 0 -12 -42 +503 - 24 = 3 0.25 Vy= -3 -19 +1.25 V3 – Node ☺ at nodal Apply analysis vy-13 vy - 2 = 0 + - 4 CS Scanned with CamScanner

24-13 + y24 = 8 [divide - 23 +504 = by 4) - 4 4 23 + 5 VU = 8 4 + 용 -0.25 13 + 1.25 24 = four equation the write all + (1.25) V + (-0.2') Vz + (-o2') x2 + (-) V3 + () Vy = 0 (-0.2), (0.2) : + (O) V + (ODVy = 5 (-0.23 Vy = -3 (-1.25 ) V3 + ® (-1)v + (0 ) v2 2 (1029) Vy + (-0.25) V3 ☺ (o)v + (0 ) Vz form write in Matrix vi -10 -0.2 5 1.25 292 -3 0.2 -002 13 1.25 -0.25 -] 24: 0 -0.25 CS Scanked with CamScanner