- Please Solve As Soon As Possible Thank You 1 (141.42 KiB) Viewed 32 times
1. Method # 1: Circuit equation Solver - Show the rest of computation for both nodes v1 and v2 to obtain their values. Applying KCL to node 1 of circuit in figure 3.0 we can obtain the following equations: (2) v1-0 v1 - v2 10 6 Applying KCL to node 2 of circuit in figure 3.0 we can obtain the following equations: i2 + 14 = i1 + i5 (4) v1 - v2 10 2. Method # 2: Matrix Method: show the rest of computations to find V1 and V2 Matrix Method: i1 = 12 +13 ans = 2= We can also use the matrix method to solve for v1 and v2. To solve for v1 and v2 using the Cramer's rule or matrix inversion we have to represent equations 3 and 5 in the following format: v1 [ 12, 12] (+v1-²=2 (6) 10 ² ( 1² + 1² ) = −² (7) Equations (6) and (7) can be represented in the matrix format in the following manner: + - v2 02-0 (5) 6 10 +4=2+ 1 10 (3) 3. Enter the following MATLAB code to verify the results shown in part 1 above. >> e1=¹ (v1-0)/6+ (v1-v2)/10 =2¹; >> e2= (v1-v2)/10 +4=2+ (v2-0)/6¹; >> s-solve (e1,e2, 'v1', 'V2'); >> [s.v1, s.v2] 1 10 + [²] = [²₂] - G+ +²)]