Question

.2 K3 R39.1 k2 6 v 1.1 kn FIG. 8.120 Problems 18 and 23. $19. For the network in Fig. 8.121 a. Write the equations necessary to solve for the branch currents b. By substitution of Kirchhoff's current law, reduce the c. Rewrite the equations in a format that can be solved d. Solve for the branch current through the resistor R3. set to three equations. using third-order determinants. R3 R5 4 2 3Ω FIG. 8.121 Problems 19, 24, and 35 SECTION 8.4 Mesh Analysis (General Approach) 20. a. Using the general approach to mesh analysis, determine

Answer 1

By KVL in i_1 -10 + 2i_1 + 1i_1 - 1i_2 = 0 3i_1 - i_2 = 10 ---(1) By KVL in i_2 4i_2 + 5i_2 - 5i_3 = 0 ---(2) By KVL in i_3 3i_3 + 6 5i_3 - 5i_2 = 0 -5i_2 + 8i_3 = -6 ---(3) \r\n equation (1) 3i_1 - i_2 + 0i_3 = 10 equation (2) -i_1 + 10i_2 -5i_3 = 0 equation (3) 0i_1 -5i_2 + 8i_3 = -6 From (1), (2), & (3) equations [Third order determinant]Vertical-bar 3 - 1 0 -1 10 -5 0 -5 8 Vertical-bar [i_1 i_2 i_3] = [10 0 -6] Branch current through R_3 (i.e., i_2) i_2 = Delta_2/Delta = Vertical-bar 3 -1 0 10 0 -6 0 -5 8 Vertical-bar / Vertical-bar 3 -1 0 -1 10 -5 0 -5 8 Vertical-bari_2 = 3[-30] - 10[-8] + 0/3[55] + 1[-8] + 0 i_2 = -30 + 80/165 - 8 = 0.318A