Question

Question 2 Using Superposition theorem, solve for the current through R3 in the Figure below. ls1 = 100 mA, Vs1 = 28 R1 = 680 Ω, R2 = 220 Ω, R3 = 330 Ω, R4-494 Ω, Enter positive for down direction and negative for up direction. Enter your results in mA. R3 R2 Is1 R1 R4

Answer 1

Case (1): When current source (I_a_1) is only Active we have to replace voltage circuit as short circuit By applying KCL at node 'v' we get V/680 + v/220 + v/824 = 100 m A 997 v/77044 = 100 times 10^- 3 therefore V = 13.832 v i_I = V/824 = 13.832/824 = 0.01678 Therefore i_I = 16.786 m A Case (ii): - When voltage source (V_S_1) is only active we have to replace current as open circuit. By applying KCL at node 'V_a' we get V_a/680 + v_a + 28/220 + V_a/824 = 0 (7.23 times 10^- 3) V_a = -0.127 V_a = -17.603 V therefore i_v = V_a/824 = -17.603/824 = -0.02136 therefore i_v = -21.36 m A i = i_I + i_v = 16.786 m A + (- 21.36 m A) = -4.577 times 10^- 3 = -4.577 m A