3I2 + 2I1 2 (3) I4= − … From equations (1) and (2), we obtain = 2(2I3 + 4I4) + I3 I4 = 4I3 + 8I4 + I3 I4 3I3 = 9I4 − 3I4 = + (4) I3 − … Putting equation (4) in equation (1), we obtain = 2I2 + I3 I4 4I4 = 2I2 − = 2I4 (5) I2 − … It is evident from the given figure that, = + (6) I1 I3 I2 … Putting equation (6) in equation (1), we obtain 3I2 +2(I3 + 2 I2) I4= − 5I + 2I3 (7) I4= − 2 … 2 Putting equations (4) and (5) in equation (7), we obtain + 3 2 I4) I4) I4= 5(−2 2(− − 10I 6I4 2 I4= − − − 4 17I = − 2 4 Equation (4) reduces to I3 = − 3(I4) In branch BC = In branch CD = In branch AD= In branch BD = Total current =
3I2 + 2I1 − I4= 2 … (3) From equations (1) and (2), we obtain I3 = 2(2I3 + 4I4) + I4 I3 = 4I3 + 8I4 + I4 − 3I3 = 9I4 − 3I4 = + I3 … (4) Putting equation (4) in equation (1), we obtain I3 = 2I2 + I4 − 4I4 = 2I2 I2 = − 2I4 … (5) It is evident from the given figure that, I1 = I3 + I2 … (6) Putting equation (6) in equation (1), we obtain 3I2 +2(I3 + I2) − I4= 2 5I 2 + 2I3 − I4= 2 … (7) Putting equations (4) and (5) in equation (7), we obtain 5(−2 I4) + 2(− 3 I4) − I4= 2 − 10I 4 − 6I4 − I4= 2 17I 4 = − 2 Equation (4) reduces to I3 = − 3(I4) In branch BC = In branch CD = In branch AD= In branch BD = Total current =
857054491workshop_for_the_month_of_july.docx