Determine the current in each branch of the network shown in fig 3.30:

Current flowing through various branches of the circuit is represented in the given figure. I1= Current flowing through the outer circuit I2= Current flowing through branch AB I3= Current flowing through branch AD I2−I4= Current flowing through branch BC I3+I4= Current flowing through branch CD I4= Current flowing through branch BD For the closed circuit ABDA, potential is zero i.e., 10I2+ 5I4− 5I3= 0 2I2+I4−I3= 0 I3= 2I2+I4… (1) For the closed circuit BCDB, potential is zero i.e., 5(I2−I4) − 10(I3+I4) − 5I4= 0 5I2+ 5I4− 10I3− 10I4− 5I4= 0 5I2− 10I3− 20I4= 0 I2= 2I3+ 4I4… (2) For the closed circuit ABCFEA, potential is zero i.e., −10 + 10 (I1) + 10(I2) + 5(I2−I4) = 0 10 = 15I2+ 10I1− 5I4 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 5I2+ 2I3−I4= 2 … (7) Putting equations (4) and (5) in equation (7), we obtain 5(−2I4) + 2(− 3I4)−I4= 2 − 10I4− 6I4−I4= 2 17I4= − 2 Equation (4) reduces to I3= − 3(I4) Therefore, current in branch In branch BC = In branch CD = In branch AD In branch BD = Total current = read less