For the network shown in the figure the value of the current i is:

Step-by-step solution

Since the network is balanced, wheatstone's bridge. Therefore no current flows through the arm AC . Here R₁ and R₂ are in series R₁ ^ =R₁+R₂=4+2=6 R₃ and R₄ are in series R^*=R₃+R₄=6+3=9 Now R^ and R^ are in parallel array l 1 R_ e q = 1 R^ + 1 R^n = 1 6 + 1 9 = 5 18 R_ e q = 16 5 Now from V =I r_ e q I = V R_ e q = 5 18 ~V array