Find the impedance of the circuit shown SolutionIts a series

Find the impedance of the circuit shown.

Its a series circuit

Apply KCL in ckt : v = i/(j3) + i/j5 + i(2) +i(j4) +i/j7

= i( 1/j3 +1/5j +2 +4j +1/7j )

V/i = Z = 1/j3 +1/5j +2 +4j +1/7j

= j/3j^2 + j/5j^2 +4j + j/7j^2 +2

= 2 - j/5 +4j -j/7 -j/3

= 2 + j ( -1/5 +4 -1/7 -1/3)

Z = 2 + j3.32

Z= 3.87<58.93deg