How are J-factors used to solve RLC circuit problems

Question

10 volt Ac circuit at 60 HZ,       R= 200 Ω   Xc= 150 Ω  and XL= 300Ω

Answer 1

Impedence 

 

Z = R +j wL - j/wC

 

Where Z is impedence of circuit in Ohms

R = resistor value in Ohm

 X_L = Inductive reactance in Ohms = wL

X_{C  = Capacitive Reactance  in Ohms =wC}

+j means leads in case of Inductane and - j before X_C means , current leads the voltage

 the value of j = (-1)^1/2

and j² = -1

Z = R +jwL -j/wC or

 

   =  R + jX_{L -} j X_c  in j operator form, which means

 

Z² modulus = ( R² + (xL -XC )² = 200x200 + ( 300-200 )² = 62500

  Gives Z = 250 Ohm as Impedence 

Current = V/Z= 0.04 Amp