An inductance of ( 200 ) mH , a capacitance of ( 10⁻³ ) F and a resistance of 10 are connected in series with an AC source 220 ~V , 50 ~Hz . The phase angle of the circuit is

Step-by-step solution

The phase angle ( ) between I and V is given by = X_ L -X_ C R where, X_ L =2 f L aligned &=2 50 [ 200 10⁻³ ]=20 X_ C &= 1 2 f C = 1 2 50 10⁻³ =10 aligned and R=10 Substituting values of X_ L , X_ C and R is Eq. (i), we get = 20-10 10 =1 = 4 = 4 The phase angle of the circuit is 4 .