An inductance of `(200)/(pi) mH` a capacitance of `(10^(-3))/(pi)` and a resistance of `10 Omega` are connected in series with an `AC` source of `220 V, 50Hz`. The phase angle of the circuit is

Options

A. `pi/6`
B. `pi/4`
C. `pi/2`
D. `pi/3`

Answer

B

Explanation

Here `L = (200)/(pi) m H = (200 xx 10^(-3))/(pi) H = (0.2)/(pi) H`
`C = (10^(-3))/(pi) F, R = 10 Omega`
`V_(rms) = 220V,v = 50Hz`
The inductive reactance is
`X_L = omega L = 2pi v L = 2pi xx 50 xx (0.2)/(pi) = 20 Omega`
The capacitive reactance is
`X_C = (1)/(omega C) = (1)/(2pi v C) = (pi)/(2pi xx 50 xx 10^(-3)) = 10 Omega`
` tan phi = ((X_L- X_C))/( R) = (20-10)/(10) = 1, phi = pi/4` s