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

Options

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

Answer

B

Explanation

Phase difference between E and I = `theta`
`therefore tan theta = (X_(L) - X_(C))/(R)`
Now `X_(L) = 2pifL = 2pi xx 50 xx ((200)/(pi)xx10^(-3)) = 20Omega`
`X_(C) = (1)/(2pifC) = (1 pi)/(2pi xx 50 xx 10^(-3)) = 10Omega`
`R = 10Omega`
`therefore tan theta = (20-10)/(10) = (10)/(10) = 1 = tan (pi)/(4)`
`therefore theta = (pi/4)`.The current will lag by `(pi/4)`