A capacitance of `((10^(-3))/(2pi))` F and an inductance of `((100)/(pi))` mH and a resistance of `10 Omega` are connected in series with an AC voltage source of 220 V, 50 Hz. The phase angle of the circuit is ``60^(@)`

Options

A. `60^(@)`
B. `30^(@)`
C. `45^(@)`
D. `90^(@)`

Answer

c

Explanation

Capacitance, `C = (10^(-3))/(2 pi) F`
Inductance, `L = ((100)/(pi)) mH = (100)/(pi) xx 10^(-3) H`
Resistance, `R = 10 Omega`
voltage of source = 220 volt
frequency of source , f = 50 Hz
Phase angle `phi` is given by
`tan phi = (x_(L) - x_(c ))/(R )`
Where, `X_(L)` = inductive impedance `= omega L`
`= 2 pi fL`
`:. X_(L) = 2 pi xx 50 xx (100)/(pi) xx 10^(-3)`
`= 10 Omega`
and `X_(C )` = capactive impedance
`= (1)/(omega C) = (1)/(2 pi//C)`
`= (1 xx 2 pi)/(2 pi xx 50 xx 10^(-5)) Omega = (100)/(5) = 20 Omega`
`:.` phase angle,
`tan phi = (X_(c ) - x_(L))/(R ) = (20 - 10)/(10) = 1` or `phi = 45^(@)`