In an electrical circuit `R,L,C` and an `AC` voltage source are all connected in series. When `L` is removed from the circuit, the phase difference between the voltage and the current in the circuit is `pi//3`. If instead, `C` is removed from the circuit, difference the phase difference is again `pi//3`. The power factor of the circuit is

Options

A. `1/2`
B. `(1)/(sqrt2)`
C. 1
D. `(sqrt3)/(2)`

Answer

C

Explanation

When L is removed , the phase difference between the voltage and current is
`tan phi_1 = (X_C)/(R ) , tan pi/3 = (X_C)/( R) " or " R tan 60^@ " or " X_C = sqrt3 R`
When C is removed, the phase difference between the voltage and current is
`tan phi_2 = (X_L)/( R)`
`tan pi/3 = (X_L)/( R) " or " X_L = R tan 60^@ = sqrt3 R`
As `X_L = X_C` ,the seris LCR circuit is in resonance . Impedance of the circuit,
` Z = sqrt(R^2 + (X_L - X_C)^2) = R " " (because X_L = X_C)`
Power factor ,`cos phi = R/Z = R/R = 1`