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 $$\frac{\pi }{3}.$$ If instead, $$C$$ is removed from the circuit, the phase difference is again $$\frac{\pi }{3}.$$ The power factor of the circuit is

Solution :
As we know that phase difference for $$L,C,R$$  series circuit is given by
$$\tan \phi = \frac{{{X_L} - {X_C}}}{R}$$
When $$L$$ is removed $$\tan \frac{\pi }{3} = \frac{{{X_C}}}{R}$$
$$\eqalign{ & \sqrt 3 = \frac{{{X_C}}}{R} \cr & \Rightarrow {X_C} = R\sqrt 3 \cr} $$
When $$C$$ is removed, $$\tan \frac{\pi }{3} = \sqrt 3 = \frac{{{X_L}}}{R}$$
$$ \Rightarrow {X_L} = R\sqrt 3 $$
Hence, in resonant circuit
$$\tan \phi = \frac{{\sqrt 3 R - \sqrt 3 R}}{R} = 0$$
So, $$\phi = 0$$
Power factor, $$\cos \phi = 1$$
It is the condition of resonance, therefore, phase difference between voltage and current is zero and power factor, $$\cos \phi = 1.$$