In an electrical circuit $R, L, C$ and $A C$ 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, the phase difference is again $\pi / 3$. The power factor of the circuit is

• A

  $1 / 2$
• B

  $1 \sqrt{2}$
• Correct answer
  C

  1
• D

  $\frac{\sqrt{3}}{2}$

Solution

Here, phase difference $\tan \phi=\frac{X_L-X_C}{R} \Rightarrow \tan \frac{\pi}{3}=\frac{X_L-X_C}{R}$ When $L$ is removed, $\tan \frac{\pi}{3}=\frac{X_C}{R}=\sqrt{3}$
Similarly, when $C$ is removed $\begin{aligned} \tan \frac{\pi}{3} & =\frac{X_L}{R}=\sqrt{3} \\ \Rightarrow \quad X_L & =\sqrt{3} R \end{aligned}$
$\begin{array}{lc} \text { Now, } & \tan \phi=0 \\ \Rightarrow & \phi=0^{\circ} \end{array}$ $\therefore$ Power factor, $\cos \phi=\cos 0^{\circ}=1$

Hence, the correct answer is option: 'C'.