Question

In a series AC RLC circuit, when the inductor is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\pi /3$. If instead, the capacitor is removed from the circuit, the phase difference is again $\pi /3$. The power factor of the circuit is -

• A. $1$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $1$

If the inductor is removed, then the circuit become an AC RC circuit.

So, power factor,

$\cos \varphi =\frac{R}{Z}$

$\Rightarrow \cos \left(\pi /3\right)=\frac{R}{\sqrt{R^2+X_C^2}}$

$\Rightarrow \frac{1}{2}=\frac{R}{\sqrt{R^2+X_C^2}}$

$\Rightarrow X_C=\sqrt{3}R...\left(1\right)$

Similarly, if the capacitor is removed, then the circuit become an AC RL circuit.

So, power factor,

$\cos \varphi =\frac{R}{Z}$

$\Rightarrow \cos \left(\pi /3\right)=\frac{R}{\sqrt{R^2+X_L^2}}$

$\Rightarrow \frac{1}{2}=\frac{R}{\sqrt{R^2+X_L^2}}$

$\Rightarrow X_L=\sqrt{3}R...\left(2\right)$

Now, the power factor of the AC RLC circuit,

$\cos \varphi =\frac{R}{Z}$

$\Rightarrow \cos \varphi =\frac{R}{\sqrt{R^2+(X_C-X_L{)}^2}}$

$\Rightarrow \cos \varphi =\frac{R}{\sqrt{R^2+(\sqrt{3}R-\sqrt{3}R{)}^2}}\left[\text{From (1) and (2)}\right]$

$\Rightarrow \cos \varphi =1$

Hence, option $\left(A\right)$ is the correct answer.