Question

A series of $\text{LCR}$ circuit is connected to an $\text{AC}$ voltage source. When $L$ is removed from the circuit, the phase difference between current and voltage is $\frac{\pi }{3}$. If instead $C$ is removed from the circuit, the phase difference is again $\frac{\pi }{3}$ between current and voltage. The power factor of the circuit is:

• A. $1.0$
• B. $-1.0$
• C. zero
• D. $0.5$

Answer 1

The correct option is A $1.0$

When $L$ is removed from $LCR$ circuit, then
$\tan \theta =\frac{X_C}{R}\Rightarrow \tan (\frac{\pi }{3})=\frac{X_C}{R}=\sqrt{3}$

$\Rightarrow X_C=\sqrt{3}R$

And when $C$ is removed then
$\tan \theta =\frac{X_L}{R}\Rightarrow \tan (\frac{\pi }{3})=\frac{X_L}{R}=\sqrt{3}$

$\Rightarrow X_L=\sqrt{3}R$

From above relation, we get

$X_L=X_C=\sqrt{3}R$


Since, $X_L=X_C$

So impedance $Z$ of the given circuit will be

$Z=\sqrt{R^2+(X_L-X_C{)}^2}=R$

Power factor of the circuit is

$\text{Power factor}=\cos \theta =\frac{R}{Z}=\frac{R}{R}=1$

Hence, option $\left(a\right)$ is correct.