Question

A series $LCR$ circuit is connected to an ac voltage source. When $L$ is removed from circuit, the phase difference between current and voltage is $\frac{\pi }{4}$. If instead of $L,C$ is removed from the circuit, the phase difference between current and voltage is again $\frac{\pi }{4}$. The power factor of the original circuit is

• A. $0.707$
• B. $0.5$
• C. $1.0$
• D. $0.785$

Answer 1

The correct option is C $1.0$

When $L$ is removed, phasor will be as follows


$\frac{I{X}_C}{IR}=\tan \left(\frac{\pi }{4}\right)$
$\therefore X_C=R$

When $C$ is removed, phasor will be as follows

$\frac{I{X}_c}{IR}=\tan \left(\frac{\pi }{4}\right)\therefore X_L=R$

$X_L=X_C$

Hence, the given circuit is in resonance.
$\therefore$Power factor of $LCR$ circuit$=\cos 0^{\circ }=1$