A series LCR circuit is connected to an 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 :

Question

3 Answers

kritika · Answer 1 · 2021-12-29T05:56:13+0000

Solution:
When

$L$ is removed,

$\tan \phi=\frac{\left|X_{C}\right|}{R}$

$\Rightarrow \tan \frac{\pi}{3}=\frac{X_{c}}{R} \ldots(i)$
When

$C$ is removed,

$\tan \phi=\frac{\left|X_{L}\right|}{R}$

$\Rightarrow \tan \frac{\pi}{3}=\frac{X_{L}}{R} \ldots$

$\Rightarrow \tan \frac{\pi}{3}=\frac{X_{L}}{R} \ldots$ From

$(i)$ and

$(i i)$ ,
Since

$X_{L}=X_{C}$ , the circuit is in resonance

$Z=R$
Power factor

$=\cos \phi=\frac{R}{Z}=1$