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

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

3 Answers

kritika · Answer 1 · 2021-12-23T09:38:46+0000

Correct answer is (4) $1.0$
When $\mathrm{L}$ removed $\tan \phi=\frac{X_{C}}{R}$
When $L$ removed $\tan \phi=\frac{X_{L}}{R}$
$$
\begin{aligned}
&\frac{X_{C}}{R}=\frac{X_{L}}{R} \Rightarrow \text { Resonance } \\
&Z=R
\end{aligned}
$$
$$
\cos \phi=\frac{\mathrm{R}}{Z}=\frac{\mathrm{R}}{\mathrm{R}}=1
$$

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