A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance R = 80$\Omega$, an inductor of inductive reactance $\mathrm{X_L=70\Omega}$, and a capacitor of capacitive reactance $\mathrm{X_C=130\Omega}$. The power factor of circuit is $\frac{x}{10}$. The value of $x$ is :
Answer (integer)
8
Solution
$$
\begin{aligned}
& \cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2}} \\\\
& \cos \phi=\frac{80}{\sqrt{(80)^2+(60)^2}} \\\\
& \cos \phi=\frac{80}{100} \Rightarrow \frac{8}{10}
\end{aligned}
$$<br/><br/>
So, $x=8$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
This question is part of PrepWiser's free JEE Main question bank. 115 more solved questions on Alternating Current are available — start with the harder ones if your accuracy is >70%.