JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 24)
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
8
Explanation
$$
\begin{aligned}
& \cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2}} \\\\
& \cos \phi=\frac{80}{\sqrt{(80)^2+(60)^2}} \\\\
& \cos \phi=\frac{80}{100} \Rightarrow \frac{8}{10}
\end{aligned}
$$
So, $x=8$
So, $x=8$
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