In a series LCR circuit, the inductive reactance $left(X_{L}ight)$ is $10 Omega$ and the

Question:

In a series LCR circuit, the inductive reactance $\left(X_{L}\right)$ is $10 \Omega$ and the

capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)$ is $4 \Omega$. The resistance $(\mathrm{R})$ in the circuit is $6 \Omega$.

The power factor of the circuit is:

  1. (1) $\frac{1}{2}$

  2. (2) $\frac{1}{2 \sqrt{2}}$

  3. (3) $\frac{1}{\sqrt{2}}$

  4. (4) $\frac{\sqrt{3}}{2}$


Correct Option: 3,

Solution:

(3)

We know that power factor is $\cos \phi$,

$\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}} \ldots(1)$

$\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$

$(\omega L-1 / \omega C)$

$\Rightarrow Z=\sqrt{6^{2}+(10-4)^{2}}$

$\Rightarrow \mathrm{Z}=6 \sqrt{2} \mid \cos \phi=\frac{6}{6 \sqrt{2}}$

$\cos \phi=\frac{1}{\sqrt{2}}$

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