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(C) The power factor of an $LCR$ circuit is given by $\cos \phi = \frac{R}{Z}$.The impedance $Z$ of the series $LCR$ circuit is calculated as:$Z = \sq

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$\frac{1}{\sqrt{2}}$

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(C) The power factor of an $LCR$ circuit is given by $\cos \phi = \frac{R}{Z}$.The impedance $Z$ of the series $LCR$ circuit is calculated as:$Z = \sqrt{R^2 + (X_L - X_C)^2}$Given values are $R = 6\, \Omega$,$X_L = 10\, \Omega$,and $X_C = 4\, \Omega$.Substituting these values:$Z = \sqrt{6^2 + (10 - 4)^2}$$Z = \sqrt{36 + 6^2}$$Z = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\, \Omega$Now,calculating the power factor:$\cos \phi = \frac{R}{Z} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$

In a series $LCR$ circuit,the inductive reactance $(X_{L})$ is $10\, \Omega$ and the capacitive reactance $(X_{C})$ is $4\, \Omega$. The resistance $(R)$ in the circuit is $6\, \Omega$. The power factor of the circuit is :

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