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(C) The power factor of an $LCR$ series circuit is given by $\cos \phi = \frac{R}{Z}$.First, we calculate the impedance $Z$ of the circuit:$Z = \sqrt{

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$6$

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(C) The power factor of an $LCR$ series circuit is given by $\cos \phi = \frac{R}{Z}$.First, we calculate the impedance $Z$ of the circuit:$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$Given $R = 60 \ \Omega$, $X_{L} = 70 \ \Omega$, and $X_{C} = 150 \ \Omega$.$Z = \sqrt{60^{2} + (70 - 150)^{2}}$$Z = \sqrt{60^{2} + (-80)^{2}}$$Z = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \ \Omega$.Now, the power factor is:$\cos \phi = \frac{R}{Z} = \frac{60}{100} = \frac{6}{10}$.Comparing this with the given power factor $\frac{\alpha}{10}$, we get $\alpha = 6$.

For the series $LCR$ circuit connected with a $220 \ V$, $50 \ Hz$ a.c. source as shown in the figure, the power factor is $\frac{\alpha}{10}$. The value of $\alpha$ is . . . . . .

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