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(C) The power factor of an $L-R$ series circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $\cos \phi = \frac{\sqrt{3}

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$\frac{1}{2 \sqrt{3} \pi} \ H$

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(C) The power factor of an $L-R$ series circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $\cos \phi = \frac{\sqrt{3}}{2}$,this implies $\phi = \frac{\pi}{6}$.The phase angle is also given by $	an \phi = \frac{X_L}{R}$.Since $	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$,we have $\frac{X_L}{R} = \frac{1}{\sqrt{3}}$.Substituting $R = 50 \ \Omega$,we get $X_L = \frac{50}{\sqrt{3}} \ \Omega$.We know $X_L = 2 \pi f L$. Given $f = 50 \ Hz$,we have $\frac{50}{\sqrt{3}} = 2 \pi (50) L$.Solving for $L$: $L = \frac{50}{\sqrt{3} 	imes 100 \pi} = \frac{1}{2 \sqrt{3} \pi} \ H$.

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