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(B) The power factor of an $LR$ series circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $R = 450 \Omega$,$\cos \phi

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$4 	ext{ H}$

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(B) The power factor of an $LR$ series circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $R = 450 \Omega$,$\cos \phi = 0.6$,and $f = \frac{75}{\pi} 	ext{ Hz}$.Since $\cos \phi = 0.6 = \frac{3}{5}$,we have $\frac{R}{Z} = \frac{3}{5}$.This implies $\frac{R^2}{R^2 + X_L^2} = \frac{9}{25}$.$25R^2 = 9R^2 + 9X_L^2 \implies 16R^2 = 9X_L^2$.Taking the square root,$4R = 3X_L \implies X_L = \frac{4}{3}R$.Substituting $R = 450 \Omega$,$X_L = \frac{4}{3} 	imes 450 = 600 \Omega$.We know $X_L = 2\pi f L$,so $600 = 2\pi 	imes \frac{75}{\pi} 	imes L$.$600 = 150 	imes L$.$L = \frac{600}{150} = 4 	ext{ H}$.

$A$ resistor of $450 \Omega$ and an inductor are connected in series to an ac source of frequency $\frac{75}{\pi} \text{ Hz}$. If the power factor of the circuit is $0.6$,then the inductance connected in the circuit is:

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