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(D) The power factor of an $LR$ series circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (\omega L)^2}}$.Given at $f_1 = 50 \ Hz$,$\

Vedclass · Accepted Answer

$0.5$

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(D) The power factor of an $LR$ series circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (\omega L)^2}}$.Given at $f_1 = 50 \ Hz$,$\cos \phi_1 = \frac{\sqrt{3}}{2}$.Thus,$\frac{R}{\sqrt{R^2 + (2\pi f_1 L)^2}} = \frac{\sqrt{3}}{2}$.Squaring both sides: $\frac{R^2}{R^2 + (2\pi f_1 L)^2} = \frac{3}{4}$.$4R^2 = 3R^2 + 3(2\pi f_1 L)^2$,which implies $R^2 = 3(2\pi f_1 L)^2$,so $R = \sqrt{3}(2\pi f_1 L)$.When the frequency is increased by $200 \%$,the new frequency $f_2 = f_1 + 200\% 	ext{ of } f_1 = f_1 + 2f_1 = 3f_1 = 150 \ Hz$.The new power factor is $\cos \phi_2 = \frac{R}{\sqrt{R^2 + (2\pi f_2 L)^2}} = \frac{R}{\sqrt{R^2 + (2\pi (3f_1) L)^2}}$.Substituting $R = \sqrt{3}(2\pi f_1 L)$:$\cos \phi_2 = \frac{\sqrt{3}(2\pi f_1 L)}{\sqrt{(\sqrt{3}(2\pi f_1 L))^2 + (3(2\pi f_1 L))^2}} = \frac{\sqrt{3}}{\sqrt{3 + 9}} = \frac{\sqrt{3}}{\sqrt{12}} = \frac{\sqrt{3}}{2\sqrt{3}} = 0.5$.

An inductor and a resistor are connected in series to an ac source of variable frequency. When the frequency of the applied ac is $50 \ Hz$,the power factor of the circuit is $\frac{\sqrt{3}}{2}$. If the frequency of the ac is increased by $200 \%$,the power factor of the circuit is . . . . . . .

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