The power factor of a CR circuit is frac{1}{s | vedclass

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(B) The power factor of a $CR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C} = \frac{1}

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$\frac{1}{\sqrt{5}}$

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(B) The power factor of a $CR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.Given $\cos \phi_1 = \frac{1}{\sqrt{2}}$,this implies $\frac{R}{Z} = \frac{1}{\sqrt{2}}$,so $R = X_{C1} = \frac{1}{2\pi f_1 C}$.When the frequency is halved,$f_2 = \frac{f_1}{2}$.The new capacitive reactance is $X_{C2} = \frac{1}{2\pi f_2 C} = \frac{1}{2\pi (f_1/2) C} = 2 X_{C1} = 2R$.The new power factor is $\cos \phi_2 = \frac{R}{\sqrt{R^2 + X_{C2}^2}} = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.

The power factor of a $CR$ circuit is $\frac{1}{\sqrt{2}}$. If the frequency of an $AC$ signal is halved,then the power factor of the circuit will become:

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