In a series LR circuit with X_L = R, the powe | vedclass

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Question

(A) The power factor in an $AC$ circuit is given by $\cos \phi = \frac{R}{Z}$.For the initial $LR$ circuit, $Z_1 = \sqrt{R^2 + X_L^2}$. Given $X_L = R

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$1 : \sqrt{2}$

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(A) The power factor in an $AC$ circuit is given by $\cos \phi = \frac{R}{Z}$.For the initial $LR$ circuit, $Z_1 = \sqrt{R^2 + X_L^2}$. Given $X_L = R$, we have $Z_1 = \sqrt{R^2 + R^2} = R\sqrt{2}$.Thus, $P_1 = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$.When a capacitor with $X_C = X_L$ is added in series, the circuit becomes an $LCR$ circuit at resonance.At resonance, $X_L - X_C = 0$, so the impedance $Z_2 = \sqrt{R^2 + (X_L - X_C)^2} = R$.The new power factor $P_2 = \frac{R}{Z_2} = \frac{R}{R} = 1$.The ratio $\frac{P_1}{P_2} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}$.

In a series $LR$ circuit with $X_L = R$, the power factor is $P_1$. If a capacitor of capacitance $C$ with $X_C = X_L$ is added to the circuit, the power factor becomes $P_2$. The ratio of $P_1$ to $P_2$ will be:

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