For a series LCR circuit,R = X_L = 2X_C. The | vedclass

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(B) Given: $R = X_L = 2X_C$.From the given relation,we have $X_L = R$ and $X_C = R/2$.The impedance $Z$ of a series $LCR$ circuit is given by:$Z = \sq

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$\frac{\sqrt{5} R}{2}, 	an^{-1}(1/2)$

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(B) Given: $R = X_L = 2X_C$.From the given relation,we have $X_L = R$ and $X_C = R/2$.The impedance $Z$ of a series $LCR$ circuit is given by:$Z = \sqrt{R^2 + (X_L - X_C)^2}$Substituting the values:$Z = \sqrt{R^2 + (R - R/2)^2} = \sqrt{R^2 + (R/2)^2} = \sqrt{R^2 + R^2/4} = \sqrt{5R^2/4} = \frac{\sqrt{5}R}{2}$.The phase difference $\phi$ is given by:$	an \phi = \frac{X_L - X_C}{R}$Substituting the values:$	an \phi = \frac{R - R/2}{R} = \frac{R/2}{R} = 1/2$.Therefore,$\phi = 	an^{-1}(1/2)$.Thus,the impedance is $\frac{\sqrt{5}R}{2}$ and the phase difference is $	an^{-1}(1/2)$.

For a series $LCR$ circuit,$R = X_L = 2X_C$. The impedance of the circuit and the phase difference between $V$ and $I$ respectively will be:

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