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(A) At the resonant angular frequency $\omega_0$, the capacitive reactance $X_C$ and inductive reactance $X_L$ are equal, so $X_C = X_L$.Given that $X

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$1/4$

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(A) At the resonant angular frequency $\omega_0$, the capacitive reactance $X_C$ and inductive reactance $X_L$ are equal, so $X_C = X_L$.Given that $X_C = 1/(\omega_0 C)$ and $X_L = \omega_0 L$, at $\omega_0$ we have $1/(\omega_0 C) = \omega_0 L$.When the angular frequency is doubled to $\omega' = 2\omega_0$, the new capacitive reactance is $X_C' = 1/(2\omega_0 C) = X_C / 2$.The new inductive reactance is $X_L' = (2\omega_0) L = 2 X_L$.Since $X_C = X_L$, the ratio of the new capacitive reactance to the new inductive reactance is $X_C' / X_L' = (X_C / 2) / (2 X_L) = 1/4$.

In an $LCR$ circuit, at a particular angular frequency $\omega_0$, the capacitive reactance and inductive reactance are the same. If the angular frequency is doubled to $2\omega_0$, what will be the ratio of the reactance of the capacitor to that of the inductor?

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