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I read in my high-school book that in a RLC circuit for frequencies lower than $f_{res}$, $X_C>X_L$ and for higher frequencies $X_L>X_C$, where $X_C$ is the resistance of a capacitor and $X_L$ the resistance of the inductor. I know that $2\pi f_{res}=\omega_{res}$ and $ \omega_{res}L=\frac{1}{\omega_{res}C}$, but I do not know how to prove how is $X_L>X_C$ for $f<f_{res}$ and $X_C>X_L$ for $f>f_{res}$. I think I need calculus to prove it but I am not sure and I cannot find any proof on the internet.

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  • $\begingroup$ You don't need calculus to prove this, ypu can prove it by using $\omega_{\rm res}=\frac{1}{\sqrt{LC}}$. Also, $X_C$ and $X_L$ are the reactance of the capacitor and the inductor (respectively), not the resistance. $\endgroup$ – user258881 Jun 14 at 6:22
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if $\omega$ is the frequency of an ac signal, then the inductive and capacitive reactances are given by

$X_C = \frac{1}{C\omega}$

$X_L = L\omega$

You know that at resonant frequency, $\omega_0$, both $X_L$ and $X_C$ are equal. Now if we increase $\omega$, $X_C$ decreases and $X_L$ Increases from the same value. Opposite is true when we decrease $\omega$ from resonant value Hope this helps.

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For a series RLC circuit, the resonant frequency occurs when $X_{C}=X_{L}$. This corresponds to the minimum impedance of the circuit, equal to $R$. It is due to the fact that the impedance of the inductor and capacitor are 180$^0$ out of phase and effectively cancel each other.

At lower frequencies the capacitor impedance increases since, as you already know, its impedance varies inversely with frequency. At the same time the impedance of the inductor decreases at lower frequency since it varies linearly with frequency. For these reasons, at frequencies lower than the resonant frequency $X_{C}>X_{L}$ and at frequencies above resonant frequency $X_{L}>X_{C}$.

Hope this helps.

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