Problem — Reactances of an RLC as omega varies

Problem

In a series RLC circuit driven at angular frequency $\omega$, explain why, as $\omega$ increases, the inductive reactance $X_L$ grows while the capacitive reactance $X_C$ decreases. What is the circuit’s behaviour in the limits $\omega\to 0$ and $\omega\to\infty$?

Solution

The two reactances depend on the angular frequency in opposite ways: $X_L=\omega L,\qquad X_C=\frac{1}{\omega C}$

Inductive reactance $X_L=\omega L$. Grows linearly with $\omega$. The inductor opposes changes in current: the faster the oscillations (high $\omega$), the stronger the self-induced emf that opposes them. At high frequencies the inductor tends to block the current.

Capacitive reactance $X_C=1/(\omega C)$. Decreases with $\omega$. The capacitor charges and discharges: at low frequency it has time to charge fully and blocks the current (in DC it is an open circuit); at high frequency it charges and discharges so quickly that it lets the current pass almost freely.

Limit $\omega\to 0$ (DC): $X_C=\frac{1}{\omega C}\to\infty,\qquad X_L=\omega L\to 0$ $X_C$ and $R$ dominate: the capacitor behaves as an open circuit and blocks the current. The current is limited by $C$ (and by $R$).

Limit $\omega\to\infty$ (high frequencies): $X_L=\omega L\to\infty,\qquad X_C=\frac{1}{\omega C}\to 0$ $X_L$ dominates: the inductor behaves as an open circuit and blocks the current.

In between, at the resonant angular frequency $\omega_0=1/\sqrt{LC}$ we have $X_L=X_C$: the two reactances cancel, the impedance reduces to $R$ alone, and the current is maximum.

$\ev{\omega\to 0:\ \text{blocks } C;\quad \omega\to\infty:\ \text{blocks } L;\quad \omega_0=1/\sqrt{LC}:\ X_L=X_C}$

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Topics: Electromagnetic induction Concepts: Alternating current · Resonance