Reactances and impedance

Under the square root of the current formula there appears a combination of $R$, $\omega L$ and $1/(\omega C)$. These three terms all have the same dimensions (ohms) and play analogous roles: they define how much each element opposes the passage of an alternating current of angular frequency $\omega$. We call them resistance $R$, inductive reactance and capacitive reactance:

Reactances

$\ev{X_L = \omega L \qquad X_C = \frac{1}{\omega C}}$

These are “pseudo-resistances” that depend on frequency: $X_L$ grows with $\omega$ (an inductor blocks high frequencies), $X_C$ decreases with $\omega$ (a capacitor blocks direct current and lets high frequency through).

For a series RLC circuit fed by a sinusoidal generator, the impedance $Z$ is defined as the ratio between the voltage and current amplitudes:

Series RLC impedance

$\ev{Z = \frac{V_0}{I_0} = \sqrt{R^2 + (X_L - X_C)^2}}$

measured in ohms. Ohm’s law extends to alternating current in the form $V_0 = Z\,I_0$ (for RMS values $V_\text{eff} = Z\,I_\text{eff}$). The phase shift $\varphi$ between voltage and current is

$\tan\varphi = \frac{X_L - X_C}{R}$

and is positive when the inductor dominates (current lagging behind voltage) and negative when the capacitor dominates (current leading voltage). At resonance $X_L = X_C$: the reactances cancel, the impedance reduces to $R$ alone, and $\varphi = 0$.

Key formula

$Z = \sqrt{R^2 + (X_L - X_C)^2}, \qquad \text{resonance: } X_L = X_C \Rightarrow \omega_0 = \frac{1}{\sqrt{LC}},\ Z_{\min} = R$

Example — RLC on the domestic mains

A circuit with $R = 50\;\Omega$, $L = 0{,}10$ H and $C = 20\;\mu\text{F}$ is fed from the Italian mains ($V_\text{eff} = 230$ V, $f = 50$ Hz, $\omega = 2\pi\cdot 50 \approx 314$ rad/s).

• $X_L = \omega L = 314\cdot 0{,}10 \approx 31{,}4\;\Omega$.
• $X_C = 1/(\omega C) = 1/(314\cdot 2\cdot 10^{-5}) \approx 159\;\Omega$.
• $Z = \sqrt{50^2 + (31{,}4 - 159)^2} \approx \sqrt{2500 + 16\,280} \approx 137\;\Omega$.
• $I_\text{eff} = V_\text{eff}/Z = 230/137 \approx 1{,}68$ A.
• $\tan\varphi = (31{,}4 - 159)/50 \approx -2{,}55$, so $\varphi \approx -68^\circ$ (leading current: $C$ dominates).

The resonance frequency would be $f_0 = 1/(2\pi\sqrt{LC}) \approx 113$ Hz, well above the mains’ 50 Hz: this is why at 50 Hz the circuit is “capacitive”.

Collegamenti

Argomenti: Electromagnetic induction Concetti: Alternating current · Resonance · Ohm’s law

Esercizi collegati: Problem — Reactances of an RLC as omega varies · Problem — Impedance and phase shift of a series RLC · Problem — Ranking RMS current in RLC