Alternating Current

In a series LCR circuit driven by a source $V = V_0\sin(\omega t)$, each component opposes the changing current differently. The inductor’s opposition (inductive reactance) and the capacitor’s opposition (capacitive reactance) both depend on frequency, unlike a resistor’s R:

VL leads the current by 90°, VC lags it by 90°, so the two are 180° out of phase with each other. Combining all three voltage phasors (VR in phase with I, VL and VC opposing each other) gives the total impedance:

Z is smallest exactly when the squared term vanishes, i.e. when $X_L = X_C$. Setting $\omega L = \dfrac{1}{\omega C}$ and solving for ω gives the resonance condition:

At $\omega=\omega_0$, $Z=R$ (its minimum), so $I=V/R$ is as large as this circuit can ever drive it.

Take the exact circuit in the simulation above: L = 0.5 H, C = 2 × 10⁻⁶ F, R = 10 Ω, driven by a source of amplitude V = 10 V. Resonance sits at $\omega_0 = 1/\sqrt{LC} = 1000$ rad/s. There, $X_L = \omega_0 L = 500\ \Omega$ and $X_C = 1/(\omega_0 C) = 500\ \Omega$ exactly, so $Z = R = 10\ \Omega$ and $I = V/R = 1.0$ A.

Now compute the voltage across just the inductor: $V_L = IX_L = (1.0)(500) = 500$ V. The voltage across just the capacitor: $V_C = IX_C = (1.0)(500) = 500$ V. The source is 10 V. Each of VL and VC is fifty times that, sitting inside a circuit whose only external supply is 10 V.

This is not a contradiction of Kirchhoff’s voltage law, it’s a direct consequence of phase. VL and VC are 180° out of phase with each other (one leads the current by 90°, the other lags it by 90°), so as phasors they point in opposite directions and very nearly cancel: $V_L - V_C = 0$ at resonance exactly. The phasor sum of VR, VL and VC still equals V at every instant, it’s just that VL and VC are fighting each other, not adding onto VR, so they’re free to each individually grow far beyond the source.

The ratio of this internal voltage to the source voltage has a name: the quality factor, $Q = X_L/R = (1/R)\sqrt{L/C}$ at resonance. Here $Q = 500/10 = 50$, matching the 50× magnification seen above exactly. This is real and dangerous enough that resonant LC circuits in practice need components rated well above the nominal supply voltage, and it’s also precisely why a radio tuner, built as a high-Q LC circuit, can isolate one station from its neighbours so sharply.