Series RLC Resonance
The one point where the reactances cancel
The graph's horizontal axis is frequency on a logarithmic scale, and the vertical axis is current relative to its maximum. Raise the frequency from low to high: at low frequencies the capacitor blocks the way, at high frequencies the inductor does, and the current stays small. Find the resonant frequency ω_0, where the two reactances cancel exactly and the current rises highest.
Inductor and capacitor are opposites
The inductive reactance X_L = ωL grows in proportion to frequency, while the capacitive reactance X_C = 1/ωC shrinks in inverse proportion. On top of that, the two point in opposite directions in the complex plane (+j and −j). So in series the two reactances do not add but subtract, and the net reactance is X_L − X_C.
The resonant frequency ω_0 = 1/√(LC)
The resonance point is the frequency where the net reactance becomes 0, that is, where X_L = X_C. Solving ωL = 1/ωC gives ω_0 = 1/√(LC). At this frequency the imaginary part of the impedance vanishes, leaving only the resistance R, so the impedance is smallest; apply the same voltage and the current V/R flows at its maximum. The circuit opens up most widely at that frequency.
Q sets the sharpness of the peak
The smaller the resistance, the taller and narrower the resonance peak, and its sharpness is measured by the quality factor Q = ω_0L/R. A large Q gives a sharp selectivity that passes only a narrow band around the resonant frequency, while a small Q passes a broad range. Picking out the station you want on a radio is exactly this selectivity at work. This plot fixes Q at 4 and scales the peak height to 1.
Back to the first screen
At low frequencies the capacitor blocked the way and at high frequencies the inductor did, so the current was small. At ω_0 = 1/√(LC), where the two reactances become exactly equal, they cancelled, the impedance fell to the resistance R, and the current shot up into a sharp peak. What set the height and width of that peak was the quality factor Q.