With L = 10 mH and C = 10 μF, what is the resonant frequency?

F₀ = 1/(2π√(LC)) = 1/(2π√(0.01 × 10⁻⁵)) ≈ 503 Hz.

At resonance with R = 50 Ω, L = 10 mH, C = 10 μF, what is the quality factor Q?

Q = (1/R)√(L/C) = (1/50)√(0.01/10⁻⁵) = (1/50)(31.6) ≈ 0.63.

What happens to the resonant frequency if you double both L and C?

F₀ = 1/(2π√(LC)). Doubling both: f₀' = 1/(2π√(4LC)) = f₀/2. It halves.

Pro 🔒~25 min

RLC Circuit

Resonance, impedance, and transient response in AC circuits

How it works

An RLC circuit consists of a resistor R, inductor L, and capacitor C in series with an AC voltage source V₀sin(ωt). The impedance Z determines the current amplitude I₀ = V₀/Z. At resonance (ω₀ = 1/√(LC)), the inductive reactance XL = ωL exactly cancels the capacitive reactance XC = 1/(ωC), leaving Z = R (minimum impedance, maximum current). The phase angle φ = arctan((XL - XC)/R) indicates whether current leads or lags voltage. The quality factor Q = ω₀L/R = 1/(R√(C/L)) measures the sharpness of the resonance peak. In transient response (no driving source), the circuit exhibits damped oscillations. The damping ratio ζ = R/(2√(L/C)) determines the behavior: underdamped (ζ < 1, oscillating decay), critically damped (ζ = 1, fastest non-oscillating decay), or overdamped (ζ > 1, slow exponential decay).

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Step-by-step

Set R, L, and C values, then adjust the driving frequency.
Watch the current and voltage waveforms in real time.
The impedance curve shows the resonance peak — try to match the driving frequency to f₀ for maximum current.
The phasor diagram shows the phase relationship between V and I.
Toggle to transient mode to see damped oscillations.

Key formulas

$Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$ Impedance of a series RLC circuit: depends on R, inductive reactance ωL, and capacitive reactance 1/(ωC)
$f_0 = \frac{1}{2\pi\sqrt{LC}}$ Resonant frequency: where inductive and capacitive reactances cancel (XL = XC)
$Q = \frac{1}{R}\sqrt{\frac{L}{C}}$ Quality factor: higher Q means sharper resonance peak and lower energy loss per cycle

Frequently asked questions

With L = 10 mH and C = 10 μF, what is the resonant frequency?: F₀ = 1/(2π√(LC)) = 1/(2π√(0.01 × 10⁻⁵)) ≈ 503 Hz.
At resonance with R = 50 Ω, L = 10 mH, C = 10 μF, what is the quality factor Q?: Q = (1/R)√(L/C) = (1/50)√(0.01/10⁻⁵) = (1/50)(31.6) ≈ 0.63.
What happens to the resonant frequency if you double both L and C?: F₀ = 1/(2π√(LC)). Doubling both: f₀' = 1/(2π√(4LC)) = f₀/2. It halves.