RLC Circuit Simulator
Simulate a series RLC circuit driven by AC. See voltage, current, and impedance. Find resonance frequency by adjusting R, L, and C.
Series RLC Circuit
A series RLC circuit contains a resistor, inductor, and capacitor in series driven by an AC source. At resonance, the inductive and capacitive reactances cancel, leaving only resistance, giving maximum current.
Impedance and Reactance
Inductive reactance XL = ωL increases with frequency. Capacitive reactance XC = 1/(ωC) decreases with frequency. Total impedance Z = √(R² + (XL − XC)²). At resonance XL = XC and Z = R is minimum.
Key Variables
| Symbol | Name | Unit | Description |
|---|---|---|---|
| R | Resistance | Ω | Dissipates energy as heat |
| L | Inductance | mH | Stores energy in magnetic field |
| C | Capacitance | μF | Stores energy in electric field |
| f | Drive frequency | Hz | Frequency of the AC source |
| f₀ | Resonant frequency | Hz | f₀ = 1/(2π√LC), maximum current here |
| Z | Impedance | Ω | Total AC 'resistance': Z = √(R² + (XL-XC)²) |
| φ | Phase angle | ° | Phase difference between V and I |
| Q | Quality factor | — | Sharpness of resonance peak: Q = f₀L/(R·2π·f₀/2π) = (1/R)√(L/C) |
Phase Angle
φ = arctan((XL − XC)/R). When φ > 0 (XL > XC), voltage leads current — the circuit is inductive. When φ < 0, current leads voltage — the circuit is capacitive. At resonance φ = 0.
Key Formulas
| Formula | Description | Notes |
|---|---|---|
| XL = ωL = 2πfL | Inductive reactance | Increases linearly with f |
| XC = 1/(ωC) = 1/(2πfC) | Capacitive reactance | Decreases with f |
| Z = √(R²+(XL-XC)²) | Series impedance | Minimum at resonance (Z = R) |
| f₀ = 1/(2π√LC) | Resonant frequency | XL = XC at this frequency |
| I₀ = V₀/Z | Peak current | Maximum when Z is minimum |
| Q = (1/R)√(L/C) | Quality factor | High Q = sharp, narrow resonance |
Frequently Asked Questions
What happens at resonance in a series RLC circuit?
At resonance XL = XC, so Z = R (minimum impedance), current is maximum, and the voltage and current are in phase (φ = 0).
What is quality factor Q?
Q = f₀/(f₂-f₁) where f₁, f₂ are the half-power (−3 dB) frequencies. High Q means a sharper resonance peak and more selective frequency response. Q = (1/R)√(L/C).
Why does current lead voltage in a capacitive circuit?
For a capacitor, I = C dV/dt. If V = V₀ sin(ωt), then I = CV₀ω cos(ωt) = CV₀ω sin(ωt + 90°). Current peaks 90° before voltage — current leads. When XC > XL, the circuit is net capacitive.
What happens to impedance at very high and very low frequencies?
At very low frequencies, XC → ∞ (capacitor blocks DC), so Z → ∞ and I → 0. At very high frequencies, XL → ∞ (inductor blocks high-f AC), so Z → ∞ and I → 0. Maximum current (minimum Z) occurs at resonance.