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Series Resonance

CI

CircuitSim

·

Created Oct 3, 2026

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Updated Oct 3, 2026

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Details

Components

4


Schematic

Yes


Simulation

AC sweep


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About this circuit

Sweep a series RLC circuit through resonance. Inductive and capacitive reactance cancel, maximizing current and the resistor voltage.

Circuit notes

Series Resonance

Sweep a series RLC circuit through resonance. Inductive and capacitive reactance cancel, maximizing current and the resistor voltage. L1 is 10 mH and C1 is approximately 2.533 µF, setting the resonant frequency to 1 kHz. R1 = 10 Ω represents series loss and makes the resonance finite.

Resonance condition

f0=12πLC=1 kHzXL(f0)=∣XC(f0)∣≈62.83 Ω\begin{aligned} f_0&=\frac{1}{2\pi\sqrt{LC}}=1\,\mathrm{kHz}\\ X_L(f_0)&=|X_C(f_0)|\approx62.83\,\Omega \end{aligned}

The same current passes through L1, C1 and R1. Their reactive voltage drops cancel at resonance, leaving only the resistor in the total impedance.

Z=R+j(ωL−1ωC)I(f0)=1 V10 Ω=100 mAVR(f0)=1 VQ=ω0LR≈6.28\begin{aligned} Z&=R+j\left(\omega L-\frac{1}{\omega C}\right)\\ I(f_0)&=\frac{1\,\mathrm V}{10\,\Omega}=100\,\mathrm{mA}\\ V_R(f_0)&=1\,\mathrm V\\ Q&=\frac{\omega_0L}{R}\approx6.28 \end{aligned}

Read the circuit and response

V1 supplies AC analysis magnitude = 1 V. The green VR probe reads the grounded resistor voltage, and the blue I probe measures series current.

The example opens in AC sweep, from 100 Hz to 10 kHz, with 100 points per decade and a logarithmic frequency axis. The chart shows magnitude and phase. Magnitude in dB is 20log⁡10(∣V∣/1 V)20\log_{10}(|V|/1\,\mathrm V) for voltage or 20log⁡10(∣I∣/1 A)20\log_{10}(|I|/1\,\mathrm A) for current; a 1 V voltage peak is 0 dB. These current dB values are referenced to one ampere, not voltage gain. Open Table to inspect numeric frequency samples, including 1 kHz.

At resonance VR is 1 V and I is 100 mA. Their phases are zero relative to the source. Below resonance the circuit is capacitive; above resonance it is inductive. AC sweep uses the source’s AC analysis magnitude/phase fields. The transient waveform Frequency and Phase fields do not control this sweep.

Try two changes

  1. Double C1 → Capacitance to approximately 5.066 µF (5.0661 µF). Resonance moves down to f0/2≈707.1 Hzf_0/\sqrt{2}\approx707.1\,\mathrm{Hz}. Restore C1 to 2.533 µF.
  2. Double R1 → Resistance to 20 Ω. The peak current halves to 50 mA, Q halves and the response broadens. The VR peak remains 1 V at resonance. Restore R1 to 10 Ω afterward.

Model assumptions

This is small-signal sinusoidal steady-state analysis of ideal linear L and C with explicit resistance. It does not show switch-on ringing, saturation, breakdown, winding/ESR frequency dependence or parasitic resonances. Series resonance minimizes driving-point impedance.

Components used

Components used
4 components
NameManufacturerMPNCountType

AC Voltage Source

Generic

—

1

Official

Capacitor

Generic

—

1

Official

Inductor

Generic

—

1

Official

Resistor

Generic

—

1

Official

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