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RC, RL, RLC circuit simulator

Pick a circuit and a source, set R, L and C: the time response, oscilloscope, Bode plot and calculations update live. For high school and university physics.

Circuit and source

Circuit diagram

Transient response

Click on a graph to place cursors A and B.

Calculations

Export:

Sinusoidal steady state: Bode plot

How to use the simulator

  1. Pick the circuit (series RC, RL, RLC, LC, parallel RLC) or a ready-made one: capacitor charging and discharging, current build-up, underdamped, critical and overdamped responses, filters, resonance, tank circuit…
  2. Set the components with the fields (choice of units) or the logarithmic sliders, plus the source's internal resistance and the coil's resistance.
  3. Pick the source: step, square, sine or triangle, with the initial conditions.
  4. Read the results: curves, oscilloscope (two channels, XY mode), energies, phase portrait, Bode plot, calculations with numbers. Cursors measure times and voltages.

How it is computed

The transient response comes from integrating the circuit's differential equations with the 4th-order Runge-Kutta method, with a step much smaller than the time constant, so curves are valid for any source. The Bode plot uses the complex impedances $\underline{Z}_R = R$, $\underline{Z}_L = r + \mathrm{j}L\omega$ and $\underline{Z}_C = 1/(\mathrm{j}C\omega)$; the gain is $G_{\mathrm{dB}} = 20\log|\underline{H}|$.

Frequently asked questions

How do I find the time constant on the curve?

The tangent at the origin meets the asymptote at $t = \tau$, when the voltage has covered 63% of its change. Tick “Tangent at the origin and τ”, or place the cursors.

Why is the voltage resonance not exactly at $f_0$?

Across the capacitor, the peak is at $f_r = f_0\sqrt{1 - 1/(2Q^2)}$, which only exists if $Q > 1/\sqrt{2}$. The current resonance (voltage across R) is always at $f_0$.

Are my settings sent anywhere?

No: everything runs in your browser. The share link contains the circuit itself.

Guided activities

1. Charging a capacitor (RC)

With R = 1 kΩ and C = 1 µF, the time constant is τ = RC = 1 ms. Under a 5 V step, the capacitor voltage reaches 63% of 5 V (3.16 V) after τ, and 99% after 5τ = 5 ms: the usual length of the transient. Find τ three ways: tangent at the origin, reading 63% with the cursors, and the displayed calculation. Double R: τ doubles.

2. The three regimes of a series RLC

With L = 10 mH and C = 1 µF, the natural frequency is f₀ = 1/(2π√(LC)) ≈ 1,592 Hz, and the critical resistance R = 2√(L/C) = 200 Ω.

The phase portrait and energy curves show energy swapping between the capacitor (electric) and the coil (magnetic), and dissipating in the resistor.

3. RC low-pass filter

With R = 1 kΩ and C = 1 µF, the cut-off frequency is f_c = 1/(2πRC) ≈ 159 Hz. On the Bode plot the gain is −3 dB there, then falls by 20 dB per decade; the phase goes from 0 to −90°, with −45° at f_c. Feed a square wave at 50 Hz then 2 kHz to the oscilloscope: at high frequency the filter “integrates” the signal into a triangle.

4. Resonance and quality factor

For a series RLC, the quality factor is Q = (1/R)√(L/C). With the values above and R = 20 Ω, Q = 5: current resonance is sharp and the bandwidth is f₀/Q ≈ 318 Hz. Increase R: the curve flattens. Compare with the capacitor voltage, whose peak shifts below f₀.

Common lab mistakes

Exporting for a lab report

Curves export as SVG or PNG, data as CSV (time domain and Bode) for a spreadsheet or Python, and calculations as LaTeX. The share link keeps the circuit: handy for setting students an exercise.

Guide written and checked by Erwan Hautot · updated 3 November 2026.