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RLC Circuit Calculator

Solve rlc circuit problems with clear steps, notation, and a final check.

Ω

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RLC Circuit Calculator explained

The short version

  • An RLC circuit has a resistor, an inductor and a capacitor, and it can ring like a struck bell.
  • The resonant frequency f₀ = 1/(2π√(LC)) depends only on L and C. The resistor decides how fast the ringing dies.
  • Compare R against 2√(L/C): below it the circuit rings, above it the circuit just sags back.

The formula this page uses

f₀ = 1 / (2π√(LC)) ζ = R / (2√(L/C)) Q = (1/R)·√(L/C) bandwidth = f₀ / Q

What each part means

SymbolWhat it means
R — ResistanceIn ohms. It is the only part that removes energy, so it controls the damping.
L — InductanceIn henries. 100 mH is 0.1 H.
C — CapacitanceIn farads. 1 µF is 0.000001 F.
ζ — Damping ratioA plain number. Below 1 the circuit oscillates, at 1 it is critically damped, above 1 it is overdamped.

Show your work: a full example

  1. The circuita series RLC with R = 100 Ω, L = 0.1 H and C = 1 µF
  2. Product of L and C0.1 × 0.000001 = 0.0000001, and √0.0000001 = 0.0003162
  3. Resonant frequencyf₀ = 1 ÷ (2π × 0.0003162) = 503.3 Hz
  4. Critical resistance2√(L/C) = 2√(0.1 ÷ 0.000001) = 2√100000 = 2 × 316.23 = 632.46 Ω
  5. CompareR = 100 Ω is well below 632.46 Ω, so the circuit is underdamped and will ring
  6. Quality factorQ = 316.23 ÷ 100 = 3.162, and the damping ratio is ζ = 100 ÷ 632.46 = 0.158
  7. Bandwidth of the resonance peak503.3 ÷ 3.162 = 159.2 Hz wide

A second, different case

  1. A different case: keep L and C, raise the resistanceR = 2000 Ω, L = 0.1 H, C = 1 µF
  2. f₀ is unchangedstill 503.3 Hz, because R never appears in that formula
  3. Critical resistance is unchanged toostill 632.46 Ω
  4. Compare2000 Ω is above 632.46 Ω, so this circuit is overdamped
  5. Damping ratioζ = 2000 ÷ 632.46 = 3.162, more than three times critical
  6. Quality factorQ = 316.23 ÷ 2000 = 0.158, so there is no ringing at all
  7. Set R exactly to 632.46 Ω insteadζ = 1 exactly, which is critical damping: the fastest return with no overshoot
Copy-ready example

a series RLC with R = 100 Ω, L = 0.1 H and C = 1 µF

The circuit

Damping in a series RLC with L = 0.1 H and C = 1 µF, so f₀ = 503.3 Hz and the critical resistance is 632.46 Ω

Rζ = R / 632.46QNameWhat you would see
0 Ω (ideal)0infiniteundampedrings forever at 503.3 Hz
100 Ω0.1583.162underdampedseveral visible cycles, then settles
400 Ω0.6320.791underdampedone overshoot, then settles
632.46 Ω1.0000.5critically dampedfastest settle, no overshoot at all
2000 Ω3.1620.158overdampedslow sag back, no overshoot

Three mistakes to check for

What students writeWhy it's wrongDo this instead
f₀ = 1/(2π√(LC)) with L = 100 and C = 1Millihenries and microfarads were left unconverted, which shifts the answer by many orders of magnitude.Use base units: L = 0.1 H and C = 0.000001 F, giving 503.3 Hz.
Expecting R to change the resonant frequencyR appears in the damping and the bandwidth, never in f₀ itself.Both worked examples above resonate at 503.3 Hz even though one has 100 Ω and the other 2000 Ω.
Assuming a bigger R means more ringingR is the loss term, so more of it means the energy drains faster and the ringing dies sooner.At 100 Ω the circuit rings; at 2000 Ω it does not oscillate at all.

Questions about the RLC Circuit Calculator

What does resonance actually mean here?

At f₀ the inductor's reactance and the capacitor's reactance are equal and opposite, so they cancel. A series circuit is then left with only R, which is why its current peaks sharply at 503.3 Hz.

What is Q in plain words?

It is how sharp and how long-lived the ringing is. Q = 3.162 means the resonance peak is 159 Hz wide out of 503 Hz, and the circuit rings for roughly three cycles before fading.

Why is critical damping useful if it does not ring?

Because it is the fastest possible return to rest without ever overshooting. Car suspensions and instrument needles are tuned near ζ = 1 for exactly that reason: you want the bump gone quickly, not bouncing.

How does a parallel RLC differ from a series one?

The resonant frequency formula is identical, but the roles of R invert. In series, a small R gives a high Q; in parallel, a large R gives a high Q, because the resistor is now a leak across the tank rather than a brake in the loop.

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