RLC Resonance Analyser

Set R, L, and C values to see the resonance frequency, Q factor, and impedance curve.

Starting values — RLC Resonance Analyser

Resistance R (Ω)	100
Inductance L (H)	0.01
Capacitance C (F)	0.00000253

Change any control above and the results recompute from these.

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Lesson

The theory — RLC Resonance Analyser

A series RLC circuit is at resonance at the one frequency where the inductor’s reactance and the capacitor’s reactance are equal and opposite, and so cancel exactly. At that frequency the circuit behaves as though only the resistor were there.

What each symbol means

R: resistance, in ohms. The only element that dissipates energy, and therefore the only thing that limits Q.
L: inductance, in henries. Its reactance X_L = 2πfL rises with frequency.
C: capacitance, in farads. Its reactance X_C = 1/(2πfC) falls with frequency — so the two curves cross at exactly one point.
Q: the quality factor — how sharp the resonance is. It also fixes the bandwidth, BW = f₀/Q.

Where the formula comes from

Resonance is by definition the frequency at which the two reactances match: X_L = X_C, that is 2πf₀L = 1/(2πf₀C).
Multiply both sides by 2πf₀C and divide by L, which leaves (2πf₀)² = 1/(LC).
Take the square root and divide by 2π: f₀ = 1/(2π√(LC)). Note what is absent — R does not appear at all, so resistance changes how sharp the peak is but never where it sits.

How to read what you see

Eight rows, each printed alongside the expression that produced it, so the arithmetic can be checked rather than trusted. The first four are the ones the derivation above reaches: resonant frequency f₀, angular frequency ω₀, quality factor Q = (1/R)√(L/C), and bandwidth BW = f₀/Q. At the default components — R = 100 Ω, L = 10 mH, C = 2.53 µF — they read 1.00 kHz, 6287 rad/s, 0.6287 and 1.59 kHz. The four below restate the same circuit from other angles: the two 3 dB frequencies at 483 Hz and 2.07 kHz, the damping ratio ζ = 0.7953, and the minimum series impedance — 100 Ω, which is R itself, because at resonance the reactances cancel and nothing else is left. The diagram states where the output is taken: across R.

Assumes: Ideal components in sinusoidal steady state — a pure resistance, a pure inductance and a pure capacitance, driven long enough that transients have died away. Nothing here describes what the circuit does in the first few cycles after switch-on.
Breaks when: The default values are barely resonant at all, and the readout says so if you read the two frequencies together: Q = 0.6287 gives a bandwidth of 1.59 kHz, which is wider than the 1.00 kHz centre frequency itself — there is no peak worth the name. Sharpness needs R small compared with √(L/C). Real inductors also carry their own winding resistance, which adds to R and caps the Q any physical build can reach.

Q counts how many times the energy sloshes before it becomes heat 🖖

Q is a general measure of stored energy divided by energy lost per cycle. In a series RLC circuit, low resistance lets energy shuttle many times between the inductor magnetic field and capacitor electric field before it becomes heat. The same idea appears in atomic clocks: a narrow spectral linewidth means an enormous Q, so the oscillator can distinguish extremely small frequency errors.

What resonance really means 🖖

Every inductor–capacitor pair has a natural frequency, f₀ = 1/(2π√(LC)), where energy sloshes back and forth like a child pumping a swing. Drive the circuit at f₀ and the reactances of L and C cancel, leaving only the resistance — so the current peaks sharply. Notice that f₀ depends only on L and C: turning the resistance knob sharpens or broadens the peak, but never shifts it.

Voltage that outgrows its source 🖖

At resonance the source only "sees" the resistance, yet the voltage across the inductor and the capacitor each swell to Q times the input — and, being 180° out of phase, they cancel exactly. Drive a Q = 50 circuit with 10 V and the capacitor sees 500 V. This voltage magnification is how spark-gap transmitters and Tesla coils reach huge voltages, and it can quietly destroy a capacitor rated only for the input.

Practice

Check yourself

Predict the answer first, then use the controls above to find out. Reveal only after you have committed to a guess — that is what makes it practice.

The defaults are L = 10 mH and C = 2.53 µF, giving f₀ = 1.00 kHz and Q = 0.6287. Double the inductance to 0.02 and halve the capacitance to 1.265e-6. Predict what each does.

Show answer

f₀ does not move — still 1.00 kHz — while Q doubles to 1.257 and the bandwidth halves from 1.59 kHz to 796 Hz. f₀ depends on the product LC, which you left unchanged; Q = (1/R)√(L/C) depends on the ratio, which you quadrupled. One pair of components, two independent knobs: trade L against C to sharpen the peak without moving it, or scale them together to move it without changing its shape.
Put the components back and drop the resistance from 100 to 10. Six of the eight rows change. Which two do not?

Show answer

f₀ and ω₀ — R appears nowhere in f₀ = 1/(2π√(LC)). Everything else moves: Q from 0.6287 to 6.287, bandwidth from 1.59 kHz to 159 Hz, the damping ratio from 0.7953 to 0.07953, and the two 3 dB frequencies closing in around the unmoved peak — 924 Hz and 1.08 kHz in place of 483 Hz and 2.07 kHz. The minimum series impedance follows R exactly, 100 to 10.0, because at resonance the two reactances cancel and only the resistor is left.

Problem solved in full

Resonant frequency of 100 Ω, 10 mH, 2.53 µF in series 6 steps
R = 100 Ω, L = 10 mH, C = 2.53 µF in series. Find the resonant frequency and the Q — then say what a Q below 1 actually means for this circuit.
1. The reactances of the inductor and capacitor have opposite signs, so at one frequency they cancel exactly and the impedance is purely resistive. That frequency is the definition of resonance.
  
  $Z (ω) = R + j (ω L - \frac{1}{ω C}); ℑ Z = 0 ⟹ ω_{0} = \frac{1}{L C}$
2. Substitute. The product LC is 2.53 × 10⁻⁸, and its square root is what everything else hangs on.
  
  $ω_{0} = \frac{1}{0.01 \times 2.53 \times 1 0 ^{- 6}} = \frac{1}{1.5906 \times 1 0 ^{- 4}} = 6287 rad/s, f_{0} = \frac{ω _{0}}{2 π} = 1001 Hz$
3. Q compares the reactance at resonance with the resistance. The convenient form √(L/C)/R needs no frequency at all, and ζ is just its reciprocal halved.
  
  $Q = \frac{ω _{0} L}{R} = \frac{1}{R} \frac{L}{C} = \frac{1}{100} 3953 = 0.6287, ζ = \frac{1}{2 Q} = 0.7953$
4. Bandwidth is the centre frequency divided by Q, and the two −3 dB corners are not symmetric about it — they are symmetric in the logarithm, so their geometric mean is f₀.
  
  $BW = \frac{f _{0}}{Q} = 1592 Hz; f_{1, 2} = f_{0} (1 + \frac{1}{4 Q ^{2}} \mp \frac{1}{2 Q}) = 483 and 2074 Hz$
5. Now read Q as energy rather than as bandwidth. The two definitions agree, and the energy one is what tells you whether anything will ring.
  
  $Q = 2 π \frac{energy stored}{energy lost per cycle} < 1 ⟹ it loses more each cycle than it holds$
6. Finally, work backwards from a real requirement to the resistance it implies.
  
  $AM at 1 MHz, 9 kHz spacing : Q = \frac{f _{0}}{BW} = 111 ⟹ R = \frac{1}{Q} \frac{L}{C} = 0.57 Ω$
Answer

Q = 0.63, which means the circuit loses more energy per cycle than it stores. That is the physical reading of Q: 2π times stored over lost-per-cycle. Below 1 there is nothing left to ring with, and the 1.59 kHz bandwidth around a 1.00 kHz centre says the same thing in the frequency domain — the passband is wider than the frequency it is centred on. This is a filter, not a resonator, and the resistor is why. Keep the same L and C and the same 1 kHz centre, and asking for the Q ≈ 111 that separating 9 kHz-spaced AM stations demands needs the total series resistance down to 0.57 Ω, which is less than the resistance of the coil's own wire. That is the real reason radio front ends use tuned transformers rather than a resistor, an inductor and a capacitor in a line.

$f_{0} = 1.00 kHz, Q = 0.6287, BW = 1.59 kHz ⟹ a filter, not a resonator$

Learning path

Resistance, then reactance

References (1)

Series resonance, the quality factor and bandwidth as set out in a standard circuits text: J. W. Nilsson & S. A. Riedel, Electric Circuits, 10th ed. Pearson/Prentice Hall, 2015. ISBN 978-0-13-376003-3.

By the 42Math team · Updated 2026-07-13

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

AM radio - Q = 31.62 sounds sharp until you read the bandwidth beside it: 31.8 kHz, on a 1.01 MHz carrier. American AM stations sit 10 kHz apart and European ones 9 kHz, so this circuit passes three neighbours at once. One tuned LC is not a radio — it is why receivers mix down to a fixed intermediate frequency and do the real filtering there.
Audio filter - The half-power points read 483 Hz and 2.07 kHz, and 1.00 kHz is not halfway between them — the midpoint is 1,277 Hz. It is the geometric mean: √(483 × 2,070) = 999.9, the 1.00 kHz the panel prints. A resonant peak sits at the geometric centre of its own passband, which is why filter responses are drawn on a log axis and why an octave, not a hertz, is the natural unit here.
High-Q (sharp) - High-Q circuit: sharp resonance peak at ~5 kHz
Overdamped - Q = 0.01000 and the damping ratio reads 50.00 — fifty times critically damped. The bandwidth comes to 159 kHz around a centre of 1.59 kHz, so the passband is a hundred times wider than the frequency it is nominally centred on, and the lower half-power point falls at 15.9 Hz. Nothing here resonates; R has swamped the circuit.