Resonance: the frequency where a circuit stops fighting itself
How a radio picks one station out of hundreds arriving at the antenna.
Start here: pushing a child on a swing
You are pushing someone on a swing. If you push at random moments, you fight the swing as often as you help it, and it barely moves.
But push at exactly the right rhythm — a small push each time it comes back to you — and the swing goes higher and higher. You are not pushing harder. You are pushing in time.
Every swing has one natural rhythm where this works. Push at that rate and the effect builds; push at any other rate and it fights you. Circuits have exactly the same property, and the frequency where it happens is called resonance.
What is doing the "fighting" in a circuit
Two components resist alternating current, and they do it in opposite ways:
↔️ Opposite reactions
An inductorFights fast changes hardest. The higher the frequency, the more it resists.
A capacitorFights slow changes hardest. The lower the frequency, the more it resists.
One gets harder as frequency rises; the other gets easier. So somewhere in the middle there is exactly one frequency where they are equal — and because they oppose each other, they cancel completely.
At resonance, with L = 0.1 H and C = 10 µF:
frequency f₀ = 159 Hz
inductor's opposition X_L = 100 Ω
capacitor's opposition X_C = 100 Ω
They are equal and opposite → they cancel exactly.
What is left? Only the resistance.
That is the whole idea. At resonance the L and the C vanish from the circuit's point of view, and it behaves as if only the resistor were there. The current is then simply V/R — the largest it can possibly be.
Why a radio needs this
Your antenna picks up every station at once — dozens of signals arriving together. Somehow the radio must respond to one and ignore the rest.
A resonant circuit does exactly that. At its resonant frequency it passes current freely; at every other frequency the L and C no longer cancel, so they block. Tuning the dial changes C, which moves the resonant frequency, which changes which station gets through.
💡 That is literally what "tuning" means — you are changing the circuit's natural rhythm until it matches the station you want, exactly as you would change your pushing rhythm to match a different swing.
How fussy is it? The Q-factor
Some swings are fussy about rhythm and some are forgiving. Circuits are the same, and the number describing it is Q.
🎯 What Q tells you
High QVery fussy. Responds strongly to its own frequency and rejects everything else sharply. Good for a radio that must separate two nearby stations.
Low QForgiving. Responds to a broad range of frequencies. Good when you want a whole band rather than one channel.
The range it responds to is the bandwidth, and the relationship is as simple as it sounds: bandwidth = f₀ / Q. Higher Q, narrower band, fussier circuit.
💡 Resistance is what lowers Q. A resistor absorbs energy on every cycle, which is like pushing the swing while someone drags their feet — the build-up is damped. Less resistance means higher Q and a sharper peak, which is why radio tuning circuits use coils with as little resistance as possible.
An inductor's reactance rises with frequency; a capacitor's falls. Put them together and there is exactly one frequency where the two are equal and opposite — they cancel completely, and the circuit behaves as if neither were there. That frequency is resonance, and selecting it is how every tuner ever built works.
Note this is the same ω₀ that appeared in the R-L-C transient topic — a resonant circuit and a ringing transient are the same physics seen in the frequency and time domains.
Series resonance — minimum impedance
Z = R + j(X_L − X_C)
At resonance the reactances cancel:
Z = R ← MINIMUM, purely resistive
I = V/R ← MAXIMUM current
Power factor = 1 (voltage and current in phase)
Called an ACCEPTOR circuit — it accepts the resonant
frequency and passes it easily.
Parallel resonance — maximum impedance
At resonance:
Z = L/(CR) ← MAXIMUM (also called dynamic impedance)
I = MINIMUM from the source
Power factor = 1
Called a REJECTOR circuit — it blocks the resonant
frequency, presenting a huge impedance to it.
Series resonance gives minimum impedance and maximum current. Parallel resonance gives maximum impedance and minimum current. Exactly opposite, and a question will absolutely test whether you know which is which. Mnemonic: in series the current has only one path so it peaks; in parallel the two branch currents circulate against each other, so the source barely supplies any.
Q-factor and bandwidth
f₀ = 1 / (2π√(LC)) With L = 10 mH, C = 100 nF, R = 20 Ω.
Change R and watch f₀ stay exactly where it is — resistance does not move the resonant frequency, it only changes Q and therefore how sharp the peak is. Only L and C decide where resonance happens.
Quality factor (series):
Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C)
Bandwidth:
BW = f₂ − f₁ = f₀/Q (Hz)
Half-power frequencies (where I falls to 0.707 of peak,
i.e. power halves):
f₁ = f₀ − BW/2 f₂ = f₀ + BW/2 (approximately)
Voltage magnification at series resonance:
V_L = V_C = Q × V_supply ← can far exceed the supply!
High Q means a narrow bandwidth — sharper selectivity. That's exactly what a radio tuner needs: high Q to reject the station 100 kHz away. But high Q in a series circuit also means the voltage across L and C is Q times the supply voltage, which can destroy components. A Q of 100 on a 10 V supply puts 1000 V across the capacitor.
Worked numerical 1 — series resonance, full analysis
A series circuit has R = 10 Ω, L = 0.1 H, C = 10 µF, fed by 20 V. Find f₀, current at resonance, Q, bandwidth, and the voltage across L.
Resonant frequency:
f₀ = 1/(2π√(LC)) = 1/(2π√(0.1 × 10 × 10⁻⁶))
= 1/(2π√(10⁻⁶)) = 1/(2π × 10⁻³)
= 159.15 Hz
At resonance Z = R = 10 Ω, so:
I = V/R = 20/10 = 2 A ← maximum
Check the reactances cancel:
ω₀ = 2π × 159.15 = 1000 rad/s
X_L = ω₀L = 1000 × 0.1 = 100 Ω
X_C = 1/(ω₀C) = 1/(1000 × 10⁻⁵) = 100 Ω ✔ equal
Q-factor:
Q = ω₀L/R = 1000 × 0.1/10 = 10
Bandwidth:
BW = f₀/Q = 159.15/10 = 15.92 Hz
f₁ ≈ 159.15 − 7.96 = 151.2 Hz
f₂ ≈ 159.15 + 7.96 = 167.1 Hz
Voltage across the inductor:
V_L = I × X_L = 2 × 100 = 200 V
= Q × V = 10 × 20 = 200 V ✔
Note: 200 V across L from a 20 V supply — ten times the
input. This is voltage magnification, and it is real.
Worked numerical 2 — parallel resonance
A coil of R = 5 Ω and L = 0.2 H is in parallel with C = 20 µF across a 100 V supply. Find f₀ and the dynamic impedance.
f₀ = 1/(2π√(LC)) (R is small, so this approximation holds)
= 1/(2π√(0.2 × 20 × 10⁻⁶))
= 1/(2π√(4 × 10⁻⁶)) = 1/(2π × 2 × 10⁻³)
= 79.6 Hz
Dynamic (maximum) impedance:
Z = L/(CR) = 0.2/(20 × 10⁻⁶ × 5)
= 0.2/(10⁻⁴) = 2000 Ω
Source current at resonance — minimum:
I = V/Z = 100/2000 = 0.05 A = 50 mA
Compare: away from resonance the impedance is far lower
and the source current far higher. The circuit is
"rejecting" 79.6 Hz.
💡 Exam angle: an 8–10 mark favourite. The reliable marks are f₀ = 1/(2π√(LC)), then the series/parallel contrast (min Z & max I versus max Z & min I), then Q and BW = f₀/Q. Always verify X_L = X_C at your computed f₀ — it takes one line and proves your frequency is right before you build the rest of the answer on it.