Resonance in LCR circuits

Key ideas

In a series LCR circuit, the inductor and capacitor voltages are 180° apart (one leads the current by 90°, the other lags), so their phasors directly oppose.
As frequency rises, $X_L = \omega L$ grows while $X_C = \dfrac{1}{\omega C}$ shrinks. At one special frequency they are equal — and cancel completely:

$X_L = X_C \Rightarrow f_0 = \frac{1}{2\pi\sqrt{LC}}$

At this resonant frequency:
- the impedance is at its minimum — just $Z = R$
- the current is at its maximum — $I = V/R$
- the supply voltage and current are in phase
- large (equal and opposite) voltages can exist across L and C individually, even bigger than the supply — they cancel each other, not the physics.
Off resonance the circuit is reactive again: below $f_0$ the capacitor dominates ( $X_C > X_L$ ); above, the inductor dominates.
Application: a radio tuner is an LCR circuit — adjusting $C$ moves $f_0$ so only the station at that frequency drives a large current.

A graph of current against frequency for a series LCR circuit, showing the current rising to a sharp maximum at the resonant frequency f-zero, where the inductive and capacitive reactances cancel, and falling away on either side.

Worked ExampleTuning an LCR circuit

A series circuit has $L = 0.50$ H and $C = 20$ µF. Find the resonant frequency, and the rms current at resonance when the supply is $12$ V rms and $R = 60$ Ω.

Step 1 — Resonant frequency

$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{(0.50)(20\times10^{-6})}} = \frac{1}{2\pi(3.16\times10^{-3})} = 50 \text{ Hz}$

Step 2 — At resonance the reactances cancel, so $Z = R$

$I_{rms} = \frac{V_{rms}}{R} = \frac{12}{60} = 0.20 \text{ A}$

Tips

To raise $f_0$ , make $L$ or $C$ smaller ( $f_0 \propto 1/\sqrt{LC}$ ) — check your answer moves the right way before boxing it.
Explain resonance in terms of phasor cancellation ( $V_L$ and $V_C$ opposite, equal at $f_0$ ) — "the reactances cancel" plus the phasor reason is what earns Excellence.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A series LCR circuit has $L = 0.10$ H and $C = 1.0$ µF.

Show that its resonant frequency is about $500$ Hz.

Merit

A series LCR circuit with $R = 25$ Ω is driven at its resonant frequency by a $5.0$ V rms supply.

Calculate the current, and state the phase relationship between the supply voltage and the current.

Excellence

Explain why the current in a series LCR circuit peaks at the resonant frequency, referring to the phasors of the inductor and capacitor voltages and the frequency dependence of the two reactances.

Resonance in LCR circuits

Covers AS91526

Key ideas

In a series LCR circuit, the inductor and capacitor voltages are 180° apart (one leads the current by 90°, the other lags), so their phasors directly oppose.
As frequency rises, $X_L = \omega L$ grows while $X_C = \dfrac{1}{\omega C}$ shrinks. At one special frequency they are equal — and cancel completely:

$X_L = X_C \Rightarrow f_0 = \frac{1}{2\pi\sqrt{LC}}$

At this resonant frequency:
- the impedance is at its minimum — just $Z = R$
- the current is at its maximum — $I = V/R$
- the supply voltage and current are in phase
- large (equal and opposite) voltages can exist across L and C individually, even bigger than the supply — they cancel each other, not the physics.
Off resonance the circuit is reactive again: below $f_0$ the capacitor dominates ( $X_C > X_L$ ); above, the inductor dominates.
Application: a radio tuner is an LCR circuit — adjusting $C$ moves $f_0$ so only the station at that frequency drives a large current.

Worked ExampleTuning an LCR circuit

A series circuit has $L = 0.50$ H and $C = 20$ µF. Find the resonant frequency, and the rms current at resonance when the supply is $12$ V rms and $R = 60$ Ω.

Step 1 — Resonant frequency

$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{(0.50)(20\times10^{-6})}} = \frac{1}{2\pi(3.16\times10^{-3})} = 50 \text{ Hz}$

Step 2 — At resonance the reactances cancel, so $Z = R$

$I_{rms} = \frac{V_{rms}}{R} = \frac{12}{60} = 0.20 \text{ A}$

Tips

To raise $f_0$ , make $L$ or $C$ smaller ( $f_0 \propto 1/\sqrt{LC}$ ) — check your answer moves the right way before boxing it.
Explain resonance in terms of phasor cancellation ( $V_L$ and $V_C$ opposite, equal at $f_0$ ) — "the reactances cancel" plus the phasor reason is what earns Excellence.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A series LCR circuit has $L = 0.10$ H and $C = 1.0$ µF.

Show that its resonant frequency is about $500$ Hz.

Merit

A series LCR circuit with $R = 25$ Ω is driven at its resonant frequency by a $5.0$ V rms supply.

Calculate the current, and state the phase relationship between the supply voltage and the current.

Excellence

Explain why the current in a series LCR circuit peaks at the resonant frequency, referring to the phasors of the inductor and capacitor voltages and the frequency dependence of the two reactances.

Resonance in LCR circuits

Key ideas

Step 1 — Resonant frequency

Step 2 — At resonance the reactances cancel, so Z=RZ = RZ=R

Tips

Test yourself

Practice by grade

Resonance in LCR circuits

Key ideas

Step 1 — Resonant frequency

Step 2 — At resonance the reactances cancel, so Z=RZ = RZ=R

Tips

Test yourself

Practice by grade

Step 2 — At resonance the reactances cancel, so $Z = R$

Step 2 — At resonance the reactances cancel, so $Z = R$