MHT-CET Physics · AC Circuits

Resonance in Series LCR Circuits

At the frequency where X_L = X_C, a series LCR circuit behaves as a pure resistor: impedance is least (Z = R), current is greatest and in phase with the voltage, and f₀ = 1/2π√(LC).

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Why this matters

32 PYQs, five HARD. Three shapes: the resonant frequency and what changes it (only L and C — never R), what happens at resonance (Z least, current greatest, zero phase), and the large voltages across L and C at resonance, measured by the quality factor.

Concept 1 of 3: The Resonant Frequency

X_L rises with frequency and X_C falls; they cross at one frequency, set only by L and C. Change L or C and f₀ moves as 1/√(LC); change R and it does not move at all.

Definition

$\omega_0 = \dfrac{1}{\sqrt{LC}}$ , $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$ . Independent of $R$ .
$L \to 3L$ , $C \to 6C$ : $f \to \dfrac{f}{\sqrt{18}} = \dfrac{f}{3\sqrt{2}}$ . Keep $f_0$ when $C \to 3C$ : $L \to \dfrac{L}{3}$ .
Frequency doubled, keep resonance: $LC$ must fall 4 times — e.g. both halved.
Capacitance for resonance (or 'voltage and current in phase', 'maximum current'): $C = \dfrac{1}{\omega^2L} = \dfrac{1}{4\pi^2f^2L}$ .
Combined elements: series inductors add, parallel capacitors add — then use $L_{\text{eq}}C_{\text{eq}}$ .

Resonant frequency

f_0 = \frac{1}{2\pi\sqrt{LC}}

Worked example

L = 0.4

H and

C = 10\,\mu

F in a series circuit. Resonant frequency?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q27Moderate

Example 1 · AC Circuits · Resonance in Series LCR Circuits

A series resonant circuit consists of inductor 'L' of negligible resistance and a capacitor 'C' which produces resonant frequency 'f'. If L is changed to 3L and 'C' is changed to 6C, the resonant frequency will become.

'Increased by 3C' is 4C

'C increased BY 3C' makes 4C; 'changed TO 3C' makes 3C. The two readings land on different options,

\frac{f}{2\sqrt{3}}

and

\frac{f}{3}

Drill 13 more on the resonant frequency

Concept 2 of 3: What Happens at Resonance

At resonance the two reactances cancel, so the circuit looks like its resistor alone. Sweep the frequency upward and the impedance falls to R and rises again — so the current, and a series bulb's brightness, rise to a peak and fall.

Definition

$Z_{\min} = R$ , $I_{\max} = \dfrac{V}{R}$ , $\phi = 0$ , power factor 1.
$Z$ against $f$ : large at low $f$ ( $X_C$ ), least at $f_0$ , large again at high $f$ ( $X_L$ ).
$V_L = V_C$ and they cancel: the whole supply appears across $R$ .
A circuit with L and C can never pass more current than R alone at the same voltage: $Z \ge R$ .
PARALLEL LC at resonance: current from the source least, voltage across the pair greatest.

At resonance

X_L = X_C, \quad Z = R, \quad \phi = 0

Worked example

A series LCR circuit with

R = 40\,\Omega

is at resonance on 200 V. Impedance, current and phase?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q21Moderate

Example 2 · AC Circuits · Resonance in Series LCR Circuits

In LCR series circuit if the frequency is increased, the impedance of the circuit

Impedance zero at resonance

The reactances cancel, the resistance does not:

Z = R

, never zero. 'At resonance, impedance is zero' is a planted false statement.

Drill 10 more on what happens at resonance

Concept 3 of 3: Voltages Across L and C at Resonance, and the Quality Factor

At resonance the current is limited only by R, so it can be large, and I × X_L across the coil can be far larger than the supply voltage. The quality factor Q = X_L/R says how many times larger — and how sharp the resonance is.

Definition

$I = \dfrac{V}{R}$ ; $V_L = V_C = IX_L$ . 0.1 V, $2\,\Omega$ , $X_L = 500\,\Omega$ ⇒ 25 V across the coil, 250 times the supply.
$Q = \dfrac{V_L}{V_R} = \dfrac{\omega_0L}{R} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ ; bandwidth $= \dfrac{R}{L}$ rad/s.
From measured voltages: $L = \dfrac{V_LR}{V_R\omega}$ , $C = \dfrac{V_R}{V_L\omega R}$ .
'Ratio of energies in L and C at maximum current': the papers use $\dfrac{\frac{1}{2}LI^2}{\frac{1}{2}CV^2}$ with $V$ the applied voltage, giving $\dfrac{L}{CR^2}$ .

Quality factor

Q = \frac{V_L}{V_R} = \frac{1}{R}\sqrt{\frac{L}{C}}

Worked example

At resonance

X_L = X_C = 200\,\Omega

R = 10\,\Omega

, supply 5 V. Current, voltage across L, and Q?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift I · Q6Moderate

Example 3 · AC Circuits · Resonance in Series LCR Circuits

In series LCR resonant circuit,

R = 800\Omega

C = 2\mu\text{ }F

and voltage across resistance is 200 V . The angular frequency is

250rad/s

. At resonance, the voltage across the inductance is

Capping V_L at the supply voltage

Series resonance MAGNIFIES the voltage on L and C. 0.1 V applied can put 25 V across the coil; the answer below the supply voltage is there for students who assume otherwise.

Drill 6 more on voltages across l and c at resonance, and the quality factor

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The Resonant Frequency
Resonant frequency
$f_0 = \frac{1}{2\pi\sqrt{LC}}$
What Happens at Resonance
At resonance
$X_L = X_C, \quad Z = R, \quad \phi = 0$
Voltages Across L and C at Resonance, and the Quality Factor
Quality factor
$Q = \frac{V_L}{V_R} = \frac{1}{R}\sqrt{\frac{L}{C}}$

Watch out for (3)

'Increased by 3C' is 4C
→ The Resonant Frequency
Impedance zero at resonance
→ What Happens at Resonance
Capping V_L at the supply voltage
→ Voltages Across L and C at Resonance, and the Quality Factor

Test yourself on AC Circuits

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