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Resonance in a Series LCR Circuit

Resonance in a series LCR circuit occurs when inductive reactance equals capacitive reactance, making impedance minimum and current maximum.

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Student-friendly explanation

At resonance, XL = XC. Therefore the net reactance XL - XC becomes zero and the circuit behaves like a pure resistor. The impedance becomes Z = R, which is its minimum value, so current reaches its maximum value Imax = Vrms/R. The resonant angular frequency is omega0 = 1/sqrt(LC). A sharp resonance means current changes greatly for a small change in frequency, and this sharpness is measured using the quality factor Q.

How to write this in exams

  1. 1

    Start with the exact idea

    Resonance in a series LCR circuit occurs when inductive reactance equals capacitive reactance, making impedance minimum and current maximum.

  2. 2

    Then show how to use it

    Set XL = XC. Substitute omega L = 1/(omega C). Solve for omega0. Convert to f0 if required. At resonance, take Z = R and calculate current using RMS voltage. Add the physical meaning: current maximum and circuit purely resistive.

  3. 3

    Add one concrete example

    Radio tuning uses resonance. A receiver selects a particular frequency when its LC circuit resonates with the frequency of the desired signal.

  4. 4

    Avoid this incomplete answer

    Using omega0 = sqrt(LC) is wrong because the correct relation is the reciprocal square root, omega0 = 1/sqrt(LC).

Definition

Resonance in a series LCR circuit occurs when inductive reactance equals capacitive reactance, making impedance minimum and current maximum.

Example

Radio tuning uses resonance. A receiver selects a particular frequency when its LC circuit resonates with the frequency of the desired signal.

Rule to remember

Condition: XL = XC. Since omega L = 1/(omega C), omega0 = 1/sqrt(LC). Resonant frequency f0 = 1/(2 pi sqrt(LC)). At resonance, Z = R, phi = 0, and power factor = 1. Q-factor for a series circuit may be written as Q = omega0 L/R = 1/(omega0 C R). L is in henry, C in farad, R in ohm, omega0 in rad s^-1, and f0 in hertz.

Memory hook

At resonance, L and C cancel their reactance, leaving only R to control current.

Examples and method

Worked example

A series LCR circuit has L = 0.5 H and C = 8 microfarad. omega0 = 1/sqrt(LC) = 1/sqrt(0.5 x 8 x 10^-6) = 1/sqrt(4 x 10^-6) = 1/(2 x 10^-3) = 500 rad s^-1. If R = 50 ohm and Vrms = 100 V, current at resonance is I = V/R = 100/50 = 2 A.

Method to apply

Set XL = XC. Substitute omega L = 1/(omega C). Solve for omega0. Convert to f0 if required. At resonance, take Z = R and calculate current using RMS voltage. Add the physical meaning: current maximum and circuit purely resistive.

Diagram support

A resonance curve of current versus frequency is useful. The x-axis is frequency, the y-axis is current, and the peak occurs at f0. A sharper peak represents larger Q. If a phasor is drawn at resonance, VL and VC are equal and opposite.

How CBSE asks it

Exams ask for resonant frequency, condition for resonance, maximum current, power factor at resonance, Q-factor meaning, or interpretation of a current-frequency graph.

Avoid common mistakes

Common confusion

Students sometimes say resonance means voltage is maximum across the source. In a series LCR circuit, resonance mainly means current is maximum because impedance is minimum.

Common wrong answer

Using omega0 = sqrt(LC) is wrong because the correct relation is the reciprocal square root, omega0 = 1/sqrt(LC).

Exam tip

In resonance questions, immediately write XL = XC, Z = R, phi = 0, and omega0 = 1/sqrt(LC). These four statements cover most scoring points.

Quick check

Why is current maximum at resonance in a series LCR circuit?

Current is maximum at resonance because XL equals XC, so the net reactance becomes zero. The impedance reduces to R, its minimum value, and therefore I = Vrms/R becomes maximum.

Answer writing and exam use

1-mark answer

Resonance in a series LCR circuit occurs when inductive reactance equals capacitive reactance, making impedance minimum and current maximum.

2-mark answer

Resonance in a series LCR circuit occurs when inductive reactance equals capacitive reactance, making impedance minimum and current maximum. Condition: XL = XC. Since omega L = 1/(omega C), omega0 = 1/sqrt(LC). Resonant frequency f0 = 1/(2 pi sqrt(LC)). At resonance, Z = R, phi = 0, and power factor = 1. Q-factor for a series circuit may be written as Q = omega0 L/R = 1/(omega0 C R). L is in henry, C in farad, R in ohm, omega0 in rad s^-1, and f0 in hertz. Radio tuning uses resonance. A receiver selects a particular frequency when its LC circuit resonates with the frequency of the desired signal.

3-mark answer

At resonance, XL = XC. Therefore the net reactance XL - XC becomes zero and the circuit behaves like a pure resistor. The impedance becomes Z = R, which is its minimum value, so current reaches its maximum value Imax = Vrms/R. The resonant angular frequency is omega0 = 1/sqrt(LC). A sharp resonance means current changes greatly for a small change in frequency, and this sharpness is measured using the quality factor Q. Condition: XL = XC. Since omega L = 1/(omega C), omega0 = 1/sqrt(LC). Resonant frequency f0 = 1/(2 pi sqrt(LC)). At resonance, Z = R, phi = 0, and power factor = 1. Q-factor for a series circuit may be written as Q = omega0 L/R = 1/(omega0 C R). L is in henry, C in farad, R in ohm, omega0 in rad s^-1, and f0 in hertz. A series LCR circuit has L = 0.5 H and C = 8 microfarad. omega0 = 1/sqrt(LC) = 1/sqrt(0.5 x 8 x 10^-6) = 1/sqrt(4 x 10^-6) = 1/(2 x 10^-3) = 500 rad s^-1. If R = 50 ohm and Vrms = 100 V, current at resonance is I = V/R = 100/50 = 2 A. Exams ask for resonant frequency, condition for resonance, maximum current, power factor at resonance, Q-factor meaning, or interpretation of a current-frequency graph. Using omega0 = sqrt(LC) is wrong because the correct relation is the reciprocal square root, omega0 = 1/sqrt(LC).
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