In a series LCR circuit, the peak current $I_{max}$ is plotted against the driving angular frequency $\omega$. The plot is a resonance (bell-shaped) curve: the vertical axis $I_{max}$ is marked in amperes at $0.5$ and $1.0$, and the horizontal axis $\omega$ is marked in rad/s at $0.5,\ 1.0,\ 1.5$ and $2.0$. Starting near zero, the current rises to a peak of about $1.0$ A close to $\omega_0\approx 1.2$ rad/s and then falls back towards zero. Find the bandwidth of this resonance curve and indicate the half-power points on it.

In a series LCR circuit, the peak current Imax is plotted against the driving angular frequency ω. The plot is

NCERT Exemplar · Q12

I_{max}

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Concept understanding — Resonance in AC Circuits

Resonance in AC Circuits

A series circuit containing a resistor $R$ , an inductor $L$ and a capacitor $C$ driven by an AC source exhibits resonance — a sharp condition at which the circuit responds most strongly.

The Competing Reactances

In a series RLC circuit the inductor and capacitor oppose the current in opposite senses. Their reactances are

X_L = \omega L, \qquad X_C = \frac{1}{\omega C}

where $\omega = 2\pi f$ is the angular frequency. As frequency rises, $X_L$ grows while $X_C$ shrinks. The total impedance is

Z = \sqrt{R^2 + (X_L - X_C)^2}

The Resonance Condition

At one special frequency the two reactances become exactly equal and cancel:

X_L = X_C \;\Rightarrow\; \omega_0 L = \frac{1}{\omega_0 C} \;\Rightarrow\; \omega_0 = \frac{1}{\sqrt{LC}}

The corresponding resonant frequency is

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

At this frequency the impedance falls to its minimum, $Z = R$ (purely resistive), so the current reaches its maximum value

I_{\max} = \frac{V_{\text{rms}}}{R}

Because the reactances cancel, the source voltage and current are exactly in phase — the power factor is 1 at resonance.

Physical Picture

At resonance energy sloshes back and forth entirely between the inductor's magnetic field and the capacitor's electric field, cycle after cycle. The source only has to make up the small amount of energy lost as heat in $R$ . This is the electrical analogue of a swing pushed at its natural frequency: a small periodic drive builds a large oscillation.

Sharpness and the Q-factor

How sharply the current peaks around $f_0$ is measured by the quality factor:

Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}

A large $Q$ (small $R$ ) gives a tall, narrow resonance curve — the circuit is highly selective, responding to a very narrow band of frequencies. A small $Q$ gives a broad, flat peak.

Why It Matters …

Why this formula?

Resonance in AC Circuits: Why the Key Formulas Hold

Resonance in an AC circuit occurs when the inductive reactance ( $X_L$ ) and capacitive reactance ( $X_C$ ) exactly cancel each other out. Let's build the understanding step-by-step.

1. The Core Condition for Resonance

Consider a series RLC circuit (resistor $R$ , inductor $L$ , capacitor $C$ ) driven by an AC voltage source $V = V_0 \sin(\omega t)$ .

The total impedance $Z$ of the series combination is:

Z = R + j(X_L - X_C)

where:

$X_L = \omega L$ (inductive reactance)
$X_C = \frac{1}{\omega C}$ (capacitive reactance)
$j = \sqrt{-1}$

Why resonance happens:

The circuit "wants" to let maximum current flow. The opposition to current comes from both resistance and reactance. But reactance can be negative (capacitive) or positive (inductive). When they are equal in magnitude but opposite in sign, they cancel:

X_L = X_C

This is the fundamental condition — not a formula to memorize, but a logical consequence of impedance minimization.

2. Deriving the Resonant Frequency

From $X_L = X_C$ :

\omega L = \frac{1}{\omega C}

Multiply both sides by $\omega$ :

\omega^2 L C = 1

Thus:

\omega_0 = \frac{1}{\sqrt{LC}}

Since $\omega = 2\pi f$ , the resonant frequency in hertz is:

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

Why this makes sense:

A larger $L$ or $C$ means the circuit takes longer to "oscillate" — lower frequency.
A smaller $L$ or $C$ means faster oscillations — higher frequency.
The product $LC$ controls the natural time scale of the circuit.

3. What Happens at Resonance — Key Consequences

(a) Impedance is Minimum (Purely Resistive)

At resonance, $X_L - X_C = 0$ , so:

Z = R + j(0) = R

Why: The reactive parts cancel, leaving only the resistance. The circuit behaves like a pure resistor.

(b) Current is Maximum

From Ohm's law for AC:

I = \frac{V}{Z}

At resonance, $Z = R$ (minimum possible), so current is maximum:

I_{\text{max}} = \frac{V}{R}

Why: The opposition to current is smallest when reactance cancels.

(c) Voltage Across L and C Can Be Very Large

The voltage across the inductor:

V_L = I \cdot X_L = \frac{V}{R} \cdot \omega_0 L

The voltage across the capacitor:

V_C = I \cdot X_C = \frac{V}{R} \cdot \frac{1}{\omega_0 C}

Since $X_L = X_C$ at resonance, $V_L = V_C$ in magnitude, but they are 180° out of phase — they cancel each other in the loop.

Why this is important:

If $R$ is small, $V_L$ and $V_C$ can be many times larger than the source voltage $V$ . This is called voltage magnification — a key concept for tuned circuits and filters.

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Bandwidth of a resonance curve is the frequency interval between the two half-power points, where the current drops to $I_{max}/\sqrt2$ . With $I_{max}=1.0$ A the half-power level is $\approx0.71$ A; the two frequencies at that level are about $1.0$ and $1.4$ rad/s, so the bandwidth is roughly $0.4$ rad/s.

Concept: bandwidth and half-power points

At resonance ( $\omega_0$ ) the series LCR current is maximum, $I_{max}$ . The half-power points $\omega_1$ and $\omega_2$ are the frequencies on either side of $\omega_0$ where the power delivered is half the maximum. Since power $\propto I^2$ , half power means

I=\frac{I_{max}}{\sqrt2}\approx0.707\,I_{max}.

The bandwidth is $\Delta\omega=\omega_2-\omega_1$ .

Reading the curve

The peak height is $I_{max}=1.0$ A, occurring near $\omega_0\approx1.2$ rad/s.
The half-power current is $\dfrac{1.0}{\sqrt2}\approx0.71$ A.
Draw a horizontal line at $I=0.71$ A. It cuts the rising side of the curve at $\omega_1\approx1.0$ rad/s (right at the printed $1.0$ gridline) and the falling side at $\omega_2\approx1.4$ rad/s (a little short of the printed $1.5$ gridline). These are the two points to mark on the graph. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

GUJCET 2026Set x1 markMCQ
$\mu$
›Reveal solutionSolution

$\omega=1/\sqrt{LC}\approx1100$ rad/s.

$LC=(27\times10^{-3})(30\times10^{-6})=8.1\times10^{-7},\qquad \sqrt{LC}=9.0\times10^{-4}.$ …
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GUJCET 2025Set 031 markMCQ
$Q. In which of the following AC circuit, we get the value of power factor 1 at resonance condition? (A) LCR series circuit (B) CR series circuit (C) Only inductor (L) circuit (D) LR series circuit$
›Reveal solutionSolution
[!TLDR]

An LCR series circuit has power factor 1 at resonance, because the reactances cancel and the impedance is purely resistive.

Concept

Power factor $\cos\phi=\dfrac{R}{Z}$ equals 1 only when the net reactance is zero. Resonance ( $X_L=X_C$ ) is defined only for a circuit containing both L and C.

Solution

In a series LCR circuit, $Z=\sqrt{R^2+(X_L-X_C)^2}$ .

At resonance $X_L=X_C$ , so $Z=R$ (purely resistive). …
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GUJCET 2024Set 131 markMCQ
$\infty$
›Reveal solutionSolution

At resonance the reactances cancel; impedance $=R$ , so power factor $=1$ . …

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GUJCET 2023Set 091 markMCQ
$\Omega$
›Reveal solutionSolution

In a series RL circuit the phase angle is $\phi=\tan^{-1}(X_L/R)$ .

Concept: Inductive reactance
$X_L=2\pi fL=2\pi(50)(25.48\times10^{-3})\approx8\ \Omega.$
Since $X_L=R=8\ \Omega$ , …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Which of the following combination should be selected for better tuning of an LCR a.c. circuit used for communication?$
(a) R = 15 ohm, L = 3.5 H, C = 30 microF
(b) R = 25 ohm, L = 2.5 H, C = 45 microF
(c) R = 20 ohm, L = 1.5 H, C = 35 microF
(d) R = 25 ohm, L = 1.5 H, C = 45 microF
›Reveal solutionSolution
Sharp tuning means high Q = (1/R) sqrt(L/C); computing Q for each set, option (a) (low R, high L/C) gives the largest value.

Good (sharp) tuning requires a high quality factor: Q = (1/R) sqrt(L/C).

Approximate Q for each:

(1/15) sqrt(3.5/30e-6) = (1/15)(342) approximately 23.

(1/25) sqrt(2.5/45e-6) = (1/25)(236) approximately 9.4.

(1/20) sqrt(1.5/35e-6) = (1/20)(207) approximately 10.4. …
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GUJCET 2022Set 171 markMCQ
$\mu$
›Reveal solutionSolution
Free (undamped) LC oscillations have angular frequency $\omega = 1/\sqrt{LC}$ .

Concept. A charged capacitor discharging through an inductor exchanges energy between the electric and magnetic fields, oscillating at the natural frequency $\omega = \dfrac{1}{\sqrt{LC}}$ .

Steps.

$L = 16\text{ mH} = 16\times10^{-3}$ H, $C = 10\ \mu\text{F} = 10\times10^{-6}$ F. …
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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. At resonance, the value of the power factor is ______.$
(a) infinity
(b) 1
(c) 0
(d) 0.5
›Reveal solutionSolution

At resonance, the reactive parts cancel and the circuit behaves as pure resistance, giving unity power factor.

Power factor is $\cos\phi = \dfrac{R}{Z}$ , where $Z = \sqrt{R^2 + (X_L - X_C)^2}$ .

…

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GUJCET 2021Set 151 markMCQ
$L = 25$
›Reveal solutionSolution

Series resonance frequency is $f_0=\dfrac{1}{2\pi\sqrt{LC}}$ .

Concept:

$LC=(25\times10^{-3})(62.5\times10^{-6})=1.5625\times10^{-6},\quad \sqrt{LC}=1.25\times10^{-3}.$ …
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GUJCET 2021Set 151 markMCQ
$L = 2$
›Reveal solutionSolution

The quality factor of a series LCR circuit is $Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ .

Concept. $Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ .

Solution. $L=2$ H, $C=18\times10^{-6}$ F, $R=10\,\Omega$ . …
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GUJCET 2020Set 071 markMCQ
$R = 3\,\Omega$
›Reveal solutionSolution

At resonance the reactances cancel and impedance equals the resistance.

Concept: For a series LCR circuit, $Z=\sqrt{R^2+(X_L-X_C)^2}$ . At resonance $X_L=X_C$ , …
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GUJCET 2019Set 131 markMCQ
$10\Omega$
›Reveal solutionSolution

The quality factor $Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}=30$ .

Concept: For a series L-C-R resonant circuit the quality factor is $Q=\dfrac{\omega_0 L}{R}=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ .

Steps: …
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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. For L-C-R A.C. circuit resonance frequency is 600 Hz and frequencies at half power points are 550 Hz and 650 Hz. What will be the Q-factor?$
(a) 1/6
(b) 1/3
(c) 6
(d) 3
›Reveal solutionSolution

Q equals resonance frequency divided by the bandwidth between half-power points: 600/(650-550) = 6.

The half-power (3 dB) points are at 550 Hz and 650 Hz, so the bandwidth is: …

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Concept: bandwidth and half-power points

Reading the curve

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