JEE Mains Physics · Alternating Current

Resonance, Quality Factor and Bandwidth

A series LCR circuit resonates when its two reactances are equal, at ω₀ = 1/√(LC); there the impedance is just R, the current and power are largest, and the quality factor ω₀L/R sets how sharp the peak is.

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Why this matters

Thirty-one PYQs, fifteen of them asking for a number, the highest share in the chapter. Sixteen find the resonant frequency or the L or C that produces it. Ten ask what the current, impedance and power do at resonance. Five use the quality factor or the bandwidth. Most are one formula and a careful power of ten.

Concept 1 of 3: The resonant frequency

As the frequency rises, the inductor's reactance climbs and the capacitor's falls. At one frequency they are equal and cancel. That is resonance. Setting ωL = 1/ωC gives ω₀ = 1/√(LC), and the resistance plays no part in it. Questions rarely say 'resonance'; they say maximum current, current in phase with the emf, maximum power or minimum impedance, and all of these mean the same thing.

Definition

Resonance: $X_L = X_C$ , so $\omega_0 = \dfrac{1}{\sqrt{LC}}$ and $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$ . R does not appear.
Signs of resonance: maximum current, maximum average power, minimum impedance (Z = R), current in phase with the emf, power factor 1.
Missing part: $L = \dfrac{1}{\omega_0^{2}C}$ or $C = \dfrac{1}{\omega_0^{2}L}$ . If $X_L$ at the source frequency is given, the capacitor must have $X_C = X_L$ , so $C = \dfrac{1}{\omega X_L}$ .
At resonance each reactance equals $\omega_0L = \sqrt{\dfrac{L}{C}}$ .
To raise $f_0$ , make LC smaller. A capacitor added in series lowers the total C. To keep $f_0$ fixed, LC must stay fixed.
A long line spreads its capacitance: total C = (capacitance per km) × length.
Two tests on the same circuit: removing C and finding a lag of $45^{\circ}$ gives $X_L = R$ ; removing L and finding a lead of $45^{\circ}$ gives $X_C = R$ . If both hold, $X_L = X_C$ and the full circuit is at resonance.

Resonant frequency

\omega_0 = \frac{1}{\sqrt{LC}} \qquad f_0 = \frac{1}{2\pi\sqrt{LC}}

Worked example

A series circuit has L = 0.5 H, C =

8\ \mu\text{F}

and R =

10\ \Omega

. Find

\omega_0

f_0

and the reactance of each part at resonance.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q23Moderate

Example 1 · Alternating Current · Resonance, Quality Factor and Bandwidth

A series LCR circuit with R

= 20\Omega,\text{ }L = 1.6H

and

C = 40\mu\text{ }F

is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is

\_\_\_\_

\Omega

ω or f

\dfrac{1}{\sqrt{LC}}

is ω in rad/s. For hertz, divide by 2π. Check which one the blank asks for.

Putting R into the formula

The resistance changes how tall and wide the resonance peak is, never where it sits.

Losing a power of ten

Convert first:

1\ \mu\text{F} = 10^{-6}\ \text{F}

1\ \text{nF} = 10^{-9}\ \text{F}

1\ \text{mH} = 10^{-3}\ \text{H}

. Most wrong answers here are off by a power of ten.

Drill 15 more on the resonant frequency

Concept 2 of 3: Current, impedance and power at resonance

At resonance the reactances cancel, so the circuit behaves as if only R were there. The current is V/R, it is in step with the voltage, and the power is V²/R. L and C decide where resonance happens; only R decides how big the current gets. Below resonance the capacitor's reactance is larger, so the circuit is capacitive; above it, inductive.

Definition

At resonance $Z = R$ : $I_0 = \dfrac{V_0}{R}$ , $I_{rms} = \dfrac{V_{rms}}{R}$ , $\cos\phi = 1$ , $P = \dfrac{V_{rms}^{2}}{R}$ .
The resonant current goes as $\dfrac{1}{R}$ : halving R doubles it. Changing L or C (with LC kept fixed) does not change it.
Current against ω: it rises, peaks at $\omega_0$ and falls. Left of $\omega_0$ : capacitive ( $X_C > X_L$ ), current leads. Right of it: inductive, current lags.
The same R alone across the same supply draws V/R. A series LCR with that R can only match it, at resonance; anywhere else it draws less.
At resonance $V_L = V_C = IX_L$ . These can be far larger than the supply voltage; they cancel each other.
Ideal L and C in PARALLEL at resonance: infinite impedance, so the line current is zero.
Resonance needs both an L and a C. A circuit with only one of them cannot resonate.

At resonance

Z = R \qquad I_0 = \frac{V_0}{R} \qquad P_{max} = \frac{V_{rms}^{2}}{R}

Worked example

R =

25\ \Omega

, L = 0.2 H and C =

20\ \mu\text{F}

are in series with a 150 V (rms) variable-frequency supply. At resonance find the rms and peak current, the rms voltage across L, and the power.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q25Moderate

Example 2 · Alternating Current · Resonance, Quality Factor and Bandwidth

An inductor of

0.5mH

, a capacitor of

20\mu F

and resistance of

20\Omega

are connected in series with a

220\text{ }V

ac source. If the current is in phase with the emf, the amplitude of current of the circuit is

\sqrt{x}

A. The value of

x

is-

Forgetting √2 on the amplitude

A supply quoted as '220 V' is rms. At resonance V/R is then the rms current; the amplitude is √2 times that.

Using L and C to find the resonant current

Once the circuit is at resonance, L and C have cancelled. The current depends only on V and R.

Reading the I–ω curve the wrong way round

On the low-frequency side

X_C

is large, so the circuit is capacitive. On the high side

X_L

wins.

Drill 9 more on current, impedance and power at resonance

Concept 3 of 3: Quality factor and bandwidth

The resonance peak can be sharp or broad. Its width is measured between the two frequencies where the current falls to 1/√2 of its peak, which is where the power halves. That width, the bandwidth, is R/L. The quality factor compares the resonant frequency with the width: a small R gives a narrow, tall peak and a large Q.

Definition

$Q = \dfrac{\omega_0L}{R} = \dfrac{1}{\omega_0CR} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ . It has no unit.
Half-power frequencies $\omega_1, \omega_2$ : the current is $\dfrac{I_{max}}{\sqrt{2}}$ and the power is half the peak.
Bandwidth $\Delta\omega = \omega_2 - \omega_1 = \dfrac{R}{L}$ , and $Q = \dfrac{\omega_0}{\Delta\omega}$ . For a sharp peak $\omega_0 \approx \dfrac{\omega_1 + \omega_2}{2}$ .
Raising R widens the bandwidth and lowers Q; $\omega_0$ stays put.
Scaling with C fixed: $Q \propto \dfrac{\sqrt{L}}{R}$ . L made 4 times gives Q twice; R made 3 times gives Q a third.
$\dfrac{Q}{\Delta\omega} = \dfrac{\omega_0L^{2}}{R^{2}}$ , with the unit of time.

Quality factor and bandwidth

Q = \frac{\omega_0L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}} \qquad \Delta\omega = \frac{R}{L} = \frac{\omega_0}{Q}

Worked example

R =

20\ \Omega

, L = 0.4 H and C =

10\ \mu\text{F}

are in series. Find

\omega_0

, Q, the bandwidth and the two half-power frequencies.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q111Moderate

Example 3 · Alternating Current · Resonance, Quality Factor and Bandwidth

A series combination of resistor of resistance 100

\Omega

, inductor of inductance

1H

and capacitor of capacitance

6.25\mu F

is connected to an ac source. The quality factor of the circuit will be___

Half power is not half current

At the edges of the band the CURRENT is 1/√2 of its peak. Power goes as I², so the power is half.

Q with f₀ instead of ω₀

Q = ω₀L/R uses the angular frequency. Using f₀ makes Q too small by a factor of 2π.

Bandwidth in rad/s or in Hz

R/L is a bandwidth in rad/s. In hertz it is R/(2πL).

Drill 4 more on quality factor and bandwidth

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
$\omega_0 = \frac{1}{\sqrt{LC}} \qquad f_0 = \frac{1}{2\pi\sqrt{LC}}$
Current, impedance and power at resonance
At resonance
$Z = R \qquad I_0 = \frac{V_0}{R} \qquad P_{max} = \frac{V_{rms}^{2}}{R}$
Quality factor and bandwidth
Quality factor and bandwidth
$Q = \frac{\omega_0L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}} \qquad \Delta\omega = \frac{R}{L} = \frac{\omega_0}{Q}$

Watch out for (9)

ω or f
→ The resonant frequency
Putting R into the formula
→ The resonant frequency
Losing a power of ten
→ The resonant frequency
Forgetting √2 on the amplitude
→ Current, impedance and power at resonance
Using L and C to find the resonant current
→ Current, impedance and power at resonance
Reading the I–ω curve the wrong way round
→ Current, impedance and power at resonance
Half power is not half current
→ Quality factor and bandwidth
Q with f₀ instead of ω₀
→ Quality factor and bandwidth
Bandwidth in rad/s or in Hz
→ Quality factor and bandwidth

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