PhysicsISC Class 12 · Alternating Current

Resonance in a Series LCR Circuit: Resonant Frequency, Q Factor and Bandwidth

Updated 15 October 2026 · Fact-checked

A series LCR circuit is at resonance when the inductive and capacitive reactances are equal, $X_L = X_C$. Then the impedance falls to its minimum value Z = R, the current is maximum and in phase with the voltage, and the resonant frequency is $f_0 = \frac{1}{2\pi\sqrt{LC}}$. The power factor at resonance is 1.

Where X_L and X_C cross at f₀, the current is greatest; below f₀ the circuit is capacitive and above f₀ it is inductive.

Illustration drawn for QuestionBuddy.

Understand Resonance in Series LCR Circuits

What resonance means. In a series LCR circuit $Z = R^{2} + (X_{L} - X_{C})^{2}$ , where $X_{L} = ω L$ rises with frequency and $X_{C} = \frac{1}{ω C}$ falls with frequency. At one particular frequency the two are equal, the reactive term vanishes and the circuit behaves as a pure resistance. This is electrical resonance.

Resonant frequency. Setting $X_{L} = X_{C}$ :
$ω_{0} L = \frac{1}{ω _{0} C} \Rightarrow ω_{0}^{2} = \frac{1}{L C} \Rightarrow ω_{0} = \frac{1}{L C}, f_{0} = \frac{1}{2 π L C} .$
Note that $f_{0}$ depends only on L and C, not on R. It is the same frequency at which a charged capacitor and an inductor exchange energy in an LC circuit ( $ω_{0} = \frac{1}{L C}$ ), so the driving source is "in step" with the circuit's natural oscillation.

What happens at resonance.

$X_{L} = X_{C}$ , so $Z = Z_{min} = R$ .
Current is maximum: $I_{r m s} = \frac{V _{r m s}}{R}$ .
$tan ϕ = 0$ : voltage and current are in phase; power factor $cos ϕ = 1$ .
Power drawn from the source is maximum: $P = \frac{V _{r m s}^{2}}{R}$ .
$V_{L}$ and $V_{C}$ are equal in size and opposite in phase, so they cancel each other; the whole supply voltage appears across R. Each of $V_{L}$ and $V_{C}$ can be many times the supply voltage.

Away from resonance. Below $f_{0}$ , $X_{C} > X_{L}$ : the circuit is capacitive (current leads). Above $f_{0}$ , $X_{L} > X_{C}$ : the circuit is inductive (current lags).

Graphs (asked in ISC 2024, Q18(i)(b), 2 marks).

Z against f: very large at low f (capacitor dominates), falls to a minimum equal to R at $f_{0}$ , then rises again almost linearly at high f (inductor dominates). The curve never touches the f-axis, because its lowest value is R, not zero.
I against f: zero at f = 0 (a capacitor blocks d.c.), rises to a peak $\frac{V}{R}$ at $f_{0}$ , then falls. A smaller R gives a taller, sharper peak; $f_{0}$ does not move.

Sharpness: Q factor and bandwidth (no derivation in the syllabus). The bandwidth is the range of frequencies between the two half-power points, where the power is half its maximum, that is where the current is $\frac{I _{ma x}}{2}$ . For a series circuit the bandwidth is $Δ ω = \frac{R}{L}$ (or $Δ f = \frac{R}{2 π L}$ ). The quality factor
$Q = \frac{ω _{0}}{Δ ω} = \frac{ω _{0} L}{R} = \frac{1}{ω _{0} C R} = \frac{1}{R} \frac{L}{C}$
is a pure number. A high Q means a narrow band and a sharp response: the circuit picks out one frequency strongly. At resonance $V_{L} = V_{C} = Q \times V$ , so Q is also the voltage magnification of the circuit.

Why both L and C are needed. Resonance needs two reactances that change in opposite ways with frequency. An RL or RC circuit has only one reactance, so its impedance never reaches a minimum equal to R at a finite non-zero frequency: there is no resonance.

Uses. Tuning circuits of radio and television receivers (a variable capacitor shifts $f_{0}$ to the wanted station), metal detectors and oscillators.

Resonance in Series LCR Circuits: formulas and key facts

Resonance condition: $X_{L} = X_{C}, ω_{0} L = \frac{1}{ω _{0} C}$; Reactive terms cancel.
Resonant frequency: $ω_{0} = \frac{1}{L C}, f_{0} = \frac{1}{2 π L C}$; Independent of R.
At resonance: $Z = R, I_{ma x} = \frac{V}{R}, cos ϕ = 1$; V and I in phase.
Maximum power: $P_{ma x} = \frac{V _{r m s}^{2}}{R}$; Watts.
Quality factor: $Q = \frac{ω _{0} L}{R} = \frac{1}{R} \frac{L}{C}$; Dimensionless; no derivation needed.
Bandwidth: $Δ ω = \frac{R}{L}, Q = \frac{ω _{0}}{Δ ω} = \frac{f _{0}}{Δ f}$; Between half-power points.
Voltage across L or C: $V_{L} = V_{C} = Q V$; Only at resonance.

How to solve Resonance in Series LCR Circuits questions

Most resonance numericals follow one chain: f₀ from L and C, then Z = R, then current, power and Q.

1Convert units first: mH to H, µF to F, pF to F.
2Find $ω_{0} = \frac{1}{L C}$ , then $f_{0} = \frac{ω _{0}}{2 π}$ . State which one the question wants.
3At resonance put Z = R: $I = \frac{V}{R}$ and $P = \frac{V ^{2}}{R}$ (r.m.s. values).
4For voltages across L or C use $V_{L} = I X_{L} = I ω_{0} L$ , or $V_{L} = Q V$ .
5For sharpness: $Q = \frac{ω _{0} L}{R}$ , bandwidth $Δ ω = \frac{R}{L}$ or $Δ f = \frac{f _{0}}{Q}$ .
6To find C (or L) for a given $f_{0}$ , rearrange: $C = \frac{1}{4 π ^{2} f _{0}^{2} L}$ .
7For a frequency off resonance, go back to the full $Z = R^{2} + (X_{L} - X_{C})^{2}$ and say whether the circuit is inductive or capacitive.

Resonance in Series LCR Circuits in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Tuning a radio

Suppose a medium-wave radio has a tuning coil of 250 µH and you want to listen to a station broadcasting at 1 MHz.

What capacitance must the variable capacitor be set to?

Try it first, then show the solutionSolution

At resonance $f_{0} = \frac{1}{2 π L C}$ , so $C = \frac{1}{4 π ^{2} f _{0}^{2} L}$ .
$C = \frac{1}{4 π ^{2} \times ( 1 0 ^{6} ) ^{2} \times 250 \times 1 0 ^{- 6}} \approx 1.01 \times 1 0^{- 10}$ F ≈ 101 pF.
At this setting the 1 MHz signal drives a large current; other stations, off resonance, meet a large impedance.

Answer: About 101 pF.

Memory hook: Turning the dial changes C, and C moves f₀ onto the station.

Dangerous voltage magnification

Imagine a technician who connects a coil and a capacitor, each of reactance 230 Ω at 50 Hz, in series with a 46 Ω resistor across the 230 V (r.m.s.) mains, thinking nothing in the circuit can exceed 230 V.

Find the current and the voltage across the capacitor.

Try it first, then show the solutionSolution

$X_{L} = X_{C}$ , so the circuit is at resonance and Z = R = 46 Ω.
$I = \frac{230}{46} = 5$ A.
$V_{C} = I X_{C} = 5 \times 230 = 1150$ V, five times the supply (Q = 5).

Answer: 5 A; 1150 V across the capacitor, which could break down its insulation.

Memory hook: At resonance, L and C swing big but cancel; the meter on the supply hides it.

Tricks to remember Resonance in Series LCR Circuits

Short memory aids. Say each one aloud once, then use it on a question the same day.

Equal reactances, least impedance
$X_{L} = X_{C}$ ⇒ Z = R ⇒ I max ⇒ cos φ = 1.
Use it when: Four facts in one chain.
One over two pi root LC
Say it as a rhythm: "one by two-pi root-L-C".
Use it when: Resonant frequency.
Below is C, above is L
Below f₀ the capacitor wins (leading current); above f₀ the inductor wins (lagging current).
Use it when: Off-resonance behaviour.
Small R, sharp peak
$Q = \frac{ω _{0} L}{R}$ : halve R and the peak gets twice as sharp; f₀ stays put.
Use it when: Graph and Q questions.

Resonance in Series LCR Circuits: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Power factor at resonance · 1 mark · MCQBoard level

At resonance, the power factor of a series LCR circuit is:
(a) 0
(b) 0.5
(c) 0.707
(d) 1

Try it first, then show the solutionSolution

At resonance Z = R, so $cos ϕ = \frac{R}{Z} = 1$ .

Answer: (d).

Example 2: Resonant frequency and current · 2 marks · Short answerBoard level

A 0.4 H inductor, a 25 µF capacitor and a 20 Ω resistor are in series across a 100 V (r.m.s.) variable-frequency source. Find the resonant frequency and the current at resonance.

Try it first, then show the solutionSolution

$L C = 0.4 \times 25 \times 1 0^{- 6} = 1 0^{- 5}$ ; $f_{0} = \frac{1}{2 π 1 0 ^{- 5}} \approx 50.3$ Hz. (1)
At resonance Z = R = 20 Ω, so $I = \frac{100}{20} = 5$ A. (1)

Answer: ≈ 50.3 Hz; 5 A.

Example 3: Derivation and graphs · 4 marks · Long answerHarder board level

(a) Obtain the condition for resonance in a series LCR circuit and an expression for the resonant frequency. (b) Sketch graphs of impedance and current against frequency and mark $f_{0}$ .

Try it first, then show the solutionSolution

$Z = R^{2} + (X_{L} - X_{C})^{2}$ is least when $X_{L} = X_{C}$ : then Z = R and the current $\frac{V}{R}$ is greatest. This is resonance. (1)
$ω_{0} L = \frac{1}{ω _{0} C}$ gives $ω_{0} = \frac{1}{L C}$ and $f_{0} = \frac{1}{2 π L C}$ . (1)
Z-f graph (f on x-axis): falls from very large values, minimum Z = R at $f_{0}$ (curve does not touch the axis), then rises. (1)
I-f graph: starts at zero, peak $\frac{V}{R}$ at $f_{0}$ , then falls; label both axes and mark $f_{0}$ . (1)

Answer: $X_{L} = X_{C}$ ; $f_{0} = \frac{1}{2 π L C}$ ; Z has a minimum R and I a maximum at $f_{0}$ .

Example 4: Q factor, bandwidth and voltage magnification · 4 marks · Long answerHarder board level

A series circuit has L = 50 mH, C = 2 µF and R = 5 Ω on a 10 V (r.m.s.) source tuned to resonance. Find (a) $ω_{0}$ and $f_{0}$ , (b) the Q factor, (c) the bandwidth in hertz, (d) the voltage across the inductor.

Try it first, then show the solutionSolution

(a) $ω_{0} = \frac{1}{0.05 \times 2 \times 1 0 ^{- 6}} = \frac{1}{1 0 ^{- 7}} \approx 3162$ rad/s; $f_{0} \approx 503$ Hz. (1)
(b) $Q = \frac{ω _{0} L}{R} = \frac{3162 \times 0.05}{5} \approx 31.6$ . (1)
(c) $Δ ω = \frac{R}{L} = 100$ rad/s, so $Δ f = \frac{100}{2 π} \approx 15.9$ Hz (check: $\frac{503}{31.6} \approx 15.9$ Hz). (1)
(d) $I = \frac{10}{5} = 2$ A; $V_{L} = I ω_{0} L = 2 \times 158.1 \approx 316$ V $= Q V$ . (1)

Answer: (a) ≈ 3162 rad/s, ≈ 503 Hz; (b) ≈ 31.6; (c) ≈ 15.9 Hz; (d) ≈ 316 V.

Example 5: Choosing a capacitor for resonance · 3 marks · Short answerHOTS / competency

A coil of inductance 0.2 H and resistance 10 Ω is to be brought to resonance on a 50 Hz supply by adding a capacitor in series. Find the capacitance. If the frequency is then raised to 60 Hz, is the circuit inductive or capacitive? Why?

Try it first, then show the solutionSolution

$C = \frac{1}{4 π ^{2} f _{0}^{2} L} = \frac{1}{4 π ^{2} \times 2500 \times 0.2} \approx 50.7 \times 1 0^{- 6}$ F ≈ 50.7 µF. (1)
At 60 Hz (above $f_{0}$ ), $X_{L}$ rises and $X_{C}$ falls, so $X_{L} > X_{C}$ . (1)
The circuit is inductive: the current lags the voltage. (1)

Answer: ≈ 50.7 µF; inductive above resonance.

Example 6: Tuning range of a receiver · 4 marks · Case-basedHOTS / competency

Suppose the tuning circuit of a receiver has a fixed coil of 200 µH and a variable capacitor, and it must cover 600 kHz to 1500 kHz.
(i) Write the condition that selects one station. (1 mark)
(ii) Find the capacitance for 600 kHz. (1 mark)
(iii) Find the capacitance for 1500 kHz. (1 mark)
(iv) Show that the ratio of the two capacitances equals the square of the frequency ratio. (1 mark)

Try it first, then show the solutionSolution

(i) Resonance: the station frequency must equal $f_{0} = \frac{1}{2 π L C}$ .
(ii) $C = \frac{1}{4 π ^{2} ( 6 \times 1 0 ^{5} ) ^{2} ( 2 \times 1 0 ^{- 4} )} \approx 352$ pF.
(iii) $C = \frac{1}{4 π ^{2} ( 1.5 \times 1 0 ^{6} ) ^{2} ( 2 \times 1 0 ^{- 4} )} \approx 56.3$ pF.
(iv) $C \propto \frac{1}{f _{0}^{2}}$ , so $\frac{C _{1}}{C _{2}} = (\frac{1500}{600})^{2} = 6.25$ ; and $\frac{352}{56.3} \approx 6.25$ .

Answer: (i) $f = f_{0}$ ; (ii) ≈ 352 pF; (iii) ≈ 56.3 pF; (iv) ratio 6.25 both ways.

Where students lose marks in Resonance in Series LCR Circuits

The slips examiners see most often, and what to write instead.

Mistake: Taking Z = R in a circuit that is not at resonance
The ISC 2024 examiners reported candidates who treated an LR coil as an LCR circuit and put Z = R.
Wrong working
Coil of R = 300 Ω, L = 1 H on 50 Hz: "Z = R = 300 Ω".
Right working
$X_{L} \approx 314$ Ω, $Z = 30 0^{2} + 31 4^{2} \approx 434$ Ω.
Do this: Use Z = R only when $X_{L} = X_{C}$ ; I = V/R only at resonance.
Usually costs: 2 marks.
Mistake: Z-f graph drawn carelessly
In 2024 graphs had $f_{0}$ not marked, axes interchanged or unlabelled, the wrong shape, or the curve touching the axis.
Do this: Put f on the x-axis and Z on the y-axis; minimum at $f_{0}$ equal to R, above the axis; mark $f_{0}$ .
Usually costs: 1 to 2 marks.
Mistake: Mixing ω₀ and f₀
Answers often give $\frac{1}{L C}$ in hertz.
Wrong working
$f_{0} = \frac{1}{L C} = 500$ Hz.
Right working
$ω_{0} = 500$ rad/s, $f_{0} = \frac{500}{2 π} \approx 79.6$ Hz.
Do this: Write the symbol and the unit every time.
Usually costs: 1 mark.
Mistake: Forgetting unit prefixes
µF and mH left unconverted give answers wrong by powers of ten.
Do this: Convert to F and H on the first line.
Usually costs: 1 mark.

Exam tips

State the resonance condition $X_{L} = X_{C}$ before writing $f_{0}$ .
In graphs, label axes, mark $f_{0}$ and show the minimum Z = R above the axis.
Q and bandwidth need no derivation, but quote the formula you use.
Give units: rad/s for ω, Hz for f, Ω for Z; Q has no unit.

Practice questions on Resonance in Series LCR Circuits

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
At resonance in a series LCR circuit, the impedance is:
(a) zero
(b) equal to R
(c) maximum
(d) equal to $X_{L} + X_{C}$
Hint
Reactances cancel.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
Find the resonant angular frequency and frequency of a series circuit with L = 0.5 H and C = 8 µF.
Hint
$L C = 4 \times 1 0^{- 6}$ .
Answer
$ω_{0} = 500$ rad/s; $f_{0} \approx 79.6$ Hz.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): An RL circuit connected to a variable-frequency source cannot show resonance.
Reason (R): Resonance needs two reactances that vary in opposite ways with frequency.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Only L is present.
Answer
(a).
Q4 · 1 mark · Short answerHarder board level
The capacitance in a resonant series LCR circuit is doubled. By what factor does the resonant frequency change?
Hint
$f_{0} \propto \frac{1}{C}$ .
Answer
It becomes $\frac{1}{2}$ times (about 0.71 times) the old value.
Q5 · 2 marks · Short answerHarder board level
A series circuit has R = 55 Ω and $X_{L} = X_{C} = 110$ Ω on a 220 V (r.m.s.) supply. Find the current, the power drawn and the voltage across the inductor.
Hint
Resonance: Z = R.
Answer
4 A; 880 W; 440 V (twice the supply).
Q6 · 3 marks · Short answerHarder board level
For L = 0.8 H, C = 5 µF and R = 40 Ω in series, find $ω_{0}$ , the Q factor and the bandwidth in hertz.
Hint
$Δ ω = \frac{R}{L}$ .
Answer
$ω_{0} = 500$ rad/s; Q = 10; $Δ ω = 50$ rad/s, $Δ f \approx 7.96$ Hz.
Q7 · 3 marks · Short answerHOTS / competency
A series circuit with R = 200 Ω resonates at 1 kHz, where $X_{L} = X_{C} = 100$ Ω. The same supply voltage is now applied at 2 kHz. Find the new impedance, say whether the circuit is inductive or capacitive, and find the ratio of the new current to the current at resonance.
Hint
$X_{L}$ doubles and $X_{C}$ halves.
Answer
$X_{L} = 200$ Ω, $X_{C} = 50$ Ω, $Z = 20 0^{2} + 15 0^{2} = 250$ Ω; inductive; current ratio $\frac{200}{250} = 0.8$ .
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student builds a series circuit with L = 5 mH, C = 0.5 µF and R = 10 Ω and drives it from a 5 V (r.m.s.) signal generator.
(i) Find $ω_{0}$ and $f_{0}$ . (1 mark)
(ii) Find the current at resonance. (1 mark)
(iii) Find Q and the voltage across the capacitor at resonance. (1 mark)
(iv) Find the bandwidth in hertz and say what replacing R by 5 Ω would do to the I-f curve. (1 mark)
Hint
$L C = 2.5 \times 1 0^{- 9}$ .
Answer
(i) $ω_{0} = 2 \times 1 0^{4}$ rad/s, $f_{0} \approx 3183$ Hz. (ii) 0.5 A. (iii) Q = 10; $V_{C} = 50$ V. (iv) $Δ ω = 2000$ rad/s, $Δ f \approx 318$ Hz; with 5 Ω the peak doubles in height and becomes sharper (half the bandwidth), at the same $f_{0}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Condition for resonance?Show answer
$X_{L} = X_{C}$ .
2.Resonant frequency formula?Show answer
$f_{0} = \frac{1}{2 π L C}$ .
3.Impedance at resonance?Show answer
Z = R (minimum).
4.L = 1 mH, C = 1 mF: ω₀?Show answer
1000 rad/s.
5.Does R change f₀?Show answer
No; it changes only the height and sharpness of the peak.
6.Q factor in terms of L, C and R?Show answer
$Q = \frac{1}{R} \frac{L}{C}$ .

Resonance in Series LCR Circuits: frequently asked questions

Why is the current maximum at resonance?

Because the impedance is least. The reactances of L and C cancel, leaving only R to limit the current, so I = \frac{V}{R}.

How can the voltage across L be larger than the supply?

At resonance V_L and V_C are equal and opposite, so they cancel in the total. Each can be Q times the supply voltage while their sum is zero.

What is the difference between ω₀ and f₀?

\omega_0 = \frac{1}{\sqrt{LC}} is the angular frequency in rad/s; f_0 = \frac{\omega_0}{2\pi} is the frequency in hertz.

What does the ISC 2027 syllabus need on resonance?

The condition X_L = X_C, Z = R as minimum, maximum power, f_0 = \frac{1}{2\pi\sqrt{LC}}, graphs of Z and I against f, and the idea of bandwidth and Q factor without derivation.

How has resonance been asked recently?

ISC 2024 Question 18(i)(b) (2 marks) asked for a labelled graph of Z against f for a series LCR circuit with f_0 marked.