LC resonance, Q factor & bandwidth: practice problems with step-by-step solutions

Free LC resonance, Q factor & bandwidth practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: natural frequency, Q and energy, series and parallel RLC, design, losses and decay.

Free practice problemEasy
C = ?L = 4.7 mH

An LC circuit with L = 4.7 mH must resonate at 50 kHz. Find the capacitance C in nF.

L = 4.7 mHf0 = 50 kHz
nF
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What you practise

  1. Easynatural frequency, Q and energy
  2. Mediumseries and parallel RLC
  3. Harddesign, losses and decay

How to solve LC resonance, Q factor & bandwidth problems

  1. Find the resonance frequency first: ω₀ = 1/√(LC) and f₀ = ω₀/(2π).
  2. Decide whether the circuit is series or parallel: series Q = (1/R)√(L/C), parallel Q = R√(C/L).
  3. Use BW = f₀/Q for the bandwidth, or the shortcuts R/(2πL) (series) and 1/(2πRC) (parallel).
  4. For energy questions, use that the total energy ½CV₀² stays constant in an ideal LC circuit and swings between C and L every quarter period.

Common mistakes

  • Mixing up ω₀ (rad/s) and f₀ (Hz): they differ by 2π.
  • Using the series formula for Q in a parallel circuit, where a larger R means a higher Q.
  • Placing f₀ halfway between f₁ and f₂: it is their geometric mean √(f₁f₂).
  • Forgetting to convert µH, nF and pF to H and F.

Key formulas

Natural frequency
ω0 = 1/√(LC), f0 = 1/(2π√(LC))
Energy
W = ½CV0² = ½Cv² + ½Li²
Peak current
Imax = V0√(C/L)
Quality factor
Q = 2π·(energy stored)/(energy lost per cycle)
Series RLC
Q = (1/R)√(L/C), BW = R/(2πL)
Parallel RLC
Q = R√(C/L), BW = 1/(2πRC)
Bandwidth
BW = f2 − f1 = f0/Q, f0 = √(f1f2)
Half-power frequencies
f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q))
Tank with coil resistance r
RD = L/(C·r) = Q²·r
Decay of the amplitude
e−Rt/(2L)

Learn LC resonance, Q factor & bandwidth from scratch

Before this: AC impedance & phasors

1What happens in an LC circuit

Charge a capacitor and connect it to an inductor. The capacitor drives a current through the coil. The current builds up a magnetic field, and when the capacitor is empty the field keeps the current going, so the capacitor charges the other way. Then everything runs back.

C+−Li

The energy swings between the electric field of C (½Cv²) and the magnetic field of L (½Li²), like a pendulum swings between height and speed. Without resistance the total energy ½CV0² stays the same.

Self-check At the moment the capacitor voltage is zero, where is the energy of an ideal LC circuit?
  1. All in the capacitor
  2. All in the inductor, and the current is at its peak
  3. Half in C and half in L
  4. It has been lost as heat

Answer: All in the inductor, and the current is at its peak. With v = 0 the capacitor holds ½Cv² = 0, so all the energy is ½Li² in the inductor: the current is at its maximum. An ideal circuit has no resistance, so nothing is lost.

2The natural frequency

The circuit oscillates at the frequency where the reactances of L and C are equal, ωL = 1/(ωC):

ω0 = 1/√(LC) f0 = ω0/(2π) = 1/(2π√(LC))

With L = 10 mH and C = 1 µF: LC = 10⁻⁸ s², ω0 = 10 000 rad/s and f0 = 1592 Hz. The energy moves from C to L in a quarter of the period T = 1/f0.

Energy also gives the peak current: ½CV0² = ½LImax², so Imax = V0√(C/L).

Self-check What happens to f0 when you double C?
  1. It doubles
  2. It falls to f0/√2, about 0.71·f0
  3. It halves
  4. It rises by a factor √2

Answer: It falls to f0/√2, about 0.71·f0. f0 is proportional to 1/√C, so doubling C divides it by √2. It halves only when C is four times larger.

3Q factor and bandwidth

Real circuits lose energy in resistance. The quality factor Q compares the stored energy with the loss: Q = 2π·(energy stored)/(energy lost per cycle). A high Q gives a sharp resonance peak.

f|I|0.707f0​f1​f2​BW

The bandwidth is the width of the peak between the half-power frequencies f1 and f2, where the current is 0.707 of its peak:

BW = f2 − f1 = f0/Q f0 = √(f1·f2)
  • Series RLC: Q = (1/R)·√(L/C) and BW = R/(2πL). A small series resistance gives a high Q.
  • Parallel RLC: Q = R·√(C/L) and BW = 1/(2πRC). A large parallel resistance gives a high Q.

4The method step by step

  1. Convert L and C to H and F, then find ω0 = 1/√(LC) and f0 = ω0/(2π).
  2. Decide whether R is in series or in parallel, and use the matching formula for Q.
  3. Find the bandwidth BW = f0/Q.
  4. For the half-power frequencies use f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q)); for a high Q, f0 ∓ BW/2 is close.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A series RLC circuit has R = 10 Ω, L = 10 mH and C = 1 µF. Find its resonance frequency, its quality factor, its bandwidth and its half-power frequencies.

1

Look at the circuit

RLC

All three parts carry the same current, so this is a series circuit.

2

Resonance frequency

f0 = 1/(2π√(0.01·10⁻⁶)) = 1/(2π·10⁻⁴) = 1592 Hz
3

Quality factor

Q = (1/10)·√(0.01/10⁻⁶) = (1/10)·100 = 10
4

Bandwidth

BW = f0/Q = 1592/10 = 159.2 Hz

Check: R/(2πL) = 10/(2π·0.01) = 159.2 Hz.

5

Half-power frequencies

f1,2 = 1592·(√(1 + 1/400) ∓ 0.05) = 1514 Hz and 1673 Hz

Check: 1673 − 1514 = 159 Hz, and √(1514·1673) = 1592 Hz.

Answer

f0 = 1.59 kHz, Q = 10, BW = 159 Hz, f1 = 1.51 kHz and f2 = 1.67 kHz.

6Common mistakes

  • Mixing up ω0 in rad/s and f0 in Hz: they differ by 2π.
  • Using the series formula for Q in a parallel circuit.
  • Placing f0 halfway between f1 and f2: it is their geometric mean.
  • Forgetting to convert mH, µH, nF and pF to H and F.
Self-check A parallel RLC circuit has R = 10 kΩ, L = 10 mH and C = 1 µF. What is its Q?
  1. 0.01
  2. 10 000
  3. 100
  4. 1

Answer: 100. In parallel, Q = R·√(C/L) = 10 000·√(10⁻⁴) = 100. The series formula (1/R)·√(L/C) gives 0.01, and leaving out the square root gives 1.

7How exam questions look

  • Easy: natural frequency, Q and energy. You find f0 or the C for a given f0, the bandwidth from Q or Q from the bandwidth, the stored energy or the peak current.
  • Medium: series and parallel RLC. You find Q or the bandwidth of a series or parallel circuit, the exact half-power frequency, the time for the energy to move from C to L, or the energy in L at a given capacitor voltage.
  • Hard: design, losses and decay. You size R for a bandwidth or C for a filter, find the energy lost per cycle, the dynamic resistance of a tank, the number of cycles until the amplitude halves, or Q from the two half-power frequencies.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
C = ?L = 47 µH

An LC circuit with L = 47 µH must resonate at 500 kHz. Find the capacitance C in nF.

L = 47 µHf0 = 500 kHz

Find C (nF)

Show hints
  1. f₀ = 1/(2π√(LC)).
  2. C = 1/(ω₀²·L).
  3. ω₀ = 2π·500000 = 3.142 × 10⁶ rad/s.
Show the answer and the step-by-step solution

LC resonance: a charged capacitor connected to an inductor drives a current that charges it the other way: the energy swings between the electric field of C (½Cv²) and the magnetic field of L (½Li²) at the natural frequency f0 = 1/(2π√(LC)). Without resistance the total energy ½CV0² stays constant, so the peak current is Imax = V0√(C/L). One period is T = 1/f0; the energy moves from C to L in a quarter of it.

Angular frequency

ω₀ = 2π·500000 = 3.142 × 10⁶ rad/s

Capacitance

C = 1/(ω₀²L) = 1/(9.87 × 10¹²·4.7 × 10⁻⁵) = 2.156 × 10⁻⁹ F = 2.156 nF

Answer

C = 2.156 nF
Example 2Medium
V2 Ω2.2 mH220 nF

A series RLC circuit has R = 2 Ω, L = 2.2 mH and C = 220 nF. Find its bandwidth BW in Hz.

R = 2 ΩL = 2.2 mHC = 220 nF

Find BW (Hz)

Show hints
  1. The bandwidth is f₀/Q.
  2. For series RLC this simplifies to BW = R/(2πL): C drops out.
  3. R/L = 909.1 rad/s.
Show the answer and the step-by-step solution

Q factor and bandwidth: in a resonant circuit Q = 2π·(energy stored)/(energy lost per cycle). For series RLC Q = ω0L/R = (1/R)√(L/C); for parallel RLC Q = R/(ω0L) = R√(C/L). The bandwidth between the half-power frequencies f1 and f2 (where the power is half and the current or voltage 1/√2 of its peak) is BW = f2 − f1 = f0/Q, which is R/(2πL) in series and 1/(2πRC) in parallel. The resonance frequency is the geometric mean f0 = √(f1f2), and f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q)). Losses also make the oscillation decay: in series RLC the amplitude falls as e−Rt/(2L).

Bandwidth in rad/s

R/L = 2/0.0022 = 909.1 rad/s

Bandwidth in Hz

BW = 909.1/(2π) = 144.7 Hz = 144.7 Hz

Answer

BW = 144.7 Hz

With f₀ = 7.234 kHz this gives Q = 50.

Example 3Hard
C = 10 nFr = 0.5 ΩL = 47 µH

A parallel tank circuit is made of a 10 nF capacitor and a 47 µH coil with a winding resistance of 0.5 Ω. At resonance it behaves like a resistor, its dynamic resistance RD = L/(C·r). Find RD in kΩ.

L = 47 µHC = 10 nFr = 0.5 Ω

Find RD (kΩ)

Show hints
  1. The tank draws only a small current at resonance, so its impedance is large.
  2. RD = L/(C·r) = Q²·r.
  3. L/C = 4700 Ω².
Show the answer and the step-by-step solution

Q factor and bandwidth: in a resonant circuit Q = 2π·(energy stored)/(energy lost per cycle). For series RLC Q = ω0L/R = (1/R)√(L/C); for parallel RLC Q = R/(ω0L) = R√(C/L). The bandwidth between the half-power frequencies f1 and f2 (where the power is half and the current or voltage 1/√2 of its peak) is BW = f2 − f1 = f0/Q, which is R/(2πL) in series and 1/(2πRC) in parallel. The resonance frequency is the geometric mean f0 = √(f1f2), and f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q)). Losses also make the oscillation decay: in series RLC the amplitude falls as e−Rt/(2L).

Dynamic resistance

RD = 4.7 × 10⁻⁵/(1 × 10⁻⁸·0.5) = 9400 Ω = 9.4 kΩ

Check with Q

Q = (1/r)·√(L/C) = 137.1, Q²·r = 9400 Ω

Answer

RD = 9.4 kΩ

Questions and answers

What is LC resonance, Q factor & bandwidth?

Charge a capacitor and connect it to an inductor. The capacitor drives a current through the coil. The current builds up a magnetic field, and when the capacitor is empty the field keeps the current going, so the capacitor charges the other way. Then everything runs back.

The energy swings between the electric field of C (½Cv²) and the magnetic field of L (½Li²), like a pendulum swings between height and speed. Without resistance the total energy ½CV0² stays the same.

How do I solve LC resonance, Q factor & bandwidth problems step by step?
  1. Find the resonance frequency first: ω₀ = 1/√(LC) and f₀ = ω₀/(2π).
  2. Decide whether the circuit is series or parallel: series Q = (1/R)√(L/C), parallel Q = R√(C/L).
  3. Use BW = f₀/Q for the bandwidth, or the shortcuts R/(2πL) (series) and 1/(2πRC) (parallel).
  4. For energy questions, use that the total energy ½CV₀² stays constant in an ideal LC circuit and swings between C and L every quarter period.
What are common mistakes in LC resonance, Q factor & bandwidth?
  • Mixing up ω₀ (rad/s) and f₀ (Hz): they differ by 2π.
  • Using the series formula for Q in a parallel circuit, where a larger R means a higher Q.
  • Placing f₀ halfway between f₁ and f₂: it is their geometric mean √(f₁f₂).
  • Forgetting to convert µH, nF and pF to H and F.
What do the LC resonance, Q factor & bandwidth exercises cover?
  • Easy: natural frequency, Q and energy
  • Medium: series and parallel RLC
  • Hard: design, losses and decay

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

AC impedance & phasors

Calculators

Resonant frequency calculator (LC and RLC)

Courses with this topic

Circuit Analysis 2: AC circuits

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Printable worksheet

LC resonance, Q factor & bandwidth worksheet with answers
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