LC resonance filters (A-level only) — AQA A-Level Physics
Teaching note checked against the official specification.
LC resonance filters are circuits containing an inductor (L) and capacitor (C) that selectively pass or block signals at a specific resonant frequency, f0 = 1/(2*pi*sqrt(LC)). At resonance, the reactances of the inductor and capacitor are equal and opposite, causing the impedance to be purely resistive and either maximising (series) or minimising (parallel) the current, which is exploited in tuning circuits and filters.
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LC resonance filters (A-level only) explained
LC resonance filters (A-level only) covers 4 specification points, including Analogy between LC circuit and mass–spring… and Inductance as mass analogy..
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Open a point below for its explanation and revision guidance.
Analogy between LC circuit and mass–spring system.
An LC circuit and a mass–spring system both oscillate by exchanging energy between two stores. In the mechanical system, energy shifts between the kinetic store of the moving mass and the elastic potential store of the spring. In the electrical system, energy shifts between the magnetic field of the inductor and the electric field of the capacitor. Mass corresponds to inductance, as both resist changes in motion or current. The spring constant corresponds to the reciprocal of capacitance. The analogy helps derive the resonant angular frequency ω = 1/√(LC) by comparing it with ω = √(k/m). Damping by friction or resistance reduces the amplitude of oscillation in both systems. Voltage maps to force, charge to displacement, and current to velocity.
Inductance as mass analogy.
In an LC circuit, the inductor opposes changes in current, just as mass opposes changes in velocity. Energy stored in the magnetic field of an inductor, ½LI², plays the role of kinetic energy, ½mv². A larger inductance L means a slower build-up or decay of current for a given potential difference, mirroring how a larger mass accelerates less under the same force. This analogy helps explain why the angular frequency of free oscillations is ω = 1/√(LC): the inductor behaves like the inertial mass and the capacitor like the spring. When analysing a circuit, treat current as the analogue of velocity and charge as the analogue of displacement, so the inductor's back e.m.f. corresponds to inertia.
Capacitance as spring analogy.
In an LC circuit, the capacitor stores energy in its electric field and provides the restoring effect that drives oscillations, just as a spring stores elastic potential energy and provides the restoring force. The energy stored, ½CV² or ½Q²/C, is analogous to ½kx². A larger capacitance C means more charge is stored for a given potential difference, so the oscillation is slower, mirroring how a softer spring (smaller k) gives a longer period. Together with the inductance-as-mass analogy, this explains why the angular frequency is ω = 1/√(LC). When analysing the circuit, treat charge as displacement and current as velocity, so the capacitor's voltage corresponds to the spring's restoring force.
Energy (voltage) response curve.
An energy or voltage response curve shows how the output of an LC filter varies with driving frequency. In a series LC circuit, current peaks at the resonant frequency f₀, while the voltage across the capacitor or inductor peaks at frequencies slightly away from f₀. In a parallel LC circuit, the impedance is maximum at resonance, so the voltage across the parallel combination peaks at f₀. The curve is often plotted as output voltage against frequency, with a sharp maximum or minimum at f₀. The width of the peak relates to the quality factor Q: a higher Q gives a narrower, sharper resonance. You should be able to sketch the general shape, label the resonant frequency, and explain how the curve changes when L, C or resistance changes.
Your focus
- Describe the energy transfers in an LC circuit and in a mass–spring system.
- Compare the roles of inductance, capacitance, mass and spring constant in the two systems.
- Explain how the resonant frequency depends on the properties of each system.
Show all 12 objectives
- Describe how an inductor opposes changes in current in terms of the mass analogy.
- Relate the energy stored in an inductor to kinetic energy in a mechanical oscillator.
- Use the analogy to predict how changing L affects the frequency of free oscillations.
- Describe how a capacitor provides a restoring effect in an LC circuit using the spring analogy.
- Relate the energy stored in a capacitor to elastic potential energy in a mechanical oscillator.
- Use the analogy to predict how changing C affects the frequency of free oscillations.
- Sketch and interpret an energy or voltage response curve for an LC filter.
- Identify the resonant frequency and describe how the curve shape depends on resistance and quality factor.
- Explain the difference between series and parallel LC circuit responses at resonance.
Which command word is this?
You must use the correct formula, substitute values with units, and give the answer to an appropriate number of significant figures. Show all steps of your working.
LC resonance filters (A-level only) exam tips
Topic Overview
LC resonance filters are a fundamental application of alternating current theory in A-level Physics. They combine an inductor and a capacitor to selectively pass or attenuate signals at a particular frequency, known as the resonant frequency. This topic extends your understanding of reactance, impedance, and phasor diagrams, and it is essential for explaining how tuning circuits in radios and filters in electronics work.
In the AQA A-Level specification, this topic is examined in Paper 2 and requires you to analyse both series and parallel LC circuits. You must be able to calculate resonant frequency, explain the phase relationships between voltage and current, and describe how the circuit's impedance changes with frequency. Mastery of this topic also supports your practical work on AC circuits and reinforces key mathematical skills such as using the formula f0 = 1/(2*pi*sqrt(LC)).
Key Concepts
- →Resonance occurs when the inductive reactance equals the capacitive reactance: XL = XC, which gives the resonant frequency f0 = 1/(2*pi*sqrt(LC)).
- →In a series LC circuit, impedance is minimum (equal to resistance R) at resonance, so current is maximum.
- →In a parallel LC circuit, impedance is maximum at resonance, so total current is minimum; the circuit acts as a band-stop filter.
- →At resonance, the voltages across the inductor and capacitor are equal in magnitude but opposite in phase, so they cancel in a series circuit.
- →The sharpness of resonance (Q-factor) depends on resistance; lower resistance gives a sharper, more selective resonance.
Marking Points
- States that both systems exchange energy between two stores and oscillate at a natural frequency.
- Matches the mass to the inductance, because mass opposes changes in velocity and inductance opposes changes in current.
- Matches the spring to the capacitor, because the spring stores elastic potential energy and the capacitor stores electric field energy.
- Compares the resonant angular frequency of the LC circuit, ω = 1/√(LC), with that of the mass–spring system, ω = √(k/m).
- Explains that increasing inductance or mass lowers the natural frequency, while increasing the reciprocal of capacitance or the spring constant raises it.
- Describes damping in the mechanical system as friction or air resistance and in the electrical system as resistance, both reducing oscillation amplitude.
- States that an inductor opposes change of current, analogous to mass opposing change of velocity.
- Links magnetic field energy ½LI² to kinetic energy ½mv².
- Explains that larger L gives slower current change for a given p.d., as larger mass gives smaller acceleration for a given force.
- Uses the analogy to justify ω = 1/√(LC) for free oscillations in an ideal LC circuit.
- Identifies current as the analogue of velocity and charge as the analogue of displacement.
- States that a capacitor stores energy in its electric field, analogous to a spring storing elastic potential energy.
- Links the energy expressions ½Q²/C or ½CV² to ½kx².
- Explains that a larger capacitance gives slower oscillations for a given inductance, as a softer spring gives a longer period.
- Uses the analogy to justify ω = 1/√(LC) for free oscillations in an ideal LC circuit.
- Identifies charge as the analogue of displacement and current as the analogue of velocity.
- Sketch a graph with frequency on the horizontal axis and output voltage or energy on the vertical axis.
- Identify the resonant frequency f₀ as the frequency at which the response is maximum or minimum, depending on circuit type.
- Describe the shape of the curve: a sharp peak for a high-Q circuit and a broader peak for a low-Q circuit.
- Explain that at resonance the reactances cancel, so impedance is minimum in a series circuit and maximum in a parallel circuit.
- Relate the peak width to energy loss and damping, noting that increased resistance broadens the response.
- Use the curve to determine the resonant frequency or to compare the selectivity of two filter designs.
Examiner Tips
- 💡Draw the two systems side by side and label the corresponding quantities before writing your explanation.
- 💡Use the equations ω = 1/√(LC) and ω = √(k/m) to justify each part of the analogy rather than relying on memory alone.
- 💡When discussing energy, name the electric field store, magnetic field store, kinetic store and elastic potential store explicitly.
- 💡Check that any comparison you make is consistent with the equations, especially the inverse relationship between capacitance and frequency.
- 💡Use the analogy to explain why increasing L lowers the resonant frequency.
- 💡When deriving ω, compare the energy expressions ½LI² and ½mv² to show the parallel.
- 💡State clearly that the analogy is between electrical and mechanical oscillators, not a physical identity.
- 💡Use the energy expressions to show the parallel between ½Q²/C and ½kx².
- 💡Explain how changing C affects the resonant frequency of an LC circuit.
- 💡State that the analogy is between electrical and mechanical oscillators, not a literal equivalence.
- 💡Label both axes with quantity and unit, and mark f₀ on the frequency axis.
- 💡Use a ruler to draw smooth curves and make the peak or trough clearly visible.
- 💡When comparing two curves, state which has the higher Q and justify it by referring to peak width.
- 💡If asked to explain the curve, link the shape to the frequency-dependent reactances X_L and X_C.
- 💡Always specify whether the circuit is series or parallel when discussing resonance effects, as the behaviour of impedance and current is opposite.
- 💡When calculating resonant frequency, show your working clearly and give your answer to an appropriate number of significant figures, usually 2 or 3.
- 💡Use phasor diagrams or graphs to illustrate how the reactances and impedance vary with frequency; this demonstrates deeper understanding and can gain explanation marks.
Common Mistakes
- Matching capacitance directly to the spring constant; correct by matching capacitance to the reciprocal of the spring constant, since ω = 1/√(LC) and ω = √(k/m).
- Claiming that the capacitor stores magnetic energy; correct by stating that the capacitor stores energy in its electric field and the inductor stores energy in its magnetic field.
- Ignoring resistance or friction when describing ideal oscillations; correct by noting that resistance in the LC circuit and friction in the mass–spring system cause damping.
- Confusing the roles of current, charge and voltage with velocity, displacement and force; correct by treating current as analogous to velocity, charge as analogous to displacement, and voltage as analogous to force.
- Treating inductance as a source of energy rather than a store; correction: the magnetic field stores energy and returns it to the circuit.
- Confusing the mass analogy with resistance; correction: resistance dissipates energy, whereas mass and inductance store it.
- Assuming the analogy means the inductor has literal mass; correction: it is a mathematical and conceptual correspondence only.
- Confusing capacitance with resistance; correction: capacitance stores energy, whereas resistance dissipates it.
- Thinking the spring analogy means the capacitor physically stretches; correction: it is a conceptual correspondence between restoring mechanisms.
- Assuming a larger C always increases frequency; correction: increasing C decreases ω because ω = 1/√(LC).
- Drawing a curve that peaks at zero frequency instead of at the resonant frequency; correct this by marking f₀ clearly on the frequency axis.
- Confusing the current response with the voltage response; correct this by checking whether the question asks about current or voltage across a specific component.
- Assuming the curve is symmetrical for all circuits; correct this by noting that practical resonance curves can be asymmetric when resistance is significant.
- Ignoring the effect of resistance on peak height and width; correct this by stating that greater resistance reduces the peak and broadens the curve.
- Students often think that resonance occurs when impedance is maximum in both series and parallel circuits. Correction: In series, impedance is minimum; in parallel, impedance is maximum.
- Students may believe that the resonant frequency depends on resistance. Correction: f0 depends only on L and C; resistance affects the bandwidth and peak current but not the resonant frequency.
- Students sometimes confuse the phase relationship between current and voltage in inductors and capacitors. Correction: In an inductor, current lags voltage by 90 degrees; in a capacitor, current leads voltage by 90 degrees.
Revision Plan
- 1Day 1-2: Revise the basics of AC circuits, focusing on reactance and impedance. Practice calculating XL and XC for different frequencies.
- 2Day 3-4: Learn the conditions for resonance in series and parallel LC circuits. Derive the resonant frequency formula from XL = XC and memorise it.
- 3Day 5-6: Work through calculations of resonant frequency, impedance, and current at resonance. Use past paper questions to apply the formula.
- 4Day 7-8: Study the filtering action of LC circuits. Draw graphs of impedance versus frequency for both series and parallel circuits and explain the shape.
- 5Day 9-10: Complete a full exam-style question on LC resonance filters, including explanations and calculations. Review mark schemes to identify common pitfalls.
Exam Question Types
- 📋Calculation of resonant frequency from given L and C values, often combined with finding current or voltage at resonance.
- 📋Explanation of how a series or parallel LC circuit acts as a filter, including reference to impedance and phase cancellation.
- 📋Analysis of graphs showing how current or impedance varies with frequency, requiring identification of resonant frequency and bandwidth.
- 📋Practical question on tuning a radio circuit, asking why a variable capacitor is used and how resonance selects a station.
Command Word Expectations (AQA)
You must use the correct formula, substitute values with units, and give the answer to an appropriate number of significant figures. Show all steps of your working.
You must provide a detailed reasoning that links cause and effect, using correct physics terminology. For example, explain why impedance is minimum at resonance in a series circuit by referring to the cancellation of reactances.
You must give a clear account of the behaviour or process, including key details such as the phase relationships or the variation of impedance with frequency, without necessarily giving reasons.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A series LC circuit has an inductance of 50 mH and a capacitance of 20 microF. Calculate the resonant frequency. If the resistance is 10 ohms, calculate the current at resonance when connected to a 12 V AC source.
- 1.Step 1: Identify given values: L = 50 mH = 0.050 H, C = 20 microF = 20 x 10^-6 F, R = 10 ohms, V = 12 V.
- 2.Step 2: Use the resonant frequency formula: f0 = 1/(2*pi*sqrt(LC)). Calculate LC = 0.050 * 20 x 10^-6 = 1.0 x 10^-6. sqrt(LC) = 1.0 x 10^-3. So f0 = 1/(2*pi*1.0 x 10^-3) = 159.15 Hz.
- 3.Step 3: At resonance, impedance Z = R = 10 ohms. Use Ohm's law: I = V/Z = 12/10 = 1.2 A.
Question: Explain how a parallel LC circuit can be used as a filter to block a specific frequency. Include the behaviour of currents and impedance at resonance.
- 1.Step 1: In a parallel LC circuit, the inductor and capacitor are connected across the same source. At resonance, the reactances are equal: XL = XC.
- 2.Step 2: The current through the inductor lags the voltage by 90 degrees, while the current through the capacitor leads by 90 degrees. These currents are equal in magnitude but opposite in phase, so they cancel when summed at the junction.
- 3.Step 3: The total current drawn from the source is therefore minimum (ideally zero if no resistance), meaning the impedance of the parallel combination is maximum at resonance. Thus, the circuit presents a high impedance to the resonant frequency, effectively blocking it from passing to the output.
- 4.Step 4: Frequencies far from resonance see a low impedance because one reactance dominates, so they pass through easily. This makes the parallel LC circuit a notch filter or band-stop filter.