ARCHETYPE: Tuning & Resonance Problems

At an AM radio's antenna, every station in the city arrives at once — a dozen carriers stacked on one wire, none stronger than a few millivolts. Turn the knob and one station swells out of the pile while its neighbours vanish. No computation, no software: the knob is a variable capacitor, and the physics is one formula from the resonance module, $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$ , being solved for $C$ by your fingers. This archetype is the exam gym built on that gesture. Everything it uses was derived earlier — the resonance condition, $Q = \omega_0L/R$ , the bandwidth rule $\Delta f = f_0/Q$ and the magnification $V = QV_{\text{in}}$ belong to Resonance & the Q-Factor; the pipeline discipline to the LCR Numericals Ladder next door. What this module drills is the problem grammar boards and entrance exams write in: park a circuit on a target frequency (solve for $C$ or $L$ ); cover a band with a component's swing (the tuning-range problem, with its square-root sting); predict which way $f_0$ jumps when a knob moves; and price the circuit's pickiness — bandwidth, selectivity, magnification, the trio that decides whether your radio hears one station or three. The setting is deliberately concrete — the classic AM receiver with its 240 μH antenna coil and air-gang capacitor — because that is the exact circuit examiners keep re-costuming, and its historical designers faced, in earnest, the very numbers you are about to push.

Key takeaways

Tuning problems invert the resonance formula: to park $f_0$ on a station, $C = \dfrac{1}{(2\pi f_0)^2 L}$ (or the same solved for $L$ ).
The tuning-range rule carries a square root: a capacitor swing $C_{\min}$ – $C_{\max}$ covers a frequency ratio $\dfrac{f_{\max}}{f_{\min}} = \sqrt{\dfrac{C_{\max}}{C_{\min}}}$ — a 9:1 capacitor sweep buys only a 3:1 band.
Knob predictions all run through $f_0 \propto 1/\sqrt{LC}$ : double $C$ → $f_0/\sqrt2$ ; iron core in → $L$ up → $f_0$ down; and $R$ never moves $f_0$ at all.
Pickiness is priced by $Q = \omega_0L/R$ : bandwidth $\Delta f = f_0/Q$ (used, not derived), and neighbouring stations are rejected when they fall outside $\Delta f$ .
The same $Q$ multiplies the antenna's millivolts: $V_C = QV_{\text{in}}$ at resonance — the free amplification that made early radio possible.
Key idea: every tuning problem is $f_0 = 1/2\pi\sqrt{LC}$ read in some direction — solve it for the unknown, respect the square root, and price the sharpness with $Q$ .

The core insight

Tuning problems are the resonance formula worked in all directions: solve $f_0 = 1/2\pi\sqrt{LC}$ for the component that sets the station, remember that frequency scales as the inverse square root of $LC$ (so component ratios enter under a root), and use $Q = \omega_0L/R$ — as bandwidth $\Delta f = f_0/Q$ , selectivity against neighbouring frequencies, and magnification $V_C = QV_{\text{in}}$ — to price how sharply the circuit keeps what it tuned.

Where this lives today

The AM gang capacitor has retired to hobby kits, but its arithmetic never stopped working. Modern radios tune by varactor — a diode whose junction capacitance varies with an applied DC voltage — so the knob became a control voltage while the governing equation stayed $f_0 = 1/2\pi\sqrt{LC}$ verbatim; a phone's RF front end still hangs tuned LC networks (spiral inductors and finger capacitors printed on silicon) on every antenna path. The selectivity window became a profession: channel filters for Wi-Fi and LTE are cascades of resonators whose collective $Q$ -versus-bandwidth trade is Move 4 compounded. The gym's physics also escaped radio. RFID tags are LC loops tuned to a reader's 13.56 MHz; wireless chargers tune coil pairs to share power efficiently; metal detectors ring a tuned loop and listen for the $f_0$ shift when nearby metal changes $L$ — the knob table's iron-core row, weaponized. Within this track, the chapter's remaining stations are the transformer-chain drill (ARCHETYPE: Transformer & Line-Loss Numericals) and the speed-craft technique module (Phasor & Limit Shortcuts in AC) — and the charges sloshing in every tuned antenna are the doorway to the electromagnetic-waves chapter, where what the antenna launches finally gets its own physics.

The canonical setup: a receiver's front end

The gym's fixed apparatus, worth knowing as a circuit with a story: an antenna feeds every broadcast at once, as tiny EMFs of a few millivolts, into a series loop of one coil (the standard AM antenna coil, $L = 240\ \mathrm{\mu H}$ , wound on a ferrite rod) and one variable capacitor — interleaved metal vanes whose overlap changes as the knob rotates, classically from about 40 pF open to a few hundred pF meshed. The loop's small residual resistance $R$ (mostly the coil's copper) completes a series LCR circuit driven simultaneously at a dozen frequencies. The resonance module says what happens: for the one carrier matching $f_0 = 1/2\pi\sqrt{LC}$ , the reactances cancel, the loop responds with maximum current, and the capacitor's voltage builds to $Q$ times the antenna's whisper. Every other station, parked off the peak, drives almost nothing. Rotating the knob rewrites $C$ , slides $f_0$ along the dial, and hands the peak to a different station: the resonance curve, steered. (This series "acceptor" front end is the honest HS model of a receiver's first stage; what a real radio adds beyond it is engineering the exam does not touch.)

Figure

The tuning gym's apparatus — one coil, one variable capacitor, and a dial that steers the resonance peak

The master moves

Move 1 — park the peak. Given $L$ and a target station $f_0$ , invert the resonance formula:

C = \frac{1}{(2\pi f_0)^2 L} \qquad \text{or} \qquad L = \frac{1}{(2\pi f_0)^2 C}

where: everything in SI before the arithmetic — hertz, henry, farad — with radio answers landing in picofarads. Anchor: parking the 240 μH coil on 1000 kHz takes $C = 1/\big((2\pi \times 10^6)^2 \times 2.4\times10^{-4}\big) = 105.5$ pF. This single inversion, in one costume or another, is the most-set resonance question on every board carrying the topic.

Move 2 — cover the band. A capacitor swings from $C_{\min}$ to $C_{\max}$ ; what band is covered? Since $f_0 \propto 1/\sqrt{C}$ at fixed $L$ : highest frequency at smallest capacitance, and

\frac{f_{\max}}{f_{\min}} = \sqrt{\frac{C_{\max}}{C_{\min}}}

The square root is the sting: a generous 9:1 capacitance sweep covers only a $\sqrt{9} = 3$ :1 frequency band. This is a real design fact, not an exam artifice — the AM band (roughly 530–1700 kHz, ratio 3.2) is almost exactly what one classic air-gang capacitor's ~10:1 swing can span, which is why the historic "365 pF" variable capacitor became a standard part: with the 240 μH coil, the band's endpoints demand $C = 375.7$ pF at 530 kHz and 36.5 pF at 1700 kHz — recompute them from Move 1 and you have re-derived a component catalogue from physics.

Derivation AThe tuning range rule and why the square root halves the band

The goal is the gym's one general result — $f_{\max}/f_{\min} = \sqrt{C_{\max}/C_{\min}}$ — derived from the resonance formula so that its square root stops being a memorized surprise. Nothing new about resonance is proved; the pattern is Resonance & the Q-Factor's formula interrogated at the two ends of a component's travel, which is the archetype's master solution shape. The strategy: write $f_0$ at both capacitor extremes, divide, and watch what survives.

Step 1. Fix the coil, swing the capacitor. At the fully-meshed end, $C = C_{\max}$ and the circuit tunes its lowest frequency; at the open end, $C = C_{\min}$ , its highest — because $C$ sits in the denominator of $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$ : less capacitance, a stiffer circuit, a faster ring. (This pairing — small $C$ , high $f$ — is itself a marked step in range problems.)

Step 2. Write both endpoint frequencies and divide:

\frac{f_{\max}}{f_{\min}} = \frac{1/2\pi\sqrt{LC_{\min}}}{1/2\pi\sqrt{LC_{\max}}} = \sqrt{\frac{C_{\max}}{C_{\min}}}

where: everything else — the $2\pi$ , the fixed $L$ — cancels in the ratio. Why the root: frequency depends on the product $LC$ through a square root, so any component ratio reaches the frequency as its root — the exponent is halved, always.

Step 3. Read the design consequences, both directions. Forward: a 9:1 capacitor swing covers a $\sqrt9 = 3$ :1 band. Backward (the design question): to cover a band of frequency ratio $k$ , the component must swing by $k^2$ — the AM band's 3.2:1 demands a $10.3$ :1 capacitance sweep, which is why the classic air-gang capacitor was built with almost exactly that travel.

Step 4. Note what the ratio cannot tell you: the band's placement. Both endpoints scale together as $1/\sqrt{L}$ , so changing the coil slides the whole band without changing its ratio — the split of labour exploited in the worked design example below (capacitor sets the ratio, coil sets the address). Verified: with $L = 240\ \mathrm{\mu H}$ — $f(360\ \mathrm{pF}) = 541$ kHz and $f(40\ \mathrm{pF}) = 1624$ kHz, ratio $3.000 = \sqrt{360/40}$ ✓; AM endpoints $f(375.7\ \mathrm{pF}) = 530$ kHz, $f(36.5\ \mathrm{pF}) = 1700$ kHz, ratio $3.21 = \sqrt{10.3}$ ✓.

Move 3 — predict the jump (the six standard knobs). All shift questions run through $f_0 \propto 1/\sqrt{LC}$ ; take the factor applied to $LC$ and answer with its inverse square root:

the move	effect	$f_0$ becomes
double $C$ (mesh the vanes deeper)	$LC \times 2$	$f_0/\sqrt2 \approx 0.71 f_0$
quadruple $C$	$LC \times 4$	$f_0/2$
slide an iron/ferrite core into the coil	$L$ up (say ×4)	$f_0$ down (÷2)
withdraw the core	$L$ down	$f_0$ up
halve $C$ and double $L$	$LC$ unchanged	$f_0$ unchanged — the classic bait
change $R$ (any amount)	$LC$ untouched	$f_0$ unchanged; only height/sharpness move

Move 4 — price the pickiness ( $Q$ , bandwidth, selectivity, magnification). The quality factor $Q = \omega_0L/R$ prices the peak's shape (all faces built in Resonance & the Q-Factor; the bandwidth rule is used exactly as the ISC syllabus stipulates — no derivation). Bandwidth: $\Delta f = f_0/Q$ , the window the circuit genuinely accepts. Selectivity is bandwidth read against the dial — the exam's usual standard is $\Delta f \leq$ the station spacing, so separating stations 20 kHz apart at $f_0 = 1$ MHz needs $Q \geq 50$ . Magnification: at resonance $V_C = QV_{\text{in}}$ — with $Q = 100$ , the antenna's 5 mV becomes 0.5 V, enough for a diode detector, and the reason a crystal radio needs no battery. Note the tension the best problems exploit: raising $Q$ sharpens and magnifies, but a real AM signal carries audio sidebands ±5 kHz wide, so a $Q$ much above $f_0/(10\ \mathrm{kHz})$ starts eating the music itself — designers live inside a $Q$ window, and JEE-flavoured stems ask for both edges of it.

Figure

The selectivity window — rejection pushes Q up, the sidebands cap it from above

Limiting cases

case	result	meaning
$C \to C/4$	$f_0$ doubles	inverse square root, always
$C_{\max}/C_{\min} = 9$	band ratio 3	the root halves the exponent
$R$ doubled	$f_0$ same; $\Delta f$ doubled; $V_C = QV$ halved	$R$ owns the shape, never the position
$Q \to \infty$ (ideal)	zero bandwidth, infinite magnification	unbuildable — and would clip the sidebands anyway
ferrite core fully in	$L$ max ⇒ $f_0$ min	why sliding the core tunes cheap radios
two stations, spacing $< \Delta f$	both heard at once	insufficient selectivity: raise $Q$ or accept the mix

Common trapTo double the tuned frequency, halve the capacitance

Halving $C$ raises $f_0$ by $\sqrt2$ only — a 41% rise, not a doubling; reaching $2f_0$ demands $C/4$ . The inverse-square-root law is the single most reliable mark-harvester in this archetype, because linear intuition ("component halves, frequency doubles") feels so natural. It bites in all four costumes: the knob question (double $C$ gives $f_0/\sqrt2$ , not $f_0/2$ ); the range question (a 10:1 capacitor sweep covers a 3.16:1 band, not 10:1); the design question (to sweep 530→1700 kHz, a 3.2:1 frequency ratio, the capacitance must swing by $3.2^2 = 10.3$ :1 — square the frequency ratio to get the component ratio); and the combined-knob bait ( $L$ doubled and $C$ halved leaves $f_0$ exactly where it was, since only the product $LC$ matters). The two-second discipline that defuses every variant: write the factor applied to the product $LC$ , then answer with its inverse square root — never operate on $L$ or $C$ 's factor directly. And keep $R$ out of it entirely: no resistance change ever moves $f_0$ ; $R$ redraws the peak's height and width, not its address.

Worked examples

Board level 1 — park the peak, then jump it (ISC/NEET pattern, 3 marks)

A receiver's tuning circuit has a 240 μH coil. (a) What capacitance tunes it to 1000 kHz? (b) If the capacitance is then doubled, to what frequency does the circuit retune?

Step 1. Part (a): invert the resonance formula. $C = \dfrac{1}{(2\pi f_0)^2 L} = \dfrac{1}{(2\pi \times 10^6)^2 \times 240\times10^{-6}}$ . Compute the bracket once: $(2\pi \times 10^6)^2 = 3.948\times10^{13}$ . Then $C = \dfrac{1}{3.948\times10^{13} \times 2.4\times10^{-4}} = \boxed{105.5\ \mathrm{pF}}$ .

Step 2. Part (b): no recomputation — the product $LC$ doubles, so $f_0 \to f_0/\sqrt2 = 1000/1.414 = \boxed{707\ \mathrm{kHz}}$ . One proportionality sentence (" $f_0 \propto 1/\sqrt{C}$ at fixed $L$ ") carries the mark.

Board level 2 — the pickiness audit (ISC pattern, 3 marks)

The circuit of part (a), tuned to 1000 kHz, has total series resistance 15.1 Ω. Find (a) the quality factor, (b) the bandwidth, (c) the voltage across the capacitor when the antenna injects 5.0 mV at resonance.

Step 1. $Q = \dfrac{\omega_0 L}{R} = \dfrac{2\pi \times 10^6 \times 240\times10^{-6}}{15.1} = \dfrac{1508}{15.1} = \boxed{100}$ (the numerator is $X_L$ at resonance — 1508 Ω of reactance against 15 Ω of resistance is what "high $Q$ " looks like in components).

Step 2. Bandwidth: $\Delta f = \dfrac{f_0}{Q} = \dfrac{1000\ \mathrm{kHz}}{100} = \boxed{10\ \mathrm{kHz}}$ — the circuit listens to a 10 kHz window around the dial setting. (Rule used as the syllabus directs: stated, not derived.)

Step 3. Magnification: $V_C = QV_{\text{in}} = 100 \times 5.0\ \mathrm{mV} = \boxed{0.50\ \mathrm{V}}$ — a hundredfold, battery-free amplification; this is why early crystal sets worked at all.

Competitive level 1 — the band-coverage design (JEE-Main pattern, ~3 minutes)

A variable capacitor swings from 40 pF to 360 pF across a 240 μH coil. (a) Find the frequency band covered. (b) A designer wants the same 3:1 band shifted to start at 1.0 MHz; what inductance is needed with the same capacitor?

Step 1. Endpoints by Move 1 — highest $f$ at smallest $C$ : $f_{\max} = \dfrac{1}{2\pi\sqrt{240\times10^{-6} \times 40\times10^{-12}}} = 1624$ kHz; $f_{\min} = \dfrac{1}{2\pi\sqrt{240\times10^{-6} \times 360\times10^{-12}}} = 541$ kHz. Band: $\boxed{541\ \mathrm{kHz\ to\ 1624\ kHz}}$ . Audit with the range rule: $\sqrt{360/40} = 3.00 = 1624/541$ ✓.

Step 2. Part (b): the band's ratio is fixed by the capacitor swing alone (3:1 whatever $L$ is); $L$ only slides the whole band. Demand $f_{\min} = 1.0$ MHz at $C_{\max} = 360$ pF: $L = \dfrac{1}{(2\pi \times 10^6)^2 \times 360\times10^{-12}} = \dfrac{1}{3.948\times10^{13} \times 3.6\times10^{-10}} = \boxed{70.4\ \mathrm{\mu H}}$ — a smaller coil for a higher band, as $f_0 \propto 1/\sqrt{L}$ predicts. Timing note: 3 minutes. Compute one endpoint honestly, then get the other from the range rule — never two full inversions. In (b), recognizing that the capacitor fixes the ratio and $L$ fixes the placement turns a two-equation problem into one line.

Competitive level 2 — the selectivity window (JEE-Advanced flavour, ~4 minutes)

A tuned circuit at $f_0 = 1.0$ MHz ( $L = 240\ \mathrm{\mu H}$ ) must (i) reject a station 20 kHz away and (ii) pass its own station's audio sidebands, which extend 5 kHz either side of the carrier. Taking "reject" to require the bandwidth at most the station spacing and "pass" to require the bandwidth at least the full sideband width, find (a) the allowed range of Q, (b) the corresponding range of series resistance, (c) the capacitor voltage per millivolt of antenna signal at the largest allowed Q.

Step 1. Part (a): rejection needs $\Delta f = f_0/Q \leq 20$ kHz $\Rightarrow Q \geq \dfrac{10^6}{2\times10^4} = 50$ . Passing the ±5 kHz sidebands needs $\Delta f \geq 10$ kHz $\Rightarrow Q \leq \dfrac{10^6}{10^4} = 100$ . Window: $\boxed{50 \leq Q \leq 100}$ — sharp enough to be alone, broad enough to keep the music.

Step 2. Part (b): $Q = \omega_0L/R$ with $\omega_0 L = 2\pi \times 10^6 \times 240\times10^{-6} = 1508\ \Omega$ , so $R = 1508/Q$ : from $R = 1508/100 = 15.1\ \Omega$ up to $R = 1508/50 = 30.2\ \Omega$ . Allowed: $\boxed{15.1\ \Omega \leq R \leq 30.2\ \Omega}$ — the coil's copper, not the capacitor, is what the designer actually controls.

Step 3. Part (c): at $Q = 100$ , $V_C = QV_{\text{in}} = \boxed{0.10\ \mathrm{V\ per\ mV}}$ of antenna signal — and this is the top of the window precisely because pushing the magnification higher would narrow $\Delta f$ below the sidebands and clip the audio. Timing note: 4 minutes. The problem is two inequalities and one component formula — the difficulty is bookkeeping which requirement bounds Q from which side (rejection pushes Q up; fidelity caps it). Write both inequalities before computing either. The design tension itself — selectivity versus fidelity — is a genuine engineering trade-off, and stems that ask for "the largest Q consistent with..." are testing whether you saw both walls of the window.

Exam traps

The square root, four ways. Component ratios enter $f_0$ under a square root; frequency ratios demand the component ratio squared. Halve $C$ : $f_0 \times 1.41$ , not doubled. Cover 3:1 in frequency: swing $C$ by 9:1.

Highest frequency, smallest capacitance. The dial's high end is the capacitor's open end. Range problems that pair $f_{\max}$ with $C_{\max}$ invert the entire answer.

$R$ never tunes. No resistance change moves $f_0$ — $R$ trades peak height and bandwidth only. "The resistance is doubled; find the new resonant frequency" is a free mark: unchanged.

Only the product matters. $L$ doubled with $C$ halved leaves $f_0$ fixed. Any knob question should start by computing the factor on $LC$ .

Q at resonance only. $Q = \omega_0L/R$ uses the resonant $\omega_0$ . And $\Delta f = f_0/Q$ is a stated rule at this level (ISC says "no derivation") — use it, cite it, don't manufacture a proof.

Picofarad hygiene. Radio numbers live at $10^{-12}$ F and $10^{-6}$ H; a slipped prefix moves the answer six orders of magnitude. Anchor: 240 μH with ~100 pF resonates near 1 MHz — memorize that landmark and sanity-check against it.

Boards and depth. This gym is ISC and competitive-exam territory. ISC 861 names the kit explicitly — electrical resonance, "bandwidth of signals and Q factor (no derivation)", $f_0 = 1/2\pi\sqrt{LC}$ with maximum power at resonance — which is exactly the four master moves at board register. JEE-Main sets Moves 1–3 as 2-minute MCQs (parking, jumping, range) and increasingly Move 4; JEE-Advanced builds multi-part stems like the selectivity window, leaning on the LC/LCR entries in its syllabus. NEET keeps to Move 1 and the knob table at single-concept depth. CBSE's rationalized syllabus carries resonance itself but not sharpness/Q — a CBSE student needs Moves 1–3 and should read Move 4 as enrichment. AP Physics 2 has no AC on its syllabus at all, and IB (2023 onward) dropped AC circuits entirely — for those boards this module is how-radio-works reading, not examinable material.

Origins · 1 minGuglielmo Marconi and how this was discovered

Guglielmo Marconi

1874–1937 · London → the transatlantic stations

Marconi is the man who turned this module's gym into an industry — by patenting the tuning. His first wireless systems (1896 onward) were untuned: spark transmitters splattering energy across the spectrum, receivers hearing everything at once — workable for one link, chaos for two. The fix, filed in London in 1900 as British patent No. 7777 — "four sevens", wireless telegraphy's most famous number — was resonant circuits at both ends: transmitter and receiver each built around an LC loop set to the same $1/2\pi\sqrt{LC}$ , so that stations on different tunings could share the air, each receiver accepting its partner and rejecting the rest. That is precisely this module's selectivity arithmetic, deployed as a business model; a year later his tuned apparatus caught the three dots of "S" across the Atlantic. Honesty requires the asterisks: Lodge had patented syntonic tuning in 1897 (his priority story lives in the induction chapter's timeline), Tesla held earlier resonance patents, and the four-sevens patent was litigated for decades — in 1943 the US Supreme Court, adjudicating the American counterparts posthumously, restored key priority to Lodge and Tesla. Marconi's unassailable achievement sits beside the litigation: he engineered, financed and scaled tuned wireless into a global system, sharing the 1909 Nobel Prize. Hertz used resonance to detect waves; Marconi used it to organize an airspace.

History

1897

— Lodge patents syntonic tuning: transmitter and receiver deliberately set to one frequency

1900

— Marconi's patent No. 7777 makes tuned LC circuits at both ends the architecture of practical wireless

1901

— tuned apparatus receives the transatlantic "S" at St John's, Newfoundland

1912

— the Titanic disaster forces regulation of the airwaves: assigned frequencies make selectivity a legal requirement

1943

— the US Supreme Court, revisiting the tuning patents, restores priority to Lodge and Tesla

Going further

Texts. SL Arora, ISC Physics XII, electrical resonance, bandwidth and Q · HC Verma Vol. 2, Ch. 39 (resonance problems) · NCERT Physics Class 12, Ch. 7 (the tuning example) · Aitken, Syntony and Spark, for the Lodge–Marconi tuning war.

Provenance. All results standard; every number recomputed. A verifier should recompute: $C(1000\ \mathrm{kHz}, 240\ \mathrm{\mu H}) = 105.5$ pF; doubled $C \to 707$ kHz; $Q = 1508/15.1 = 100$ , $\Delta f = 10$ kHz, $V_C = 0.50$ V from 5 mV; band 40–360 pF → 1624–541 kHz, ratio 3.00 = $\sqrt{9}$ ✓; $L$ for 1.0 MHz at 360 pF = 70.4 μH; window $50 \leq Q \leq 100$ , $R \in [15.1, 30.2]\ \Omega$ ; AM endpoints 375.7 pF and 36.5 pF ✓.

Prerequisites. Resonance & the Q-Factor (every law this gym applies) · ARCHETYPE: The LCR Numericals Ladder (the pipeline discipline). Leads to. ARCHETYPE: Transformer & Line-Loss Numericals and TECHNIQUE: Phasor & Limit Shortcuts in AC (the chapter's remaining gym stations, coming up) · Alternating-Current Power & Resonance (UG deep dive, incl. the receiver end-to-end).

Why does the receiver respond only to the tuned station when every station drives it at once?

Superposition plus a very prejudiced load. All the carriers coexist on the antenna, each driving the LC loop at its own frequency, and the circuit responds to each independently. But the size of each response is set by the impedance at that frequency: the tuned station sees the collapsed $Z = R$ — perhaps 15 Ω — while a station 100 kHz off-peak faces the uncancelled reactance difference, often tens of kilohms. Same millivolts in, responses different by a factor of a thousand: the off-tune stations are not absent, just microscopically represented, and the detector stage never notices them. Turning the knob doesn't silence anyone — it reassigns which station gets the 15 Ω welcome.

Why do tuning circuits vary the capacitor rather than the coil?

Mechanical honesty: a smoothly variable capacitor is easy to build (rotate one set of vanes into another — air is the dielectric, the law is geometry), while smoothly varying an inductance means moving a core or switching taps, lumpier and lossier. So the classic architecture fixes $L$ (a well-made coil with stable $Q$ ) and puts the whole swing on the rotor — accepting the square-root tax that a 10:1 sweep covers only 3.16:1 of band, conveniently just the AM band's ratio. Cheap pocket radios did the opposite for coarse tuning — sliding a ferrite core (the knob table's $L$ row) — and modern sets replaced the rotor with a varactor diode, a voltage-controlled capacitance. Every variant obeys the same inversion of $f_0 = 1/2\pi\sqrt{LC}$ ; only which symbol the mechanism grips has changed.

Higher Q seems all-upside — sharper AND more magnification. What actually stops designers cranking it up?

The signal itself has width. A broadcast is not a bare carrier: the audio rides in sidebands several kilohertz either side (±5 kHz for standard AM), and the receiver must accept that whole slice or the music arrives with its treble amputated. Bandwidth $\Delta f = f_0/Q$ therefore has a floor — at 1 MHz, keeping ±5 kHz demands $Q \leq 100$ — while neighbour rejection sets the ceiling from the other side: the window of the last worked example. Physics adds its own cap: $Q = \omega_0L/R$ is limited by the coil's unavoidable copper resistance, so kilohertz-sharp filtering at megahertz carriers is unreachable with one LC pair. Real receivers escape architecturally — multiple tuned stages, and the superheterodyne trick of shifting every station to one fixed frequency where a single optimized filter discriminates. That story is engineering; the window arithmetic motivating it is this module's.

Is the Q·V magnification really free amplification? Where's the catch?

It is real voltage — a crystal radio lights headphones with no battery, powered entirely by the transmitted wave — but not a free lunch in power. The circuit takes the antenna's tiny power and, over many cycles, piles energy into the LC slosh until the capacitor's swing reaches $Q$ times the drive; nothing is created, only accumulated and concentrated at one frequency. Three catches. It works only in a whisker-thin band ( $\Delta f = f_0/Q$ — the same $Q$ that magnifies also narrows). It is voltage magnification, not power gain — draw significant current from the capacitor and you load the circuit, drop its effective $Q$ , and the arithmetic deflates. And it takes $\sim Q$ cycles to build, which is why extreme- $Q$ resonators respond sluggishly. Radio uses the trick where it is strong: lifting a millivolt signal to detector level, at one frequency, drawing almost nothing.

Does the same tuning physics work for FM radio and higher bands?

The formula transfers verbatim; only the component sizes shrink. FM broadcast sits near 100 MHz — a hundred times the AM band — and since $f_0 \propto 1/\sqrt{LC}$ , the product $LC$ must fall by $100^2 = 10^4$ : FM front ends use coils of a fraction of a microhenry (often just a few turns of bare wire, or a printed track) and capacitors of a few tens of picofarads. The same scaling continues upward — the 13.56 MHz of an NFC card, the gigahertz paths in a phone — until, in the microwave regime, lumped coils and capacitors blur into cavity and stripline resonators whose geometry is their $L$ and $C$ . Two HS-honest footnotes: FM's detection (frequency modulation) needs circuitry beyond this chapter, and at 100 MHz the stray picofarads of the wiring itself matter, which is why RF construction looks so fussy. But the exam-facing core — park $1/2\pi\sqrt{LC}$ on the carrier, budget $Q$ against bandwidth — is unchanged from 530 kHz to 5 GHz.

How do I sanity-check picofarad answers without redoing the whole computation?

Carry one landmark and scale from it. The landmark this module hands you: 240 μH with 105 pF resonates at 1000 kHz. Every radio-range answer should be reachable from it by the square-root law in a couple of mental steps. Need 500 kHz (half the frequency)? Four times the capacitance: ~420 pF. Need 2 MHz? A quarter: ~26 pF. Different coil — say 60 μH, a quarter of the landmark's? That doubles every frequency at the same $C$ , or equivalently quarters the needed $C$ at the same frequency. If your computed answer sits orders of magnitude off the scaled landmark, a prefix died somewhere in the $(2\pi f)^2 L$ denominator — the usual suspects being μ→ $10^{-6}$ and p→ $10^{-12}$ . This is the same anchor-value discipline the reactance modules built (100 μF ↔ 32 Ω at mains), promoted to radio frequencies.