In short

Connect a resistor R, an inductor L, and a capacitor C in series across a sinusoidal source v(t) = V_0\sin(\omega t). Because the three elements are in series, the same current i(t) = I_0\sin(\omega t - \phi) flows through all three. The voltage across the resistor is in phase with i; the voltage across the inductor leads i by 90°; the voltage across the capacitor lags i by 90°. Add the three as phasors and the source voltage amplitude satisfies

\boxed{\;V_0 = I_0\,Z,\qquad Z = \sqrt{R^2 + (X_L - X_C)^2}\;,\qquad \tan\phi = \dfrac{X_L - X_C}{R}\;}

where X_L = \omega L and X_C = 1/(\omega C) are the reactances.

Resonance happens at the frequency where X_L = X_C, i.e.

\boxed{\;\omega_0 = \dfrac{1}{\sqrt{LC}},\qquad f_0 = \dfrac{1}{2\pi\sqrt{LC}}\;}

At \omega = \omega_0 the inductive and capacitive voltages cancel exactly, the impedance collapses to Z = R, and the current amplitude I_0 = V_0/R is at its maximum. The phase angle \phi is zero — the circuit behaves like a pure resistor.

Quality factor Q = \omega_0 L/R = 1/(\omega_0 RC) measures how sharp the resonance peak is. The bandwidth (the width of the frequency band where the power is at least half its peak) is \Delta\omega = \omega_0/Q. A high Q gives a narrow, selective peak — what you want in an AM radio picking out Akashvani Delhi (819 kHz) from Akashvani Mumbai (1053 kHz) on a crowded band.

Take an old transistor radio in Lucknow. Turn the tuning dial slowly. For most positions you hear noise, hiss, faint overlapping voices — the broadcast band is a crowded soup of signals, all arriving at the antenna at the same instant, all different frequencies. Then at one particular spot on the dial the static gives way to a single clean station: "Yeh Akashvani hai." Nudge the dial a millimetre further and it is gone — hiss again. The radio was doing nothing magical. All it did was pick one frequency out of the chorus and ignore the rest. Behind that dial is an LCR series circuit whose resonant frequency, adjusted by rotating a tiny variable capacitor, matches the carrier frequency of exactly one station.

This chapter is about that circuit. Three elements in a row — a resistor, an inductor, a capacitor — driven by one sinusoidal source. The physics is richer than it looks. Across all three the same current flows, but the voltage across each has its own phase. The net impedance depends on frequency in a non-monotonic way: high at low and high frequencies, minimum at one special frequency. And that minimum — the resonance — is the mechanism behind every tuning circuit, every MRI scanner's RF coil, every spectrum analyser, and every satellite transmitter in ISRO's downlink stack.

The three voltages across R, L, C — recall from the previous chapter

Before stacking the elements in series, remember what each one does on its own when a current i(t) = I_0\sin(\omega t - \phi) flows through it. (The reason for choosing the current as the reference phasor, instead of the source voltage, will become clear in a moment — the current is common to all three series elements.)

Resistor. Ohm's law holds instantaneously:

v_R(t) = i(t)\,R = I_0 R\sin(\omega t - \phi)

The voltage across R is in phase with the current; its peak is V_{R,0} = I_0 R.

Inductor. From v_L = L\,di/dt:

v_L(t) = L\,\frac{d}{dt}[I_0\sin(\omega t - \phi)] = I_0\omega L\cos(\omega t - \phi) = I_0\omega L\sin(\omega t - \phi + \tfrac{\pi}{2})

Why: derivative of sine is cosine; \cos\theta = \sin(\theta + \pi/2). So v_L leads the current by 90°. Peak: V_{L,0} = I_0\omega L = I_0 X_L with X_L = \omega L.

Capacitor. Charge is q = Cv_C, so current is i = dq/dt = C\,dv_C/dt, and

v_C(t) = \frac{1}{C}\int i\,dt = \frac{1}{C}\int I_0\sin(\omega t - \phi)\,dt = -\frac{I_0}{\omega C}\cos(\omega t - \phi) = \frac{I_0}{\omega C}\sin(\omega t - \phi - \tfrac{\pi}{2})

Why: integral of sine is -\cos/\omega; -\cos\theta = \sin(\theta - \pi/2). So v_C lags the current by 90°. Peak: V_{C,0} = I_0/(\omega C) = I_0 X_C with X_C = 1/\omega C.

So with the current as reference, the three instantaneous voltages are

v_R \text{ in phase with } i,\qquad v_L \text{ leads } i \text{ by } 90°,\qquad v_C \text{ lags } i \text{ by } 90°

Three sinusoids of the same frequency but different phases. Adding them as functions of time looks ugly — but adding them as phasors turns the problem into geometry.

Building the impedance — phasor addition

Kirchhoff's voltage law around the series loop gives, at every instant,

v(t) = v_R(t) + v_L(t) + v_C(t)

Three sinusoids of the same frequency add to another sinusoid of that frequency — the only question is what the amplitude and phase of the sum are. The cleanest way to compute them is to represent each sinusoid as a phasor: an arrow of length equal to the peak value, rotating anti-clockwise at angular frequency \omega. At any instant the actual voltage or current is the vertical projection of the arrow. Because all three rotate together at the same \omega, their relative orientations are fixed, and we can draw a frozen snapshot.

Put the current phasor \vec{I} along the positive x-axis (our reference direction). Then:

The source voltage phasor \vec{V} = \vec{V_R} + \vec{V_L} + \vec{V_C}. The inductor and capacitor phasors point in opposite directions along y, so they partially cancel, leaving a net reactive component I_0(X_L - X_C) along +y (or -y if X_C > X_L).

Phasor diagram for LCR series circuit Current phasor along the positive x-axis. Resistor voltage phasor is along the positive x-axis, in phase with current. Inductor voltage phasor points along positive y-axis, 90 degrees ahead. Capacitor voltage phasor points along negative y-axis, 90 degrees behind. Their difference is a net reactive voltage. The source voltage is the vector sum of resistor and net reactive components, at phase angle phi above the x-axis. Re (x) Im (y) V_R = I₀R I (reference) V_L = I₀X_L V_C = I₀X_C V_L − V_C V₀ = I₀Z φ
Frozen phasor snapshot with the current along the positive $x$-axis. $V_R$ is in phase with $I$; $V_L$ is 90° ahead; $V_C$ is 90° behind. The cancellation $V_L - V_C$ leaves a net vertical arrow, and the source voltage is the diagonal from the origin to the tip of the sum — of length $I_0 Z$, at angle $\phi$ above the current.

Compute the length of \vec{V} from its two perpendicular components — Pythagoras:

Step 1. Components of the source phasor.

V_x = I_0 R,\qquad V_y = I_0(X_L - X_C)

Step 2. Magnitude.

V_0 = \sqrt{V_x^2 + V_y^2} = \sqrt{(I_0 R)^2 + [I_0(X_L - X_C)]^2} = I_0\sqrt{R^2 + (X_L - X_C)^2}

Step 3. Define the impedance Z as the peak voltage per peak current.

\boxed{\;Z \;=\; \frac{V_0}{I_0} \;=\; \sqrt{R^2 + (X_L - X_C)^2}\;}

Why: we want Ohm's law to hold for amplitudes. V_0 = I_0 Z generalises V = IR by replacing the resistance with an impedance that folds in the reactive drop.

Step 4. The phase angle \phi between source voltage and current is the angle of \vec{V} above the x-axis:

\boxed{\;\tan\phi \;=\; \frac{V_y}{V_x} \;=\; \frac{X_L - X_C}{R}\;}

Why: \phi is the phase by which the source voltage leads the current. If X_L > X_C (inductive), \phi > 0, current lags voltage. If X_L < X_C (capacitive), \phi < 0, current leads voltage. If X_L = X_C, \phi = 0, they are in phase — this is resonance.

Step 5. Write the time-domain solution.

The current that satisfies Kirchhoff's loop with source v(t) = V_0\sin(\omega t) is

i(t) \;=\; \frac{V_0}{Z}\sin(\omega t - \phi)

That is the complete steady-state response of a series LCR circuit to a sinusoidal drive.

Impedance vs frequency — the resonance peak

The formula Z(\omega) = \sqrt{R^2 + (\omega L - 1/\omega C)^2} hides a beautiful non-monotonic shape. At very low frequencies \omega \to 0, X_C = 1/\omega C \to \infty and Z \to \infty — the capacitor blocks DC, so no current flows. At very high frequencies \omega \to \infty, X_L = \omega L \to \infty and Z \to \infty — the inductor blocks high-frequency AC. In between, there is one frequency where X_L = X_C exactly, and the two reactances cancel each other:

\omega L = \frac{1}{\omega C} \;\Longrightarrow\; \omega^2 = \frac{1}{LC} \;\Longrightarrow\; \boxed{\;\omega_0 = \frac{1}{\sqrt{LC}}\;}

or equivalently f_0 = 1/(2\pi\sqrt{LC}). At this resonant frequency the impedance collapses to its minimum value Z_\min = R — just the resistor — and the current amplitude I_0 = V_0/R is at its maximum.

Interactive: LCR current amplitude as a function of frequency A Lorentzian-shaped curve shows the current amplitude I0 as a function of driving angular frequency, peaking at omega equals omega_0. A draggable vertical line lets you scan frequency and read off the current and impedance. At resonance the current is maximum and the impedance equals R. driving angular frequency ω / ω₀ current amplitude I₀R / V₀ 1 (ω₀) 2 3 0 0.5 1.0 ½-power: I/I_max = 1/√2 Q = 5 drag the red point along the axis
The current amplitude $I_0 R / V_0 = R/Z(\omega)$ plotted against the normalised driving frequency $\omega/\omega_0$ for $Q = 5$. The peak is at $\omega = \omega_0$. The dashed horizontal line marks $1/\sqrt{2} \approx 0.707$ — the half-power level. The two frequencies where the curve crosses this line define the bandwidth $\Delta\omega$. Drag the red point to explore.

Two things to notice about the shape.

  1. The peak height is V_0/R, exactly. No matter how large or small L and C are, at \omega_0 the impedance is Z = R, so I_0 = V_0/R. The peak height depends only on R and V_0.

  2. The peak width depends on the ratio of R to the reactances at resonance. A circuit with small R has a sharp, narrow peak (the denominator changes quickly as \omega moves off \omega_0). A circuit with large R has a broad peak — the reactive terms dominate only far from resonance.

This sharpness is captured by the quality factor Q.

The quality factor Q — how sharp is sharp?

Define Q at resonance as the ratio of reactance to resistance:

Q \;=\; \frac{\omega_0 L}{R} \;=\; \frac{1}{\omega_0 R C} \;=\; \frac{1}{R}\sqrt{\frac{L}{C}}

Why: at \omega_0, \omega_0 L = 1/(\omega_0 C) so both give the same number; substituting \omega_0 = 1/\sqrt{LC} into \omega_0 L gives L/\sqrt{LC} = \sqrt{L/C}. The three expressions are identically equal.

A typical AM-radio tank circuit has Q \sim 80–200 (very selective). A loudspeaker crossover network might have Q \sim 0.7 (deliberately broad). An MRI RF coil at 1.5 T and 63.87 MHz has Q \sim 200–400.

The physical meaning of Q

Q tells you two equivalent things:

Bandwidth — the width of the resonance peak

Define the bandwidth \Delta\omega as the width of the frequency range over which the power dissipated in R is at least half its peak. Power dissipated is P = I_\text{rms}^2 R, so half-power means I/I_\max = 1/\sqrt{2}, i.e. Z = R\sqrt{2}.

Setting Z^2 = 2R^2 and using Z^2 = R^2 + (X_L - X_C)^2:

(X_L - X_C)^2 = R^2 \;\Longrightarrow\; \omega L - \frac{1}{\omega C} = \pm R

Step 1. Take the +R branch (upper half-power frequency \omega_2 > \omega_0).

\omega_2 L - \frac{1}{\omega_2 C} = R \;\Longrightarrow\; \omega_2^2 LC - 1 = R C\omega_2 \;\Longrightarrow\; \omega_2^2 - \frac{R}{L}\omega_2 - \omega_0^2 = 0

Why: multiply through by \omega_2/L; use 1/LC = \omega_0^2.

Step 2. Solve the quadratic for \omega_2 using the quadratic formula.

\omega_2 = \frac{R/L + \sqrt{(R/L)^2 + 4\omega_0^2}}{2}

Step 3. Similarly the -R branch gives

\omega_1 = \frac{-R/L + \sqrt{(R/L)^2 + 4\omega_0^2}}{2}

Step 4. Subtract.

\Delta\omega \;=\; \omega_2 - \omega_1 \;=\; \frac{R}{L}

Why: the two square-root terms are identical and cancel, leaving just R/L. The bandwidth in angular frequency is R/L — independent of the square roots, independent of C.

Step 5. Divide by \omega_0 to get the dimensionless width.

\frac{\Delta\omega}{\omega_0} \;=\; \frac{R/L}{\omega_0} \;=\; \frac{R}{\omega_0 L} \;=\; \frac{1}{Q}

So

\boxed{\;\Delta\omega \;=\; \frac{\omega_0}{Q}\;}

High Q means narrow bandwidth means high selectivity. This is the equation that makes radio tuning possible.

Worked examples

Example 1: Tuning into Akashvani Delhi (AM, 819 kHz)

An AM receiver in Delhi has a tank-circuit inductor L = 200\ \mu\text{H}. A variable tuning capacitor C is set to resonate at Akashvani Delhi's carrier frequency f_0 = 819 kHz. The effective series resistance is R = 8.0\ \Omega.

Find (a) the required capacitance C, (b) the quality factor Q, (c) the bandwidth \Delta f, and (d) the peak voltage across the inductor when a 10 mV carrier is picked up at resonance.

Series LCR tank-circuit schematic Schematic of a series loop containing a resistor R = 8 ohms, an inductor L = 200 microhenries, and a variable capacitor C tuned to 819 kHz, driven by an AC source labelled antenna signal 10 millivolts at 819 kilohertz. antenna (10 mV, 819 kHz) R = 8 Ω L = 200 μH C = ?
A simple series LCR tuning circuit. Find the capacitance that resonates with $L = 200\ \mu\text{H}$ at 819 kHz.

Step 1. From f_0 = 1/(2\pi\sqrt{LC}), solve for C.

C \;=\; \frac{1}{(2\pi f_0)^2 L}

Why: square both sides of 2\pi f_0 = 1/\sqrt{LC}, rearrange.

Step 2. Plug in numbers.

\omega_0 = 2\pi\times 819\times 10^3 \;=\; 5.146\times 10^6\text{ rad/s}
\omega_0^2 = (5.146\times 10^6)^2 = 2.648\times 10^{13}\text{ rad}^2/\text{s}^2
C = \frac{1}{2.648\times 10^{13}\times 200\times 10^{-6}} = \frac{1}{5.297\times 10^9} \approx 1.89\times 10^{-10}\text{ F}
\boxed{\;C \;\approx\; 189\text{ pF}\;}

Why: inside a typical Indian pocket-radio tuning gang, the variable capacitor ranges from about 20 pF to 360 pF — 189 pF sits comfortably in the middle of that range.

Step 3. Quality factor.

Q \;=\; \frac{\omega_0 L}{R} \;=\; \frac{5.146\times 10^6\times 200\times 10^{-6}}{8.0} \;=\; \frac{1029}{8.0} \;\approx\; 129

Step 4. Bandwidth.

\Delta f \;=\; \frac{\Delta\omega}{2\pi} \;=\; \frac{\omega_0}{2\pi Q} \;=\; \frac{f_0}{Q} \;=\; \frac{819000}{129} \;\approx\; 6.35\text{ kHz}

Why: \Delta\omega = \omega_0/Q in angular frequency; divide by 2\pi to get hertz. The All India Radio AM channels are spaced every 9 kHz, and the 6.35 kHz bandwidth here sits within that spacing — the radio passes one channel and attenuates its neighbours.

Step 5. Peak voltage across the inductor at resonance.

At resonance, current amplitude is I_0 = V_0/R = 0.010/8.0 = 1.25 mA. Voltage across the inductor:

V_{L,0} = I_0 X_L = I_0\omega_0 L = 1.25\times 10^{-3}\times 5.146\times 10^6\times 200\times 10^{-6}
= 1.25\times 10^{-3}\times 1029 \approx 1.29\text{ V}

Why: V_{L,0} = Q V_0 = 129\times 10 mV = 1.29 V. A 10 mV antenna signal becomes 1.29 V across the inductor — a 129-fold step-up that boosts the tiny picked-up signal into something the next stage can amplify without adding noise. This is why resonant circuits are used in front of every receiver.

Result. C \approx 189 pF; Q \approx 129; \Delta f \approx 6.35 kHz; V_{L,0} \approx 1.29 V.

What this shows. The little variable capacitor in an AM radio is doing precision engineering: one particular setting matches exactly one station, rejects the ones 9 kHz above and below it by a factor of nearly \sqrt{2} on the first adjacent channel, and voltage-amplifies the desired carrier by Q = 129 before any transistor has touched it. All three functions — tuning, selectivity, pre-amplification — happen in the same one-loop LCR circuit.

Example 2: A 230 V, 50 Hz mains circuit with $R, L, C$ in series

A series LCR circuit is connected across the 230 V, 50 Hz Indian mains. The components are R = 100\ \Omega, L = 0.50 H, C = 20\ \mu\text{F}. Find (a) the reactances X_L and X_C, (b) the impedance Z, (c) the RMS current, (d) the phase angle \phi, and (e) the resonant frequency f_0.

Phasor diagram at 50 Hz for R=100 Ω, L=0.50 H, C=20 μF Phasor diagram with current on the x-axis. V_R of magnitude 100 I0 along x. V_L of magnitude 157 I0 along positive y. V_C of magnitude 159 I0 along negative y. Their difference is a tiny negative arrow; source voltage lies almost exactly along x. V_R = 100 I₀ V_L = 157 I₀ V_C = 159 I₀ V₀ ≈ 100 I₀ nearly at resonance — the two reactances almost cancel
Phasor diagram at 50 Hz. $V_L$ and $V_C$ nearly cancel because $X_L \approx X_C$ — the circuit is operating close to its resonant frequency.

Step 1. Reactances at \omega = 2\pi\times 50 = 314.16 rad/s.

X_L = \omega L = 314.16\times 0.50 = 157.1\ \Omega
X_C = \frac{1}{\omega C} = \frac{1}{314.16\times 20\times 10^{-6}} = \frac{1}{6.283\times 10^{-3}} = 159.2\ \Omega

Why: X_L and X_C are almost equal — a clue that 50 Hz is close to \omega_0/2\pi for this particular L and C.

Step 2. Impedance.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + (157.1 - 159.2)^2} = \sqrt{10000 + 4.41} \approx 100.02\ \Omega

Why: the reactive term (X_L - X_C) = -2.1\ \Omega is tiny compared to R = 100\ \Omega, so its square adds only 4.41 to the 10000 already there. The impedance is essentially R.

Step 3. RMS current.

I_\text{rms} = \frac{V_\text{rms}}{Z} = \frac{230}{100.02} \approx 2.30\text{ A}

Step 4. Phase angle.

\tan\phi = \frac{X_L - X_C}{R} = \frac{-2.1}{100} = -0.021 \;\Longrightarrow\; \phi \approx -1.2°

Why: negative \phi means the current leads the voltage slightly — the circuit is marginally capacitive because X_C is marginally greater than X_L at 50 Hz.

Step 5. Resonant frequency.

\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.50\times 20\times 10^{-6}}} = \frac{1}{\sqrt{10^{-5}}} = \frac{1}{3.162\times 10^{-3}} \approx 316.2\text{ rad/s}
f_0 = \frac{\omega_0}{2\pi} = \frac{316.2}{6.283} \approx 50.3\text{ Hz}

Why: the designer chose L and C so that resonance falls at 50.3 Hz — essentially at the grid frequency. That is the principle of a grid-connected series-LCR filter used in UPS systems and solar inverter output stages.

Result. X_L = 157.1\ \Omega; X_C = 159.2\ \Omega; Z \approx 100\ \Omega; I_\text{rms} \approx 2.30 A; \phi \approx -1.2°; f_0 \approx 50.3 Hz.

What this shows. A circuit tuned near resonance on the mains behaves almost like a pure resistor — V and I nearly in phase, Z close to R. A small mistuning (50 vs 50.3 Hz) produces a tiny but non-zero phase angle. Industrial filters and PF-correction banks at Tata Steel, Jindal, or any large ISRO test facility exploit exactly this near-resonant condition to deliver real power cleanly into the load.

Common confusions

You have the series LCR, the impedance formula, the resonance condition, and the Q factor. What follows is JEE-Advanced territory: the differential-equation derivation, the complex-impedance formalism, the parallel LCR (anti-resonance), and why resonance is mathematically the same as a damped driven oscillator.

The LCR differential equation

Starting from Kirchhoff's voltage law around the loop, with q the charge on the capacitor and i = dq/dt the loop current:

L\frac{di}{dt} + Ri + \frac{q}{C} = V_0\sin(\omega t)

Substitute i = dq/dt:

L\frac{d^2 q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = V_0\sin(\omega t)

This is a second-order linear ODE with a sinusoidal driving term — structurally identical to the damped-driven harmonic oscillator

m\ddot{x} + b\dot{x} + kx = F_0\sin(\omega t)

under the correspondence m \leftrightarrow L, b \leftrightarrow R, k \leftrightarrow 1/C, x \leftrightarrow q, F_0 \leftrightarrow V_0. Every result from the mechanical driven oscillator translates directly. The natural frequency \omega_0 = \sqrt{k/m} = 1/\sqrt{LC}; the damping ratio \zeta = b/(2\sqrt{km}) = R/2\sqrt{C/L} = 1/(2Q).

The steady-state solution is q(t) = Q_0\sin(\omega t - \phi - \pi/2) with Q_0 = V_0/(\omega Z) — differentiating once gives the i(t) = (V_0/Z)\sin(\omega t - \phi) we already found by phasors.

Complex impedance

The phasor method is the polar-form version of a more powerful trick: write voltages and currents as complex exponentials, v(t) = \operatorname{Re}[V_0 e^{j\omega t}]. Then the complex impedances are

Z_R = R,\qquad Z_L = j\omega L,\qquad Z_C = \frac{1}{j\omega C} = -\frac{j}{\omega C}

In series, complex impedances add (as in a DC circuit):

Z = R + j\omega L - \frac{j}{\omega C} = R + j\left(\omega L - \frac{1}{\omega C}\right)

Magnitude and phase:

|Z| = \sqrt{R^2 + (\omega L - 1/\omega C)^2},\qquad \arg(Z) = \arctan\frac{\omega L - 1/\omega C}{R}

The magnitude is the impedance we derived; the argument is the phase angle \phi. The complex formalism reduces all AC circuit problems — no matter how complicated — to the same arithmetic you already know for DC circuits, with the single replacement that resistances become complex impedances.

Parallel LCR — anti-resonance

Put R, L, C in parallel across a sinusoidal source. The admittances (reciprocals of impedances) add:

Y = \frac{1}{R} + \frac{1}{j\omega L} + j\omega C = \frac{1}{R} + j\left(\omega C - \frac{1}{\omega L}\right)

Resonance occurs when the imaginary part vanishes: \omega C = 1/(\omega L), i.e. the same \omega_0 = 1/\sqrt{LC}. But now |Y| is minimum at \omega_0, so |Z| = 1/|Y| is maximum, and the current from the source is minimum. This is anti-resonance or a rejector circuit — the opposite selectivity. The tuned plate circuit in a valve AM transmitter is parallel-LCR: at the carrier frequency it presents a huge impedance to the plate, forcing all the RF power into the aerial rather than back through the tube.

Inside the parallel LCR tank at \omega_0, a large current sloshes between L and C (the "tank current"), but only a small current flows from the source — because L and C together form a closed loop that the source does not need to feed. Energy oscillates between magnetic (in L) and electric (in C) at \omega_0, and the source only replenishes what R dissipates.

Why f_0 = 1/(2\pi\sqrt{LC}) is a deep fact, not a formula to memorise

The product LC has units of (second)². This is the only combination of L (henries = V·s/A), C (farads = A·s/V), and nothing else that gives a time — and dimensional analysis forces the natural time scale of any LC circuit to be \sqrt{LC}. The 1/(2\pi) factor is the geometric consequence of the solution being a sinusoid whose period is 2\pi times the time-scale. Once you recognise that \sqrt{LC} is the natural unit of time for a circuit with only L and C, the formula is inevitable — not something to memorise.

Applications across the spectrum

Where f_0 Role of the resonance
AM radio tuner (Akashvani) 500 kHz – 1.6 MHz Selects one station from many
FM radio tuner 88 – 108 MHz Same, narrower channels
MRI RF coil at 1.5 T 63.87 MHz (Larmor, ¹H) Matches the spin precession frequency for maximum energy transfer
MRI at 3 T 127.74 MHz Stronger signal because the Boltzmann-split population difference grows with B_0
ISRO S-band telemetry 2.2 – 2.3 GHz Narrow-band receiver front-ends
GMRT (Pune) low-band 150 MHz Radio astronomy receivers tuned to deep-space continuum emission
Induction hob (Butterfly Smart) 20 – 40 kHz Drives the cooking pan via a resonant inverter at high efficiency
Wireless phone charger 100 – 200 kHz Resonant coupling between coils for maximum power transfer

Each application needs its own trade-off between bandwidth (how much signal must get through) and selectivity (how much interference to reject) — and the design knob is Q.

The L_\text{rms} that matters at resonance

In many textbook problems you are asked to find V_L and V_C individually at resonance and notice they equal QV_0. That is correct and instructive. But be aware: V_L and V_C are not in phase with the source. You cannot add their magnitudes to the source voltage; you have to add them as phasors, which gives zero contribution from V_L + V_C in the loop (because they cancel). A high-voltage reading across L alone does not mean the circuit is storing more energy than the source can supply — it means energy is sloshing between L and C, and the oscilloscope across L captures the reactive-storage voltage, not a net energy injection.

Why Q measures stored energy over dissipated energy

At resonance, current amplitude is I_0 = V_0/R. Peak energy stored in the inductor is

U_L^{\max} = \tfrac{1}{2}L I_0^2

Energy dissipated in R over one full period T = 2\pi/\omega_0 is

E_\text{diss} = \tfrac{1}{2}I_0^2 R T = \frac{\pi I_0^2 R}{\omega_0}

The ratio:

\frac{2\pi U_L^{\max}}{E_\text{diss}} = \frac{2\pi\cdot\tfrac{1}{2}L I_0^2}{\pi I_0^2 R/\omega_0} = \frac{\omega_0 L}{R} = Q

So Q = 2\pi\times(\text{peak stored energy})/(\text{energy lost per cycle}). A Q = 100 resonator loses 2\pi/100 \approx 6\% of its stored energy per cycle — that is why it rings for about Q/2\pi cycles after the drive stops, a useful rule of thumb.

Where this leads next