AC Power and Resonance

Two questions run this article, and both are worth money. First: when the voltage and current in an AC circuit are out of step — as the phasor article showed they usually are — what do you actually pay for? A factory can draw sixty amperes all day and be billed as if it used far less energy than the cables suggest, because part of that current is borrowed and returned every cycle without ever becoming heat or work. Utilities charge for the discrepancy; engineers install banks of capacitors to erase it. Second: how does a circuit choose a frequency? Turn an old radio dial and one station swells out of a spectrum crowded with hundreds, all present at the antenna simultaneously. The series RLC of the last article holds the answer: at exactly one frequency the inductor's and capacitor's oppositions cancel, the impedance collapses to its bare-resistance minimum, and the circuit responds enormously to that frequency while shrugging at all others. The sharpness of that preference is measured by a single dimensionless number, the quality factor $Q$ — and the same $Q$ that sharpens the tuning also quietly multiplies voltages, so a millivolt whisper from an antenna can become a half-volt signal with no amplifier in sight. You have the tools ready: impedance and phase from the phasor article, the natural frequency $\omega_0 = 1/\sqrt{LC}$ from the oscillations article, and the mechanics track's resonance theory waiting to be cashed through the dictionary. What this article adds is the power bookkeeping and the resonance anatomy — the two pillars of every AC power system and every tuned receiver on Earth.

Key takeaways

Instantaneous AC power pulses at twice the line frequency; its average is $\bar P = V_{\mathrm{rms}} I_{\mathrm{rms}}\cos\varphi$ , where $\varphi$ is the voltage–current phase angle.
The rms value $V_0/\sqrt2$ is defined by heating equivalence: an rms ampere produces exactly the heat of a DC ampere. Mains "230 V" is rms; the peak is 325 V.
$\cos\varphi$ is the power factor: reactive elements shuttle energy back and forth (borrow-and-return) without dissipating, inflating the current for the same delivered power.
A series RLC driven at $\omega_0 = 1/\sqrt{LC}$ has minimum impedance $Z = R$ : current, and power delivery, peak there — resonance.
The quality factor $Q = \omega_0 L/R$ measures the resonance three equivalent ways: reactance-to-resistance ratio, frequency-to-bandwidth ratio $\omega_0/\Delta\omega$ , and $2\pi$ times stored-to-dissipated-per-cycle energy.
At resonance the capacitor and inductor voltages are each $Q$ times the source voltage — free, passive voltage magnification.
KEY IDEA: only the in-phase part of the current carries average power ( $\bar P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$ ), and a series RLC concentrates its response into a band of width $\omega_0/Q$ around resonance.

The core insight

Average AC power is $\bar P = V_{\mathrm{rms}} I_{\mathrm{rms}}\cos\varphi$ — reactance moves current but no net energy — and the driven series RLC responds resonantly at $\omega_0 = 1/\sqrt{LC}$ with sharpness $Q = \omega_0 L/R = \omega_0/\Delta\omega$ , magnifying both current and element voltages by the same factor.

Where this lives today

The grid's reactive-power problem has been re-invented by electronics. Every phone charger and server power supply is a rectifier — a nonlinear load whose current comes in gulps near the voltage peaks — so regulators (IEC 61000-3-2) now force anything above 75 W to include active power-factor-correction circuitry: a switching converter that shapes the input current into a sinusoid in phase with the mains, a real-time, silicon implementation of Worked example 2. At grid scale the capacitor bank has evolved into the STATCOM — an inverter that synthesizes leading or lagging current on demand, holding voltage steady on lines dominated by wind and solar farms. Resonance meanwhile became a wireless power channel: a Qi phone pad and an electric-toothbrush base are loosely coupled coils that would transfer almost nothing — except both sides are tuned to the same ~100–130 kHz, and the resonant magnification recovers the efficiency that geometry throws away. The same $Q$ bookkeeping, pushed to superconducting microwave circuits with $Q > 10^6$ , sets how long a qubit's readout resonator holds a photon — the energy-per-cycle reading of Derivation B, now counted photon by photon.

The physical picture

Watch the power flow into a purely resistive load — a kettle — over one mains cycle. The instantaneous power $p = vi$ is always positive (voltage and current reverse together), but it pulses: zero a hundred times a second, peaking in between. The kettle doesn't care; its thermal mass averages the pulses into steady heat. Now watch a load with phase lag — a motor, say, at $\varphi = 60°$ . For most of each half-cycle power flows in, but there are intervals where $v$ and $i$ have opposite signs and the power flows backwards, out of the motor and into the grid. The motor's magnetic field is returning energy it borrowed a few milliseconds earlier.

Figure

Instantaneous power over two cycles: a resistive load versus a load with 60 degrees of phase lag

The averages tell the economic story: same voltage, same current amplitude, but the lagging load's mean power is only $\cos 60° = 0.5$ of the resistive load's. The other half of the current amplitude is circulating current — real amperes heating real cables, delivering nothing. That is why "power factor" appears on industrial electricity bills, and why the fix (Worked example 2) is a capacitor: an element whose borrowing is timed opposite to the motor's, so the two trade energy locally instead of hauling it through kilometres of grid. The resonance picture is the second half of the story. Sweep the driving frequency across a series RLC and the current traces a peaked curve: starved at low frequency by the capacitor, starved at high frequency by the inductor, released at $\omega_0$ where the two cancel. The peak's height is set by $R$ alone; its width is set by $R$ too — smaller resistance means a taller and narrower peak, with the product fixed. One number, $Q$ , captures the shape.

Figure

The resonance curve: current versus frequency for two resistances, same L and C — computed

The power ledger

Multiply the driven circuit's voltage and current — $v = V_0\cos\omega t$ , $i = I_0\cos(\omega t - \varphi)$ from the phasor article — and average. The product-to-sum identity gives

p(t) = V_0 I_0 \cos\omega t\,\cos(\omega t - \varphi) = \frac{V_0 I_0}{2}\big[\cos\varphi + \cos(2\omega t - \varphi)\big]

where: the second term oscillates at $2\omega$ — the hundred-per-second pulsing of the hero figure — and averages to zero over any whole cycle, while the first term is a constant floor set by the phase angle. The steady, billable part is the floor:

\bar P = \frac{V_0 I_0}{2}\cos\varphi = V_{\mathrm{rms}}\, I_{\mathrm{rms}}\cos\varphi

where: $\cos\varphi$ is the power factor and the rms values absorb the factor of $\tfrac12$ . Read it physically: only the component of current in phase with the voltage delivers energy on average; the quadrature component sloshes energy in and out at $2\omega$ with zero net. A purely reactive load ( $\varphi = \pm90°$ ) draws current, heats the supply cables, trips breakers — and delivers nothing.

Derivation ARoot-mean-square values, from the heating they cause

We want to justify the $\sqrt2$ and the very definition of "rms," not just quote them. The physical anchor is heating: define the effective AC current as the DC current that would cook a resistor at the same average rate — this is how AC and DC were compared in the 1890s, calorimeter against calorimeter, and it is still the operational meaning of every AC meter reading. The strategy is to compute the average of $i^2$ over a cycle honestly, then package the result.

Step 1. In a resistor the instantaneous heating is $p = i^2 R$ with $i = I_0\cos\omega t$ , so the average rate is $\bar p = R\,\overline{I_0^2\cos^2\omega t} = I_0^2 R\,\overline{\cos^2\omega t}$ .

Step 2. Average the squared cosine. Over a full period, $\cos^2$ spends as much time above $\tfrac12$ as below it — the identity $\cos^2\theta = \tfrac12(1 + \cos 2\theta)$ shows the oscillating half averages away:

\overline{\cos^2\omega t} = \frac12 \quad\Longrightarrow\quad \bar p = \frac{I_0^2 R}{2}

Step 3. Match to DC. A steady current $I_{\mathrm{eff}}$ heats at $I_{\mathrm{eff}}^2 R$ . Equating:

I_{\mathrm{rms}} = \frac{I_0}{\sqrt2}, \qquad V_{\mathrm{rms}} = \frac{V_0}{\sqrt2}

Why "root mean square"? The recipe is read right-to-left: square the signal, take the mean, take the root. For a sine the answer is the peak over $\sqrt2$ ; for other waveforms (square, triangle, audio) the recipe still applies but the factor differs — which is why cheap "average-responding" meters mis-read non-sinusoidal currents.

Step 4. Cash the convention. Mains "230 V" is rms, so the peak is $230\sqrt2 = 325\ \mathrm{V}$ — the number insulation must survive. And with both $V$ and $I$ in rms, the resistive power formulas keep their DC form exactly: $\bar P = V_{\mathrm{rms}}I_{\mathrm{rms}} = I_{\mathrm{rms}}^2 R = V_{\mathrm{rms}}^2/R$ — the reason the rms convention won. Verified: $\overline{\cos^2} = 1/2$ by the double-angle identity and by direct numerical average; $230\sqrt2 = 325.3\ \mathrm{V}$ ; a 2.0 kW kettle on 230 V draws $I = P/V = 8.70\ \mathrm{A}$ rms, peak $12.3\ \mathrm{A}$ .

Engineering bookkeeping splits $V_{\mathrm{rms}}I_{\mathrm{rms}}$ into a right triangle: real power $P = VI\cos\varphi$ (watts, does work), reactive power $Q_r = VI\sin\varphi$ (volt-amperes-reactive, "var" — the borrowed flow), and their hypotenuse the apparent power $S = VI$ (volt-amperes, what the cables and transformer must carry). The triangle is just the impedance triangle rescaled, and it is the language of Worked example 2: utilities bill industry for low $\cos\varphi$ because their copper carries $S$ while the customer only uses $P$ .

Resonance: the circuit that picks a frequency

Drive the series RLC at variable frequency with fixed amplitude $V_0$ . The phasor article's result gives the current amplitude

I_0(\omega) = \frac{V_0}{\sqrt{R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2}}

where: everything frequency-dependent sits in the net reactance. The denominator is smallest — bare $R$ — when $\omega L = 1/\omega C$ , i.e. at $\omega_0 = 1/\sqrt{LC}$ : the driven circuit responds maximally at exactly the frequency the undriven circuit rings at (aq-em-lc-rlc-oscillations). At that point the phase is zero and the source, seeing pure resistance, delivers maximum power $V_{\mathrm{rms}}^2/R$ ; at half-power the delivered power is down by 2, which is the origin of the "−3 dB" bandwidth convention. Through the oscillator dictionary this curve is the same driven-oscillator response anatomized in the mechanics track's Resonance and Resonance Curve and Bandwidth — every theorem there (peak shift with heavy damping, Lorentzian shape near the peak) transfers without re-derivation.

Go deeper · Derivation B · 2 minQ three ways: reactance ratio, bandwidth, energy

Derivation BQ three ways: reactance ratio, bandwidth, energy

We want the quality factor $Q$ to emerge as one number with three independent readings — because in practice it is used in all three: designers compute $\omega_0 L/R$ , measurers read $\omega_0/\Delta\omega$ off a spectrum, and physicists reason with the energy ratio. The strategy: define the half-power bandwidth honestly by solving for the two half-power frequencies exactly, then audit the energy books in the resonant steady state. The satisfying surprise is that the bandwidth result $\Delta\omega = R/L$ is exact, not a narrow-band approximation.

Step 1. Half-power condition. Power goes as $I_0^2$ , so half power means $I_0 = I_{\max}/\sqrt2$ , i.e. the denominator of $I_0(\omega)$ equals $\sqrt2\,R$ :

\left(\omega L - \frac{1}{\omega C}\right)^2 = R^2 \quad\Longrightarrow\quad \omega L - \frac{1}{\omega C} = \pm R

Step 2. Solve exactly. Each sign gives a quadratic, $\omega^2 \mp (R/L)\,\omega - \omega_0^2 = 0$ , whose positive roots are the half-power edges:

\omega_{1,2} = \mp\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \omega_0^2}

Subtracting: $\Delta\omega = \omega_2 - \omega_1 = R/L$ exactly, and multiplying: $\omega_1\omega_2 = \omega_0^2$ exactly — the edges sit at equal ratios from $\omega_0$ , not equal distances (the curve is symmetric on a log axis). Why is the width $R/L$ and not something involving $C$ ? $R/L$ is the ringdown decay rate of the free circuit (envelope $e^{-Rt/2L}$ , energy $e^{-Rt/L}$ ): a resonance's width in frequency is its decay rate in time. Sharp filters ring long; this reciprocity is the same theorem as a spectral line's natural width, met at the field level in the radiation chapters.

Step 3. First two readings. Define $Q \equiv \omega_0/\Delta\omega$ (the measurer's reading). Then

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

— the designer's reading: the reactance of either element at resonance, over the resistance.

Step 4. Energy reading. In the resonant steady state with current $i = I_0\cos\omega_0 t$ , the stored energy $\tfrac12 Li^2 + q^2/2C$ is constant at $\tfrac12 L I_0^2$ (the slosh audit of the oscillations article: each store fills as the other empties). The dissipation per cycle is $\bar P\,T = \tfrac12 I_0^2 R \cdot 2\pi/\omega_0$ . Their ratio:

2\pi\,\frac{\text{energy stored}}{\text{energy dissipated per cycle}} = 2\pi\,\frac{\tfrac12 L I_0^2}{\tfrac12 I_0^2 R\,(2\pi/\omega_0)} = \frac{\omega_0 L}{R} = Q

Why does this reading matter? It is the definition that generalizes beyond circuits — to cavities, mechanical resonators, atoms — anywhere "stored" and "lost per cycle" make sense. A $Q$ of a million (a quartz crystal) means the oscillation is a millionfold better at storing than losing.

Step 5. The magnification corollary. At resonance the current is $I_0 = V_0/R$ , so the capacitor's peak voltage is $I_0 X_C = (V_0/R)(1/\omega_0 C) = Q V_0$ — and likewise $V_L = QV_0$ , the two in opposite phase, cancelling in the loop sum while each stands $Q$ times taller than the source. A 10 mV antenna signal across a $Q = 50$ tuned circuit presents 0.5 V to the detector: passive, free amplification, and the reason tuning capacitors carry voltage ratings far above the signals they handle. Verified: for $R = 25\ \Omega$ , $L = 30\ \mathrm{mH}$ , $C = 8\ \mu\mathrm{F}$ — the resonance-curve figure — the exact roots give $\Delta f = 132.6\ \mathrm{Hz} = R/2\pi L$ and $f_1 f_2 = f_0^2$ to machine precision; $Q = \omega_0 L/R = 2.45 = f_0/\Delta f = 325/133$ . ✓

Go deeper · Derivation C · 2 minTuning a radio: the resonant receiver, end to end

Derivation CTuning a radio: the resonant receiver, end to end

We want the flagship application computed honestly: an AM broadcast receiver that must sweep 530–1700 kHz with one knob and reject a station 10 kHz away. The strategy is to size the tuning capacitor from the resonance condition, then confront selectivity with the exact response formula — including the honest disappointment that launched a whole architecture (the superheterodyne). The numbers are the historical ones: a standard 240 μH ferrite antenna coil and the classic variable capacitor.

Step 1. Size the capacitor range. Resonance requires $C = 1/\omega_0^2 L = 1/(2\pi f)^2 L$ . With $L = 240\ \mu\mathrm{H}$ :

C(530\ \mathrm{kHz}) = \frac{1}{(2\pi \times 5.3\times10^5)^2 (2.4\times10^{-4})} = 376\ \mathrm{pF}, \qquad C(1700\ \mathrm{kHz}) = 36.5\ \mathrm{pF}

Why does a 10:1 capacitance swing cover only a 3.2:1 frequency band? Because $f \propto 1/\sqrt{C}$ — the square root compresses the dial. The classic "365 pF" air-gang variable capacitor exists precisely because of this calculation: its 10–365 pF swing (plus strays) spans the AM band with one shaft.

Step 2. Selectivity at the band's middle. Stations sit 10 kHz apart (9 kHz in some regions). Near resonance the exact response ratio is

\frac{I(\omega)}{I(\omega_0)} = \frac{1}{\sqrt{1 + Q^2\left(\frac{\omega}{\omega_0} - \frac{\omega_0}{\omega}\right)^2}} \approx \frac{1}{\sqrt{1 + \left(2Q\,\delta\right)^2}}, \qquad \delta = \frac{\Delta f}{f_0}

For a good LC front end $Q \approx 50$ : at $f_0 = 1000$ kHz an interferer at 1010 kHz has $2Q\delta = 2(50)(0.01) = 1$ , response $1/\sqrt2$ — the neighbour sits exactly at the half-power edge, attenuated a mere 3 dB. Even $Q = 100$ manages only $1/\sqrt5 = 0.45$ (−7 dB). Why can't we just build Q = 5000? Coil losses set a floor on $R$ ; worse, the bandwidth must stay wide enough for the program: AM audio needs ±5 kHz, so $Q > f_0/10\ \mathrm{kHz} = 100$ at the top of the band would start eating the music itself. Single-tuned selectivity is structurally mediocre.

Step 3. The honest resolution. Real receivers since Armstrong (1918) convert every station to one fixed intermediate frequency (455 kHz), where a cascade of tuned stages — each contributing its $Q$ — builds the sharp skirt once and for all: the superheterodyne. The front-end LC of Step 1 survives in every one of them, now doing the coarse job it is actually good at.

Step 4. The magnification bonus. The antenna induces perhaps 10 mV in the coil; with $Q = 50$ the tuned circuit presents $QV = 0.5\ \mathrm{V}$ of the chosen station to the detector diode — the passive gain that made crystal sets (no battery, no amplifier, one diode, one earphone) genuinely work, powered entirely by the transmitter. Verified: capacitor endpoints 375.7 pF and 36.5 pF recomputed from $C = 1/(2\pi f)^2L$ ; response ratios $1/\sqrt{1+(2Q\delta)^2}$ = 0.707 at $Q=50$ and 0.447 at $Q=100$ for 10 kHz at 1 MHz; $50 \times 10\ \mathrm{mV} = 0.5\ \mathrm{V}$ .

A one-paragraph mirror completes the picture: put $L$ and $C$ in parallel across a source and everything inverts. At $\omega_0$ the tank's impedance is maximum (ideally infinite; in practice $\approx Q^2 R$ with $R$ the coil resistance): current circulates vigorously between $L$ and $C$ — the free oscillation, sustained — while the source supplies only the losses. Series resonance selects a frequency by conducting it; parallel resonance selects by refusing to load it. Oscillators, RF amplifier loads, and induction-heating tanks are parallel-resonant; the mathematics is the same triangle with impedance and admittance swapping roles.

Limiting cases

case	result	meaning
$\varphi = 0$ (resistive / at resonance)	$\bar P = V_{\mathrm{rms}}I_{\mathrm{rms}}$ , pf $= 1$	every ampere works; DC formulas apply verbatim
$\varphi = \pm90°$ (pure reactance)	$\bar P = 0$	current flows, cables heat, nothing delivered
$\omega \ll \omega_0$	capacitor starves the loop; current leads	capacitive regime
$\omega \gg \omega_0$	inductor starves the loop; current lags	inductive regime
$R \to 0$	$Q \to \infty$ : response narrows to a spike, $V_C/V_s \to \infty$	the ideal filter — and a voltage hazard
$Q \lesssim 1/\sqrt2$	no usable peak; curve slumps	overdamped driving — matches the ringless regime of the oscillations article
$Q$ large	$\Delta\omega = \omega_0/Q$ , edges at equal ratios ( $\omega_1\omega_2 = \omega_0^2$ )	narrow-band filter; symmetric on a log axis

Common trap“Power factor 0.7 means the motor only uses 70% of the energy supplied — the rest is wasted inside it”

Nothing is consumed twice and nothing vanishes inside the motor: a 10 kW motor at power factor 0.7 uses exactly 10 kW. What the low power factor means is that delivering those 10 kW requires $10/0.7 = 14.3$ kVA of apparent power — the motor draws 62 A instead of the 43.5 A a unity-power-factor load would need, returning the surplus energy to the grid every half cycle. The waste is in the delivery infrastructure: cable and transformer heating scale as $I^2$ , so the utility's losses rise by $(1/0.7)^2 \approx 2$ while the customer's meter (which integrates real power) never sees it. That is why utilities meter and penalize reactive power for industry, and why the cure — Worked example 2's capacitor bank — happens at the factory, not the power station. The reciprocal mistake also bites: "kVA" on a transformer nameplate is not "kW available"; at pf 0.8 a 100 kVA transformer tops out at 80 kW.

Worked examples

Example 1 — the kettle and the motor (confidence builder). A 2.0 kW kettle and a 2.0 kW induction motor (power factor 0.7 lagging) each run on 230 V rms mains. Find each device's current.

Step 1. Kettle: resistive, $\cos\varphi = 1$ , so $I = P/V = 2000/230 = \mathbf{8.70\ A}$ .

Step 2. Motor: $I = P/(V\cos\varphi) = 2000/(230 \times 0.7) = \mathbf{12.4\ A}$ — 43% more current for the same power and the same electricity bill.

Step 3. Read the consequence: the motor's supply wiring dissipates $(12.4/8.70)^2 = 2.0$ times the heat of the kettle's for identical delivered energy. Multiply by a factory of motors and the utility's interest in $\cos\varphi$ stops being academic.

Example 2 — designing a power-factor correction bank (exam level). A workshop's machines draw $P = 10\ \mathrm{kW}$ at power factor 0.70 lagging from 230 V, 50 Hz mains. Size the parallel capacitor that raises the plant's power factor to 0.95, and find the supply-current saving.

Step 1. Before: $S_1 = P/0.70 = 14.3\ \mathrm{kVA}$ , supply current $I_1 = 14{,}300/230 = 62.1\ \mathrm{A}$ , reactive power $Q_1 = P\tan(\arccos 0.70) = 10\,\tan 45.6° = 10.2\ \mathrm{kvar}$ (lagging — the machines' magnetization).

Step 2. Target: at pf 0.95, $Q_2 = P\tan(\arccos 0.95) = 10\,\tan 18.2° = 3.29\ \mathrm{kvar}$ . The capacitor must supply the difference locally: $\Delta Q = Q_1 - Q_2 = 6.92\ \mathrm{kvar}$ .

Step 3. Size the capacitor. A capacitor across the mains sources reactive power $Q_C = V^2/X_C = \omega C V^2$ , so

C = \frac{\Delta Q}{\omega V^2} = \frac{6920}{(2\pi \times 50)(230)^2} = \mathbf{416\ \mu F}

Step 4. After: $I_2 = P/(V \times 0.95) = 10{,}000/218.5 = \mathbf{45.8\ A}$ — a 26% current reduction with zero change in delivered power. The machines still run on their own lagging current; it now circulates between them and the capacitor bank across the room instead of through the substation. (Design note: never overshoot to pf = 1 exactly — a fully resonant plant plus the grid's inductance is a tuned circuit, and switching transients would ring it.)

Origins · 1 minNikola Tesla and how this was discovered

Nikola Tesla

1856–1943 · Smiljan, Budapest, New York

This track's magnetostatics chapter credits Tesla's rotating field geometry — the induction-motor insight itself, told in Magnetic Field of Steady Currents. This article inherits the other half of his 1888 argument: polyphase AC as a power-delivery system. His AIEE lecture "A New System of Alternate Current Motors and Transformers" laid out, with phasor-free 1880s reasoning, why multiple currents at fixed phase offsets deliver constant total power (the $2\omega$ pulsing of our hero figure cancels between phases — a three-phase machine hums where a single-phase one buzzes) and why the whole scheme lives or dies on the phase bookkeeping this chapter formalizes. Licensed to Westinghouse, the system won the generation-and-transmission war (that industrial story belongs to the Transformers article); what belongs here is Tesla the resonance engineer: his magnifying transmitter was a deliberately driven high- $Q$ resonant transformer, exploiting exactly the $QV$ voltage magnification of this article to reach millions of volts from mains-fed primaries — the same physics, scaled until it made lightning.

History

1888

— Tesla's AIEE lecture presents the polyphase AC system; Westinghouse licenses the patents within months.

1891

— Lauffen–Frankfurt demonstration transmits three-phase power 175 km at 75% efficiency, ending the DC-vs-AC engineering argument.

1893

— Westinghouse lights the Chicago World's Fair with polyphase AC; Kennelly and Steinmetz present the complex-impedance methods that make such systems calculable.

1918

— Armstrong's superheterodyne receiver concedes the selectivity limits of single-tuned circuits and builds sharpness at a fixed intermediate frequency instead.

1920

s — Utilities begin metering reactive power; the capacitor bank becomes standard industrial equipment.

Going further

The natural next step is aq-em-transformers, where rms power bookkeeping meets Faraday's law and becomes the architecture of the grid — including why the war Tesla's system won was really an argument about $I^2R$ . The phasor machinery underneath came from aq-em-ac-circuits-phasors; the resonant frequency and the ringdown from aq-em-lc-rlc-oscillations; the dissipation element from aq-em-current-resistance-resistivity. On the mechanics side, the resonance curve's full anatomy — peak shift at low $Q$ , phase walk through 90°, the Lorentzian approximation — is developed in Resonance and Resonance Curve and Bandwidth, and transfers here symbol for symbol through the oscillator dictionary. Provenance: Authored fresh for the SuperSTEM E&M track's circuits chapter (recon-§d gap heal); sources: v3 circuit pilots (rewritten to UG register) + legacy wiki circuit articles (narrative) — no AllQFT base exists for this topic.

If reactive power does no work, why do utilities bother billing for it?

Because delivering it costs real money even though consuming it costs nothing. The reactive component of current is indistinguishable from working current in every wire, breaker, and transformer winding between the power station and the load: it occupies ampacity and causes $I^2R$ losses that the utility, not the customer, pays for. A plant at pf 0.7 forces the grid to build and cool infrastructure for 43% more current than its energy use justifies. Billing kvar-hours (or penalizing low pf) simply hands that cost to the party who can fix it cheapest — with a local capacitor bank.

Why is average power V·I·cosφ and not just V·I?

Write both signals and multiply: the product of two sinusoids at the same frequency splits into a constant term $\tfrac12 V_0I_0\cos\varphi$ and a term at $2\omega$ that averages to zero. Only the overlap of the current with the in-phase part of the voltage survives averaging. Geometrically: decompose the current phasor into a component along the voltage ( $I\cos\varphi$ , the working part) and one at right angles ( $I\sin\varphi$ , the borrowed part); the right-angle component's energy flow reverses sign four times per cycle and cancels exactly.

Is the resonance curve of a circuit really the same as a mechanical resonance, or just similar-looking?

The same, exactly, through the dictionary of the oscillations article: $L\ddot q + R\dot q + q/C = V_0\cos\omega t$ maps term by term onto $m\ddot x + b\dot x + kx = F_0\cos\omega t$ . The current amplitude here corresponds to the velocity amplitude there — which is why the electrical current peaks exactly at $\omega_0$ , just as mechanical velocity resonance does, while the mechanical displacement peak sits slightly below $\omega_0$ . If you plot capacitor charge instead of current, you recover the displacement curve, shifted peak and all. One equation, two costumes.

What physically limits how high Q can be made?

Whatever loss you cannot remove. In circuits: the coil's wire resistance (including its rise with frequency from the skin effect), core losses, capacitor dielectric loss, and — often decisive — the energy the circuit usefully delivers, since a loaded resonator's effective $R$ includes the load. Air-cored LC circuits manage $Q$ of tens to a few hundred; quartz crystals reach $10^4$ – $10^6$ by storing energy mechanically in a nearly lossless lattice; superconducting cavities exceed $10^9$ . And a broadcast receiver deliberately keeps $Q$ finite: bandwidth is $f_0/Q$ , and a filter narrower than the signal's sidebands destroys the very program it selects.

How can the capacitor voltage exceed the source voltage without violating energy conservation?

Voltage is not energy. At resonance the inductor and capacitor voltages are equal, opposite, and each $Q$ times the source — their sum in the loop is zero, so Kirchhoff is satisfied, and the source only ever supplies the resistor's dissipation. The tall element voltages were built up over many cycles: the source deposits a little energy each cycle and the resonator accumulates it (that is the energy reading of $Q$ ), exactly as small pushes accumulate into a child's high swing. Nothing is created; it is stored, gradually, and sloshed.

Why does AC power pulse at twice the mains frequency, and does anything notice?

Power is a product of two signals at $\omega$ , and products of sinusoids contain the sum frequency: $2\omega$ . Physically, power delivery peaks twice per cycle (both when $v$ and $i$ are both positive and when both are negative). Plenty of things notice: single-phase motors and transformers hum at 100/120 Hz (twice the mains frequency — a diagnostic worth remembering), incandescent filaments flicker slightly at $2f$ , and LED drivers need storage capacitors precisely to bridge the hundred-times-a-second power nulls. Three-phase systems exist largely because their three staggered $2\omega$ pulsations sum to a constant — Tesla's original selling point.

For a square wave or a dimmer-chopped waveform, is P = Vrms·Irms·cosφ still right?

Not as written — that formula is for single-frequency sinusoids. The honest general statement is $\bar P = \overline{v(t)\,i(t)}$ , averaged over a period. For non-sinusoidal waveforms you either compute that average directly or decompose into Fourier components: each harmonic delivers $V_n I_n \cos\varphi_n$ independently, and cross-frequency terms average to zero. This is why "true-rms" meters matter for chopped waveforms, and why engineers define a distortion power factor: a rectifier can draw current perfectly in phase yet still have pf < 1 because its current harmonics carry rms without carrying power.