Power in AC Circuits & Power Factor

An electric kettle and an induction motor sit on the same 230 V mains. Each consumes exactly 2.0 kilowatts — the electricity meter swears to it. Yet an ammeter tells a different story: the kettle draws 8.7 amperes, the motor 12.4. Same voltage, same watts, forty percent more current. Somewhere in the motor's circuit, nearly four extra amperes are flowing that deliver nothing — real current, heating real wires, billed to nobody. This module is the account book that explains them. The chapter so far has established a strange regime: inductors and capacitors host full-sized voltages and currents while consuming, on average, zero energy — every joule borrowed is returned. So in a general AC circuit, some of the current is doing work and some is merely shuttling energy back and forth. The exact split is set by the phase angle $\varphi$ that the impedance triangle handed us, and the headline result is one clean formula:

P = V_{\mathrm{rms}}\, I_{\mathrm{rms}} \cos\varphi

The factor $\cos\varphi$ — the power factor — is the fraction of the volt-ampere product that becomes real work. The rest of the current, $I\sin\varphi$ , is the famous wattless current: present, measurable, wire-heating, and energetically sterile. One product-to-sum identity derives all of it, the choke coil's free-lunch claim from earlier in the chapter finally gets its arithmetic, and the kettle-versus-motor mystery resolves into the single number utilities care about most.

Key takeaways

Instantaneous power in AC is $p = vi$ , computed with both waves' timing respected: $p = V_0I_0\sin\omega t\,\sin(\omega t - \varphi)$ .
Averaged over a cycle this gives $P = \tfrac{1}{2}V_0I_0\cos\varphi = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$ — the only AC power formula you need.
$\cos\varphi = R/Z$ is the power factor: 1 for a pure resistor, 0 for a pure inductor or capacitor, in between for real circuits.
Only the resistor eats energy: equivalently $P = I_{\mathrm{rms}}^2R$ — the two power formulas always agree.
The current component $I\sin\varphi$ is wattless: it flows, it heats supply cables, it delivers no net energy — the choke coil's current is entirely of this kind.
Key idea: $P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$ — the phase angle splits the current into a working part ( $I\cos\varphi$ ) and a wattless part ( $I\sin\varphi$ ), and only the working part shows up on the meter.

The core insight

The average power delivered to any AC circuit is $P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$ , where the power factor $\cos\varphi = R/Z$ measures what fraction of the current works in phase with the voltage; the quadrature remainder $I\sin\varphi$ is wattless — real current that borrows and returns energy without ever delivering any, dissipating nothing itself but taxing every cable that carries it.

Where this lives today

Power factor never left the bill. Every industrial electricity tariff in India carries a power-factor clause — typically a penalty below 0.9 and a small rebate near unity — and every factory's electrical room answers it with racks of correction capacitors, switching in and out as motor load changes. The physics is this module's decomposition run in reverse: buy leading wattless current to cancel the motors' lagging wattless current before it reaches the utility's cables. The modern twist is that the loads have changed shape. A phone charger or LED driver draws current in narrow gulps rather than a lagging sine, spoiling the power factor by waveform rather than phase — which is why regulations now force anything above a few watts to include "power factor correction" circuitry on board, and why the label on a laptop brick quotes VA and W as separate numbers. Data centres, with megawatts behind their walls, employ engineers whose entire brief is the gap between those two numbers. Meanwhile the wattless current's one genuine use survives at the top of the chapter's next module: utilities deliberately circulate reactive current to hold up transmission-line voltage. The grid that must carry all these amperes — working and wattless alike — at continental scale is The Grid: Why Transmit at High Voltage, next door.

The physical picture

Watch the energy flow into a purely resistive load — the kettle — over one mains cycle. The instantaneous power $p = vi$ pulses (it is zero whenever the waves cross zero, a hundred times a second on 50 Hz mains) but it is never negative: voltage and current reverse together, so their product is always ≥ 0. Every milliwatt that leaves the supply arrives in the water. The kettle's thermal mass smooths the pulses into steady heat, and the average works out to exactly $V_{\mathrm{rms}}I_{\mathrm{rms}}$ — which is precisely what rms values were invented to make true (AC Signals: Peak, Average & RMS). Now watch a load whose current lags by, say, $\varphi = 60°$ — a motor, a choke, any circuit with a net reactance. For most of each half-cycle, $v$ and $i$ have the same sign: power flows in. But there are now intervals where the lagging current still runs positive after the voltage has gone negative (and vice versa) — the product turns negative, and power flows backwards, out of the load and into the supply. The load's magnetic field is returning energy it borrowed a few milliseconds earlier. The net delivery is what survives after these refunds, and the bigger the phase gap, the deeper and longer the refund intervals: at $\varphi = 90°$ — the pure inductor and pure capacitor of the earlier modules — the backward lobes exactly cancel the forward ones and the average is zero, the quarter-cycle deposit-and-refund ledger both those modules kept.

Figure

Instantaneous power over one cycle — a resistive load versus a load lagging by 60 degrees

The phasor decomposition on the right of the figure is the module's mental model: split the current arrow into a component along the voltage ( $I\cos\varphi$ — in phase, doing steady work) and a component at right angles ( $I\sin\varphi$ — pure borrowing and returning). The average power is the voltage times the working component only. Everything below is that sentence made rigorous.

The flagship derivation: averaging the power

Derivation AFrom instantaneous product to the power factor formula

The goal is $P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$ , derived honestly from the instantaneous product with nothing beyond one trigonometric identity and one symmetry fact this chapter already owns. The strategy: write both waves with the phase gap in place, multiply, convert the awkward product into a sum with the product-to-sum identity, and then let the full-cycle average kill every oscillating term. Watch which term survives — it is the one that does not oscillate at all.

Step 1. Set up the general case. Let the supply voltage be $v = V_0\sin\omega t$ and the current $i = I_0\sin(\omega t - \varphi)$ — a current lagging by the circuit's phase angle $\varphi$ (for a leading current, $\varphi$ is negative; nothing below changes). The instantaneous power is the plain product:

p = vi = V_0I_0\,\sin\omega t\,\sin(\omega t - \varphi)

Step 2. Convert the product to a sum. The identity $\sin A\sin B = \tfrac{1}{2}[\cos(A-B) - \cos(A+B)]$ , with $A = \omega t$ and $B = \omega t - \varphi$ , gives

p = \frac{V_0I_0}{2}\big[\cos\varphi - \cos(2\omega t - \varphi)\big]

Why this is the decisive move: the messy product of two waves has split into a constant term, $\cos\varphi$ , and a pure oscillation at double frequency, $\cos(2\omega t - \varphi)$ . The physics of the figure is already visible — power pulses at $2\omega$ around a steady offset, and the offset is set by the phase gap.

Step 3. Average over one full cycle. The full-cycle average of any sinusoid is zero — the symmetry fact established back in AC Signals: Peak, Average & RMS: equal areas above and below its mean. The $\cos(2\omega t - \varphi)$ term therefore averages to nothing, and only the constant survives:

\bar{p} = \frac{V_0I_0}{2}\cos\varphi

Step 4. Dress it in rms clothing. Since $V_0 = \sqrt{2}\,V_{\mathrm{rms}}$ and $I_0 = \sqrt{2}\,I_{\mathrm{rms}}$ , the half absorbs the two $\sqrt2$ 's:

P = V_{\mathrm{rms}}\,I_{\mathrm{rms}}\cos\varphi

where: $\cos\varphi$ is the power factor, read straight off the impedance triangle of the Series LCR module: adjacent over hypotenuse, $\cos\varphi = R/Z$ . The three sanity checks: pure resistor, $\varphi = 0$ , $P = V_{\mathrm{rms}}I_{\mathrm{rms}}$ — the resistive law recovered ✓. Pure inductor or capacitor, $\varphi = \pm90°$ , $P = 0$ — both modules' zero-power ledgers recovered ✓. And substituting $V_{\mathrm{rms}} = I_{\mathrm{rms}}Z$ with $\cos\varphi = R/Z$ gives $P = I_{\mathrm{rms}}^2 Z \cdot (R/Z) = I_{\mathrm{rms}}^2R$ — the statement that only the resistance ever eats ✓. Verified: numerically averaging $\sin\omega t\sin(\omega t - 60°)$ over one cycle gives 0.2500 = ½cos 60° ✓; the identity check ½[cos 60° − cos(2ωt − 60°)] reproduces the product pointwise ✓; for the anchor circuit below, both power routes agree to the watt.

The formula family is worth displaying as a family, because examiners rotate freely among its members:

P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi = I_{\mathrm{rms}}^2R = \frac{V_{R,\mathrm{rms}}^2}{R}

All three say the same thing: find the resistor's share, because nobody else is eating. The product $V_{\mathrm{rms}}I_{\mathrm{rms}}$ without the cosine also has a name — the apparent power, measured in volt-amperes (VA) — and it is what the supply infrastructure must be built to carry, whether or not the load makes it useful.

Wattless current — the choke's free lunch, audited

Split the current as in the figure: $I\cos\varphi$ in phase with the voltage, $I\sin\varphi$ in quadrature. The in-phase component delivers all the power; the quadrature component delivers, over any full cycle, exactly none — it is wattless current (the syllabus term; "idle current" in older books). It is not fictitious: it flows through every wire, trips breakers, heats cables by $I^2R$ in the cables — it merely does no work at the load. Now the choke coil's promise from AC Through an Inductor can be cashed with arithmetic instead of a ledger. A choke in series with a lamp limits the current by adding reactance rather than resistance. For the ideal choke itself, $\varphi = 90°$ : its power factor is $\cos 90° = 0$ , its entire current is wattless, and its power consumption is $P = VI\cos 90° = 0$ — current limiting with no heat bill at the choke, which is exactly why tube-light ballasts are chokes and not resistors. (A real choke's thin winding resistance spoils this slightly; "almost wattless" is the honest engineering verdict.) The same split explains the opening mystery — and why utilities police the power factor of industry. To deliver a fixed power $P$ at fixed voltage $V$ , a load needs current $I = P/(V\cos\varphi)$ : the worse the power factor, the more current for the same watts. The kettle ( $\cos\varphi = 1$ ) takes 8.7 A; the motor ( $\cos\varphi = 0.7$ ) takes 12.4 A for the same 2.0 kW, and cables sized for it. Since cable heating goes as $I^2$ , the delivery losses for the motor run $(1/0.7)^2 \approx 2$ times the kettle's — losses the utility pays for but the customer's watt-hour meter never records. Multiply by a city of motors and the interest stops being academic: industrial tariffs bill low power factor explicitly, and factories install banks of capacitors — elements whose leading wattless current cancels the motors' lagging wattless current locally, so the two trade energy across the factory floor instead of hauling it through the grid. (Sizing such a bank is a lovely calculation, but it is UG fare; and the full grid-scale economics of moving power at high voltage is the next module's whole story.) One last dividend from $P = VI\cos\varphi$ : at resonance, the previous module showed $\varphi = 0$ and $I = V/R$ — both factors peak together, so power delivered is maximum at resonance, the syllabus point that ties the two modules into one sentence.

Figure

The power triangle — apparent power splits into real and reactive parts by the same angle φ

Limiting cases

case	power factor	consequence
pure resistor	$\cos\varphi = 1$	$P = V_{\mathrm{rms}}I_{\mathrm{rms}}$ ; all current works
pure inductor	$0$ ( $\varphi = +90°$ )	$P = 0$ ; all current wattless — the choke
pure capacitor	$0$ ( $\varphi = -90°$ )	$P = 0$ ; wattless but leading
series LCR at resonance	$1$ ( $Z = R$ )	maximum power; circuit resistive
general series LCR	$\cos\varphi = R/Z$	read it off the impedance triangle
same $P$ , halved $\cos\varphi$	current doubles	cable heating quadruples — the utility's grievance

Common trapA device with power factor 0.7 wastes 30 percent of the energy it consumes

It wastes none of it — the meter reading and the physics both say the 2 kW motor uses exactly 2 kW. What $\cos\varphi = 0.7$ means is that delivering those honest watts requires $1/0.7 = 1.43$ times the current a resistive load would need, because a wattless component rides along with the working one, borrowed and returned every half-cycle. The waste, such as it is, happens in the supply infrastructure: every cable, transformer and fuse between the power station and the motor carries the full 12.4 A and heats as its square, roughly doubling the delivery losses compared with a unity-power-factor load. Three linked mark-losers hide here. First: "power consumed" questions must use $P = VI\cos\varphi$ , never $P = VI$ — the product $VI$ is apparent power (volt-amperes), a capacity rating, not a consumption. Second: the reciprocal trap — a transformer or generator plated "100 kVA" delivers at most 80 kW into a 0.8-power-factor load; kVA is not kW. Third: "wattless" never means "harmless" — the wattless amperes are real in every conductor they traverse, which is exactly why utilities meter and penalize them for industry. Nothing is consumed twice, nothing vanishes in the motor; the bill for a bad power factor is paid in copper, upstream.

Worked examples

Board level — one circuit, both power routes (CBSE/ISC pattern, 3 marks)

A series circuit with $R = 40\ \Omega$ and $X_L = 30\ \Omega$ is connected to a 230 V, 50 Hz supply. Find the power factor and the power dissipated, and verify by a second route.

Step 1. Impedance: $Z = \sqrt{40^2 + 30^2} = 50\ \Omega$ (the 3-4-5 family). Current: $I = 230/50 = 4.6$ A.

Step 2. Power factor from the triangle: $\cos\varphi = R/Z = 40/50 = \boxed{0.8}$ (lagging, since the reactance is inductive — say so for the qualifying mark).

Step 3. Power: $P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi = 230 \times 4.6 \times 0.8 = \boxed{846.4\ \mathrm{W}}$ .

Step 4. Verify via the resistor's share: $P = I^2R = (4.6)^2 \times 40 = 21.16 \times 40 = 846.4$ W ✓. Two routes, one answer — writing the check earns the "verify" mark and catches slips for free.

Competitive level — the kettle and the motor, audited to the ampere (JEE-Main/NEET pattern, ~3 minutes)

A 2.0 kW kettle (power factor 1) and a 2.0 kW induction motor (power factor 0.7 lagging) each run on 230 V rms mains. Find (a) each device's current, (b) the wattless component of the motor's current, (c) the ratio of supply-cable heating for the two devices.

Step 1. Kettle: $I = P/(V\cos\varphi) = 2000/(230 \times 1) = \boxed{8.70\ \mathrm{A}}$ — all of it working.

Step 2. Motor: $I = 2000/(230 \times 0.7) = \boxed{12.42\ \mathrm{A}}$ . Same watts, forty-three percent more amperes — the current the power factor formula predicts and the opening mystery reported.

Step 3. Wattless component: $\sin\varphi = \sqrt{1 - 0.7^2} = \sqrt{0.51} = 0.714$ , so $I\sin\varphi = 12.42 \times 0.714 = \boxed{8.87\ \mathrm{A}}$ — nearly as large as the kettle's entire current, flowing and delivering nothing. (Check the decomposition: working component $12.42 \times 0.7 = 8.70$ A — exactly the kettle's, as it must be for equal watts at equal voltage; and $\sqrt{8.70^2 + 8.87^2} = 12.42$ ✓.)

Step 4. Cable heating goes as $I^2$ : ratio $= (12.42/8.70)^2 = (1/0.7)^2 = \boxed{2.04}$ . The motor's cables run twice as hot for the same delivered energy — the utility's grievance, quantified. Timing note: 3 minutes. The efficient order is watts → current (divide by $V\cos\varphi$ in one step), then trig for the split. The elegant check in (c) — the heating ratio is just $1/\cos^2\varphi$ , no currents needed — is worth internalizing: many MCQs ask only for that ratio, and it is a five-second answer.

Exam traps

$P = VI$ is not a formula. In AC it is the apparent power. Real power always carries the cosine: $P = VI\cos\varphi$ . The only time the bare product is right is $\varphi = 0$ — and then you should say why (resistive, or at resonance).

The power factor comes from the triangle. $\cos\varphi = R/Z$ — adjacent over hypotenuse. Not $R/X$ , not $X/Z$ (that's $\sin\varphi$ , the wattless fraction). One glance at the triangle prevents all three confusions.

$I^2R$ , never $I^2Z$ or $I^2X$ . Only resistance dissipates. $I^2Z$ is the apparent power again in disguise; $I^2X$ is the peak rate of energy borrowing, not a consumption. If a power answer needs a reactance in it, the answer is wrong.

"Wattless" ≠ "currentless." The quadrature component is physically present in every wire — it heats cables, sizes fuses and trips breakers. What it never does is deliver net energy to the load.

kVA versus kW. Equipment ratings in kVA state the volt-ampere capacity; the deliverable watts are kVA × power factor. Reading "100 kVA" as "100 kW" is a one-mark gift to the examiner.

Power at resonance. Maximum — both because $I$ peaks and because $\cos\varphi$ hits 1. "At resonance the power factor is unity" is a free mark whenever it applies.

Boards and depth. CBSE/NEET: "power in AC circuits; wattless current" is the syllabus line verbatim — the derivation of $\tfrac12 V_0I_0\cos\varphi$ , the power factor, and the wattless split. ISC: adds the explicit three-member formula family $P = \tfrac12 V_0I_0\cos\varphi = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi = I_{\mathrm{rms}}^2R$ and the choke coil "analytical only" — the $\cos\varphi = 0$ , $P = 0$ argument made here. JEE-Main: formula-family numericals and the kVA/kW discrimination. JEE-Advanced: power audits inside full LCR problems. AP Physics 2 has no AC on its syllabus at all, and IB (2023 onward) dropped AC circuits entirely — this module is CBSE/ISC/JEE/NEET territory.

Origins · 1 minElihu Thomson and how this was discovered

Elihu Thomson

1853–1937 · Lynn, Massachusetts → General Electric

Thomson is the engineer who made $\cos\varphi$ a line item on a bill. A Philadelphia high-school teacher turned inventor — nearly 700 patents by the end — he co-founded the Thomson-Houston company and built it into the Edison empire's great AC rival. His problem was the one this module derives: on alternating current, volts times amperes is not what a customer consumes, and in the 1880s nobody could meter the difference — utilities were billing guesses. Thomson's recording wattmeter (1889) settled it: an instrument whose deflection tracked the true product $vi$ averaged over every cycle — mechanically computing $V I\cos\varphi$ decades before anyone wrote it on a syllabus — and logging watt-hours a clerk could read monthly. Honest AC billing made the AC business viable: if you cannot meter it, you cannot sell it. In 1892 Thomson-Houston merged with Edison General Electric to form General Electric — the AC camp effectively absorbing the DC one — and Thomson spent his later decades as GE's elder statesman and twice acting president of MIT. The kettle-versus-motor audit of this module is his instrument's daily work, performed a billion times a night on every metered wall in the world.

History

1882

— Edison's Pearl Street station bills DC customers by lamp-count and primitive meters; AC has no honest meter at all

1888

— polyphase AC systems make phase angles an unavoidable commercial reality

1889

— Elihu Thomson's recording wattmeter measures true average power, cos φ included, and makes AC billable

1892

— Thomson-Houston and Edison General Electric merge into GE; the metered AC grid becomes the industry standard

1920

s — utilities begin billing industrial customers explicitly for poor power factor; capacitor-bank correction becomes routine engineering

Going further

Texts. NCERT Physics Class 12, Ch. 7 (§7.9) · HC Verma Vol. 2, Ch. 39 · SL Arora, ISC Physics XII, power in a.c. circuits, choke coil · Thomson's wattmeter: T. Hughes, Networks of Power, for the metering war.

Provenance. All results standard. A verifier should recompute: numerical average of $\sin\omega t\sin(\omega t - 60°)$ over a cycle $= 0.250 = \tfrac12\cos 60°$ ✓; board circuit $Z = 50$ , $I = 4.6$ A, $\cos\varphi = 0.8$ , $P = 230 \times 4.6 \times 0.8 = 846.4$ W $= 4.6^2 \times 40$ ✓; kettle 8.70 A, motor 12.42 A, wattless $12.42 \times \sqrt{0.51} = 8.87$ A, working $12.42 \times 0.7 = 8.70$ A, $\sqrt{8.70^2 + 8.87^2} = 12.42$ ✓, heating ratio $(1/0.7)^2 = 2.04$ ✓.

Prerequisites. AC Signals: Peak, Average & RMS (rms values and the zero-average symmetry) · The Series LCR Circuit & Impedance (the triangle whose cosine this is). Leads to. LC Oscillations (the pure borrowing, undriven) · The Grid: Why Transmit at High Voltage (the amperes-versus-watts economics at continental scale) · ARCHETYPE: The LCR Numericals Ladder (power questions drilled in context) · Alternating-Current Power & Resonance (UG deep dive).

If the wattless current delivers no energy, why do utilities care so much about it?

Because delivery is priced in amperes, not watts. Every conductor between the generator and the load — transmission line, transformer winding, street cable, fuse — must be thick enough to carry the total current and cool enough to survive its $I^2R$ heating, and the wattless component inflates that current without adding a single billable watt-hour. A factory at power factor 0.7 forces the utility to build and cool infrastructure for 43% more current than its consumption justifies, and to eat roughly double the delivery losses. The customer's meter integrates real power only, so without a separate penalty clause the utility pays for the copper and the heat. That is the entire economics: wattless current is free to the load, expensive to the wire. It is also why the cure — the factory's own capacitor bank — works: it moves the borrowing local, so the sterile amperes circulate across the factory floor instead of through fifty kilometres of grid.

Why does the derivation give exactly half of V₀I₀cos φ — where does the half come from?

From the same place as every rms half in this chapter: a sinusoid spends its life mostly below its peak. The instantaneous power splits into a constant part plus a double-frequency wobble; the wobble averages away, and the constant part is $\tfrac12 V_0I_0\cos\varphi$ because the product of two sines averages to half the product of their peaks (when in phase — the $\cos\varphi$ handles the timing gap). The rms machinery then absorbs the half elegantly: $V_0I_0/2 = (\sqrt2 V_{\mathrm{rms}})(\sqrt2 I_{\mathrm{rms}})/2 = V_{\mathrm{rms}}I_{\mathrm{rms}}$ . This is precisely why rms values exist: they are the DC-equivalent numbers under which AC power formulas shed their halves and look like Ohm's-law bookkeeping again. If a power formula you have written contains both a peak value and no half (or an rms value and a stray half), a factor of 2 has escaped — audit it.

Does the power really flow backwards during part of the cycle, physically?

Yes — measurably and usefully. During the intervals where $v$ and $i$ have opposite signs, the load's stored field (magnetic in a motor's windings, electric in a capacitor) is discharging back through the supply wires: the load is momentarily a generator. On the grid this is routine traffic; the supply absorbs the returned energy and redelivers it a few milliseconds later. You can see the bookkeeping in the figure's negative lobes, and an oscilloscope across a real motor shows them plainly. What never flows backwards is the average: as long as $\varphi < 90°$ the forward lobes outweigh the refunds, and the net movement is source-to-load. At exactly 90° the tug-of-war ties — the pure reactance — and beyond 90° would need the "load" to be a net source, which is what a generator is. The refund traffic is also why single-phase motors physically hum at double the mains frequency: the power pulsation at $2\omega$ in the derivation is a mechanical vibration you can hear.

How do I find the power factor if I'm given the circuit components rather than φ?

Build the triangle, take adjacent over hypotenuse — no inverse-trig required. From $R$ , $X_L$ , $X_C$ : form $Z = \sqrt{R^2 + (X_L - X_C)^2}$ and read $\cos\varphi = R/Z$ directly. Most exam circuits are 3-4-5 or 5-12-13 triangles in disguise, so the cosine is a clean fraction (0.8, 0.6, 12/13...) — if you find yourself computing $\varphi$ itself in degrees and then taking its cosine, you are doing two unnecessary steps and inviting rounding. The same triangle also hands you $\sin\varphi = (X_L - X_C)/Z$ for wattless-current parts, and its sign tells lagging (inductive, positive) versus leading (capacitive, negative) — one construction, all four facts. Reserve actual angle-finding for questions that explicitly ask "by what angle does the current lag?"

Is the wattless current the same thing as the current in a choke coil?

In the ideal choke, exactly — the choke is the wattless idea built as a product. An ideal inductor's current lags its voltage by the full 90°, so its working component $I\cos 90°$ is zero and everything it draws is quadrature: current that limits the lamp, sizes the wires, and consumes nothing. That is the entire design brief of the tube-light ballast — current limiting whose "resistance" refuses to heat. In a general circuit the wattless current is the component $I\sin\varphi$ of the total, not a separate current you could touch: one physical current flows, and the split into working and wattless parts is bookkeeping against the voltage's timing. The bookkeeping is still physical enough to act on — the factory capacitor bank cancels precisely that component — but no ammeter reads $I\sin\varphi$ alone; you compute it from the meter's total and the phase.

Why is the power maximum at resonance? Two different modules seem to claim credit

Both are right, and multiplying them is the point. Power is (current) × (voltage) × (power factor). At resonance the impedance bottoms out at $R$ , so the current peaks at $V/R$ — that is the resonance module's contribution. And the phase angle collapses to zero, so $\cos\varphi$ peaks at 1 — this module's contribution. $P = VI\cos\varphi$ takes both peaks at once: $P_{\max} = V \cdot (V/R) \cdot 1 = V^2/R$ , delivered entirely to the resistance. Away from $f_0$ both factors sag together — less current, and a smaller working fraction of it — which is why the power-versus-frequency curve is even sharper than the current curve (it is essentially its square). The examiner's phrasing "power delivered to the circuit by the source is maximum at resonance" is thus two modules' headlines multiplied into one sentence.