Transformers

Between a generating station and your phone charger, electricity changes its voltage a dozen times — 20 kV out of the alternator, 400 kV across the countryside, 11 kV down your street, 230 V at the wall, 5 V into the USB port. The machine performing each change has no moving parts, no exotic materials, and an efficiency that embarrasses every other energy converter humans build: the transformer — two coils of wire sharing one loop of iron, obeying Faraday's law in stereo. This article is the payoff of the induction chapter you have already climbed. Mutual Inductance built the coupling machinery — flux from one circuit threading another, the coefficient $k$ , the ideal limit $k \to 1$ where all flux is shared. Here we live in that limit and ask the device questions: why the voltage ratio is exactly the turns ratio, why stepping voltage up steps current down, why that one fact decided the most famous engineering war of the nineteenth century, and how a box of iron and copper pretends to be a different resistor than the one you connected — the impedance trick hi-fi amplifiers and radio antennas both live by. Mastering it buys you the grid: you can read the nameplate on any substation, price the $I^2R$ bill that AC was adopted to dodge, and see why the same argument, run through modern power electronics, is now bringing high-voltage direct current back. One device, one law — the reason electricity conquered distance.

Key takeaways

A transformer is two windings sharing one core flux. Faraday's law per turn gives the voltage ratio = turns ratio: $V_2/V_1 = N_2/N_1$ .
It transfers power, never creates it: $V_1I_1 = V_2I_2$ , so stepping voltage up steps current down by the same factor — enforced by mmf balance $N_1I_1 \approx N_2I_2$ .
That current reduction is why AC won the grid: line loss $I^2R$ falls as $1/V^2$ at fixed delivered power. 100 MW through a 10 Ω line at 400 kV loses 0.625%; at 10 kV the same line would burn more than you send.
Seen from the primary, a load masquerades as $Z_{\rm in} = (N_1/N_2)^2 Z_L$ — the impedance transformation that matches amplifiers to speakers and antennas to cables.
Real cores pay an eddy bill (fought with laminations — Eddy Currents) and a hysteresis bill (the loop area — Hysteresis), yet big units clear 99%.
KEY IDEA: shared flux converts voltage ratios into turns ratios, and power conservation converts voltage into current headroom — which turns $I^2R$ from a dealbreaker into a rounding error.

The core insight

Two windings on a common core see the same $\mathrm{d}\Phi/\mathrm{d}t$ per turn, so their voltages stand in the ratio of their turn counts while their currents stand in the inverse ratio; a transformer converts voltage into current headroom at constant power, which is the entire economic case for the AC grid.

Where this lives today

The transformer's empire is splitting into two kingdoms. The classical one stands: a hundred million distribution transformers hum on poles worldwide, grain-oriented steel at 50/60 Hz, many over 99% efficient, some in service for half a century. The insurgent kingdom runs on the EMF equation's $fNA$ trade: switch-mode converters chop DC into 100 kHz–1 MHz square waves precisely so their ferrite transformers can be gram-scale — why a 65 W laptop charger outperforms a 2 kg linear brick from 1990. And at the largest scales the wheel has turned full circle: ±800 kV HVDC interties carry hydropower across China and under seas — DC rehabilitated not against the transformer's physics but by it, since every HVDC terminal is built of transformers feeding semiconductor valves. The turns ratio, 190 years old, remains the cheapest factor of a thousand in all of engineering.

The physical picture

Wind two coils on the same closed loop of iron — say 100 turns on the left leg, 1000 on the right — and drive the left one with alternating current. The iron does what the magnetization chapter trained it to do: multiplies the winding's flux and, being a closed magnetic circuit with no air gap, channels essentially every flux line around the loop and through both windings. Whatever $\Phi(t)$ threads one turn of the left coil threads each turn of the right coil too. Now Faraday's law reads the same script to both windings: every turn, left or right, sees the same $\mathrm{d}\Phi/\mathrm{d}t$ and develops the same volts-per-turn, so a winding's total voltage is its head count times that common figure. Ten times the turns, ten times the voltage — no adjustment, pure counting. That is the whole secret of the voltage ratio.

Figure

One core flux read by two windings — volts per turn is common property, so voltage counts turns

The current story is the mirror image, enforced by the iron itself. Load the secondary and its current, by Lenz's sign, tries to kill the core flux. But the primary voltage — pinned to the mains — insists on its $\mathrm{d}\Phi/\mathrm{d}t$ , so the primary draws exactly enough extra current to restore what the secondary removed. The core is a flux thermostat (Mutual Inductance's image, inherited whole): the set-point is dictated by the primary voltage, and the windings' magnetomotive forces settle the bill between them, $N_1 I_1 \approx N_2 I_2$ . More voltage on one side means less current on that side, watt for watt. One honest subtlety before the mathematics: the power that crosses a transformer never travels through the iron. The core carries flux; the energy flows through the electromagnetic field around the windings, exactly as The Poynting Vector's coaxial audit showed for a cable — wires and core are guides, not pipes.

The ideal transformer, minted

The whole device rests on one fact, certified upstream: a closed, high-permeability core in the $k \to 1$ limit shares all its flux between windings. Coupling coefficients and the bound $M^2 \le L_1 L_2$ belong to the theorems of Mutual Inductance; the induction law itself is Faraday's Law of Induction. Here, we apply both.

Derivation ATurns ratio and current ratio from one shared flux

We derive the two nameplate equations of the ideal transformer — $V_2/V_1 = N_2/N_1$ and $N_1I_1 = N_2I_2$ — from shared core flux plus Faraday's law, keeping honest books about the approximation each needs (perfect coupling for the first; negligible magnetizing current for the second). The strategy: write the per-turn EMF the common flux dictates, count turns, then let mmf balance fix the currents. Watch where the ideality assumptions enter — they enter separately, which is why a lightly loaded real transformer can obey the voltage ratio beautifully while flunking the current ratio.

Step 1. Let $\Phi(t)$ be the flux through one turn anywhere on the core — a single symbol because the closed, high- $\mu$ core channels the same flux through every turn of both windings (the $k \to 1$ limit; leakage flux is the correction, deferred to the non-ideal ledger below). With the primary connected to a source $V_1(t)$ and winding resistance negligible, Kirchhoff's loop rule for the primary demands that the source be balanced by the winding's induced EMF:

V_1(t) = N_1 \frac{\mathrm{d}\Phi}{\mathrm{d}t}

where: $N_1$ is the primary turn count and $N_1\,\mathrm{d}\Phi/\mathrm{d}t$ is Faraday's law applied turn by turn. Read it as a constraint on the core: the mains voltage dictates $\mathrm{d}\Phi/\mathrm{d}t$ ; the iron has no say. Why write the loop rule this way? With zero winding resistance there is nothing else in the primary loop — so the flux waveform is slaved to the applied voltage. This is the flux-thermostat set-point.

Step 2. The secondary winding watches the same $\Phi(t)$ and develops its open-circuit voltage by the same law, $V_2(t) = N_2\,\mathrm{d}\Phi/\mathrm{d}t$ . Divide the two Faraday equations and $\mathrm{d}\Phi/\mathrm{d}t$ — the only quantity the device actually manufactures — cancels out:

\frac{V_2}{V_1} = \frac{N_2}{N_1}

where: the ratio holds instant by instant for any waveform, not just sinusoids. A 1:10 transformer delivers ten times the voltage at every moment — which is also why it faithfully transforms distortion: square-ish input, square-ish output. Why does nothing about the core appear? Both windings read the same flux; the core's size and material set how feasible that flux is (Step 4) but drop out of the ratio. The ratio is pure counting.

Step 3. Now load the secondary so $I_2$ flows. Ampère's law for $H$ around the core loop (Ampère's Law for H's magnetic-circuit form) gives $\oint \mathbf{H}\cdot\mathrm{d}\boldsymbol\ell = N_1I_1 - N_2I_2$ , the minus sign because the secondary current, by Lenz, opposes the flux. For a high-permeability core at finite $B$ , $H = B/\mu$ is nearly zero, so the magnetomotive forces must cancel:

N_1 I_1 \;\approx\; N_2 I_2

where: the small residual $N_1 I_{\rm mag} = \oint \mathbf{H}\cdot\mathrm{d}\boldsymbol\ell$ is the magnetizing current — the current the primary draws even with the secondary open, just to push the set-point flux around the iron. The ideal-transformer statement is precisely the claim $I_{\rm mag} \ll I_1$ : honest at full load on a good core, false near no-load. Why does this enforce power conservation? Multiply the two boxed results: $V_2I_2 = V_1I_1$ . Nothing is stored (at $k=1$ the coupled energy has a null mode — Mutual Inductance) and nothing created; the transformer is a ratio machine, not a source.

Step 4. Finally, price the flux itself. For a sinusoid $\Phi = \Phi_{\max}\sin\omega t$ , Step 1 gives peak voltage $N_1\omega\Phi_{\max}$ , hence the RMS EMF equation $V_{\rm rms} = \frac{2\pi}{\sqrt 2} f N B_{\max} A \approx 4.44\, f N B_{\max} A$ , with $A$ the core cross-section. This equation sizes transformers: iron saturates near $B_{\max} \approx 1.5$ – $1.8$ T, so at fixed voltage the product $fNA$ is spoken for. A 200-turn winding on a $100\ \mathrm{cm^2}$ core at 50 Hz and 1.5 T supports $666\ \mathrm{V}$ RMS — and this single scaling is why 50 Hz grid transformers are tonnes of iron while a 100 kHz USB-charger transformer fits on a fingertip. Note: multiplying the two ratio equations gives $V_1I_1 = V_2I_2$ exactly, confirming the energy audit; the EMF-equation constant is $\pi\sqrt{2} = 4.4429$ , and the 666 V example follows directly from the formula.

The caveat worth carrying everywhere: the current ratio silently assumed negligible magnetizing current. Unload the secondary and $I_2 = 0$ , yet the primary still draws $I_{\rm mag}$ — a mostly reactive trickle keeping the flux circulating. Where it stops being small — saturation, inrush — is flagged in the warning below.

Why AC won: the transmission-loss theorem

Here is the calculation that built the twentieth century's largest machine. A transmission line has resistance $R$ ; you must push power $P$ through it to a distant city. The line current follows from the delivery voltage, $I = P/V$ (power factor taken as unity — AC Power and Resonance owns that bookkeeping), and the line then eats

P_{\rm loss} = I^2 R = \frac{P^2 R}{V^2}

where: the inverse-square dependence on $V$ is the entire story — voltage costs nothing to raise (turns ratio!) while loss falls with its square. Doubling $V$ quarters the heating; a forty-fold step-up cuts it by 1600.

Go deeper · Derivation B · 1 minThe 1600-to-1 verdict on a real line

Derivation BThe 1600-to-1 verdict on a real line

We put honest numbers into $P_{\rm loss} = P^2R/V^2$ for one line and two candidate voltages, because arithmetic — not rhetoric — ended the war of the currents. Take $P = 100$ MW to deliver (a small city) through a line of total resistance $R = 10\ \Omega$ (roughly a hundred kilometres of practical conductor), comparing 10 kV against 400 kV. Watch for the low-voltage case to fail qualitatively, not just expensively — the formula will demand more heat than there is power, the honest signal that the scheme is impossible rather than merely wasteful.

Step 1. At $V = 400\ \mathrm{kV}$ : the line current is $I = P/V = 10^8/(4\times10^5) = 250\ \mathrm{A}$ . The line's bill is $P_{\rm loss} = I^2R = (250)^2 \times 10 = 625\ \mathrm{kW}$ — just 0.625% of the shipment. A rounding error; the copper stays cool.

Step 2. At $V = 10\ \mathrm{kV}$ : the same delivery would need $I = 10^8/10^4 = 10{,}000\ \mathrm{A}$ , and the formula charges $P_{\rm loss} = (10^4)^2\times 10 = 1000\ \mathrm{MW}$ — ten times the power being sent. Why is an impossible number the right answer? Because the premise fails: forcing 100 MW through this line at 10 kV would drop $IR = 100$ kV across the line — more than the source itself. The absurdity is the physics speaking: at low voltage this line cannot carry the load at all. The realistic 10 kV options are all bad — thicker copper (cost scales with conductor mass), shorter lines (a power station every couple of kilometres: literally Edison's Pearl Street architecture), or less power.

Step 3. The loss ratio between the two voltages is $(400/10)^2 = 1600$ exactly, independent of the line — bought entirely by two static machines at its ends, each better than 99% efficient. Direct current in 1890 had no such machine: DC conversion meant rotating motor-generator sets, costly and lossy at every node. The turns ratio was the technology; the $1/V^2$ was the argument. The arithmetic: $250^2 \times 10 = 6.25\times10^5$ W and $(10^4)^2 \times 10 = 10^9$ W; the loss ratio equals $(V_2/V_1)^2 = 1600$ .

Figure

The grid ladder — every rung is a transformer, and the high rung is where distance is cheap

The story attached to this arithmetic deserves its numbers. In 1886, in Great Barrington, Massachusetts, William Stanley took a 500 V alternator, stepped it up to 3,000 V, sent it about 1,200 metres down Main Street, and stepped it back down to light shop lamps at 100 V — the first practical AC distribution system, rungs and all. The publicity war that followed ended on arithmetic: when the Niagara Adams plant's 5,000-horsepower (3.7 MW) two-phase alternators sent power 26 miles to Buffalo in 1896, stepped up to 11,000 V for the journey, the argument was over everywhere on Earth. One modern honesty: the 1890s verdict condemned 1890s DC conversion technology, not DC itself. Semiconductor power electronics now transform DC voltage efficiently, and the longest hauls run HVDC at ±800 kV (the FAQ below tells that story). The $1/V^2$ theorem never changed sides; only the machinery for climbing the voltage ladder did.

The impedance looking glass

A transformer changes what a load looks like: with voltage scaled one way and current the other, any impedance hung on the secondary appears from the primary rescaled by the turns ratio squared.

Go deeper · Derivation C · 1 minA load seen through a transformer is the load times the turns ratio squared

Derivation CA load seen through a transformer is the load times the turns ratio squared

We derive $Z_{\rm in} = (N_1/N_2)^2 Z_L$ — the impedance-transformation theorem — directly from the two ideal-transformer ratios, then spend it on the matching problem it was born for. The route is two substitutions long; the physics to watch is that voltage and current scale in opposite directions, so their ratio scales by the square. This is a genuinely AC theorem: it inherits the ideal ratios, which need changing flux.

Step 1. Hang $Z_L$ on the secondary: $V_2 = Z_L I_2$ (amplitudes or phasors; the theorem carries phase intact). The primary-side observer measures $Z_{\rm in} = V_1/I_1$ .

Step 2. Substitute the ideal ratios $V_1 = (N_1/N_2) V_2$ and $I_1 = (N_2/N_1) I_2$ :

Z_{\rm in} = \frac{V_1}{I_1} = \frac{(N_1/N_2)\,V_2}{(N_2/N_1)\,I_2} = \left(\frac{N_1}{N_2}\right)^{2} Z_L

where: the square arises because both rescalings help — a step-down transformer ( $N_1 > N_2$ ) shows the primary a magnified load, and the magnification applies to resistance and reactance alike, phase untouched. Why would anyone want to disguise a load? Because maximum power transfer from a source of internal resistance $r$ demands a load matched to $r$ — the theorem derived in EMF and Internal Resistance. When the actual load is wrong, a transformer fixes the disguise instead of the load.

Step 3. The classic worked match: a valve audio amplifier behaves like a source with $r \approx 5\ \mathrm{k\Omega}$ ; a loudspeaker is $Z_L = 8\ \Omega$ — mismatched by 625. Interpose a transformer with $(N_1/N_2)^2 = 5000/8 = 625$ , i.e. $N_1/N_2 = 25$ : the amplifier sees $625\times8 = 5\ \mathrm{k\Omega}$ — matched — while the speaker receives the power at its own low impedance. Every valve amplifier shipped with this iron; the same theorem at radio frequency matches antennas to feedlines. The numbers: $\sqrt{5000/8} = 25.0$ exactly, and $(25)^2 \times 8\ \Omega = 5000\ \Omega$ ; phases carry through unchanged since both ratios are real.

The non-ideal ledger

Real transformers leak a little at four audited doors — each owned in depth elsewhere in the track, recalled here by name and priced.

Copper (winding) loss. The windings have resistance; the bill is $I^2R_{\rm winding}$ per side, scaling with load squared — why nameplate efficiency is quoted at rated load.

Eddy-current core loss. The core is itself a conductor bathed in changing flux, hosting the bulk whirlpools of Eddy Currents; that article's lamination theorem — $N$ insulated sheets cut the loss by $N^2$ , thin-sheet dissipation scaling as $f^2 B_{\max}^2 d^2/\rho$ — is why every core is a stack of varnished laminations, and why high-frequency units abandon iron for ferrites.

Hysteresis core loss. Each cycle traces the $B$ – $H$ loop once and pays the loop's area in heat — Hysteresis's loss bill, scaling as $f\,B_{\max}^{1.6}$ by Steinmetz's law (that article's own object). Grid cores use grain-oriented silicon steel with deliberately skinny loops.

Leakage flux. Not all flux links both windings; the stray fraction acts as a small series inductance per side, $k$ slipping just below 1 — Mutual Inductance's territory. Leakage even has a use: it limits fault currents, and designers sometimes build it in on purpose. Add the doors up and the scorecard still astonishes: distribution transformers clear 98–99%, large grid units 99.5% — the most efficient energy converters in common use.

Limiting cases

case	result	meaning
$N_1 = N_2$	$V_2 = V_1$ , $Z_{\rm in} = Z_L$	isolation transformer: no ratio change, just galvanic separation
secondary open ( $I_2 = 0$ )	primary draws only $I_{\rm mag}$	flux thermostat idles; voltage ratio still exact
DC drive ( $\omega \to 0$ )	$\mathrm{d}\Phi/\mathrm{d}t \to 0$ : no secondary voltage; primary current climbs to saturation	transformers are AC-only devices; DC input is a short circuit through winding resistance
$\mu \to \infty$ , $k \to 1$ , lossless	ideal transformer exactly	both nameplate ratios exact; magnetizing current zero
$f$ raised at fixed $V$	required $\Phi_{\max} \propto 1/f$	EMF equation: higher frequency shrinks the iron — the switch-mode power-supply strategy
load $Z_L \to 0$ (short)	$I$ limited only by leakage + winding resistance	fault regime: leakage inductance is the designed-in safety margin

Common trapA transformer that raises voltage gives you more power

It never does. The two nameplate equations multiply to $V_1I_1 = V_2I_2$ : whatever factor the voltage gains, the available current loses, exactly. A step-up transformer is a gearbox, not an engine — a bicycle in high gear moves faster per pedal stroke only by demanding proportionally more force. The reverse misreading causes real accidents: a step-down transformer is a current amplifier (exactly how a spot welder works — a couple of turns of massive busbar as secondary, thousands of amps at a volt or two), so "only 2 volts" is not "only" anything. And the DC corner ruins hardware: with no $\mathrm{d}\Phi/\mathrm{d}t$ to generate back-EMF, the primary is a low-resistance coil across a source — it saturates, overheats, and burns. Transformers trade voltage for current in an AC world; they mint nothing.

Worked examples

1 · The pole transformer on your street (confidence builder)

A neighbourhood transformer steps the 11 kV distribution line down to 230 V for homes. (a) What turns ratio does it need? (b) When the houses on it draw 4.6 kW (unity power factor), what current flows in each winding, ideal-transformer books?

Step 1. Turns ratio from the voltage ratio: $N_1/N_2 = V_1/V_2 = 11{,}000/230 = 47.8$ — about 48 primary turns per secondary turn.

Step 2. Secondary current: $I_2 = P/V_2 = 4600/230 = \mathbf{20\ A}$ .

Step 3. Primary current two ways — power books, $I_1 = P/V_1 = 4600/11{,}000 = \mathbf{0.418\ A}$ , or mmf balance, $I_1 = 20/47.8 = 0.418$ A. The books agree, as Derivation A promised. The moral: the same 4.6 kW is 20 A on the house side and under half an amp on the line side — the street's thin high line and your panel's thick cables are both sized by the transformer.

2 · Sizing the iron, then auditing the ladder (exam level)

(a) A 50 Hz transformer must accept 666 V RMS on a 200-turn primary without driving its core past $B_{\max} = 1.5$ T. What core cross-section does it need? (b) The full grid path runs generator → step-up → line → step-down → homes, with each transformer 99.0% efficient and the 400 kV line of Derivation B losing 0.625%. What fraction of the generated 100 MW reaches the city?

Step 1. Invert the EMF equation for the area: $A = V_{\rm rms}/(4.44\,f\,N\,B_{\max}) = 666/(4.44 \times 50 \times 200 \times 1.5) = \mathbf{0.010\ m^2}$ — a core face of $100\ \mathrm{cm^2}$ , ten centimetres on a side. (Run the same voltage at 100 kHz and $A$ shrinks two-thousand-fold: the switch-mode argument, in one line.)

Step 2. Chain the efficiencies multiplicatively: $\eta_{\rm total} = 0.990 \times (1 - 0.00625) \times 0.990 = 0.974$ .

Step 3. Delivered power: $0.974 \times 100\ \mathrm{MW} = \mathbf{97.4\ MW}$ . The end-to-end bill for crossing a hundred kilometres is under 3% — of which the line, the part that covers distance, is only a quarter. The transformers made distance the cheap part. Check: without the ladder, Derivation B showed the line cannot even carry the load — in 1890 the 2.6% was an improvement on impossible.

The engineer who bet the company

Origins · 1 minGeorge Westinghouse and how this was discovered

George Westinghouse

1846–1914 · Pittsburgh — the industrialist who bought the turns ratio

Westinghouse was not a physicist; he was the rare industrialist who could read a physics argument and stake a fortune on it. Already rich from the railway air brake, he saw in 1885 what Edison would not concede for a decade: Faraday's law, embodied in Gaulard and Gibbs's crude "secondary generators," made voltage a commodity — and $I^2R$ made high voltage the only way electricity would ever leave the neighbourhood of its power station. He imported the European transformers, hired William Stanley to perfect a closed-core design, and bankrolled the Great Barrington demonstration. Edison's counterattack was theatrical — staged electrocutions, the "Westinghoused" smear; Westinghouse's rebuttal was arithmetic delivered at scale: a quarter-million lamps at the 1893 Chicago World's Fair, then the Niagara contract with Tesla's polyphase patents. He lost his company in the 1907 panic and died in 1914 — before the grid he gambled on became the largest machine ever built, running on his bet that two coils and an iron ring beat a power station on every street corner.

History

1831

— Faraday's induction ring: two windings on an iron torus — structurally the first transformer, read as a discovery about flux

1885

— Zipernowsky, Bláthy and Déri patent the closed-core "ZBD" transformer and coin the parallel-connected distribution scheme; Stanley refines the design for Westinghouse

1886

— Great Barrington: first practical AC system — 500 V stepped to 3,000 V, transmitted down Main Street, stepped down to 100 V for shop lighting

1893

— Westinghouse AC lights the Chicago World's Columbian Exposition; the safety propaganda war effectively ends

1896

— Niagara's Adams plant transmits power 26 miles to Buffalo at 11,000 V — the template for every grid since

1954

— Sweden's Gotland link opens the modern HVDC era: power electronics finally give direct current its own voltage ladder

Going further

Read next, in this chapter: this article and The Hall Effect close the circuits chapter; back-references worth revisiting are AC Power and Resonance (RMS and power-factor bookkeeping) and EMF and Internal Resistance (the matching theorem the impedance looking-glass serves).

Read next, upstream: Mutual Inductance (the coupling machinery and the $k \to 1$ limit) · Faraday's Law of Induction (the law read per-turn here) · Eddy Currents (the lamination theorem) · Hysteresis (the loop-area bill and Steinmetz's law) · The Poynting Vector (why the power crosses in the field, not the iron).

Beyond the track: three-phase power and per-unit system analysis (power engineering); switch-mode converters and planar magnetics (power electronics); HVDC converter stations. 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.

Why do transformers only work with AC?

Because the only thing a winding can hand its neighbour is a changing flux. Faraday's law pays volts per turn in proportion to dΦ/dt; steady DC, after its switch-on transient, holds the flux constant and the secondary voltage falls to zero. Worse, the primary loses its back-EMF, becomes a bare low-resistance coil across the source, saturates the core, and cooks. AC is not a convenience here; it is the operating principle. (DC-DC converters do use transformers — by first chopping the DC into high-frequency AC, which is conceding the point.)

If the secondary is open and no power flows, why does the primary draw any current at all?

That residual draw is the magnetizing current — the small current that pushes the set-point flux around the core against its finite permeability (exactly the term Ampère's law for H kept before Derivation A idealized it away). It is mostly reactive: current and voltage sit 90° apart, so the energy shuttles in and out of the core twice per cycle rather than being consumed — the ledger Faraday's Law of Induction audits in its own transformer FAQ. What little real power an idle transformer eats is the core's hysteresis and eddy bill — why utilities care about "no-load loss" on devices energized around the clock.

Where exactly does the energy cross from primary to secondary — through the iron?

No. The iron carries flux, which carries no power by itself — an idling transformer has full flux and transfers nothing. The energy travels through the electromagnetic field around and between the windings, the same accounting The Poynting Vector performs for a coaxial cable (all the power in the gap, none inside the metal). Core and copper are the guides that shape the field, not the pipe the energy uses.

What is the actual difference between a transformer and just two coils near each other?

Only the coupling grade. Two loose coils have k = M/√(L₁L₂) well below 1 — most primary flux misses the secondary, and the device behaves like Mutual Inductance's general coupled pair. A transformer is the engineered k → 1 endpoint: a closed high-μ core channels essentially all flux through both windings, collapsing the coupled equations to the two clean nameplate ratios. Wireless phone chargers sit deliberately in between (k ≈ 0.3–0.6 through the air gap), which is why they need resonance tricks to move power a mains transformer moves with brute flux.

Why do transformers hum, and why at twice the mains frequency?

Magnetostriction: iron shortens and lengthens by parts per million as it magnetizes, at every flux peak — positive or negative alike — so the core flexes twice per electrical cycle: a 100 Hz fundamental on 50 Hz mains (120 Hz in the Americas), plus harmonics from the effect's nonlinearity. Loose laminations and tank walls act as sounding boards. The hum is an acoustic tax, not a meaningful energy loss; utilities specify noise limits for neighbours, not efficiency.

Is a spot welder really just a transformer? Thousands of amps sounds like something more exotic.

It is exactly a step-down transformer taken to its logical extreme: hundreds of primary turns, and a secondary of one to three turns of massive copper busbar. Step the voltage down by 100 and the available current steps up by 100 — mmf balance is indifferent to which direction you exploit it. A welder delivering 5,000 A at 2 V moves the same 10 kW that 43 A carries at 230 V; it has traded all its voltage headroom for current, because melting a spot of steel is a job for I²R concentrated in a milliohm contact.

If AC won the war, why do modern mega-projects use HVDC — did the physics change?

The physics never changed; the machinery did. The 1890s case for AC was really a case for the transformer — then the only efficient voltage changer in existence, and it needs AC. Semiconductor power electronics finally gave DC its own voltage ladder, and with conversion available at both ends, DC wins specific niches on merit: no reactive current charging a long submarine cable every cycle, no need to synchronize the grids it joins, full use of the insulation (no √2 peak-to-RMS penalty). AC still owns distribution because transformers remain cheaper and more reliable than converter stations at every street corner.