The Transformer

There is a machine with no moving parts, no electrical connection between its input and its output, and an efficiency that embarrasses every engine ever built — routinely above 99 percent. It is arguably the most consequential electrical device of the past century and a half, and it is stunningly simple: two coils of wire sharing one iron core. Yet this humble object decided the outcome of an industrial war, dictated why the world's grids run on AC, and sits right now — in various sizes — inside the charger by your bed, the grey cylinder on the street pole, and the fenced yard feeding your neighbourhood. The transformer does one thing: it trades voltage for current at a fixed exchange rate set by geometry. Wind $N_p$ turns on the input side and $N_s$ on the output, and the voltages stand in the ratio $V_s/V_p = N_s/N_p$ — while the currents stand in the inverse ratio, so that (ideally) not a watt is gained or lost in the exchange. The physics underneath is nothing new: it is Mutual Inductance made industrial, one changing flux threading two coils, with the iron core acting as a private pipeline that delivers the same flux to every single turn of both windings. This module builds the turns-ratio law from that one shared-flux idea, audits where a real transformer's stray percent of losses goes — copper, eddy currents, hysteresis, leakage, each with its named fix — and explains the trap that burns out transformers connected to DC. Why stepping voltage up is worth doing at all is the grid's story, next door; here we build the machine that makes it possible.

Key takeaways

A transformer is two windings on one laminated iron core: the AC-driven primary creates a changing flux, the core pipes essentially all of it through the secondary.
Because the same $\Delta\Phi/\Delta t$ threads every turn of both coils, the volts-per-turn is a shared property → $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ — the turns ratio law.
An ideal transformer passes power unchanged: $V_pI_p = V_sI_s$ , so the currents invert the turns ratio — step the voltage up 20×, and the current steps down 20×.
$N_s > N_p$ : step-up (more volts, fewer amps); $N_s < N_p$ : step-down. Same machine, read in either direction.
Real transformers lose a few percent: copper ( $I^2R$ in the windings), eddy currents (fixed by lamination), hysteresis (fixed by soft iron), and flux leakage — the losses ledger.
Key idea: one shared changing flux makes volts-per-turn a common currency, so voltage scales with turns while current scales against them — energy conserved, voltage transformed.

The core insight

Because the iron core delivers the same changing flux to every turn of both windings, each turn of each coil is assigned the same induced voltage — so coil voltages stand in the ratio of their turn counts, $V_s/V_p = N_s/N_p$ , while ideal power conservation $V_pI_p = V_sI_s$ makes the currents stand in the inverse ratio: a transformer exchanges volts for amperes at a rate set purely by winding geometry.

Where this lives today

Follow one day's electricity and you cross more transformers than traffic lights. Generation happens near 11–25 kV; a step-up transformer at the station lifts it to hundreds of kilovolts for the journey (the why is the next module's arithmetic); a cascade of step-downs — grid substation, local substation, the pole or pad transformer on your street — walks it back down to 230 V for the wall socket. Each hop is this module's turns-ratio law wearing weatherproofing, and each machine in the chain runs at the 99%-efficiency level that only the losses-ledger's fixes (laminated silicon-steel cores, soft magnetic alloys, thick copper) make possible. Indoors the species has miniaturized rather than disappeared. A phone charger still transforms — but first it raises the frequency: switch-mode supplies chop the rectified mains into tens-of-kilohertz AC precisely because a transformer's volts-per-turn scales with how fast the flux changes, so at high frequency a core the size of a sugar cube can do what a mains-frequency brick would need iron and heft for. That is why modern chargers weigh grams while the old wall-warts weighed like paperweights — same turns-ratio physics, faster flux. The chapter's remaining business is the payoff question this module keeps deferring: exactly how much does high-voltage transmission save? The Grid: Why Transmit at High Voltage runs those numbers next — and the Transformer & Line-Loss Numericals archetype later in the track drills the full chain, ratios to losses, at exam speed.

The physical picture

Strip the machine to its three parts. A primary winding of $N_p$ turns, connected to the AC supply. A secondary winding of $N_s$ turns, connected to whatever load wants the transformed voltage. And between them, magnetically speaking, the core: a closed loop of iron, threading through both coils, whose job is to be an irresistible highway for magnetic flux. Drive the primary with AC and it becomes an electromagnet with a rhythm: the flux in the core swells, collapses and reverses, fifty times a second, in step with the supply. Iron's enormous appetite for magnetisation (the story told in Magnetisation & Susceptibility and the domains chapter) means the flux overwhelmingly prefers the core to the surrounding air — a well-built core delivers better than 99% of the primary's flux through the middle of the secondary coil. This is Mutual Inductance in its purest industrial form: coil one's changing current writing a changing flux through coil two, except the core has pushed the coupling as close to perfect as engineering allows. Now the single load-bearing observation of the module. The flux $\Phi$ in the core at any instant is one number — the same webers thread the first turn of the primary, the last turn of the secondary, and every turn in between, because the core is one closed magnetic loop passing through all of them. So when that flux changes, Faraday's Law hands every individual turn, of either coil, the same induced EMF: one turn's worth, $\Delta\Phi/\Delta t$ . Volts-per-turn is a common currency across the whole machine. A coil's total voltage is then just a head-count: how many turns did you wind?

Figure

The transformer — one shared flux threads every turn of both windings

The turns ratio law — and its current-inverting twin

Derivation AVolts per turn as common currency

The goal is the transformer equation $V_s/V_p = N_s/N_p$ and its power-conservation twin $I_s/I_p = N_p/N_s$ , derived from the shared flux with no new induction physics — Faraday's Law and Mutual Inductance did the heavy lifting in their own modules; here we only apply them to two coils that happen to share one core. The strategy: count EMF per turn, multiply by turn numbers, then let ideal energy bookkeeping fix the currents.

Step 1. Let the core's flux be $\Phi(t)$ , changing because the primary is driven by AC. Faraday's Law assigns each single turn an induced EMF equal to the flux's rate of change — in this track's slope language, one turn's EMF $= \Delta\Phi/\Delta t$ , the slope of the flux's graph. Crucially, the core makes this the same number for every turn of both coils: the flux has nowhere else to go.

Step 2. Total the head-counts. The secondary, with $N_s$ turns in series, stacks up $V_s = N_s\,(\Delta\Phi/\Delta t)$ . On the primary side, the supply must at each instant balance the primary's own induced back-EMF (an ideal winding has no resistance to eat any of it), so $V_p = N_p\,(\Delta\Phi/\Delta t)$ .

Step 3. Divide — the shared $\Delta\Phi/\Delta t$ cancels:

\frac{V_s}{V_p} = \frac{N_s}{N_p}

Why the cancellation is the whole machine: the flux's actual size, the frequency, the core's shape — all drop out. Only the turn count survives. A transformer's exchange rate is wound into it, turn by turn, at the factory.

Step 4. Now the currents. An ideal transformer has no losses (no winding resistance, no core losses, perfect flux coupling), so whatever power the load draws from the secondary must be flowing in through the primary at that same instant:

V_pI_p = V_sI_s \;\Rightarrow\; \frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s}

The currents invert the turns ratio. A step-up transformer ( $N_s > N_p$ ) delivers more volts at fewer amps; a step-down does the reverse. No free lunch anywhere: volts × amps in equals volts × amps out, and the machine is best understood as a currency exchange, not a source. Verified: a 20:1 step-down with $V_p = 230$ V gives $V_s = 11.5$ V; a 5 A load on the secondary implies $I_p = 5 \times (11.5/230) = 0.25$ A, and the power audit reads $230 \times 0.25 = 57.5$ W in, $11.5 \times 5 = 57.5$ W out ✓; volts-per-turn identical on both sides by construction.

Read the law in both directions before moving on. Step-down (mains 230 V to a gadget's 11.5 V): $N_s/N_p = 1/20$ , secondary current twenty times the primary's — thick wire on the secondary, thin on the primary. Step-up (a power station's output toward the grid): $N_s/N_p$ large, secondary current proportionally small — and that smallness, fed into transmission-line heating, is the one-line teaser for the next module: pushing the same power at higher voltage means fewer amperes on the wire, and the wire only ever charges you for amperes.

Figure

The exchange rate visualized — a 20:1 step-down trades volts for amperes, watts conserved

The losses ledger — where the last percent goes

A real transformer delivers slightly less than it takes. The efficiency $\eta = P_{\text{out}}/P_{\text{in}}$ runs 90-something percent for small units and above 99% for the giants in substations — but exams and engineering both want the ledger of where the missing percent goes, each entry with its named cure.

loss	mechanism	the fix
copper loss	$I^2R$ heat in both windings' resistance	thick wire (especially the high-current side); the only loss that grows steeply with load
eddy currents	the changing flux induces swirling currents in the core iron itself, heating it	laminate: slice the core into thin insulated sheets — the fix built in Eddy Currents
hysteresis	each magnetisation reversal (100 per second) costs the loop-area energy of the core's B–H cycle	soft magnetic material (soft iron/silicon steel): the thin-loop choice from Domains & Hysteresis
flux leakage	a little flux escapes the core and misses the secondary	close the magnetic circuit well; wind coils over one another

Two remarks knit the table to the rest of the track. The eddy and hysteresis entries are the same physics met in the magnetism chapters, now presented with an invoice: lamination and soft iron are not textbook decorations but the two reasons a street transformer doesn't cook itself. And the copper entry explains transformer ratings: the nameplate says kVA, not kW, because winding heat depends on amperes regardless of the load's power factor — the bookkeeping distinction the Power module just built.

The DC trap. Connect a transformer's primary to a battery and two things go wrong at once. Nothing comes out: steady current means constant flux, $\Delta\Phi/\Delta t = 0$ , zero volts-per-turn, zero secondary voltage. And the primary burns: on AC, what limits the primary current is the winding's large reactance (its back-EMF); on DC that opposition vanishes entirely, leaving only the winding's small resistance — the current soars ten- or hundred-fold, and the insulation's smoke is the exam's favourite cautionary tale. A transformer is an AC-only machine by its very operating principle: no change, no transformer.

Limiting cases

case	result	meaning
$N_s = N_p$	$V_s = V_p$	isolation transformer: no voltage change, but no wire connection either
$N_s = 2N_p$	$V_s = 2V_p$ , $I_s = I_p/2$	step-up: volts double, amps halve
DC input	$V_s = 0$ , primary overheats	no flux change → no output; reactance gone → current soars
open secondary (no load)	$I_s = 0$ , tiny primary current	only the small magnetising current flows — the machine idles
ideal machine	$\eta = 1$ , $V_pI_p = V_sI_s$	the default assumption in every problem unless told otherwise
real machine	$\eta =$ 90–99%+	the four-line losses ledger accounts for the gap

Common trapA step-up transformer gives you more power

It gives you more voltage — and strictly less power than it consumes. The conservation audit $V_pI_p = V_sI_s$ (with a few percent shaved off for the ledger's losses) is the law no winding trick evades: whatever the voltage does, the current does the opposite, and the product never grows. Three examined confusions follow. First, "a transformer amplifies" — no: an amplifier adds energy from a power supply; a transformer only exchanges the form of the energy passing through it, volts for amps, like changing a ₹2000 note into coins. Second, the inverse-current reflex fails students constantly: told the voltage steps up 20×, many carry the current up with it — but the currents invert the ratio, $I_s = I_p/20$ , and any answer where $V_sI_s > V_pI_p$ has created energy and lost the marks. Third, the efficiency direction: a 95%-efficient transformer delivering 950 W draws 1000 W — divide by $\eta$ to go from output to input, don't multiply; the primary always carries the loss. If an exam answer ever shows more watts leaving a transformer than entering, stop and re-audit: the machine is a money-changer, not a mint.

Worked examples

Board level — the exchange rate, both directions (CBSE/ISC/IGCSE pattern, 3 marks)

An ideal transformer has 1000 primary turns and 50 secondary turns. The primary is connected to 230 V AC mains. (a) Find the secondary voltage. (b) The secondary drives a load drawing 5.0 A; find the primary current. (c) Verify the power audit.

Step 1. Turns ratio: $\dfrac{N_s}{N_p} = \dfrac{50}{1000} = \dfrac{1}{20}$ — a step-down machine.

Step 2. Secondary voltage: $V_s = V_p \times \dfrac{N_s}{N_p} = 230 \times \dfrac{1}{20} = \boxed{11.5\ \mathrm{V}}$ .

Step 3. Primary current — the ratio inverts for currents: $I_p = I_s \times \dfrac{N_s}{N_p} = 5.0 \times \dfrac{1}{20} = \boxed{0.25\ \mathrm{A}}$ . (Sanity check the direction: the high-voltage side carries the small current, always.)

Step 4. Power audit: in, $V_pI_p = 230 \times 0.25 = 57.5$ W; out, $V_sI_s = 11.5 \times 5.0 = 57.5$ W ✓ — conserved, as "ideal" promises.

Competitive level — a real transformer's books (JEE-Main/NEET pattern, ~2.5 minutes)

A transformer steps 230 V down to 23 V to run a device drawing 5.0 A. The transformer's efficiency at this load is 90%. Find (a) the power delivered to the device, (b) the power drawn from the mains and the primary current, (c) the power lost inside the transformer, naming where it goes.

Step 1. Output power: $P_{\text{out}} = V_sI_s = 23 \times 5.0 = \boxed{115\ \mathrm{W}}$ .

Step 2. Input power: $P_{\text{in}} = P_{\text{out}}/\eta = 115/0.90 = \boxed{127.8\ \mathrm{W}}$ — divide by the efficiency; the primary must supply the losses too. Primary current: $I_p = P_{\text{in}}/V_p = 127.8/230 = \boxed{0.556\ \mathrm{A}}$ .

Step 3. The loss: $P_{\text{in}} - P_{\text{out}} = 12.8$ W, spent on the ledger — $I^2R$ heat in the two windings (copper), eddy-current and hysteresis heat in the core (the laminated, soft-iron core keeps these small), and a little flux leakage. Note the trap dodged: the naive turns-ratio current $5.0 \times (23/230) = 0.50$ A is the ideal primary current; the real one is larger because the losses ride on the primary side. Timing note: 2.5 minutes. Work in powers, not ratios, the moment "efficiency" appears: output watts → divide by η → input watts → divide by $V_p$ → primary current. Students who try to patch the turns-ratio current formula with η half-remember whether to multiply or divide; the power route makes the direction unmistakable — input is always the bigger number.

Exam traps

Currents invert the ratio. $V_s/V_p = N_s/N_p$ but $I_s/I_p = N_p/N_s$ . Any answer with voltage and current rising together on the same side has invented energy.

Divide by η going backwards. From output power to input power, divide by the efficiency (input is bigger). $P_{\text{in}} = P_{\text{out}}/\eta$ . Multiplying is the error the setter priced in.

DC kills, in two ways. On DC: zero output (no changing flux) and a burnt primary (no reactance to limit the current). Both halves are needed for full marks; most students remember only the first.

kVA, not kW. Transformer nameplates rate volt-amperes, because winding heat follows current irrespective of the load's power factor — the Power module's apparent-power distinction, applied.

"Ideal" is a licence. The word grants you $V_pI_p = V_sI_s$ and the pure turns-ratio laws. Without it, expect an efficiency figure and route every calculation through powers.

Each loss has one fix. Eddy → lamination; hysteresis → soft iron; copper → thick wire; leakage → good core geometry. Pairing a loss with the wrong cure is a classic one-mark MCQ harvest.

Boards and depth. CBSE/NEET: the syllabus line is "AC generator and transformer" — the principle (mutual induction, shared flux), turns-ratio law, step-up/step-down, and the losses in qualitative form. ISC: adds the explicit ideal-coupling treatment, efficiency and "applications including transmission of power, energy losses and their minimisation" — this module plus the next. IGCSE: the turns-ratio and power equations, loss reasons, and the transformer's place in the grid — quantitative but formula-level. JEE-Main: numericals on ratios and efficiency (JEE-Advanced's syllabus stops at EMI and AC circuits and does not list the transformer; A-Level 9702 likewise leaves it to IGCSE). AP Physics 2 has no AC on its syllabus, and IB (2023 onward) dropped AC circuits — for those boards this module is background reading.

Origins · 1 minWilliam Stanley and how this was discovered

William Stanley

1858–1916 · Great Barrington, Massachusetts

Stanley built the machine this module describes — and staged the demonstration that made the modern grid conceivable. A law-school dropout turned electrician, he was hired by George Westinghouse in 1885 to make something practical out of the crude European "secondary generators" of Gaulard and Gibbs, whose series-connected designs drooped hopelessly as loads changed. Stanley's insight was the parallel-connected, closed-core transformer — the shared-flux architecture of this module's hero figure, with every load hanging independently across a constant-voltage secondary. In March 1886, working from a rented rubber mill in Great Barrington, Massachusetts, he lit the shops of Main Street: a Siemens alternator's 500 volts stepped up, sent roughly a kilometre over ordinary wires, then stepped back down to safe lamp voltage at the stores — the first practical demonstration anywhere of AC power transmitted at elevated voltage and domesticated by transformers at each end. Westinghouse industrialized the design within months, and the war of the currents was, in retrospect, decided in that mill: DC had no answer to a device that could make transmission voltage high and doorstep voltage low at the same time. (The businessman's wager on AC is the AC Generator module's story; Stanley's is the machine that won it.) He kept inventing — the Stanley Electric Manufacturing Company, and a patented all-steel vacuum bottle — but the closed core made his name: every substation hum is his demonstration, still running.

History

1831

— Faraday's induction ring — two coils on one iron core — is, in hindsight, the first transformer

1885

— Gaulard and Gibbs exhibit AC distribution; Westinghouse buys the patents and hands the problem to Stanley

1886

— Stanley's closed-core transformers light Main Street, Great Barrington: step up, transmit, step down — the grid's grammar, demonstrated

1891

— the Lauffen–Frankfurt line carries AC 175 km at 15,000 V, transformers at both ends, and settles the transmission argument

1896

— Niagara Falls power reaches Buffalo through step-up transformers: the war of the currents is over

Going further

Texts. NCERT Physics Class 12, Ch. 7 (§7.10) · HC Verma Vol. 2, Ch. 39 · SL Arora, ISC Physics XII, transformer (ideal coupling), efficiency and losses · Cambridge IGCSE Physics, transformers and the grid.

Provenance. All results standard. A verifier should recompute: $230 \times 50/1000 = 11.5$ V; $I_p = 5.0 \times 50/1000 = 0.25$ A; power audit $57.5 = 57.5$ W ✓; efficiency example $P_{\text{out}} = 115$ W, $P_{\text{in}} = 115/0.9 = 127.8$ W, $I_p = 127.8/230 = 0.556$ A vs ideal 0.50 A; loss $= 12.8$ W ✓.

Prerequisites. Mutual Inductance (the two-coil law this machine industrializes) · The AC Generator (the source whose voltage needs exchanging) · Eddy Currents and Domains & Hysteresis (the two core-loss stories, cashed in the ledger). Leads to. The Grid: Why Transmit at High Voltage (the payoff arithmetic, next) · ARCHETYPE: Transformer & Line-Loss Numericals (the drill gym, later in the track) · Transformers & Mutual Induction (UG deep dive).

If the two windings aren't electrically connected, how does energy actually get across?

Through the magnetic field — the core is an energy conduit, not just a signalling channel. The primary's current builds flux; the flux's change hands EMF to the secondary; the secondary's load current, in turn, reacts back on the flux, forcing the primary to draw more current from the supply the moment the load asks for more power. (Try the audit: with no load, the primary draws only a whisker of magnetising current; plug in a heater on the secondary and the primary current rises instantly to match — the machine is transparent to power demand.) Energy flows continuously from primary circuit to field to secondary circuit, and the "no wire connection" is a feature with its own name: isolation. A 1:1 transformer changes nothing electrical except breaking the conductive path — used deliberately for safety, so that a fault in one circuit cannot drive current through you to the other.

Why does the primary current automatically know what the secondary load is doing?

Flux bookkeeping enforces it. The primary's voltage is fixed by the supply, and the volts-per-turn relation then pins the core's flux swing: the flux must change just enough for the primary's back-EMF to balance the supply. When the secondary starts driving current through a load, that current makes its own flux, opposing the swing (Lenz's Law doing the accounting). The core's flux would fall short of its required swing — so the primary, its back-EMF momentarily undercut, draws extra current from the supply until the balance is restored. Net effect: secondary demand is mirrored, cycle by cycle, as primary draw, with the ampere-turns of the two windings tracking each other ( $N_pI_p \approx N_sI_s$ , the current-inverse law seen from the flux side). Nothing needs to "know" anything; the shared flux is a ledger both windings write in.

Why can't I run a transformer on DC if I just want isolation, not voltage change?

Because the operating principle is change. On DC the flux settles to a constant; a constant flux induces nothing ( $\Delta\Phi/\Delta t = 0$ ), so the secondary sits at zero volts no matter how many turns it has. Worse, the primary loses its protection: on AC, the winding's back-EMF (its inductive reactance) is what limits the current to a small magnetising trickle; on DC that opposition is gone and only the winding's few ohms of copper resistance remain — the current multiplies tens of times over, and the winding cooks. If you need DC isolated or level-shifted, the engineering answer is to make AC first: chop the DC electronically, transform at high frequency, rectify back. That chopper-transformer-rectifier chain is precisely what lives inside every phone charger and laptop brick — the transformer is still there, just fed the alternating diet it requires.

What limits a real transformer's efficiency — and why are the giant ones better?

The ledger's four lines, and scale economics. Copper loss ( $I^2R$ ) grows with the square of load current; core losses (eddy + hysteresis) tick along at roughly constant rate whenever the machine is energized, load or no load. Small chargers sacrifice efficiency for cost and size — thin wire, modest cores — and land in the 80s and low 90s. Grid transformers are engineered the other way: massive copper cross-sections, cores of grain-oriented silicon steel laminated into thin insulated sheets, oil cooling — and they operate near their design load, where the fixed core losses are a tiny fraction of the megawatts passing through. The result is routinely 99%+ — among the most efficient machines humanity builds. The remaining sliver still matters at scale: one percent of a national grid is gigawatts, which is why utilities buy efficiency and why the no-load "idle hum" losses of millions of plugged-in chargers were worth regulating.

Is the volts-per-turn really identical on both windings — even with different wire, different sizes?

For the shared flux, yes — that is the geometry doing the work. Each turn, wherever it sits and whatever its wire gauge, encircles the same core and therefore the same $\Phi(t)$ ; Faraday's Law prices EMF per turn at $\Delta\Phi/\Delta t$ with no interest in the copper's thickness. Wire gauge matters to the losses, not the law: the winding carrying more current (the low-voltage side) needs thicker wire to keep its $I^2R$ down, which is why a step-down transformer's secondary looks stubby and fat next to its slender primary. The small print is flux leakage: the few tenths of a percent of flux that stray outside the core skip some turns, making the real ratio deviate slightly from $N_s/N_p$ . Engineering shrinks the leak with closed cores and interleaved windings, and exam problems ignore it under the word "ideal."

Which way does a transformer work — can I use the same one to step up or down?

Either way — the machine has no preferred direction. "Primary" simply names the winding you connect to the source; feed the 50-turn side and the 1000-turn side becomes a step-up output, feed the 1000-turn side and you have a step-down. Both windings sit on the same flux and obey the same volts-per-turn arithmetic. The practical caveats are ratings, not physics: each winding's insulation is built for its intended voltage and its wire gauge for its intended current, so driving a winding far beyond its design numbers risks the copper or the insulation even though the ratio law holds. Bench supplies and lab demonstrations reverse small transformers routinely; the grid's giants are likewise the same species of machine at the step-up (station) and step-down (substation) ends of a line — the direction of energy flow is chosen by where you put the source, not by the transformer.