Transformers

Somewhere between the power plant on the edge of your city and the switch on your wall, the electricity you use has been squeezed and stretched in voltage several times over — up to hundreds of thousands of volts for the cross-country run, then back down, step by step, to the safe 220 V that reaches your socket. Every one of those conversions is done by the same humble device: two coils of copper wire wound on a shared iron core, with no motor, no moving part, and — if it is well built — almost no wasted energy. That device is the transformer, and it is one of the quiet reasons electricity can be generated far from where it is used at all. The idea traces back to Michael Faraday's discovery that a changing magnetic flux induces an EMF in a nearby coil. A transformer does nothing more exotic than harness that single fact twice: an alternating current in one coil (the primary) creates a changing flux in the shared core, and that same changing flux induces an EMF in the second coil (the secondary). Because both coils feel the same rate of change of flux, the ratio of their induced voltages is fixed purely by how many times each is wound — the turns ratio. Wind more turns on the output side and you step voltage up; wind fewer and you step it down. This article follows the CBSE-prescribed thread exactly: how the voltage relation $V_s/V_p = N_s/N_p$ falls out of Faraday's law, why an ideal (lossless) transformer must trade current for voltage so that power is conserved, and why that trade-off is the entire justification for high-voltage power transmission — sending the same power at higher voltage and lower current slashes the $I^2R$ heating losses in the long cables in between. We will also see, briefly and qualitatively, why a real transformer falls a little short of ideal.

Key takeaways

A transformer transfers AC electrical energy between two coils — primary ( $N_p$ turns) and secondary ( $N_s$ turns) — linked by a shared, changing magnetic flux through a laminated iron core.
Voltage relation: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ — output-to-input voltage ratio equals the turns ratio.
$N_s > N_p$ gives a step-up transformer ( $V_s > V_p$ ); $N_s < N_p$ gives a step-down transformer ( $V_s < V_p$ ).
For an ideal (lossless) transformer, input power equals output power at every instant: $V_pI_p = V_sI_s$ , so $\dfrac{I_s}{I_p} = \dfrac{N_p}{N_s}$ — current changes inversely to voltage.
This inverse trade-off is why power is transmitted at very high voltage and low current over long lines: since line loss goes as $I^2R$ , cutting the current cuts the loss far more than raising the voltage costs.
Real transformers lose a little energy to resistive (copper) heating, hysteresis in the core, and eddy currents — all mentioned only qualitatively at this level, never calculated.
A transformer only works on AC, never on steady DC, because a constant current makes a constant (non-changing) flux, and Faraday's law needs a changing flux to induce any EMF at all.
Key idea: because both coils of a transformer share the same changing flux, their induced voltages scale with their turns count, $V_s/V_p = N_s/N_p$ , and an ideal transformer conserves power ( $V_pI_p = V_sI_s$ ), so raising voltage always means lowering current by the same factor — the trade-off that makes long-distance AC transmission efficient.

The core insight

A transformer uses a shared, changing magnetic flux to make $V_s/V_p = N_s/N_p$ , and because an ideal transformer conserves power ( $V_pI_p = V_sI_s$ ), stepping voltage up always steps current down by the same factor, which is exactly why power is transmitted at high voltage and low current to minimise $I^2R$ line losses.

Where it fits in the syllabus

Transformers close the Alternating Current unit in CBSE Class 12, sitting right after the AC generator in the prescribed list — both are applications of electromagnetic induction to AC systems, drawing on Faraday's law and mutual inductance from the previous chapter. The syllabus keeps this topic tightly bounded: only the principle, working, and the step-up/step-down turns-ratio relation $V_s/V_p = N_s/N_p$ together with the ideal-transformer power relation $V_pI_p = V_sI_s$ are examinable. Transformer losses — hysteresis, eddy currents, copper (resistive) loss — are explicitly to be mentioned qualitatively at most; there is no efficiency formula, no numerical loss calculation, and no derivation of core-loss mechanisms expected at board level. This article honours that boundary throughout: every formula used below is one of these two relations or a direct algebraic consequence of them.

The physical picture

A transformer has two separate windings — the primary, with $N_p$ turns, connected to the AC input supply, and the secondary, with $N_s$ turns, connected to the output load — wound on a common laminated soft-iron core. The core's high permeability concentrates almost all the magnetic flux produced by the primary's current so that essentially the same flux $\phi_B$ threads every single turn of both coils, primary and secondary alike. Lamination (thin, insulated iron sheets rather than one solid block) is used specifically to choke off eddy currents circulating within the core itself, one of the real-transformer loss mechanisms mentioned below. Because each turn of a coil independently experiences the same rate of change of flux, Faraday's law gives the EMF induced in each winding as proportional to its own number of turns:

\varepsilon_p = -N_p\frac{d\phi_B}{dt}, \qquad \varepsilon_s = -N_s\frac{d\phi_B}{dt}

where: $\varepsilon_p,\varepsilon_s$ are the instantaneous EMFs induced in the primary and secondary respectively, $N_p,N_s$ are their turns counts, and $d\phi_B/dt$ is the (common) rate of change of magnetic flux through one turn of the core. The physical reading is simple: the flux does not care which coil it is threading, so whichever coil has more turns collects a proportionally larger EMF — voltage transformation is nothing but a turns-counting trick riding on one shared, changing flux. For an ideal transformer — no winding resistance, no flux leakage out of the core, no core losses — the terminal voltage across the primary equals its back-EMF in magnitude, and the terminal voltage across the secondary equals its induced EMF. Dividing the two induced-EMF expressions above (the common $d\phi_B/dt$ cancels) gives the central transformer relation:

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

where: $V_p$ is the RMS (or peak) voltage applied to the primary, $V_s$ is the RMS (or peak) voltage delivered by the secondary, and $N_s/N_p$ is the turns ratio. If $N_s > N_p$ , more secondary turns mean $V_s > V_p$ — a step-up transformer, the kind used at power plants to raise voltage before long-distance transmission. If $N_s < N_p$ , fewer secondary turns mean $V_s < V_p$ — a step-down transformer, used at substations and in household adapters to bring voltage back down to safe, usable levels.

Figure

A transformer core diagram showing a primary coil with fewer turns on the left connected to an AC source and a secondary coil with more turns on the right connected to a load, joined by a common laminated iron core carrying the shared changing flux, illustrating a step up transformer

For an ideal transformer, no energy is created or destroyed, so the power drawn from the source equals the power delivered to the load at every instant:

V_pI_p = V_sI_s

where: $I_p,I_s$ are the RMS currents in the primary and secondary. Combined with the voltage relation, this immediately gives the current relation:

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

The physical reading is the trade-off at the heart of the whole topic: whatever a transformer gains in voltage, it loses proportionally in current, and vice versa. A step-up transformer raises voltage but lowers current; a step-down transformer does the opposite. Neither transformer creates or destroys power — it only reshapes the voltage-current split.

Voltage up, current down: why this makes long-distance transmission efficient

Derivation AWhy stepping voltage up before transmission cuts power loss

The goal here is to show, starting only from the ideal-transformer power relation $V_pI_p = V_sI_s$ and the resistive-heating formula $P_{loss} = I^2R$ , that delivering a fixed amount of power over a long transmission line at a higher voltage and correspondingly lower current reduces the power wasted as heat in the line — the entire practical justification for building step-up transformers at power stations and step-down transformers at the receiving end, without needing any transformer-efficiency numbers.

Step 1. Fix the power that must reach the far end of a transmission line at some value $P$ . From the ideal power relation applied to the sending end, $P = VI$ , where $V$ is the line (transmission) voltage and $I$ is the line current carrying that power. Since $P$ is fixed by the load's needs, $V$ and $I$ cannot both be chosen freely — raising one forces the other down in the same proportion, exactly the inverse relationship a step-up transformer enforces.

Step 2. Write the resistive power lost in the transmission line itself, of fixed resistance $R$ , as $P_{loss} = I^2R$ . This loss depends on the square of the current, not on the voltage at all — the line resistance heats up purely because current flows through it, regardless of what voltage that current happens to be riding on.

Step 3. Substitute $I = P/V$ from Step 1 into the loss expression: $P_{loss} = \left(\dfrac{P}{V}\right)^2 R = \dfrac{P^2R}{V^2}$ . For a fixed power $P$ to be delivered and a fixed line resistance $R$ , the loss falls as the square of the transmission voltage. Doubling the transmission voltage, for the same delivered power, quarters the line loss; a step-up transformer that raises voltage tenfold cuts the loss to one-hundredth.

Step 4. Read the practical conclusion. A power station cannot simply choose to send a small current at a small voltage and still deliver the same power over a long, resistive line without paying dearly in $I^2R$ heat. The transformer's ability to trade voltage for current for free (in the ideal case) is therefore exploited deliberately: step the voltage up to hundreds of kilovolts for the cross-country run, keeping the current — and hence the heat lost in the cables — small, then step it back down through a cascade of step-down transformers to a safe delivery voltage at homes and factories.

Worked examples

Board level - step-down transformer for household supply

A step-down transformer converts a $2200\ \text{V}$ (RMS) transmission-line voltage to $220\ \text{V}$ for household use. The primary coil has $5000$ turns, and the secondary supplies a current of $10\ \text{A}$ to the load. Assume an ideal transformer.

Step 1. Find the turns ratio from the voltage relation: $\dfrac{N_s}{N_p} = \dfrac{V_s}{V_p} = \dfrac{220}{2200} = 0.1$ .

Step 2. Find the number of secondary turns: $N_s = 0.1\times N_p = 0.1\times5000 = 500$ turns.

Step 3. Find the primary current from the ideal power relation: $V_pI_p = V_sI_s \implies I_p = \dfrac{V_sI_s}{V_p} = \dfrac{220\times10}{2200} = 1\ \text{A}$ .

The transformer steps 2200 V down to 220 V (turns ratio 10:1) while stepping current up from 1 A to 10 A — voltage and current move in exactly opposite directions, as the ideal power relation requires.

Competitive level - line loss comparison with and without step-up transmission

Timing note: aim to finish this in under three minutes — it is two independent applications of $P_{loss}=I^2R$ with a ratio at the end, no new physics beyond the derivation above. A power station must deliver $P = 100\ \text{kW}$ of power to a town over a transmission line of total resistance $R = 20\ \Omega$ . Compare the power lost in the line if the power is sent (a) directly at $V_1 = 220\ \text{V}$ , versus (b) after a step-up transformer raises it to $V_2 = 11{,}000\ \text{V}$ for transmission (with a step-down transformer restoring $220\ \text{V}$ at the receiving end).

Step 1. Case (a): find the line current at $220\ \text{V}$ using $P = V_1I_1$ : $I_1 = \dfrac{P}{V_1} = \dfrac{100{,}000}{220} = 454.5\ \text{A}$ .

Step 2. Find the line loss in case (a): $P_{loss,1} = I_1^2R = (454.5)^2\times20 = 2.07\times10^{7}\ \text{W} = 20{,}700\ \text{kW}$ . This absurdly large loss (over $200$ times the power actually delivered) shows that transmitting bulk power at ordinary domestic voltage is simply impossible in practice.

Step 3. Case (b): find the line current at $11{,}000\ \text{V}$ : $I_2 = \dfrac{P}{V_2} = \dfrac{100{,}000}{11{,}000} = 9.09\ \text{A}$ .

Step 4. Find the line loss in case (b): $P_{loss,2} = I_2^2R = (9.09)^2\times20 = 1.65\times10^{3}\ \text{W} = 1.65\ \text{kW}$ , only about $1.65\%$ of the delivered power.

Stepping the transmission voltage up by a factor of $50$ (from 220 V to 11,000 V) cuts the line loss by a factor of $50^2=2500$ , from an unworkable 20,700 kW down to a modest 1.65 kW — the quantitative reason every grid uses high-voltage transmission with transformers at both ends.

Common traps

Believing a transformer can increase power — an ideal transformer conserves power ( $V_pI_p=V_sI_s$ ); it only trades voltage against current, never creating extra energy from nothing.
Assuming a step-up transformer, which raises voltage, also raises current — current actually falls in a step-up transformer and rises in a step-down transformer; the two ratios are always reciprocals of each other, never equal.
Confusing the turns ratio with the current ratio directly — the turns ratio $N_s/N_p$ equals the voltage ratio $V_s/V_p$ , while the current ratio $I_s/I_p$ is its reciprocal, $N_p/N_s$ .
Thinking a transformer works on steady DC — Faraday's law needs a changing flux to induce any EMF; a constant DC current produces a constant flux and induces nothing, so an ordinary transformer core simply does not respond to DC at all.
Writing $P_{loss}=I^2R$ using the secondary-side current when the question asks about line (transmission-cable) loss, or vice versa — always identify which current actually flows through the resistive segment being asked about.

Common trapHigh transmission voltage lowers line loss because it lowers current, not because voltage itself "loses less"

The single most common wrong explanation for high-voltage transmission is some vague claim that "higher voltage travels better" or "voltage doesn't get lost as easily as current." That is not the mechanism. Resistive line loss depends only on $I^2R$ — it has no direct voltage term at all. The real chain of reasoning is: for a fixed power to be delivered, $P=VI$ forces current down whenever voltage is raised (by a step-up transformer), and it is that lower current, squared, which shrinks the $I^2R$ heat dissipated in the cable. State the mechanism as "raising the transmission voltage lowers the current needed to deliver the same power, and since line loss goes as current squared, the loss falls sharply" — not simply "high voltage loses less."

You should be able to…

Describe the construction and working principle of a transformer, including why the core is laminated and why the input must be AC, not DC.
State and apply the voltage relation $V_s/V_p = N_s/N_p$ and the ideal power relation $V_pI_p = V_sI_s$ to find any one unknown among $V_p,V_s,N_p,N_s,I_p,I_s$ .
Distinguish step-up from step-down transformers by their turns ratio and correctly state which of voltage or current increases and which decreases in each case.
Derive, from $P=VI$ and $P_{loss}=I^2R$ , why raising the transmission voltage reduces resistive line loss for a fixed delivered power, and compute that loss numerically.
Explain qualitatively (without formulas) the three sources of energy loss in a real transformer: copper (resistive) loss, hysteresis loss, and eddy-current loss.

Ask the tutor: if you want to work through a mixed turns-ratio-and-transmission-loss problem step by step, or want the full numeric comparison of $I^2R$ loss at different transmission voltages for your own chosen numbers, ask the AI tutor to walk through it with you.

Why can't we just use a bigger wire instead of transformers to cut transmission loss?

A thicker wire does have lower resistance, which does reduce $I^2R$ loss, but only up to a point — copper or aluminium cable that thick, strung for hundreds of kilometres, would be prohibitively heavy, expensive, and hard to support on towers. Raising the transmission voltage with a transformer, by contrast, is essentially free once you have built the transformer, and its effect on loss (loss falls as the square of voltage, from the derivation above) is far more dramatic than any practical wire upgrade could achieve. In practice grids use both a reasonably good conductor and very high transmission voltages, but the transformer's voltage-for-current trade is doing most of the work.

Does a transformer's turns ratio have to be a whole number, like exactly 10:1?

No — the turns ratio is just $N_s/N_p$ , and while real transformers are built with whole numbers of physical turns (you cannot wind half a loop of wire), the ratio itself can come out to any value once you divide, such as 0.1 or 3.7. In board problems you will sometimes be given the ratio directly (like 1:20) rather than the individual turns counts, and you use it exactly the same way in $V_s/V_p=N_s/N_p$ .

If a transformer is not perfectly efficient, does that break the power relation VpIp = VsIs used in problems?

The relation $V_pI_p=V_sI_s$ is only exactly true for an ideal transformer — the CBSE syllabus tells you to assume this idealisation in every numerical problem unless a specific efficiency is given. Real transformers do lose a small amount of energy to resistive heating in the windings, hysteresis in the core material, and eddy currents circulating inside the core, so a real secondary delivers very slightly less power than the primary supplies. But the syllabus keeps this at a qualitative mention only — no efficiency formula or loss numbers are examinable, so in every worked problem you should treat the transformer as ideal and use $V_pI_p=V_sI_s$ without hesitation.

Why does the iron core have to be laminated instead of a solid block?

A solid iron core, sitting inside a changing magnetic field, would itself act like a giant short-circuited secondary coil — the changing flux induces swirling "eddy currents" directly within the metal, which heat the core and waste energy (one of the three real-transformer loss mechanisms mentioned in this chapter, alongside copper loss and hysteresis loss). Slicing the core into thin sheets, each coated with a thin insulating layer, breaks up the paths available to these eddy currents and confines them to tiny loops within each sheet, dramatically reducing the wasted heat. This lamination detail is mentioned qualitatively in the syllabus and is a favourite one-line "why" question in exams.