Transformers

Electricity is generated at power plants, often far from the towns that need it, and the device that makes the long journey down the wires efficient — stepping voltage up for transmission and back down for safe household use — is the transformer. It works entirely on electromagnetic induction, with no moving parts and no direct electrical connection between its two coils, the changing magnetic flux in a shared iron core doing all the work. This section closes the electromagnetic-induction thread of the ICSE Class 10 Electricity and Magnetism chapter. You will pin down the transformer's principle and construction, the turns-ratio and power relations that let you solve any transformer numerical, and the Joule's-law reasoning that explains why power grids run at very high voltage. Real losses (eddy currents, hysteresis) are mentioned only qualitatively; three-phase systems and grid engineering stay out of scope.

The core insight

A transformer changes AC voltage by mutual induction between two coils sharing an iron core, with $V_s/V_p = N_s/N_p$ fixing the voltage change and $V_pI_p = V_sI_s$ fixing the matching trade-off in current, which is exactly why stepping voltage up before transmission cuts $I^2R$ line losses.

The core idea

A transformer is a static AC device with two coils — the primary ( $N_p$ turns) and the secondary ( $N_s$ turns) — wound on a common soft-iron core. The coils are not electrically connected; energy passes between them purely through the changing magnetic field in the core.

Why it needs AC. An alternating current in the primary produces a continuously changing magnetic flux in the core. Because the secondary coil links the same core, that changing flux induces an alternating EMF in it too, by Faraday's law of electromagnetic induction. A steady DC primary current would produce a constant flux — no change, no induced EMF, no output. This is the one fact examiners test most often.

Turns ratio fixes the voltage. Each turn of the primary and each turn of the secondary encloses the same changing flux, so the EMF induced per turn is identical on both sides. Total EMF is therefore proportional to the number of turns, giving the ideal-transformer relation

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

where: $V_p, V_s$ are the primary and secondary voltages, and $N_p, N_s$ are the number of turns in the primary and secondary coils.

A step-up transformer has $N_s > N_p$ , so $V_s > V_p$ : it raises voltage.
A step-down transformer has $N_s < N_p$ , so $V_s < V_p$ : it lowers voltage.

Power is conserved (ideally). With no losses, the electrical power entering the primary equals the power leaving the secondary:

V_p I_p = V_s I_s

where: $I_p, I_s$ are the primary and secondary currents. Whenever voltage is stepped up, current is stepped down by the same factor, and vice versa — a transformer trades voltage for current, it never creates extra power.

Figure

A transformer core showing a primary coil with fewer turns and a secondary coil with more turns linked by the same changing magnetic flux, illustrating a step up in voltage

Derivation ATurns ratio from equal induced EMF per turn

Show that the ratio of secondary to primary voltage in an ideal transformer equals the ratio of secondary to primary turns, starting only from the fact that both coils enclose the same changing magnetic flux produced in their shared iron core.

Step 1. By Faraday's law, the EMF induced in a coil equals the number of turns times the rate of change of flux through one turn: $e = -N\dfrac{d\phi}{dt}$ , where $\phi$ is the flux linking a single turn.

Step 2. In an ideal transformer, the entire flux $\phi$ produced by the primary is confined to the soft-iron core and links every turn of both coils equally — no flux leaks out. So the same $\dfrac{d\phi}{dt}$ appears in the EMF equation for both coils.

Step 3. The EMF induced in the primary is $e_p = N_p\dfrac{d\phi}{dt}$ , and the EMF induced in the secondary is $e_s = N_s\dfrac{d\phi}{dt}$ . Taking the ratio, the common factor $\dfrac{d\phi}{dt}$ cancels: $\dfrac{e_s}{e_p} = \dfrac{N_s}{N_p}$ .

Step 4. For an ideal transformer with negligible coil resistance, the terminal voltages equal the induced EMFs, so $V_p \approx e_p$ and $V_s \approx e_s$ . Substituting gives the turns-ratio relation, $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ , exactly as stated above — the voltage change is set entirely by the ratio of turns, nothing else.

Worked example

A transformer has $500$ turns in its primary coil and $2500$ turns in its secondary coil, and the primary is connected to a $220\ \text{V}$ AC supply. Find the secondary voltage, classify the transformer, and find the primary current if the secondary delivers $2\ \text{A}$ to its load.

Step 1. Write the turns-ratio relation and substitute the known values: $\dfrac{V_s}{220} = \dfrac{N_s}{N_p} = \dfrac{2500}{500} = 5$ .

Step 2. Solve for $V_s$ : $V_s = 220 \times 5 = 1100\ \text{V}$ .

Step 3. Classify the transformer. Since $N_s > N_p$ and $V_s > V_p$ , this is a step-up transformer.

Step 4. Use the ideal power relation, $V_pI_p = V_sI_s$ , to find the primary current given $I_s = 2\ \text{A}$ : $220 \times I_p = 1100 \times 2$ , so $I_p = \dfrac{2200}{220} = 10\ \text{A}$ . So the secondary voltage is $1100\ \text{V}$ , this is a step-up transformer, and the primary current is $10\ \text{A}$ — note current is stepped down by the same factor of $5$ that voltage is stepped up, keeping input power ( $220 \times 10 = 2200\ \text{W}$ ) equal to output power ( $1100 \times 2 = 2200\ \text{W}$ ).

Competitive stretch

A step-down transformer converts $6600\ \text{V}$ to $220\ \text{V}$ with a primary current of $2\ \text{A}$ . Why is high-voltage transmission chosen in the first place? Power lines have resistance $R$ , and heat lost obeys Joule's law, $H = I^2Rt$ . To deliver a fixed power $P = VI$ , a higher transmission voltage $V$ means a proportionally lower current $I$ , and since loss depends on $I^2$ , even a modest current reduction (say, to one-thirtieth, as $6600\ \text{V}$ steps down to $220\ \text{V}$ ) cuts transmission losses by a factor of $30^2 = 900$ . Here $I_s = \dfrac{6600 \times 2}{220} = 60\ \text{A}$ at the low-voltage end, confirming current rises exactly as voltage falls.

Common traps

Believing a transformer can run on DC — it needs a continuously changing flux to induce EMF; a steady DC current gives no induced secondary EMF at all.
Applying the turns ratio to current — current is inversely related to the turns ratio via $V_pI_p = V_sI_s$ , not directly, so $I_s/I_p \ne N_s/N_p$ .
Thinking a step-up transformer "creates" power — an ideal transformer only conserves power; voltage rises only at the cost of a proportional drop in current.
Saying "high voltage reduces loss" without the current link — losses fall because higher voltage permits lower current for the same delivered power, and $I^2R$ depends on the square of that current.

Ask the tutor: Want to try a few more step-up or step-down numericals, or have the high-voltage-transmission reasoning walked through again with different numbers?

Does the iron core actually carry any voltage or current itself?

No. The core is not part of either electrical circuit — it carries no current and has no voltage across it. Its only job is to guide the magnetic flux efficiently from the primary coil to the secondary coil, so that as much of the changing flux as possible links both coils. That is why it is made of soft iron, a material that is easy to magnetise and demagnetise as the AC alternates.

Why do textbooks say the transformer relations hold only for an "ideal" transformer?

A real transformer loses a little energy as heat, from the coils' own resistance and from eddy currents and hysteresis in the core, so its output power is always slightly less than its input power. The "ideal" assumption of zero losses lets us use the clean equalities $V_s/V_p = N_s/N_p$ and $V_pI_p = V_sI_s$ for numericals; at this level you only need to know that real losses exist, not calculate them.

If stepping up voltage is so efficient, why not transmit at even higher voltages than the grid already uses?

Very high voltages need thicker insulation, taller towers, and more expensive equipment to prevent sparking and breakdown, so engineers pick a transmission voltage that balances lower $I^2R$ losses against these rising equipment costs. This trade-off (not covered further at ICSE level) is why grids use very high but still finite voltages, stepped back down close to homes for safety.

Transformers

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads