Transformers

The transformer is one of the most elegant and economically important devices in electrical engineering. Without transformers, the high-voltage transmission of electricity over hundreds of kilometres — which is how modern national grids deliver power efficiently — would be impossible. Without them, alternating current (AC) would have no advantage over direct current (DC), and Edison might have won the "War of the Currents" against Tesla and Westinghouse. A transformer is, at heart, a direct application of Faraday's law from the previous lesson: two coils linked by a common magnetic circuit, with one coil's changing current inducing an e.m.f. in the other. This lesson concludes OCR H556 Module 6.3.

Spec mapping (OCR H556 Module 6.3 — electromagnetism). This lesson covers the structure of a simple iron-cored transformer (primary, secondary, soft-iron core); derivation of the transformer equation $N_p/N_s = V_p/V_s$ from Faraday's law applied to both coils linked by a common flux; the ideal-transformer power conservation $V_p I_p = V_s I_s$ giving $I_s/I_p = N_p/N_s$ ; qualitative understanding of eddy-current and hysteresis losses (laminated core, soft iron); the role of transformers in high-voltage transmission for the National Grid; and the requirement that transformers operate on AC (DC produces no flux change and no induced e.m.f.).

Transformer Structure

A simple transformer has three components:

A primary coil of $N_p$ turns, connected to the AC input.
A secondary coil of $N_s$ turns, connected to the output load.
A laminated soft-iron core linking the two coils, which channels the magnetic flux so that almost all of the flux generated by the primary links the secondary.

The soft-iron core is chosen because:

Soft iron is easily magnetised and demagnetised (high permeability, low remanence), so it follows the rapidly changing field from the primary efficiently and with minimal hysteresis loss.
Laminating the core (thin insulated sheets stacked together, instead of a solid block) breaks up the closed paths available for eddy currents induced in the iron itself. Eddy currents would otherwise circulate in the bulk iron, dissipating energy as ohmic heat by Lenz's law applied to the iron.
The core ensures that virtually all of the flux from the primary also links the secondary — without it, only a small fraction would be captured and the transformer would be inefficient.

OCR does not expect a quantitative treatment of eddy-current power or hysteresis loops, but does expect qualitative awareness of why the soft-iron core is laminated.

How It Works — Deriving the Transformer Equation

When an alternating voltage $V_p$ is applied to the primary, an alternating current flows, which produces an alternating flux $\Phi$ in the core. By Faraday's law, this changing flux induces an e.m.f. in each turn of both coils equal to $-\mathrm{d}\Phi/\mathrm{d}t$ .

Since both coils link the same core flux:

$V_p = -N_p\,\frac{\mathrm{d}\Phi}{\mathrm{d}t}, \qquad V_s = -N_s\,\frac{\mathrm{d}\Phi}{\mathrm{d}t}.$

(Here $V_p$ is identified with the back-e.m.f. in the primary — for an ideal transformer with no resistive losses, the applied $V_p$ equals minus the back-e.m.f., and similarly the induced $V_s$ equals the open-circuit secondary voltage.) Dividing the two equations and cancelling the common $\mathrm{d}\Phi/\mathrm{d}t$ :

$\frac{N_p}{N_s} = \frac{V_p}{V_s}, \qquad \text{equivalently} \qquad \frac{V_s}{V_p} = \frac{N_s}{N_p}.$

This is the transformer equation. The ratio of output to input voltage equals the ratio of secondary to primary turns — the turns ratio.

Step-up vs Step-down

flowchart TD
    A[Compare N_p and N_s] --> B{N_s > N_p?}
    B -->|yes| C[Step-up: V_s > V_p]
    B -->|no| D{N_s < N_p?}
    D -->|yes| E[Step-down: V_s < V_p]
    D -->|no| F[1:1 — isolation transformer]
    C --> G[I_s smaller<br/>by same ratio]
    E --> H[I_s larger<br/>by same ratio]

Step-up transformer: $N_s > N_p$ , so $V_s > V_p$ . Used for long-distance transmission, where high voltage minimises $I^2 R$ losses.
Step-down transformer: $N_s < N_p$ , so $V_s < V_p$ . Used for distribution to homes (final stage of the National Grid) and for small electronics (mains 230 V down to 12 V or 5 V).
1:1 isolation transformer ( $N_p = N_s$ , $V_p = V_s$ ): does not change voltage, but isolates secondary from primary — useful for safety (e.g. medical equipment).

A $1$ : $10$ step-up transformer takes $230$ V in and gives $2300$ V out. A $100$ : $1$ step-down transformer takes $11\,000$ V in and gives $110$ V out.

The Ideal Transformer and Power Conservation

An ideal transformer is one in which (a) all the flux from the primary links the secondary, and (b) there are no losses of any kind (no copper losses, no iron losses, no leakage flux). For such a transformer, the power delivered to the secondary equals the power drawn from the primary:

$P_p = P_s \quad \Rightarrow \quad V_p I_p = V_s I_s.$

Rearranging:

$\frac{V_s}{V_p} = \frac{I_p}{I_s}.$

Combining with the turns-ratio equation gives the full transformer relationship that OCR examines:

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

Note that the currents ratio is inverted compared with the voltages and turns. If you step voltage up by a factor of $10$ , you step current down by a factor of $10$ . Power is conserved in the ideal case.

Exam tip. OCR transformer questions almost always boil down to the chain $N_p/N_s = V_p/V_s = I_s/I_p$ plus ideal power conservation $V_p I_p = V_s I_s$ . Learn these by heart.

Why Transmit at High Voltage?

This current-step-down is the entire reason the National Grid uses high-voltage transmission. The power lost as heat in the transmission cables is

$P_{\text{loss}} = I^2 R,$

where $R$ is the (fixed) resistance of the cables and $I$ is the current flowing through them. Halving $I$ (by doubling the voltage) reduces the power loss by a factor of four. In the UK, the power is generated at $\sim 25$ kV, stepped up to $275$ kV or $400$ kV for long-distance transmission, and stepped down in stages (through $33$ kV substations and $11$ kV transformers) to $230$ V at the home. The combined effect is a reduction of transmission losses from roughly $50$ % (if everything were at $25$ kV) to less than $3$ %.

Worked Example 1 — Voltage and Current Ratio

A transformer with $1200$ primary turns and $50$ secondary turns steps down $240$ V AC for a model train layout. The train draws a current of $2.0$ A. Assuming ideal behaviour, calculate (a) the secondary voltage, (b) the current drawn from the primary.

(a) Secondary voltage.

$\frac{V_s}{V_p} = \frac{N_s}{N_p} \quad \Rightarrow \quad V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{50}{1200} = 10\ \text{V}.$

(b) Primary current (using power conservation):

$V_p I_p = V_s I_s \quad \Rightarrow \quad I_p = \frac{V_s I_s}{V_p} = \frac{10 \times 2.0}{240} \approx 0.083\ \text{A}.$

Or directly from the turns ratio:

$\frac{I_p}{I_s} = \frac{N_s}{N_p} = \frac{50}{1200} \quad \Rightarrow \quad I_p = 2.0 \times \frac{50}{1200} \approx 0.083\ \text{A}.$

Both methods give $I_p \approx 83$ mA at $240$ V — exactly the same $20$ W of power that is delivered to the secondary at $10$ V, $2.0$ A.

Worked Example 2 — National Grid Transmission

A power station generates $100$ MW at $25$ kV. A transformer steps this up to $400$ kV for transmission along a line of total resistance $10$ $\Omega$ . Calculate (a) the transmission current and power loss at $400$ kV, (b) the transmission current and power loss if the voltage were instead $25$ kV, and (c) the ratio.

(a) At $400$ kV.

$I_{400} = \frac{P}{V} = \frac{100 \times 10^6}{400 \times 10^3} = 250\ \text{A}$

$P_{\text{loss}} = I^2 R = (250)^2 \times 10 = 6.25 \times 10^5\ \text{W} = 625\ \text{kW}.$

Only $0.625$ % of the generated power is lost — acceptable by engineering standards.

(b) At $25$ kV.

$I_{25} = \frac{100 \times 10^6}{25 \times 10^3} = 4000\ \text{A}$

$P_{\text{loss}} = (4000)^2 \times 10 = 1.6 \times 10^8\ \text{W} = 160\ \text{MW}.$

We would lose more than the entire generated output to heat in the cables. Clearly impossible.

(c) Ratio of losses.

$\frac{P_{\text{loss}}(25)}{P_{\text{loss}}(400)} = \left(\frac{I_{25}}{I_{400}}\right)^2 = \left(\frac{4000}{250}\right)^2 = 16^2 = 256.$

Transmitting at $400$ kV instead of $25$ kV reduces losses by a factor of $256$ . This is the economic reason for the National Grid's high-voltage transmission.

Worked Example 3 — A Real-World 11 kV → 230 V Substation

A distribution-substation transformer steps $11\,000$ V down to $230$ V for a row of houses. The street loadings draw a total of $50$ A at $230$ V.

(a) Find the turns ratio.

$\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{11\,000}{230} \approx 47.8.$

So $N_p/N_s \approx 48$ : $1$ .

(b) Find the primary current (ideal transformer):

$I_p = I_s \times \frac{N_s}{N_p} = 50 \times \frac{1}{48} \approx 1.04\ \text{A}.$

$P_p = V_p I_p = (11\,000)(1.04) \approx 11.4\ \text{kW}, \quad P_s = V_s I_s = (230)(50) = 11.5\ \text{kW}.$

Equal within rounding error — power is conserved in the ideal case.

(d) If the transformer is $98$ % efficient (a typical real value), the primary current is slightly higher:

$P_p = P_s / 0.98 = 11.5/0.98 \approx 11.73\ \text{kW}, \quad I_p = 11.73 \times 10^3 / 11\,000 \approx 1.07\ \text{A}.$

About $230$ W is dissipated in the core and windings.

AC Is Essential

Transformers work only with alternating current. A DC current in the primary would produce a constant flux, $\mathrm{d}\Phi/\mathrm{d}t = 0$ , and no e.m.f. would be induced in the secondary. Transformers are the reason the whole world standardised on AC rather than DC in the late 19th century — you simply cannot step DC up and down the way you can step AC up and down.

Modern electronic DC-DC converters achieve the same trick by rapidly switching a DC supply into a pulsed form that contains enough $\mathrm{d}I/\mathrm{d}t$ to make a transformer work — typically at $20$ kHz to $1$ MHz. The result: laptop power supplies and phone chargers are essentially miniature switched-mode transformers with very small ferrite cores (because higher frequency reduces the required $A$ and $N$ ).

What OCR Does NOT Require

To keep the topic manageable, OCR H556 deliberately excludes several advanced transformer topics that are covered by other specifications:

Rotating-coil e.m.f. derivation ( $\varepsilon = NBA\omega\sin\omega t$ ). OCR treats rotating coils only qualitatively.
RMS and peak values for AC, including $V_{\text{rms}} = V_0/\sqrt{2}$ . These belong to other modules.
Eddy-current calculations. OCR expects qualitative awareness only.
Hysteresis loops and core losses. Qualitative only.
Transformer efficiency formulae beyond "power in $=$ power out (ideal)".

For OCR you can safely skip the numerical RMS and eddy-current questions — they will not appear in your exam.

Synoptic Links

Module 6.3 (electromagnetic induction, previous lesson). Both transformer coils obey Faraday's law $\varepsilon = -N\,\mathrm{d}\Phi/\mathrm{d}t$ ; the transformer equation is just the ratio of two Faraday's-law statements with a common flux.
Module 6.3 (Lenz's law). Eddy-current losses in the core are dissipation associated with the back-reaction described by Lenz's law: the changing core flux induces eddy currents whose own field opposes the change, doing ohmic work in the iron.
Module 4 (electricity). $P = VI$ for power, $P = I^2 R$ for resistive dissipation in cables. These are the algebraic backbone of the high-voltage-transmission argument.
Module 4 (resistivity). Cable resistance $R = \rho L/A$ : high-voltage transmission allows the use of thinner cables (smaller $A$ , larger $R$ ) while keeping losses acceptable.
Engineering — switched-mode power supplies, induction cookers, wireless charging. All use transformer physics at frequencies from $50$ Hz to $\sim$ MHz.
Economics of energy distribution. The Tesla-Edison War of the Currents (1880s-90s) was settled by the discovery of practical transformers; AC won because transformers cannot work with DC at any reasonable scale.

Specimen question modelled on the OCR H556 paper format

Question (9 marks): A transformer in a school physics laboratory is to be used to step a $230$ V AC supply down to $12$ V to power a low-voltage lamp drawing $3.0$ A. The primary winding has $1150$ turns.

(a) Calculate the number of turns required on the secondary, assuming an ideal transformer. [2]

(b) Calculate the current drawn from the primary. [2]

Transformers

Transformers

Transformer Structure

How It Works — Deriving the Transformer Equation

Step-up vs Step-down

The Ideal Transformer and Power Conservation

Why Transmit at High Voltage?

Worked Example 1 — Voltage and Current Ratio

Worked Example 2 — National Grid Transmission

Worked Example 3 — A Real-World 11 kV → 230 V Substation

AC Is Essential

What OCR Does NOT Require

Synoptic Links

Specimen question modelled on the OCR H556 paper format

More in Physics