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.

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Transformer Structure

A simple transformer has three components:

1. A primary coil of N_p turns, connected to the AC input.
2. A secondary coil of N_s turns, connected to the output load.
3. A laminated soft-iron core (or more exotic magnetic material) linking the two coils, which channels the magnetic flux so that almost all of the flux generated by the primary links the secondary.

graph TD
    subgraph transformer[Iron-cored transformer]
        PI[Primary input<br/>V_p, I_p] --> PC[Primary coil N_p turns]
        PC -->|flux in core| SC[Secondary coil N_s turns]
        SC --> SO[Secondary output<br/>V_s, I_s]
    end

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 without hysteresis losses.
• Laminating the core (thin insulated sheets stacked together) reduces eddy currents, although at A-Level for OCR you only need a qualitative understanding of this.

OCR does not expect a quantitative treatment of transformer power losses, eddy currents or hysteresis.

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How It Works

When an alternating voltage V_p is applied to the primary, an alternating current flows, which produces an alternating flux Φ in the core. By Faraday's law, this changing flux induces an e.m.f. in each turn of both coils equal to −dΦ/dt. Since both coils link the same core flux:

V_p = −N_p (dΦ/dt)      (magnitude of primary back-EMF)
V_s = −N_s (dΦ/dt)      (magnitude of secondary induced EMF)

Dividing:

V_s / V_p = 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

• Step-up transformer: N_s > N_p, so V_s > V_p. Raises voltage (typically for long-distance transmission).
• Step-down transformer: N_s < N_p, so V_s < V_p. Lowers voltage (typically for distribution to homes).

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.

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The Ideal Transformer and Power Conservation

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

P_p = P_s
V_p I_p = V_s I_s

Rearranging,

V_s / V_p = I_p / I_s

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

V_s / V_p = N_s / N_p = I_p / I_s

Note the currents ratio is inverted compared with voltages. If you step voltage up by a factor of 10, you step current down by a factor of 10.

Why Transmit at High Voltage?

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

P_loss = I² 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 ~25 kV, stepped up to 275 kV or 400 kV for long-distance transmission, and stepped down in stages to 230 V at the home. The combined effect is a reduction of transmission losses from about 50% (if everything were at 25 kV) to less than 3%.

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Worked Example — 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

V_s / V_p = N_s / N_p
V_s = V_p × N_s / N_p = 240 × 50 / 1200 = 10 V

(b) Primary current

V_p I_p = V_s I_s
I_p = V_s I_s / V_p = 10 × 2.0 / 240 = 0.083 A

Alternatively using the turns ratio directly:

I_p / I_s = N_s / N_p = 50/1200
I_p = 2.0 × 50/1200 = 0.083 A

The primary draws 83 mA at 240 V — the same 20 W of power that is delivered to the secondary at 10 V 2 A.

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Worked Example — 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 Ω. 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 = P/V = 100 × 10⁶ / 400 × 10³ = 250 A
P_loss = I² R = 250² × 10 = 625 × 10³ W = 625 kW

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

(b) At 25 kV

I_25 = P/V = 100 × 10⁶ / 25 × 10³ = 4000 A
P_loss = 4000² × 10 = 1.6 × 10⁸ W = 160 MW