Transformers and AC Power Transmission

Transformers are the unsung workhorse of the modern electrical grid. Every kilowatt-hour delivered to a UK home has passed through at least three transformers — at the power station, at the supergrid substation, and at the local 11 kV distribution substation. AQA specification 3.7.5 expects you to understand the transformer equation, the construction features that achieve high efficiency, the power-conservation relation V_p I_p = V_s I_s for the ideal case, the loss mechanisms in the real device, and the engineering rationale for high-voltage long-distance transmission via the National Grid.

Spec mapping: AQA 7408 A-Level Physics, Section 3.7.5 (Transformers and AC power transmission). This lesson develops the ideal-transformer turns-ratio equation V_p/V_s = N_p/N_s = I_s/I_p, the four principal loss mechanisms in real transformers and their engineering remedies (laminated core, soft iron, thick copper windings, closed core), the use of step-up and step-down transformers in the National Grid, the I²R cable-loss argument for high transmission voltages, and a worked comparison of cable losses at 11 kV vs 400 kV. Refer to the official AQA 7408 specification document for the authoritative wording.

Synoptic links: (i) Electromagnetic induction and Faraday's law (3.7.5) — the transformer works because the primary's alternating current generates a changing flux in the iron core, and Faraday's law applied to the secondary winding produces an induced EMF in the ratio of turns. (ii) AC waveforms and rms values (3.7.5) — transmission losses are calculated using rms current values, and the power-conservation relation involves rms quantities. (iii) Energy and power (3.5.1) — the entire National Grid is a study in efficient energy transfer over a distance, with each transformer trading current for voltage to minimise I²R cable losses.

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The Ideal Transformer

A transformer transfers electrical energy from one circuit to another through electromagnetic induction. Two coils are wound on a shared iron core:

• The primary coil (Np turns) carries the input alternating current I_p at voltage V_p.
• The secondary coil (Ns turns) delivers the output current I_s at voltage V_s.

graph LR
    A["AC source V_p"] --> B["Primary coil (Np turns)"]
    B --> C["Iron core (laminated, soft iron)"]
    C --> D["Secondary coil (Ns turns)"]
    D --> E["Load: V_s = V_p × Ns/Np"]

    style B fill:#3498db,color:#fff
    style D fill:#27ae60,color:#fff
    style C fill:#95a5a6,color:#fff

How the Transformer Works

1. The alternating current in the primary creates an alternating magnetic flux Φ(t) in the iron core.

2. The iron core's high magnetic permeability channels nearly all the flux through the core (rather than dissipating it as stray field), so the same Φ(t) threads both coils.

3. By Faraday's law applied to each coil, the EMFs induced are proportional to the number of turns:

   ε_p = −Np dΦ/dt and ε_s = −Ns dΦ/dt

4. Dividing one by the other:

   ε_s / ε_p = N_s / N_p

In an ideal transformer, the primary back-EMF ε_p equals the applied voltage V_p (no resistive loss in the primary winding), and the secondary EMF ε_s equals the secondary terminal voltage V_s (no resistive loss in the secondary). This gives the standard form:

V_s / V_p = N_s / N_p

• Step-up transformer (Ns > Np): V_s > V_p; voltage increases, current decreases.
• Step-down transformer (Ns < Np): V_s < V_p; voltage decreases, current increases.
• Isolation transformer (Ns = Np): V_s = V_p; provides electrical isolation but no voltage change.

Power Conservation

An ideal transformer is 100% efficient — all the input power appears at the output:

V_p I_p = V_s I_s

Combining with the turns-ratio equation:

I_s / I_p = N_p / N_s = V_p / V_s

So the current ratio is the inverse of the turns ratio. A 10:1 step-up transformer (Ns = 10 Np, so V_s = 10 V_p) has I_s = I_p/10 — the secondary current is one-tenth of the primary current.

Worked Example 1 — Step-up Transformer

Question: A power-station transformer steps up an alternating supply from 25 kV (rms) to 400 kV (rms). If the primary current is 800 A (rms), calculate the turns ratio and the secondary current, assuming the transformer is ideal.

Solution:

Turns ratio: N_s / N_p = V_s / V_p = 400 / 25 = 16

Secondary current: I_s = I_p × (V_p / V_s) = 800 × (25 / 400) = 50 A (rms)

Check: V_p I_p = 25 000 × 800 = 2.0 × 10⁷ W = 20 MW V_s I_s = 400 000 × 50 = 2.0 × 10⁷ W = 20 MW ✓

The same 20 MW of power is delivered, but at 16× the voltage and 1/16 the current.

Worked Example 2 — Step-down Transformer

Question: A local distribution transformer steps down 11 kV (rms) to the domestic 230 V (rms) supply. The transformer has 5000 turns on the primary. Calculate the number of turns on the secondary, and the primary current when the secondary delivers 100 A (rms) to consumers.

Solution:

N_s / N_p = V_s / V_p = 230 / 11 000 = 0.0209 N_s = 5000 × 0.0209 = 104.5 ≈ 105 turns

I_p = I_s × (V_s / V_p) = 100 × (230 / 11 000) = 2.09 A (rms)

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Real Transformer Construction

To approach ideal behaviour as closely as practical, real transformers use specific construction features that minimise each of the four principal loss mechanisms.

1. Laminated Iron Core

The iron core is built from thin (~0.3 mm) iron sheets separated by insulating varnish, stacked together. The lamination orientation is parallel to the flux direction.

Why: The changing flux in a solid iron core would induce eddy currents (Lenz-law loops) within the core itself, dissipating energy as I²R heat in the iron. Lamination breaks these loops into small isolated regions, drastically reducing the eddy current path length and the energy lost.

2. Soft Iron Core Material

The core is made of soft iron (or modern silicon-steel laminates) rather than hard ferromagnetic material like steel for permanent magnets.

Why: As the AC current cycles, the magnetic domains in the core repeatedly realign. Each domain reorientation dissipates energy (hysteresis loss), with the loss per cycle proportional to the area enclosed by the B-H hysteresis loop. Soft iron has a thin hysteresis loop, so the loss per cycle is small. Hard magnetic materials (high coercivity) would have wide loops and high hysteresis losses.

3. Thick Copper Windings

The primary and secondary coils are wound from thick copper conductors to minimise the resistance R of each winding.

Why: The I²R copper loss (Joule heating) in each winding is proportional to its resistance. Doubling the cross-sectional area of the copper halves R and halves the copper loss for given current. For grid-scale transformers, the copper alone can mass tens of tonnes.

4. Closed Magnetic Circuit (Shell or Core Construction)

The core is shaped to form a complete closed magnetic loop linking the primary and secondary coils, with both windings sharing the same flux path.

Why: A closed core ensures that virtually all the flux from the primary links the secondary, with minimal flux leakage into surrounding air. Flux leakage is energy that reaches the secondary as a reduced flux linkage and therefore lower induced EMF. A typical commercial transformer achieves 99%+ of flux linkage.

Efficiency

The efficiency of a real transformer is:

η = (output power / input power) × 100% = (V_s I_s / V_p I_p) × 100%

Modern grid-scale transformers achieve 99% to 99.5% efficiency. The 1% loss appears as heat in the core (eddy currents + hysteresis) and the windings (copper losses), and is removed by oil-circulation cooling. The largest grid transformers are immersed in tanks of insulating mineral oil, which both insulates the high-voltage windings and circulates heat to external radiator fins.

Worked Example 3 — Real Transformer Efficiency

Question: A transformer has a primary input of 6.0 kW at 240 V (rms). The secondary delivers 24 A at 240 V to a resistive load. Calculate the secondary power, the efficiency of the transformer, and the total power loss in the transformer.

Solution:

P_p (input) = 6000 W P_s (output) = V_s × I_s = 240 × 24 = 5760 W

η = P_s / P_p × 100% = 5760 / 6000 × 100% = 96%

Power loss = P_p − P_s = 6000 − 5760 = 240 W

This loss is the sum of all the loss mechanisms — copper loss in both windings, eddy currents in the core, hysteresis loss in the core, and any flux leakage.

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AC Power Transmission and the National Grid

The electrical power generated at a station must be delivered to consumers over distances of tens or hundreds of kilometres. The cables connecting station to consumer have resistance R, and the I²R loss in the cables is the dominant energy loss in the transmission system.

Why High Transmission Voltage?

For a power station delivering power P to a load over cables of total resistance R:

I = P / V (current required at transmission voltage V)

P_loss = I²R = (P/V)² R = P²R/V²

The cable loss is proportional to V⁻². Doubling V cuts the loss by a factor of 4. Increasing V by a factor of 10 cuts the loss by a factor of 100. The transformer makes this trick possible: stepping up to high voltage at the station and stepping back down near the consumer.

UK National Grid Voltage Hierarchy

The UK National Grid uses a hierarchical voltage structure to minimise losses while keeping local-distribution voltages safe and practical:

Tier	Voltage	Use
Super-grid	400 kV (some 275 kV)	Long-distance transmission between regions
Primary distribution	132 kV / 66 kV / 33 kV	Regional distribution to substations
Secondary distribution	11 kV	Local distribution to street substations
Domestic	230 V	Inside homes and businesses

These are well-established round figures for the UK grid. A power station generates at ~25 kV, steps up to 400 kV for the super-grid, and a sequence of step-down transformers reduces this back through 132 kV → 33 kV → 11 kV → 230 V over the journey to a domestic socket. Each transformer's losses are small (~1%), but the cumulative chain still delivers ~92% to the consumer.

Worked Example 4 — Comparing Transmission at 11 kV and 400 kV

Question: A 50 MW power station is connected to a substation 60 km away via aluminium transmission cables with a resistance of 0.020 Ω per km (so total cable resistance = 1.2 Ω). Compare the cable power loss when transmitting at (a) 11 kV and (b) 400 kV. Express each loss as a percentage of the transmitted power.

Solution:

(a) At 11 kV: I = P/V = 50 × 10⁶ / 11 × 10³ = 4545 A P_loss = I²R = (4545)² × 1.2 = 2.066 × 10⁷ × 1.2 = 2.48 × 10⁷ W = 24.8 MW Fractional loss = 24.8/50 = 49.6% — nearly half the generated power lost as cable heat!

(b) At 400 kV: I = P/V = 50 × 10⁶ / 400 × 10³ = 125 A P_loss = I²R = (125)² × 1.2 = 15 625 × 1.2 = 1.875 × 10⁴ W = 18.75 kW Fractional loss = 0.01875/50 = 0.038% — less than one-tenth of one per cent!

Ratio of loss at 11 kV to 400 kV: 24.8 MW / 0.01875 MW = 1322 Square of voltage ratio: (400/11)² = 1322 ✓

The factor-of-1322 reduction in cable loss is precisely the V² scaling that makes the National Grid feasible. Without high-voltage transmission, half the generated power would be wasted as heat in the cables.

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Practical Limits on Transmission Voltage

If higher voltage is better, why doesn't the UK transmit at 1 MV or 10 MV? Three practical constraints set the upper limit at 400 kV:

1. Insulation costs — air-gap and porcelain-insulator clearances scale with voltage. Doubling the voltage requires roughly doubling the inter-conductor spacing and the insulator length. This drives up the cost of the pylons, the substations, and the insulators themselves.

2. Corona discharge — at very high field strengths, the air around the conductors ionises and forms a faintly glowing plasma sheath. This loses energy continuously (an audible humming and crackling at substations is corona) and produces radio interference. Modern bundled conductors (two or four parallel cables for each phase) reduce the local field strength and mitigate corona, but it remains a practical upper limit.