Transformers and Power Transmission

Transformers are devices that change the voltage of an alternating current supply. They are among the most important practical applications of electromagnetic induction and are essential to the efficient transmission of electrical power across the National Grid.

How a Transformer Works

A transformer consists of two coils of wire (the primary and secondary) wound on a shared iron core.

An alternating current in the primary coil creates a changing magnetic field
The iron core channels this changing magnetic flux to the secondary coil
By Faraday’s law, the changing flux linkage through the secondary coil induces an alternating EMF

The transformer only works with alternating current because a steady (DC) current produces a constant magnetic field — there is no change in flux and therefore no induced EMF.

Why Use an Iron Core?

Iron has a very high permeability — it concentrates the magnetic field lines and ensures that almost all the flux from the primary coil passes through the secondary coil. Without the core, much of the flux would spread out into the surrounding air and be lost. A well-designed core ensures flux linkage between the coils is nearly 100%.

The Transformer Equation

For an ideal transformer:

Vₛ / Vₚ = Nₛ / Nₚ

where:

Vₛ = secondary voltage
Vₚ = primary voltage
Nₛ = number of turns on the secondary coil
Nₚ = number of turns on the primary coil

This comes directly from Faraday’s law: the same rate of change of flux passes through both coils, so the induced EMF per turn is the same in both. The total EMF is proportional to the number of turns.

Step-Up and Step-Down Transformers

Step-up transformer: Nₛ > Nₚ, so Vₛ > Vₚ (voltage is increased)
Step-down transformer: Nₛ < Nₚ, so Vₛ < Vₚ (voltage is decreased)

Worked Example 1

A transformer has 2000 turns on the primary coil and 100 turns on the secondary coil. The primary voltage is 230 V. What is the secondary voltage?

Vₛ = Vₚ × (Nₛ / Nₚ) = 230 × (100 / 2000) = 230 × 0.05 = 11.5 V

This is a step-down transformer (turns ratio 20:1), used for low-voltage applications like phone chargers.

Worked Example 2: Designing a Transformer

A transformer must step up from 230 V to 11.5 kV. If the primary coil has 200 turns, how many turns are needed on the secondary?

Nₛ = Nₚ × Vₛ / Vₚ = 200 × 11 500 / 230 = 200 × 50 = 10 000 turns

Power in an Ideal Transformer

In an ideal transformer, no energy is lost. The power input equals the power output:

VₚIp = VₛIₛ

This means that if the voltage is stepped up, the current must be stepped down by the same factor (and vice versa):

Iₛ / Iₚ = Nₚ / Nₛ = Vₚ / Vₛ

Worked Example 3

A step-up transformer increases the voltage from 230 V to 11 500 V. The current in the primary coil is 10 A. What is the current in the secondary coil (assuming 100% efficiency)?

VₚIₚ = VₛIₛ

Iₛ = VₚIₚ / Vₛ = 230 × 10 / 11 500 = 0.20 A

The voltage increased by a factor of 50, so the current decreased by a factor of 50.

Transformer Efficiency

Real transformers are not 100% efficient. The efficiency is:

Efficiency = (VₛIₛ / VₚIₚ) × 100%

or equivalently:

Efficiency = (power output / power input) × 100%

Good power transformers achieve efficiencies of 95–99%.

Sources of Energy Loss

Loss	Cause	Reduction Method	Typical Impact
Eddy currents in the core	Changing flux induces currents in the conducting iron core, dissipating energy as heat	Use a laminated core (thin iron sheets insulated from each other)	Largest loss if not controlled
Resistance of the coils (copper loss)	Current flowing through the wire produces I²R heating	Use thick, low-resistance copper wire	Increases with load current
Hysteresis	Energy is needed to repeatedly magnetise and demagnetise the core	Use soft iron (easily magnetised/demagnetised, narrow hysteresis loop)	Constant at given frequency
Flux leakage	Not all the flux from the primary reaches the secondary	Wind coils on top of each other; use a closed (toroidal) core	Small in well-designed transformers

flowchart TD
    A[Electrical Power Input] --> B[Primary Coil]
    B --> C{Iron Core}
    C --> D[Secondary Coil]
    D --> E[Useful Power Output]
    B --> F[Copper Loss: I squared R heating in coils]
    C --> G[Eddy Current Loss: currents in core produce heat]
    C --> H[Hysteresis Loss: energy to magnetise and demagnetise core]
    C --> I[Flux Leakage: not all flux reaches secondary]

Worked Example 4: Efficiency Calculation

A transformer has a primary voltage of 230 V and a primary current of 2.0 A. The secondary delivers 12 V at 35 A. Calculate the efficiency.

Input power: P_in = 230 × 2.0 = 460 W

Output power: P_out = 12 × 35 = 420 W

Efficiency = (420 / 460) × 100% = 91.3%

Power lost = 460 − 420 = 40 W (dissipated as heat in the core and coils)

Power Transmission and the National Grid

Electrical power from power stations must be transmitted over long distances to consumers. The cables have resistance, so some energy is lost as heat. The key relationship is:

P_loss = I²R

where I is the current in the cables and R is their resistance.

Why High Voltage?

If a power station generates power P at voltage V:

P = IV, so I = P / V

Substituting into the power loss equation:

P_loss = (P / V)² × R = P²R / V²

This shows that the power lost is inversely proportional to the square of the transmission voltage. Doubling the voltage reduces the power loss to one quarter.

The National Grid Voltage Stages

Stage	Typical Voltage	Purpose
Power station output	25 kV	Generated at moderate voltage
Step-up for transmission	400 kV (or 275 kV)	Long-distance overhead lines
Step-down for distribution	33 kV	Regional distribution
Step-down for local supply	11 kV	Local substations
Step-down for consumers	230 V	Domestic supply

Worked Example 5

A power station transmits 500 MW of power. The transmission cables have a total resistance of 10 Ω. Compare the power loss at 25 kV and 400 kV.

At 25 kV: I = P / V = 500 × 10⁶ / 25 000 = 20 000 A P_loss = I²R = (20 000)² × 10 = 4.0 × 10⁹ W = 4000 MW

This is eight times the power being transmitted — clearly impossible. The cables would melt.

At 400 kV: I = P / V = 500 × 10⁶ / 400 000 = 1250 A P_loss = I²R = (1250)² × 10 = 1.56 × 10⁷ W = 15.6 MW

This is only 3.1% of the power transmitted — acceptable.

The voltage was increased by a factor of 16 (400/25), so the power loss decreased by a factor of 16² = 256.

Worked Example 6: Minimum Transmission Voltage

A power company must transmit 50 MW of power over cables with resistance 8.0 Ω. The maximum acceptable power loss is 2% of the transmitted power. What is the minimum transmission voltage?

Maximum power loss = 0.02 × 50 × 10⁶ = 1.0 × 10⁶ W

P_loss = P²R / V²

V² = P²R / P_loss = (50 × 10⁶)² × 8.0 / (1.0 × 10⁶) = 2.5 × 10¹⁵ × 8.0 / 10⁶ = 2.0 × 10¹⁰

V = √(2.0 × 10¹⁰) = 1.41 × 10⁵ V ≈ 141 kV

Summary of the National Grid System

Power station → Step-up transformer → High-voltage transmission lines → Step-down transformer → Consumers

The transformers are the crucial components. Without them, either the transmission losses would be enormous (at low voltage) or the voltage would be dangerously high at the consumer end (at high voltage). Transformers allow us to have the best of both worlds.

Real-World Applications

Phone chargers: A small step-down transformer (or increasingly, a switched-mode power supply) reduces the 230 V mains to about 5 V for charging. Switched-mode supplies operate at high frequency (50–500 kHz), allowing much smaller and lighter transformers.

Welding transformers: A step-down transformer reduces the voltage but massively increases the current (to hundreds of amps). The high current melts the metal at the welding point. The transformer is designed with very few secondary turns and thick copper wire to minimise resistance.

High-voltage DC (HVDC) transmission: For very long distances (hundreds of km), DC transmission is more efficient than AC because it avoids inductive and capacitive losses in the cables. AC is converted to DC for transmission, then back to AC at the receiving end. The conversion uses power electronics rather than transformers.

Common Exam Mistakes

Trying to use a transformer with DC: Transformers ONLY work with AC. A constant current produces a constant flux, giving zero induced EMF.
Confusing primary and secondary: The primary is connected to the input (power source). The secondary is connected to the output (load).
Forgetting that P_loss = I²R, not P = IV for the cables: The power lost in the cables depends on the current through them and their resistance, not on the voltage across the load.
Not squaring V in the transmission loss formula: P_loss = P²R/V². Doubling V reduces loss by a factor of 4, not 2.

Key Points

Transformer equation: Vₛ/Vₚ = Nₛ/Nₚ
Ideal transformer: VₚIₚ = VₛIₛ (power in = power out)
Energy losses: eddy currents (laminated core), resistance (thick wire), hysteresis (soft iron), flux leakage
Power loss in cables: P_loss = I²R = P²R/V²
High-voltage transmission reduces current, dramatically reducing I²R losses
Doubling the transmission voltage reduces power loss to one quarter

A-Level Deep Dive: Transformers and Power Transmission

Spec mapping

Edexcel 9PH0 Topic 7 — Electric and Magnetic Fields includes the operation of the ideal transformer, the use of Vₛ/Vₚ = Nₛ/Nₚ and IₚVₚ = IₛVₛ for an ideal transformer, the reasons for energy losses in real transformers (eddy currents, resistive heating, hysteresis, flux leakage), and the use of step-up transformers in the national grid to minimise I²R transmission losses (refer to the official specification document for exact wording). Although the transformer sub-strand sits inside the broader fields topic, it is genuinely synoptic: it depends on EM induction (Topic 7 earlier sub-strand), draws on AC theory introduced in Topic 4 (Electric Circuits), and feeds forward into Topic 11 (Nuclear and Particle Physics) only loosely — its main exam home is Paper 2 (Section A: Topics 6–7) with possible carry-over into Paper 3 synoptic items. The Edexcel formula booklet does list the transformer equation but does not list P = I²R rearranged into the transmission-loss form P_loss = P²R/V²; that algebraic step must be reproduced by the candidate.

Worked example with full mark scheme

Question (8 marks):

A power station generates 2.0 MW of electrical power at 5.0 kV. This is transmitted through a cable of total resistance 4.0 Ω to a substation 30 km away. An ideal step-up transformer with primary coil of 200 turns is used at the power station; the secondary coil has 4000 turns.

(a) Calculate the secondary voltage and the current in the transmission cable. (3)

(b) Calculate the power dissipated in the cable as heat, and the percentage of generated power lost in transmission. (3)

(c) Without using a step-up transformer (i.e. transmitting at 5.0 kV directly), calculate the power dissipated in the same cable. Comment on the result. (2)

Solution with mark scheme:

(a) Step 1 — apply the ideal transformer equation.

\begin{aligned} V_{s} / V_{p} = N_{s} / N_{p} \\ V_{s} = V_{p} \times (N_{s} / N_{p}) = 5000 \times (4000 / 200) = 5000 \times 20 = 100 000 V = 100 kV \end{aligned}

M1 — correct substitution into the turns-ratio equation. Common error: inverting the ratio and writing Vₚ/Vₛ = Nₛ/Nₚ, which gives 250 V (a step-down result) and contradicts the description "step-up". The unit of the answer must be stated.

A1 — correct Vₛ = 100 kV (or 1.0 × 10⁵ V).

Step 2 — find the current in the transmission cable using power conservation.

For an ideal transformer, PₚIₚ = PₛIₛ so power transmitted at the secondary is the same 2.0 MW. The cable carries the secondary current:

I_{s} = P / V_{s} = 2.0 \times 10^{6} / 1.0 \times 10^{5} = 20 A

B1 — correct cable current Iₛ = 20 A. Examiners reward students who explicitly state "ideal transformer, so power in = power out" before computing — this earns the implicit method mark even if the arithmetic later slips.

(b) Step 1 — power dissipated in the cable.

P_{\text{loss}} = I^{2}R = (20)^{2} \times 4.0 = 400 \times 4.0 = 1600 W = 1.6 kW

M1 — correct use of P = I²R with the cable current (not the generator current).

A1 — correct P_loss = 1.6 kW.

Step 2 — percentage loss.

$\%\text{ loss} = \dfrac{1600}{2\,000\,000} \times 100 = 0.080\,\%$

B1 — correct percentage to 2 s.f.

\begin{aligned} I = P / V = 2.0 \times 10^{6} / 5.0 \times 10^{3} = 400 A \\ P_{\text{loss}} = (400)^{2} \times 4.0 = 160 000 \times 4.0 = 640 000 W = 640 kW \end{aligned}

M1 — correct re-computation of current at the lower voltage and substitution into I²R.

A1 — correct P_loss = 640 kW, with comment that this is 32% of generated power (i.e. 400× the loss at 100 kV) — confirming why high-voltage transmission is essential. The factor of 400 arises because current scales as 1/V and dissipation scales as I², so P_loss ∝ 1/V².

Total: 8 marks (M3 A3 B2).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A national grid operator transmits 500 MW over a 200 km line of total resistance 8.0 Ω.

(a) Show that, when transmitted at 400 kV, the percentage power loss is approximately 2.5%. (3)

(b) The operator is considering upgrading the line to 800 kV (HVDC alternative aside). Determine the new percentage power loss and comment on the practical implication. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2) — current calculation: I = P/V = 5.0 × 10⁸ / 4.0 × 10⁵ = 1250 A.
M1 (AO2) — power-loss calculation: P_loss = I²R = (1250)² × 8.0 = 1.25 × 10⁷ W = 12.5 MW.
A1 (AO1) — percentage loss = 12.5 / 500 × 100 = 2.5%, matching the printed value.

(b)

M1 (AO2) — recognising the 1/V² scaling: doubling V quarters the loss, so percentage loss ≈ 0.625%. Equivalent full computation also accepted: I = 625 A, P_loss = (625)² × 8.0 = 3.125 MW, percentage = 0.625%.
A1 (AO1) — correct percentage to 2–3 s.f.
B1 (AO3) — practical comment: doubling the voltage quarters I²R losses but requires more expensive insulation, taller pylons, larger right-of-way, and worsens corona-discharge losses — so the gain in efficiency must be weighed against capital and environmental cost.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. This is a typical Paper 2 transmission item: AO2 dominates because the question demands chained substitution; AO3 enters only in the evaluative comment.

Synoptic links

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