Physics › Magnetic fields › Transformers

Transformers

Two coils on a shared iron core make a transformer: the alternating flux links both, and the turns ratio Ns/Np sets the voltage ratio Vs/Vp. It works only on ac, because induction needs a changing flux. Winding resistance, eddy currents, flux leakage and repeatedly magnetising the core lower the efficiency, and the grid transmits at high voltage because cable loss goes as I²R.

Transformers

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Year 13AQA 3.7.5.6OCR A 6.3.3(f)(i)EDEXCEL 125

Builds on Electromagnetic induction: Faraday and Lenz and Alternating currents and Series, parallel and power.

IN THIS TOPIC

  • Explain transformer operation through alternating flux and induced emf.
  • Use the turns-ratio equation, and say why a transformer needs ac.
  • Name the four causes of inefficiency, and use the efficiency equation.
  • Calculate transmission-line power losses, and explain the high-voltage grid.

COMMON MISCONCEPTION

A transformer can step up power.

A transformer cannot step up power: stepping the voltage up steps the current down in the same ratio, and a real transformer delivers slightly less power than it takes in.

Coupled by a shared flux

The transformer: change or nothing (animated figure)
Describe the drawing

Alternating current in the five-turn primary drives a magnetic flux round the iron core, pulsing and reversing with the supply. The eight-turn secondary links that same ever-changing flux, so an emf appears across it, a quarter cycle out of step because emf follows the rate of change, not the flux itself. Stop the change and the secondary dies: transformers run on a.c. or not at all. More secondary turns than primary makes this one step up the voltage.

FIG. 1Alternating current in the five-turn primary drives a magnetic flux round the iron core, pulsing and reversing with the supply. The eight-turn secondary links that same ever-changing flux, so an emf appears across it, a quarter cycle out of step because emf follows the rate of change, not the flux itself. Stop the change and the secondary emf falls to zero: a transformer needs changing flux, so steady dc gives no sustained secondary emf, only a brief one at switch-on and switch-off. More secondary turns than primary makes this one step up the voltage.

A transformer is two coils wound on one iron core. Alternating current in the primary drives an alternating flux round the core, and the core delivers that same changing flux through every turn of the secondary, where Faraday's law induces the same emf in each turn. The two voltages are therefore in the ratio of the numbers of turns.

NsNp=VsVp\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{p}}IN THE AQA DATA BOOKLET

More secondary turns step up the voltage, and fewer step it down. The mechanism also explains why a transformer works on ac only: it needs changing flux. A steady dc primary current makes a steady flux, and a steady flux induces no emf.

WORKED EXAMPLE

Designing a charger's transformer

A charger steps 230 V down to 12 V. Its primary has 1150 turns. Find the secondary turn count.

The turns ratio equals the voltage ratio, so Ns/Np = Vs/Vp = 12/230.

Ns = 1150 × 12/230 = 60 turns.

The flux through each turn is the same in both coils, so each coil's voltage is proportional to its number of turns.

Efficiency and the causes of loss

An ideal transformer transfers all of the input power to the output. With output power IsVs and input power IpVp, the efficiency is

efficiency=IsVsIpVp\mathrm{efficiency} = \frac{I_{s}V_{s}}{I_{p}V_{p}}IN THE AQA DATA BOOKLET
Two cores side by side: a solid block with one large amber eddy-current swirl, and a stack of thin insulated laminations where only small loops survive, wasting far less energy as heat.
FIG. 2The changing flux induces eddy currents in the core itself. Building the core from thin insulated laminations restricts those current loops and reduces the energy dissipated.

Real transformers reach efficiencies above 95%, and the shortfall has four causes. The windings have resistance and their temperature rises as current flows.

The changing flux induces eddy currents in the iron core, which raise its temperature; building the core from thin insulated laminations restricts those loops and is the standard remedy. A small amount of flux leaks and does not link the secondary, and some energy is used in re-magnetising the core each cycle. All four causes carry credit. Stepping the voltage up steps the current down by at least the same factor, because the output power cannot exceed the input power.

GUIDED PRACTICE

An efficiency from four meters

The charger above draws 0.52 A at 230 V and delivers 9.0 A at 12 V. Find its efficiency.

Show the working

Input is 230 × 0.52 = 120 W, and output is 12 × 9.0 = 108 W.

Efficiency = 108/120 = 0.90, or 90%. The remaining 12 W raises the temperature of the windings and the core, which is why a charger becomes warm in use.

Power loss in transmission lines

A power station feeds a step-up transformer, a long transmission line carries high voltage and small current, and a step-down transformer feeds the homes: since cable loss goes as current squared, twenty-five times the voltage means six hundred and twenty-five times less waste.
FIG. 3Step up, transmit at high voltage and small current, step down: cable loss goes as the current squared.

Transmission cables have resistance, and the power dissipated in them is P = I2R, so the loss depends on the square of the current. For a given transmitted power, P = IV, so increasing the voltage reduces the current in proportion, and the squared dependence means a moderate increase in voltage gives a large reduction in loss: twenty-five times the voltage gives 625 times less power dissipated in the same cables. That is the reason for the national grid design: step up to hundreds of kilovolts at the power station, transmit at small current, and step down near the consumer. Mains electricity is ac because transformers require a changing flux.

ASSESSMENT FOCUS

  • Give the operation as a chain: alternating primary current, alternating flux in the core, changing flux linkage in the secondary, induced emf. Four steps, in that order.
  • Transformers work on ac only, because induction requires changing flux. The dc case carries a mark of its own: steady current gives steady flux and therefore no induced emf.
  • The four causes of inefficiency form a list answer: winding resistance, eddy currents that laminations reduce, flux leakage, and the energy used in repeatedly magnetising the core.
  • Give the grid argument in symbols: for fixed P = IV, a higher V means a lower I, and the cable loss I2R falls as the square. State the square explicitly.
  • In efficiency calculations, keep the primary and secondary quantities separate and check every subscript.
  • AQA prints both equations on its data sheet without their conditions, so state them: the turns ratio assumes that all the flux links both coils, and real cores leak a little, which is one reason a measured secondary voltage is slightly below the calculated value.

CHECK YOURSELF

A station sends 10 MW down cables of total resistance 5.0 Ω. Find the power lost when transmitting at 25 kV, and at 400 kV, as a percentage of the power sent each time.

Show a hint

Find the current from P = IV first, then the loss from I²R.

Show the answer

At 25 kV the current is I = P/V = 107/(2.5 × 104) = 400 A, so the loss is I2R = 4002 × 5.0 = 8.0 × 105 W, or 0.80 MW, which is 8.0% of the power sent.

At 400 kV the current falls to 25 A, so the loss is 252 × 5.0 = 3.1 × 103 W, or 3.1 kW, which is 0.031% of the power sent.

Sixteen times the voltage gives 256 times less loss, because the loss depends on the square of the current. This is why transmission lines operate at hundreds of kilovolts.

The turns ratio sets the voltage ratio, and the output power cannot exceed the input power.

For a given transmitted power, increasing the voltage reduces the current. This reduces cable losses because P_loss = I²R.

WORKBOOK

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Question workbook

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Answer book

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  • Explain transformer operation through alternating flux and induced emf.
  • Use the turns-ratio equation, and say why a transformer needs ac.
  • Name the four causes of inefficiency, and use the efficiency equation.
  • Calculate transmission-line power losses, and explain the high-voltage grid.

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