Physics › Magnetic fields › Transformers

Transformers

Two coils share a core, and Faraday's law does the rest: the turns ratio sets the voltage ratio. Real transformers lose a little power as heat in the windings and the core, and the national grid depends on one consequence, that transmitting at high voltage means small current and small losses in the cables.

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

Builds on Electromagnetic induction: Faraday and Lenz and Alternating currents.

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.

Coupled by a shared flux

The transformer: change or nothing (animated figure)one shared flux, always changingprimary a.c.secondary emffive turnseight turns
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 d.c. gives no sustained secondary emf, only a blip 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 an emf in each turn. More turns collect more emf, and the voltages sit in the turns ratio.

NsNp=VsVp\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{p}}ON THE AQA DATA SHEET

More secondary turns step up the voltage, and fewer step it down. The mechanism also explains the one absolute restriction. A transformer needs changing flux, so it works on ac only. A steady dc primary current makes a steady flux, and a steady flux induces nothing whatsoever.

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.

Sixty against eleven hundred and fifty. The flux through each turn is identical, so the voltage divides itself among however many turns each side happens to have.

Where the power leaks

An ideal transformer passes power through untouched, and with the output power IsVs and input power IpVp, the report card is

efficiency=IsVsIpVp\mathrm{efficiency} = \frac{I_{s}V_{s}}{I_{p}V_{p}}ON THE AQA DATA SHEET
Why cores are laminated: a solid core lets large eddy currents swirl and heat it; thin insulated sheets cut the loops smallsolid core: large eddy currentslaminated: loops cut smallthe changing flux induces currents in the core itself
FIG. 2The changing flux induces eddy currents in the core itself; laminations cut the loops small and starve the loss.

Real transformers reach efficiencies in the high nineties of per cent, and the shortfall has nameable causes. The windings have resistance and warm up as current flows.

The changing flux induces eddy currents in the iron core itself, swirling charge that heats the metal; building the core from thin insulated laminations cuts those loops small and is the standard fix. A little flux leaks, missing the secondary, and a little energy is spent repeatedly re-magnetising the core each cycle. Whenever a question asks, name all four, since each carries its own mark. Stepping the voltage up steps the current down by at least the same factor, because the power out can never exceed the power in.

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 missing tenth warms the windings and the core. Chargers run warm for that reason alone, and the four causes of loss above are worth naming by heart.

The grid's one big idea

Why the grid runs at hundreds of kilovolts: for the same power, raising the voltage lowers the current, and cable loss falls as the current squaredstationstepuphigh V, small Istepdownhomesloss = I²R: current is the villain, so the grid starves it25× the volts, the same power, 625× less lost
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 they waste is P = I2R, so the current, squared, is what dominates the loss. For a fixed power delivered, P = IV, so raising the voltage lowers the current in proportion, and the squared dependence turns a modest voltage increase into a dramatic collapse in losses. Twenty-five times the voltage means six hundred and twenty-five times less power lost in the same cables. That is the logic of the national grid. Step up to hundreds of kilovolts at the power station, cross the country at small current, step back down near the user. Mains electricity is ac in the first place because transformers exist.

ASSESSMENT FOCUS

  • The operation answer is a Faraday chain. Alternating primary current, alternating core flux, changing flux linkage in the secondary, induced emf. Four links, in that order.
  • Transformers work on ac only, because induction needs changing flux. The dc case scores as a mark of its own, with steady flux and therefore no emf.
  • The four inefficiency causes make a list question. Winding resistance, eddy currents that laminations reduce, flux leakage, and the energy of repeatedly magnetising the core.
  • Grid questions want the argument in symbols. Fixed P = IV, so higher V means lower I, and the cable loss I2R falls as the square. Name that square explicitly.
  • In efficiency calculations keep primary and secondary quantities strictly apart; the subscripts are where the marks hide.
  • Both equations here are printed on the AQA data sheet, and neither comes with its caveat printed alongside. Supply it yourself. The turns ratio assumes every line of flux links both coils, and real cores leak a little, which is one reason a measured secondary voltage sits slightly low.

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

Current first from P = IV; then the loss is that current squared times R.

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, a mere 0.031%.

Sixteen times the voltage, 256 times less loss. That is the square at work, and the reason pylons carry hundreds of kilovolts.

Turns set the voltage ratio; power only ever passes through.

The grid starves I²R with high volts, small current and tiny loss.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 59 KB · print edition · white background

Question workbook

PDF · 59 KB · screen edition · navy

Answer book

PDF · 36 KB · fully worked · navy
18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the transformers questions page.

5 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.

  • 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.

Open the full revision checklist to track your progress across the whole unit.