The transformer — stepping voltage up or down — Physics

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Three voltages, one wire

A large power station's generator produces electricity at about 25 000 V. The pylons that carry it across the country run at 275 000 V or 400 000 V. The socket in your wall gives 230 V in many countries. The same electricity passes through all three.

Something must change the voltage, up and then down again, and do it with almost no waste. The generator lecture ended by saying that this device needs a current that keeps changing, which is one reason the world's power grids use alternating current (a.c.). This lecture explains that device, the transformer, and why the voltage on the pylons is so high.

The idea behind it is not new to you. You met it in the Faraday's law lecture, on 29 August 1831, when Faraday wound two coils on one iron ring.

Two coils, one core

A transformer is two coils of wire wound on the same iron core. The primary coil is connected to an a.c. supply. The secondary coil is connected to whatever needs the electricity, called the load. The two coils are not joined by any wire.

A transformer: two coils on one iron core. The alternating current in the primary makes a changing flux, which the core carries round through the secondary. The coils are not joined by any wire. Here the secondary has more turns, so this is a step-up transformer.

Follow the energy step by step. The alternating p.d. across the primary drives an alternating current through it. That current makes a magnetic field in the iron, which grows, shrinks and reverses with the current. Iron carries magnetic flux far better than air, so almost all the flux goes round the core and through the secondary coil. The flux through the secondary keeps changing, so by Faraday's law an e.m.f. is induced in it. A changing current in one coil induces an e.m.f. in another: this is called mutual induction.

Now the generator lecture's promise. What if the primary were connected to a battery instead? A steady current makes a steady flux. A steady flux is not changing, so no e.m.f. is induced in the secondary. Before you read on, predict what a voltmeter on the secondary would show.

Predict — a transformer on a battery

A transformer's primary is connected to a 12 V battery through a switch. A voltmeter is connected across the secondary. The switch is closed, left closed for a minute, then opened. What does the voltmeter show?

This box explains what a voltmeter on the secondary shows when the primary is connected to a battery.

The voltmeter flicks briefly when the switch is closed, reads zero while the current is steady, and flicks the other way when the switch is opened. An e.m.f. is induced only while the flux is changing: as the current starts and as it stops. This is exactly what Faraday saw with his ring in 1831.

So a transformer on d.c. gives only a brief flick at switching on and off. Faraday's ring of 1831 did exactly this, and it was, in effect, the first transformer. To give a steady output, the flux must change all the time. That is what a.c. does: it grows, shrinks and reverses many times a second.

Why the voltage depends on the turns

This is the hard part of the lecture, so take it slowly. Suppose every bit of flux in the core passes through every turn of both coils. Then, at any instant, the flux through one turn of the primary changes at exactly the same rate as the flux through one turn of the secondary. By Faraday's law, each turn has the same e.m.f. as every other turn, on either coil.

The turns of a coil are joined one after another, in series, so their e.m.f.s add. A coil with twice as many turns has twice the e.m.f. If the e.m.f. of one turn is $e$ , then the secondary's e.m.f. is $N_{s} e$ . In an ideal transformer, the coils have no resistance, so the p.d. across each coil equals its e.m.f. The primary's p.d. is then $N_{p} e$ . Divide one by the other, and $e$ cancels:

\frac{V _{s}}{V _{p}} = \frac{N _{s} e}{N _{p} e} = \frac{N _{s}}{N _{p}}

The transformer equation

For an ideal transformer, the ratio of the p.d.s equals the ratio of the numbers of turns:

$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$

A step-up transformer has more turns on the secondary, so $V_{s} > V_{p}$ . A step-down transformer has fewer, so $V_{s} < V_{p}$ .

That is it. That is the whole idea. Every turn gets the same share of e.m.f., so a coil's p.d. is set by how many turns it has.

One note on the numbers. The p.d. of an a.c. supply rises and falls, so "230 V" is a kind of average, chosen so that it heats a resistor as much as 230 V of d.c. would. It is called the r.m.s. value, and the Pre-University lectures explain it. Here, all p.d.s and currents are these values, and the equations work for them.

More voltage means less current

Can a step-up transformer give you more energy than you put in? No. A transformer has no source of energy of its own. In an ideal transformer, the power out equals the power in:

V_{p} I_{p} = V_{s} I_{s}

So if the p.d. goes up by a factor of 10, the current must go down by a factor of 10. Combining this with the transformer equation gives $\frac{I _{s}}{I _{p}} = \frac{N _{p}}{N _{s}}$ : the current ratio is the turns ratio upside down.

Lenz's law shows how the primary "knows" what the secondary needs. When the load draws a current from the secondary, that current makes its own flux, which opposes the change in the core's flux. To keep the core's flux changing as the supply demands, the primary must draw more current. A transformer with nothing connected draws almost nothing. Connect a load, and the primary's current rises to supply it, as the generator lecture found for a generator under load.

Stop and think

A student says: "A step-up transformer turns 230 V into 23 000 V, so it multiplies the power by 100." What is wrong?

Show the answer

The p.d. goes up by 100, but the current goes down by 100, so the power stays the same (in an ideal transformer) or falls slightly (in a real one). A transformer changes how the power is carried, high p.d. with low current or low p.d. with high current, not how much power there is. Energy is conserved.

Real transformers

Real transformers are not perfect, but they come close. Three things waste energy. The coils have resistance, so the currents heat them. Some flux escapes the core. And the changing flux induces eddy currents in the iron core itself, the same eddy currents you met in the Lenz's law lecture, which heat the core. To cut eddy currents, the core is built from thin iron sheets, called laminations, each insulated from the next. The insulation blocks the loops in which eddy currents would flow. With these steps, large transformers waste only a very small fraction of the power that passes through them.

Why the pylons carry such high voltages

Transmission lines are long, so they have resistance. The current in them heats them, and the power wasted is

P_{lost} = I^{2} R

where $I$ is the current in the line and $R$ is the line's resistance. For a given power $P$ sent at p.d. $V$ , the current is $I = P / V$ . Double the p.d., and the current halves. Because the loss depends on $I^{2}$ , halving the current cuts the loss to a quarter. This is why the voltage is stepped up before transmission and stepped down again near the users.

Try it on a model line below. It sends 500 MW from a power station through a line of total resistance 5.0 Ω. Choose the voltage at which the power is sent.

Real-world model — choose the transmission voltage

Choose a transmission voltage. Watch the current in the line, the power wasted as heat, and what reaches the far end.

This table shows a model line that sends 500 MW through a total resistance of 5.0 Ω at different voltages. The model leaves out the line's other effects and assumes the power and voltage are measured where the line starts.

Voltage	Current (A)	Power wasted (MW)	Share wasted
25 kV	20 000	impossible: the line would need 100 kV across it
66 kV	≈ 7576	≈ 287	≈ 57%
132 kV	≈ 3788	≈ 71.7	≈ 14%
275 kV	≈ 1818	≈ 16.5	≈ 3.3%
400 kV	1250	≈ 7.8	≈ 1.6%

Raising the voltage from 132 kV to 400 kV, about three times, cuts the wasted power by about nine times.

You should have seen two things. Each step up in voltage cuts the waste sharply, because the loss depends on the square of the current. And at 25 kV the model breaks completely. A current of 20 000 A through 5.0 Ω needs a p.d. of 100 kV across the line itself, more than the 25 kV supplied. The power simply cannot be sent that way. This is why power stations step their output up to hundreds of kilovolts at once. Real networks show the result: in 2018–19, transmission losses on the England and Wales network, which runs at 275 kV and 400 kV, were about 1.8% of the electricity carried (source: National Grid Electricity Transmission, transmission losses report 2018/19).

Worked examples

Worked Example 1 — A step-down transformer

A transformer has 500 turns on the primary and 50 on the secondary. The primary is connected to a 230 V a.c. supply. Find the secondary p.d. If the secondary supplies 4.0 A, find the primary current. Assume the transformer is ideal.

Think first. Fewer turns on the secondary, so this steps down. The current must step up by the same factor.

$V_{s} = V_{p} \times \frac{N _{s}}{N _{p}} = 230 \times \frac{50}{500} = 23$ V.

Power in = power out: $230 \times I_{p} = 23 \times 4.0 = 92$ W, so $I_{p} = 92 \div 230 = 0.40$ A.

Check. The turns ratio is 10, and $4.0 \div 0.40 = 10$ . ✓ The smaller current is on the higher-voltage side. ✓

Worked Example 2 — From the generator to the pylons

A power station's generator gives 25 kV. A transformer steps it up to 400 kV. The primary has 1000 turns. How many turns does the secondary need? If 500 MW passes through, what are the two currents?

Think first. The p.d. must rise by a factor of $400 \div 25 = 16$ , so the turns must too.

$N_{s} = 1000 \times 16 = 16000$ turns. On the generator side, $I_{p} = 500 \times 1 0^{6} \div 25000 = 20000$ A. On the line side, $I_{s} = 500 \times 1 0^{6} \div 400000 = 1250$ A.

Check. $20000 \div 1250 = 16$ , the turns ratio. ✓ The huge 20 000 A flows only through the short, very thick connections inside the power station, never through a long line.

Worked Example 3 — Comparing two line voltages

A line of total resistance 8.0 Ω carries 200 MW. Compare the power wasted if it is sent at 132 kV and at 400 kV.

Think first. Find the current first, from $I = P / V$ . Then use $I^{2} R$ .

At 132 kV: $I = 200 \times 1 0^{6} \div 132000 \approx 1515$ A, so $P_{lost} \approx 151 5^{2} \times 8.0 \approx 18.4$ MW, about 9.2%. At 400 kV: $I = 500$ A, so $P_{lost} = 50 0^{2} \times 8.0 = 2.0$ MW, which is 1.0%.

Check. The voltage rose by $400 \div 132 \approx 3.03$ times, so the loss should fall by about $3.0 3^{2} \approx 9.2$ times: $18.4 \div 2.0 = 9.2$ . ✓

Three traps

Trap 1 — A transformer on d.c.

A steady current gives a steady flux, and a steady flux induces nothing. A transformer connected to a battery gives no output, apart from a brief flick at switching on and off. Transformers need a.c.

Trap 2 — Flipping the wrong ratio

The p.d.s follow the turns: $V_{s} / V_{p} = N_{s} / N_{p}$ . The currents go the other way: $I_{s} / I_{p} = N_{p} / N_{s}$ . Check with the rule "the high-voltage side carries the small current".

Trap 3 — Using the line voltage in V²/R

The power wasted in a line is $I^{2} R$ , with $I$ the current in the line. Writing $V^{2} / R$ with the transmission voltage gives $40000 0^{2} \div 5.0 = 32000$ MW, which is nonsense. In $V^{2} / R$ , the $V$ must be the p.d. across the resistance itself, which is the small drop along the line, not the voltage it is sent at.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A transformer has 920 turns on its primary and 40 on its secondary, and is connected to 230 V a.c. (a) Find the secondary p.d. (b) It is used in a lamp that needs 24 W. Find the secondary and primary currents, assuming it is ideal.

Hint 1
Use the turns ratio for the p.d.

Hint 2
Current = power ÷ p.d., on each side.

Full answer
(a) $V_{s} = 230 \times 40 \div 920 = 10$ V. (b) $I_{s} = 24 \div 10 = 2.4$ A, and $I_{p} = 24 \div 230 \approx 0.104$ A. Check: $2.4 \div 0.104 \approx 23$ , and $920 \div 40 = 23$ . ✓
A transformer must give 6.0 V from a 230 V supply. Its primary has 1150 turns. (a) How many turns are needed on the secondary? (b) A student connects it to a 230 V d.c. supply by mistake. What will the secondary give?

Hint 1
$N_{s} = N_{p} \times V_{s} \div V_{p}$ .

Hint 2
(b) Is the flux changing?

Full answer
(a) $N_{s} = 1150 \times 6.0 \div 230 = 30$ turns. (b) Nothing steady: a brief flick at switch-on, then zero, because the flux stops changing. The primary, which has little resistance, may also draw a large current and overheat, because there is no changing flux to limit it. Check: $230 \div 1150 = 0.2$ V per turn, and $30 \times 0.2 = 6.0$ V. ✓
A transformer's secondary supplies 12 V at 4.0 A. The transformer is 95% efficient and its primary is connected to 230 V. Find (a) the output power, (b) the input power, (c) the primary current, and (d) the power wasted.

Hint 1
Efficiency = useful power out ÷ power in.

Hint 2
Power in = power out ÷ 0.95.

Full answer
(a) $12 \times 4.0 = 48$ W. (b) $48 \div 0.95 \approx 50.5$ W. (c) $50.5 \div 230 \approx 0.22$ A. (d) $50.5 - 48 \approx 2.5$ W, mostly as heat in the coils and the core. Check: the primary current is a little more than the ideal $48 \div 230 \approx 0.21$ A, because it also supplies the waste. ✓
A stretch. Using the model line of this lecture (500 MW, 5.0 Ω), find (a) the current and the power wasted at 275 kV, and (b) the voltage drop along the line, $I R$ . (c) Explain why the 25 kV case in the model is impossible, and (d) suggest one other way, besides raising the voltage, to reduce the waste.

Hint 1
$I = P / V$ , then $I^{2} R$ and $I R$ .

Hint 2
(d) Look at $I^{2} R$ . Which other quantity could be made smaller, and how?

Full answer
(a) $I = 500 \times 1 0^{6} \div 275000 \approx 1818$ A, and $I^{2} R \approx 181 8^{2} \times 5.0 \approx 16.5$ MW, about 3.3%. (b) $I R \approx 1818 \times 5.0 \approx 9.1$ kV, a small part of 275 kV. (c) At 25 kV the current would be 20 000 A, and the drop along the line would be $20000 \times 5.0 = 100$ kV. A line cannot lose more p.d. than it is given, so this current, and this power, are impossible. (d) Reduce the resistance: use thicker wires, or several wires side by side, or a metal with lower resistivity. Check: $16.5 \div 500 = 0.033$ , and $9.1 \div 275 \approx 0.033$ : the share of power wasted equals the share of p.d. dropped. ✓

The bigger idea

The transformer decided what kind of electricity the world uses. In the early 1880s, Lucien Gaulard and John Dixon Gibbs built early transformers in Europe. In 1884–85, Ottó Bláthy, Miksa Déri and Károly Zipernowsky in Budapest made the closed iron core that modern transformers use. In 1886, William Stanley used step-up and step-down transformers to light the town of Great Barrington, in the United States (source: Edison Tech Center, "The History of the Transformer"). Systems built on direct current had no easy way to change their voltage, so they could not send power far. Alternating current, with transformers, could.

Today some very long links do use direct current, at very high voltages, with electronic converters at each end. But almost every home is still fed through a chain of transformers, from the generator's 25 kV, up to the pylons and back down to the socket. With this lecture, the induction unit is complete: a changing flux makes an e.m.f. (Faraday), in the direction that opposes the change (Lenz), in a turning coil (the generator) and between two coils on one core (the transformer).

Key concepts in this lesson

A transformer is two coils on one iron core; an alternating current in the primary makes a changing flux that induces an e.m.f. in the secondary (mutual induction).
A transformer needs a.c., because a steady current makes a steady flux, which induces nothing.
For an ideal transformer, $V_{s} / V_{p} = N_{s} / N_{p}$ , because every turn has the same e.m.f.
An ideal transformer conserves power, $V_{p} I_{p} = V_{s} I_{s}$ , so stepping the p.d. up steps the current down by the same factor.
Laminated cores reduce eddy currents, and real transformers waste only a small fraction of the power.
Power is sent at high voltage because the waste in a line is $I^{2} R$ , and a higher voltage needs a smaller current.