Physics · 4 Electricity and magnetism

4.5.6 The transformer

Before you start

The difference between d.c. and a.c.: alternating current keeps reversing direction (4.2.2).
$P = I V$ and $V = I R$ , and their rearrangements (4.2.4, 4.2.5).
A current in a coil produces a magnetic field, and reversing the current reverses the field (4.5.3).
A changing magnetic field linking a coil induces an e.m.f. in it (4.5.1).
Soft iron is easily magnetised and demagnetised; steel keeps its magnetism (4.1).

Syllabus objectives covered (8)

Describe the construction of a simple transformer with a soft-iron core, as used for voltage transformations
Use the terms primary, secondary, step-up and step-down
Recall and use the equation Vp / Vs = Np / Ns where p and s refer to primary and secondary
Describe the use of transformers in high-voltage transmission of electricity
State the advantages of high-voltage transmission
Explain the principle of operation of a simple iron-cored transformer
Recall and use the equation for 100% efficiency in a transformer IpVp = IsVs where p and s refer to primary and secondary
Recall and use the equation P = I²R to explain why power losses in cables are smaller when the voltage is greater

Part 1 of 7

What a transformer is made of

A transformer changes the size of an alternating voltage. It can take the $230$ V mains and give $12$ V for a lamp, or take a low voltage and give a much higher one. It has no moving parts.
It is built from three parts. The first is a core of soft iron. In a simple transformer the core is usually a closed loop, often drawn as a square ring.
The second and third parts are two coils of insulated wire, each wound around the core. In the simple transformer they sit on opposite sides of the ring. One coil is connected to the a.c. supply. The other is connected to whatever needs the new voltage.
Notice what is not there: no wire joins the two coils. They are electrically separate from each other and from the core. Whatever reaches the second coil gets there through the iron, as a magnetic field. The section on how a transformer works shows how.
The wire is insulated so that the current goes round every turn of the coil. If bare turns touched each other or the core, the current would take a short cut and those turns would not count.
The core is soft iron because soft iron is easily magnetised and easily demagnetised. The core has to be magnetised one way, then the other, every time the current reverses. Steel keeps its magnetism, which is why it is used for permanent magnets, so it is the wrong material here.
apparatus diagram — not drawn yet
A closed square ring of soft iron; on the left side a coil of a few turns of wire connected to an a.c. supply symbol, labelled 'primary coil, Np turns' and 'input Vp'; on the right side a coil of more turns connected to a lamp, labelled 'secondary coil, Ns turns' and 'output Vs'; the ring labelled 'soft-iron core'; no wire joining the two coils.
The two coils share the iron core, but no wire connects one coil to the other.
So a description of the construction needs three things: a soft-iron core, a primary coil and a secondary coil, both of insulated wire and both wound on the core.

Part 2 of 7

Primary, secondary, step-up and step-down

The coil connected to the supply is the primary coil. It is the input. Its voltage is $V_{p}$ and its number of turns is $N_{p}$ .
The coil connected to the load (the lamp, motor or cable that uses the output) is the secondary coil. It is the output. Its voltage is $V_{s}$ and its number of turns is $N_{s}$ .
A step-up transformer makes the output voltage bigger than the input voltage: $V_{s} > V_{p}$ . It does this by having more turns on the secondary than on the primary: $N_{s} > N_{p}$ .
A step-down transformer makes the output voltage smaller: $V_{s} < V_{p}$ . It has fewer turns on the secondary than on the primary: $N_{s} < N_{p}$ .
To decide which kind you have, compare the turns or compare the voltages. Do not go by which coil is drawn on the left. The primary is the coil connected to the supply, wherever it is drawn.

Part 3 of 7

The turns-ratio equation

The two voltages are in the same ratio as the two numbers of turns. Paper 4 does not give you this equation. You have to recall it.
$\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}$
Read it like this: on both sides, primary is on top and secondary is underneath. Each voltage goes with the turns of its own coil.
A quick example. A transformer has $100$ turns on its primary and $500$ on its secondary. The secondary has $5$ times as many turns, so it gives $5$ times the voltage: $12$ V in gives $60$ V out.
To find the secondary voltage, flip both fractions so that $V_{s}$ is on top. That is allowed, because when two fractions are equal, their upside-down versions are equal too. Then multiply both sides by $V_{p}$ .
$V_{s} = V_{p} \times \frac{N_{s}}{N_{p}}$
In the same way, the number of turns on the secondary is:
$N_{s} = N_{p} \times \frac{V_{s}}{V_{p}}$
Always check the answer against step-up or step-down. If the secondary has fewer turns, your output voltage must come out smaller than the input. If it does not, you have turned the ratio upside down.
A number of turns has no unit. The two voltages can be in V or in kV, but they must both be in the same unit, because the equation compares one with the other.

Part 4 of 7

How a transformer works

The explanation is a chain of four links. Each link comes from something you already know about magnetism and induction.
1. Alternating current in the primary. The a.c. supply drives a current in the primary coil. This current keeps changing in size and keeps reversing direction.
2. A changing magnetic field in the core. A current in a coil makes a magnetic field. Because the current keeps changing, the field changes too: it grows, falls and reverses in step with the current. The soft-iron core is magnetised first one way, then the other.
3. The core carries the field through the secondary. The core is a closed loop of iron, so the changing magnetic field made by the primary also passes through the secondary coil. The core links the two coils magnetically.
4. An e.m.f. is induced in the secondary. A changing magnetic field linking a coil induces an e.m.f. in that coil. So an alternating e.m.f. appears across the secondary, at the same frequency as the supply. If the secondary is part of a complete circuit, an alternating current flows in it.
Every turn of the secondary has the same changing field through it, so every turn adds the same e.m.f. More turns on the secondary means a bigger total e.m.f. That is where the turns ratio comes from.
Why d.c. does not work. A steady direct current makes a steady magnetic field. A steady field is not changing, so no e.m.f. is induced in the secondary. There is only a brief pulse at the moment the current is switched on or off, because only then does the field change.
The word that matters is changing. 'The primary makes a magnetic field which induces a voltage in the secondary' misses it, because a magnetic field that does not change induces nothing.

Part 5 of 7

Power in, power out

A transformer cannot create energy. If a transformer is 100% efficient, all the power that goes into the primary comes out of the secondary.
Electrical power is $P = I V$ , so the power in is $I_{p} V_{p}$ and the power out is $I_{s} V_{s}$ . Make them equal and you have the equation you must recall:
$I_{p} V_{p} = I_{s} V_{s}$
Here is what it means. Whatever the transformer does to the voltage, it does the opposite to the current. Step the voltage up $5$ times and the current comes out $5$ times smaller.
Check it with the transformer from the turns-ratio example. With $12$ V and $2.0$ A going in, the input power is $24$ W. The output is $60$ V, so the output current is $24 \div 60 = 0.40$ A. And $60 \times 0.40$ is $24$ W again.
So a step-up transformer does not give you more power. It gives a bigger voltage and a smaller current.
Real transformers are not quite 100% efficient: the coils and the core warm up, so some energy is wasted as heat. Use $I_{p} V_{p} = I_{s} V_{s}$ when a question says the transformer is 100% efficient.

Part 6 of 7

Transformers in electricity transmission

Electricity travels from power stations to homes along transmission cables, which can be hundreds of kilometres long.
At the power station, a step-up transformer raises the voltage to a very high value before the electricity goes into the transmission cables. These voltages can be hundreds of thousands of volts.
Near the places where the electricity is used, step-down transformers lower the voltage again, in one or more stages, down to the mains voltage supplied to homes (for example $230$ V).
apparatus diagram — not drawn yet
A left-to-right block diagram: a box labelled 'power station (generator)', an arrow to a box labelled 'step-up transformer', then a long pair of cables on pylons labelled 'high-voltage transmission cables (small current)', then a box labelled 'step-down transformer', then an arrow to a house labelled 'homes (mains voltage)'.
Voltage goes up for the long journey and comes down again before it reaches the homes.
Transformers only work with a.c., and that is one reason the mains supply is a.c.
Why go to the trouble? For a given power, a higher voltage means a smaller current in the cables, because $P = I V$ .
Advantage 1. A smaller current heats the cables less. So less energy is wasted as heat in the cables, and more of the power reaches the consumers.
Advantage 2. A smaller current can be carried by thinner cables, and thinner cables are cheaper and lighter.
In an answer, give the cause and the effect together: 'high voltage means a small current, so less power is wasted heating the cables'. 'High voltage is more efficient' alone does not say why.

Part 7 of 7

Why a higher voltage wastes less power: P = I²R

The Extended objective wants the reason with an equation. Start from two equations you already know: $P = I V$ and $V = I R$ .
Put $V = I R$ into $P = I V$ . The power turned into heat in a resistance $R$ that carries a current $I$ is:
$P = I \times I R = I^{2} R$
Apply this to the cables. Their resistance $R$ is fixed by their material, length and thickness. So the power wasted in them depends on the square of the current.
The power station must deliver a certain power $P$ . From $P = I V$ , the current needed is $I = P / V$ . Raise the voltage and the same power needs a smaller current.
Now put the two ideas together. Make the voltage $10$ times bigger: the current becomes $10$ times smaller, and the wasted power $I^{2} R$ becomes $10^{2} = 100$ times smaller.
The trap here is to use $P = V^{2} / R$ with the transmission voltage. That would make a bigger voltage waste more power, which is wrong. The p.d. across the cables themselves is only the small drop along their length, not the transmission voltage. Work with the current in the cables and use $I^{2} R$ .

Worked examples

Example 1. A doorbell needs $6.0$ V. It runs from the $230$ V a.c. mains through a transformer whose primary coil has $920$ turns. Calculate the number of turns on the secondary coil, and state whether the transformer is step-up or step-down.

Show the working

Write down the equation you have recalled.
$\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}$
The unknown is $N_{s}$ . Flip both fractions to put it on top, then multiply both sides by $N_{p}$ .
$N_{s} = N_{p} \times \frac{V_{s}}{V_{p}}$
Substitute the values.
$N_{s} = 920 \times \frac{6.0}{230} = 24$
Check: the output voltage is smaller than the input, and $24$ turns is fewer than $920$ . Fewer turns on the secondary means step-down. The two facts agree, so the ratio is the right way up.

Answer: $24$ turns. It is a step-down transformer, because the secondary has fewer turns than the primary and the output voltage is lower than the input.

Example 2. A transformer is 100% efficient. Its primary coil has $50$ turns and is connected to a $12$ V a.c. supply. The output voltage is $240$ V. (a) Calculate the number of turns on the secondary coil. (b) The current in the primary coil is $2.5$ A. Calculate the current in the secondary coil.

Show the working

(a) The unknown is $N_{s}$ , so use the rearranged turns equation.
$N_{s} = N_{p} \times \frac{V_{s}}{V_{p}} = 50 \times \frac{240}{12} = 1000$
That is more turns than the primary has, which fits: the voltage goes up, so this is a step-up transformer.
(b) 100% efficient means power in equals power out, so use $I_{p} V_{p} = I_{s} V_{s}$ . Divide both sides by $V_{s}$ to get $I_{s}$ on its own.
$I_{s} = \frac{I_{p} V_{p}}{V_{s}} = \frac{2.5 \times 12}{240} = 0.125 A$
Check that it makes sense. The voltage went up $20$ times, so the current must come down $20$ times: $2.5 \div 20 = 0.125$ A. Power in is $2.5 \times 12 = 30$ W, and power out is $240 \times 0.125 = 30$ W.

Answer: (a) $1000$ turns. (b) $0.125$ A (about $0.13$ A).

Example 3. A power station sends $500$ kW of power along cables with a total resistance of $4.0$ Ω. Calculate the power wasted as heat in the cables when the power is sent at (a) $10$ kV and (b) $100$ kV. Comment on the result.

Show the working

Convert to base units first: $500$ kW $= 500 000$ W, $10$ kV $= 10 000$ V and $100$ kV $= 100 000$ V.
(a) Find the current in the cables from $P = I V$ .
$I = \frac{P}{V} = \frac{500 000}{10 000} = 50 A$
Then find the heating in the cables with $P = I^{2} R$ , using the cable resistance.
$P_{loss} = I^{2} R = 50^{2} \times 4.0 = 10 000 W$
(b) Same power, ten times the voltage.
$I = \frac{500 000}{100 000} = 5.0 A$
Square the new current and multiply by the same resistance.
$P_{loss} = {5.0}^{2} \times 4.0 = 100 W$
Comment: ten times the voltage gave one tenth of the current and one hundredth of the wasted power. At $10$ kV, $10$ kW of the $500$ kW is wasted ( $2$ %). At $100$ kV, only $100$ W is wasted ( $0.02$ %).

Answer: (a) $10 000$ W ( $10$ kW). (b) $100$ W. Raising the voltage $10$ times makes the current $10$ times smaller and the power wasted in the cables $100$ times smaller.

Example 4. Explain why a transformer gives an output voltage when its primary coil is connected to an a.c. supply, but not when the primary is connected to a battery.

Show the working

Start with the input. The a.c. supply makes a current in the primary coil that keeps changing in size and direction.
That current makes a magnetic field in the soft-iron core. The field changes too, and it reverses every time the current reverses.
The core carries this changing field through the secondary coil.
A changing magnetic field linking the secondary induces an alternating e.m.f. across it. That e.m.f. is the output voltage.
Now the battery. It gives a steady direct current, which makes a steady magnetic field in the core.
A steady field is not changing, so it induces no e.m.f. in the secondary, apart from a brief pulse at the moment the battery is connected or disconnected.

Answer: With a.c., the changing current in the primary makes a changing magnetic field in the core. The core carries this field through the secondary, and the changing field induces an alternating e.m.f. in the secondary. A battery gives a steady current and so a steady field, which induces no e.m.f., so there is no output.

Where people go wrong

Mistake: Writing the turns equation upside down on one side, as $V_{p} / V_{s} = N_{s} / N_{p}$ .

Instead: Each voltage goes with the turns of its own coil: $V_{p} / V_{s} = N_{p} / N_{s}$ .

With the ratio upside down, a step-down transformer comes out giving a bigger voltage. Checking your answer against step-up or step-down catches this.

Mistake: Thinking a step-up transformer gives out more power than it takes in.

Instead: It steps the voltage up and the current down. At 100% efficiency, $I_{p} V_{p} = I_{s} V_{s}$ , so power out equals power in.

A transformer cannot create energy. A real one gives out slightly less power than it takes in, because some energy heats the coils and the core.

Mistake: Thinking a transformer works with a battery (d.c.).

Instead: A steady current makes a steady field, and a steady field induces no e.m.f. The primary needs a.c.

Induction needs a changing magnetic field. Only a.c. gives a field that changes all the time.

Mistake: Saying that current flows from the primary coil to the secondary coil through the core.

Instead: The coils are not connected by any wire, and they are insulated from the core. Only the changing magnetic field passes from one coil to the other.

The e.m.f. in the secondary is induced, not supplied along a wire. That is the whole principle of the transformer.

Mistake: Using $P = V^{2} / R$ with the transmission voltage and concluding that a higher voltage wastes more power.

Instead: Find the current from $I = P / V$ , then use $P = I^{2} R$ for the cables.

The p.d. across the cables is only the small drop along their length, not the transmission voltage, so $V^{2} / R$ with the transmission voltage gives the wrong quantity.

Check yourself

What is the core of a transformer made of, and why is that material chosen?
Answer
Soft iron, because it is easily magnetised and demagnetised, so its magnetism can reverse each time the current reverses. It carries the magnetic field from the primary coil through the secondary coil.
A transformer has $50$ turns on its primary coil and $1000$ turns on its secondary coil. Is it step-up or step-down, and by what factor does it change the voltage?
Answer
Step-up, because the secondary has more turns. $1000 \div 50 = 20$ , so the output voltage is $20$ times the input voltage.
A transformer has $600$ turns on its primary and $30$ turns on its secondary. The primary is connected to $240$ V a.c. Calculate the output voltage.
Answer
$V_{s} = V_{p} \times N_{s} / N_{p} = 240 \times 30 / 600 = 12$ V.
Why does a transformer need an alternating supply?
Answer
An e.m.f. is induced in the secondary only when the magnetic field through it changes. A.c. makes a field that changes all the time; d.c. makes a steady field that induces no e.m.f.
A 100% efficient transformer steps $230$ V down to $23$ V. The current in the secondary is $4.0$ A. Calculate the current in the primary.
Answer
$I_{p} = I_{s} V_{s} / V_{p} = 4.0 \times 23 / 230 = 0.40$ A.
Where in the transmission system are step-up transformers used, and where are step-down transformers used?
Answer
Step-up transformers at the power station, to raise the voltage before transmission. Step-down transformers near the consumers, to lower it to the mains voltage.
State two advantages of transmitting electricity at a high voltage.
Answer
The current is smaller, so less power is wasted as heat in the cables. Thinner, cheaper cables can be used.
The same power is transmitted at double the voltage through the same cables. What happens to the current, and what happens to the power wasted in the cables?
Answer
The current halves (from $P = I V$ ). The power wasted, $I^{2} R$ , becomes ${(1 / 2)}^{2} = 1 / 4$ of what it was.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A transformer is used to run a $12$ V lamp from the $230$ V a.c. mains supply.

(a) Describe the construction of a simple transformer. 2 marks No calculator

Mark scheme

B1 a core made of soft iron
B1 two coils of (insulated) wire, primary and secondary, wound on the core

(b) The primary coil has $2300$ turns. Calculate the number of turns on the secondary coil. 2 marks

Mark scheme

C1 Vp / Vs = Np / Ns or any correct rearrangement used, e.g. Ns = 2300 × 12 / 230, or any equivalent form
A1 120 (no unit needed)

(c) State whether this is a step-up or a step-down transformer. 1 mark No calculator

Mark scheme

B1 step-down; ignore any reason given

Question 2. The primary coil of a transformer is connected to a $230$ V a.c. supply. The secondary coil gives an output of $46$ V to a heater. The transformer is 100% efficient.

(a) Explain how the current in the primary coil produces a voltage across the secondary coil. 3 marks No calculator

Mark scheme

B1 the alternating current in the primary produces a changing / alternating magnetic field in the core
B1 the (soft-iron) core carries / links this field through the secondary coil
B1 the changing field induces an (alternating) e.m.f. / voltage in the secondary; insufficient: a field that is not described as changing anywhere in the answer

(b) The current in the secondary coil is $5.0$ A. Calculate the current in the primary coil. 3 marks

Mark scheme

C1 IpVp = IsVs or Ip / Is = Vs / Vp, or any equivalent form
C1 Ip = (5.0 × 46) / 230
A1 1.0 A; unit required

Question 3. A power station supplies $1.2$ MW of electrical power to a town through transmission cables with a total resistance of $5.0$ Ω.

(a) Describe how transformers are used between the power station and the homes in the town. 2 marks No calculator

Mark scheme

B1 a step-up transformer at the power station increases the voltage for transmission
B1 step-down transformer(s) near the town / homes decrease the voltage (to the mains value)

(b) The power is transmitted at $20$ kV. Calculate the current in the cables. 2 marks

Mark scheme

C1 I = P / V, e.g. 1 200 000 / 20 000
A1 60 A; unit required

(c) Calculate the power wasted as heat in the cables. 2 marks

Mark scheme

C1 P = I²R used, e.g. 60² × 5.0
A1 18 000 W (18 kW); unit required; ecf from (b)

(d) The same power is now transmitted at $400$ kV. Explain, using $P = I^{2} R$ , why less power is wasted in the cables. 2 marks No calculator

Mark scheme

B1 for the same power, a higher voltage gives a smaller current (I = P / V), e.g. 3.0 A
B1 power wasted = I²R with R unchanged, so a smaller current gives a much smaller loss, e.g. 45 W, owtte

(e) State one other advantage of transmitting electricity at a high voltage. 1 mark No calculator

Mark scheme

B1 thinner (so cheaper or lighter) cables can be used; insufficient: less power wasted, which repeats (d)

How this is examined

Calculating the power wasted in transmission cables, and explaining why a higher voltage reduces it: a run of short Calculate parts (the current from $P = I V$ , then the heating from $P = I^{2} R$ ) followed by an Explain part about raising the voltage $I = P / V$ with kW and kV converted to W and V; $P = I^{2} R$ using the cable resistance, with the answer in W; then the chain: same power at a higher voltage needs a smaller current, the loss is $I^{2} R$ with $R$ unchanged, so the loss is much smaller. Paper 4 gives neither equation, so both are recalled.
Explaining how a changing current in one coil produces a voltage across the other: a three-mark Explain part, which can be followed by a part asking why the transformer does not work with d.c. Three links: alternating current in the primary makes a changing magnetic field in the core; the core carries the field through the secondary; the changing field induces an alternating e.m.f. in the secondary. The word changing (or alternating) must describe the field. For d.c.: steady current, steady field, so no e.m.f. is induced.
Using the 100% efficiency equation to find the current in the other coil: a three-mark Calculate part that says the transformer is 100% efficient $I_{p} V_{p} = I_{s} V_{s}$ recalled and written down, which a C mark can credit if the final answer is wrong; the values substituted; the answer with the unit A. Check: if the voltage is stepped up, the current must come out smaller.
Finding a number of turns or a voltage from the turns ratio: a two-mark Calculate part that gives three of $V_{p}$ , $V_{s}$ , $N_{p}$ and $N_{s}$ $V_{p} / V_{s} = N_{p} / N_{s}$ recalled and rearranged for the unknown, and a correct answer: a voltage with the unit V, or a number of turns with no unit. A quick check that a step-down transformer gives a smaller voltage.
Describing what a transformer is made of and naming the two kinds: a short Describe part, labels to add to a drawing, or a State part asking whether a transformer is step-up or step-down A soft-iron core, and two coils of insulated wire (primary and secondary) wound on it, the primary connected to the a.c. supply. Step-up: more turns on the secondary, $V_{s} > V_{p}$ . Step-down: fewer turns on the secondary, $V_{s} < V_{p}$ .
Where the transformers are in the transmission system, and what high voltage gains: a two-mark Describe part, or a State part asking for the advantages A step-up transformer at the power station and a step-down transformer near the consumers. Advantages: a smaller current, so less power is wasted as heat in the cables; thinner, cheaper cables can be used.

Words to know

Transformer: A device with two coils wound on a soft-iron core that changes the size of an alternating voltage.
Primary coil: The coil connected to the a.c. supply: the input. Its voltage is $V_{p}$ and its number of turns is $N_{p}$ .
Secondary coil: The coil that gives the output to the load. Its voltage is $V_{s}$ and its number of turns is $N_{s}$ .
Step-up transformer: A transformer with more turns on the secondary than on the primary, so the output voltage is bigger than the input voltage.
Step-down transformer: A transformer with fewer turns on the secondary than on the primary, so the output voltage is smaller than the input voltage.
Soft iron: Iron that is easily magnetised and easily demagnetised, used for the core so that its magnetism can reverse with the current.
100% efficient transformer: An ideal transformer in which power out equals power in: $I_{p} V_{p} = I_{s} V_{s}$ .
High-voltage transmission: Sending electricity along cables at a very high voltage, so that the current, and the power wasted as heat in the cables, are small.

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