Lesson video

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Hello, my name's Dr.

George and I'm going to be helping you with this lesson on transformers, which is part of the unit: mains electricity.

The outcome for the lesson is I can describe how a transformer works and solve problems using transformer equations.

These are the key words for the lesson.

You can come back to this slide anytime if you need to remind yourself of the meanings.

The lesson has three parts called, how transformers work, transformer efficiency, and transformer calculations.

So transformers change the size of potential difference.

That's what they're for.

And they do this with AC, alternating current.

This diagram shows the parts of the transformer.

And it's a step up transformer and we can see that because it has more turns of wire on its secondary coil than on its primary coil.

And that means the p.

across the secondary coil will be larger than the p.

applied across the primary coil.

What the transformer can't do though is transfer more energy from the secondary than is transferred to it by the primary because of conservation of energy.

So a question for you, which of the following can a transformer increase? And when I ask a question, I'll wait five seconds, but if you need longer, press pause and press play when you're ready.

And the correct answer is potential difference only.

It can't increase energy, it can't make energy out of nowhere, and so it can't increase power, energy transferred per second either.

So a transformer simply has two coils of insulated wire each wrapped around a soft iron core.

Soft iron means pure ion that's easily magnetised and demagnetized, and there's no direct electrical connection between the coils because they're insulated.

And so also no current can flow between the coil and the core.

And alternating p.

is applied across the primary coil and that causes the soft iron core to become magnetised because of coil of insulated wire around a soft iron core is an electromagnet.

You have to use an alternating p.

A direct p.

wouldn't work.

The alternating p.

across the primary coil causes an alternating current and that makes the soft iron core become magnetised in one particular direction.

We could represent it like this, say with north pole at the top and south pole at the bottom.

And then when the p.

direction reverses, the magnetic field direction reverses.

So now we have south pole at the top and north pole at the bottom.

And because main's frequency is 50 hertz, this cycle repeats 50 times each second.

But making the magnetic field flip direction by having an alternating p.

across the primary coil has an equivalent effect to pushing a magnet in and out of the secondary coil 50 times each second.

'Cause the secondary coil is experiencing a changing magnetic field at 50 hertz.

If we actually did this, we could connect a microammmeter to the coil, we could detect the induced current caused by the potential difference induced across the coil.

And when a current is induced by a changing magnetic field around a conductor, that's called the generator effect, Watch the ammeter.

This current switches direction each time the magnet moves in or moves out.

If we simply had a direct potential difference across the primary coil, so if it didn't change direction, then when we first turned it on, that would be like pushing the magnet into the coil and we would get an induced current, but then we'd get nothing more 'cause there'd be no further change to the p.

d, so no further change to the magnetic field.

The generator effect only works with a moving magnetic field, a magnetic field that is changing.

You can think of magnetic field lines cutting across the wires and that's what causes an induced p.

and current.

So now, when is a p.

induced across the secondary coil of a transformer? Press pause while you read these four options.

There may be more than one correct one.

And press play when you have your answer ready.

And the first statement is correct.

When a direct p.

across the primary coil is turned on or off because then for a very short time there's a changing p.

and so there's a changing magnetic field.

But also when an alternating p.

across the primary coil is turned on or off, because again, there'll be a change to the p.

while it's turning on just briefly and while it's turning off.

And so there'll be a changing magnetic field.

And finally, what we normally do with a transformer is we have an alternating p.

across the primary coil and it's left turned on.

But because the p.

keeps switching direction, the magnetic field keeps switching direction and then a p.

is induced across the secondary coil.

So the only one of these that won't work is if a direct p.

across the primary coil is simply left turned on.

There's no change to the magnetic field, so there are no field lines cutting the secondary coil and no induced p.

across the secondary.

Well done, if you picked out those correct statements.

And a slightly longer task for you, can you describe how an alternating p.

is induced across the secondary coil of a transformer? Press pause while you're writing out your answer and press play when you're ready to check it.

There are different ways of wording the answer, but here are key points that your answer could include.

An alternating p.

across the primary coil pushes current through it and it becomes an electromagnet.

This creates a magnetic field around the soft iron core that appears inside the secondary coil.

As it is formed, the moving magnetic field induces a p.

across the secondary coil, swapping the direction of the p.

across the primary coil, turns the magnetic field round, which induces a p.

across the secondary coil in the opposite direction.

This cycle repeats 50 times each second for 50 hertz alternating p.

So compare your answer with this one and look for points that you've made that are the same as these.

And check whether you said anything that wasn't quite right or whether you missed anything out.

Now let's move on to the second part of the lesson, transformer efficiency.

Having a soft iron core shaped as a complete loop like this, make sure that the magnetic field inside the secondary coil is as strong as it can be.

So the iron core, iron is a magnetise material, strengthens a magnetic field and helps it extend as far as the secondary coil where it's still strong.

And having this core reduces the amount of energy that's dissipated, that is spreads out as heat by the magnetic field.

But we make the soft iron core not out of one solid piece of iron, but of thin sheets of soft iron, with a very thin layer of electrical insulation between each one.

And we describe the core as laminated, made of sheets.

And the insulating layers stop induced current flowing through the core itself because the core is a conductor and it's a conductor in a changing magnetic field when the transformer is working.

And so p.

can be induced across the core itself, making currents flow in the core.

Those currents would be called eddy currents and they would cause heating of the core and so dissipate energy.

That would be a waste of energy because then that energy wouldn't go into the current in the secondary coil, which is where we want it.

Also, the separate parts of the core are securely bolted together to minimise vibrations caused by the changing magnetic field.

Sometimes a transformer will hum and that's a vibration and that also leads to energy dissipation.

So which of the following prevents some energy dissipating from a working transformer? Press pause while you think about this and press play when you're ready.

So B is a correct answer.

Laminating the soft iron core to reduce eddy currents in the core.

C, securely bolting separate parts of the soft iron core together to reduce vibration of the core.

And also D, shaping the soft iron core as a complete loop.

A isn't really a correct answer to this question because if we didn't insulate the wire of the coils, then current would simply flowed directly from one coil to the other via the core and this device wouldn't even work as a transformer.

So thanks to all these measures, a transformer dissipates very little energy overall.

And in fact, in calculations we often assume that a transformer has an efficiency of 100%.

So the useful output energy transfer is the same as the input energy transfer.

This makes calculations simpler and gives answers that are still very close to the actual values.

And so when we're assuming that efficiency is 100% for a transformer, we can say that the power input is the same as the power output.

And for an electrical device we can express power as current times potential difference.

So this equation expresses power input equals power output for a transformer.

The P doesn't represent a quantity, it's simply a label on the current and potential difference on the left to show that they belong to the primary coil.

And the S on the right is a label representing secondary coil.

The secondary potential difference depends on two things, the size of the potential difference across the primary coil and the design of the transformer.

So here we have a transformer with two circuits connected to it.

There's an input circuit on the left with potential difference VP, and we must be using the transformer because that's not the p.

that we want in the circuit on the right.

So we use this transformer and we have potential difference VS across the circuit on the right.

The secondary current mainly depends on the secondary p.

and the resistance of the secondary circuit.

You could think of the secondary coil as acting like an AC power supply for the circuit on the right.

The primary current drawn from the power supply on the left is then decided by this relationship, power input equals power output.

Now I'm going to show you an example calculation question and then I'll ask you to try one.

What current is drawn from a 230 volt power supply to a transformer if the output specifications are 12.

0 volts and 250 milliamps? So that means that the circuit attached to the secondary coil needs 12.

0 volts and 250 milliamps of current.

Now when we use an equation, we use the standard units for the quantities.

So let's convert the output current into amps.

So 250 milliamps, that's 250 thousandths of an amp is 0.

250 amps.

And now we could write the equation we're going to use, power input equals power output, substituting the values that we know.

The transformer draws current from the 230 volt power supply.

So that must be the primary potential difference.

And then the output is for the circuit connected to the secondary.

We can start rearranging this by doing the right hand multiplication and then dividing both sides by 230.

Alternatively, you could start by rearranging the equation in symbols to make the primary current the subject and then substitute in the numbers.

When we do the calculation, we get this on the calculator, but we can't really know that many significant figures in the answer because we only had three significant figures in the measurements we used to calculate it.

When we round that to three significant figures, we could either write 0.

0130 amps, or we could write as milliamps, 13.

0 milliamps.

And we should write that final zero because it is the third significant figure.

We know it's a zero and not some other digit.

And now here's a question for you.

What current is drawn from a 230 volt power supply to a transformer if the output is 24.

0 volts and 800 milliamps? So let's take a look at the work solution.

Again, we should convert the output current into the standard unit, amps.

Write down the equation we're going to use, power input equals power output, substitute in the values or rearrange the equation first if you prefer.

Then rearrange and do the calculation.

And you could write your answer in amps or given that the question used milliamps, you could write your answer in milliamps and to three significant figures, that's 83.

5 milliamps.

Well done if you've got that right.

And now a couple more questions for you.

I'll leave you to read these.

So press pause while you write your answers and press play when you're ready to check them.

Let's take a look at the worked answers.

So you have to calculate the current drawn from the power supply when the current in the secondary circuit is 0.

40 amps and you're given the potential differences for the primary and secondary coils.

So as usual, we use the equation for power in equals power out, substitute the values into the right places.

And when we do the calculation, we get a primary current of 3.

8 amps, two significant figures there because we only knew two significant figures in the secondary current, which we used in the calculation.

And then you were asked to calculate the current drawn from the power supply when two more lamps are connected in parallel to the secondary circuit and the secondary current increases to 1.

20 amps.

And that is what you would expect to happen to the current if you added two identical lamps in parallel with the first, it's tripled because there's the same current in each branch of 0.

40 amps.

So again, write the equation, substitute in the values, and now we have primary current of 11.

5 amps.

So by changing what's in the secondary circuit, we've changed the current drawn by the primary coil.

You could say we're demanding higher power for the secondary circuit.

And now let's take a look at a different kind of transformer calculation in the third part of this lesson.

The strength of electromagnet that's made when a current flows through the primary coil is determined by the p.

across it.

If you increase the p.

across a coil, you get a stronger magnetic field because you get a larger current.

We could represent the magnetic field like this because the magnetic field around a current carrying coil is very similar to the magnetic field around a bar magnet.

And the strength of this electromagnet is equal in size to the strength of the electromagnet that appears in the secondary coil when an induced current appears in that coil.

If we made the secondary coil identical to the primary coil, the induced p.

across the secondary would be equal to the p.

across the primary.

That wouldn't be very useful because it wouldn't step up or step down the potential difference.

Transformer would just give us the same potential difference out as we put in.

But if we double, the number of turns on the secondary coil we'll double the induced potential difference across it.

So if the secondary coil has twice the number of turns of wire as the primary coil, the p.

across it will be double the p.

that's across the primary.

So that would be a kind of step up transformer that is doubling the p.

So what effect would putting four times the number of turns on the secondary coil of a transformer have on the p.

across it? And the answer is A, it would make it four times bigger.

So the number of turns on the secondary is proportional to the p.

we get across the secondary if we don't change anything else.

If the number of turns on the secondary coil is three times greater than the number of turns on the primary coil, the p.

across the secondary is three times the p.

across the primary.

And it goes the other way too.

If the number of turns on the secondary is two times smaller, it's half the number of turns on the primary coil, the p.

across the secondary will be half the p.

across the primary and that would be a step down transformer.

The ratio of the number of turns on each coil is equal to the ratio of the p.

ds.

We can write that using fractions like this.

So the primary potential difference over the secondary potential difference equals the number of turns in the primary coil over the number of turns in the secondary coil.

Two questions again, I'll show you how to do the first one.

There are 60 turns on the primary coil of a transformer and 240 turns on the secondary coil.

How did the p.

across the secondary coil compare to the p.

across the primary coil? So we need to use this relationship we've just learned and we can substitute in the numbers of turns on the two coils, 60 over 240, primary over secondary and we could simplify that to a quarter.

So that tells us that the p.

across the secondary is four times bigger than the p.

across the primary 'cause VP over VS is one over four.

So now a question for you.

There are 2,400 turns on the primary coil of a transformer and 120 turns on the secondary coil.

If the p.

across the primary coil is 230 volts, what is the p.

across the secondary coil? So again, we use this relationship.

At this time, we have to do a calculation rather than just show a ratio so we can substitute in the three values that we know, the number of turns on each coil and the potential difference across the primary.

There are different ways of rearranging this.

You could start by calculating 2,400 over 120, which is 20.

Another possible step is when you have an equation that is simply a fraction on each side, you can invert both of those fractions, you can turn them upside down.

When you rearrange, you should end up with VS is 11.

5 volts.

If you haven't got that, it might be your rearranging that has gone wrong.

Think about using the same skills that you use in maths when you rearrange a formula.

And now here's the final task for this lesson.

Three questions for you.

And when you answer these, show all of your working out.

So press pause when you do this.

And when you press play, I'll show you the work solutions.

Let's start with question one.

There are 88 turns on the primary coil of a transformer that has a p.

of 12 volts across it.

How many turns are needed on the secondary coil for it to have a p.

of 66 volts across it? So we use the same equation that the ratio of p.

ds is the same as the ratio of numbers of turns.

Substitute in the values we know.

And then we need to rearrange this to make the number of turns on the secondary the subject.

We get the answer 484 turns.

Now a transformer for a laptop changes mains p.

d, that's 230 volts, into the 19.

5 volt supply that it needs.

How many turns are needed on the secondary coil if there are 4,600 turns on the primary coil? Substitute in again, rearranged to make the number of turns on the secondary the subject and we get 390 turns.

And for most laptops you will see that there's a sort of black box partway along the lead that connects to the mains and inside that is a transformer.

The transformer in question two supplies a current at 3.

33 amp to the laptop.

How much current does the transformer draw from the main supply? This time we need to use the power equation.

Power in equals power out because we're trying to find a current so we can substitute in the values that we know for the primary and secondary, rearrange to find the current in the primary.

And when we round, we should round to three significant figures because the quantities that went into calculating this, we knew to three significant figures and we get 0.

282 amps.

Well done if you're getting some or all of these right.

And now we've reached the end of this lesson.

So here's a summary.

A transformer is a device that can change the p.

of an AC supply.

The efficiency of a transformer is close to 100%.

Power input equals power output.

IP VP equals IS VS.

The current drawn from the supply depends on the current in the secondary circuit.

The ratio of turns on each coil is equal to the ratio of p.

s across each one.

VP over VS equals NP over NS.

Well done for working through this lesson And I hope to see you again in a future lesson.

Bye for now.

I've finished the video