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Hello, my name's Dr.

George, and this lesson is called Power of an Appliance or Electric Circuit.

It's part of the unit, Mains Electricity.

The outcome of this lesson is I can describe how the power of an appliance or electric circuit depends on current and potential difference.

I'll be using these keywords during the lesson.

And remember, power is the amount of energy transferred each second.

If you want to remind yourself of the meanings of these words, come back to this slide anytime.

The lesson has three parts.

They're called calculating power, power of mains appliances, and the correct fuse.

Let's start.

It turns out that it always requires the same amount of energy to heat the same amount of water by the same number of degrees.

And we can use that when thinking about power.

So the same amount of energy is always needed to heat one litre of water from 20 to a 100 degrees, which is the sort of thing a kettle does.

This kettle on the left has a power of 1,500 watts and it boils a litre of water in four minutes.

This kettle on the right has a power of 2,000 watts and it boils a litre of water in only three minutes.

So a more powerful kettle will heat up the water more quickly.

That must mean that the more powerful kettle is supplying the energy needed more quickly.

To heat two times the amount of water from 20 degrees C to a 100 degrees C, each kettle would need to transfer two times the amount of energy.

So on the left, four megajoules of energy, that's 4 million joules are needed to heat this water from 20 degrees to a 100 degrees C.

And on the right with twice as much water.

Eight megajoules of energy are needed to heat it from 20 to a 100 degrees C.

So how long would it take a careful to heat water from 20 to a 100 degrees C if it contained two times as much water as before? When I ask a question, I'll wait five seconds, but you may need more time.

In which case, press pause and press play when you have your answer ready.

The correct answer is D.

It would take twice as much time to heat twice as much water.

Now power is equal to the amount of energy transferred each second.

So power can be calculated by dividing the amount of energy transferred by the time it takes.

We can write the equation in symbols as P equals E over T.

For this equation to work, we use power in watts, energy in jewels, and time in seconds.

Now, out of these four, which two equations are correct? The first correct equation is B, power is energy divided by time.

That's what we saw on the previous slide.

But C is also correct if we multiply both sides of the equation in B, by time we get energy equals power times time.

In this circuit there's a cell and a lamp, and the lamp is transferring energy to the surroundings by emitting light.

This lamp has a power of three watts and that means it's transferring three jewels of energy to the surroundings each second in the form of light and perhaps some heat.

Now if we add another lamp in parallel with the first, we find that doesn't change the brightness of the first lamp and the second one has the same brightness as that.

So if they have the same brightness as before, they must each have the same power as before.

They're transferring the same amount of energy into light per second.

So if the power of the first lamp was three watts, the combined power of these two in parallel must be six watts.

And these three in parallel still equally bright, the power is nine watts.

But we also know that when we add identical bulbs in parallel, it increases the current through the cell.

Let's say this is two amps.

If we add another identical bulb in parallel, the current is now four amps.

There's two amps in each of these branches containing a bulb, and that adds up to four amps where the amateur is and through the cell.

And if we add another amp in parallel, we have a total current through the cell of six amps.

So now I want you to think about this true or false, the power of an electric circuit is proportional to the current, and when I say current, I mean the current in the cell.

And what is the evidence for your answer? So why do you think that? Press pause while you think about this and press play when you're ready with your answer.

The correct answer is that it's true.

How could you know that from what you've just seen? Well, we saw that when identical bulbs are added in parallel, they each have the same p.

d.

Because they have the same p.

d.

as the p.

d.

across the cell and they each have the same current.

They have equal brightness and that must be because they have the same power.

So the total power is proportional to the number of bulbs.

If you double the number of bulbs, you double the total power.

But we also saw that the cell current is proportional to the number of bulbs.

If you double the number of bulbs, you double the current in the cell.

So if the power and the current are both proportional to the number of bulbs, they must be proportional to each other as well.

So the power of an electric circuit is proportional to the current.

If we add bulbs in series with each other, they will become less bright and so we'll be changing the power of each bulb.

But that won't happen if we add a cell each time we add a bulb.

So in the first circuit, if we have power three watts, we've got equally bright bulbs again, so total power is six watts, and we've doubled the p.

d.

by doubling the number of cells.

And the current is the same.

We've doubled the p.

d.

, which determines the push on the electrons, but we've also doubled the resistance.

So we get the same current.

` If we now add another cell and another bulb in series.

We've got three times the power that we had at the start, three times the total p.

d.

and the same current again.

So thinking about that, is this true or false? The power of an electric circuit is proportional to the potential difference and I mean the total potential difference across the cell or battery.

And how do you know? What is the evidence for your answer? Press pause while you think about this.

And this statement is also true.

And one way to explain how we know from what we've seen on the previous slides.

If the number of cells always equals the number of bulbs in series, which is what we were looking at before.

The current stays the same, but the total p.

d.

is proportional to the number of cells, the number of cells doubles, the total p.

d.

doubles.

The total power is also proportional to that number because the power is proportional to the number of bulbs we have.

So the power must be proportional to the p.

d.

since both the power and the p.

d.

are proportional to the number of cells and bulbs.

Well done if you realise that.

So we found that the power in an electric circuit, the total power is proportional to the current in the cell or battery and it's also proportional to the potential difference across the cell or battery.

So we can summarise these relationships with the equation power is current times potential difference.

Or P equals I times V.

And this is a very useful equation when we're thinking about electric circuits.

And it's actually true for any individual component in a circuit that its power, the energy it transfers per second equals the current through it times the potential difference across it.

But this equation only works if we use standard units.

We have power measured in watts, current in amps, and potential difference in volts.

And now here's a worked example.

I'll show you how to do this and then I'll ask you a question.

If the current through an electric motor is 2.

4 amps and the p.

d.

across it is 6.

0 volts, what is the power of the motor? Well, we could start by writing down the equation we're going to use.

We could also write down the quantities that we know.

So we have I equals 2.

4 amps and V equals 6.

0 volts.

Substitute the quantities into the equation and we find that the power is 14.

4 watts, but we should really round that to two significant figures because we only knew I and V to two significant figures.

So we get 14 watts.

And now here's one for you to try.

If the p.

d.

across a washing machine is 230 volts and the current through it is 2.

4 amps, what is the power of the washing machine? Press pause when you do this and press play when you're ready to check your answer.

Again, we can use the equation P equals I times V.

Substitute in the current in p.

d.

We get 552 watts, but we only knew the current to two significant figures, so we can only really know our answer to two significant figures.

And that's 550 watts.

Well done if you got that.

Now here's a longer written task for you.

A group of pupils want to model the power of an electric circuit.

They use a loop of rope and you can see in the picture that there are two hands moving the rope representing the battery.

There's a hand gripping the rope, representing the lamp, and there's a current represented by the movement of the rope.

So the questions are how could they use the model to show how energy is transferred by an electric circuit? How could they use the model to show the effect of increasing current on the power of a circuit? And how can they use the model to show the effect of increasing p.

d.

on the power of a circuit? Press pause while you write down your answers.

And when you're ready, press play and I'll show you some example answers.

So here are some ways you could have written these answers.

So first of all, showing how energy is transferred.

The pupil modelling the battery pulls the rope around the circuit representing current.

Through the grip of the pupil, who is the lamp.

Friction from the rope heats up their hands and then dissipates into the surroundings.

Similar to how energy is transferred in a real electric circuit.

And then how to show the effect of increase in current.

The pupil modelling the battery can increase current by pulling the rope more quickly.

The larger current heats the hands of the lamp more quickly showing that energy is transferred more quickly.

And finally, how to show the effect of increasing the p.

d.

Adding a second lamp to the rope loop and pulling the rope around at the same speed, same current, will transfer energy twice as quick.

To do this, the battery will need to push two times harder with a p.

d.

that is twice as big.

Well done if you've got some of these points into your answers.

And now we'll move on to the second part of the lesson.

Power of main's appliances.

In a house, the p.

d.

across a main circuit is always 230 volts.

Here we have a circuit that includes a kettle and in parallel with it a lamp.

And because all main circuits are wired as parallel circuits, the p.

d.

across each appliance is always the same.

It's always 230 volts.

So the p.

d.

across the kettle here is 230 volts and so is the p.

d.

across the lamp.

But their powers are different.

This kettle has a power of 1,500 watts and the lamp only 40 watts.

And we've seen that power is current times potential difference.

So as the p.

d.

across the kettle is the same as the p.

d.

across the lamp, the current through the kettle must be greater because that's the only way it could have a higher power.

And so another pair of questions, I'll show you how to do the first one.

What is the current through a 1,500 watt kettle that is plugged into the mains? So although we're not told it, we can write down the p.

d.

across the kettle, mains p.

d.

is 230 volts.

We could also write down the power 1,500 watts.

And then we use this equation P equals I times V.

You can either rearrange the equation in symbols first or you can substitute in the values and then rearrange.

You're going to need to rearrange one way or another to find I.

Here we can divide both sides by 230 volts.

Calculate I we get 6.

5217, et cetera amps shown on the calculator, but we can't really know the answer to that many significant figures.

We can give three significant figures here because we only have the p.

d.

to three significant figures.

And now here's a question for you.

What's the current through a 40 watt lump that is plugged into the mains? Press pause when you write your answer And press play when you're finished.

Again, we can write down the p.

d.

straight away.

We're talking about a main circuit, so it's 230 volts, P equals I times V.

Substitute the values, rearrange to find the current.

And we get this answer on the calculator, which here is rounded to two significant figures because we only knew the power to two significant figures at most.

So we can see that it's true what we saw on the previous slide.

These two appliances have the same p.

d.

across them, but since they have very different powers, they must have very different currents.

If they had the same current and the same p.

d.

, it'd have to have the same power.

And so the current in the kettle is much larger than the current in the lower power lamp.

The current through the kettle and the lamp add up to give the current through the main circuit.

So we add these two together and we get 6.

69 amps.

Plugging in too many appliances can cause a very large current in a main circuit.

Because every time you plug in an appliance it's in parallel with whatever else is switched on.

And so it adds extra current that goes into that new branch.

If the current exceeds the size of a circuit breaker and the consumer box in the house, the whole circuit will be turned off.

The circuit breaker will detect that the current is too high and it will switch off the circuit.

That's a safety precaution to make sure that nothing gets so hot that wires melt or even start a fire.

Now here's an example question.

What is the current through a mains circuit when two 1,500 watt kettles and three 40 watt lamps are plugged into it? We could start with an initial calculation of the total power.

So two times the kettle and three times the lamp power, 3,120 watts.

Now that we have the power, we can use P equals IV.

Because we know V, it's mains it's 230 volts.

So substituting in rearranging to find the current, we get 13 point something amps.

And if we round two significant figures, 14 amps.

Now here's one for you.

What is the current to remain circuit when four 1,500 watt kettles are plugged into it? Press pause while you write down your working.

Again, let's start by working out the total power.

It's 6,000 watts and substituting into P equals I times V.

Again, it's 230 volts, it's mains.

Rearranging to find the current.

To two significant figures, we get 26 amps.

That's essentially a large current for a main circuit.

Now four questions and for each calculation, show all your working out, show the equation you're using and the quantities you're using and give each answer to the correct number of significant figures.

Basing that on how many significant figures are in the quantities that you use to calculate your answer.

Press pause while you do this and press play when you're ready to check your answers.

So here are the worked solutions.

The p.

d.

across the 2,000 watt mains kettle with a current of 8.

7 amps, 230 volts, again, it's mains.

Calculate the power of a main's electric kettle that has a current of 9.

2 amps through it.

P equals I times V.

Again, it's 230 volts p.

d.

substituting in.

And we should round to two significant figures because we know the current to two significant figures.

So 2,100 watts.

The current used by a main lamp with a power of 11 watts, again rearranging the equation.

P equals I times V.

And we should write our answer to two significant figures because we know the power to two significant figures.

So the current is only 0.

048 amps.

And the current used by an electric oven with power 3,500 watts.

And electric oven will often be the highest power device in a household.

So again, rearranging to find the current using 230 volts, we find that the current is 15 amps to two significant figures because that power of 3,500 watts may only be written to two significant figures if it's rounded to the nearest a 100.

Well done if you're getting these right.

Now for the last part of the lesson, the correct fuse.

It can happen inside an appliance that a wire comes loose and makes a connection where it shouldn't.

And that can cause a short circuit.

That's a circuit in which there's very little resistance.

So imagine that this wire has come loose and it's actually inside this main's appliance.

The current no longer has to pass through the main resistance of the appliance.

It's just passing through a low resistance wire.

And so we get a very large current, alternating current because main's provides AC.

And since power is current times potential difference.

And the p.

d.

is still 230 volts but we have a large current, then there'll be a large power.

And all that power is doing that energy transfer is causing heating of the wires in the main circuit and inside the appliance where it's flowing.

That could cause a fire if things get very hot.

So the faulty appliance needs to turn off as soon as possible and that's the point of a circuit breaker or a fuse.

It's to switch off the current if it gets dangerously high.

In the plug inside every appliance there's a fuse that contains a thin wire and that wire melts breaking the circuit and turning off the electricity.

If the current becomes too large.

There are three sizes of fuse available for plugs, three amp, five amp, and 13 amp.

And for each of these fuses, the wire melts if a current larger than the size of the fuse flows through it.

So if you have a three amp fuse, the wire inside it will melt and break the circuit if the current is greater than three amps.

The correct fuse to use depends on the current used by the appliance when it's working normally.

For example, which size fuse would allow a 2,000 watt kettle to work properly.

I'll show you how to do this and then I'll give you a question to try.

So we're looking to find the current in the kettle so that we can decide what's the best fuse.

So we'll use P equals I times V, it's main's voltage, 230 volts, rearranged to find the current.

Divide the power by the p.

d.

and we get 8.

69 or so amps.

That means we have to use a 13 amp fuse because that will allow that much current to flow without melting.

A three amp or five amp fuse would melt as soon as this kettle started working normally.

So that would not be appropriate.

And now a question for you.

Which size fuse would allow a 60 watt lamp to work? Three amps, five amps or 13 amps.

Press pause while you do the calculation And decide which fuse.

Press play when you're ready.

So we start by working out the current in the lamp.

We use P equals I times V, substitute in the power and the main voltage.

Rearrange and the current is about 0.

26 amps.

We would have to use a three amp fuse, so that allows any current up to three amps to flow.

A five amp or 13 amp views would allow this current to flow, but they would also allow a much larger current than the lamp should have when it's working normally.

An appliance that uses less than three amps should be fitted with a three amp fuse.

If there's a short circuit, the three amp fuse will melt faster than a five amp or 13 amp fuse.

So the three amp fuse is more sensitive to lower currents.

If you use a 13 amp fuse with this lamp, then 13 amps could flow through the appliance before it gets switched off, before the wire inside the fuse breaks.

And that could damage the lamp.

Which of these sizes of fuse should be fitted to a television that works normally with a current of 3.

2 amps? The correct answer is five amps.

3.

2 amps is just a little bit more than three amps.

So the three amp views would melt when the TV was working normally, and we don't want that.

So we need to choose the next one up.

And that's the five amp fuse.

Now here's a table of appliances and their powers.

They're all going to be used in main circuits.

So can you calculate missing values for current and state which type of fuse should be used for each of these appliances? And you're choosing from three amp, five amp, and 13 amp each time.

Press pause while you do this and press play when you're ready to check your answers.

And here are the answers.

Did you realise that none of the three standard sizes of views are going to work for an electric oven? The wire inside them will melt no matter which views you use.

So electric ovens are usually plugged into a special oven socket that's connected to a 20 amp circuit breaker in the consumer box.

So now we've reached the end of the lesson and here's a summary.

Power is equal to the amount of energy transferred each second, P equals E divided by T.

And in electric circuits, P equals I times V.

Power, P is measured in watts.

Energy, E is measured in joules.

Time, T is measured in seconds.

Current, I is measured in amps And p.

d.

, V is measured in volts.

Fuses that are fitted in the plugs of main's appliances contain a thin wire that melts if the current becomes larger than the size of the fuse because of a short circuit.

Fuses usually come in three sizes, three amp, five amp, and 13 amp.

The smallest fuse possible is used, but it needs to be greater than the current the appliance uses when it is working.

So well done for working through this lesson.

I hope you found it interesting and useful.

and you will need to use this knowledge if you ever have to decide what size of fuse to use.

I hope to see you again in the future lesson.

Bye for now.