Hello, my name's Dr.

George.

And this lesson is called Analysing Series Circuits.

It's part of the unit, Electric Fields and Circuit Calculations.

The outcome for the lesson which I'll be helping you achieve by the end is I can use circuit rules and the equation I = V/R to analyse series circuits.

Here are the key words for the lesson.

I'll introduce them as we go along, but the slide is here in case you want to come back and check the meanings anytime for yourself.

There are three parts to the lesson called current in series circuits, current in p.

d.

in series circuits, and analysing series circuits.

Let's get started.

Here's a representation of an electric current flowing around a series circuit.

A series circuit is a circuit that only has a single loop, and current is the flow of electrons around the wires and components.

And current in a series circuit is the same everywhere.

It's the same in the cell as it is in the wires or in this lamp.

So if you measured the amount of charge passing a point per second, which is the current, it would be the same no matter which point in this circuit you chose.

Now zooming into what's happening in the metal wires, a metal consists of a lattice of ions.

A lattice is a regular arrangement as you see here.

And ions are atoms that have lost one or more of their electrons.

And also in the metal are those electrons that have escaped from their atoms and they're loosely held.

They stay within the metal, but they're able to drift through the lattice.

And that's why a metal is a good conductor because it's possible to make those electrons move in one direction.

And then you have a current.

By the way, this model doesn't show the charges.

There would actually be negative charges on the electrons and the ions are positively charged.

So to make the electrons move, we have a cell.

And in a cell, positive charge builds up on one side, and negative charge builds up on the other.

And on a lot of cells that you can buy, you'll actually see a little positive and negative symbol on opposite ends.

And when you have charge building up in one place, there's an electric field around it.

That's an area of influence that exerts forces on other charges.

And it's this field that pushes electrons around a circuit.

And in fact, when you connect a cell to a circuit, the field isn't just close to the cell.

There's an electric field all around that circuit.

So that field exerts a force on all the charges in a metal.

And it makes the free electrons, wherever they are, all move at the same time in the same direction.

The field also exerts a force in the opposite direction on the positively charged ions, but they're fixed in place in the lattice.

So they don't move.

So looking at this series circuit, we have an ammeter there which measures current.

And we've said that current is the same everywhere.

So if we put an ammeter here, we'll measure the same current, and also here, or here, and here.

So now look at this circuit.

And can you say what is the current through bulb X? With these short questions, I'll wait five seconds.

But if you need longer, press Pause.

And press Play when you have your answer ready.

And the correct answer is 0.

15 amps because the current through bulb X will be the same as the current where the ammeter is measuring it next to the cell.

And now a slightly longer question.

Have a look at this circuit.

And I'd like you to explain what will happen to the current in three different situations.

So what if the ammeter is moved to another position? What if another identical resistor is added in series? And what if another identical cell is added in series and correctly aligned? So I mean it's in the same direction as the cell that's already there.

So press Pause while you're working out your answers.

And press Play when you're ready to check them.

So here are the answers to the questions.

Moving the ammeter to another position in series will not affect the current.

It's the same everywhere in the circuit.

Adding another resistor will double the resistance in the circuit, and that will halve the current everywhere in the circuit.

Adding another cell will double the strength of the electric field, so double the push on the electrons, and therefore double the current in the circuit.

So well done if you got those.

And now for the second part of this lesson, current and p.

d.

in series circuits.

The potential difference, or p.

d.

, of a cell is a measure of the size of the push that it gives to charges.

But we can think of it as well as a measure of the size of the electric field since that is what pushes the charge around the circuit.

Here's a cell with a p.

d.

of 1.

5 volts.

Two cells in series with each other, like this, have a total p.

d.

of 3.

0 volts.

It's just the sum of the p.

ds.

across the individual cells.

The electric field pushing the charge is a bit like the slope of a ramp, which causes a force on a ball.

If you put a ball at the top of a ramp, it wants to roll down.

And if you put a charge in an electric field, it also experiences a force and wants to move in a particular direction.

And if we increase the p.

d.

by adding another cell, we increase the size of the electric field and we increase the size of the force.

Now, of course, what you see here isn't what's really happening in the circuit.

We don't really have balls rolling down slopes.

That's a model to help us think about what's going on.

Now, look at this circuit.

The push on the electrons due to this 1.

5 volt cell causes a large current in this circuit.

And that current is the same everywhere.

It's a series circuit.

Now, if we increase the resistance, how difficult it is for current to flow measured in ohms? If we increase it by adding another resistor, the field can't push any harder.

There's still only a 1.

5 volt cell.

And so the current will decrease.

And that decreased current is the same everywhere in the circuit.

So a question about that.

What will happen if a resistor is added in series to the circuit shown? Look at all the options.

Think carefully.

Press Pause if you need more time.

And press Play when you've decided your answer.

And the correct answer is the current will decrease.

We've done nothing to change the push from the cell.

It's the same cell as before.

But we've increased the total resistance in the circuit, so that will decrease the current.

If we instead add another cell, that will increase the p.

d.

And that will increase the size of the electric field.

So now we've doubled the p.

ds.

three volts, and it increases how hard the electrons are pushed.

And so the current increases.

Question, what will happen if another cell is added in series to the circuit shown? Press Pause, and press Play when you've chosen your answer.

And if we add another cell, the push will increase.

We'll now have a battery more than one cell connected in series.

And because of that, the current in the circuit will increase.

So well done if you spotted both of those.

Now, look at this voltmeter here.

It seems to be measuring the potential difference across the cell.

But if we change the positions where we connect the voltmeter, to here or to here, the reading on the voltmeter doesn't change.

Here, it looks as though we're measuring the potential difference across the bulb.

And we are.

But the p.

d.

is the same as when we measured it across the cell.

Here's a representation of that.

Have a look at the reading on the voltmeter.

And as the positions of the contacts change, the reading doesn't change.

Here's the voltmeter measuring the p.

d.

across the cell.

And let's switch that lamp for three components.

It doesn't change the voltmeter reading.

The cell still has the same voltage.

And what we're now measuring is the total potential difference across the components, which equals the potential difference across the cell.

Now, if we focus in on one of those components, this resistor, the potential difference across the resistor alone is determined by its resistance or how hard it is to push current through it.

A higher resistance requires a higher p.

d.

to push the same current through it.

Look carefully about that, and then I have a question for you.

In this circuit, which component has the highest p.

d.

across it? Press Pause if you need longer than five seconds to think.

The correct answer is the motor because it has the highest resistance.

We know that in a serious circuit, the current is the same everywhere.

The only way you can get the same current through the motor, which has the highest resistance, is to have a larger p.

d.

across it.

Slightly different question.

Which of these four things has the highest p.

d.

across it? Did you realise that it's the cell.

The p.

d.

across the cell is the sum of the p.

ds.

across the resistor, lamp, and motor.

And now some more circuit questions for you to think about.

In each of these questions, I'd like you to write down the missing p.

d.

values from the circuits.

So press Pause for as long as you need, and press Play when you have your answers ready.

Now, let's take a look at the answers.

Question one, there's only a cell and a lamp in the circuit.

The p.

d.

across the both will be the same.

Question two, the p.

d.

across the resistor and motor should add up to equal the p.

d.

across the cell, 1.

5 volts.

So the resistor must have 0.

6 volts across it.

In question three, the potential differences across the three components should add to equal the potential difference across the cell.

And the sum of those three is 3.

0 volts.

And in question four, we have two cells and three components.

We should find that the potential differences across the three components add up to equal the sum of the potential differences across the cells.

So that's three volts in total across the cells and three volts in total across the three components.

Well done if you got most or all of those right.

And if you didn't, you might want to go back and watch some of the earlier slides again.

And now for part three of the lesson, analysing series circuits.

This equation relates current, p.

d.

, and resistance for any component in a circuit.

So the current through the component equals the p.

d.

across the component divided by the component resistance.

And we can write this equation in symbols as I = V/R, where I is current measured in amps, V is potential difference measured in volts, and R is resistance measured in ohms. So now can you match the symbol to the correct unit? Here are the answers.

I is current measured in amperes, sometimes shortened to amps.

V is potential difference measured in volts.

And R is resistance measured in ohms. And the symbol for the ohm is a Greek letter.

Let's try a calculation.

Calculate the current through a bulb if the voltage across it is 12.

0 volts and that has a resistance of 40.

0 ohms. So it's a good idea to start by writing down what you know and what you're trying to find out.

We're trying to find the current.

And so let's write down the equation with current as the subject.

And substituting in the values 12 divided by 40, we'll get an answer of 0.

30 amps.

Now, I'd like you to try one.

Calculate the current through a motor if the voltage across it is 10.

0 volts and it has a resistance of 25.

0 ohms. Press Pause while you write down your working, and press Play when you're ready.

Again, let's start by writing down what we know and what we're trying to find out.

And the same arrangement of the equation we're trying to work out at current, 10 divided by 25, 0.

40 amps.

So well done if you got that.

Now, what if we want to calculate a p.

d.

? We need to make V the subject of the equation.

And to do that, we can multiply both sides by R.

And on the right-hand side, we have divide by R times R, which overall has no effect.

So we can cross those out, and we get I x R = V, or we could write it the other way around as V = I x R.

And our question, using that arrangement, what is the p.

d.

across the resistor if it has a resistance of 38.

0 ohms and a current of 0.

40 amps passes through it? Again, let's write down what we know and what we want to find.

And we'll write down the equation in the arrangement that we need, V = I x R.

Substituting the values, and we find that the potential difference is 15.

2 volts across this resistor.

Now, one for you to try.

What is the p.

d.

across a thin wire if it has a resistance of 26.

0 ohms and a current of 0.

50 amps passes through it? Press Pause while you write out your answer, and press Play when you're ready to check it.

So again, we write down what we know and what we're trying to find.

We write down the equation in the arrangement that we need.

We substitute in the values.

And we calculate our answer.

13.

0 volts.

And of course there's a third way that we could arrange the equation.

We can make R the subject.

We'll do that in two steps.

Multiply both sides by R again.

And now we get I x R = V.

And now divide both sides by I.

And if we do that, we end up multiplying by I and dividing by I on the left.

And overall that has no effect.

We could say those operations cancel out.

So we can cross that out, and we're left with R = V/I.

And a question, what is the resistance of a motor if the voltage across it is 24.

0 volts and the current through it is 0.

70 amps? Write down what you know and what you're trying to find.

Write down the equation in the form that you want it.

Substitute in the values and calculate.

34.

3 ohms. And one for you.

What is the resistance of a thermistor if the voltage across it is 8.

0 volts and the current through it is 0.

20 ohms? Press Pause, and press Play when you're ready to check your answer.

So first of all, write down the quantities that you know and the quantity you're trying to find.

Write down the equation, substitute in the values, and calculate.

40.

0 ohms. So as we've seen, current, p.

d.

, and resistance are related by these equations, which really are all one equation just arranged in different ways.

And if you can rearrange them, then you only need to learn one of them.

And the techniques that you use for rearranging are exactly the same as what you would do if you had to rearrange a formula in a maths lesson.

Which equation here is correctly arranged to find the missing quantity in the circuit shown? And the answer, I = V/R, because the missing quantity here, the one that we aren't told is the current in the circuit.

All three of these arrangements are actually correct, but only B gives us what we need.

And now, here's some questions for you where you'll need to calculate V or I or R in different situations.

So write out your solutions.

And when you're ready to check your answers, press Play.

Now, I'm going to show you the solutions step by step.

Question one, a light bulb has resistance of 2.

5 ohms. What is the current through the light bulb when the p.

d.

across it is 7.

5 volts? We write down the quantities we're given, V and R.

We write down the equation in the arrangement that we need, I = V/R, because we want to calculate current.

And we substitute in the values for V and R, and we get 3.

0 amps.

And 3.

0 is a suitable number of significant figures to write because the quantities we use to calculate it were given to two significant figures.

Now, question two, the p.

d.

across a motor is 12 volts.

If its resistance is 24 ohms, what is the current through it? So again, we've been given V and R, and we want to calculate I.

Write down the equation, substitute the values, and the answer is 0.

5 ohms. If you remember a different arrangement of the equation from when you need in a question, you can write down your rearranging if you need to.

Question three, a light bulb has a current of 0.

50 amps flowing through it when it is connected to a 4.

5 volt battery.

And I think we can assume here that there's nothing else in the circuit.

What is the resistance of the light bulb? So this time, we write down V and I, which the question tells us, and we want to calculate R.

So we use the arrangement R = V/I.

Substitute in the numbers, and we get 9.

0 ohms. Question four, a buzzer, just another type of component that you can use in the circuit.

A buzzer has a resistance of 7.

5 ohms. What p.

d.

is needed to make a current of 0.

20 amps flow through it? So this time, we're given the resistance R, and we're given the current I, and we need to find the p.

d.

, V.

So we use the arrangement of the equation V = I x R.

So 0.

20 amps x 7.

5 ohms, we get 1.

5 volts.

And, finally, question five.

Two 1.

5 volt cells are connected in series and push a current of 0.

3 amps through a motor.

What is the resistance of the motor? And no other components are mentioned here.

So it must be just the motor in series with these two cells.

This time we're not directly given the potential difference.

We need to work it out.

And it's going to be the sum of the p.

ds.

across the two cells, which gives 3.

0 volts.

So we can do that little calculation at the beginning.

We are given the current, 0.

3 amps, and we want to find the resistance R.

So we use R = V/I.

So that's 3.

0 volts divided by 0.

3 amps, which is 10 ohms. So that's the resistance of the resistor.

So well done if you're getting the hang of this.

And if you've got any of these wrong, think about whether you got them wrong just because you made a slip and you understand what went wrong, or did they go wrong because you didn't understand what you were doing, in which case you might want to go back and watch some of the earlier slides.

And now we've come to the end of the lesson.

So I'll give you a summary.

Current in a series circuit is the same everywhere in the circuit.

The p.

d.

of the battery equals that across the components in the circuit.

So the sum of the p.

ds.

across the components equals the p.

d.

across the battery.

The current in a series circuit depends on the p.

d.

and the resistance in the circuit.

According to the equation, current equals p.

d.

divided by resistance.

This equation can be rearranged so that either p.

d.

or R is the subject of the equation.

If two of the three quantities, I, V, or R, are known, the remaining one can be algebraically found.

So well done for working through this lesson.

I hope you found that you've improved at rearranging the equation and using it in the right ways, and aim to see you again in a feature lesson.

Bye for now.